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insert  into `chapter`(`id`,`subjectId`,`pharseId`,`gradeId`,`editionId`,`chapter`,`unit`,`section`,`knowledgeId`,`chapterOrder`,`unitOrder`,`sectionOrder`) values (30957,9,2,402,21,'第9章 化学与社会发展','9.3 环境污染的防治','',50345,9,3,0);
insert  into `chapter`(`id`,`subjectId`,`pharseId`,`gradeId`,`editionId`,`chapter`,`unit`,`section`,`knowledgeId`,`chapterOrder`,`unitOrder`,`sectionOrder`) values (30958,9,2,402,21,'第9章 化学与社会发展','9.3 环境污染的防治','',50346,9,3,0);
insert  into `chapter`(`id`,`subjectId`,`pharseId`,`gradeId`,`editionId`,`chapter`,`unit`,`section`,`knowledgeId`,`chapterOrder`,`unitOrder`,`sectionOrder`) values (30959,9,2,402,21,'第9章 化学与社会发展','9.3 环境污染的防治','',50347,9,3,0);
insert  into `chapter`(`id`,`subjectId`,`pharseId`,`gradeId`,`editionId`,`chapter`,`unit`,`section`,`knowledgeId`,`chapterOrder`,`unitOrder`,`sectionOrder`) values (30960,9,2,402,21,'第9章 化学与社会发展','9.3 环境污染的防治','',50348,9,3,0);
insert  into `chapter`(`id`,`subjectId`,`pharseId`,`gradeId`,`editionId`,`chapter`,`unit`,`section`,`knowledgeId`,`chapterOrder`,`unitOrder`,`sectionOrder`) values (30961,9,2,402,21,'第9章 化学与社会发展','9.3 环境污染的防治','',50349,9,3,0);

/*Table structure for table `edition` */

DROP TABLE IF EXISTS `edition`;

CREATE TABLE `edition` (
  `Id` int(11) NOT NULL,
  `Name` varchar(255) DEFAULT NULL,
  PRIMARY KEY (`Id`)
) ENGINE=MyISAM DEFAULT CHARSET=utf8;

/*Data for the table `edition` */

insert  into `edition`(`Id`,`Name`) values (8,'粤教版');
insert  into `edition`(`Id`,`Name`) values (21,'沪教版');
insert  into `edition`(`Id`,`Name`) values (59,'人教新版');
insert  into `edition`(`Id`,`Name`) values (62,'北京课改新版');
insert  into `edition`(`Id`,`Name`) values (64,'鲁教五四新版');
insert  into `edition`(`Id`,`Name`) values (80,'鲁教新版');
insert  into `edition`(`Id`,`Name`) values (83,'上海新版');

/*Table structure for table `grade` */

DROP TABLE IF EXISTS `grade`;

CREATE TABLE `grade` (
  `gradeId` int(11) NOT NULL,
  `gradeName` varchar(255) DEFAULT NULL,
  PRIMARY KEY (`gradeId`)
) ENGINE=MyISAM DEFAULT CHARSET=utf8;

/*Data for the table `grade` */

insert  into `grade`(`gradeId`,`gradeName`) values (110,'一年级');
insert  into `grade`(`gradeId`,`gradeName`) values (111,'一年级上');
insert  into `grade`(`gradeId`,`gradeName`) values (112,'一年级下');
insert  into `grade`(`gradeId`,`gradeName`) values (120,'二年级');
insert  into `grade`(`gradeId`,`gradeName`) values (121,'二年级上');
insert  into `grade`(`gradeId`,`gradeName`) values (122,'二年级下');
insert  into `grade`(`gradeId`,`gradeName`) values (130,'三年级');
insert  into `grade`(`gradeId`,`gradeName`) values (131,'三年级上');
insert  into `grade`(`gradeId`,`gradeName`) values (132,'三年级下');
insert  into `grade`(`gradeId`,`gradeName`) values (140,'四年级');
insert  into `grade`(`gradeId`,`gradeName`) values (141,'四年级上');
insert  into `grade`(`gradeId`,`gradeName`) values (142,'四年级下');
insert  into `grade`(`gradeId`,`gradeName`) values (150,'五年级');
insert  into `grade`(`gradeId`,`gradeName`) values (151,'五年级上');
insert  into `grade`(`gradeId`,`gradeName`) values (152,'五年级下');
insert  into `grade`(`gradeId`,`gradeName`) values (160,'六年级');
insert  into `grade`(`gradeId`,`gradeName`) values (161,'六年级上');
insert  into `grade`(`gradeId`,`gradeName`) values (162,'六年级下');
insert  into `grade`(`gradeId`,`gradeName`) values (200,'七年级');
insert  into `grade`(`gradeId`,`gradeName`) values (201,'七年级上');
insert  into `grade`(`gradeId`,`gradeName`) values (202,'七年级下');
insert  into `grade`(`gradeId`,`gradeName`) values (300,'八年级');
insert  into `grade`(`gradeId`,`gradeName`) values (301,'八年级上');
insert  into `grade`(`gradeId`,`gradeName`) values (302,'八年级下');
insert  into `grade`(`gradeId`,`gradeName`) values (400,'九年级');
insert  into `grade`(`gradeId`,`gradeName`) values (401,'九年级上');
insert  into `grade`(`gradeId`,`gradeName`) values (402,'九年级下');
insert  into `grade`(`gradeId`,`gradeName`) values (500,'高一');
insert  into `grade`(`gradeId`,`gradeName`) values (501,'高一上');
insert  into `grade`(`gradeId`,`gradeName`) values (502,'高一下');
insert  into `grade`(`gradeId`,`gradeName`) values (600,'高二');
insert  into `grade`(`gradeId`,`gradeName`) values (601,'高二上');
insert  into `grade`(`gradeId`,`gradeName`) values (602,'高二下');
insert  into `grade`(`gradeId`,`gradeName`) values (700,'高三');
insert  into `grade`(`gradeId`,`gradeName`) values (701,'高三上');
insert  into `grade`(`gradeId`,`gradeName`) values (702,'高三下');
insert  into `grade`(`gradeId`,`gradeName`) values (10100,'必修1');
insert  into `grade`(`gradeId`,`gradeName`) values (10200,'必修2');
insert  into `grade`(`gradeId`,`gradeName`) values (10300,'必修3');
insert  into `grade`(`gradeId`,`gradeName`) values (10400,'必修4');
insert  into `grade`(`gradeId`,`gradeName`) values (10500,'必修5');
insert  into `grade`(`gradeId`,`gradeName`) values (20100,'选修1');
insert  into `grade`(`gradeId`,`gradeName`) values (20101,'选修1-1');
insert  into `grade`(`gradeId`,`gradeName`) values (20102,'选修1-2');
insert  into `grade`(`gradeId`,`gradeName`) values (20200,'选修2');
insert  into `grade`(`gradeId`,`gradeName`) values (20201,'选修2-1');
insert  into `grade`(`gradeId`,`gradeName`) values (20202,'选修2-2');
insert  into `grade`(`gradeId`,`gradeName`) values (20203,'选修2-3');
insert  into `grade`(`gradeId`,`gradeName`) values (20300,'选修3');
insert  into `grade`(`gradeId`,`gradeName`) values (20301,'选修3-1');
insert  into `grade`(`gradeId`,`gradeName`) values (20302,'选修3-2');
insert  into `grade`(`gradeId`,`gradeName`) values (20303,'选修3-3');
insert  into `grade`(`gradeId`,`gradeName`) values (20304,'选修3-4');
insert  into `grade`(`gradeId`,`gradeName`) values (20305,'选修3-5');
insert  into `grade`(`gradeId`,`gradeName`) values (20400,'选修4');
insert  into `grade`(`gradeId`,`gradeName`) values (20401,'选修4-1');
insert  into `grade`(`gradeId`,`gradeName`) values (20402,'选修4-2');
insert  into `grade`(`gradeId`,`gradeName`) values (20403,'选修4-3');
insert  into `grade`(`gradeId`,`gradeName`) values (20404,'选修4-4');
insert  into `grade`(`gradeId`,`gradeName`) values (20405,'选修4-5');
insert  into `grade`(`gradeId`,`gradeName`) values (20406,'选修4-6');
insert  into `grade`(`gradeId`,`gradeName`) values (20407,'选修4-7');
insert  into `grade`(`gradeId`,`gradeName`) values (20500,'选修5');
insert  into `grade`(`gradeId`,`gradeName`) values (20600,'选修6');
insert  into `grade`(`gradeId`,`gradeName`) values (20700,'选修7');
insert  into `grade`(`gradeId`,`gradeName`) values (20800,'选修8');
insert  into `grade`(`gradeId`,`gradeName`) values (20900,'选修');
insert  into `grade`(`gradeId`,`gradeName`) values (30100,'选择性必修1');
insert  into `grade`(`gradeId`,`gradeName`) values (30200,'选择性必修2');
insert  into `grade`(`gradeId`,`gradeName`) values (30300,'选择性必修3');

/*Table structure for table `knowledge_basic` */

DROP TABLE IF EXISTS `knowledge_basic`;

CREATE TABLE `knowledge_basic` (
  `id` int(9) NOT NULL AUTO_INCREMENT,
  `knowledgeName` varchar(200) DEFAULT NULL COMMENT '知识点名称',
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  `pharseId` int(11) DEFAULT NULL COMMENT '学历Id',
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  UNIQUE KEY `index_md5` (`md5`)
) ENGINE=MyISAM AUTO_INCREMENT=109009 DEFAULT CHARSET=utf8;

/*Data for the table `knowledge_basic` */

insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50002,'对科学探究的理解',9,2,'F67FE69D3660B4D35A731817B538B21D');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50003,'科学探究的意义',9,2,'0CCE9D48EB96FDF93FBAE8640D547B8E');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50004,'猜想与事实验证',9,2,'171AB172EFB24344684ECA5B04ABFFCA');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50005,'科学探究的基本方法I',9,2,'E7BEB1DCF073B1D1E700FB02ECCAF064');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50006,'科学探究的基本环节I',9,2,'2432FC2EFE99899B0ECFF8ADE0211E7D');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50007,'科学探究能力',9,2,'1CC41F4AB8528178818A29B9EF5FABBB');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50008,'测定空气里氧气含量的探究',9,2,'79C3489E2392AFD26733D285DEE3ABD0');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50009,'制取气体的反应原理的探究',9,2,'F0547ECD4E64A31E247C34B64547F812');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50010,'质量守恒定律的实验探究',9,2,'3E53AE683F8E8C84221DB763B30FE907');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50011,'金属活动性的探究',9,2,'FBBBADB6D1A15C0C924C73B0B0A4B7CB');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50012,'探究酸碱的主要性质',9,2,'9BA86C2987B9321A45B4DBF1EFF6BB4A');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50013,'药品是否变质的探究',9,2,'8EC41A3E649625A55CEAFC35F6FA45E8');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50014,'缺失标签的药品成分的探究',9,2,'87B0CBA64000C51C883F57274C04519C');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50015,'实验探究物质的性质或变化规律',9,2,'F7C3C4088DFE80933E84CA084FA3524A');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50016,'探究金属锈蚀的条件',9,2,'5A5A84625F44B7E7345B4EA6FDE06627');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50017,'燃烧的条件与灭火原理探究',9,2,'85DBDB1CBB78B9BE83CCEDD468732E0A');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50018,'影响化学反应速率的因素探究',9,2,'16D37A42180158171D57E1CC8122B415');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50019,'实验探究物质变化的条件和影响物质变化的因素',9,2,'C3A8217C9D3A5D9C5E76A77D8F4A8FDE');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50020,'食品干燥剂、保鲜剂和真空包装的成分探究',9,2,'4E2598D3FA41AE72D1927B81328DBD51');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50021,'化肥有效成分含量的探究',9,2,'81B993DAE9D5735B0714C325C526AEE5');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50022,'味精中食盐含量的探究',9,2,'D3D2A1A264FEB84BD8BA9D0557AAFCA8');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50023,'实验探究物质的组成成分以及含量',9,2,'D5E390212EA61535B492B740102DF78A');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50024,'物质除杂或净化的探究',9,2,'CF50B28EF624912FF106C57CA9BE41DC');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50025,'气体制取装置的探究',9,2,'A4EACDF08E8FDA83C7784C8FD21F7811');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50026,'实验分析与处理能力',9,2,'9A84AF5408986FAAB11F648A07867D84');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50027,'实验数据处理或者误差分析的探究',9,2,'31BD7CC9213175D709FCFA2EEB4B202A');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50028,'实验操作注意事项的探究',9,2,'2D084A4ACD512E6314D6E8AE111B8205');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50029,'实验步骤的探究',9,2,'2A12B41ADEEDC754B55EC468D1A41D09');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50030,'化学实验方案设计与评价',9,2,'33702A9C691C0F5AAAC103D7DD1952EB');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50031,'化学竞赛',9,2,'D081111DBDEE3C687D1439B444D64004');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50032,'基本的实验技能',9,2,'DFDC9E0C03A33349408E99F28D07F899');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50033,'用于加热的仪器',9,2,'0610027C7B4268080E7C1C5F04AF05A7');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50034,'测量容器-量筒',9,2,'CB07ACCC409BBB4C0ADC6AFB26CF351B');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50035,'称量器-托盘天平',9,2,'A1F3A4E959C66A4DD4F330F13FF4D808');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50036,'加热器皿-酒精灯',9,2,'9F75E281CBE6072BD91A286E64FB6F0D');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50037,'挟持器-铁夹、试管夹、坩埚钳',9,2,'9BF3F8E2F454487987A4888544F9E1BE');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50038,'分离物质的仪器',9,2,'664A26F366B9EF4988631E95AF9B366D');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50039,'量气装置',9,2,'F19EA2AD04C46F33134D405510650A60');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50040,'固体药品的取用',9,2,'3B4421D0AB0E43C65932C51FB58F593F');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50041,'液体药品的取用',9,2,'CA355F31B8E517ABC70BF477CA77F4CE');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50042,'给试管里的固体加热',9,2,'AFA8024DE2C03966E71D6F94A93B6B93');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50043,'给试管里的液体加热',9,2,'CE9E053A63F6A8AED199BED09F1E498E');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50044,'物质的溶解',9,2,'3CF419E05D85881157B758A01C6EF399');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50045,'浓硫酸的性质及浓硫酸的稀释',9,2,'AF7994B458C40E4A18EC60F5E622E522');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50046,'一定溶质质量分数的溶液的配制',9,2,'C3BEB22D8BB8A4B874FD7BB8A8914643');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50047,'混合物的分离方法',9,2,'5F9F76D679371D223DEEDA050BDC9D85');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50048,'过滤的原理、方法及其应用',9,2,'218171BD4087237ACDCC6D3846B9CDA5');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50049,'结晶的原理、方法及其应用',9,2,'D38AAD5D5676BE87EAF6ADE964CAFF4F');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50050,'蒸发与蒸馏操作',9,2,'8BA23D23CE49F63D802D34B1BCEEBFE1');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50051,'仪器的装配或连接',9,2,'BDBF65C0985144843465CF6C4785094B');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50052,'检查装置的气密性',9,2,'D007BEAADEE2CF8A702432F742825E70');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50053,'玻璃仪器的洗涤',9,2,'C1E624FB0970AAAE461D24D6B6AEB0E9');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50054,'常见的意外事故的处理方法',9,2,'4C4D7D721CF9A48F1EDB88443BD1E863');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50055,'实验室制取气体的思路',9,2,'9512192A18D42848EEF03F3F0A3DFBF6');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50056,'常用气体的发生装置和收集装置与选取方法',9,2,'602295B7058619BF89A19BFD6438B320');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50057,'常用气体的收集方法',9,2,'922DCFDCC5792CD2364CBCCE9B00C96E');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50058,'常见气体的检验与除杂方法',9,2,'9E64CBA5E1C5B2C14D7667A735F62E00');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50059,'气体的净化（除杂）',9,2,'CFEEDCDD5E287BEF4B583158A12363F1');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50060,'气体的干燥（除水）',9,2,'CF164E398BEBD2384CBBBFE73FA72FCF');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50061,'酸碱指示剂的使用',9,2,'E2F964E176EFB40969652E3249023645');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50062,'溶液的酸碱度测定',9,2,'02DD0428A167BDE5E5B544CC1AAE3F74');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50063,'证明盐酸和可溶性盐酸盐',9,2,'451BF16FFF6233CCA8D9AD69B31C33B9');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50064,'证明硫酸和可溶性硫酸盐',9,2,'9165D83699E3D014274F4737D169A2A0');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50065,'证明碳酸盐',9,2,'E1544EE9437A8407BA1A88AD797FD3E2');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50066,'证明铵盐',9,2,'6BD1D12CB7B4668D39BD7B9D01A98BAD');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50067,'常用仪器的名称和选用',9,2,'F8411C202A5CF1E26A0EB50979857D44');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50068,'化学的基本常识',9,2,'9FA0BB67BA754C1B4E5921C3BC837CEB');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50069,'化学的历史发展过程',9,2,'47F1ADC470BAAF1874D79DC4144AF95F');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50070,'化学的研究领域',9,2,'CB306AAB7CDB746C2996F02A7B7D1CE4');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50071,'化学的用途',9,2,'491EF963C5AC9A9136F60C8CB2188636');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50072,'绿色化学II',9,2,'0A3CB8D6A1D544B17D920FF2D36B8C18');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50073,'蜡烛燃烧实验',9,2,'C20168038C0E10C441F2C118AC418EE7');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50074,'吸入空气与呼出气体的比较',9,2,'DDECC260427B448F88F296B1EF62B8A8');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50075,'学习化学的重要途径及学习方法',9,2,'4C3860ECA356F2A5EE47B2BFC2A9A93E');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50076,'有关化学之最',9,2,'C6D16D558CA9F98AD8E8FCE9B6DED577');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50077,'化学常识',9,2,'EA407C6C8898C300A60B1F84D8F15D80');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50078,'身边的化学物质',9,2,'219F81CC6C7826D1BB55686944865323');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50079,'地球周围的空气',9,2,'D5253BC36FA24D25543DCC144F93B3DC');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50080,'空气的成分及各成分的体积分数',9,2,'BB07992B31827AFC0B94591957936729');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50081,'空气组成的测定',9,2,'1FE73DFBA86C3B53A3597C79E251ECB2');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50082,'空气对人类生活的重要作用',9,2,'D2803A212B8CA2EA925E3C7242B0B84F');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50083,'空气的污染及其危害',9,2,'1BE3614EC5D67A9FE3FD389516F369EA');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50084,'防治空气污染的措施',9,2,'CBBFD74D4CB5AAF24C9E4A10D9F056D9');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50085,'目前环境污染问题',9,2,'B9F879D2C97FE4E050A153E142F0D9BA');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50086,'氧气的物理性质',9,2,'AEFB4027BC1463273DEAF0B90B53A694');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50087,'氧气的化学性质',9,2,'1AA75EAD4FB22C7A6F25C91B5C48727E');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50088,'氧气的用途',9,2,'9D9B6DF568499E7A2B1E4091229913F3');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50089,'氧气与碳、磷、硫、铁等物质的反应现象',9,2,'7CBD2A56F3BD5796FE663F8FF3F39FEE');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50090,'氧气的工业制法',9,2,'8D3FA11102EE509EBF012560CE3DD396');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50091,'实验室制取氧气的反应原理',9,2,'90C42A8DA5F45FDC451FD59001C6355D');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50092,'氧气的制取装置',9,2,'90F6A7F5CDD8F8F753CDCFAF51B207EF');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50093,'氧气的收集方法',9,2,'B30BEE5EC24F87C4A4E28163E78C8B22');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50094,'氧气的检验和验满',9,2,'6B9A66AFA26A38E7D9127E2CBDEC35A7');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50095,'制取氧气的操作步骤和注意点',9,2,'062B476579A41DAEA439432AC779B391');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50096,'自然界中的氧循环',9,2,'C3F2CD4BAD8360893F1CE25F71F75037');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50097,'自然界中的碳循环',9,2,'0D45AF329C08F0C1A96B274414684A37');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50098,'催化剂的特点与催化作用',9,2,'63AE3E64663572042ECDD0EF493C16E1');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50099,'常见气体的用途',9,2,'3478EBC519F2D13B79FC22803F47B765');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50100,'二氧化碳的实验室制法',9,2,'5B5C37F4490273A893DD9A89D14CA062');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50101,'二氧化碳的检验和验满',9,2,'86A2E69C428AAFB50B38A78F62D35B8D');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50102,'制取二氧化碳的操作步骤和注意点',9,2,'1A04AE79DF0720DE2B40A38B48E5CDD2');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50103,'二氧化碳的工业制法',9,2,'F7F8EB12E0F61A9321597157C0D61791');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50104,'二氧化碳的物理性质',9,2,'0FFE862DE5E1E5F4C37E267F87743EC4');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50105,'二氧化碳的化学性质',9,2,'3D57962739FD6CE041A88150F92290B8');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50106,'二氧化碳的用途',9,2,'9D512E28838BD0483C473D80A285D4E8');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50107,'二氧化碳对环境的影响',9,2,'B796D0E168FEC2FECF1F71B40343933F');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50108,'一氧化碳的物理性质',9,2,'4F4ED24EBD74E0C83FB1DCA8E065D4CA');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50109,'一氧化碳的化学性质',9,2,'906B92B2E09A38A0DC5933B4943E87A0');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50110,'一氧化碳的毒性',9,2,'1608D514520AB9C8F76DECB1D579F099');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50111,'探究氧气的性质',9,2,'27A463FA0B44830511E5ED817924AFDD');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50112,'探究二氧化碳的性质',9,2,'49C965EEB9807E1F10E4E809CFFEE8FB');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50113,'水与常见的溶液',9,2,'E382852F04D7D7D131BC0FA2C4CF8CC9');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50114,'电解水实验',9,2,'5A8BE7163B9B6D457877516EF0C8C257');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50115,'水的组成',9,2,'729D75BF4AE9760E4F1F3522025CEF25');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50116,'水的性质和应用',9,2,'D5CB15B7ECEEB8965BB6937B1F34CFC1');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50117,'水的合成与氢气的燃烧',9,2,'41F3CBA68FFB2F2A2181B08ABB545F89');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50118,'水的净化',9,2,'2F2D48064F39B473234D59790C132658');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50119,'自来水的生产过程与净化方法',9,2,'29D051A2EB4F0FDD54196648F21C5BB6');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50120,'硬水与软水',9,2,'512D92D7A986CBE30B2D82A804E3F009');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50121,'水资源状况',9,2,'8AF16A79FFAA4B864EF189A5D7449082');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50122,'水资源的污染与防治',9,2,'AE9A5410BBFB439D970FE33802E86836');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50123,'溶解现象与溶解原理',9,2,'53414900E2F2C28C88F5BB5BA754C49B');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50124,'常见的溶剂',9,2,'89240B73186F95914E88BE3C281807C7');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50125,'溶液的概念、组成及其特点',9,2,'39795321717AE630A8DA44715D783FF7');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50126,'溶液、溶质和溶剂的相互关系与判断',9,2,'496D691D99A11C0C98A1611AD4B1D52E');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50127,'影响溶解快慢的因素',9,2,'9C3508E704AA16945CF79B05AF57848C');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50128,'溶解时的吸热或放热现象',9,2,'EE15C3FF282DB9CFC0557A4117C5EDD0');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50129,'悬浊液、乳浊液的概念及其与溶液的区别',9,2,'5958C80AABBCD2A448BE9C36F0DFB50C');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50130,'乳化现象与乳化作用',9,2,'EEF050E9354CE1EBC48096D9B0345EC5');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50131,'饱和溶液和不饱和溶液',9,2,'825D2EA00913A4AA9BDF8E72CD2EA810');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50132,'饱和溶液和不饱和溶液相互转变的方法',9,2,'83C733DC5D84598104562151CC1C79B3');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50133,'浓溶液、稀溶液跟饱和溶液、不饱和溶液的关系',9,2,'3E450E4AF4A57D230A480691C7B26A04');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50134,'固体溶解度的概念',9,2,'A82EDD56D1EF3F81E94FEFAC1757FF8F');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50135,'固体溶解度的影响因素',9,2,'38C37B881D026580EF22A9E4677F8096');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50136,'固体溶解度曲线及其作用',9,2,'61C3EA9FB954034C3C481FD1E767E23E');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50137,'气体溶解度的影响因素',9,2,'224C6AA84B57848656F7DFD11CD67D34');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50138,'晶体和结晶的概念与现象',9,2,'F3D24E8D7784DB1ED4EC0B98C85E1DF8');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50139,'物质的溶解性及影响溶解性的因素',9,2,'EA3840C049893EF0051EDF8BB6FA23F8');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50140,'溶质的质量分数',9,2,'E623FF4F394D554C5A1D095987994377');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50141,'用水稀释改变浓度的方法',9,2,'00EDCA0EAC577434EBE1B76715E828EC');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50142,'溶质的质量分数、溶解性和溶解度的关系',9,2,'0820738EC28CCC2B06A82D833D0BA1E8');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50143,'有关溶质质量分数的简单计算',9,2,'0341610337A295BE1B50F527A701BC75');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50144,'氢气的制取和检验',9,2,'6ABE352B723CEBDC0B6B9E28D8FF40AA');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50145,'金属与金属矿物',9,2,'C25EB64A3EB4FDBEE9C8EC96EFCC53F6');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50146,'常见金属的特性及其应用',9,2,'AA92AE094EA15D74A793A7976DAF3768');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50147,'金属的物理性质及用途',9,2,'C8306DD6F56237F32FD2F50C8BB7A566');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50148,'合金与合金的性质',9,2,'2C1D21E96A6D80A69A395B6FF1956367');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50149,'常见的金属和非金属的区分',9,2,'1BE70C69CC6DE24490AC34B68BD64A0C');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50150,'金属的化学性质',9,2,'A13F3BA20B13B210C59FAF8017D236C8');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50151,'金属活动性顺序及其应用',9,2,'21AB1E26E44F941996510754C733A7F2');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50152,'金属元素的存在及常见的金属矿物',9,2,'64B856008505D7261B5DF826362AE3B8');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50153,'探究金属铜的冶炼原理',9,2,'FD5B92CBB6B0596AB3CC6B8CE5A94668');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50154,'金属材料及其应用',9,2,'FCDAE5ACB17CD164A212D497D591B2BA');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50155,'一氧化碳还原氧化铁',9,2,'111D10C784B62405E0CE2F721CCC1ACC');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50156,'铁的冶炼',9,2,'9800A941E051517633763A1DE894BDD2');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50157,'碳、一氧化碳、氢气还原氧化铜实验',9,2,'A38B49209472B0772D3937C7EC229295');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50158,'常见金属的冶炼方法',9,2,'7F35B00DA0F2864D5BD499A013BD8A63');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50159,'生铁和钢',9,2,'5ECF438E3BA6480CC3A438114D08F6C5');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50160,'含杂质物质的化学反应的有关计算',9,2,'0E6A5AC64D865D6AC07C2FCE64D7B1BA');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50161,'金属锈蚀的条件及其防护',9,2,'0B75652361ACFB1ADB97925881B5F088');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50162,'铁锈的主要成分',9,2,'74A026491D88FB1A903FA60BCDB5FED2');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50163,'金属资源的保护',9,2,'F807D346EA8DCA278C97B975E1F81580');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50164,'废弃金属对环境的污染',9,2,'B8E4FFCBA1F8184A704C00FE3BA67999');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50165,'金属的回收利用及其重要性',9,2,'0DB1BF9A0326038FE66D8A95AF74919B');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50166,'生石灰的性质与用途',9,2,'E7CC4E909F739405272E79CFA00367F3');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50167,'碳酸钙、生石灰、熟石灰之间的转化',9,2,'E2F5BEBF17A82CDE2F8F86FEE5A83E22');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50168,'生活中的常见化合物',9,2,'BA91C53141A643C6E46492FC64D8EEBB');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50169,'酸碱指示剂及其性质',9,2,'679B4E067AE5950E56C31CED459A841D');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50170,'酸的物理性质及用途',9,2,'AE5314A8DA1E6872EC5B45D207E97671');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50171,'酸的化学性质',9,2,'92438C080AF2AEA5EAB3352F8CC07875');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50172,'常见碱的特性和用途',9,2,'CE6771B7C1F859F5F56119792AE868D4');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50173,'碱的化学性质',9,2,'004978819088D183A8F9841C6D2AA432');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50174,'溶液的导电性及其原理分析',9,2,'F6C15B2D371681BA5250DD75C528842E');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50175,'中和反应及其应用',9,2,'F33D24234403AB352C4FCAAA86D9AEDF');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50176,'酸碱溶液的稀释',9,2,'9D89F19778A174ED652FD947B1BF751B');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50177,'溶液的酸碱性与pH值的关系',9,2,'753CFB115E2DB521CD069A1BA64A1AA1');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50178,'溶液的酸碱性测定',9,2,'58391E8604E4FC624FDB8114D42D1B65');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50179,'酸碱性对生命活动和农作物生长的影响',9,2,'7001D40712221DE0D30FE64A620B1A65');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50180,'酸雨的产生、危害及防治',9,2,'16F60ABADA0C6E46AEC7BE29A16D05F1');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50181,'酸碱盐的应用',9,2,'9A2B936B4A7A1E12B42F1D7D9DAFFEB7');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50182,'氯化钠与粗盐提纯',9,2,'E47BD79901F90F2CC539E45C205B4800');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50183,'碳酸钠、碳酸氢钠与碳酸钙',9,2,'82AC5EF22D6C457C4B4B0BA73F0ED930');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50184,'常用盐的用途',9,2,'9504C1AF6B347E373FAC1BACEABF46EF');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50185,'盐的化学性质',9,2,'322CD5382C6C967C00DDCB78643884B3');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50186,'复分解反应及其发生的条件',9,2,'22764440B2C9F2FB903CE8957A2AA01F');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50187,'常见化肥的种类和作用',9,2,'D8CBEB88C52AACB0C5087B874D44A229');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50188,'化肥的简易鉴别',9,2,'B7545F7B47B6048A7B6409BC345D2A9C');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50189,'铵态氮肥的检验',9,2,'378D72809ABD14A5B67378228E9304E6');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50190,'施用化肥对环境的影响',9,2,'A933845584061F71DCAF5044998C7980');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50191,'醋酸的性质及醋酸的含量测定',9,2,'84650374B6526AEAB39B489F2622673A');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50192,'海水晒盐的原理和过程',9,2,'99E8052E09A34539245A1C7AF9617FE3');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50193,'纯碱的制取',9,2,'75E785CC9A3945B6E57DB6E924DB2F28');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50194,'酸碱盐的溶解性',9,2,'5DCA9332AC63DE3F172FFBBD16178338');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50195,'离子或物质的共存问题',9,2,'03250FF73B1415D706A3BBDCAAB0F475');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50196,'空气中常见酸碱盐的质量或性质变化及贮存法',9,2,'6F0AB8AED38B46D8E7B80C4F917408F2');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50197,'根据浓硫酸或烧碱的性质确定所能干燥的气体',9,2,'BC80062EA0785AA09FEFF108DEA55ED2');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50198,'酸、碱、盐的鉴别',9,2,'64FDC1CCEF34439FD0ABE59A5FAE6869');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50199,'常见离子的检验方法及现象',9,2,'71A8117EB8BB945AFDCD7551E28A6B53');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50200,'物质构成的奥秘',9,2,'D0B416A4CCAC7CA256E9E0B8D137EC0B');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50201,'化学物质的多样性',9,2,'DF3C83CABEAB2F43401DC6A7006C2D2A');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50202,'物质的三态及其转化',9,2,'7493594CEFAF0A08A47135EA019AC24A');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50203,'物质的简单分类',9,2,'D05854F0A3EB6B2B5A879183BEE5B54C');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50204,'从组成上识别氧化物',9,2,'0E8970A411A619A5812F91322963EF99');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50205,'纯净物和混合物的概念',9,2,'9E2A64855E05B663520CCBC2581B7998');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50206,'纯净物和混合物的判别',9,2,'C28D7FCAB728585C954BFC481EDB2F43');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50207,'单质和化合物的概念',9,2,'6FB46146D54B9533B525DEDC9308DB3F');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50208,'单质和化合物的判别',9,2,'2B040BF76E436C1916A75A7B376ACD9E');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50209,'氧化物、酸、碱和盐的概念',9,2,'F768624B1F592EA6C9B2D6B7174E38F2');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50210,'常见的氧化物、酸、碱和盐的判别',9,2,'63CDCAF145863D57408D00BAFBE9EEE9');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50211,'有机物的特征、分类及聚合物的特性',9,2,'09A263C7643FEEE92B371E16CC0632E2');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50212,'甲烷、乙醇等常见有机物的性质和用途',9,2,'C36342AACC12BDE610850D6BA7300A3E');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50213,'有机物与无机物的区别',9,2,'BE5BC41C9CE398D8E5324B70CF066DAA');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50214,'物质的多样性及其原因',9,2,'4AD38ECF2884A47A5EFA657F3358E635');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50215,'物质的鉴别、推断',9,2,'2C40DE1F33E9CCB8DFFD2F3CC6A0C903');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50216,'物质的相互转化和制备',9,2,'B264C38FE3C4092EFAB48AB1CE030329');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50217,'微粒构成物质',9,2,'86B0D735B14AFC1F1C64986EF01AFC22');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50218,'物质的微粒性',9,2,'F88ED1BAA2D998584A6D7A861B7A2055');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50219,'分子、原子、离子、元素与物质之间的关系',9,2,'22052ED6DB7561A2A8ABC3E50D68F829');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50220,'微粒观点及模型图的应用',9,2,'0F29A3092AA170140432866B7C7E0AD9');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50221,'原子的定义与构成',9,2,'7149244DE63A0BAA9BD91FF165CFAF79');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50222,'分子和原子的区别和联系',9,2,'9407C90C49BF04F8B59677647EF56522');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50223,'原子和离子的相互转化',9,2,'A1E95B7D953B84C8F0C03A8A933F1169');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50224,'核外电子在化学反应中的作用',9,2,'1F5B519CDE67AC0D0FCAB419AA3048A4');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50225,'原子结构示意图与离子结构示意图',9,2,'755E7BABE12F5F5B55756943FAB23A2C');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50226,'利用分子与原子的性质分析和解决问题',9,2,'AB7902A82898C827AE39F10473CD9888');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50227,'分子的定义与分子的特性',9,2,'F1EC6F11FDCA959B087365BFF349DA2A');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50228,'原子的有关数量计算',9,2,'82EEB7C49F8ACC2D9B7CE8287EA8FEA0');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50229,'认识化学元素',9,2,'602852260C81B93A01693165F1AE9DD7');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50230,'元素的概念',9,2,'9643EA653FA4C97E6C0B01346AB1AD6E');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50231,'地壳中元素的分布与含量',9,2,'347D36720BAFA2296C3C9EEB2099AA7A');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50232,'元素的符号及其意义',9,2,'53D3EEAF4D7A0111EDEBC65644961762');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50233,'元素的简单分类',9,2,'16396618378B70ABE31493EAD61CA631');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50234,'元素周期表的特点及其应用',9,2,'E898C81260902D7597BA9196736A9B52');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50235,'碳单质的物理性质及用途',9,2,'00DED208A2A2A932217ED8CC9C13C1D6');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50236,'碳的化学性质',9,2,'A4CD7B365B8116DC933969A17D33C9E8');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50237,'物质的元素组成',9,2,'78F9C02A11B92FAECDFC0B1E52FAD018');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50238,'物质的构成和含量分析',9,2,'DE288FE869DBBBDD91C4F8F89886F647');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50239,'碳元素组成的单质',9,2,'C0F43F92577E6A9E94982DFDA4565340');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50240,'氧元素组成的单质',9,2,'3BC8E52202B74E5845EA52EE395623E1');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50241,'元素在化学变化过程中的特点',9,2,'CB9337BC8A0CDE6EE31A94A201F0333D');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50242,'同素异形体和同素异形现象',9,2,'CCEBCEC8775C71CA0C3E559870894DC4');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50243,'物质组成的表示',9,2,'0FB8280EA2D4C3DEA9DB1C1C7FFA4581');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50244,'化学式的书写及意义',9,2,'34DC54D96373B2263B664F8DD1E628AC');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50245,'化合价的概念',9,2,'3F4718C3FE9E06F834777E39835E3DFD');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50246,'常见元素与常见原子团的化合价',9,2,'B988BCE88D7AE76FEF0F2E0F3E809C0F');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50247,'化合价与离子表示方法上的异同点',9,2,'58CA3758D5E6DAEA6802485A22F4D1ED');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50248,'化合价规律和原则',9,2,'72342F16DA6DA710F86B49BFAD61DFF6');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50249,'有关元素化合价的计算',9,2,'72FF247339A16A5E0850F7801504EAD1');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50250,'相对原子质量的概念及其计算方法',9,2,'E090DC0D50BEC3A36EB5D6371FD6030E');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50251,'相对分子质量的概念及其计算',9,2,'54D22F8632EB479849F7BAD14F024CE5');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50252,'元素质量比的计算',9,2,'96F6441CBED455698217B98F989F0D90');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50253,'元素的质量分数计算',9,2,'04B9894BEC8FB548D80DAC08563DB359');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50254,'化合物中某元素的质量计算',9,2,'293A3DDBA6D1A6BB1F2B2DC364CFA5F2');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50255,'混合物中某元素的质量计算',9,2,'35B07AF7D3FF07F8D3A4403E3C302808');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50256,'物质组成的综合计算',9,2,'8A913F4AB05BDAF0E0211F5914910C01');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50257,'标签上标示的物质成分及其含量',9,2,'F77CF08EE99FA7701632A59DB043B925');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50258,'化学符号及其周围数字的意义',9,2,'A9ED0CDD23029E0981337FF9A45F6321');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50259,'有关化学式的计算和推断',9,2,'AEA1D6F516DAC44D20B2943846C1A8EB');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50260,'物质的化学变化',9,2,'915084E1490F0B0903E69748E0CEACD5');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50261,'化学变化的基本特征I',9,2,'F16FAF5D680D7B88E2E157C1C137C497');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50262,'化学变化的基本特征II',9,2,'82BE72A97FDBBDACEED45092AC9D3520');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50263,'物理变化的特点',9,2,'AD0E54281F5A23CF8D1120DADD57A727');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50264,'化学变化和物理变化的判别',9,2,'9A4BEE6584F98D9C0CEA9EAB4E89DFA2');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50265,'化学性质与物理性质的差别及应用',9,2,'F9480B6E092115185F23EAA12D5E3FAC');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50266,'物质发生化学变化时的能量变化',9,2,'AC0E9478875811DAA1DA261D75DF0F52');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50267,'化学的研究对象及其发展',9,2,'59843BFB2CA080A23BCBBC5278FA6982');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50268,'反应现象和本质的联系',9,2,'D82C780D2A2961EA0E0D62A13C024C81');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50269,'化学反应的实质',9,2,'70E1A3C032F8B2A08D1EE55D5CEB23EF');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50270,'认识几种化学反应',9,2,'B7D5B31DAA4C9F6CC67772A4DAF5AEEA');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50271,'化合反应及其应用',9,2,'E785FE017D50ED8E46E6ED13F19A3CDF');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50272,'分解反应及其应用',9,2,'E12E5A207C1C51BDF37F1D829294669A');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50273,'置换反应及其应用',9,2,'E0E5486446ED62CE06D20E36F5768613');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50274,'复分解反应及其应用',9,2,'6DFDE8AA3DD913DD25BD717FAEE1BF79');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50275,'氧化反应',9,2,'137B81B130A9003977A2F513032A738A');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50276,'复分解反应的条件与实质',9,2,'48997E34BFDE15452D4F69DE04B87F57');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50277,'金属活动性顺序与置换反应',9,2,'35BB97F361BCCA162714C1C41FCFA775');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50278,'酸、碱性废水的处理',9,2,'21F13A5B34B547642228FD4E9FC56A15');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50279,'光合作用与呼吸作用',9,2,'D1B8C0F3BAA7B05C8CEA547E188A5A7C');
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insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50281,'还原反应',9,2,'6C50A85BD85A359AE0225A9D9A11AF0C');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50283,'质量守恒定律及其应用',9,2,'54778FAEBFB990828EC9D581A3A5C7CB');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50284,'化学方程式的概念、读法和含义',9,2,'17C4EF97B946C1798E10E7F9086802E2');
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insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50286,'常见化学反应中的质量关系',9,2,'2625E71721BF8A56E5B453FF7B74748E');
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insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50288,'根据化学反应方程式的计算',9,2,'DBAF53621CCB49960FD0A32093C38BE2');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50289,'化学与社会发展',9,2,'89DA44BC2F3EF61B413FE4976FD21114');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50290,'化学与能源、资源利用',9,2,'1B3CE0EEBEA29CFC50AD439503CBEDDF');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50291,'燃烧与燃烧的条件',9,2,'1BA7439A02D9539077510205DD05123F');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50292,'完全燃烧与不完全燃烧',9,2,'1530FF634B23607909249284AC7EDA45');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50293,'常用燃料的使用与其对环境的影响',9,2,'E57942BBFD46919C7D2C7B73BF09FB1C');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50294,'灭火的原理和方法',9,2,'4C320A31119CD5D8114FC81267C53F23');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50295,'防范爆炸的措施',9,2,'2C6AC01AC0D8EA3AE36555A9D181F9E5');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50296,'燃烧、爆炸、缓慢氧化与自燃',9,2,'28CE99F02D994EC5227307D28DB232E5');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50297,'化石燃料及其综合利用',9,2,'DC12F8DB498B3F86414155559FD0B476');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50298,'石油的组成',9,2,'89C0590CD1499525B298FE68F263ED07');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50299,'石油加工的产物',9,2,'BFE897BE4CFAC69AB90DAB87238A7C57');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50300,'原油泄漏对生态环境的危害及其处理',9,2,'9E08261C0D1FF04F85381FA86FCE1087');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50301,'保护水资源和节约用水',9,2,'12BB430BE526CEBB26B7248683B51FAB');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50302,'海洋中的资源',9,2,'FFE367E29927354B95526F42FAB1E71A');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50303,'对海洋资源的合理开发与利用',9,2,'C2A2C749A957B42D69460247E0482569');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50304,'资源综合利用和新能源开发',9,2,'3DB85D1C2F2E830EED7246C33FA19EDD');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50305,'燃烧和爆炸实验',9,2,'99426C3199C2B5E7EDF93918D715C120');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50306,'易燃物和易爆物安全知识',9,2,'6209D3873549A30E8E72459469532FD4');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50307,'几种常见的与化学有关的图标',9,2,'06A0C7649829DE337BF494D3CE6C9850');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50308,'几种常用的灭火器',9,2,'D757BC02ACAD5F1A8CE151D0948F8AE9');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50309,'氢气的物理性质',9,2,'0F3DF30B426F47754822336CE604D5CD');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50310,'氢气的化学性质与燃烧实验',9,2,'86C7AD1C5ECC2DE984715F2977F43D6E');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50311,'氢气的爆鸣实验',9,2,'9750F0D8E0E5365485FDCA09BE800CED');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50312,'氢气、一氧化碳、甲烷等可燃气体的验纯',9,2,'7754747050D51E40650737204A3C02D5');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50313,'氢气的用途和氢能的优缺点',9,2,'187F9F36EA07EBAFDFA037EC925DA73A');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50314,'常见能源的种类、能源的分类',9,2,'19887FF0F3B04E8D42B028A30F412FBA');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50315,'常见的化学合成材料',9,2,'A980F899A57B8BA9CE6BFF1C9FCE9456');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50316,'塑料及其应用',9,2,'F88C967C823AF458243BD26B225798E0');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50317,'塑料制品的回收、再生与降解',9,2,'5B4EE033BC747A6A5C345DDF1FE4030A');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50318,'合成橡胶及其应用',9,2,'4D51348A67E197AEEFE296D4BC379D24');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50319,'合成纤维及其应用',9,2,'3A1BCAE6012B71EE86F1367F1DE77423');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50320,'塑料制品使用的安全',9,2,'4D22D72019911546884011BACB8CA059');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50321,'白色污染与防治',9,2,'5A88ADE3643E2BA1D804F2247B52E630');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50322,'合成材料的使用及其对人和环境的影响',9,2,'DFDA491FA263FB10E68EA6A7DE049A39');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50323,'新材料的开发与社会发展的密切关系',9,2,'ED4531C827B20FC93E120B756EC0E5AD');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50324,'有机高分子材料的分类及鉴别',9,2,'FAC45B9EE8435AD8572BCD6D71F2A8CF');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50325,'复合材料、纳米材料',9,2,'8ABA2FFF73615C43D750822F4C0C497E');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50326,'棉纤维、羊毛纤维和合成纤维的鉴别',9,2,'3DCD852CECD4A1991BD42A43C7C03D56');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50327,'化学物质与健康',9,2,'2E7CB9DA2FA1A111AB1FE0CBEBA97A50');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50328,'人体的元素组成与元素对人体健康的重要作用',9,2,'A26D87D40DE6192BCF1909654B74C9EB');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50329,'生命活动与六大营养素',9,2,'64A690142AC2ADFA6C55D4FE8C43C364');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50330,'亚硝酸钠、甲醛等化学品的性质与人体健康',9,2,'8939D08AE0823B1488A204F4B09365A2');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50331,'食品、药品与健康食品中的有机营养素',9,2,'8F2CEC0B8F27C169C1F03A26C05446BA');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50332,'均衡营养与健康',9,2,'411D393824B67DB8BAD71A4FB00D5773');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50333,'矿物质与微量元素',9,2,'EECEAB35F5E0B21C53A7C5868D1B08CB');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50334,'加碘盐的检验',9,2,'2CE217D7731633423261C5A17249F180');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50335,'常见中毒途径及预防方法',9,2,'05B9022BB8E1260A26AAD5E8C65560BB');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50336,'烟的危害性及防治',9,2,'823643E2144F9166803244882BA1B655');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50337,'毒品的危害性及预防',9,2,'CA227F3B7209F2B8BA93BAD35C925FA4');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50338,'药品的分类',9,2,'EA5D180498A00AB2C3246396A7052AD2');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50339,'家庭常备药品',9,2,'0478ACFF2DA0846C01408D49FA268E71');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50340,'鉴别淀粉、葡萄糖的方法与蛋白质的性质',9,2,'4CD436BF13296D674D046C80F7E7ECDE');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50341,'微量元素、维生素与健康的关系及摄取方法',9,2,'83FC2CC3304417AC0ABF4A258507EE67');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50342,'保护好我们的环境',9,2,'63F9212841B4B9338FCB63EB72380985');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50343,'三废处理的必要性和一般原则',9,2,'0667F552D10337B4D16F6A154F399AF9');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50344,'常见污染物的来源、危害及治理',9,2,'596525FB85C296245F55562F755DE682');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50345,'生活污水和工业废液的危害',9,2,'2A137E2F08657A148B2F3EDD0924F3C0');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50346,'富营养化污染与含磷洗衣粉的禁用',9,2,'1A01F836B0C2494B9E95700F520F7DF4');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50347,'臭氧空洞和臭氧层保护',9,2,'EC3F02C3F47663F8D884A1C6A8A54FA4');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50348,'合理使用化肥、农药对保护环境的重要意义',9,2,'5D498AA1F7978898EF8E9C4F87572D37');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50349,'化学在环境监测与环境保护中的重要作用',9,2,'8CF2191BDCF8CD1B8C58E20D6502A774');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (50350,'无土栽培I',9,2,'067E1DDD683AC1F3C40F854448EAF2A8');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (107057,'自然界中的氮循环',9,2,'EBE611B005232D6284896A9A0AA6D4EB');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (108416,'科学探究',9,2,'987A11324A278EF679E24102BA30D426');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (108434,'化学变化的基本特征',9,2,'2FA2E93D006CDC0C2CFF0DE945E438B9');
insert  into `knowledge_basic`(`id`,`knowledgeName`,`subjectId`,`pharseId`,`md5`) values (108436,'质量守恒定律',9,2,'441272188E0474D1E23973AAB9A2AF0C');

/*Table structure for table `knowledge_level` */

DROP TABLE IF EXISTS `knowledge_level`;

CREATE TABLE `knowledge_level` (
  `id` int(11) DEFAULT NULL,
  `subjectId` int(11) DEFAULT NULL,
  `pharseId` int(11) DEFAULT NULL,
  `knowledge_level_1` int(11) DEFAULT NULL,
  `knowledge_level_2` int(11) DEFAULT NULL,
  `knowledge_level_3` int(11) DEFAULT NULL,
  `knowledgeId` int(11) DEFAULT NULL,
  `level1` varchar(100) DEFAULT NULL,
  `level2` varchar(100) DEFAULT NULL,
  `level3` varchar(100) DEFAULT NULL,
  UNIQUE KEY `index` (`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`)
) ENGINE=MyISAM DEFAULT CHARSET=utf8;

/*Data for the table `knowledge_level` */

insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9284,9,2,50078,50079,50080,50080,'身边的化学物质','地球周围的空气','空气的成分及各成分的体积分数');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9285,9,2,50078,50079,50081,50081,'身边的化学物质','地球周围的空气','空气组成的测定');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9286,9,2,50078,50079,50082,50082,'身边的化学物质','地球周围的空气','空气对人类生活的重要作用');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9287,9,2,50078,50079,50083,50083,'身边的化学物质','地球周围的空气','空气的污染及其危害');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9288,9,2,50078,50079,50084,50084,'身边的化学物质','地球周围的空气','防治空气污染的措施');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9289,9,2,50078,50079,50085,50085,'身边的化学物质','地球周围的空气','目前环境污染问题');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9290,9,2,50078,50079,50086,50086,'身边的化学物质','地球周围的空气','氧气的物理性质');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9291,9,2,50078,50079,50087,50087,'身边的化学物质','地球周围的空气','氧气的化学性质');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9292,9,2,50078,50079,50088,50088,'身边的化学物质','地球周围的空气','氧气的用途');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9293,9,2,50078,50079,50089,50089,'身边的化学物质','地球周围的空气','氧气与碳、磷、硫、铁等物质的反应现象');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9294,9,2,50078,50079,50090,50090,'身边的化学物质','地球周围的空气','氧气的工业制法');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9295,9,2,50078,50079,50091,50091,'身边的化学物质','地球周围的空气','实验室制取氧气的反应原理');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9296,9,2,50078,50079,50092,50092,'身边的化学物质','地球周围的空气','氧气的制取装置');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9297,9,2,50078,50079,50093,50093,'身边的化学物质','地球周围的空气','氧气的收集方法');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9298,9,2,50078,50079,50094,50094,'身边的化学物质','地球周围的空气','氧气的检验和验满');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9299,9,2,50078,50079,50095,50095,'身边的化学物质','地球周围的空气','制取氧气的操作步骤和注意点');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9300,9,2,50078,50079,50096,50096,'身边的化学物质','地球周围的空气','自然界中的氧循环');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9301,9,2,50078,50079,50097,50097,'身边的化学物质','地球周围的空气','自然界中的碳循环');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9302,9,2,50078,50079,50098,50098,'身边的化学物质','地球周围的空气','催化剂的特点与催化作用');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9303,9,2,50078,50079,50099,50099,'身边的化学物质','地球周围的空气','常见气体的用途');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9304,9,2,50078,50079,50100,50100,'身边的化学物质','地球周围的空气','二氧化碳的实验室制法');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9305,9,2,50078,50079,50101,50101,'身边的化学物质','地球周围的空气','二氧化碳的检验和验满');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9306,9,2,50078,50079,50102,50102,'身边的化学物质','地球周围的空气','制取二氧化碳的操作步骤和注意点');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9307,9,2,50078,50079,50103,50103,'身边的化学物质','地球周围的空气','二氧化碳的工业制法');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9308,9,2,50078,50079,50104,50104,'身边的化学物质','地球周围的空气','二氧化碳的物理性质');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9309,9,2,50078,50079,50105,50105,'身边的化学物质','地球周围的空气','二氧化碳的化学性质');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9310,9,2,50078,50079,50106,50106,'身边的化学物质','地球周围的空气','二氧化碳的用途');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9311,9,2,50078,50079,50107,50107,'身边的化学物质','地球周围的空气','二氧化碳对环境的影响');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9312,9,2,50078,50079,50108,50108,'身边的化学物质','地球周围的空气','一氧化碳的物理性质');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9313,9,2,50078,50079,50109,50109,'身边的化学物质','地球周围的空气','一氧化碳的化学性质');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9314,9,2,50078,50079,50110,50110,'身边的化学物质','地球周围的空气','一氧化碳的毒性');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9315,9,2,50078,50079,50111,50111,'身边的化学物质','地球周围的空气','探究氧气的性质');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9316,9,2,50078,50079,50112,50112,'身边的化学物质','地球周围的空气','探究二氧化碳的性质');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9317,9,2,50078,50079,107057,107057,'身边的化学物质','地球周围的空气','自然界中的氮循环');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9318,9,2,50078,50113,50114,50114,'身边的化学物质','水与常见的溶液','电解水实验');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9319,9,2,50078,50113,50115,50115,'身边的化学物质','水与常见的溶液','水的组成');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9320,9,2,50078,50113,50116,50116,'身边的化学物质','水与常见的溶液','水的性质和应用');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9321,9,2,50078,50113,50117,50117,'身边的化学物质','水与常见的溶液','水的合成与氢气的燃烧');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9322,9,2,50078,50113,50118,50118,'身边的化学物质','水与常见的溶液','水的净化');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9323,9,2,50078,50113,50119,50119,'身边的化学物质','水与常见的溶液','自来水的生产过程与净化方法');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9324,9,2,50078,50113,50120,50120,'身边的化学物质','水与常见的溶液','硬水与软水');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9325,9,2,50078,50113,50121,50121,'身边的化学物质','水与常见的溶液','水资源状况');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9326,9,2,50078,50113,50122,50122,'身边的化学物质','水与常见的溶液','水资源的污染与防治');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9327,9,2,50078,50113,50123,50123,'身边的化学物质','水与常见的溶液','溶解现象与溶解原理');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9328,9,2,50078,50113,50124,50124,'身边的化学物质','水与常见的溶液','常见的溶剂');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9329,9,2,50078,50113,50125,50125,'身边的化学物质','水与常见的溶液','溶液的概念、组成及其特点');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9330,9,2,50078,50113,50126,50126,'身边的化学物质','水与常见的溶液','溶液、溶质和溶剂的相互关系与判断');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9331,9,2,50078,50113,50127,50127,'身边的化学物质','水与常见的溶液','影响溶解快慢的因素');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9332,9,2,50078,50113,50128,50128,'身边的化学物质','水与常见的溶液','溶解时的吸热或放热现象');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9333,9,2,50078,50113,50129,50129,'身边的化学物质','水与常见的溶液','悬浊液、乳浊液的概念及其与溶液的区别');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9334,9,2,50078,50113,50130,50130,'身边的化学物质','水与常见的溶液','乳化现象与乳化作用');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9335,9,2,50078,50113,50131,50131,'身边的化学物质','水与常见的溶液','饱和溶液和不饱和溶液');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9336,9,2,50078,50113,50132,50132,'身边的化学物质','水与常见的溶液','饱和溶液和不饱和溶液相互转变的方法');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9337,9,2,50078,50113,50133,50133,'身边的化学物质','水与常见的溶液','浓溶液、稀溶液跟饱和溶液、不饱和溶液的关系');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9338,9,2,50078,50113,50134,50134,'身边的化学物质','水与常见的溶液','固体溶解度的概念');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9339,9,2,50078,50113,50135,50135,'身边的化学物质','水与常见的溶液','固体溶解度的影响因素');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9340,9,2,50078,50113,50136,50136,'身边的化学物质','水与常见的溶液','固体溶解度曲线及其作用');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9341,9,2,50078,50113,50137,50137,'身边的化学物质','水与常见的溶液','气体溶解度的影响因素');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9342,9,2,50078,50113,50138,50138,'身边的化学物质','水与常见的溶液','晶体和结晶的概念与现象');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9343,9,2,50078,50113,50139,50139,'身边的化学物质','水与常见的溶液','物质的溶解性及影响溶解性的因素');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9344,9,2,50078,50113,50140,50140,'身边的化学物质','水与常见的溶液','溶质的质量分数');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9345,9,2,50078,50113,50141,50141,'身边的化学物质','水与常见的溶液','用水稀释改变浓度的方法');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9346,9,2,50078,50113,50142,50142,'身边的化学物质','水与常见的溶液','溶质的质量分数、溶解性和溶解度的关系');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9347,9,2,50078,50113,50143,50143,'身边的化学物质','水与常见的溶液','有关溶质质量分数的简单计算');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9348,9,2,50078,50113,50144,50144,'身边的化学物质','水与常见的溶液','氢气的制取和检验');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9349,9,2,50078,50145,50146,50146,'身边的化学物质','金属与金属矿物','常见金属的特性及其应用');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9350,9,2,50078,50145,50147,50147,'身边的化学物质','金属与金属矿物','金属的物理性质及用途');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9351,9,2,50078,50145,50148,50148,'身边的化学物质','金属与金属矿物','合金与合金的性质');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9352,9,2,50078,50145,50149,50149,'身边的化学物质','金属与金属矿物','常见的金属和非金属的区分');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9353,9,2,50078,50145,50150,50150,'身边的化学物质','金属与金属矿物','金属的化学性质');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9354,9,2,50078,50145,50151,50151,'身边的化学物质','金属与金属矿物','金属活动性顺序及其应用');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9355,9,2,50078,50145,50152,50152,'身边的化学物质','金属与金属矿物','金属元素的存在及常见的金属矿物');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9356,9,2,50078,50145,50153,50153,'身边的化学物质','金属与金属矿物','探究金属铜的冶炼原理');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9357,9,2,50078,50145,50154,50154,'身边的化学物质','金属与金属矿物','金属材料及其应用');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9358,9,2,50078,50145,50155,50155,'身边的化学物质','金属与金属矿物','一氧化碳还原氧化铁');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9359,9,2,50078,50145,50156,50156,'身边的化学物质','金属与金属矿物','铁的冶炼');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9360,9,2,50078,50145,50157,50157,'身边的化学物质','金属与金属矿物','碳、一氧化碳、氢气还原氧化铜实验');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9361,9,2,50078,50145,50158,50158,'身边的化学物质','金属与金属矿物','常见金属的冶炼方法');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9362,9,2,50078,50145,50159,50159,'身边的化学物质','金属与金属矿物','生铁和钢');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9363,9,2,50078,50145,50160,50160,'身边的化学物质','金属与金属矿物','含杂质物质的化学反应的有关计算');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9364,9,2,50078,50145,50161,50161,'身边的化学物质','金属与金属矿物','金属锈蚀的条件及其防护');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9365,9,2,50078,50145,50162,50162,'身边的化学物质','金属与金属矿物','铁锈的主要成分');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9366,9,2,50078,50145,50163,50163,'身边的化学物质','金属与金属矿物','金属资源的保护');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9367,9,2,50078,50145,50164,50164,'身边的化学物质','金属与金属矿物','废弃金属对环境的污染');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9368,9,2,50078,50145,50165,50165,'身边的化学物质','金属与金属矿物','金属的回收利用及其重要性');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9369,9,2,50078,50145,50166,50166,'身边的化学物质','金属与金属矿物','生石灰的性质与用途');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9370,9,2,50078,50145,50167,50167,'身边的化学物质','金属与金属矿物','碳酸钙、生石灰、熟石灰之间的转化');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9371,9,2,50078,50168,50169,50169,'身边的化学物质','生活中的常见化合物','酸碱指示剂及其性质');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9372,9,2,50078,50168,50170,50170,'身边的化学物质','生活中的常见化合物','酸的物理性质及用途');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9373,9,2,50078,50168,50171,50171,'身边的化学物质','生活中的常见化合物','酸的化学性质');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9374,9,2,50078,50168,50172,50172,'身边的化学物质','生活中的常见化合物','常见碱的特性和用途');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9375,9,2,50078,50168,50173,50173,'身边的化学物质','生活中的常见化合物','碱的化学性质');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9376,9,2,50078,50168,50174,50174,'身边的化学物质','生活中的常见化合物','溶液的导电性及其原理分析');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9377,9,2,50078,50168,50175,50175,'身边的化学物质','生活中的常见化合物','中和反应及其应用');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9378,9,2,50078,50168,50176,50176,'身边的化学物质','生活中的常见化合物','酸碱溶液的稀释');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9379,9,2,50078,50168,50177,50177,'身边的化学物质','生活中的常见化合物','溶液的酸碱性与pH值的关系');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9380,9,2,50078,50168,50178,50178,'身边的化学物质','生活中的常见化合物','溶液的酸碱性测定');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9381,9,2,50078,50168,50179,50179,'身边的化学物质','生活中的常见化合物','酸碱性对生命活动和农作物生长的影响');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9382,9,2,50078,50168,50180,50180,'身边的化学物质','生活中的常见化合物','酸雨的产生、危害及防治');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9383,9,2,50078,50168,50181,50181,'身边的化学物质','生活中的常见化合物','酸碱盐的应用');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9384,9,2,50078,50168,50182,50182,'身边的化学物质','生活中的常见化合物','氯化钠与粗盐提纯');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9385,9,2,50078,50168,50183,50183,'身边的化学物质','生活中的常见化合物','碳酸钠、碳酸氢钠与碳酸钙');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9386,9,2,50078,50168,50184,50184,'身边的化学物质','生活中的常见化合物','常用盐的用途');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9387,9,2,50078,50168,50185,50185,'身边的化学物质','生活中的常见化合物','盐的化学性质');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9388,9,2,50078,50168,50186,50186,'身边的化学物质','生活中的常见化合物','复分解反应及其发生的条件');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9389,9,2,50078,50168,50187,50187,'身边的化学物质','生活中的常见化合物','常见化肥的种类和作用');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9390,9,2,50078,50168,50188,50188,'身边的化学物质','生活中的常见化合物','化肥的简易鉴别');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9391,9,2,50078,50168,50189,50189,'身边的化学物质','生活中的常见化合物','铵态氮肥的检验');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9392,9,2,50078,50168,50190,50190,'身边的化学物质','生活中的常见化合物','施用化肥对环境的影响');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9393,9,2,50078,50168,50191,50191,'身边的化学物质','生活中的常见化合物','醋酸的性质及醋酸的含量测定');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9394,9,2,50078,50168,50192,50192,'身边的化学物质','生活中的常见化合物','海水晒盐的原理和过程');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9395,9,2,50078,50168,50193,50193,'身边的化学物质','生活中的常见化合物','纯碱的制取');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9396,9,2,50078,50168,50194,50194,'身边的化学物质','生活中的常见化合物','酸碱盐的溶解性');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9397,9,2,50078,50168,50195,50195,'身边的化学物质','生活中的常见化合物','离子或物质的共存问题');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9398,9,2,50078,50168,50196,50196,'身边的化学物质','生活中的常见化合物','空气中常见酸碱盐的质量或性质变化及贮存法');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9399,9,2,50078,50168,50197,50197,'身边的化学物质','生活中的常见化合物','根据浓硫酸或烧碱的性质确定所能干燥的气体');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9400,9,2,50078,50168,50198,50198,'身边的化学物质','生活中的常见化合物','酸、碱、盐的鉴别');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9401,9,2,50078,50168,50199,50199,'身边的化学物质','生活中的常见化合物','常见离子的检验方法及现象');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9402,9,2,50200,50201,50202,50202,'物质构成的奥秘','化学物质的多样性','物质的三态及其转化');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9403,9,2,50200,50201,50203,50203,'物质构成的奥秘','化学物质的多样性','物质的简单分类');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9404,9,2,50200,50201,50204,50204,'物质构成的奥秘','化学物质的多样性','从组成上识别氧化物');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9405,9,2,50200,50201,50205,50205,'物质构成的奥秘','化学物质的多样性','纯净物和混合物的概念');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9406,9,2,50200,50201,50206,50206,'物质构成的奥秘','化学物质的多样性','纯净物和混合物的判别');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9407,9,2,50200,50201,50207,50207,'物质构成的奥秘','化学物质的多样性','单质和化合物的概念');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9408,9,2,50200,50201,50208,50208,'物质构成的奥秘','化学物质的多样性','单质和化合物的判别');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9409,9,2,50200,50201,50209,50209,'物质构成的奥秘','化学物质的多样性','氧化物、酸、碱和盐的概念');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9410,9,2,50200,50201,50210,50210,'物质构成的奥秘','化学物质的多样性','常见的氧化物、酸、碱和盐的判别');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9411,9,2,50200,50201,50211,50211,'物质构成的奥秘','化学物质的多样性','有机物的特征、分类及聚合物的特性');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9412,9,2,50200,50201,50212,50212,'物质构成的奥秘','化学物质的多样性','甲烷、乙醇等常见有机物的性质和用途');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9413,9,2,50200,50201,50213,50213,'物质构成的奥秘','化学物质的多样性','有机物与无机物的区别');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9414,9,2,50200,50201,50214,50214,'物质构成的奥秘','化学物质的多样性','物质的多样性及其原因');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9415,9,2,50200,50201,50215,50215,'物质构成的奥秘','化学物质的多样性','物质的鉴别、推断');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9416,9,2,50200,50201,50216,50216,'物质构成的奥秘','化学物质的多样性','物质的相互转化和制备');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9417,9,2,50200,50217,50218,50218,'物质构成的奥秘','微粒构成物质','物质的微粒性');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9418,9,2,50200,50217,50219,50219,'物质构成的奥秘','微粒构成物质','分子、原子、离子、元素与物质之间的关系');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9419,9,2,50200,50217,50220,50220,'物质构成的奥秘','微粒构成物质','微粒观点及模型图的应用');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9420,9,2,50200,50217,50221,50221,'物质构成的奥秘','微粒构成物质','原子的定义与构成');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9421,9,2,50200,50217,50222,50222,'物质构成的奥秘','微粒构成物质','分子和原子的区别和联系');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9422,9,2,50200,50217,50223,50223,'物质构成的奥秘','微粒构成物质','原子和离子的相互转化');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9423,9,2,50200,50217,50224,50224,'物质构成的奥秘','微粒构成物质','核外电子在化学反应中的作用');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9424,9,2,50200,50217,50225,50225,'物质构成的奥秘','微粒构成物质','原子结构示意图与离子结构示意图');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9425,9,2,50200,50217,50226,50226,'物质构成的奥秘','微粒构成物质','利用分子与原子的性质分析和解决问题');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9426,9,2,50200,50217,50227,50227,'物质构成的奥秘','微粒构成物质','分子的定义与分子的特性');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9427,9,2,50200,50217,50228,50228,'物质构成的奥秘','微粒构成物质','原子的有关数量计算');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9428,9,2,50200,50229,50230,50230,'物质构成的奥秘','认识化学元素','元素的概念');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9429,9,2,50200,50229,50231,50231,'物质构成的奥秘','认识化学元素','地壳中元素的分布与含量');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9430,9,2,50200,50229,50232,50232,'物质构成的奥秘','认识化学元素','元素的符号及其意义');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9431,9,2,50200,50229,50233,50233,'物质构成的奥秘','认识化学元素','元素的简单分类');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9432,9,2,50200,50229,50234,50234,'物质构成的奥秘','认识化学元素','元素周期表的特点及其应用');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9433,9,2,50200,50229,50235,50235,'物质构成的奥秘','认识化学元素','碳单质的物理性质及用途');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9434,9,2,50200,50229,50236,50236,'物质构成的奥秘','认识化学元素','碳的化学性质');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9435,9,2,50200,50229,50237,50237,'物质构成的奥秘','认识化学元素','物质的元素组成');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9436,9,2,50200,50229,50238,50238,'物质构成的奥秘','认识化学元素','物质的构成和含量分析');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9437,9,2,50200,50229,50239,50239,'物质构成的奥秘','认识化学元素','碳元素组成的单质');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9438,9,2,50200,50229,50240,50240,'物质构成的奥秘','认识化学元素','氧元素组成的单质');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9439,9,2,50200,50229,50241,50241,'物质构成的奥秘','认识化学元素','元素在化学变化过程中的特点');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9440,9,2,50200,50229,50242,50242,'物质构成的奥秘','认识化学元素','同素异形体和同素异形现象');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9441,9,2,50200,50243,50244,50244,'物质构成的奥秘','物质组成的表示','化学式的书写及意义');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9442,9,2,50200,50243,50245,50245,'物质构成的奥秘','物质组成的表示','化合价的概念');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9443,9,2,50200,50243,50246,50246,'物质构成的奥秘','物质组成的表示','常见元素与常见原子团的化合价');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9444,9,2,50200,50243,50247,50247,'物质构成的奥秘','物质组成的表示','化合价与离子表示方法上的异同点');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9445,9,2,50200,50243,50248,50248,'物质构成的奥秘','物质组成的表示','化合价规律和原则');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9446,9,2,50200,50243,50249,50249,'物质构成的奥秘','物质组成的表示','有关元素化合价的计算');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9447,9,2,50200,50243,50250,50250,'物质构成的奥秘','物质组成的表示','相对原子质量的概念及其计算方法');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9448,9,2,50200,50243,50251,50251,'物质构成的奥秘','物质组成的表示','相对分子质量的概念及其计算');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9449,9,2,50200,50243,50252,50252,'物质构成的奥秘','物质组成的表示','元素质量比的计算');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9450,9,2,50200,50243,50253,50253,'物质构成的奥秘','物质组成的表示','元素的质量分数计算');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9451,9,2,50200,50243,50254,50254,'物质构成的奥秘','物质组成的表示','化合物中某元素的质量计算');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9452,9,2,50200,50243,50255,50255,'物质构成的奥秘','物质组成的表示','混合物中某元素的质量计算');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9453,9,2,50200,50243,50256,50256,'物质构成的奥秘','物质组成的表示','物质组成的综合计算');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9454,9,2,50200,50243,50257,50257,'物质构成的奥秘','物质组成的表示','标签上标示的物质成分及其含量');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9455,9,2,50200,50243,50258,50258,'物质构成的奥秘','物质组成的表示','化学符号及其周围数字的意义');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9456,9,2,50200,50243,50259,50259,'物质构成的奥秘','物质组成的表示','有关化学式的计算和推断');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9457,9,2,50260,50270,50271,50271,'物质的化学变化','认识几种化学反应','化合反应及其应用');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9458,9,2,50260,50270,50272,50272,'物质的化学变化','认识几种化学反应','分解反应及其应用');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9459,9,2,50260,50270,50273,50273,'物质的化学变化','认识几种化学反应','置换反应及其应用');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9460,9,2,50260,50270,50274,50274,'物质的化学变化','认识几种化学反应','复分解反应及其应用');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9461,9,2,50260,50270,50275,50275,'物质的化学变化','认识几种化学反应','氧化反应');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9462,9,2,50260,50270,50276,50276,'物质的化学变化','认识几种化学反应','复分解反应的条件与实质');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9463,9,2,50260,50270,50277,50277,'物质的化学变化','认识几种化学反应','金属活动性顺序与置换反应');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9464,9,2,50260,50270,50278,50278,'物质的化学变化','认识几种化学反应','酸、碱性废水的处理');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9465,9,2,50260,50270,50279,50279,'物质的化学变化','认识几种化学反应','光合作用与呼吸作用');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9466,9,2,50260,50270,50280,50280,'物质的化学变化','认识几种化学反应','反应类型的判定');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9467,9,2,50260,50270,50281,50281,'物质的化学变化','认识几种化学反应','还原反应');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9468,9,2,50260,108434,50261,50261,'物质的化学变化','化学变化的基本特征','化学变化的基本特征I');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9469,9,2,50260,108434,50262,50262,'物质的化学变化','化学变化的基本特征','化学变化的基本特征II');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9470,9,2,50260,108434,50263,50263,'物质的化学变化','化学变化的基本特征','物理变化的特点');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9471,9,2,50260,108434,50264,50264,'物质的化学变化','化学变化的基本特征','化学变化和物理变化的判别');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9472,9,2,50260,108434,50265,50265,'物质的化学变化','化学变化的基本特征','化学性质与物理性质的差别及应用');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9473,9,2,50260,108434,50266,50266,'物质的化学变化','化学变化的基本特征','物质发生化学变化时的能量变化');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9474,9,2,50260,108434,50267,50267,'物质的化学变化','化学变化的基本特征','化学的研究对象及其发展');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9475,9,2,50260,108434,50268,50268,'物质的化学变化','化学变化的基本特征','反应现象和本质的联系');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9476,9,2,50260,108434,50269,50269,'物质的化学变化','化学变化的基本特征','化学反应的实质');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9477,9,2,50260,108434,108434,108434,'物质的化学变化','化学变化的基本特征','化学变化的基本特征');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9478,9,2,50260,108436,50283,50283,'物质的化学变化','质量守恒定律','质量守恒定律及其应用');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9479,9,2,50260,108436,50284,50284,'物质的化学变化','质量守恒定律','化学方程式的概念、读法和含义');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9480,9,2,50260,108436,50285,50285,'物质的化学变化','质量守恒定律','化学方程式的配平');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9481,9,2,50260,108436,50286,50286,'物质的化学变化','质量守恒定律','常见化学反应中的质量关系');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9482,9,2,50260,108436,50287,50287,'物质的化学变化','质量守恒定律','书写化学方程式、文字表达式、电离方程式');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9483,9,2,50260,108436,50288,50288,'物质的化学变化','质量守恒定律','根据化学反应方程式的计算');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9484,9,2,50289,50290,50291,50291,'化学与社会发展','化学与能源、资源利用','燃烧与燃烧的条件');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9485,9,2,50289,50290,50292,50292,'化学与社会发展','化学与能源、资源利用','完全燃烧与不完全燃烧');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9486,9,2,50289,50290,50293,50293,'化学与社会发展','化学与能源、资源利用','常用燃料的使用与其对环境的影响');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9487,9,2,50289,50290,50294,50294,'化学与社会发展','化学与能源、资源利用','灭火的原理和方法');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9488,9,2,50289,50290,50295,50295,'化学与社会发展','化学与能源、资源利用','防范爆炸的措施');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9489,9,2,50289,50290,50296,50296,'化学与社会发展','化学与能源、资源利用','燃烧、爆炸、缓慢氧化与自燃');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9490,9,2,50289,50290,50297,50297,'化学与社会发展','化学与能源、资源利用','化石燃料及其综合利用');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9491,9,2,50289,50290,50298,50298,'化学与社会发展','化学与能源、资源利用','石油的组成');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9492,9,2,50289,50290,50299,50299,'化学与社会发展','化学与能源、资源利用','石油加工的产物');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9493,9,2,50289,50290,50300,50300,'化学与社会发展','化学与能源、资源利用','原油泄漏对生态环境的危害及其处理');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9494,9,2,50289,50290,50301,50301,'化学与社会发展','化学与能源、资源利用','保护水资源和节约用水');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9495,9,2,50289,50290,50302,50302,'化学与社会发展','化学与能源、资源利用','海洋中的资源');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9496,9,2,50289,50290,50303,50303,'化学与社会发展','化学与能源、资源利用','对海洋资源的合理开发与利用');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9497,9,2,50289,50290,50304,50304,'化学与社会发展','化学与能源、资源利用','资源综合利用和新能源开发');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9498,9,2,50289,50290,50305,50305,'化学与社会发展','化学与能源、资源利用','燃烧和爆炸实验');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9499,9,2,50289,50290,50306,50306,'化学与社会发展','化学与能源、资源利用','易燃物和易爆物安全知识');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9500,9,2,50289,50290,50307,50307,'化学与社会发展','化学与能源、资源利用','几种常见的与化学有关的图标');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9501,9,2,50289,50290,50308,50308,'化学与社会发展','化学与能源、资源利用','几种常用的灭火器');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9502,9,2,50289,50290,50309,50309,'化学与社会发展','化学与能源、资源利用','氢气的物理性质');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9503,9,2,50289,50290,50310,50310,'化学与社会发展','化学与能源、资源利用','氢气的化学性质与燃烧实验');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9504,9,2,50289,50290,50311,50311,'化学与社会发展','化学与能源、资源利用','氢气的爆鸣实验');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9505,9,2,50289,50290,50312,50312,'化学与社会发展','化学与能源、资源利用','氢气、一氧化碳、甲烷等可燃气体的验纯');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9506,9,2,50289,50290,50313,50313,'化学与社会发展','化学与能源、资源利用','氢气的用途和氢能的优缺点');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9507,9,2,50289,50290,50314,50314,'化学与社会发展','化学与能源、资源利用','常见能源的种类、能源的分类');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9508,9,2,50289,50315,50316,50316,'化学与社会发展','常见的化学合成材料','塑料及其应用');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9509,9,2,50289,50315,50317,50317,'化学与社会发展','常见的化学合成材料','塑料制品的回收、再生与降解');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9510,9,2,50289,50315,50318,50318,'化学与社会发展','常见的化学合成材料','合成橡胶及其应用');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9511,9,2,50289,50315,50319,50319,'化学与社会发展','常见的化学合成材料','合成纤维及其应用');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9512,9,2,50289,50315,50320,50320,'化学与社会发展','常见的化学合成材料','塑料制品使用的安全');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9513,9,2,50289,50315,50321,50321,'化学与社会发展','常见的化学合成材料','白色污染与防治');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9514,9,2,50289,50315,50322,50322,'化学与社会发展','常见的化学合成材料','合成材料的使用及其对人和环境的影响');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9515,9,2,50289,50315,50323,50323,'化学与社会发展','常见的化学合成材料','新材料的开发与社会发展的密切关系');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9516,9,2,50289,50315,50324,50324,'化学与社会发展','常见的化学合成材料','有机高分子材料的分类及鉴别');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9517,9,2,50289,50315,50325,50325,'化学与社会发展','常见的化学合成材料','复合材料、纳米材料');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9518,9,2,50289,50315,50326,50326,'化学与社会发展','常见的化学合成材料','棉纤维、羊毛纤维和合成纤维的鉴别');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9519,9,2,50289,50327,50328,50328,'化学与社会发展','化学物质与健康','人体的元素组成与元素对人体健康的重要作用');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9520,9,2,50289,50327,50329,50329,'化学与社会发展','化学物质与健康','生命活动与六大营养素');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9521,9,2,50289,50327,50330,50330,'化学与社会发展','化学物质与健康','亚硝酸钠、甲醛等化学品的性质与人体健康');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9522,9,2,50289,50327,50331,50331,'化学与社会发展','化学物质与健康','食品、药品与健康食品中的有机营养素');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9523,9,2,50289,50327,50332,50332,'化学与社会发展','化学物质与健康','均衡营养与健康');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9524,9,2,50289,50327,50333,50333,'化学与社会发展','化学物质与健康','矿物质与微量元素');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9525,9,2,50289,50327,50334,50334,'化学与社会发展','化学物质与健康','加碘盐的检验');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9526,9,2,50289,50327,50335,50335,'化学与社会发展','化学物质与健康','常见中毒途径及预防方法');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9527,9,2,50289,50327,50336,50336,'化学与社会发展','化学物质与健康','烟的危害性及防治');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9528,9,2,50289,50327,50337,50337,'化学与社会发展','化学物质与健康','毒品的危害性及预防');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9529,9,2,50289,50327,50338,50338,'化学与社会发展','化学物质与健康','药品的分类');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9530,9,2,50289,50327,50339,50339,'化学与社会发展','化学物质与健康','家庭常备药品');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9531,9,2,50289,50327,50340,50340,'化学与社会发展','化学物质与健康','鉴别淀粉、葡萄糖的方法与蛋白质的性质');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9532,9,2,50289,50327,50341,50341,'化学与社会发展','化学物质与健康','微量元素、维生素与健康的关系及摄取方法');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9533,9,2,50289,50342,0,0,'化学与社会发展','保护好我们的环境','无土栽培');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9534,9,2,50289,50342,50343,50343,'化学与社会发展','保护好我们的环境','三废处理的必要性和一般原则');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9535,9,2,50289,50342,50344,50344,'化学与社会发展','保护好我们的环境','常见污染物的来源、危害及治理');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9536,9,2,50289,50342,50345,50345,'化学与社会发展','保护好我们的环境','生活污水和工业废液的危害');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9537,9,2,50289,50342,50346,50346,'化学与社会发展','保护好我们的环境','富营养化污染与含磷洗衣粉的禁用');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9538,9,2,50289,50342,50347,50347,'化学与社会发展','保护好我们的环境','臭氧空洞和臭氧层保护');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9539,9,2,50289,50342,50348,50348,'化学与社会发展','保护好我们的环境','合理使用化肥、农药对保护环境的重要意义');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9540,9,2,50289,50342,50349,50349,'化学与社会发展','保护好我们的环境','化学在环境监测与环境保护中的重要作用');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9541,9,2,50289,50342,50350,50350,'化学与社会发展','保护好我们的环境','无土栽培I');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9542,9,2,108416,50002,0,0,'科学探究','对科学探究的理解','科学探究的基本方法');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9543,9,2,108416,50002,0,0,'科学探究','对科学探究的理解','科学探究的基本环节');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9544,9,2,108416,50002,50003,50003,'科学探究','对科学探究的理解','科学探究的意义');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9545,9,2,108416,50002,50004,50004,'科学探究','对科学探究的理解','猜想与事实验证');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9546,9,2,108416,50002,50005,50005,'科学探究','对科学探究的理解','科学探究的基本方法I');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9547,9,2,108416,50002,50006,50006,'科学探究','对科学探究的理解','科学探究的基本环节I');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9548,9,2,108416,50007,50008,50008,'科学探究','科学探究能力','测定空气里氧气含量的探究');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9549,9,2,108416,50007,50009,50009,'科学探究','科学探究能力','制取气体的反应原理的探究');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9550,9,2,108416,50007,50010,50010,'科学探究','科学探究能力','质量守恒定律的实验探究');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9551,9,2,108416,50007,50011,50011,'科学探究','科学探究能力','金属活动性的探究');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9552,9,2,108416,50007,50012,50012,'科学探究','科学探究能力','探究酸碱的主要性质');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9553,9,2,108416,50007,50013,50013,'科学探究','科学探究能力','药品是否变质的探究');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9554,9,2,108416,50007,50014,50014,'科学探究','科学探究能力','缺失标签的药品成分的探究');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9555,9,2,108416,50007,50015,50015,'科学探究','科学探究能力','实验探究物质的性质或变化规律');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9556,9,2,108416,50007,50016,50016,'科学探究','科学探究能力','探究金属锈蚀的条件');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9557,9,2,108416,50007,50017,50017,'科学探究','科学探究能力','燃烧的条件与灭火原理探究');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9558,9,2,108416,50007,50018,50018,'科学探究','科学探究能力','影响化学反应速率的因素探究');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9559,9,2,108416,50007,50019,50019,'科学探究','科学探究能力','实验探究物质变化的条件和影响物质变化的因素');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9560,9,2,108416,50007,50020,50020,'科学探究','科学探究能力','食品干燥剂、保鲜剂和真空包装的成分探究');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9561,9,2,108416,50007,50021,50021,'科学探究','科学探究能力','化肥有效成分含量的探究');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9562,9,2,108416,50007,50022,50022,'科学探究','科学探究能力','味精中食盐含量的探究');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9563,9,2,108416,50007,50023,50023,'科学探究','科学探究能力','实验探究物质的组成成分以及含量');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9564,9,2,108416,50007,50024,50024,'科学探究','科学探究能力','物质除杂或净化的探究');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9565,9,2,108416,50007,50025,50025,'科学探究','科学探究能力','气体制取装置的探究');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9566,9,2,108416,50026,50027,50027,'科学探究','实验分析与处理能力','实验数据处理或者误差分析的探究');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9567,9,2,108416,50026,50028,50028,'科学探究','实验分析与处理能力','实验操作注意事项的探究');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9568,9,2,108416,50026,50029,50029,'科学探究','实验分析与处理能力','实验步骤的探究');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9569,9,2,108416,50026,50030,50030,'科学探究','实验分析与处理能力','化学实验方案设计与评价');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9570,9,2,108416,50026,50031,50031,'科学探究','实验分析与处理能力','化学竞赛');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9571,9,2,108416,50032,50033,50033,'科学探究','基本的实验技能','用于加热的仪器');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9572,9,2,108416,50032,50034,50034,'科学探究','基本的实验技能','测量容器-量筒');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9573,9,2,108416,50032,50035,50035,'科学探究','基本的实验技能','称量器-托盘天平');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9574,9,2,108416,50032,50036,50036,'科学探究','基本的实验技能','加热器皿-酒精灯');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9575,9,2,108416,50032,50037,50037,'科学探究','基本的实验技能','挟持器-铁夹、试管夹、坩埚钳');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9576,9,2,108416,50032,50038,50038,'科学探究','基本的实验技能','分离物质的仪器');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9577,9,2,108416,50032,50039,50039,'科学探究','基本的实验技能','量气装置');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9578,9,2,108416,50032,50040,50040,'科学探究','基本的实验技能','固体药品的取用');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9579,9,2,108416,50032,50041,50041,'科学探究','基本的实验技能','液体药品的取用');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9580,9,2,108416,50032,50042,50042,'科学探究','基本的实验技能','给试管里的固体加热');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9581,9,2,108416,50032,50043,50043,'科学探究','基本的实验技能','给试管里的液体加热');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9582,9,2,108416,50032,50044,50044,'科学探究','基本的实验技能','物质的溶解');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9583,9,2,108416,50032,50045,50045,'科学探究','基本的实验技能','浓硫酸的性质及浓硫酸的稀释');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9584,9,2,108416,50032,50046,50046,'科学探究','基本的实验技能','一定溶质质量分数的溶液的配制');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9585,9,2,108416,50032,50047,50047,'科学探究','基本的实验技能','混合物的分离方法');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9586,9,2,108416,50032,50048,50048,'科学探究','基本的实验技能','过滤的原理、方法及其应用');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9587,9,2,108416,50032,50049,50049,'科学探究','基本的实验技能','结晶的原理、方法及其应用');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9588,9,2,108416,50032,50050,50050,'科学探究','基本的实验技能','蒸发与蒸馏操作');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9589,9,2,108416,50032,50051,50051,'科学探究','基本的实验技能','仪器的装配或连接');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9590,9,2,108416,50032,50052,50052,'科学探究','基本的实验技能','检查装置的气密性');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9591,9,2,108416,50032,50053,50053,'科学探究','基本的实验技能','玻璃仪器的洗涤');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9592,9,2,108416,50032,50054,50054,'科学探究','基本的实验技能','常见的意外事故的处理方法');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9593,9,2,108416,50032,50055,50055,'科学探究','基本的实验技能','实验室制取气体的思路');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9594,9,2,108416,50032,50056,50056,'科学探究','基本的实验技能','常用气体的发生装置和收集装置与选取方法');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9595,9,2,108416,50032,50057,50057,'科学探究','基本的实验技能','常用气体的收集方法');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9596,9,2,108416,50032,50058,50058,'科学探究','基本的实验技能','常见气体的检验与除杂方法');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9597,9,2,108416,50032,50059,50059,'科学探究','基本的实验技能','气体的净化（除杂）');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9598,9,2,108416,50032,50060,50060,'科学探究','基本的实验技能','气体的干燥（除水）');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9599,9,2,108416,50032,50061,50061,'科学探究','基本的实验技能','酸碱指示剂的使用');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9600,9,2,108416,50032,50062,50062,'科学探究','基本的实验技能','溶液的酸碱度测定');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9601,9,2,108416,50032,50063,50063,'科学探究','基本的实验技能','证明盐酸和可溶性盐酸盐');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9602,9,2,108416,50032,50064,50064,'科学探究','基本的实验技能','证明硫酸和可溶性硫酸盐');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9603,9,2,108416,50032,50065,50065,'科学探究','基本的实验技能','证明碳酸盐');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9604,9,2,108416,50032,50066,50066,'科学探究','基本的实验技能','证明铵盐');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9605,9,2,108416,50032,50067,50067,'科学探究','基本的实验技能','常用仪器的名称和选用');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9606,9,2,108416,50068,0,0,'科学探究','化学的基本常识','绿色化学');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9607,9,2,108416,50068,50069,50069,'科学探究','化学的基本常识','化学的历史发展过程');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9608,9,2,108416,50068,50070,50070,'科学探究','化学的基本常识','化学的研究领域');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9609,9,2,108416,50068,50071,50071,'科学探究','化学的基本常识','化学的用途');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9610,9,2,108416,50068,50072,50072,'科学探究','化学的基本常识','绿色化学II');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9611,9,2,108416,50068,50073,50073,'科学探究','化学的基本常识','蜡烛燃烧实验');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9612,9,2,108416,50068,50074,50074,'科学探究','化学的基本常识','吸入空气与呼出气体的比较');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9613,9,2,108416,50068,50075,50075,'科学探究','化学的基本常识','学习化学的重要途径及学习方法');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9614,9,2,108416,50068,50076,50076,'科学探究','化学的基本常识','有关化学之最');
insert  into `knowledge_level`(`id`,`subjectId`,`pharseId`,`knowledge_level_1`,`knowledge_level_2`,`knowledge_level_3`,`knowledgeId`,`level1`,`level2`,`level3`) values (9615,9,2,108416,50068,50077,50077,'科学探究','化学的基本常识','化学常识');

/*Table structure for table `pharse` */

DROP TABLE IF EXISTS `pharse`;

CREATE TABLE `pharse` (
  `pharseId` int(11) NOT NULL,
  `pharseName` varchar(255) DEFAULT NULL,
  PRIMARY KEY (`pharseId`)
) ENGINE=MyISAM DEFAULT CHARSET=utf8;

/*Data for the table `pharse` */

insert  into `pharse`(`pharseId`,`pharseName`) values (1,'小学');
insert  into `pharse`(`pharseId`,`pharseName`) values (2,'初中');
insert  into `pharse`(`pharseId`,`pharseName`) values (3,'高中');

/*Table structure for table `pharsegrade` */

DROP TABLE IF EXISTS `pharsegrade`;

CREATE TABLE `pharsegrade` (
  `id` int(10) unsigned NOT NULL AUTO_INCREMENT,
  `pharseId` int(11) DEFAULT NULL,
  `gradeId` varchar(255) DEFAULT NULL,
  PRIMARY KEY (`id`)
) ENGINE=MyISAM AUTO_INCREMENT=60 DEFAULT CHARSET=utf8;

/*Data for the table `pharsegrade` */

insert  into `pharsegrade`(`id`,`pharseId`,`gradeId`) values (1,1,'151');
insert  into `pharsegrade`(`id`,`pharseId`,`gradeId`) values (2,1,'152');
insert  into `pharsegrade`(`id`,`pharseId`,`gradeId`) values (3,1,'161');
insert  into `pharsegrade`(`id`,`pharseId`,`gradeId`) values (4,1,'162');
insert  into `pharsegrade`(`id`,`pharseId`,`gradeId`) values (5,2,'201');
insert  into `pharsegrade`(`id`,`pharseId`,`gradeId`) values (6,2,'202');
insert  into `pharsegrade`(`id`,`pharseId`,`gradeId`) values (7,2,'301');
insert  into `pharsegrade`(`id`,`pharseId`,`gradeId`) values (8,2,'302');
insert  into `pharsegrade`(`id`,`pharseId`,`gradeId`) values (9,2,'401');
insert  into `pharsegrade`(`id`,`pharseId`,`gradeId`) values (10,2,'402');
insert  into `pharsegrade`(`id`,`pharseId`,`gradeId`) values (11,3,'10100');
insert  into `pharsegrade`(`id`,`pharseId`,`gradeId`) values (12,3,'10200');
insert  into `pharsegrade`(`id`,`pharseId`,`gradeId`) values (13,3,'10300');
insert  into `pharsegrade`(`id`,`pharseId`,`gradeId`) values (14,3,'10400');
insert  into `pharsegrade`(`id`,`pharseId`,`gradeId`) values (15,3,'10500');
insert  into `pharsegrade`(`id`,`pharseId`,`gradeId`) values (16,3,'20101');
insert  into `pharsegrade`(`id`,`pharseId`,`gradeId`) values (17,3,'20102');
insert  into `pharsegrade`(`id`,`pharseId`,`gradeId`) values (18,3,'20201');
insert  into `pharsegrade`(`id`,`pharseId`,`gradeId`) values (19,3,'20202');
insert  into `pharsegrade`(`id`,`pharseId`,`gradeId`) values (20,3,'20203');
insert  into `pharsegrade`(`id`,`pharseId`,`gradeId`) values (21,3,'20301');
insert  into `pharsegrade`(`id`,`pharseId`,`gradeId`) values (22,3,'20302');
insert  into `pharsegrade`(`id`,`pharseId`,`gradeId`) values (23,3,'20303');
insert  into `pharsegrade`(`id`,`pharseId`,`gradeId`) values (24,3,'20304');
insert  into `pharsegrade`(`id`,`pharseId`,`gradeId`) values (25,3,'20305');
insert  into `pharsegrade`(`id`,`pharseId`,`gradeId`) values (26,3,'20401');
insert  into `pharsegrade`(`id`,`pharseId`,`gradeId`) values (27,3,'20402');
insert  into `pharsegrade`(`id`,`pharseId`,`gradeId`) values (28,3,'20403');
insert  into `pharsegrade`(`id`,`pharseId`,`gradeId`) values (29,3,'20404');
insert  into `pharsegrade`(`id`,`pharseId`,`gradeId`) values (30,3,'20405');
insert  into `pharsegrade`(`id`,`pharseId`,`gradeId`) values (31,3,'20406');
insert  into `pharsegrade`(`id`,`pharseId`,`gradeId`) values (32,3,'20407');
insert  into `pharsegrade`(`id`,`pharseId`,`gradeId`) values (33,1,'110');
insert  into `pharsegrade`(`id`,`pharseId`,`gradeId`) values (34,1,'111');
insert  into `pharsegrade`(`id`,`pharseId`,`gradeId`) values (35,1,'112');
insert  into `pharsegrade`(`id`,`pharseId`,`gradeId`) values (36,1,'120');
insert  into `pharsegrade`(`id`,`pharseId`,`gradeId`) values (37,1,'121');
insert  into `pharsegrade`(`id`,`pharseId`,`gradeId`) values (38,1,'122');
insert  into `pharsegrade`(`id`,`pharseId`,`gradeId`) values (39,1,'130');
insert  into `pharsegrade`(`id`,`pharseId`,`gradeId`) values (40,1,'131');
insert  into `pharsegrade`(`id`,`pharseId`,`gradeId`) values (41,1,'132');
insert  into `pharsegrade`(`id`,`pharseId`,`gradeId`) values (42,1,'140');
insert  into `pharsegrade`(`id`,`pharseId`,`gradeId`) values (43,1,'141');
insert  into `pharsegrade`(`id`,`pharseId`,`gradeId`) values (44,1,'142');
insert  into `pharsegrade`(`id`,`pharseId`,`gradeId`) values (45,3,'20100');
insert  into `pharsegrade`(`id`,`pharseId`,`gradeId`) values (46,3,'20200');
insert  into `pharsegrade`(`id`,`pharseId`,`gradeId`) values (47,3,'20300');
insert  into `pharsegrade`(`id`,`pharseId`,`gradeId`) values (48,3,'20400');
insert  into `pharsegrade`(`id`,`pharseId`,`gradeId`) values (49,3,'20500');
insert  into `pharsegrade`(`id`,`pharseId`,`gradeId`) values (50,3,'20600');
insert  into `pharsegrade`(`id`,`pharseId`,`gradeId`) values (51,3,'20700');
insert  into `pharsegrade`(`id`,`pharseId`,`gradeId`) values (52,1,'150');
insert  into `pharsegrade`(`id`,`pharseId`,`gradeId`) values (53,1,'160');
insert  into `pharsegrade`(`id`,`pharseId`,`gradeId`) values (54,2,'200');
insert  into `pharsegrade`(`id`,`pharseId`,`gradeId`) values (55,2,'300');
insert  into `pharsegrade`(`id`,`pharseId`,`gradeId`) values (56,2,'400');
insert  into `pharsegrade`(`id`,`pharseId`,`gradeId`) values (57,3,'500');
insert  into `pharsegrade`(`id`,`pharseId`,`gradeId`) values (58,3,'600');
insert  into `pharsegrade`(`id`,`pharseId`,`gradeId`) values (59,3,'700');

/*Table structure for table `question_knowledge_basic_id` */

DROP TABLE IF EXISTS `question_knowledge_basic_id`;

CREATE TABLE `question_knowledge_basic_id` (
  `id` int(9) NOT NULL AUTO_INCREMENT,
  `knowledge_basic_id` int(11) DEFAULT NULL COMMENT '基本知识点表ID',
  `question_id` int(9) DEFAULT NULL COMMENT '试题Id',
  PRIMARY KEY (`id`),
  UNIQUE KEY `index_questionId_kid` (`question_id`,`knowledge_basic_id`),
  KEY `index_knowledgeId` (`knowledge_basic_id`),
  KEY `index_questionId` (`question_id`)
) ENGINE=MyISAM AUTO_INCREMENT=33231035 DEFAULT CHARSET=utf8;

/*Data for the table `question_knowledge_basic_id` */

insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666809,50195,1839824);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666813,50215,1839826);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666814,50287,1839826);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666815,50041,1839828);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666816,50048,1839828);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666817,50050,1839828);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666818,50128,1839827);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666819,50130,1839827);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666820,50147,1839827);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666821,50150,1839827);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666828,50148,1839830);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666829,50147,1839830);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666830,50150,1839830);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666831,50161,1839830);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666832,50283,1839833);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666833,50114,1839832);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666834,50120,1839832);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666835,50133,1839832);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666836,50137,1839832);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666845,50019,1839834);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666846,50098,1839834);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666850,50087,1839837);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666852,50104,1839837);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666854,50105,1839837);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666856,50114,1839837);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666857,50287,1839837);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666867,50114,1839838);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666868,50087,1839838);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666869,50114,1839839);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666870,50060,1839839);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666871,50081,1839839);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666872,50173,1839839);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666873,50227,1839839);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666874,50097,1839840);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666875,50083,1839840);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666876,50107,1839840);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666877,50119,1839840);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666878,50120,1839840);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666879,50287,1839840);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666883,50185,1839841);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666884,50130,1839841);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666885,50150,1839841);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666886,50171,1839841);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666887,50210,1839841);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666888,50283,1839843);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666889,50253,1839843);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666890,50034,1839844);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666891,50045,1839844);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666892,50058,1839844);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666893,50087,1839844);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666894,50283,1839845);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666895,50052,1839846);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666896,50051,1839846);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666897,50258,1839848);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666898,50293,1839847);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666901,50195,1839850);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666902,50177,1839850);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666906,50185,1839853);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666907,50125,1839851);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666908,50258,1839854);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666909,50010,1839852);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666910,50131,1839851);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666911,50173,1839854);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666916,50023,1839856);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666917,50058,1839856);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666918,50161,1839856);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666919,50185,1839856);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666920,50287,1839856);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666922,50122,1839858);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666923,50099,1839858);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666924,50235,1839858);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666927,50195,1839857);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666929,50128,1839861);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666930,50266,1839861);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666931,50080,1839862);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666932,50099,1839862);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666933,50198,1839863);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666934,50058,1839863);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666935,50136,1839864);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666936,50049,1839864);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666937,50132,1839864);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666938,50283,1839866);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666939,50195,1839865);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666940,50169,1839865);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666941,50136,1839867);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666942,50138,1839867);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666943,50142,1839867);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666949,50111,1839869);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666950,50287,1839869);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666951,50291,1839869);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666952,50220,1839871);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666953,50023,1839870);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666954,50288,1839872);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666955,50204,1839871);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666956,50058,1839870);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666957,50143,1839872);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666958,50185,1839870);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666959,50287,1839870);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666960,50116,1839873);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666961,50206,1839873);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666962,50227,1839873);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666963,50100,1839875);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666964,50058,1839875);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666965,50104,1839875);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666966,50105,1839875);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666967,50287,1839875);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666968,50222,1839877);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666969,50219,1839877);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666970,50221,1839877);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666971,50122,1839874);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666972,50114,1839874);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666973,50120,1839874);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666974,50206,1839874);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666975,50230,1839874);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666976,50288,1839876);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666977,50143,1839876);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666978,50136,1839878);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666979,50142,1839878);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666980,50169,1839879);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666981,50088,1839879);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666982,50106,1839879);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666983,50128,1839879);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666984,50147,1839879);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666985,50235,1839879);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666986,50329,1839879);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666987,50019,1839881);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666988,50094,1839881);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666989,50098,1839881);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666990,50171,1839881);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666991,50287,1839881);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666992,50024,1839882);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666993,50244,1839880);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666994,50058,1839882);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666995,50185,1839882);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666996,50288,1839883);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666997,50143,1839883);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666998,50150,1839883);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2666999,50185,1839884);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667000,50138,1839884);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667001,50150,1839884);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667002,50177,1839884);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667003,50215,1839885);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667004,50280,1839885);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667005,50287,1839885);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667006,50023,1839886);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667007,50058,1839886);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667008,50265,1839886);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667009,50234,1839887);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667010,50287,1839887);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667019,50023,1839889);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667026,50036,1839891);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667028,50062,1839891);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667029,50312,1839891);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667032,50128,1839893);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667033,50138,1839893);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667034,50100,1839894);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667035,50091,1839894);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667036,50101,1839894);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667037,50287,1839894);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667038,50215,1839895);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667039,50210,1839895);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667040,50287,1839895);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667041,50159,1839897);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667042,50166,1839897);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667043,50184,1839897);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667044,50166,1839898);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667045,50287,1839898);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667046,50175,1839896);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667047,50177,1839896);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667048,50291,1839899);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667049,50100,1839900);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667050,50098,1839900);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667051,50104,1839900);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667052,50105,1839900);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667053,50287,1839900);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667059,50171,1839902);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667060,50258,1839905);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667062,50170,1839902);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667064,50223,1839902);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667065,50227,1839902);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667066,50283,1839906);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667067,50244,1839907);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667068,50252,1839907);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667069,50253,1839907);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667070,50173,1839908);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667071,50140,1839908);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667079,50244,1839910);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667080,50264,1839912);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667081,50204,1839910);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667082,50251,1839910);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667083,50252,1839910);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667084,50253,1839910);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667085,50036,1839911);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667086,50034,1839911);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667087,50035,1839911);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667088,50043,1839911);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667089,50137,1839915);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667090,50132,1839915);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667091,50135,1839915);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667092,50138,1839915);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667094,50215,1839914);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667096,50058,1839914);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667097,50173,1839914);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667098,50326,1839914);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667099,50119,1839917);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667100,50048,1839917);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667101,50226,1839917);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667102,50230,1839917);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667103,50287,1839917);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667104,50294,1839917);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667105,50297,1839917);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667106,50013,1839916);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667107,50173,1839916);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667108,50185,1839916);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667109,50287,1839916);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667110,50048,1839918);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667111,50100,1839919);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667112,50291,1839921);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667113,50058,1839919);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667114,50173,1839921);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667115,50092,1839919);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667116,50287,1839919);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667117,50265,1839922);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667118,50175,1839920);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667120,50114,1839920);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667121,50185,1839920);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667128,50015,1839925);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667129,50098,1839925);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667130,50244,1839926);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667131,50287,1839926);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667132,50046,1839927);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667138,50092,1839929);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667139,50050,1839929);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667140,50060,1839929);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667141,50093,1839929);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667142,50098,1839929);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667146,50258,1839930);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667147,50244,1839930);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667148,50136,1839932);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667149,50049,1839932);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667150,50142,1839932);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667151,50230,1839933);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667152,50015,1839935);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667153,50058,1839935);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667154,50173,1839935);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667155,50287,1839935);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667156,50046,1839934);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667157,50143,1839937);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667158,50226,1839936);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667163,50225,1839940);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667165,50312,1839942);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667166,50045,1839942);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667167,50050,1839942);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667168,50264,1839943);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667169,50258,1839941);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667172,50008,1839945);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667173,50010,1839945);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667176,50244,1839947);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667177,50232,1839947);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667178,50258,1839949);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667179,50291,1839948);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667182,50182,1839951);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667183,50185,1839951);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667184,50278,1839954);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667185,50166,1839953);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667186,50287,1839953);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667188,50121,1839955);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667189,50080,1839955);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667190,50152,1839955);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667191,50297,1839955);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667192,50151,1839956);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667193,50136,1839956);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667194,50194,1839956);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667195,50225,1839957);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667196,50185,1839958);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667197,50143,1839958);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667198,50171,1839958);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667199,50288,1839958);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667200,50105,1839959);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667201,50058,1839959);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667202,50100,1839959);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667203,50287,1839959);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667204,50100,1839960);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667205,50092,1839960);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667206,50093,1839960);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667207,50287,1839960);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667208,50030,1839961);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667209,50081,1839961);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667210,50087,1839961);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667211,50136,1839962);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667212,50049,1839962);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667213,50142,1839962);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667214,50045,1839963);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667215,50041,1839963);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667216,50044,1839963);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667217,50048,1839963);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667218,50081,1839964);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667219,50045,1839964);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667220,50136,1839966);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667221,50142,1839966);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667222,50185,1839965);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667223,50171,1839965);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667224,50177,1839965);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667225,50155,1839964);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667226,50161,1839964);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667227,50287,1839964);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667228,50170,1839968);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667231,50105,1839969);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667236,50244,1839971);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667237,50206,1839971);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667238,50244,1839974);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667239,50120,1839972);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667240,50213,1839974);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667241,50252,1839974);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667242,50225,1839975);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667243,50288,1839977);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667244,50143,1839977);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667245,50166,1839979);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667246,50105,1839979);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667247,50185,1839979);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667248,50287,1839979);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667252,50234,1839980);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667253,50136,1839981);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667255,50225,1839983);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667256,50340,1839976);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667257,50225,1839980);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667258,50049,1839981);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667259,50287,1839980);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667260,50132,1839981);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667261,50143,1839981);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667262,50073,1839976);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667263,50128,1839976);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667264,50166,1839976);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667265,50170,1839976);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667271,50023,1839985);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667272,50220,1839987);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667273,50058,1839985);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667274,50249,1839987);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667275,50185,1839985);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667276,50280,1839987);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667278,50136,1839986);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667308,50100,1839996);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667315,50234,1839999);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667322,50023,1840002);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667334,50280,1840000);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667341,50198,1840003);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667343,50291,1840006);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667345,50310,1840006);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667358,50023,1840010);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667360,50065,1840010);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667381,50024,1840017);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667384,50295,1840019);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667386,50030,1840021);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667392,50171,1840021);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667395,50093,1840022);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667396,50287,1840022);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667397,50185,1840023);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667398,50030,1840024);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667400,50052,1840024);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667404,50101,1840024);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667406,50204,1840025);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667409,50171,1840027);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667411,50126,1840027);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667429,50050,1840031);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667432,50041,1840032);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667433,50052,1840032);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667434,50101,1840032);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667435,50181,1840033);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667438,50331,1840033);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667439,50161,1840035);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667544,50119,1840073);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667550,50244,1840073);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667554,50283,1840073);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667555,50105,1840074);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667556,50089,1840074);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667557,50150,1840074);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667561,50023,1840078);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667562,50058,1840078);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667563,50185,1840078);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667564,50193,1840078);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667565,50287,1840078);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667567,50100,1840079);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667568,50058,1840079);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667569,50060,1840079);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667570,50287,1840079);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667577,50130,1840081);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667578,50015,1840082);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667579,50160,1840081);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667580,50287,1840081);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667584,50234,1840085);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667585,50225,1840085);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667594,50176,1840088);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667595,50098,1840088);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667596,50150,1840088);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667597,50177,1840088);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667598,50120,1840089);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2667606,50023,1840093);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671371,50081,1841409);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671410,50091,1841421);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671411,50100,1841421);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671418,50185,1841422);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671430,50016,1841425);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671431,50010,1841425);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671432,50112,1841425);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671433,50287,1841425);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671451,50257,1841432);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671453,50249,1841432);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671454,50328,1841432);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671455,50161,1841434);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671457,50147,1841434);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671459,50150,1841434);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671461,50160,1841434);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671463,50287,1841434);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671477,50130,1841440);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671478,50099,1841440);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671479,50175,1841440);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671480,50330,1841440);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671523,50116,1841456);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671524,50048,1841456);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671525,50124,1841456);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671526,50043,1841455);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671606,50220,1841484);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671626,50232,1841489);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671720,50304,1841529);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671737,50302,1841539);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671739,50156,1841539);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671740,50192,1841539);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671770,50232,1841551);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671771,50244,1841551);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671779,50303,1841556);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671780,50140,1841556);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671781,50177,1841556);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671782,50185,1841556);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671783,50193,1841556);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671786,50154,1841554);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671787,50110,1841554);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671794,50045,1841559);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671796,50057,1841559);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671801,50150,1841561);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671802,50098,1841561);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671803,50176,1841561);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671818,50057,1841566);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671856,50225,1841579);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671887,50030,1841591);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671888,50058,1841591);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671889,50134,1841591);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671890,50291,1841591);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671898,50030,1841595);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671899,50073,1841595);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671900,50087,1841595);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671902,50291,1841595);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671911,50087,1841599);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671912,50139,1841599);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671913,50013,1841597);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671914,50173,1841597);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671916,50287,1841597);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671921,50234,1841602);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671922,50175,1841602);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671923,50220,1841602);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671924,50224,1841602);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671925,50225,1841602);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671927,50216,1841604);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671928,50128,1841604);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671929,50136,1841604);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671930,50140,1841604);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671931,50142,1841604);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671932,50167,1841604);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671933,50185,1841604);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671940,50295,1841609);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671941,50291,1841609);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671942,50294,1841609);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671947,50023,1841607);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671949,50199,1841607);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2671964,50017,1841612);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672026,50185,1841637);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672030,50189,1841637);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672032,50199,1841637);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672053,50151,1841648);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672064,50161,1841650);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672066,50291,1841650);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672090,50227,1841656);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672093,50283,1841656);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672118,50228,1841660);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672120,50118,1841660);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672121,50150,1841660);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672122,50161,1841660);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672123,50287,1841660);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672128,50116,1841665);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672131,50114,1841665);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672133,50115,1841665);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672135,50204,1841665);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672137,50030,1841669);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672138,50150,1841669);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672139,50167,1841669);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672140,50185,1841669);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672152,50109,1841675);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672153,50104,1841675);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672154,50105,1841675);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672155,50107,1841675);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672156,50108,1841675);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672157,50110,1841675);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672173,50017,1841684);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672174,50287,1841684);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672188,50284,1841688);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672193,50062,1841693);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672194,50044,1841693);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672195,50045,1841693);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672196,50050,1841693);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672249,50023,1841714);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672251,50066,1841714);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672252,50098,1841714);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672253,50150,1841714);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672254,50287,1841714);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672255,50283,1841717);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672256,50050,1841717);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672257,50084,1841717);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672258,50087,1841717);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672259,50148,1841717);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672260,50173,1841717);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672261,50227,1841717);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672263,50340,1841717);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672264,50297,1841716);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672265,50248,1841716);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672290,50196,1841728);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672291,50161,1841728);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672316,50014,1841738);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672317,50094,1841738);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672318,50287,1841738);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672319,50057,1841737);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672355,50216,1841755);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672356,50048,1841755);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672357,50062,1841755);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672358,50180,1841755);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672359,50185,1841755);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672360,50280,1841755);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672361,50287,1841755);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672391,50092,1841765);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672392,50039,1841765);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672393,50093,1841765);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672394,50287,1841765);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672419,50087,1841775);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672420,50086,1841775);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672490,50182,1841804);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672492,50147,1841804);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672494,50185,1841804);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672496,50292,1841804);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672513,50196,1841810);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672514,50166,1841810);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672519,50216,1841811);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672522,50105,1841811);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672524,50109,1841811);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672527,50239,1841811);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672550,50097,1841827);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672596,50285,1841838);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672597,50239,1841839);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672598,50227,1841839);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672599,50326,1841839);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672600,50330,1841839);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672601,50030,1841840);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672602,50064,1841840);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672603,50105,1841840);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672733,50291,1841882);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672738,50176,1841887);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672741,50098,1841887);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672744,50150,1841887);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672746,50177,1841887);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672794,50244,1841905);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672796,50253,1841905);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672797,50254,1841905);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672831,50297,1841920);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672833,50293,1841920);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672834,50313,1841920);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672911,50180,1841945);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672913,50287,1841945);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672960,50299,1841963);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672963,50192,1841965);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672964,50046,1841965);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672965,50048,1841965);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672966,50140,1841965);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672967,50182,1841965);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672988,50106,1841972);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672989,50100,1841972);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2672990,50107,1841972);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673001,50235,1841976);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673002,50106,1841976);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673004,50239,1841976);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673033,50053,1841988);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673034,50184,1841987);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673035,50099,1841987);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673036,50178,1841987);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673037,50197,1841989);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673039,50062,1841989);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673040,50118,1841989);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673046,50220,1841994);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673049,50208,1841994);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673052,50269,1841994);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673070,50292,1842001);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673073,50294,1842001);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673075,50295,1842001);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673076,50308,1842001);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673102,50226,1842011);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673103,50174,1842011);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673152,50087,1842027);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673154,50081,1842027);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673155,50175,1842027);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673156,50287,1842027);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673201,50305,1842040);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673226,50023,1842053);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673228,50185,1842053);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673232,50288,1842053);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673296,50205,1842075);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673297,50204,1842075);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673298,50207,1842075);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673299,50233,1842075);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673360,50119,1842099);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673361,50062,1842099);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673362,50115,1842099);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673363,50235,1842099);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673364,50249,1842099);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673374,50011,1842105);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673398,50284,1842112);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673401,50011,1842116);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673402,50287,1842116);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673403,50080,1842115);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673404,50082,1842115);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673405,50083,1842115);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673413,50238,1842121);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673473,50275,1842144);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673474,50281,1842144);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673527,50181,1842158);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673528,50321,1842158);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673533,50220,1842159);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673534,50208,1842159);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673535,50269,1842159);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673536,50280,1842159);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673573,50027,1842172);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673574,50034,1842172);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673575,50035,1842172);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673576,50062,1842172);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673591,50130,1842180);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673592,50123,1842180);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673593,50171,1842180);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673595,50231,1842182);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673609,50187,1842188);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673610,50116,1842188);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673611,50166,1842188);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673612,50210,1842188);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673613,50244,1842188);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673642,50215,1842199);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673644,50058,1842199);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673645,50188,1842199);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673646,50198,1842199);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673647,50316,1842199);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673656,50012,1842201);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673657,50045,1842201);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673658,50125,1842203);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673659,50062,1842203);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673660,50131,1842203);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673661,50133,1842203);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673662,50135,1842203);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673663,50166,1842203);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673664,50287,1842203);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673672,50109,1842204);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673673,50104,1842204);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673674,50105,1842204);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673675,50108,1842204);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673676,50219,1842204);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673759,50275,1842239);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673760,50281,1842239);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673761,50261,1842236);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673762,50262,1842236);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673773,50120,1842243);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673776,50148,1842243);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673778,50189,1842243);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673780,50308,1842243);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673784,50158,1842246);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673785,50248,1842246);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673786,50280,1842246);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673863,50288,1842272);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673864,50193,1842272);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673962,50281,1842307);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673973,50181,1842311);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673975,50088,1842311);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2673977,50106,1842311);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2674142,50293,1842366);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2674143,50180,1842366);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2674144,50287,1842366);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2674165,50011,1842375);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2674166,50287,1842375);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2674172,50116,1842378);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2674173,50136,1842378);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2674174,50142,1842378);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2674175,50230,1842378);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2674176,50283,1842378);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2674187,50017,1842383);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2674188,50046,1842383);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2674189,50173,1842383);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2674190,50185,1842383);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2674191,50283,1842383);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2674192,50287,1842383);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2674204,50169,1842392);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2674205,50064,1842392);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2674206,50150,1842392);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2674207,50171,1842392);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2674291,50092,1842425);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2674292,50055,1842425);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2674293,50093,1842425);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2674294,50094,1842425);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2674295,50287,1842425);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2674312,50305,1842432);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2676172,50038,1843066);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2676549,50091,1843199);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2676639,50150,1843231);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2676651,50023,1843233);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2676797,50285,1843278);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2676913,50261,1843331);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2699547,50149,1851514);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2699747,50006,1851589);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2699782,50180,1851607);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2699784,50169,1851607);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2699785,50171,1851607);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2699786,50179,1851607);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2700103,50127,1851711);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2700522,50173,1851852);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2700523,50062,1851852);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2700525,50171,1851852);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2700526,50209,1851852);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2700615,50157,1851890);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2700861,50297,1851982);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2700863,50072,1851982);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2700866,50280,1851982);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2700867,50283,1851982);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2700994,50030,1852034);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2700995,50063,1852034);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2700997,50172,1852034);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2701089,50004,1852068);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2701090,50340,1852068);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2701574,50157,1852240);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2701575,50287,1852240);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2702236,50188,1852502);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2702971,50069,1852769);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2703824,50209,1853097);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2703825,50210,1853097);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2703826,50216,1853097);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2703827,50287,1853097);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2703862,50020,1853115);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2703863,50058,1853115);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2704603,50310,1853393);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2704604,50057,1853393);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2704605,50287,1853393);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2704606,50309,1853393);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2704607,50312,1853393);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2704657,50075,1853411);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2706087,50179,1853968);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2706088,50177,1853968);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2706384,50350,1854084);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2706385,50143,1854084);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2706386,50187,1854084);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2706387,50244,1854084);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2706388,50258,1854084);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2706929,50247,1854274);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2706930,50244,1854274);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2706931,50258,1854274);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2707751,50157,1854579);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2707752,50287,1854579);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2708202,50283,1854749);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2708203,50254,1854749);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2708204,50255,1854749);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2708282,50105,1854779);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2708283,50082,1854779);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2708284,50087,1854779);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2708287,50091,1854779);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2708290,50100,1854779);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2708293,50219,1854779);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2708295,50236,1854779);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2708297,50287,1854779);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2710127,50006,1855454);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2710546,50077,1855615);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2710911,50086,1855741);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2710912,50104,1855741);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2710913,50212,1855741);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2710914,50309,1855741);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2711195,50221,1855843);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2711196,50219,1855843);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2711197,50227,1855843);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2711198,50241,1855843);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2712719,50215,1856407);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2712721,50058,1856407);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2712725,50183,1856407);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2713657,50099,1856744);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2713659,50072,1856744);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2713660,50087,1856744);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2713667,50240,1856752);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2713669,50251,1856752);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2713671,50264,1856752);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2713740,50265,1856775);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2713741,50071,1856775);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2713742,50280,1856775);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2713743,50287,1856775);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2715182,50076,1857338);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2715183,50080,1857338);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2715184,50235,1857338);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2715185,50237,1857338);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2716530,50293,1857848);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2716531,50083,1857848);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2716532,50259,1857848);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2716533,50287,1857848);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2717087,50240,1858068);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2717504,50301,1858223);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2717506,50297,1858223);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2718282,50004,1858511);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2718298,50272,1858514);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2718503,50112,1858598);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2718505,50287,1858598);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2718919,50036,1858767);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2718920,50035,1858767);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2718927,50037,1858767);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2719232,50148,1858887);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2719234,50071,1858887);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2719235,50161,1858887);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2719236,50264,1858887);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2719237,50265,1858887);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2720526,50179,1859384);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2720530,50177,1859384);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2720557,50203,1859401);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2720981,50261,1859570);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2721500,50066,1859778);
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insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2721507,50335,1859778);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2721611,50227,1859828);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2721612,50105,1859828);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2721613,50242,1859828);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2721614,50283,1859828);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2721626,50261,1859835);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2721627,50262,1859835);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2721628,50263,1859835);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2721762,50076,1859896);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2721764,50080,1859896);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2721765,50147,1859896);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2721766,50231,1859896);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2721767,50328,1859896);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2721768,50184,1859897);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2721769,50235,1859897);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2721770,50324,1859897);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2722107,50006,1860027);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2723379,50067,1860532);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2723380,50037,1860532);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2723729,50070,1860664);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2723776,50114,1860683);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2723777,50117,1860683);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2723778,50280,1860683);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2723779,50287,1860683);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2724072,50028,1860800);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2724074,50035,1860800);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2724077,50087,1860800);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2724185,50263,1860846);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2724186,50261,1860846);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2724187,50262,1860846);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2725204,50028,1861253);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2725206,50034,1861253);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2725207,50052,1861253);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2725351,50259,1861311);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2726084,50147,1861597);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2726085,50149,1861597);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2726086,50154,1861597);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2728452,50221,1862549);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2728456,50055,1862549);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2728458,50085,1862549);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2728459,50141,1862549);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2728460,50159,1862549);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2728461,50223,1862549);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2728462,50227,1862549);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2728463,50248,1862549);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2728921,50118,1862745);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2728922,50117,1862745);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2729294,50246,1862885);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2729296,50248,1862885);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2729889,50241,1863109);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2729891,50237,1863109);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2754431,50259,1872987);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2754434,50299,1872987);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2754872,50004,1873165);
insert  into `question_knowledge_basic_id`(`id`,`knowledge_basic_id`,`question_id`) values (2757109,50075,1874082);

/*Table structure for table `questions` */

DROP TABLE IF EXISTS `questions`;

CREATE TABLE `questions` (
  `id` int(10) NOT NULL AUTO_INCREMENT,
  `title` text COMMENT '试题-题干',
  `option_a` text COMMENT '选项A',
  `option_b` text COMMENT '选项B',
  `option_c` text COMMENT '选项C',
  `option_d` text COMMENT '选项D',
  `option_e` text COMMENT '选项E',
  `answer1` text,
  `answer2` text COMMENT '非标准格式答案或含部分过程说明的答案',
  `parse` text COMMENT '试题解析',
  `qtpye` varchar(80) DEFAULT NULL COMMENT '试题题型',
  `diff` float(3,2) DEFAULT NULL COMMENT '试题难度，难度从0-5，越大越难',
  `md5` varchar(50) DEFAULT NULL COMMENT '试题题干的md5值',
  `subjectId` tinyint(2) DEFAULT NULL COMMENT '学科Id',
  `gradeId` int(5) DEFAULT NULL COMMENT '年级ID',
  `knowledges` varchar(225) DEFAULT NULL,
  `area` varchar(50) DEFAULT NULL COMMENT '试题区域',
  `year` int(4) DEFAULT NULL COMMENT '试题年份',
  `paperTpye` varchar(50) DEFAULT NULL COMMENT '试题类型：1，月考；2，模拟考；3，中考；4，高考；5，学业考试；6，其他',
  `source` varchar(200) DEFAULT NULL COMMENT '试题来源(试卷)',
  `isSub` tinyint(1) DEFAULT NULL COMMENT '是否有子题',
  `isNormal` tinyint(1) DEFAULT NULL COMMENT '是否常规题，如果选择题无法正常提取标准答案或者选项，有小题的答题无法正常提取小题，则isNormal为0，否则为1',
  `isKonw` tinyint(1) DEFAULT NULL COMMENT '是否匹配章节知识点，1匹配，0不匹配',
  PRIMARY KEY (`id`),
  KEY `index_qtypes` (`qtpye`),
  KEY `index_knowedges` (`knowledges`),
  KEY `index_year` (`year`),
  KEY `index_subject_fromsite` (`subjectId`,`isKonw`)
) ENGINE=MyISAM AUTO_INCREMENT=21659289 DEFAULT CHARSET=utf8;

/*Data for the table `questions` */

insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839824,'下列各组物质在水中能大量共存，并能形成无色溶液的是（　　）','NaCl、CuSO<SUB>4</SUB>、Na<SUB>2</SUB>CO<SUB>3</SUB>','Ca（OH）<SUB>2</SUB>、HCl、CaCO<SUB>3</SUB>','CaCO<SUB>3</SUB>、NaOH、KNO<SUB>3</SUB>','NaCl、Na<SUB>2</SUB>SO<SUB>4</SUB>、HCl','','D','【解答】解：A、CuSO<SUB>4</SUB>溶于水呈蓝色，故选项错误；<br />B、CaCO<SUB>3</SUB>难溶于水，不能形成溶液，CaCO<SUB>3</SUB>能够盐酸、氢氧化钙反应，不能大量共存，故此选项错误；<br />C、CaCO<SUB>3</SUB>难溶于水，不能形成溶液，故选项错误；<br />D、三者之间不反应，能大量共存，且不存在有色离子，故选项正确；<br />故选D．','【分析】若物质之间相互交换成分，能结合成沉淀的则不能得到透明的溶液；本题还要注意能得到无色透明溶液，不能含有明显有颜色的铜离子、铁离子和亚铁离子等．','选择题',3.00,'67e8e8e3306b4502ac2df9c04af119ee',9,400,'离子或物质的共存问题','',2016,'37','2016•郯城县校级一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839826,'<img  src=\"/tikuimages/9/2016/400/shoutiniao6/0f65accf-94d4-11e9-925b-b42e9921e93e_xkb26.png\" style=\"vertical-align:middle;FLOAT:right;\" />（2016•官渡区一模）已知A、B、C、D、E是初中化学中常见的五种物质．其中A、D是黑色固体，B、C、E是无色气体，B的相对分子质量小于C，D中含有使用最广泛的金属元素，它们在一定条件下的转化关系如图所示（其他反应物和生成物已略去）．<br />（1）写出下列物质的化学式：A<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；B<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）写出E转化成D的化学方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','C$###$CO$###$3Fe+2O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>Fe<SUB>3</SUB>O<SUB>4</SUB>','【解答】解：D中含有使用最广泛的金属元素，所以D中含有铁元素，D是黑色固体，含有铁的黑色固体可能是四氧化三铁或氧化亚铁，气体E能反应生成气体C和B，也能生成D，则E可能是氧气，氧气能与铁反应生成四氧化三铁，所以D是四氧化三铁，四氧化三铁能被一氧化碳还原生成铁和二氧化碳，所以C可能是二氧化碳，A是黑色固体，能生成二氧化碳，则A可能是碳，碳能反应生成一氧化碳，所以B可能是一氧化碳，二氧化碳能与碳反应生成一氧化碳，B的相对分子质量小于C，所以B是一氧化碳，C是二氧化碳，经过验证，推导正确，因此：<br />（1）A是碳，B是一氧化碳；故填：C；CO；<br />（2）E转化为D的反应是铁和氧气在点燃的条件下生成四氧化三铁，故化学方程式为：3Fe+2O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>Fe<SUB>3</SUB>O<SUB>4</SUB>．','【分析】根据A、B、C、D、E是初中化学中常见的五种物质，D中含有使用最广泛的金属元素，所以D中含有铁元素，D是黑色固体，含有铁的黑色固体可能是四氧化三铁或氧化亚铁，气体E能反应生成气体C和B，也能生成D，则E可能是氧气，氧气能与铁反应生成四氧化三铁，所以D是四氧化三铁，四氧化三铁能被一氧化碳还原生成铁和二氧化碳，所以C可能是二氧化碳，A是黑色固体，能生成二氧化碳，则A可能是碳，碳能反应生成一氧化碳，所以B可能是一氧化碳，二氧化碳能与碳反应生成一氧化碳，B的相对分子质量小于C，所以B是一氧化碳，C是二氧化碳，然后将推出的各种物质代入转化关系中验证即可．','书写',3.00,'0f6ec114f5f98a99e61cea39ddeba800',9,400,'物质的鉴别、推断,书写化学方程式、文字表达式、电离方程式','',2016,'37','2016•官渡区一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839827,'下列说法不正确的是（　　）','硝酸铵固体溶于水时放热','铜丝浸入硝酸汞溶液中，表面覆盖一层银白色的物质，说明铜比汞活泼','洗洁精具有乳化功能，可用于去除油污','用金属钨制作白炽灯丝是利用了其熔点高且能导电的性质','','A','【解答】解：A、硝酸铵溶于水时吸热，溶液温度降低，故选项说法错误；<br />B、铜丝浸入硝酸汞溶液中，表面覆盖一层银白色的物质，说明铜比汞活泼，正确；<br />C、洗洁精具有乳化功能，可用于去除油污，正确；<br />D、金属钨制作白炽灯丝是利用了其熔点高且能导电的性质，正确；<br />故选A','【分析】A、根据硝酸铵溶于水的温度变化进行分析判断；<br />B、根据金属的活动性分析；<br />C、根据乳化作用分析；<br />D、根据金属的性质分析．','选择题',3.00,'b7d375cdb795fadae36799d6349b13ee',9,400,'溶解时的吸热或放热现象,乳化现象与乳化作用,金属的物理性质及用途,金属的化学性质','太仓市',2016,'32','2016•太仓市模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839828,'下列图示实验操作中，正确的是（　　）','<img  src=\"/tikuimages/9/2016/400/shoutiniao6/0f7208de-94d4-11e9-a80e-b42e9921e93e_xkb31.png\" style=\"vertical-align:middle\" /><br />闻气味','<img  src=\"/tikuimages/9/2016/400/shoutiniao8/0f756440-94d4-11e9-a9a5-b42e9921e93e_xkb49.png\" style=\"vertical-align:middle\" /><br />取用液体药品','<img  src=\"/tikuimages/9/2016/400/shoutiniao63/0f78bfa1-94d4-11e9-8624-b42e9921e93e_xkb84.png\" style=\"vertical-align:middle\" /><br />过滤','<img  src=\"/tikuimages/9/2016/400/shoutiniao84/0f79f81e-94d4-11e9-96e6-b42e9921e93e_xkb74.png\" style=\"vertical-align:middle\" /><br />用手拿蒸发皿','','B','【解答】解：A．闻药品气味时要用扇闻的方法，故A错误；<br />B．向试管内倾倒液体时，瓶塞倒放，标签向着手心，集气瓶口紧靠试管口，符合液体的取用操作，故B正确；<br />C．过滤操作要遵循“一贴、二低、三靠”的原则，故C错误；<br />D．蒸发皿要用坩埚钳夹取，不能用手拿，故D错误．<br />故选B．','【分析】A．根据闻药品气味的方法分析；<br />B．根据液体倾倒方法分析；<br />C．根据过滤操作的注意事项分析；<br />D．根据蒸发皿的取用方法分析．','选择题',3.00,'14f6bede122c9b5960c9b102534e5e57',9,400,'液体药品的取用,过滤的原理、方法及其应用,蒸发与蒸馏操作','',2016,'32','2016•长春模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839830,'<img  src=\"/tikuimages/9/2016/400/shoutiniao30/0f836e00-94d4-11e9-a010-b42e9921e93e_xkb67.png\" style=\"vertical-align:middle;FLOAT:right\" />（2016•天津一模）铜、铁、铝是生活中常见的金属，请回答：<br />（1）黄铜是铜锌合金，其硬度比纯铜<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“大”或“小”）．<br />（2）铝块能制成铝箔是利用了铝的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>性．<br />（3）“双吸剂”的主要成分是还原铁粉，常用于食品保鲜，其原因是铁粉能吸收空气中的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>和<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（4）用等质量的锌、铁、镁三种金属分別与三份溶质质量分数相同的稀盐酸充分反应．产生氢气的质量与反应时间的关系如图所示．下列说法中正确的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填字母）．<br />A．曲线X、Y、Z分别表示锌、铁、镁的反应情况<br />B．反应后一定有剩余的金属是镁<br />C．反应后可能没有剩余的金属是锌和铁<br />D．三份盐酸均反应完．','','','','','','大$###$延展$###$水$###$氧气$###$B','【解答】解：（1）黄铜是铜锌合金，其硬度比纯铜大；<br />（2）铝块能制成铝箔是利用了铝的延展性；<br />（3）“双吸剂”的主要成分是还原铁粉，常用于食品保鲜，其原因是铁粉能吸收空气中的水和氧气；<br />（4）24份质量的镁会生成2份质量的氢气，56份质量的铁会生成2份质量的氢气，65份质量的锌会生成2份质量的氢气，结合图象可知，金属X是Mg，反应后可能已经反应完的金属是Zn、Fe，X、Y生成氢气的质量相同，则金属Mg-定有剩余，三份盐酸也会出现剩余．<br />故答案为：（1）大；（2）延展；（3）水；氧气；（4）B；','【分析】（1）根据合金的性质比组成其纯金属的性质优良进行分析；<br />（2）根据金属的物理性质进行分析；<br />（3）根据金属铁的性质分析；<br />（4）根据锌、铁、镁三种金属的活动性及质量守恒定律进行分析；','填空题',3.00,'2bb160bea297431fac9d7e3bb2e79680',9,400,'金属的物理性质及用途,合金与合金的性质,金属的化学性质,金属锈蚀的条件及其防护','',2016,'37','2016•天津一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839832,'下列关于“水”的说法，正确的是（　　）','水电解时正极得到氢气','饱和石灰水属于浓溶液','“汽水”应低温高压保存','软水一定是纯净物','','C','【解答】解：A．电解水时，正极得到的是氧气，负极得到的是氢气，故错误；<br />B．氢氧化钙微溶于水，形成的饱和溶液为稀溶液，故错误；<br />C．二氧化碳的溶解度随着温度的降低而增大，随着压强的增大而增大，所以“汽水”应低温高压保存，故正确；<br />D．软水中含有少量的钙、镁离子化合物，属于混合物，故错误．<br />故选C．','【分析】A．根据电解水的实验来分析；<br />B．根据饱和溶液与浓溶液的关系来分析；<br />C．根据影响气体溶解度的因素来分析；<br />D．根据软水的组成来分析．','选择题',3.00,'c25843449ead72c85693ceb4e887da4c',9,400,'电解水实验,硬水与软水,浓溶液、稀溶液跟饱和溶液、不饱和溶液的关系,气体溶解度的影响因素','',2016,'37','2016•重庆校级一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839833,'质量守恒定律是自然界的普遍规律．请你回答下列问题：<br />（1）化学反应过程，从微观方面说就是反应物的原子<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>成新的分子而生成新物质的过程，反应前后<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>、<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>、<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>守恒（没有发生改变）．<br />（2）由质量守恒定律可知，在“镁条燃烧生成氧化镁”的反应中，生成的氧化镁的质量与<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>的质量相等．','','','','','','重新组合$###$原子的种类$###$原子的数目$###$原子的质量$###$参加反应的镁条和氧气','【解答】解：（1）化学反应的过程就是分子分成原子，原子不再分，而是重新组合得新的分子的过程．因此，一切化学反应前后，原子的种类、原子的数目、原子的质量都没有改变．所以化学反应中存在质量守恒；<br />（2）镁条在氧气中燃烧生成氧化镁，由质量守恒定律，生成氧化镁的质量等于参加反应的镁条和氧气的质量之和．<br />故答案为：<br />（1）重新组合，原子的种类、原子的数目、原子的质量．<br />（2）参加反应的镁条和氧气．','【分析】（1）从微观和宏观角度分析质量守恒定律，化学反应前后原子的种类、数目和质量都不发生改变；参加反应的各物质质量总和等于反应后生成的各物质质量总和；<br />（2）考虑在化学反应中参加反应前各物质的质量总和等于反应后生成各物质的质量总和．','填空题',3.00,'72cb9e8e43f2a7cc4a684112047aa5aa',9,400,'质量守恒定律及其应用','',2016,'35','2016春•沂源县期中',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839834,'经过学习我们可以初步认识到：化学反应的快慢、现象、生成物等与反应物量的多少、反应条件有着密切的关系．请各举一例说明：<br />（1）使用催化剂，能改变化学反应的速率：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）反应物的量不同，生成物可能不同：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）反应物的量不同，反应现象可能不同：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（4）反应条件不同，生成物不同：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','过氧化氢溶液常温下分解速率较慢，加入少量二氧化锰，迅速产生大量气泡（合理即可）$###$碳在氧气充足时燃烧生成二氧化碳，在氧气不足时燃烧生成一氧化碳（合理即可）$###$硫在空气中燃烧发出淡蓝色火焰，而在纯氧气中燃烧发出蓝紫色火焰（合理即可）$###$常温下二氧化碳与水反应生成碳酸，在叶绿体和光照的作用下，二氧化碳与水反应生成葡萄糖和氧气（合理即可）','【解答】解：（1）将常温下分解速率较慢的过氧化氢溶液在加入少量二氧化锰，迅速产生气泡，说明了使用催化剂，能改变化学反应的速率（合理即可）．<br />（2）反应物的量不同，生成物可能不同，如碳在氧气充足时燃烧生成二氧化碳，在氧气不足时燃烧生成一氧化碳（合理即可）．<br />（3）反应物相同时，反应物的浓度不同，反应现象可能不同，如：硫在空气中燃烧发出淡蓝色火焰，而在纯氧气中燃烧发出蓝紫色火焰（合理即可）．<br />（4）反应条件不同，生成物可能不同，如常温下二氧化碳与水反应生成碳酸，在叶绿体和光照的作用下，二氧化碳与水反应生成葡萄糖和氧气（合理即可）．<br />故答案为：（1）过氧化氢溶液常温下分解速率较慢，加入少量二氧化锰，迅速产生大量气泡（合理即可）；<br />（2）硫在空气中燃烧发出淡蓝色火焰，而在纯氧气中燃烧发出蓝紫色火焰（合理即可）；<br />（3）碳在氧气充足时燃烧生成二氧化碳，在氧气不足时燃烧生成一氧化碳（合理即可）；<br />（4）常温下二氧化碳与水反应生成碳酸，在叶绿体和光照的作用下，二氧化碳与水反应生成葡萄糖和氧气（合理即可）．','【分析】（1）反应物相同时，反应条件不同，是否使用催化剂，反应速率可能不同．<br />（2）反应物的量不同，生成物可能不同，如碳在氧气中燃烧．<br />（3）反应物的量不同，反应现象不同，如铁、硫在空气或氧气中燃烧．<br />（4）反应条件不同，生成物可能不同，如反应温度不同等．','填空题',3.00,'59c02e3220e7974dc30358b6b6e75110',9,400,'实验探究物质变化的条件和影响物质变化的因素,催化剂的特点与催化作用','',2016,'32','2016•鱼台县模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839837,'下列是研究氧气、水、二氧化碳的实验．根据图示，回答问题．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao61/0f9fa88f-94d4-11e9-8693-b42e9921e93e_xkb80.png\" style=\"vertical-align:middle\" /><br />（1）A中铁丝在氧气中燃烧，反应的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）B中试管a中产生的气体为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）C实验的实验现象是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','3Fe+2O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>Fe<SUB>3</SUB>O<SUB>4</SUB>$###$O<SUB>2 </SUB>$###$蜡烛由下至上依次熄灭','【解答】解：（1）铁丝在氧气中燃烧，生成四氧化三铁，反应的化学方程式为：3Fe+2O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>&nbsp;&nbsp;Fe<SUB>3</SUB>O<SUB>4</SUB>；<br />（2）利用图所示方法，通电一段时间后，a试管中收集到的气体是氧气．<br />（3）二氧化碳的密度比空气大，不能燃烧，不支持燃烧，C实验的实验现象是蜡烛由下至上依次熄灭．<br />答案：<br />（1）3Fe+2O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>&nbsp;&nbsp;Fe<SUB>3</SUB>O<SUB>4</SUB>；<br />（2）O<SUB>2 </SUB><br />（3）蜡烛由下至上依次熄灭．','【分析】（1）根据铁丝在氧气中燃烧，生成四氧化三铁解答；<br />（2）根据电解水时，正极产生的是氧气，负极产生的是氢气，氧气和氢气的体积比约为1：2解答；<br />（3）根据二氧化碳的密度比空气大，不能燃烧，不支持燃烧解答．','书写',3.00,'1c6271adbb6ee0fe835e4e404313ed57',9,400,'氧气的化学性质,二氧化碳的物理性质,二氧化碳的化学性质,电解水实验,书写化学方程式、文字表达式、电离方程式','',2016,'37','2016•延庆县一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839838,'根据如图所示，下列说法错误的是（　　）<br /><img  src=\"/tikuimages/9/2015/400/shoutiniao91/0fa59c00-94d4-11e9-9d31-b42e9921e93e_xkb20.png\" style=\"vertical-align:middle\" />','装置A中，正极产生的是氧气，负极产生的是氢气','该反应产生的氢气和氧气的体积比应该是2：1','装置B中，集气瓶中放少量水是防止反应放出大量的热使集气瓶炸裂','将红热的铁丝插入盛氧气的集气瓶中时，要至上面下迅速插入','','D','【解答】解：A．电解水时，正极试管中收集到的气体较少是氧气，负极试管中收集到的气体较多是氢气，故正确；<br />B．由电解水图可知，通电一段时间后，正极试管中收集到的气体较少是氧气，负极试管中收集到的气体较多是氢气，氢气和氧气的体积之比是2：1，故正确；<br />C．铁丝在氧气中燃烧，在集气瓶底预先放一些水，可以防止反应放出大量的热使集气瓶炸裂，故正确；<br />D．将红热的铁丝插入盛氧气的集气瓶中时，要从上往下缓慢插入氧气瓶，否则会影响实验现象，故错误．<br />故选D．','【分析】根据电解水的实验现象、铁丝在氧气中燃烧的实验注意事项来分析解答．','选择题',3.00,'83a53a0e57a9e3e3cd6d3e3d4ecf526f',9,400,'氧气的化学性质,电解水实验','',2015,'35','2015秋•孝义市校级期中',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839839,'下列A-E是初中化学中的五个实验装置，请按要求填空：<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao23/0faffc40-94d4-11e9-8c2a-b42e9921e93e_xkb22.png\" style=\"vertical-align:middle\" /><br />（1）A实验试管2中产生起气体是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）B实验中烧杯③的作用是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）C实验中发生反应的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，集气瓶中水的作用是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（4）若用D装置除去0<SUB>2</SUB>中的水蒸气，该液体试剂为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（5）E实验中，气球的变化情况是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','氢气$###$对比$###$4P+5O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2P<SUB>2</SUB>O<SUB>5</SUB>$###$防止灼热的五氧化二磷固体接触瓶底使其受热不均而炸裂$###$浓硫酸$###$瓶内气球先变大再变小','【解答】解：（1）据图可以看出2试管中产生的气体体积大，是氢气，故填：氢气；<br />（2）因为酚酞遇碱变红，烧杯①中的酚酞变红，说明浓氨水的化学性质是呈碱性，而烧杯③是想说明酚酞在空气中不变色，起到了对比的作用，故填：对比；<br />（3）红磷燃烧生成五氧化二磷，反应的化学方程式为4P+5O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2P<SUB>2</SUB>O<SUB>5</SUB>，集气瓶中水的作用是防止灼热的五氧化二磷固体接触瓶底使其受热不均而炸裂；故填：4P+5O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2P<SUB>2</SUB>O<SUB>5</SUB>；防止灼热的五氧化二磷固体接触瓶底使其受热不均而炸裂；<br />（4）浓硫酸具有吸水性，能作干燥剂；故填：浓硫酸；<br />（5）氢氧化钠能与二氧化碳反应生成碳酸钠和水，二氧化碳被消耗，瓶内压强变小，气球变大，碳酸钠能与盐酸反应生成二氧化碳气体，瓶内压强变大，气球变小，故填：瓶内气球先变大再变小．','【分析】根据已有的知识进行分析，电解水时生成的氢气和氧气的体积比为2：1；设计实验证明分子的运动需要设计对照试验，酚酞可使碱性溶液变红；根据测定氧气的含量实验进行分析；浓硫酸具有吸水性；氢氧化钠能与二氧化碳反应生成碳酸钠和水，碳酸钠能与盐酸反应生成二氧化碳气体．','简答题',3.00,'61aee22549666141ef52089a918e6569',9,400,'气体的干燥（除水）,空气组成的测定,电解水实验,碱的化学性质,分子的定义与分子的特性','',2016,'32','2016•湘桥区模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839840,'<img  src=\"/tikuimages/9/2016/400/shoutiniao43/0fb68bf0-94d4-11e9-98cc-b42e9921e93e_xkb19.png\" style=\"vertical-align:middle;FLOAT:right\" />（2016•柳江县一模）习近平总书记说“绿水青山就是金山银山”，突显了中国领导层对环境的治理决心和可持续发展的新理念．<br />（1）树林每天能吸收二氧化碳，并释放出氧气，这是依靠植物的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>作用，这种作用有利于维持生物圈中的二氧化碳平衡．二氧化碳过量排放会引起<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）近期，中国部分城市雾霾天气严重影响了人们正常的生活，导致PM<SUB>2.5</SUB>浓度升高的人为因素有（写出其中一项）<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）自然界中的水含有许多可溶性和不溶性杂质，通过多种途径可以使水得到不同程度的净化．如图是自来水厂净化水过程示意图．<br />①吸附池中通常加入黑色固体是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />②饮用硬度大的地下水不利于人体健康．检验某地下水是硬水还是软水的方法是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />③自来水含有少量Ca（HCO<SUB>3</SUB>）<SUB>2</SUB>等可溶性化合物，烧水时Ca（HCO<SUB>3</SUB>）<SUB>2</SUB>发生分解反应，生成难溶性的碳酸钙、水和二氧化碳，这是水壶中出现水垢的原因之一．写出固体Ca（HCO<SUB>3</SUB>）<SUB>2</SUB>受热分解的化学方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','光合$###$温室效应$###$有焚烧秸杆$###$活性炭$###$取样，加入肥皂水，泡沫多的水软水，泡沫少的是硬水$###$Ca（HCO<SUB>3</SUB>）<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>CaCO<SUB>3</SUB>↓+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑','【解答】解：（1）树林每天能吸收二氧化碳，并释放出氧气，这是依靠植物的光合作用；这种作用有利于维持生物圈中的二氧化碳平衡．二氧化碳过量排放会引起<br />温室效应；<br />（2）导致PM2.5浓度升高的人为因素有焚烧秸杆，给空气质量造成很大的影响．<br />（3）①活性炭具有吸附性，可以吸附色素和异味，吸附池中通常加入黑色固体是活性炭；<br />②鉴别硬水和软水使用的是肥皂水，泡沫多的水软水，泡沫少的是硬水；<br />③碳酸氢钙受热分解生成难溶性的碳酸钙、水和二氧化碳，反应的方程式是Ca（HCO<SUB>3</SUB>）<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>CaCO<SUB>3</SUB>↓+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑<br />答案：<br />（1）光合；温室效应；<br />（2）焚烧秸杆；<br />（3）①活性炭；②取样，加入肥皂水，泡沫多的水软水，泡沫少的是硬水；③Ca（HCO<SUB>3</SUB>）<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>CaCO<SUB>3</SUB>↓+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑．','【分析】（1）根据绿色植物通过光合作用，吸收二氧化碳，释放出人类维持生命的氧．减少大气中二氧化碳的量，降低温室效应解答；<br />（2）根据致PM2.5浓度升高的原因解答；<br />（3）①根据活性炭具有吸附性，可以吸附色素和异味解答；<br />②根据区别硬水和软水使用的是肥皂水解答；<br />③根据反应物、生成物书写方程式．','书写',3.00,'0c018adfbe3095b8b445be01d1f5fd70',9,400,'空气的污染及其危害,自然界中的碳循环,二氧化碳对环境的影响,自来水的生产过程与净化方法,硬水与软水,书写化学方程式、文字表达式、电离方程式','',2016,'37','2016•柳江县一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839841,'下列叙述Ⅰ和叙述Ⅱ均正确，并且有因果关系的是（　　）<br /><table class=\"edittable\"><TBODY><TR><td width=39>选项</TD><td width=205>叙述Ⅰ</TD><td width=200>叙述Ⅱ</TD></TR><TR><td>&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp; <br />A</TD><td>Na<SUB>2</SUB>CO<SUB>3</SUB>溶液呈碱性</TD><td>Na<SUB>2</SUB>CO<SUB>3</SUB>属于碱</TD></TR><TR><td>B</TD><td>铁可以和稀硫酸反应放出H<SUB>2</SUB></TD><td>Fe属于金属</TD></TR><TR><td>C</TD><td>稀盐酸可以与多数金属氧化物反应</TD><td>稀盐酸常用来金属除锈</TD></TR><TR><td>D</TD><td>洗涤剂增大了油污在水中的溶解性</TD><td>洗涤剂能除去衣服上的油污</TD></TR></TBODY></TABLE>','A','B','C','D','','C','【解答】解：A、Na<SUB>2</SUB>CO<SUB>3</SUB>溶液呈碱性，碳酸钠是由钠离子和碳酸根离子构成的，属于盐，叙述Ⅱ错误，故选项错误．<br />B、铁可以和稀硫酸反应放出H<SUB>2</SUB>，Fe属于金属，但两者没有因果关系，故选项错误．<br />C、稀盐酸可以与多数金属氧化物反应，稀盐酸常用来金属除锈，叙述Ⅰ和叙述Ⅱ均正确，并且有因果关系，故选项正确．<br />D、洗涤剂具有乳化作用，能将大的油滴分散成细小的油滴随水冲走，而不是增大了油污在水中的溶解性，叙述Ⅰ错误，故选项错误．<br />故选：C．','【分析】A、根据碳酸钠是由钠离子和碳酸根离子构成的，进行分析判断．<br />B、根据金属的化学性质，进行分析判断．<br />C、根据酸的化学性质、用途进行分析判断．<br />D、根据洗涤剂具有乳化作用，进行分析判断．','选择题',3.00,'88c264281dc2f06fcf53123cfc022244',9,400,'乳化现象与乳化作用,金属的化学性质,酸的化学性质,盐的化学性质,常见的氧化物、酸、碱和盐的判别','',2015,'35','2015秋•周村区校级期中',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839843,'实验室用过氧化氢溶液100克与3克的二氧化锰制氧气，完全反应后剩余固液混合物99.3克，求原过氧化氢溶液中氢元素的质量分数（　　）','4.8%','10.7%','85.6%','10.2%','','B','【解答】解：过氧化氢溶液在二氧化锰的催化作用下生成水和氧气，二氧化锰作催化剂反应前后质量不变，由质量守恒定律，最终所得到的水中氢元素的质量即为原过氧化氢溶液中氢元素的质量，水中氢元素的质量为（99.3g-3g）×<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">1×2</td></tr><tr><td>1×2+16</td></tr></table>×</span>100%=10.7g．<br />原过氧化氢溶液中氢元素的质量分数为<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">10.7g</td></tr><tr><td>100g</td></tr></table>×</span>100%=10.7%．<br />故选：B．','【分析】过氧化氢溶液在二氧化锰的催化作用下生成水和氧气，二氧化锰作催化剂反应前后质量不变，由质量守恒定律，最终所得到的水中氢元素的质量即为原过氧化氢溶液中氢元素的质量，据此进行分析判断．','选择题',3.00,'1ba8d34da9d7221a2afa020c8e63f9fd',9,400,'元素的质量分数计算,质量守恒定律及其应用','',0,'37','',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839844,'如图所示的实验操作不正确的是（　　）','<img  src=\"/tikuimages/9/2016/400/shoutiniao46/0fc272d1-94d4-11e9-ad24-b42e9921e93e_xkb31.png\" style=\"vertical-align:middle\" /><br />铁丝在氧气中燃烧','<img  src=\"/tikuimages/9/2016/400/shoutiniao34/0fc38440-94d4-11e9-8bc5-b42e9921e93e_xkb86.png\" style=\"vertical-align:middle\" /><br />读取液体的体积','<img  src=\"/tikuimages/9/2016/400/shoutiniao45/0fc5f540-94d4-11e9-8725-b42e9921e93e_xkb85.png\" style=\"vertical-align:middle\" /><br />除去CO中的CO<SUB>2</SUB>','<img  src=\"/tikuimages/9/2016/400/shoutiniao29/0fc86640-94d4-11e9-9a1f-b42e9921e93e_xkb78.png\" style=\"vertical-align:middle\" /><br />稀释浓硫酸','','C','【解答】解：A、进行铁丝在纯氧中燃烧的实验操作时，要在集气瓶底预先放一些沙子或水，操作正确；<br />B、量筒的使用，视线要与凹液面的最低处保持相平，故操作正确；<br />C、除去气体中的杂质，利用洗气瓶应从长进短出．故操作错误；<br />D、稀释浓硫酸时，要把浓硫酸缓缓地沿器壁注入水中，同时用玻璃棒不断搅拌，以使热量及时地扩散，禁止把水倒入浓硫酸中，故操作正确．<br />故答案选：C','【分析】根据实验室仪器的使用方法和操作注意事项，以及气体除杂进行分析解答本题．','选择题',3.00,'481b98e244966ac5a7b72da23bb186e4',9,400,'测量容器-量筒,浓硫酸的性质及浓硫酸的稀释,常见气体的检验与除杂方法,氧气的化学性质','',2016,'32','2016•泰安模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839845,'现有化学反应2A+B═C+3D，下列说法正确的是（　　）','A和B的质量比一定等于C和D的质量比','A和B相对分子质量比为44：36，则C和D的相对发展质量的比也是44：36','若A和B各取2g使其反应，则C和D的质量总和一定等于4g','A和B各取2g恰好完全反应，则C和D的质量总和一定等于4g','','D','【解答】解：A．参加反应的AB的质量总和一定等于生成的CD的质量总和，但反应物A和B的质量比不一定等于生成物C和D的质量比，故错误．<br />B．因为A、B的化学计量数之比为2：1，C、D的化学计量数之比为1：3，所以若A和B相对分子质量比为44：36，则C和D的相对分子质量的比不会是44：36，故错误．<br />C．若A和B各取2g使其反应，若二者充分参与反应，则生成的C、D总质量为4g，若其中一种有剩余，则C、D总质量小于4g，故错误．<br />D．若A和B各取2g恰好完全反应，由质量守恒定律可知，则生成C和D的质量总和一定等于4g，故正确．<br />故选D．','【分析】根据质量守恒定律的内涵以及意义来分析解答．','选择题',3.00,'f51d055ed2b6145f7ed2ab8544e5ef41',9,400,'质量守恒定律及其应用','普兰店市',2015,'35','2015秋•普兰店市校级期中',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839846,'完成下面的实验报告．<br />实验名称：连接仪器并检査装置的气密性<br />实验目的：练习仪器的连接，并学会装罝气密性的检査方法．<br />实验步驟：①连接玻璃管和胶皮管：<br />②连接带导管的橡皮塞和试管；<br />③<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />实验现象：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />实验结论：装置的气密性良好．<br />实验反思：<br />（1）三组同学进行第③步操作时，先后顺序颠倒了，结果没有看到预期的实验现象，很多同学认为该实验失敗了，王老师说还可以通过另一个现象来判断，装置的气密性良好，该现象是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；张同学发表意见说，结合上述情况，在原实验基础上，再等一等，观察一下上述现象，更能说明装置的气密性良好．张同学的表现获得了大家一致的赞誉．<br />（2）六组的同学经过多次、准确的操作之后，均未看到预期的实验现象，最后经过交流，判定所用装置存在漏气的情况．请尝试写出一条漏气的原因<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','先把导管口放入水中，再用手握住试管$###$导管口有气泡冒出$###$导管内形成一段水柱$###$玻璃管与胶皮管没有连接好，留有较大空隙','【解答】解：检查装置气密性时，先把导管口放入水中，再用手握住试管，然后观察导管口是否有气泡冒出，如有，则证明装置气密性良好；如果装置气密性良好，握住试管的手松开一会儿后，导管内会形成一段水柱；如没有预期的现象出现，则证明装置漏气，漏气的原因可能是玻璃管与胶皮管没有连接好，留有较大空隙；或玻璃管与橡皮塞之间有较大空隙．故答案为：先把导管口放入水中，再用手握住试管；导管口是否有气泡冒出；导管内形成一段水柱；玻璃管与胶皮管没有连接好，留有较大空隙；（答案合理即可）．','【分析】根据检查装置气密性的方法分析解答；','填空题',3.00,'014ffdba922e6a439ab59a78d09caadf',9,400,'仪器的装配或连接,检查装置的气密性','',2015,'35','2015秋•周村区校级期中',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839847,'绿色交通工具是指在行驶中对环境不发生污染或污染程度很小的载客工具．下列交通工具的使用对环境污染最大的是（　　）','<img  src=\"/tikuimages/9/2016/400/shoutiniao98/0fce59ae-94d4-11e9-a14b-b42e9921e93e_xkb8.png\" style=\"vertical-align:middle\" /><br />无轨电车','<img  src=\"/tikuimages/9/2016/400/shoutiniao98/0fcf1d00-94d4-11e9-89c5-b42e9921e93e_xkb74.png\" style=\"vertical-align:middle\" /><br />柴油汽车','<img  src=\"/tikuimages/9/2016/400/shoutiniao39/0fd118cf-94d4-11e9-be42-b42e9921e93e_xkb63.png\" style=\"vertical-align:middle\" /><br />天然气汽车','<img  src=\"/tikuimages/9/2016/400/shoutiniao73/0fd27861-94d4-11e9-a714-b42e9921e93e_xkb14.png\" style=\"vertical-align:middle\" /><br />氢气动力车','','B','【解答】解：<br />A、大量使用电能不会造成环境污染，使用无轨电车不会产生污染，故正确；<br />B、柴油汽车燃烧产生一些有害气体，污染空气，故错误；<br />C、天然气汽车是比较清洁的燃料，不产生污染物，故正确；<br />D、氢气燃烧产物是水，不污染环境，故正确．<br />答案：B．','【分析】A、根据无轨电车不会产生污染解答；<br />B、柴油汽车燃烧产生一些有害气体，污染空气解答；<br />C、根据天然气汽车是比较清洁的燃料解答；<br />D、根据氢气是最洁的燃料解答．','选择题',3.00,'cfdb267bb35d7d0243e0be10527176f5',9,400,'常用燃料的使用与其对环境的影响','',2016,'37','2016•重庆校级二模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839848,'请用化学符号填空：<br />（1）4个碳原子<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（2）一氧化碳中碳元素化合价<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）食盐的阳离子<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（4）干冰的主要分成<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','4C$###$<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td style=\"font-size: 90%\">+2</td></tr><tr><td>C</td></tr></table></span>O$###$Na<SUP>+</SUP>$###$CO<SUB>2</SUB>','【解答】解：（1）原子的表示方法就是用元素符号来表示一个原子，表示多个该原子，就在其元素符号前加上相应的数字．所以4个碳原子，就可表示为：4C；<br />（2）元素化合价的表示方法：确定出化合物中所要标出的元素的化合价，然后在其化学式该元素的上方用正负号和数字表示，正负号在前，数字在后，所以一氧化碳中碳元素化合价，故可表示为：<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td style=\"font-size: 90%\">+2</td></tr><tr><td>C</td></tr></table></span>O；<br />（3）食盐是氯化钠的俗称，其阳离子为钠离子，其符号为：Na<SUP>+</SUP>；<br />（4）干冰是固态二氧化碳的俗称，其化学式为：CO<SUB>2</SUB>；<br />故答案为：（1）4C；（2）<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td style=\"font-size: 90%\">+2</td></tr><tr><td>C</td></tr></table></span>O；（3）Na<SUP>+</SUP>；（4）CO<SUB>2</SUB>；','【分析】本题考查化学用语的意义及书写，解题关键是分清化学用语所表达的对象是分子、原子、离子还是化合价，才能在化学符号前或其它位置加上适当的计量数来完整地表达其意义，并能根据物质化学式的书写规则正确书写物质的化学式，才能熟练准确的解答此类题目．','填空题',3.00,'fcd7a2ec957a82dbb73b06e91df67ccb',9,400,'化学符号及其周围数字的意义','',2016,'37','2016•鞍山一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839850,'下列各组物质在给定PH的溶液中能大量共存并形成无色溶液的是（　　）','pH=3：FeCl<SUB>3</SUB>、CuS0<SUB>4</SUB>、NaCl','pH=12：BaCl<SUB>2</SUB>、Na<SUB>2</SUB>C0<SUB>3</SUB>、Na0H','pH=1：NH<SUB>4</SUB>Cl、Na<SUB>2</SUB>S0<SUB>4</SUB>、HN0<SUB>3</SUB>','pH=4：NaOH、NaHCO<SUB>3</SUB>、K<SUB>2</SUB>SO<SUB>4</SUB>','','C','【解答】解：A、铁离子在溶液中显黄色，铜离子在溶液中显蓝色，而选项中要求形成无色溶液，不符合题意，故A错误；<br />B、碳酸根离子和钡离子会生成白色的碳酸钡沉淀，氯化钡和碳酸钠不能大量共存，故B错误；<br />C、所有的钠盐、钾盐、硝酸盐都溶于水，选项中所给物质在酸性环境中可以大量共存，故C正确；<br />D、碳酸钠在酸性环境中，会与氢离子反应生成二氧化碳气体，不能大量共存，硫酸亚铁在溶液中显浅绿色，故D错误．<br />故选：C．','【分析】A、根据铁离子在溶液中显黄色，铜离子在溶液中显蓝色进行分析；<br />B、根据碳酸根离子和钡离子会生成白色的碳酸钡沉淀进行分析；<br />C、根据所有的钠盐、钾盐、硝酸盐都溶于水进行分析；<br />D、根据碳酸根离子和氢离子会生成二氧化碳气体进行分析．','选择题',3.00,'241545f5db15a241c1f37ef9d86e04d2',9,400,'溶液的酸碱性与pH值的关系,离子或物质的共存问题','',2016,'37','2016•卧龙区一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839851,'关于溶液的说法错误的是（　　）','溶液不一定是无色的','溶液中可以含有多种溶质','溶液是均一、稳定的混合物','硝酸钾饱和溶液不能再溶解氯化钠','','D','【解答】解：A．溶液可以有颜色，例如硫酸铜溶液是蓝色，故说法正确；<br />B．溶液中溶质可以有多种，溶剂只有一种，故说法正确；<br />C．溶液是均一、稳定的混合物，故说法正确；<br />D．饱和溶液只对于某一溶质饱和，还可以溶解其它溶质，如硝酸钾的饱和溶液中仍然可以溶解氯化钠，故说法错误．<br />故选D．','【分析】A．根据溶液可以有颜色考虑；<br />B．溶液中溶质可以有多种，溶剂只有一种；<br />C．根据溶液的特点考虑；<br />D．根据饱和溶液只对于某一溶质饱和考虑．','选择题',3.00,'0814e3904ff9ca3d43bd8a8be692d5d0',9,400,'溶液的概念、组成及其特点,饱和溶液和不饱和溶液','',2016,'37','2016春•深圳月考',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839852,'<img  src=\"/tikuimages/9/2016/400/shoutiniao61/0fdc8a80-94d4-11e9-9608-b42e9921e93e_xkb55.png\" style=\"vertical-align:middle;FLOAT:right\" />（2016春•高青县期中）物质发生化学变化的前后，总质量是否发生改变？是增加、减小还是不变？小明、小强按如图的装置进行探究：<br />（1）提出问题：化学反应前后，总质量是否发生改变？<br />（2）建立假设：化学变化前后总质量不变．<br />（2）查阅资料：<br />CuSO<SUB>4</SUB>+2NaOH═Na<SUB>2</SUB>SO<SUB>4</SUB>+Cu（OH）<SUB>2</SUB>↓；<br />Ca<SUB>2</SUB>CO<SUB>3</SUB>+2HCl═CaCl<SUB>2</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑．<br />（4）设计实验：小明设计的实验装置和选用药品如A所示，小强设计的实验装置和选用药品如B所示，他们在反应前后都进行了规范的操作、准确的称量和细致的观察．<br />（5）得出结论：<br />①小明认为：在化学反应中，生成物的总质量与反应物的总质量相等；<br />小强认为：在化学反应中，生成物总质量与反应物总质量不相等．<br />②你认为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>的结论正确．<br />③导致另一个实验结论错误的原因是：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />小军认为，若将该装置进行改进也可以得出正确的结论，他改进的方法是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','小明$###$小强的实验漏掉生成物的气体的质量$###$将反应物放在密闭容器中进行','【解答】解：②小明的结论是正确的，因为在质量守恒定律中，反应物的总质量等于生成物的总质量；<br />③小强的实验漏掉生成物的气体的质量；可将反应物放在密闭容器中进行实验．<br />答案：②小明；③小强的实验漏掉生成物的气体的质量，将反应物放在密闭容器中进行．','【分析】在表达质量守恒定律时，一定不要漏掉反应物或生成物的质量，特别是气体物质，容易忽略．选择另外两种药品验证质量守恒定律时，除了两种物质要反应外，最好不生成气体物质．','填空题',3.00,'11a0811645e96efb12ea203743b9a5f3',9,400,'质量守恒定律的实验探究','',2016,'35','2016春•高青县期中',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839853,'已知：Ca（HCO<SUB>3</SUB>）<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>&nbsp;CaCO<SUB>3</SUB>+CO<SUB>2</SUB>↑+H<SUB>2</SUB>O．下列图象表示一定质量的Ca（HCO<SUB>3</SUB>）<SUB>2</SUB>固体受热过程中某些量随时间的变化趋势（该过程中CaCO<SUB>3</SUB>不分解）．其中不正确的是（　　）','<img  src=\"/tikuimages/9/2016/400/shoutiniao37/0fe208c0-94d4-11e9-83f3-b42e9921e93e_xkb89.png\" style=\"vertical-align:middle\" />','<img  src=\"/tikuimages/9/2016/400/shoutiniao80/0fe84a4f-94d4-11e9-a38a-b42e9921e93e_xkb61.png\" style=\"vertical-align:middle\" />','<img  src=\"/tikuimages/9/2016/400/shoutiniao46/0feb0970-94d4-11e9-9b5f-b42e9921e93e_xkb33.png\" style=\"vertical-align:middle\" />','<img  src=\"/tikuimages/9/2016/400/shoutiniao17/0fef2821-94d4-11e9-b4d1-b42e9921e93e_xkb92.png\" style=\"vertical-align:middle\" />','','D','【解答】解：A、刚开始加热时，温度不能达到碳酸氢钙的分解温度，因此要经过一段时间后，碳酸氢钙才开始分解，当碳酸氢钙完全分解后，容器中仍然有一定量的固体--碳酸钙，故A正确；<br />B、钙元素的质量不变，刚开始加热时，温度不能达到碳酸氢钙的分解温度，因此要经过一段时间后，碳酸氢钙才开始分解，当碳酸氢钙完全分解后，容器中仍然有一定量的固体--碳酸钙，所以钙元素的质量分数不变，增大，然后不变，故B正确；<br />C、依据质量守恒定律可知，碳元素的质量始终不变，故C正确；<br />D、加热一段时间后，产生二氧化碳，当碳酸氢钙完全分解后，二氧化碳质量不再增加，故D错误．<br />故选：D．','【分析】碳酸氢钙受热分解成碳酸钙、水和二氧化碳，根据反应情况可以判断相关方面的问题．','选择题',3.00,'259d067a1484d0bfc11d9c77a74de424',9,400,'盐的化学性质','',2016,'37','2016•房山区一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839854,'<img  src=\"/tikuimages/9/2016/400/shoutiniao33/0ff569b0-94d4-11e9-b4d3-b42e9921e93e_xkb97.png\" style=\"vertical-align:middle;FLOAT:right\" />（2016•松北区二模）分类、对比是学习化学的重要方法．<br />（1）请根据数字2的含义对下列符号进行分类：2N、N<SUB>2</SUB>、2Mg<SUP>2+</SUP>、SO<SUB>2</SUB>、2N<SUB>2</SUB>、SO<SUB>4</SUB><SUP>2-</SUP><br />分类依据是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，选出的一组符号为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）“对比实验”是化学学习中行之有效的思维方法．某化学学习小组的同学在实验室做了如下实验，请你参与并回答下列问题．<br />实验中根据塑料瓶变瘪的程度证明了<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，瓶①的作用是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','符号前面的数字均表示微粒个数为2$###$2N、2Mg<SUP>2+</SUP>，2N<SUB>2</SUB>$###$CO<SUB>2</SUB>和NaOH溶液中的溶质确实发生了反应$###$对比实验','【解答】解：（1）分类依据是：符号前面的数字均表示微粒个数为2；选出的一组符号为：2N、2Mg<SUP>2+</SUP>，2N<SUB>2</SUB>；<br />（2）根据塑料瓶变瘪的程度证明CO<SUB>2</SUB>和NaOH溶液中的溶质确实发生了反应，其中瓶①的作用是对比实验；<br />故答案为：（1）符号前面的数字均表示微粒个数为2；2N、2Mg<SUP>2+</SUP>，2N<SUB>2</SUB>；（2）CO<SUB>2</SUB>和NaOH溶液中的溶质确实发生了反应&nbsp;&nbsp;&nbsp;对比实验','【分析】（1）根据化学式表示的意义，化学式中数字的含义回答本题；<br />（2）根据二氧化碳和NaOH溶液反应生成碳酸钠和水进行解答；','填空题',3.00,'731ceb260c16071d8c3529663804f01f',9,400,'碱的化学性质,化学符号及其周围数字的意义','',2016,'37','2016•松北区二模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839856,'<img  src=\"/tikuimages/9/2016/400/shoutiniao15/0ffb360f-94d4-11e9-a03c-b42e9921e93e_xkb20.png\" style=\"vertical-align:middle;FLOAT:right\" />（2016•岑溪市一模）一天，小晨去郊游，在路上意外发现了几枚铜钱，上面有经色斑点．这些绿色斑点是怎么形成的？它的主要成分是什么？具有什么性质？围绕这些问题，她与同学们一起进行了如下探究活动：<br />【查阅资料】铜制品长时间在潮湿的空气中会生成铜绿[Cu<SUB>2</SUB>（OH）<SUB>2</SUB>CO<SUB>3</SUB>]，铜绿不稳定，受热后易分解；无水硫酸铜是一种白色固体，遇水会变蓝．<br />【猜想1】根据铜绿的成分，小晨猜测，铜变成铜绿是铜与空气中的O<SUB>2</SUB>、<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>、<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填化学式）等物质发生反应所致．<br />【猜想2】小曦还推测，铜绿受热分解的生成物为铜、氧气、水和二氧化碳．但小施认为不合理，她的理由是：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />【进行实验】为了验证铜绿受热分解的生成物，同学们从铜钱上取下适量的铜绿，进行了如下实验．<br />①连接装置如图，并检查装置的气密性．<br />②将干燥后的铜绿放入试管A中加热．<br />【解释与结论】<br />（1）A试管口要略向下倾斜，原因是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）B装置的作用是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）在实验中观察到的A装置中绿色固体逐渐变成黑色，取少量黑色固体放入另一试管中，加入稀硫酸，观察到黑色固体逐渐溶解，变成蓝色溶液．则可推测铜绿分解产物中有<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（4）在实验中还观察到C装置中无水硫酸铜变蓝色，小曦认为铜绿分解产物中有<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，但小施认为应该把BC装置调换才能得出正确的结论．她的理由是：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．根据小施的提议，同学们重新组装了仪器并进行了实验．<br />（5）在实验中，除了观察到以上现象外，同学们还观察到B装置中的澄清石灰水变浑浊，请你写出铜绿加热分解的化学方程式：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','H<SUB>2</SUB>O$###$CO<SUB>2</SUB>$###$在加热的条件下，铜能与氧气反应生成氧化铜$###$防止冷凝水倒流引起试管炸裂$###$检验铜绿分解产物中是否含有二氧化碳$###$氧化铜$###$水$###$气体先通过装置B会带出水蒸气，干扰了对水的检验$###$Cu<SUB>2</SUB>（OH）<SUB>2</SUB>CO<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2CuO+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑','【解答】解：【猜想1】根据铜绿的成分含有碳元素、氢元素和氧元素，空气中含有氢元素的是水，含有碳元素的是二氧化碳，故小晨猜测，铜变成铜绿是铜与空气中的O<SUB>2</SUB>、H<SUB>2</SUB>O和CO<SUB>2</SUB>，故填：H<SUB>2</SUB>O、CO<SUB>2</SUB>．<br />【猜想2】由于铜在加热的时候能与氧气反应生成氧化铜，故小曦还推测，铜绿受热分解的生成物为铜、氧气、水和二氧化碳的结论不合理，故填：在加热的条件下，铜能与氧气反应生成氧化铜．<br />【解释与结论】<br />（1）A试管口要略向下倾斜的目的是防止冷凝水倒流引起试管炸裂，故填：防止冷凝水倒流引起试管炸裂．<br />（2）B装置内盛有的是澄清的石灰水，其作用是检验铜绿分解产物中是否含有二氧化碳，故填：检验铜绿分解产物中是否含有二氧化碳．<br />（3）在实验中观察到的A装置中绿色固体逐渐变成黑色，取少量黑色固体放入另一试管中，加入稀硫酸，观察到黑色固体逐渐溶解，变成蓝色溶液．由此可推测铜绿分解产物中有氧化铜，故填：氧化铜．<br />（4）水能使无水硫酸铜反应变蓝，故在实验中还观察到C装置中无水硫酸铜变蓝色，小曦认为铜绿分解产物中有水，把BC装置调换才能得出正确的结论是因为气体先通过装置B会带出水蒸气，干扰了对水的检验，故填：水，气体先通过装置B会带出水蒸气，干扰了对水的检验；<br />（5）在实验中，除了观察到以上现象外，同学们还观察到B装置中的澄清石灰水变浑浊，说明碱式碳酸铜受热分解能生成氧化铜、水和二氧化碳，故填：Cu<SUB>2</SUB>（OH）<SUB>2</SUB>CO<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2CuO+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑','【分析】根据铜绿的成分，结合质量守恒定律推断铜生锈的条件，铜能与氧气加热反应生成氧化铜，根据固体加热的注意事项进行分析解答即可．','书写',3.00,'371966aadce4ad4fbb357e679f164d13',9,400,'实验探究物质的组成成分以及含量,常见气体的检验与除杂方法,金属锈蚀的条件及其防护,盐的化学性质,书写化学方程式、文字表达式、电离方程式','岑溪市',2016,'37','2016•岑溪市一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839857,'下面是小明同学对某无色溶液所含溶质的记录，你认为合理的是（　　）','NaOH、Na<SUB>2</SUB>CO<SUB>3</SUB>、NaCl','NaNO<SUB>3</SUB>、NaCl、CuCl<SUB>2</SUB>','HCl、NaOH、NaCl','Na<SUB>2</SUB>CO<SUB>3</SUB>、Ca（OH）<SUB>2</SUB>、NaCl','','A','【解答】解：A、三者之间不反应，能在同一溶液中大量共存，且不存在有色离子，故选项正确；<br />B、三者之间不反应，能在同一溶液中大量共存，但Cul<SUB>2</SUB>溶于水呈蓝色，故选项错误；<br />C、HCl、NaOH反应生成氯化钠和水，不能在同一溶液中大量共存，故选项错误；<br />D、Na<SUB>2</SUB>CO<SUB>3</SUB>、Ca（OH）<SUB>2</SUB>反应生成碳酸钙沉淀，不能在同一溶液中大量共存，故选项错误；<br />故选：A．','【分析】根据复分解反应发生的条件可知，若物质之间相互交换成分不能生成水、气体、沉淀，则不能在同一溶液中大量共存．本题还要注意是无色溶液，不能含有明显有颜色的铜离子、铁离子和亚铁离子等．','选择题',3.00,'db793b99a3c11c11858fdeaa5c66e253',9,400,'离子或物质的共存问题','',0,'37','',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839858,'下列说法不正确的是（　　）','应广泛使用含磷洗衣粉','金刚石的硬度大，可用来切割玻璃','食品包装中充入氮气以防腐','水通过“三态”变化实现在自然界的循环','','A','【解答】解：A、广泛使用含磷洗衣粉会造成水体富营养化，污染水资源，故A错误；<br />B、金刚石的硬度大，所以可用来切割玻璃，故B正确；<br />C、氮气的化学性质不活泼，所以食品包装中充入氮气以防腐，故C正确；<br />D、水是通过“三态”变化实现在自然界的循环，故D正确．<br />故选：A．','【分析】A、根据广泛使用含磷洗衣粉会造成水体富营养化进行解答；<br />B、根据金刚石的硬度大进行解答；<br />C、根据氮气的化学性质不活泼进行解答；<br />D、根据水是通过“三态”变化实现在自然界的循环进行解答．','选择题',3.00,'3cca38574523dc61f3e8a884eeec12be',9,400,'常见气体的用途,水资源的污染与防治,碳单质的物理性质及用途','',2016,'32','2016•广东模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839861,'下列变化过程中，会吸收热量的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />A&nbsp;硝酸铵溶于水&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;B&nbsp;氢氧化钠溶于水<br />C&nbsp;镁与稀盐酸反应&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;D<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','A$###$CO<SUB>2</SUB>与C反应','【解答】解：A、NH<SUB>4</SUB>NO<SUB>3</SUB>固体溶于水会吸收大量的热，使溶液温度降低．故此选项正确．<br />B、氢氧化钠溶于水放出热量，故此选项错误．<br />C、镁与盐酸反应发生化学变化放出热量，故此选项错误．<br />故选A&nbsp;&nbsp;&nbsp;&nbsp;CO<SUB>2</SUB>与C反应','【分析】吸收热量即变化时温度会降低，对各选项的反应认真分析结合实验实际解决．','填空题',3.00,'d302dd769440f5e15d1ab0c979464ef1',9,400,'溶解时的吸热或放热现象,物质发生化学变化时的能量变化','',2016,'37','2016•上饶三模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839862,'下列关于稀有气体的叙述中不正确的是（　　）','在通电时一般都能发出有色光','都是无色无味的气体','一定不能和其他物质反应','是空气中含量最少的气体','','C|D','【解答】解：A、稀有气体在通电时一般都能放出不同颜色的光，故A说法正确．<br />B、稀有气体都是无色无味的气体，故B说法正确<br />C、稀有气体化学性质不活泼一般不能和其他物质反应，但不是说稀有气体就一定不能和其他物质反应，随着科学的发展人们发现一定条件下稀有气体也能和其他物质反应，故C说法不正确．<br />D、空气中稀有气体的体积分数约为0.94%，含量大于二氧化碳，所以并不是含量最少的气体，故D说法不正确．<br />故答案选CD．','【分析】A、运用稀有气体在通电时一般都能放出不同颜色的光解答．<br />B、运用稀有气体都是无色无味的气体解答．<br />C、运用稀有气体化学性质不活泼一般很难和其他物质反应解答．<br />D、根据空气的组成解答．','多选题',3.00,'1ec29b0c326a7242eec951f224eab2ed',9,400,'空气的成分及各成分的体积分数,常见气体的用途','',0,'37','',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839863,'下列各组物质的鉴别中，所选的鉴别试剂，不正确的是（　　）<br /><table class=\"edittable\"><TBODY><TR><td width=48>　 选项</TD><td width=252>待鉴别物质</TD><td width=132>鉴别试剂</TD></TR><TR><td>　 　A　 </TD><td>氢氧化钠溶液和氢氧化钙溶液</TD><td>二氧化碳气体</TD></TR><TR><td>　B</TD><td>鉴别硫酸铵和硫酸钠固体</TD><td>氢氧化钡溶液</TD></TR><TR><td>　C</TD><td>食盐溶液和稀盐酸</TD><td>紫色石蕊</TD></TR><TR><td>　D</TD><td>氮气和二氧化碳气体</TD><td>燃着的木条</TD></TR></TBODY></TABLE>','A','B','C','D','','D','【解答】解：A、二氧化碳气体能与氢氧化钙溶液反应生成碳酸钙白色沉淀和水，能与氢氧化钠溶液反应生成碳酸钠和水，但无明显变化，可以鉴别，故选项错误．<br />B、氢氧化钡溶液能与硫酸铵反应生成硫酸钡白色沉淀、水和有刺激性气味的氨气，硫酸钠能与氢氧化钡溶液反应生成硫酸钡白色沉淀和氢氧化钠，可以鉴别，故选项错误．<br />C、食盐溶液和稀盐酸分别显中性、酸性，分别能使紫色石蕊溶液显紫色、红色，可以鉴别，故选项错误．<br />D、氮气和二氧化碳气体均不能燃烧、不能支持燃烧，均能使燃着的木条熄灭，不能鉴别，故选项正确．<br />故选：D．','【分析】根据两种物质与同种试剂反应产生的不同现象来鉴别它们，若两种物质与同种物质反应的现象相同，则无法鉴别它们．','选择题',3.00,'f049708665c545fa37319f287e392bd1',9,400,'常见气体的检验与除杂方法,酸、碱、盐的鉴别','',2016,'37','2016•青岛一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839864,'<img  src=\"/tikuimages/9/2016/400/shoutiniao50/1007921e-94d4-11e9-b5a2-b42e9921e93e_xkb67.png\" style=\"vertical-align:middle;FLOAT:right\" />（2016•孟津县一模）如图是甲、乙两种固体物质的溶解度曲线．<br />（1）由图可获得的一条信息是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（2）若乙中混有少量甲，提纯乙的方法是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（3）t<SUB>1</SUB>℃时，将不饱和的甲溶液，转化成该温度下的饱和溶液的一种方法是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，变化过程中，溶质的质量分数<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（选填“增大”、“减小”或“不变”）','','','','','','t<SUB>1</SUB>℃时，甲和乙的溶解度相等$###$蒸发结晶$###$加入溶质（或同温下蒸发溶剂）$###$增大','【解答】解：（1）通过分析溶解度曲线可知，t<SUB>1</SUB>℃时，甲和乙的溶解度相等；<br />（2）乙物质的溶解度受温度影响较大，所以乙中混有少量甲，提纯乙的方法是蒸发结晶；<br />（3）甲物质的溶解度随温度的升高而增大，t<SUB>1</SUB>℃时，将不饱和的甲溶液，转化成该温度下的饱和溶液的一种方法是加入溶质（或同温下蒸发溶剂），变化过程中，溶质的质量分数增大．<br />故答案为：（1）t<SUB>1</SUB>℃时，甲和乙的溶解度相等；<br />（2）蒸发结晶；<br />（3）加入溶质（或同温下蒸发溶剂），增大．','【分析】根据固体的溶解度曲线可以：①查出某物质在一定温度下的溶解度，从而确定物质的溶解性，②比较不同物质在同一温度下的溶解度大小，从而判断饱和溶液中溶质的质量分数的大小，③判断物质的溶解度随温度变化的变化情况，从而判断通过降温结晶还是蒸发结晶的方法达到提纯物质的目的．','填空题',3.00,'1866efd2d3bf9c35dc8ce3d7f0bbd5b3',9,400,'结晶的原理、方法及其应用,饱和溶液和不饱和溶液相互转变的方法,固体溶解度曲线及其作用','',2016,'37','2016•孟津县一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839865,'下列离子在使紫色石蕊溶液变蓝的溶液中，能大量共存的一组是（　　）','Fe<SUP>3+</SUP>、K<SUP>+</SUP>、NO<SUB>3</SUB>ˉ、C1ˉ','Na<SUP>+</SUP>、Κ<SUP>+</SUP>、NO<SUB>3</SUB>ˉ、CO<SUB>3</SUB>&nbsp;<SUP>2-</SUP>','Ag+、NH<SUB>4</SUB><SUP>+</SUP>、C1ˉ、SO<SUB>4</SUB>&nbsp;<SUP>2-</SUP>','H<SUP>+</SUP>、Na<SUP>+</SUP>、HCO<SUB>3</SUB>ˉ、Clˉ','','B','【解答】解：使紫色石蕊溶液变蓝的溶液显碱性，<br />A、氢氧根离子和铁离子结合产生氢氧化铁沉淀，不能共存，故选项错误；<br />B、离子间不能结合产生沉淀气体或水，可以共存，故选项正确；<br />C、银离子和氯离子结合产生氯化银沉淀、铵根离子和氢氧根离子结合产生氨气和水，不能共存，故选项错误；<br />D、氢离子和氢氧根离子结合产生产生水，和碳酸氢根结合产生二氧化碳和水，故选项错误；<br />故选项为：B．','【分析】根据已有的知识进行分析，碱性溶液能使石蕊试液变蓝，结合复分解反应发生的条件可知，若离子之间相互交换成分不能生成水、气体、沉淀，则能够在溶液中大量共存．','选择题',3.00,'d808dedbfaee447bd173c34690e7d0af',9,400,'酸碱指示剂及其性质,离子或物质的共存问题','',2016,'32','2016•雅安校级模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839866,'下列不能用质量守恒定律解释的是（　　）','钢铁生锈','葡萄酿酒','冰雪融化','食物腐烂','','C','【解答】解：A、钢铁生锈时生成新物质，属于化学变化，能用质量守恒定律解释；<br />B、葡萄酿酒时发生化学变化，故能用质量守恒定律来解释；<br />C、冰雪融化是物理变化，所以不能用质量守恒定律解释；<br />D、食物腐烂是化学变化，所以能用质量守恒定律解释．<br />故选C．','【分析】化学变化和物理变化的本质区别是否有新物质生成．化学反应都遵循质量守恒定律，一般物理变化不能用质量守恒定律解释．','选择题',3.00,'4507448c9d80913a1e026caf5ac71be1',9,400,'质量守恒定律及其应用','',2016,'35','2016春•沂源县期中',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839867,'如图表示是KNO<SUB>3</SUB>和NaNO<SUB>3</SUB>的溶解度曲线．下列说法正确的是（　　）<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao75/10159bde-94d4-11e9-8375-b42e9921e93e_xkb86.png\" style=\"vertical-align:middle\" />','t<SUB>1</SUB>℃时，两种溶液中溶质质量分数一定不相等','t<SUB>1</SUB>℃时，在50g水里加入15g&nbsp;KNO<SUB>3</SUB>固体，充分溶解，得到60g溶液','t<SUB>1</SUB>℃时，硝酸钾、硝酸钠两物质饱和溶液升温到t<SUB>2</SUB>℃后，溶质质量分数相等','两种饱和溶液从t<SUB>2</SUB>℃降温到t<SUB>1</SUB>℃时，析出晶体：硝酸钾一定多于硝酸钠','','B','【解答】解：A、由图可知：硝酸钾的溶解度小于硝酸钠的溶解度，若将小于或等于20g的溶质放于100g水中，则两溶液中溶质均能全部溶解，溶质的质量分数相等，故错误；<br />B、t<SUB>1</SUB>℃时硝酸钾的溶解度是20g，即100g水中最多溶解20g的硝酸钾，所以在50g水里加入15g&nbsp;KNO<SUB>3</SUB>固体，最多溶解10g，不能全部溶解，得到60g溶液，正确；<br />C、t<SUB>1</SUB>℃时硝酸钾的溶解度小于硝酸钠的溶解度，饱和时质量分数的计算式<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">溶解度</td></tr><tr><td>溶解度+100g</td></tr></table></span>×100%，即溶解度大则溶质的质量分数大，则硝酸钾的溶质质量分数小于硝酸钠的质量分数；硝酸钾、硝酸钠两物质饱和溶液升温到t<SUB>2</SUB>℃后，溶液中溶质、溶剂的质量不变，溶质的质量分数与升温前相等，所以硝酸钠的溶解度大于硝酸钾溶质质量分数，故错误；<br />D、不知两种饱和溶液的质量，无法确定两种饱和溶液从t<SUB>2</SUB>℃降温到t<SUB>1</SUB>℃时，析出晶体的质量多少，若两溶液相等，则析出晶体质量相等，故错误；<br />故选：B．','【分析】A、据该温度下两种物质的溶解度大小分析解答；<br />B、据该温度下硝酸钾的溶解度分析所加物质是否能全部溶解；<br />C、硝酸钾、硝酸钠的溶解度均随温度升高而增大，并结合升温前两物质的溶解度分析解答；<br />D、等质量的饱和溶液降低相同的温度，溶解度变化大的析出晶体多．','选择题',3.00,'5137a46ef26139aefc267e021c1e1ed3',9,400,'固体溶解度曲线及其作用,晶体和结晶的概念与现象,溶质的质量分数、溶解性和溶解度的关系','',2016,'37','2016•邗江区二模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839869,'<img  src=\"/tikuimages/9/2016/400/shoutiniao98/101a56cf-94d4-11e9-b259-b42e9921e93e_xkb35.png\" style=\"vertical-align:middle;FLOAT:right\" />（2016•高青县模拟）某化学实验兴趣小组做细铁丝在氧气中燃烧的实验，对要求铁丝绕成螺旋状的原因进行探究．<br />【查阅资料】铁丝绕成螺旋状的原因可能是：①增加与氧气的接触面积；②提高未燃部分的温度．<br />【提出问题】铁丝绕成螺旋状的原因是什么？<br />【设计实验】<br /><table class=\"edittable\"><TBODY><TR><td width=205>实验设计及操作 </TD><td width=124>实验现象 </TD><td width=198>实验结论 </TD></TR><TR><td>方案1：按图一实验，待火柴燃烧剩余<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">1</td></tr><tr><td>4</td></tr></table></span>时，伸入集气瓶．（要求：铁丝长度、粗细、集气瓶、水体积、氧气体积相同）</TD><td>①<br /><u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD><td>铁丝绕成螺旋状不是为了增加与氧气的接触面积</TD></TR><TR><td>方案2：按图二实验，待火柴燃烧剩余<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">1</td></tr><tr><td>4</td></tr></table></span>时，伸入集气瓶．（要求同上）</TD><td>②<br /><u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD><td>铁丝绕成螺旋状是为了提高未燃部分的温度</TD></TR></TBODY></TABLE>【交流与反思】<br />（1）铁丝末端系一根火柴的目的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；待火柴燃烧剩余<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">1</td></tr><tr><td>4</td></tr></table></span>时，方可伸入集气瓶，这样操作的目的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；写出细铁丝燃烧的化学方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）细铁丝比粗铁丝更容易燃烧的原因可能是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）由图二实验可知，可燃物燃烧需要的条件是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','B中铁丝剧烈燃烧$###$E中铁丝继续燃烧$###$引燃铁丝$###$防止火柴燃烧消耗氧气$###$3Fe+2O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>Fe<SUB>3</SUB>O<SUB>4</SUB>$###$与氧气接触的更充分$###$温度达到可燃物的着火点','【解答】解：【设计实验】图一中，A中铁丝不燃烧，B中铁丝燃烧；图二中，C中铁丝不燃烧，D中铁丝不燃烧，E中铁丝剧烈燃烧，故填：<table class=\"edittable\"><TBODY><TR><td width=205>实验设计及操作 </TD><td width=124>实验现象 </TD><td width=198>实验结论 </TD></TR><TR><td>方案1：按图一实验，待火柴燃烧剩余<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">1</td></tr><tr><td>4</td></tr></table></span>时，伸入集气瓶．（要求：铁丝长度、粗细、集气瓶、水体积、氧气体积相同）</TD><td>①<br />B中铁丝剧烈燃烧</TD><td>铁丝绕成螺旋状不是为了增加与氧气的接触面积</TD></TR><TR><td>方案2：按图二实验，待火柴燃烧剩余<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">1</td></tr><tr><td>4</td></tr></table></span>时，伸入集气瓶．（要求同上）</TD><td>②<br />E中铁丝继续燃烧</TD><td>铁丝绕成螺旋状是为了提高未燃部分的温度</TD></TR></TBODY></TABLE>；<br />【交流与反思】（1）铁丝末端系一根火柴的目的是引燃铁丝；待火柴燃烧剩余<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">1</td></tr><tr><td>4</td></tr></table></span>时，方可伸入集气瓶，这样操作可以防止火柴燃烧消耗氧气；细铁丝燃烧的化学方程式为3Fe+2O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>Fe<SUB>3</SUB>O<SUB>4</SUB>，故填：引燃铁丝；防止火柴燃烧消耗氧气；3Fe+2O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>Fe<SUB>3</SUB>O<SUB>4</SUB>．<br />（2）细铁丝与粗铁丝相比，与氧气接触的面积更大，有利于燃烧，故填：与氧气接触的更充分；<br />（3）由图二可以看出，可燃物要燃烧，需要温度达到可燃物的着火点，故填：温度达到可燃物的着火点．','【分析】根据铁丝在氧气中燃烧的现象、条件以及化学方程式的书写方法进行分析解答，火柴起到的是引燃的作用，据此解答．','书写',3.00,'e65c15b6d6f797647291b905cf816c76',9,400,'探究氧气的性质,书写化学方程式、文字表达式、电离方程式,燃烧与燃烧的条件','',2016,'32','2016•高青县模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839870,'将过量的Na<SUB>2</SUB>CO<SUB>3</SUB>溶液滴入到一定量CuSO<SUB>4</SUB>溶液中得到蓝色固体．某研究性学习小组对蓝色固体的成分进行了如下探究，请完成下列各题：<br />（一）猜想与假设：<br />猜想一：固体为CuCO<SUB>3</SUB>，理由：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（用化学反应方程式表示）．<br />猜想二：固体为Cu（OH）<SUB>2</SUB>，理由：Na<SUB>2</SUB>CO<SUB>3</SUB>溶液呈<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“酸”或“碱”）性．<br />猜想三：固体为Cu（OH）<SUB>2</SUB>和CuCO<SUB>3</SUB>的混合物．<br />（二）资料查阅：<br />①Cu（OH）<SUB>2</SUB>和CuCO<SUB>3</SUB>晶体均不带结晶水；　　　<br />②Cu（OH）<SUB>2</SUB>、CuCO<SUB>3</SUB>受热易分解，各生成对应的两种氧化物．<br />（三）设计与实验：<br />Ⅰ．固体的获取：<br />（1）将反应后的固、液混合物经<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>、洗涤、低温烘干得蓝色固体．<br />（2）判断固体已洗净的方法及现象<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />Ⅱ．用如图所示装置，定性探究固体的成分．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao92/101fae00-94d4-11e9-8d5d-b42e9921e93e_xkb41.png\" style=\"vertical-align:middle\" /><br />（3）若用装置A、B组合进行实验，B中无现象，则猜想<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>正确；<br />（4）小组同学将装置按　A、<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>、<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“B”、“C”）的顺序组合进行实验，验证出猜想三是正确的，实验中：B中的现象为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，C中的现象为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />结论：固体为Cu（OH）<SUB>2</SUB>和CuCO<SUB>3</SUB>的混合物．','','','','','','Na<SUB>2</SUB>CO<SUB>3</SUB>+CuSO<SUB>4</SUB>=CuCO<SUB>3</SUB>↓+Na<SUB>2</SUB>SO<SUB>4</SUB>$###$碱$###$过滤$###$取最后洗涤液少量，滴加适量的氢氧化钡，无白色沉淀产生；加稀盐酸，无气泡生成$###$二$###$C$###$B$###$澄清的石灰水变浑浊$###$白色固体变蓝','【解答】解：【猜想与假设】猜想一：因为碳酸钠和硫酸铜反应生成碳酸铜沉淀和硫酸钠，故其化学方程式为：Na<SUB>2</SUB>CO<SUB>3</SUB>+CuSO<SUB>4</SUB>=CuCO<SUB>3</SUB>↓+Na<SUB>2</SUB>SO<SUB>4</SUB>；<br />猜想二：碳酸钠是强碱弱酸盐，在水中会进行水解生成氢氧根离子，显碱性；<br />【设计与实验】（1）过滤是将不溶性物质从溶液中分离的一种操作，该题需要将固、液分离；<br />（2）固体上残留的硫酸钠、碳酸钠会与盐酸反应生成二氧化碳气体，故答案为：取最后洗涤液少量，滴加适量的氢氧化钡，无白色沉淀产生；加稀盐酸，无气泡生成；<br />（3）B中无明显现象，说明了没有二氧化碳生成，固体中也就没有碳酸铜，故答案为：二；<br />（4）要证明生成物中有水和二氧化碳，应先证明水再证明二氧化碳，若先证明二氧化碳的话，即使原气体中没有水蒸汽，气体通过石灰水后也会从石灰水中带出水蒸汽．故答案为C、B；澄清的石灰水变浑浊；白色固体变蓝．<br />故答案为：<br />（一） Na<SUB>2</SUB>CO<SUB>3</SUB>+CuSO<SUB>4</SUB>=CuCO<SUB>3</SUB>↓+Na<SUB>2</SUB>SO<SUB>4</SUB>，碱．（三）（1）过滤；（2）取最后洗涤液少量，滴加适量的氢氧化钡，无白色沉淀产生；加稀盐酸，无气泡生成．<br />（3）二；（4）C、B； 澄清的石灰水变浑浊，白色固体变蓝．','【分析】【猜想与假设】猜想一：根据碳酸钠和硫酸铜反应生成蓝色沉淀进行分析，<br />猜想二：根据氢氧根离子会与铜离子生成氢氧化铜沉淀进行分析，<br />【设计与实验】（1）根据将固体和液体进行分离的方法进行分析，<br />（2）根据澄清石灰水会与二氧化碳反应检验二氧化碳的存在进行分析，<br />（3）根据无水硫酸铜遇水变蓝进行分析，<br />（4）要分析检验水和二氧化碳的顺序．','书写',3.00,'0c495ac6f16700687491f08104d1b79a',9,400,'实验探究物质的组成成分以及含量,常见气体的检验与除杂方法,盐的化学性质,书写化学方程式、文字表达式、电离方程式','',2016,'37','2016•柳州二模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839871,'某化学反应的微观过程如图所示，下列说法错误的是（　　）<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao22/10281270-94d4-11e9-ae07-b42e9921e93e_xkb85.png\" style=\"vertical-align:middle\" />','图中共有2种氧化物','反应前后的原子种类和数目没有变','生成物会造成空气污染','参加反应的两物质的化学计量数之比为1：2','','D','【解答】解：由化学反应的微观过程图可知．各物质反应的微粒个数关系是：<img  src=\"/tikuimages/9/2016/400/shoutiniao74/102ca64f-94d4-11e9-8625-b42e9921e93e_xkb34.png\" style=\"vertical-align:middle\" /><br />由上图可知，该反应是硫化氢燃烧生成了二氧化硫和水，反应的化学方程式是：2H<SUB>2</SUB>S+3O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2SO<SUB>2</SUB>+2H<SUB>2</SUB>O．<br />A、由方程式可知，图中共有2种氧化物，故A正确；<br />B、由微粒的变化可知，反应前后的原子种类和数目没有变，故B正确；<br />C、生成物中有二氧化硫，会造成空气污染，故C正确；<br />D、由上图可知，参加反应的两物质的化学计量数之比为2：3，故D错误．<br />故选D．','【分析】观察化学反应的微观过程图，根据微粒的构成分析反应物、生成物，写出反应的化学方程式，据其意义分析判断有关的说法．','选择题',3.00,'afdbf5f6573bafca49a84623500a448c',9,400,'从组成上识别氧化物,微粒观点及模型图的应用','',2016,'32','2016•镇江模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839872,'将24g含有小苏打的胃药（杂质不溶于水，也不与酸反应）加入烧杯中，向其中加入l00g&nbsp;质量分数10.95%的稀盐酸充分反应，过滤，得到108g滤液，则该胃药中主要成分的质量分数为（　　）','80%','70%','87.5%','90%','','B','【解答】解：设反应的碳酸氢钠是质量为x<br />NaHCO<SUB>3</SUB>+HCl═NaCl+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑&nbsp;&nbsp;溶液质量增大<br />&nbsp;&nbsp;84&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 44&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 40<br />&nbsp;&nbsp;x&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 108g-100g<br /><span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">84</td></tr><tr><td>x</td></tr></table></span>=<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">40</td></tr><tr><td>108g-100g</td></tr></table></span><br />x=16.8g<br />所以该胃药中主要成分的质量分数为：<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">16.8g</td></tr><tr><td>24g</td></tr></table></span>×100%=70%．<br />故选：B．','【分析】根据反应前后物质的质量之差可以求出该反应生成的二氧化碳质量，依据化学方程式进行计算．','选择题',3.00,'f3bd38503facd1895df92e1d148c5595',9,400,'有关溶质质量分数的简单计算,根据化学反应方程式的计算','',2016,'37','2016•香坊区二模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839873,'“超临界水”因具有许多优良特质而被科学家追捧，它是指当气压和温度达到一定值时，水的液态和气态完全交融在一起的液体．下面有关“超临界水”的说法正确的是（　　）','它是混合物','它的分子之间有间隔','它的化学性质与水不同','它的一个分子由4个氢原子和2个氧原子构成','','B','【解答】解：A、超临界水只含有水一种物质，属于纯净物，错误；<br />B、超临界水的分子间有一定的间隔，正确；<br />C、超临界水是水的一种状态，不是新物质，它的化学性质与水相同，错误；<br />D、超临界水的一个分子中含有2个氢原子和一个氧原子，错误；<br />故选B．','【分析】根据题干叙述的超临界水的特点以及有关物质的微观构成的知识进行分析解答．','选择题',3.00,'74f5c4851292cbff0f6d9fbcff96734b',9,400,'水的性质和应用,纯净物和混合物的判别,分子的定义与分子的特性','',2016,'37','2016•潍坊二模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839874,'水是生命的源泉，也是不可缺少的资源．<br />（1）某矿泉水的主要矿物质成分机及含量如表，这里Ca、K、Zn、F是指<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．（填“单质、元素、分子或原子”）<br /><table class=\"edittable\"><TBODY><TR><td width=114>成分</TD><td width=114>Ca</TD><td width=114>K</TD><td width=114>Zn</TD><td width=114>F</TD></TR><TR><td>含量（mg/L）</TD><td>20</TD><td>3</TD><td>0.06</TD><td>0.02</TD></TR></TBODY></TABLE>（2）水污染日益严重，水资源的保护和合理利用已受到人们的普遍关注，请参与讨论下列有关问题：<br />（Ⅰ）自来水、冰水混合物、海水中属于纯净物的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（Ⅱ）某学校饮水处可以将自来水净化为饮用水，其中处理步骤如图所示：<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao66/1036df80-94d4-11e9-8fcc-b42e9921e93e_xkb73.png\" style=\"vertical-align:middle\" /><br />①对应的作用是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填字母序号，下同）．<br />A．杀菌消毒&nbsp;&nbsp;&nbsp; B．吸附杂质&nbsp;&nbsp;&nbsp; C．沉淀过滤&nbsp;&nbsp;&nbsp;&nbsp; D．蒸馏<br />（Ⅲ）下列做法会造成水体污染的有<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />A、工业废水直接排放&nbsp;&nbsp;&nbsp;&nbsp;B、工业废气处理后排放<br />C、禁止使用含磷洗衣粉&nbsp;&nbsp;&nbsp;&nbsp;D、大量使用化肥、农药<br />（Ⅳ）能确认水是由氧元素和氢元素组成的实验是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />A、氢气在氧气中燃烧生成水&nbsp;&nbsp;&nbsp;&nbsp;B、水的蒸发&nbsp;&nbsp;&nbsp;&nbsp;C、水的电解&nbsp;&nbsp;&nbsp;&nbsp;D、水的净化<br />（3）长期饮用硬水可能会引起体内结石，鉴别硬水和软水最简单的方法是用<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>鉴别．<br />（4）水通电分解生成最理想的能源--氢气，电解一定量的水，当其中负极产生5ml气体时，正极产生的气体体积是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>ml．','','','','','','元素$###$冰水混合物$###$B$###$AD$###$AD$###$肥皂水$###$2.5','【解答】解：（1）物质的组成常用元素来描述，某矿泉水的主要矿物质成分机及含量如表，这里Ca、K、Zn、F是指元素；<br />（2）（Ⅰ）冰水混合物是由一种物质组成的，属于纯净物．自来水、海水中常含有可溶性的杂质等，属于混合物．<br />（Ⅱ）①由于活性炭具有吸附性，所以①中的活性炭对应的作用是吸附杂质；<br />（Ⅲ）A、工业废水直接排放，会造成水体污染；<br />B、工业废气处理后排放，不会造成水体污染；<br />C、禁止使用含磷洗衣粉，能防止水的污染；<br />D、大量使用化肥、农药，会造成成水体污染．<br />（Ⅳ）A、氢气在氧气中燃烧生成水，说明了水是由氢元素和氧元素组成的；<br />B、水的蒸发是物理变化，不能说明水的组成；<br />C、水的电解生成了氢气和氧气，说明了水是由氢元素和氧元素组成的；<br />D、水的净化是除去了水中的杂质，不能说明水的组成．<br />（3）长期饮用硬水可能会引起体内结石，鉴别硬水和软水最简单的方法是用肥皂水鉴别，遇肥皂水产生的泡沫少的是硬水，遇肥皂水产生的泡沫多的是软水．<br />（4）水通电分解生成最理想的能源--氢气，电解一定量的水，当其中负极产生5ml气体时，正极产生的气体体积是2.5ml．<br />故答为：（1）元素；（2）（Ⅰ）冰水混合物．（Ⅱ）①B；（Ⅲ）AD．（Ⅳ）AD．（3）肥皂水．<br />（4）2.5．','【分析】（1）根据物质的组成分析回答；<br />（2）（Ⅰ）根据各种“水”的组成分析．<br />（Ⅱ）根据活性炭具有吸附性分析；<br />（Ⅲ）根据水体的污染源分析判断；<br />（Ⅳ）根据质量守恒定律分析；<br />（3）根据鉴别硬水和软水最简单的方法分析回答．<br />（4）根据电解水实验的现象和结论分析回答．','填空题',3.00,'d2f7d2759e4d801d052ce4c37188d2d3',9,400,'电解水实验,硬水与软水,水资源的污染与防治,纯净物和混合物的判别,元素的概念','',2016,'37','2016•红桥区一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839875,'下列是实验室制备CO<SUB>2</SUB>，并验证CO<SUB>2</SUB>性质的实验装置图，按要求回答下列问题：<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao88/10407c70-94d4-11e9-9ccb-b42e9921e93e_xkb82.png\" style=\"vertical-align:middle\" /><br />（1）关闭活塞N，打开活塞M，向装置A中加入稀盐酸，装置B处观察到的现象是：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，当观察到C处的石灰水变浑浊时，则C中发生的化学反应方程式是：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）关闭活塞M，打开活塞N，用A、D、E、F、G制取纯净干燥的CO<SUB>2</SUB>气体，并验证二氧化碳的性质，则装置E中应盛装<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>溶液，装置G中观察到的现象为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）实验室要制取干燥的氧气并验满，也可用上述装置．关闭活塞M，打开活塞N，并从A、D、E、F、G装置中，选用<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填字母编号）．','','','','','','导管口有气泡冒出，溶液变红$###$Ca（OH）<SUB>2</SUB>+CO<SUB>2</SUB>═H<SUB>2</SUB>O+CaCO<SUB>3</SUB>↓$###$浓硫酸$###$下层蜡烛先熄灭，上层蜡烛后熄灭$###$A、E、F、G','【解答】解：（1）关闭活塞N，打开活塞M，向仪A中加入稀盐酸，装置B处观察到的现象是：导管口有气泡冒出，溶液变红；二氧化碳与氢氧化钙反应生成碳酸钙和水；<br />故答案为：导管口有气泡冒出；溶液变红；Ca（OH）<SUB>2</SUB>+CO<SUB>2</SUB>═H<SUB>2</SUB>O+CaCO<SUB>3</SUB>↓；<br />（2）装置D的作用是：除去CO<SUB>2</SUB>中混有的HCl，装置E中盛装浓硫酸用于除去CO<SUB>2</SUB>气体中的水分；当观察到：G装置中矮的蜡烛熄灭说明F中已收满二氧化碳气体，下层蜡烛先熄灭，上层蜡烛后熄灭；<br />故答案为：浓硫酸；下层蜡烛先熄灭，上层蜡烛后熄灭；<br />（3）实验室要制取干燥的氧气用A、D、E、F、G中，只需选用的装置是A、E、F、G．<br />故答案为：A、E、F、G．','【分析】二氧化碳能溶于水，密度比空气的密度大，因此只能用向上排空气法收集．关闭活塞N，打开活塞M，向仪器②中加入稀盐酸，装置B处观察到的现象是：导管口有气泡冒出，溶液变红；二氧化碳与氢氧化钙反应生成碳酸钙和水；装置D的作用是：除去CO<SUB>2</SUB>中混有的HCl，装置E中盛装浓硫酸用于出去CO<SUB>2</SUB>气体中的水分；当观察到：G装置中矮的蜡烛熄灭说明F中已收满二氧化碳气体；实验室要制取干燥的氧气用A、D、E、F、G中，只需选用的装置是A、E、F、G．','书写',3.00,'8eb7bbc4b3924dc785d1d1d5f56725e3',9,400,'常见气体的检验与除杂方法,二氧化碳的实验室制法,二氧化碳的物理性质,二氧化碳的化学性质,书写化学方程式、文字表达式、电离方程式','',2016,'32','2016•邹平县模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839876,'某中学化学活动小组的同学对附近一家造纸厂排放的污水中氢氧化钠的含量用如下方法进行了测定，取80g污水于烧杯中，逐滴加入配制好的20%稀硫酸至恰好完全反应（假设污水中的其他成分不与稀硫酸反应），为减小实验误差，他们重复做了3次实验，稀硫酸的用量如下表所示：<table class=\"edittable\"><TBODY><TR><td width=142>实验次数</TD><td width=142>&nbsp;1</TD><td width=142>&nbsp;2</TD><td width=142>3&nbsp;</TD></TR><TR><td>&nbsp;加入稀硫酸的质量/g</TD><td>&nbsp;48</TD><td>&nbsp;49</TD><td>&nbsp;50</TD></TR></TBODY></TABLE>（1）实验中与污水恰好完全反应所需要的稀硫酸的质量为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>g<br />（2）计算污水中所含氢氧化钠的质量分数．','','','','','','49','【解答】解：（1）实验中与污水恰好完全反应所需要的稀硫酸的质量为（48g+49g+50g）÷3=49g．<br />（2）设污水中所含氢氧化钠的质量为x<br />2NaOH+H<SUB>2</SUB>SO<SUB>4</SUB>═Na<SUB>2</SUB>SO<SUB>4</SUB>+2H<SUB>2</SUB>O<br />80&nbsp;&nbsp;&nbsp;&nbsp; 98<br />x&nbsp;&nbsp;&nbsp;&nbsp; 49g×20%<br /><span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">80</td></tr><tr><td>98</td></tr></table>=<table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">x</td></tr><tr><td>49g×20%</td></tr></table></span>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; x=8g<br />污水中所含氢氧化钠的质量分数为<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">8g</td></tr><tr><td>80g</td></tr></table></span>×100%=10%．<br />答：（1）49；（2）污水中所含氢氧化钠的质量分数为10%．','【分析】氢氧化钠与稀硫酸反应生成硫酸钠和水，由3次实验消耗的稀硫酸的平均值，由稀硫酸中溶质的质量分数，由反应的化学方程式列式计算出消耗的氢氧化钠的质量，进而计算出所含氢氧化钠的质量分数．','计算题',3.00,'b3e9171588c555e4a55903c944eabc16',9,400,'有关溶质质量分数的简单计算,根据化学反应方程式的计算','',0,'37','',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839877,'下列对原子、分子、离子的认识，错误的是（　　）','分子由原子构成，原子是不能再分的最小粒子','相同原子可以构成不同的分子','在原子中，原子核内的质子数等于核外电子数','分子、原子、离子都是构成物质的粒子','','A','【解答】解：A、分子由原子构成，原子是化学变化中不能再分的最小粒子，错误；<br />B、相同原子可以构成不同的分子，比如氧原子能构成氧分子和臭氧分子，正确；<br />C、在原子中，原子核内的质子数等于核外电子数，正确；<br />D、分子、原子、离子都是构成物质的粒子，正确；<br />故选A．','【分析】根据已有的物质的微观构成粒子的知识进行分析解答即可．','选择题',3.00,'55cccf746fb25baaa2ca98db4996d8d4',9,400,'分子、原子、离子、元素与物质之间的关系,原子的定义与构成,分子和原子的区别和联系','',2016,'35','2016春•衡阳校级期中',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839878,'<img  src=\"/tikuimages/9/2016/400/shoutiniao50/104bee21-94d4-11e9-9471-b42e9921e93e_xkb75.png\" style=\"vertical-align:middle;FLOAT:right\" />（2016•渝北区模拟）如图为A、B、C三种物质的溶解度曲线，请回答：<br />（1）A、C两物质在<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>℃时的溶解度相等．<br />（2）t<SUB>3</SUB>℃时，30gA物质在50g水中形成饱和溶液，此时A溶液的质量分数是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．（精确到0.1%）<br />（3）现有t<SUB>3</SUB>℃时A、B、C的三种饱和溶液，分别将它们降温到t<SUB>2</SUB>℃时，溶质的质量分数由小到大的顺序是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','t<SUB>2</SUB>$###$33.3%$###$B＞A＞C','【解答】解：（1）A、C两物质的溶解度的交点对应的温度是t<SUB>2</SUB>℃；<br />（2）t<SUB>3</SUB>℃时A的溶解度是50g，所以将30g&nbsp;A物质加入到50g水中最多溶解25g，故此时A溶液的质量分数是<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">25g</td></tr><tr><td>25g+50g</td></tr></table></span>×100%=33.3%；<br />（3）AB的溶解度随温度的升高而增大，C的溶解度随温度的升高而减小，所以t<SUB>3</SUB>℃时，A、B、C的饱和溶液降温至t<SUB>2</SUB>℃时AB要析出晶体，C变为不饱和溶液，溶质的质量分数与降温前相等；据饱和时质量分数的计算式<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">溶解度</td></tr><tr><td>溶解度+100g</td></tr></table></span>×100%可知：溶解度越大质量分数也就越大，t<SUB>2</SUB>℃时B的溶解度大于A的溶解度大于t<SUB>3</SUB>℃时C的溶解度，所以溶质的质量分数由大到小的顺序是B＞A＞C；<br />故答案为：（1）t<SUB>2</SUB>；（2）33.3%；（3）B＞A＞C．','【分析】根据物质的溶解度曲线，t<SUB>2</SUB>°C时A、C两物质的溶解度交于一点；饱和溶液中溶质的质量分数为<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">溶解度</td></tr><tr><td>溶解度+100g</td></tr></table></span>×100%','填空题',3.00,'95407e869b9613bb314d5dd35f743f1f',9,400,'固体溶解度曲线及其作用,溶质的质量分数、溶解性和溶解度的关系','',2016,'32','2016•渝北区模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839879,'从下列物质中选择填空（填序号）<br />①氧气&nbsp;②石蕊溶液&nbsp;③干冰&nbsp;④石墨⑤葡萄糖⑥硝酸铵固体⑦水银<br />（1）常用的酸碱指示剂是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，（2）可用于制铅笔芯和干电池电极的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，<br />（3）能用于供给呼吸的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，（4）可用于人工降雨的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（5）溶于水后溶液温度降低的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，（6）温度计中含有的物质是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，<br />（7）人体主要的供能物质的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','②$###$④$###$①$###$③$###$⑥$###$⑦$###$⑤','【解答】解：<br />（1）石蕊溶液常用作酸碱指示剂；<br />（2）石墨质软、具有导电性，可用于制铅笔芯和干电池电极；<br />（3）氧气可以支持燃烧，帮助呼吸；<br />（4）干冰可以用于人工降雨；<br />（5）溶于水后溶液温度降低的是硝酸铵；<br />（6）温度计中含有的物质是汞；<br />（7）人体主要的供能物质的是葡萄糖．<br />答案：（1）②；（2）④；（3）①；（4）③；（5）⑥；（6）⑦；（7）⑤．','【分析】（1）根据石蕊溶液常用作酸碱指示剂解答；<br />（2）根据石墨具有导电性解答；<br />（3）根据氧气可以支持燃烧，帮助呼吸解答；<br />（4）根据干冰可以用于人工降雨解答；<br />（5）根据溶于水后溶液温度降低的是硝酸铵解答；<br />（6）根据温度计中含有的物质是汞解答；<br />（7）根据人体主要的供能物质的是葡萄糖解答．','填空题',3.00,'3c47a644616053859d40c522f36d582d',9,400,'氧气的用途,二氧化碳的用途,溶解时的吸热或放热现象,金属的物理性质及用途,酸碱指示剂及其性质,碳单质的物理性质及用途,生命活动与六大营养素','',2016,'35','2016春•平凉期中',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839880,'下列物质的化学式、名称和俗称相一致的是（　　）','NaHCO<SUB>3</SUB>&nbsp;&nbsp;碳酸氢钠&nbsp;&nbsp;苏打','C<SUB>2</SUB>H<SUB>5</SUB>OH&nbsp;&nbsp;乙醇&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;酒精','CaO&nbsp;&nbsp;&nbsp;&nbsp;氧化钙&nbsp;&nbsp;&nbsp;&nbsp;熟石灰','NaOH&nbsp;&nbsp;&nbsp;氢氧化钠&nbsp;&nbsp;&nbsp;纯碱','','A','【解答】解：A．碳酸氢钠的俗称是小苏打，其化学式为：NaHCO<SUB>3</SUB>，苏打是碳酸钠的俗称，其化学式、化学名称和俗称不相一致．<br />B．乙醇俗称是酒精，其化学式为C<SUB>2</SUB>H<SUB>5</SUB>OH，其化学式、化学名称和俗称相一致．<br />C．氧化钙的俗称是生石灰，其化学式为：CaO，熟石灰是氢氧化钙的俗称，其化学式、化学名称和俗称不相一致．<br />D．氢氧化钠俗称火碱、烧碱、苛性钠，其化学式为：NaOH，纯碱是碳酸钠的俗称，其化学式、化学名称和俗称不相一致．<br />故选B．','【分析】根据常见化学物质的名称、俗称、化学式进行分析判断即可．','选择题',3.00,'21b7da355f3cade62a5f92f02f323f12',9,400,'化学式的书写及意义','',2016,'32','2016•丹东模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839881,'科学探究是奇妙的过程，在一次实验中小文不小心把硫酸铁溶液滴加到了盛有5%H<SUB>2</SUB>O<SUB>2</SUB>溶液的试管中，立即有大量的气泡产生，硫酸铁溶液中含有三种粒子（H<SUB>2</SUB>O、SO<SUB>4</SUB><SUP>2-</SUP>、Fe<SUP>3+</SUP>），小文想知道硫酸铁溶液中的哪种粒子能使双氧水分解的速度加快，请你和小文一起通过如图1所示的两个实验完成这次探究活动，并填写空白．<br /><img  src=\"/tikuimages/9/2015/400/shoutiniao81/1055d92e-94d4-11e9-b486-b42e9921e93e_xkb7.png\" style=\"vertical-align:middle\" /><br />（1）你认为最不可能的是哪一种粒子<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，理由是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）检验实验产生的气体是否是氧气的方法是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）在实验②中加入稀硫酸后，无明显变化，说明<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>不起催化作用；<br />（4）小明如果要确定硫酸铁是催化剂，还须通过实验确认它在化学反应前后<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>不变．<br />（5）经证明硫酸铁是过氧化氢分解的催化剂，写出硫酸铁催化过氧化氢分解的化学方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（6）．小华通过实验发现，硫酸铁溶液的浓度也会影响过氧化氢分解的速率，常温下，他在a、b、c三个试管中分别加入等量5%过氧化氢的溶液，在其他条件相同时，各滴2滴5%、10%、20%的硫酸铁溶液，根据实验数据绘制了产生氧气体积和时间变化的曲线如图2，请你观察如图2回答问题：<br />①写出两条规律性结论：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />②若实验中所得的曲线如图中虚线所示，请推测其中可能的原因<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','H<SUB>2</SUB>O分子$###$过氧化氢的溶液中含有水分子$###$把带火星的木条伸入试管中，若木条复燃则是氧气$###$SO<SUB>4</SUB><SUP>2-</SUP>$###$质量和化学性质$###$2H<SUB>2</SUB>O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;硫酸铁&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O+O<SUB>2</SUB>↑$###$硫酸铁溶液的浓度越大，过氧化氢分解的速率越大；硫酸铁溶液的浓度不会影响最终产生的氧气质量$###$在其他条件相同时，过氧化氢溶液的浓度变大','【解答】解：（1）过氧化氢溶液中有水，因此是水分子所起催化作用，故不正确；故填：H<SUB>2</SUB>O分子；过氧化氢的溶液中含有水分子；<br />（2）因为氧气具有助燃行，可把一根带火星的木条伸入试管中，若木条复燃则是氧气；故填：把带火星的木条伸入试管中，若木条复燃则是氧气；<br />（3）加入硫酸后，过氧化氢溶液不会分解产生氧气，说明不是硫酸根离子所起的催化作用；故填：SO<SUB>4</SUB><SUP>2-</SUP>；<br />（4）根据催化剂的特征可知，要想确定是否是硫酸铁所起的催化作用，还需要通过实验确定其质量和化学性质没有改变才可以；故填：质量和化学性质没有改变；<br />（5）过氧化氢在硫酸铁的催化作用下分解为水和氧气；故填：2H<SUB>2</SUB>O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;硫酸铁&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O+O<SUB>2</SUB>↑；<br />（6）①根据图象可知，产生氧气的速率与硫酸铁溶液的浓度有关，浓度越大，分解速率越快，但是最终产生氧气的量相同；故填：硫酸铁溶液的浓度越大，过氧化氢分解的速率越大；硫酸铁溶液的浓度不会影响最终产生的氧气质量；<br />②反应物的浓度越大，反应速率越快，产生的氧气越多；故填：在其他条件相同时，过氧化氢溶液的浓度变大．<br />答案：<br />（1）H<SUB>2</SUB>O分子；<br />（2）把带火星的木条伸入试管中，若木条复燃则是氧气；<br />（3）SO<SUB>4</SUB><SUP>2-</SUP>；<br />（4）质量和化学性质；<br />（5）2H<SUB>2</SUB>O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;硫酸铁&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O+O<SUB>2</SUB>↑；<br />（6）①硫酸铁溶液的浓度越大，过氧化氢分解的速率越大；硫酸铁溶液的浓度不会影响最终产生的氧气质量；<br />②在其他条件相同时，过氧化氢溶液的浓度变大．','【分析】（1）根据过氧化氢溶液中含有水分子来分析；<br />（2）根据氧气具有助燃性来分析；<br />（3）根据硫酸中含有硫酸根离子来分析；<br />（4）根据催化剂的特征来分析；<br />（5）根据反应物、生成物以及反应条件来分析书写；<br />（6）根据图象信息来分析解答．','书写',3.00,'25939f5858f85966daf4f197fa5243aa',9,400,'实验探究物质变化的条件和影响物质变化的因素,氧气的检验和验满,催化剂的特点与催化作用,酸的化学性质,书写化学方程式、文字表达式、电离方程式','',2015,'37','2015秋•颍州区校级月考',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839882,'下列除杂设计（括号内为杂质）所选用试剂和操作都正确的是（　　）<br /><table class=\"edittable\"><TBODY><TR><td width=37>序号</TD><td width=152>物质</TD><td width=134>选用试剂</TD><td width=219>操作</TD></TR><TR><td>A</TD><td>KNO<SUB>3</SUB>（NaCl）</TD><td>AgNO<SUB>3</SUB>溶液</TD><td>溶解、过滤</TD></TR><TR><td>B</TD><td>MnO<SUB>2</SUB>（KCl）</TD><td>H<SUB>2</SUB>O</TD><td>溶解、过滤、洗涤、干燥</TD></TR><TR><td>C</TD><td>NaOH溶液（Na<SUB>2</SUB>CO<SUB>3</SUB>）</TD><td>稀盐酸</TD><td>加入试剂至不再产生气泡</TD></TR><TR><td>D</TD><td>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;CO<SUB>2</SUB>（CO）</TD><td>O<SUB>2</SUB></TD><td>点燃</TD></TR></TBODY></TABLE>','A','B','C','D','','B','【解答】解：A、NaCl能与AgNO<SUB>3</SUB>溶液反应生成氯化银沉淀和硝酸钠，再过滤，能除去杂质但引入了新的杂质硝酸钠，不符合除杂原则，故选项所采取的方法错误．<br />B、KCl易溶于水，KCl难溶于水，可采取加水溶解、过滤、洗涤、干燥的方法进行分离除杂，故选项所采取的方法正确．<br />C、Na<SUB>2</SUB>CO<SUB>3</SUB>和NaOH溶液均能与稀盐酸反应，不但能把杂质除去，也会把原物质除去，不符合除杂原则，故选项所采取的方法错误．<br />D、除去二氧化碳中的一氧化碳不能够通氧气点燃，这是因为除去气体中的气体杂质不能使用气体，否则会引入新的气体杂质，故选项所采取的方法错误．<br />故选：B．','【分析】根据原物质和杂质的性质选择适当的除杂剂和分离方法，所谓除杂（提纯），是指除去杂质，同时被提纯物质不得改变．除杂质题至少要满足两个条件：①加入的试剂只能与杂质反应，不能与原物质反应；②反应后不能引入新的杂质．','选择题',3.00,'7bc5d681c659d10ebdbf01cf8ad09c88',9,400,'物质除杂或净化的探究,常见气体的检验与除杂方法,盐的化学性质','',2016,'37','2016•玄武区一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839883,'有①Cu-Zn&nbsp;&nbsp;②Cu-Ag&nbsp;两种貌似黄金的合金，它们都有其特殊的用途，但一些骗子常用它们做成饰品冒充真黄金欺骗消费者．对此，化学科学有责任加以揭露．<br />（1）现有上述两种合金制成的假黄金饰品各一件，小明同学只用稀盐酸和必要的实验仪器就鉴别出其中一种饰品是假黄金，它是合金<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填写代号）；若要证明另一种饰品也是假黄金，可选用一种盐溶液来验证，这种盐可以是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（写一种盐的名称或化学式）；<br />（2）取与上述用稀盐酸鉴别出的同种合金10g，放入烧杯中，再向其中加入93.7g某稀盐酸恰好完全反后，测得烧杯内剩余物质的质量共为103.5g．<br />①上述过程中变化的质量10g，+93.7g-103.5g=0.2g是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填物质的名称或化学式）的质量；<br />②10g，该合金中铜的质量是多少？已知反应后的溶液中溶质全部溶解，求该溶液中溶质的质量分数是多少？','','','','','','①$###$硝酸银或AgNO<SUB>3</SUB>$###$氢气或H<SUB>2</SUB>','【解答】解：（1）只用稀盐酸和必要的实验仪器就鉴别出其中一种饰品是假黄金，它是铜锌合金，因为其中的锌能和稀盐酸反应生成氯化锌和氢气；<br />若要证明另一种饰品也是假黄金，可选用一种盐溶液来验证，这种盐可以是硝酸银，因为硝酸银能和铜反应生成硝酸铜和银．<br />故填：①；硝酸银或AgNO<SUB>3</SUB>．<br />（2）①上述过程中变化的质量10g，+93.7g-103.5g=0.2g是氢气或H<SUB>2</SUB>的质量．<br />故填：氢气或H<SUB>2</SUB>．<br />②设锌的质量为x，生成氯化锌质量为y，<br />Zn+2HCl═ZnCl<SUB>2</SUB>+H<SUB>2</SUB>↑，<br />65&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 136&nbsp;&nbsp;&nbsp;2<br />x&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; y&nbsp;&nbsp; 0.2g<br /><span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">65</td></tr><tr><td>x</td></tr></table></span>=<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">136</td></tr><tr><td>y</td></tr></table></span>=<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">2</td></tr><tr><td>0.2g</td></tr></table></span>，<br />x=6.5g，y=13.6g，<br />该合金中铜的质量是：10g-6.5g=3.5g，<br />该溶液中溶质的质量分数是：<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">13.6g</td></tr><tr><td>6.5g+93.7g-0.2g</td></tr></table></span>×100%=13.6%，<br />答：该合金中铜的质量是3.5g，该溶液中溶质的质量分数是13.6%．','【分析】锌能和稀盐酸反应生成氯化锌和氢气，铜不能和稀盐酸反应；<br />铜能和稀盐酸反应生成硝酸铜和银；<br />根据反应的化学方程式和提供的数据可以进行相关方面的计算．','填空题',3.00,'a0faedaf218e2f2be644c43efcfd59ef',9,400,'有关溶质质量分数的简单计算,金属的化学性质,根据化学反应方程式的计算','',2016,'35','2016春•重庆校级期中',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839884,'下列四个图象与对应叙述相符的是（　　）<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao78/105d7a51-94d4-11e9-be64-b42e9921e93e_xkb54.png\" style=\"vertical-align:middle\" /><br />①向等量的铝粉和镁粉中分别滴入质量分数相同的足量的稀盐酸<br />②将阻力的碳酸镁和氧化镁固体分别加入相同质量、相同溶质质量分数的稀盐酸中<br />③一定量的盐酸与CaCl<SUB>2</SUB>混合溶液中逐滴滴入Na<SUB>2</SUB>CO<SUB>3</SUB>溶质至过量，图中a，b之间对应的溶液的pH＜7<br />④t<SUB>1</SUB>℃时等质量的a、b、c三种物质的饱和溶液降温到t<SUB>1</SUB>℃，所得溶液中溶剂最少的是c的溶液．','①②','①③','③④','①④','','A','【解答】解：①向等量的铝粉和镁粉中分别滴入质量分数相同的足量的稀盐酸，铝生成的氢气多，故正确；<br />②将足量的碳酸镁和氧化镁固体分别加入相同质量、相同溶质质量分数的稀盐酸中，氧化镁生成的氯化镁质量大，故正确；<br />③一定量的盐酸与CaCl<SUB>2</SUB>混合溶液中逐滴滴入Na<SUB>2</SUB>CO<SUB>3</SUB>溶质至过量，oa段是碳酸钠和盐酸反应，所以图中a，b之间对应的溶液的pH=7，故错误；<br />④t<SUB>2</SUB>℃时，a物质的溶解度最大，等质量的a、b、c三种物质的饱和溶液降温到t<SUB>1</SUB>℃，所得溶液中溶剂最少的是a的溶液，故错误．<br />故选：A．','【分析】①根据向等量的铝粉和镁粉中分别滴入质量分数相同的足量的稀盐酸，铝生成的氢气多进行分析；<br />②根据将足量的碳酸镁和氧化镁固体分别加入相同质量、相同溶质质量分数的稀盐酸中，氧化镁生成的氯化镁质量大进行分析；<br />③根据一定量的盐酸与CaCl<SUB>2</SUB>混合溶液中逐滴滴入Na<SUB>2</SUB>CO<SUB>3</SUB>溶质至过量，oa段是碳酸钠和盐酸反应，所以图中a，b之间对应的溶液的pH=7进行分析；<br />④根据t<SUB>2</SUB>℃时，a物质的溶解度最大，等质量的a、b、c三种物质的饱和溶液降温到t<SUB>1</SUB>℃，所得溶液中溶剂最少的是a的溶液进行分析．','选择题',3.00,'4cea50134e4e9d3d2a7bd07d5bfb42ed',9,400,'晶体和结晶的概念与现象,金属的化学性质,溶液的酸碱性与pH值的关系,盐的化学性质','',2016,'37','2016•东河区一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839885,'有一无色液体A，在通电条件下，可以产生B和H两种单质气体．B能使带火星的木条复燃，F是植物光合作用所用的原料．其它关系如图所示，请完成下列问题．<br /><img  src=\"/tikuimages/9/2015/400/shoutiniao98/10647f30-94d4-11e9-9cca-b42e9921e93e_xkb69.png\" style=\"vertical-align:middle\" /><br />（1）写出反应 ①的基本反应类型<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）写出有关物质的化学式：A是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，D是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）写出反应④的化学方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','分解反应$###$H<SUB>2</SUB>O$###$CuO$###$H<SUB>2</SUB>+CuO<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>Cu+H<SUB>2</SUB>O','【解答】解：<br />根据图框可知A通电能生成B和H，则A为水，B和H为氢气或氧气，黑色固体G能在B在这燃烧，则B为氧气，H为氢气，G可能为碳，F可能为二氧化碳，E为红色固体，则E为铜，与氧气反应生成的黑色固体D为氧化铜，代入框图，推断合理．<br />（1）反应①为A通电能生成B和H，即水通电生成氢气和氧气，为分解反应；<br />（2）根据分析，A为水，D为氧化铜，故其化学式分别为：H<SUB>2</SUB>O，CuO；<br />（3）反应④为氢气和氧化铜能反应生成铜和水，所以其化学方程式为：H<SUB>2</SUB>+CuO<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>Cu+H<SUB>2</SUB>O．<br />故答案为：<br />（1）分解反应；（2）H<SUB>2</SUB>O，CuO；（3）H<SUB>2</SUB>+CuO<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>Cu+H<SUB>2</SUB>O．','【分析】此题为框图式物质推断题，完成此类题目，关键是找准解题突破口，根据叙述的关键，以及物质的性质和物质之间的反应，做出判断，A通电能生成B和H，则A为水，B和H为氢气或氧气，黑色固体G能在B在这燃烧，则B为氧气，H为氢气，G可能为碳，F可能为二氧化碳，E为红色固体，则E为铜，与氧气反应生成的黑色固体D为氧化铜，代入框图，推断合理．','书写',3.00,'f893265f75ea7a31d20c43e25729d3e0',9,400,'物质的鉴别、推断,反应类型的判定,书写化学方程式、文字表达式、电离方程式','',2015,'33','2015秋•甘肃校级期末',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839886,'跳跳糖是1974年开始出现在美国市场上的创新糖果，发明人是美国通用食品公司的研发科学家米切尔．跳跳糖的特征跟卖点，就是趣味小糖果颗粒在舌头上噼啪的作响!这项产品一推出就造成风行，成为小孩子的最爱．<br />有同学对跳跳糖为什么能在舌头上噼啪的作响产生疑问．<br />【提出问题】能让跳跳糖在舌头上噼啪作响的是什么？<br />【查阅资料】有人曾经做过实验呢！他们把跳跳糖放在灯下，可以看到内部有气泡，再把跳跳糖放入水中，观察到在它的表面有连续不断的气泡冒出，正是这些气泡使人有“跳”的感觉，该气体能够在高压条件下压入跳跳糖浆中，且该气体能够部分溶于水，能与食品共同保存．<br />【猜想】物质的性质决定用途，说明该物质具有<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>性质；（填写一条相关的物理或化学性质）<br />你猜想能让跳跳糖跳起来的气体是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />【验证】请你用实验证明你的猜想是否正确，并写出验证该气体的方法及现象、结论：<br /><table class=\"edittable\"><TBODY><TR><td width=225>验证方法&nbsp;</TD><td width=236>现象及结论&nbsp;</TD></TR><TR><td><u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD><td><u>&nbsp;&nbsp;&nbsp;&nbsp;</u>&nbsp;</TD></TR></TBODY></TABLE>','','','','','','能溶于水$###$二氧化碳$###$将气体通入澄清石灰水，振荡$###$石灰水变浑浊，证明该气体为二氧化碳','【解答】解：<br />【猜想】根据题中信息可知：该气体能够部分溶于水，能与食品共同保存．物质的性质决定用途，说明该物质具有能溶于水的性质；猜想能让跳跳糖跳起来的气体是二氧化碳；<br />【验证】将气体通入澄清石灰水，振荡，石灰水变浑浊，证明该气体为二氧化碳．<br />答案：<br />【猜想】能溶于水；二氧化碳；<br />【验证】：<br /><table class=\"edittable\"><TBODY><TR><td width=225>验证方法&nbsp;</TD><td width=236>现象及结论&nbsp;</TD></TR><TR><td>将气体通入澄清石灰水，振荡</TD><td>石灰水变浑浊，证明该气体为二氧化碳</TD></TR></TBODY></TABLE>','【分析】【猜想】根据题中信息分析气体性质；<br />【验证】根据验证二氧化碳，气体通入澄清石灰水，石灰水变浑浊，可以确定气体为二氧化碳解答．','填空题',3.00,'3d002a386c7222eee0f03aabbef5be8c',9,400,'实验探究物质的组成成分以及含量,常见气体的检验与除杂方法,化学性质与物理性质的差别及应用','',2014,'35','2014秋•开福区期中',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839887,'<img  src=\"/tikuimages/9/2016/400/shoutiniao41/106ce39e-94d4-11e9-b099-b42e9921e93e_xkb1.png\" style=\"vertical-align:middle;FLOAT:right;\" />（2016•安阳模拟）如下表格是元素周期表的一部分，等质量的四种元素中所含原子个数由多到少的顺序是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，硅与碳位于同一主族，同族元素性质相似，则二氧化硅与NaOH溶液反应的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','Si、P、S、Cl$###$2NaOH+SiO<SUB>2</SUB>=Na<SUB>2</SUB>SiO<SUB>3</SUB>+H<SUB>2</SUB>O','【解答】解：这四种元素的原子序数分别为11、12、13、14，均位于第三周期；原子的相对原子质量与其实际质量是成正比，由于相对原子质量的大小关系是Cl＞S＞P＞Si，真实质量也是这一关系，故等质量的各种原子中，所含原子个数由多到少的顺序是：Si＞P＞S＞Cl．<br />二氧化硅与烧碱溶液反应生成硅酸钠和水，反应的化学方程式为：2NaOH+SiO<SUB>2</SUB>═Na<SUB>2</SUB>SiO<SUB>3</SUB>+H<SUB>2</SUB>O．<br />故答案为：<br />Si＞P＞S＞Cl；&nbsp;SiO<SUB>2</SUB>+2NaOH═Na<SUB>2</SUB>SiO<SUB>3</SUB>+H<SUB>2</SUB>O．','【分析】根据11、12、13、14号元素均位于第三周期；等质量的金属，所含原子个数最多，即原子质量最小；由原子的相对原子质量与其实际质量是成正比的关系，相对原子质量最小的即是原子个数最多的．二氧化硅与烧碱溶液反应生成盐和水，写出反应的化学方程式即可．','书写',3.00,'4036414f5b0e938f8ce61effa0534b23',9,400,'元素周期表的特点及其应用,书写化学方程式、文字表达式、电离方程式','',2016,'32','2016•安阳模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839889,'宣威市生产的松花皮蛋远近闻名．将水、生石灰、纯碱、食盐按一定质量比混合而成料浆，将鲜鸭蛋粘上料浆密封一段时间即可得到松花皮蛋．某化学实验小组的同学对料浆澄清液（“料液”）的主要成分进行探究．<br />【提出问题】料液的主要成分（H<SUB>2</SUB>O除外）是什么？<br />【猜想与假设】实验小组的同学通过充分讨论，作出了如下猜想：<br />①料液中一定含有NaOH、NaCl、Ca（OH）<SUB>2</SUB>②料液中一定含有NaOH、NaCl、Na<SUB>2</SUB>CO<SUB>3</SUB><br />【对猜想的解释】<br />（1）料液中一定没有CaO的原因是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（用化学方程式表示）．<br />（2）料液中一定存在NaOH的原因是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（用化学方程式表示）．<br />【进行实验】<br />①取适量料液放入试管中，滴加2～3滴酚酞试液，料液显红色．<br />②取适量料液放入试管中，滴加少量Na<SUB>2</SUB>CO<SUB>3</SUB>溶液，料液中出现白色沉淀．<br />③取适量料液放入试管中，滴加稀盐酸直到过量，没有现象发生．<br />【实验结论】<br />（3）猜想<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>成立．<br />【交流与拓展】<br />（4）松花皮蛋味道鲜美，但直接食用会稍有涩味．如果将松花皮蛋蘸上食醋食用，则轻微涩味会被去除，味道变得更为鲜美可口．食醋可以去除松花皮蛋涩味的原因是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（5）现在有的皮蛋厂直接用烧碱（NaOH）代替生石灰和纯碱，但制得的皮蛋口感不如宣威生产的松花皮蛋．烧碱一定要密封保存，其理由是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（选序号填空）．<br />A、烧碱易吸水潮解&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;B、烧碱易吸收空气中的CO<SUB>2</SUB>而变质<br />C、烧碱具有强腐蚀性&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;D、烧碱溶于水时放出热量．','','','','','','CaO+H<SUB>2</SUB>O═Ca（OH）<SUB>2</SUB>$###$Ca（OH）<SUB>2</SUB>+Na<SUB>2</SUB>CO<SUB>3</SUB>═CaCO<SUB>3</SUB>↓+2NaOH$###$①$###$食醋中的酸（或醋酸）中和残留在皮蛋中的碱$###$AB','【解答】解：【对猜想的解释】（1）氧化钙和水反应氢氧化钙，化学方程式为：CaO+H<SUB>2</SUB>O═Ca（OH）<SUB>2</SUB>，所以料液中一定没有CaO；<br />（2）氢氧化钙和碳酸钠溶液反应生成碳酸钙沉淀和氢氧化钠，化学方程式为：Ca（OH）<SUB>2</SUB>+Na<SUB>2</SUB>CO<SUB>3</SUB>═CaCO<SUB>3</SUB>↓+2NaOH，所以料液中一定存在NaOH；<br />【实验结论】（3）碳酸钠溶液和氢氧化钙反应生成碳酸钙白色沉淀和氢氧化钠，所以取适量料液放入试管中，滴加少量Na<SUB>2</SUB>CO<SUB>3</SUB>溶液，料液中出现白色沉淀，说明一定含有氢氧化钙，所以猜想①成立；<br />（4）蛋食用时加入一些食醋，可以去除涩感，原因是酸碱起中和反应，所以食醋可以去除松花皮蛋涩味的原因是：食醋中的酸（或醋酸）中和残留在皮蛋中的碱；<br />（5）氢氧化钠固体容易吸水潮解、易吸收空气中的CO<SUB>2</SUB>而变质，所以要密封保存，故选：AB．<br />故答案为：【对猜想的解释】（1）CaO+H<SUB>2</SUB>O═Ca（OH）<SUB>2</SUB>；<br />（2）Ca（OH）<SUB>2</SUB>+Na<SUB>2</SUB>CO<SUB>3</SUB>═CaCO<SUB>3</SUB>↓+2NaOH；<br />【实验结论】（3）①；<br />（4）食醋中的酸（或醋酸）中和残留在皮蛋中的碱；<br />（5）AB．','【分析】【对猜想的解释】（1）根据氧化钙和水反应氢氧化钙进行解答；<br />（2）根据氢氧化钙和碳酸钠溶液反应生成碳酸钙沉淀和氢氧化钠进行解答；<br />【实验结论】（3）根据碳酸钠溶液和氢氧化钙反应生成碳酸钙白色沉淀和氢氧化钠进行解答；<br />（4）根据酸碱中和的原理进行解答；<br />（5）根据氢氧化钠固体容易吸水潮解、易吸收空气中的CO<SUB>2</SUB>而变质进行解答．','书写',3.00,'0ca405c9a9cae0eeba93741dfb087bf2',9,400,'实验探究物质的组成成分以及含量,生石灰的性质与用途,碱的化学性质,盐的化学性质,书写化学方程式、文字表达式、电离方程式','宣威市',2016,'37','2016•宣威市校级一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839890,'<img  src=\"/tikuimages/9/2016/400/shoutiniao57/10763270-94d4-11e9-b203-b42e9921e93e_xkb83.png\" style=\"vertical-align:middle;FLOAT:right\" />（2016•大兴区一模）粗盐通过初步的处理可除去其中的难溶性杂质．<br />（1）粗盐提纯的主要步骤为溶解、过滤、<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．在上述的三个步骤中，均要用到的一种玻璃仪器是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）某同学欲称取12g粗盐做实验，在称量过程中发现天平如图所示，则此时应进行的操作是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','蒸发$###$玻璃棒$###$继续向左盘添加粗盐，直至天平平衡','【解答】解：（1）粗盐的主要成分是氯化钠，粗盐提纯是通过溶解（把不溶物与食盐初步分离）、过滤（把不溶物彻底除去）、蒸发（食盐从溶液中分离出来而得到食盐的过程）．在溶解、过滤、蒸发操作中都要用到的玻璃仪器是玻璃棒，在溶解操作中，玻璃棒起到搅拌以加快粗盐的溶解的作用；过滤操作中，玻璃棒的作用是引流；蒸发操作中，玻璃棒起到搅拌而使液体受热均匀的作用．<br />（2）某同学欲称取12g粗盐做实验，在称量过程中发现天平如图所示，即指针偏右，说明氯化钠的质量小于砝码的质量，故应进行的操作是继续向左盘添加粗盐，直至天平平衡．<br />故答案为：（1）蒸发；玻璃棒；（2）继续向左盘添加粗盐，直至天平平衡．','【分析】（1）粗盐提纯是将粗盐中含有的泥沙等不溶物除去，据此进行分析解答．<br />（2）某同学欲称取12g粗盐做实验，在称量过程中发现天平如图所示，即指针偏右，说明药品质量小于砝码质量，据此进行分析解答．','填空题',3.00,'afd8689311ae461e33a9bd9207020f58',9,400,'称量器-托盘天平,氯化钠与粗盐提纯','',2016,'37','2016•大兴区一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839891,'化学实验操作直接影响实验结果和人身安全，下列关于实验操作的说法不正确的是（　　）','使用酒精灯时，万一洒出酒精在桌面燃烧，立即用湿布盖灭','点燃可燃性气体前必须检验纯度','称量固体药品时，右盘放砝码，左盘放药品','用pH试纸测定溶液的pH时，先用水润湿pH试纸，再将待测液滴在pH试纸上','','D','【解答】解：A、少量酒精洒在桌上并燃烧起来，立即用湿抹布扑盖，即可以降低温度，又可以隔绝空气，所以可以灭火，故正确；<br />B、可燃性气体与氧气混合点燃可能会发生爆炸，所以点燃可燃性气体前必须检验气体纯度，故正确；<br />C、用托盘天平称量固体药品时，应把称量物放在左盘，砝码放在右盘，称量时砝码用镊子夹取，故正确；<br />D、用pH试纸测定未知溶液的pH时，正确的操作方法为用玻璃棒蘸取少量待测液滴在干燥的pH试纸上，与标准比色卡对比来确定pH．不能用水湿润pH试纸，否则稀释了待测溶液，使溶液的酸碱性减弱，测定结果不准确，故错误．<br />故选D．','【分析】A、根据灭火的原理进行分析；<br />B、根据可燃性气体与氧气混合点燃可能会发生爆炸进行分析；<br />C、根据托盘天平的属于方法进行分析；<br />D、根根据用pH试纸测定未知溶液的pH的方法进行分析判断．','选择题',3.00,'65f75d56944d0dcc7dc6221c26ecb045',9,400,'称量器-托盘天平,加热器皿-酒精灯,溶液的酸碱度测定,氢气、一氧化碳、甲烷等可燃气体的验纯','',2016,'37','2016•长清区一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839893,'<img  src=\"/tikuimages/9/2016/400/shoutiniao42/1081cb30-94d4-11e9-bbcb-b42e9921e93e_xkb16.png\" style=\"vertical-align:middle;FLOAT:right\" />（2016春•中山市月考）如图1是a、b、c三种物质的溶解度曲线，回答下列问题：<br />（1）t<SUB>2</SUB>℃时，将30g&nbsp;a物质放入50g水中充分溶解得到溶液为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>溶液（填饱和或不饱和）．<br />（2）将t<SUB>1</SUB>℃时a、b、c三种物质的饱和溶液升温到t<SUB>2</SUB>℃，有晶体析出的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填字母代号）；<br />（3）将t<SUB>2</SUB>℃时，将盛有a的饱和溶液的小试管放入盛水的烧杯中（如图2），向水中加入一定量的硝酸铵固体后，试管中的现象是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','饱和$###$c$###$有晶体析出','【解答】解：<br />（1）t<SUB>2</SUB>℃时，a的溶解度为50&nbsp;g，则在该温度下，50&nbsp;g水中最多溶解25&nbsp;g&nbsp;a物质，将30&nbsp;g&nbsp;a物质加入到50&nbsp;g水中，所得溶液为饱和溶液；<br />（2）升高温度时，c物质的溶解度减小，有晶体析出，a、b物质的溶解度增大，溶液变为不饱和溶液，故所得溶液的溶质质量分数a大于c．<br />（3）硝酸铵溶于水时吸收热量，使a物质的溶解度减小，有晶体析出．<br />故答案为：（1）饱和；（2）c；（3）有晶体析出．','【分析】（1）t<SUB>2</SUB>℃时，a的溶解度为50&nbsp;g，则在该温度下，50&nbsp;g水中最多溶解25&nbsp;g&nbsp;a物质，将30&nbsp;g&nbsp;a物质加入到50&nbsp;g水中，所得溶液为饱和溶液．<br />（2）升高温度时，c物质的溶解度减小，有晶体析出，溶质质量分数减小，A物质的溶解度增大，溶液变为不饱和溶液，但溶质质量分数不变<br />（3）硝酸铵溶于水时吸收热量，使a物质的溶解度减小，有晶体析出．','填空题',3.00,'ffdd283b0ffa41e15e1c7f71940fa5ef',9,400,'溶解时的吸热或放热现象,固体溶解度曲线及其作用,晶体和结晶的概念与现象','中山市',2016,'37','2016春•中山市月考',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839894,'如图是实验室制取某些常见气体所使用的一些装置和仪器．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao98/1089e180-94d4-11e9-b2bb-b42e9921e93e_xkb68.png\" style=\"vertical-align:middle\" /><br />（1）从图1的装置中任意选择一种仪器，写出它的名称：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，简要说出它的一种用途：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）写出实验室制取氧气的一个化学方程式：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，用该方法制取氧气的发生装置可选择图1中的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填标号）．<br />（3）实验室收集二氧化碳通常选择图2中的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填标号）仪器进行组合．欲收集满一瓶二氧化碳气体，收集一段时间后，将燃着的木条放在集气瓶口，若观察到火焰熄灭，则接下来应进行的操作是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','酒精灯$###$用于加热$###$2KMnO<SUB>4</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>K<SUB>2</SUB>MnO<SUB>4</SUB>+MnO<SUB>2</SUB>+O<SUB>2</SUB>↑$###$A$###$CEF$###$将导管移出集气瓶，盖上玻璃片','【解答】（1）从图1的装置中任意选择一种仪器，写出它的名称：酒精灯；它的用途：加热；<br />（2）实验室制取氧气的一个化学方程式（书写方程式应该注意化学式、配平、条件、箭头）：<br />2KMnO<SUB>4</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>K<SUB>2</SUB>MnO<SUB>4</SUB>+MnO<SUB>2</SUB>+O<SUB>2</SUB>↑；利用高锰酸钾制氧气是固体加热，装置是A；<br />（3）实验室制取二氧化碳选择的药品是大理石和稀盐酸，收集二氧化碳通常选择图2中的CEF仪器进行组合．欲收集满一瓶二氧化碳气体，收集一段时间后，将燃着的木条放在集气瓶口，若观察到火焰熄灭，则接下来应进行的操作是将导管移出集气瓶，用玻璃片盖好，正放在桌面上．<br />故答案为：（1）酒精灯、用于加热；<br />（2）2KMnO<SUB>4</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>K<SUB>2</SUB>MnO<SUB>4</SUB>+MnO<SUB>2</SUB>+O<SUB>2</SUB>↑（或2H<SUB>2</SUB>O<SUB>2</SUB>═2H<SUB>2</SUB>O+O<SUB>2</SUB>↑）<br />A（或B）；（3）CEF、将导管移出集气瓶，盖上玻璃片．','【分析】（1）依据实验室常用仪器的认识解决此题；<br />（2）书写化学式应该注意化学式、配平、条件、箭头；利用高锰酸钾制氧气是固体加热，装置是A；<br />（3）实验室通常选择大理石和稀盐酸反应制取二氧化碳；二氧化碳的密度比空气大．','书写',3.00,'6043a7f6ffec0459ffb88331972a6ca5',9,400,'实验室制取氧气的反应原理,二氧化碳的实验室制法,二氧化碳的检验和验满,书写化学方程式、文字表达式、电离方程式','',2016,'37','2016•延庆县一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839895,'A～I是初中化学常见的物质，他们之间的相互关系如图所示，（“→”指向生成物．）其中B是红棕色粉末，A、C常温下是气体，H和I中含有一种相同元素，H不溶于稀硝酸，I是红褐色固体．请回答下面问题：<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao75/1091829e-94d4-11e9-bc60-b42e9921e93e_xkb67.png\" style=\"vertical-align:middle\" /><br />（1）写出物质的化学式，I是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，H是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）反应③的基本反应类型是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）物质F属于<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“酸”、“碱”、“盐”或“氧化物”）．<br />（4）反应①的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．反应②的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','Fe（OH）<SUB>3</SUB>$###$BaSO<SUB>4</SUB>$###$复分解反应$###$碱$###$3CO+Fe<SUB>2</SUB>O<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2Fe+3CO<SUB>2</SUB>$###$Fe<SUB>2</SUB>O<SUB>3</SUB>+3H<SUB>2</SUB>SO<SUB>4</SUB>=Fe<SUB>2</SUB>（SO<SUB>4</SUB>）<SUB>3</SUB>+3H<SUB>2</SUB>O','【解答】解：B是红棕色粉末，A、C常温下是气体，高温条件下，氧化铁和一氧化碳反应生成铁和二氧化碳，因此A是一氧化碳，B是氧化铁，C是二氧化碳，D是铁；氧化铁能够溶于酸产生G，G和F反应产生沉淀H和I，H和I中含有一种相同元素，H不溶于稀硝酸，I是红褐色固体，因此I是氢氧化铁，则E是硫酸，氧化铁和稀硫酸反应生成硫酸铁和水，硫酸铁和氢氧化钡反应生成红褐色沉淀氢氧化铁和白色沉淀硫酸钡，硫酸钡不溶于稀硝酸，带入验证，符合转化关系，因此：<br />（1）I是氢氧化铁，H是硫酸钡；故填：Fe（OH）<SUB>3</SUB>，BaSO<SUB>4</SUB>；<br />（2）反应③是硫酸铁和氢氧化钡反应生成硫酸钡沉淀和氢氧化铁沉淀，该反应属于复分解反应；故填：复分解反应；<br />（3）F是氢氧化钡，属于碱类物质；故填：碱；<br />（4）反应①是一氧化碳和氧化铁生成铁和二氧化碳的反应，反应②氧化铁和稀硫酸反应生成硫酸铁和水的反应；故填：3CO+Fe<SUB>2</SUB>O<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>&nbsp;2Fe+3CO<SUB>2</SUB>；Fe<SUB>2</SUB>O<SUB>3</SUB>+3H<SUB>2</SUB>SO<SUB>4</SUB>=Fe<SUB>2</SUB>（SO<SUB>4</SUB>）<SUB>3</SUB>+3H<SUB>2</SUB>O．','【分析】B是红棕色粉末，A、C常温下是气体，高温条件下，氧化铁和一氧化碳反应生成铁和二氧化碳，因此A是一氧化碳，B是氧化铁，C是二氧化碳，D是铁；氧化铁能够溶于酸产生G，G和F反应产生沉淀H和I，H和I中含有一种相同元素，H不溶于稀硝酸，I是红褐色固体，因此I是氢氧化铁，则E是硫酸，氧化铁和稀硫酸反应生成硫酸铁和水，硫酸铁和氢氧化钡反应生成红褐色沉淀氢氧化铁和白色沉淀硫酸钡，硫酸钡不溶于稀硝酸，据此完成相关的问题．','书写',3.00,'43935250cb66a07e7ee80980d8f9ddc8',9,400,'常见的氧化物、酸、碱和盐的判别,物质的鉴别、推断,书写化学方程式、文字表达式、电离方程式','',2016,'37','2016•龙岗区二模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839896,'<img  src=\"/tikuimages/9/2015/400/shoutiniao24/109664a1-94d4-11e9-8359-b42e9921e93e_xkb43.png\" style=\"vertical-align:middle;FLOAT:right;\" />某校化学小组在利用硫酸和氢氧化钠溶液探究酸碱中和反应时，利用数字化传感器测得烧杯中溶液pH的变化图象，如图所示．下列说法正确的是（　　）','图中c点所示溶液呈碱性','图中a点所示溶液中，含有的溶质是Na<SUB>2</SUB>S0<SUB>4</SUB>和NaOH','该实验是将氢氧化钠溶液逐滴滴入到盛有硫酸的烧杯中','由b点到c点的pH变化，是由于在发生中和反应而产生的','','B','【解答】解：<br />A、c点的pH小于7，溶液显酸性，故错误；<br />B、a点的pH大于7，溶液显碱性，溶液中的溶质是硫酸钠和氢氧化钠，故正确；<br />C、由图象可知，pH值是开始时大于7逐渐的减小到7然后小于7，可知原溶液显碱性，然后不断的加入酸性溶液，使pH减小，说明是把稀盐酸滴加到氢氧化钠溶液中，故错误；<br />D、b点到c点的pH变化，是由于酸过量而产生的，反应已停止，故错误．<br />答案：B','【分析】A、根据c点的pH小于7，溶液显酸性，据此进行分析判断．<br />B、根据a点的pH大于7，溶液显碱性，据此进行分析判断．<br />C、根据图象中pH值的变化是从大于7逐渐的减小到小于7，进行分析解答．<br />D、根据b点到c点的pH变化，是由于酸过量而产生的，反应已停止，进行分析解答．','选择题',3.00,'fa01b94eff00e7dbe886c5a5c7d7dd21',9,400,'中和反应及其应用,溶液的酸碱性与pH值的关系','',2015,'35','2015秋•周村区校级期中',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839897,'生活离不开化学，化学可以改善我们的生活，下列有六种生活中常见物质，请完成以下各题（填序号）：<br />A生石灰、B明矾、C碳酸钙、D不锈钢、E生铁、F活性炭<br />（1）用于制造医疗器械、炊具的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />（2）用作食品干燥剂的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />（3）用作补钙的盐是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />（4）净化水时可作絮凝剂的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','D$###$A$###$C$###$B','【解答】解：（1）不锈钢不易生锈，可用于制造医疗器械、炊具．<br />（2）生石灰能与水反应生成氢氧化钙，可用作食品干燥剂．<br />（3）用作补钙的盐是碳酸钙．<br />（4）明矾溶于水生成的胶状物可对杂质吸附，使杂质沉降，净化水时可作絮凝剂．<br />故答案为：（1）D；（2）A；（3）C；（4）B．','【分析】物质的性质决定物质的用途，根据不锈钢不易生锈，生石灰能与水反应生成氢氧化钙，碳酸钙含有钙元素，明矾溶于水后生成的胶状物对杂质吸附，使杂质沉降，进行分析解答．','填空题',3.00,'aaa112b932c697a9b0b4636c160061c8',9,400,'生铁和钢,生石灰的性质与用途,常用盐的用途','',0,'37','',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839898,'去年11月上旬的一则新闻报道引起了上海市市民的广泛关注：一名三岁的儿童将某种食品袋中的生石灰干燥剂撒入眼中导致一只眼睛失明，这个悲惨事件给她和她的家人带来了巨大的痛苦．<br />（1）上面的事例说明，化学物质可以造福于人类，但使用不当也会给人类带来危害，请你再举出生活中的一种类似的现象来说明这个道理：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（2）生石灰（CaO）可作于干燥剂的理由是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（用化学方程式表示）；<br />（3）看到这报道后，小静留意到，家里有一包包装略有破损的生石灰干燥剂，她想知道这包干燥剂是否还能使用，就通过如下的实验进行了验证：<br /><table class=\"edittable\"><TBODY><TR><td width=143>问题与猜想</TD><td width=143>实验步骤（验证猜想是否正确）</TD><td width=143>实验现象</TD><td width=143>实验结论</TD></TR><TR><td>小纸袋中的物质能否继续作干燥剂</TD><td>取足量小纸袋中的固体放入烧杯中，加入适量水，①触摸杯壁，②滴入无色酚酞试液</TD><td>①<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />②无色酚酞试液不变色</TD><td>生石灰已经完全变质，不能再做干燥剂</TD></TR></TBODY></TABLE>（4）变质后的物质是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','含磷洗衣粉给人带来干净卫生的同时污染了水源；手机电池给人提供电能的同时，废弃电池的重金属污染了土壤和水体；二氧化碳可以灭火，但也能产生温室效应等．$###$CaO+H<SUB>2</SUB>O═Ca（OH）<SUB>2</SUB>$###$不发热$###$碳酸钙','【解答】解：（1）事物都是“一分为二”的，化学物质也是这样，有利的同时也有害，可以从身边的化学举例子，如含磷洗衣粉给人带来干净卫生的同时污染了水源；手机电池给人提供电能的同时，废弃电池的重金属污染了土壤和水体；二氧化碳可以灭火，但也能产生温室效应等．<br />（2）氧化钙能够吸收水，生成氢氧化钙，所以氧化钙可以用作干燥剂，氧化钙和水反应的化学方程式为：<br />CaO+H<SUB>2</SUB>O=Ca（OH）<SUB>2</SUB>．<br />（3）氧化钙和水反应生成氢氧化钙，氢氧化钙与空气中的二氧化碳反应生成碳酸钙，验证生石灰是否变质，就是验证固体中是否有碳酸钙；根据碳酸钙能和盐酸反应生成能使石灰水变浑浊的气体二氧化碳；氢氧化钙属于碱，水溶液显碱性，能使酚酞试液变红色来设计实验；<br />（4）氧化钙和水反应生成氢氧化钙，氢氧化钙与空气中的二氧化碳反应生成碳酸钙．所以是碳酸钙．<br />故答案为：<br />（1）二氧化碳可以灭火，但也能产生温室效应（含磷洗衣粉给人带来干净卫生的同时污染了水源；手机电池给人提供电能的同时，废弃电池的重金属污染了土壤和水体）；<br />（2）CaO+H<SUB>2</SUB>O═Ca（OH）<SUB>2</SUB>；<br />（3）不发热；<br />（4）碳酸钙．','【分析】（1）根据辩证唯物论“一分为二”的观点，可以从身边的化学举例子，如洗衣粉、手机电池、食品添加剂、装饰材料等；<br />（2）根据反应物和生成物及其质量守恒定律可以书写化学方程式；<br />（3）氢氧化钙溶液显碱性，能使石灰水变红色；碳酸钙能和稀盐酸反应生成氯化钙、水和二氧化碳气体；<br />（4）氢氧化钙能和二氧化碳反应生成碳酸钙．','书写',3.00,'4b041019dacec8693a013403f5f40140',9,400,'生石灰的性质与用途,书写化学方程式、文字表达式、电离方程式','',2016,'37','2016•罗平县三模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839899,'（1）燃烧需要的条件为：可燃物、<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>&nbsp;和<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）最近“纸火锅”逐渐流行起来．“纸火锅”是用纸张代替金属材料做容器盛放汤料，当酒精燃烧时纸张不会燃烧．对此现象，下列解释较合理的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />A．纸张不是可燃物，不能燃烧&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;B．纸张被水浸湿，导致着火点降低<br />C．水蒸发时吸热，温度达不到纸张的着火点&nbsp;&nbsp;&nbsp;&nbsp;D．空气不充足，纸张不会燃烧．','','','','','','氧气（或空气）$###$达到可燃物的着火点$###$C','【解答】解：<br />（1）可燃物燃烧的条件是：与氧气或空气充分接触；温度达到着火点；<br />（2）<br />A、纸张具有可燃性，是可燃物，不能燃烧是因为温度没有达到纸张的着火点，故选项解释错误；<br />B、可燃物的着火点一般是不变的，纸张的着火点一般情况下不能改变，故选项解释错误．<br />C、水蒸发时吸热，导致温度达不到纸张的着火点，因此纸张不能燃烧，故选项解释正确．<br />D、纸张与空气充分接触，不能燃烧是因为温度没有达到纸张的着火点，故选项解释错误．<br />故答案为：<br />（1）氧气（或空气）；&nbsp;&nbsp;达到可燃物的着火点；<br />（2）C．','【分析】（1）根据燃烧的条件来分析解答．<br />（2）根据燃烧的条件（燃烧需要同时满足三个条件：①可燃物、②氧气或空气、③温度要达到着火点），进行分析解答．','填空题',3.00,'d7027d1408b186abf70e5b4023d7d28d',9,400,'燃烧与燃烧的条件','冷水江市',2014,'33','2014秋•冷水江市期末',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839900,'根据实验装置图，回答下列问题：<br /><img  src=\"/tikuimages/9/2015/400/shoutiniao36/10a1af40-94d4-11e9-9e85-b42e9921e93e_xkb55.png\" style=\"vertical-align:middle\" /><br />①用过氧化氢溶液和二氧化锰混合制取氧气应选用的发生装置是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，采用此装置的原因是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；其中二氧化锰在反应中起<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>作用．<br />②实验室制取CO<SUB>2</SUB>的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；收集二氧化碳气体应选择的装置是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，采用此收集方法的原因是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />③如装置F所示，点燃两支短蜡烛，然后沿烧杯内壁倾倒二氧化碳时，看到F中的实验现象是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；证明CO<SUB>2</SUB>具有<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>的性质．<br />④某同学选用G装置进行实验，看到与上一步实验F中完全相同的实验现象，他由此得出与上一步实验完全相同的结论．此结论（答“合理”或“不合理”）<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','B$###$固体与液体混合且不需加$###$催化$###$CaCO<SUB>3</SUB>+2HCl=CaCl<SUB>2</SUB>+CO<SUB>2</SUB>↑+H<SUB>2</SUB>O$###$E$###$二氧化碳的密度大于空气且能溶于水$###$蜡烛由下至上依次熄灭$###$二氧化碳的密度大于空气，不能燃烧且不支持燃烧$###$不合理','【解答】解：<br />①用过氧化氢溶液和二氧化锰混合制取氧气，不需加热，属于固液常温型，不需要加热，故选发生装置B；其中二氧化锰起催化作用；<br />②实验室用大理石和稀盐酸常温反应制取二氧化碳，反应的方程式为：CaCO<SUB>3</SUB>+2HCl=CaCl<SUB>2</SUB>+CO<SUB>2</SUB>↑+H<SUB>2</SUB>O；二氧化碳的密度比空气大且能溶于水，故只能用向上排空气法收集；<br />③装置F中，沿烧杯内壁倾倒二氧化碳时，看到的实验现象是：下面的蜡烛先熄灭，上面的蜡烛后熄灭；证明CO<SUB>2</SUB>具有的性质是：二氧化碳既不能燃烧也不能支持燃烧；二氧化碳的密度比空气小；选用G装置进行实验，实验现象是：下面的蜡烛先熄灭，上面的蜡烛后熄灭；因此得出的结论与F装置相同；比较两个发现可以发现，气体进入的方向不一样，F装置中气体从上面进入，G装置中气体直接从下面开始．所以即使看到下面的蜡烛先熄灭，也无法断定密度比空气大，故答案为：不合理．<br />故答案为：①B；固体与液体混合且不需加热；催化②CaC03+2HCl=CaCl2+H2O+CO2↑；E；二氧化碳的密度大于空气且能溶于水；③蜡烛由下至上依次熄灭；二氧化碳的密度大于空气，不能燃烧且不支持燃烧；④不合理','【分析】①根据用过氧化氢溶液和二氧化锰混合制取氧气，不需加热，属于固液常温型解答；<br />②根据制取二氧化碳的反应物的状态和反应条件选择发生装置，并据二氧化碳的密度和溶解性选择收集<br />③根据装置F中，沿烧杯内壁倾倒二氧化碳时，看到的实验现象是：下面的蜡烛先熄灭，上面的蜡烛后熄灭；证明CO<SUB>2</SUB>具有的性质是：二氧化碳既不能燃烧也不能支持燃烧；二氧化碳的密度比空气小．选用G装置进行实验，实验现象是：下面的蜡烛先熄灭，上面的蜡烛后熄灭；因此得出的结论与F装置相同解答．','书写',3.00,'41c51faa57a569069e6163163eaad6d2',9,400,'催化剂的特点与催化作用,二氧化碳的实验室制法,二氧化碳的物理性质,二氧化碳的化学性质,书写化学方程式、文字表达式、电离方程式','冷水江市',2015,'32','2015•冷水江市校级模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839902,'分析推理是化学学习中常用的思维方法．下列分析推理正确的是（　　）','O<SUB>2</SUB>和O<SUB>3</SUB>的分子构成不同，所以它们的化学性质不同','浓硫酸具有吸水性，所以浓盐酸也具有吸水性','离子是带电荷的微粒，所以带电荷的微粒一定是离子','因为酸溶液呈酸性，所以呈酸性的溶液一定是酸溶液','','A','【解答】解：A、O<SUB>2</SUB>和O<SUB>3</SUB>的分子构成不同，它们的化学性质不同，故选项推理正确．<br />B、浓硫酸具有吸水性，但浓盐酸不具有吸水性，浓盐酸具有挥发性，故选项推理错误．<br />C、离子是带电荷的微粒，但带电荷的微粒不一定是离子，也可能是质子、电子等，故选项推理错误．<br />D、酸溶液呈酸性，但呈酸性的溶液不一定是酸溶液，也可能是硫酸氢钠等盐溶液，故选项推理错误．<br />故选：A．','【分析】A、同种的分子性质相同，不同种的分子性质不同，可以简记为：“两小运间，同同不不”．<br />B、根据浓盐酸具有挥发性，进行分析判断．<br />C、根据常见的带电荷的微粒，进行分析判断．<br />D、根据呈酸性的溶液不一定是酸溶液，进行分析判断．','选择题',3.00,'02767ecee24e985d076053b1118a6709',9,400,'酸的物理性质及用途,酸的化学性质,原子和离子的相互转化,分子的定义与分子的特性','',2016,'37','2016•长清区二模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839905,'用化学用语表示：<br />（1）硫酸铁<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（2）自然界中存在最硬的物质<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（3）5个硝酸根离子<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（4）保持五氧化二磷化学性质的最小粒子<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','Fe<SUB>2</SUB>（SO<SUB>4</SUB>）<SUB>3</SUB>$###$C$###$5NO<SUB>3</SUB><SUP>-</SUP>$###$P<SUB>2</SUB>O<SUB>5</SUB>','【解答】解：（1）硫酸铁的化学式为：Fe<SUB>2</SUB>（SO<SUB>4</SUB>）<SUB>3</SUB>；<br />（2）自然界中存在最硬的物质是金刚石，其是由原子直接构成的，其化学式就是其元素符号，故为：C；<br />（3）离子的表示方法：在表示该离子的元素符号右上角，标出该离子所带的正负电荷数，数字在前，正负符号在后，带1个电荷时，1要省略．若表示多个该离子，就在其元素符号前加上相应的数字，故5个硝酸根离子可表示为：5NO<SUB>3</SUB><SUP>-</SUP>；<br />（4）保持五氧化二磷化学性质的最小粒子五氧化二磷分子，其化学式为：P<SUB>2</SUB>O<SUB>5</SUB>；<br />故答案为：（1）Fe<SUB>2</SUB>（SO<SUB>4</SUB>）<SUB>3</SUB>&nbsp;（2）C&nbsp;&nbsp;（3）5NO<SUB>3</SUB><SUP>-</SUP>&nbsp;&nbsp; （4）P<SUB>2</SUB>O<SUB>5</SUB>','【分析】本题考查化学用语的意义及书写，解题关键是分清化学用语所表达的对象是分子、原子、离子还是化合价，才能在化学符号前或其它位置加上适当的计量数来完整地表达其意义，并能根据物质化学式的书写规则正确书写物质的化学式，才能熟练准确的解答此类题目．','填空题',3.00,'0edea271519f584107b408eaa10bbace',9,400,'化学符号及其周围数字的意义','',2016,'32','2016•宜春模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839906,'今有化学反应：4XY<SUB>2</SUB>+Y<SUB>2</SUB>=2Z，则生成物Z的化学式可能用X、Y表示为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','X<SUB>2</SUB>Y<SUB>5</SUB>','【解答】解：根据所给化学方程式可知，反应前含有4个X原子，10个Y原子，根据质量守恒定律可知2个Z中应该含有10个Y原子，4个X原子，即1个Z中含有5个Y原子，2个X原子，即化学式为X<SUB>2</SUB>Y<SUB>5</SUB>．<br />故答案为：X<SUB>2</SUB>Y<SUB>5</SUB>．','【分析】根据质量守恒定律可知：反应前后元素种类不变，原子个数不变．由以上依据可推出Z的化学式．','填空题',3.00,'209663ac1082b830b890987aa6b6ad47',9,400,'质量守恒定律及其应用','',0,'37','',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839907,'乙酸乙酯（&nbsp;C<SUB>4</SUB>H<SUB>8</SUB>O<SUB>2</SUB>）常用作食品、饮料的调香剂．下列有关乙酸乙酯的叙述正确的是（　　）','由14个原子构成','其中碳元素的质量分数为41.4%','一个乙酸乙酯分子中，碳、氢、氧原子的个数比为2：4：1','其中碳、氢、氧元素的质量比为12：1：16','','C','【解答】解：A．乙酸乙酯是由乙酸乙酯分子构成的，1个乙酸乙酯分子中含有14个原子，故选项说法错误．<br />B．其中碳元素的质量分数为<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">12×4</td></tr><tr><td>12×4+1×8+16×2</td></tr></table></span>×100%≈54.5%，故选项说法错误．<br />C．一个乙酸乙酯分子是由4个碳原子、8个氢原子和2个氧原子构成的，一个乙酸乙酯分子中，碳、氢、氧原子的个数比为2：4：1，故选项说法正确．<br />D．乙酸乙酯中碳、氢、氧元素的质量比为（12×4）：（1×8）：（16×2）=6：1：4，故选项说法错误．<br />故选C．','【分析】A．根据乙酸乙酯的微观构成，进行分析判断．<br />B．根据化合物中元素的质量分数的计算方法来分析．<br />C．根据一个乙酸乙酯分子的构成进行分析判断．<br />D．根据化合物中各元素质量比=各原子的相对原子质量×原子个数之比，进行分析判断．','选择题',3.00,'7223f503301a479df1c318c3f99bd9b1',9,400,'化学式的书写及意义,元素质量比的计算,元素的质量分数计算','太仓市',2016,'32','2016•太仓市模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839908,'盛有饱和氢氧化钠溶液的烧杯在空气中放置一段较长时间，该溶液中的（　　）','氢氧化钠的质量保持不变','氢氧化钠的质量变小','氢氧化钠的质量分数增大','氢氧化钠的质量分数保持不变','','B','【解答】解：A、因为氢氧化钠和空气中的二氧化碳发生了化学反应，所以溶液中氢氧化钠质量减小，该选项说法不正确；<br />B、因为氢氧化钠和空气中的二氧化碳发生了化学反应，所以溶液中氢氧化钠质量减小，该选项说法正确；<br />C、溶液中氢氧化钠质量减小，而溶液的质量变化情况无法确定，因此不能确定氢氧化钠质量分数变化情况，该选项说法不正确；<br />D、溶液中氢氧化钠质量减小，而溶液的质量变化情况无法确定，因此不能确定氢氧化钠质量分数变化情况，该选项说法不正确．<br />故选：B．','【分析】盛有饱和氢氧化钠溶液的烧杯在空气中放置一段较长时间时，氢氧化钠能和空气中的二氧化碳反应生成碳酸钠和水，同时溶液中的水不断蒸发．','选择题',3.00,'8e6f10c909f5e6a122597c8393903c5f',9,400,'溶质的质量分数,碱的化学性质','',2016,'37','2016春•广州月考',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839910,'蒲公英，又名黄花地丁，是早春很好的一种野生蔬菜．有利尿、缓泻、退黄疸、利胆等功效．蒲公英的有效成分蒲公英甾醇的化学式为C<SUB>30</SUB>H<SUB>50</SUB>O，下列有关说法正确的是（　　）','蒲公英甾醇属于氧化物','蒲公英甾醇的相对分子质量为426','蒲公英甾醇中C、H、O三种元素的质量比为30：50：1','蒲公英甾醇中氢元素的质量分数最大','','B','【解答】解：A．由物质的化学式可知，该物质含有三种元素，不符合氧化物是由两种元素组成的特征，故不属于氧化物，故错误；<br />B．蒲公英甾醇的相对分子质量为：12×30+1×50+16=426，故正确；<br />C．蒲公英甾醇中C、H、O三种元素的质量比为（12×30）：（1×50）：16=180：25：8，故错误；<br />D．蒲公英甾醇中C、H、O三种元素的质量比为（12×30）：（1×50）：16=180：25：8，由此可见其中碳元素的质量分数最大，故错误．<br />故选B．','【分析】A．根据氧化物的概念来分析；<br />B．根据相对分子质量为构成分子的各原子的相对原子质量之和，进行分析；<br />C．根据化合物中元素质量比的计算方法来分析；<br />D．根据化合物中元素质量比的计算方法来分析．','选择题',3.00,'18d01f5d7647c1afeba73cbf6ada3713',9,400,'从组成上识别氧化物,化学式的书写及意义,相对分子质量的概念及其计算,元素质量比的计算,元素的质量分数计算','',2016,'32','2016•陕西模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839911,'下列图示实验操作中，正确的是（　　）','<img  src=\"/tikuimages/9/2015/400/shoutiniao35/10bd4d8f-94d4-11e9-8c0f-b42e9921e93e_xkb62.png\" style=\"vertical-align:middle\" />','<img  src=\"/tikuimages/9/2015/400/shoutiniao25/10c0a8f0-94d4-11e9-803b-b42e9921e93e_xkb88.png\" style=\"vertical-align:middle\" />','<img  src=\"/tikuimages/9/2015/400/shoutiniao33/10c0a8f1-94d4-11e9-a350-b42e9921e93e_xkb70.png\" style=\"vertical-align:middle\" />','<img  src=\"/tikuimages/9/2015/400/shoutiniao19/10c4eeb0-94d4-11e9-a609-b42e9921e93e_xkb7.png\" style=\"vertical-align:middle\" />','','C','【解答】解：A、点燃酒精灯时，用火柴点燃，如果用燃着的酒精灯去点燃另一个酒精灯，会引起酒精失火，造成危险，故A错误；<br />B、使用托盘天平是，注意“左物右码”，托盘天平用于粗略称量药品的质量，不能精确到0.01g，不能用托盘天平称量10.05g固体，故B错误；<br />C、给液体加热时，要用酒精灯的外焰加热，试管内液体不能超过其体积的<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">1</td></tr><tr><td>3</td></tr></table></span>，防止沸腾溅出，故C正确；<br />D、量筒量液时要注意量程的选择，应选择略大于量取液体体积的量程．故量取9.5mL液体体积应选10mL的量筒，故D错误；<br />故选C．','【分析】A、根据点燃酒精灯的方法分析；<br />B、根据托盘天平的称量能准确到0.1g，不能精确到0.01g进行分析；<br />C、根据给液体加热时的注意事项分析；<br />D、根据用量筒量液时要注意量程的选择进行分析．','选择题',3.00,'57f4c343e81a6e8328195da2ebcd05c7',9,400,'测量容器-量筒,称量器-托盘天平,加热器皿-酒精灯,给试管里的液体加热','',2015,'37','2015秋•烟台校级月考',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839912,'生活即化学，生活中有很多的变化都蕴含着化学原理，下列变化属于化学变化的是（　　）','汽油挥发','燃放烟花','活性炭吸附冰箱内的异味','分离液态空气制取氧气','','B','【解答】解：A、汽油挥发过程中只是状态发生改变，没有新物质生成，属于物理变化．<br />B、燃放烟花过程中有新物质生成，属于化学变化．<br />C、活性炭吸附冰箱内的异味过程中没有新物质生成，属于物理变化．<br />D、分离液态空气制取氧气过程中没有新物质生成，属于物理变化．<br />故选B．','【分析】化学变化是指有新物质生成的变化，物理变化是指没有新物质生成的变化，化学变化和物理变化的本质区别是否有新物质生成；据此分析判断．','选择题',3.00,'e270f381f89deb59dc859d56e3156ea1',9,400,'化学变化和物理变化的判别','',2015,'33','2015秋•云南校级期末',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839914,'下列各组物质的鉴别方法中不正确的是（　　）','用灼烧的方法鉴别羊毛和合成纤维','用盐酸鉴别氢氧化钠是否变质','用石灰水区别氧气和二氧化碳','用观察颜色的方法鉴别氧化铜和四氧化三铁','','D','【解答】解：A、灼烧羊毛和合成纤维时，有烧焦的羽毛味的是羊毛，产生刺激性气味的为合成纤维，因此，用灼烧的方法可以鉴别羊毛和合成纤维．故不符合题意；<br />B、盐酸和碳酸钠反应生成气体，故用盐酸鉴别氢氧化钠是否变质．故不符合题意；<br />C、氧气与石灰水不反应，而二氧化碳与石灰水反应，生成白色沉淀，现象明显，可以鉴别．故不符合题意；<br />D、因为氧化铜和四氧化三铁都为黑色，用观察颜色的方法无法鉴别．故D符合题意；<br />故选D．','【分析】A、考虑羊毛为蛋白质，灼烧时有烧焦的羽毛味；合成纤维含有多种元素，灼烧时产生刺激性气味；<br />B、考虑用盐酸和碳酸钠反应生成气体；<br />C、考虑二氧化碳与石灰水反应的现象，进行分析；<br />D、考虑氧化铜和四氧化三铁的颜色．','选择题',3.00,'0ff062a56a03e70d60a080d823540e60',9,400,'常见气体的检验与除杂方法,碱的化学性质,物质的鉴别、推断,棉纤维、羊毛纤维和合成纤维的鉴别','',2016,'37','2016•柳州二模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839915,'下列有关溶液的说法不正确的是（　　）','降低温度或增大压强均可以增大气体在水中的溶解度','物质的溶解度不一定都随温度升高而增大','在一定条件下，饱和溶液与不饱和溶液之间可相互转化','溶液中有晶体析出时，溶质质量减小，所以溶质的质量分数一定减小','','D','【解答】解：A、气体的溶解度随温度的升高、压强降低而增大，所以增大压强和降低温度均可增大气体在水中的溶解度，故选项说法正确．<br />B、物质的溶解度不一定都随温度升高而增大：如熟石灰随温度的升高溶解度减小；故选项说法正确．<br />C、改变温度、增加溶质的量或蒸发溶剂或增加溶剂饱和溶液与不饱和溶液之间可相互转化；故选项说法正确．<br />D、若饱和溶液通过蒸发溶剂后有晶体析出，溶质的质量分数不变，故D选项说法错误．<br />故选：D．','【分析】A、增大压强和降低温度均可增大气体在水中溶解的体积．<br />B、根据溶解度的影响因素来做题；<br />C、根据饱和溶液与不饱和溶液的相互转化关系回答；<br />D、若饱和溶液通过蒸发溶剂后有晶体析出，溶质的质量分数不变．','选择题',3.00,'a510edc3ee777a767efb5520a524f327',9,400,'饱和溶液和不饱和溶液相互转变的方法,固体溶解度的影响因素,气体溶解度的影响因素,晶体和结晶的概念与现象','',2016,'37','2016•黑龙江一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839916,'期末化学实验老师整理药品时，拿出一瓶久置的氢氧化钙粉末[Ca（OH）<SUB>2</SUB>]，他让小刘和小赵对这瓶氢氧化钙粉末的组成进行实验探究．<br />（1）提出问题：这瓶氢氧化钙是否已经生成碳酸钙（CaCO<SUB>3</SUB>）而变质？<br />（2）进行猜想：<br />A：氢氧化钙全部变为碳酸钙；B：氢氧化钙部分变为碳酸钙；C：氢氧化钙没有变质．<br />（3）设计实验方案，进行实验：如表是对猜想A进行实验探究的过程示例：<br /><table class=\"edittable\"><TBODY><TR><td width=198>实验步骤</TD><td width=198>实验现象</TD><td width=181>实验结论</TD></TR><TR><td>取样，加适量水，搅拌，过滤<br />①取少量滤液于试管中，滴入酚酞试液<br />②取少量滤渣于试管中，加入盐酸</TD><td>①滤液不变色<br />②有气泡产生</TD><td>氢氧化钙全部变为碳酸钙</TD></TR></TBODY></TABLE>请你另选择一种猜想参与探究，完成如表．<br /><table class=\"edittable\"><TBODY><TR><td width=205>实验步骤</TD><td width=205>实验现象</TD><td width=167>实验结论</TD></TR><TR><td>取样，加适量水，搅拌，过滤<br />①取少量滤液于试管中，滴入酚酞试液<br />②取少量滤渣于试管中，加入盐酸</TD><td>①<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />②<u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD><td><u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD></TR></TBODY></TABLE>（4）原理与用途：<br />①氢氧化钙俗称<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，农业上一般采用氢氧化钙改良<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>性土壤（填“酸”或“碱”）．<br />②氢氧化钙变质是由于与空气中的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>发生反应的缘故，反应的化学方程式为：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，因此氢氧化钙应<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>保存．','','','','','','滤液变为红色$###$无气泡产生（或无现象）$###$氢氧化钙没有变质$###$熟石灰（或消石灰）$###$酸$###$CO<SUB>2</SUB>（或二氧化碳）$###$CO<SUB>2</SUB>+Ca（OH）<SUB>2</SUB>═CaCO<SUB>3</SUB>↓+H<SUB>2</SUB>O$###$密封','【解答】解：（3）取样品，加适量水，搅拌，过滤，向滤液中滴加无色酚酞，若观察到变红说明含有氢氧化钙；取滤渣滴加稀盐酸，若观察到有气泡，则可说明含有碳酸钙，可得出部分变质的结论；若无气泡产生，则滤渣不含碳酸钙，则可判断氢氧化钙没有变质；<br /><table class=\"edittable\"><TBODY><TR><td width=284>实验现象</TD><td width=284>实验结论</TD></TR><TR><td>①滤液变为红色<br />②有气泡产生</TD><td>氢氧化钙部分变为碳酸钙</TD></TR><TR><td>①滤液变为红色<br />②无气泡产生（或无明显现象）</TD><td>氢氧化钙没有变质</TD></TR></TBODY></TABLE>（4）<br />①氢氧化钙又被称为熟石灰和消石灰，由于氢氧化钙属于碱，能与酸发生中和反应，因此农业上一般采用氢氧化钙改良酸性土壤；<br />②氢氧化钙变质是由于与空气中的CO<SUB>2</SUB>（或二氧化碳）发生反应的缘故，反应的化学方程式为：CO<SUB>2</SUB>+Ca（OH）<SUB>2</SUB>=CaCO<SUB>3</SUB>↓+H<SUB>2</SUB>O；因此氢氧化钙应密封保存．<br />答案：<br />（3）<br /><table class=\"edittable\"><TBODY><TR><td width=205>实验步骤</TD><td width=205>实验现象</TD><td width=167>实验结论</TD></TR><TR><td>取样，加适量水，搅拌，过滤<br />①取少量滤液于试管中，滴入酚酞试液<br />②取少量滤渣于试管中，加入盐酸</TD><td>①滤液变为红色<br />②无气泡产生（或无现象）</TD><td>氢氧化钙没有变质</TD></TR></TBODY></TABLE>（4）①熟石灰；　酸；<br />②CO<SUB>2</SUB>（或二氧化碳）；　CO<SUB>2</SUB>+Ca（OH）<SUB>2</SUB>═CaCO<SUB>3</SUB>↓+H<SUB>2</SUB>O；密封','【分析】（3）利用氢氧化钙溶液能使酚酞变红、碳酸钙能与盐酸反应放出二氧化碳检验两种物质的存在；若氢氧化化钙部分变质，则会观察到未变质的氢氧化钙可使酚酞变红、已变质形成的碳酸钙遇盐酸放出气体的现象；若氢氧化钙没有变质，则只能看到滴加酚酞变红而无气体放出的现象；<br />（4）氢氧化钙俗称熟石灰或消石灰，常用于改良酸性土壤；氢氧化钙能与二氧化碳反应生成碳酸钙而变质．','书写',3.00,'dd7d01d7e6012848fe1a79552af8457c',9,400,'药品是否变质的探究,碱的化学性质,盐的化学性质,书写化学方程式、文字表达式、电离方程式','',2016,'32','2016•揭西县模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839917,'化学与生活生产息息相关，请运用化学知识回答下列问题：<br />（1）目前市场上有：高钙奶粉、补锌口服液等，此处钙，锌是指<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（分子、原子、元素）<br />（2）自来水厂生产自来水的过程中，常加入<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，用来吸附水中一些溶解性的杂质，除去臭味．生活中使硬水软化的一种最常用方法是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />（3）海水中含有很多杂质，取一杯浑浊的海水，若要使其变得澄淸，所需要的操作方法为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />（4）在厨房里炒菜时，在客厅都能闻到菜香味的主要原因是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />（5）当森林发生火灾时，救火的措施之一是将大火蔓延线路前的一片树木砍掉，形成隔离带，其原理是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（6）目前，人们使用的燃料大多来自化石燃料，如煤、<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>、天然气等，消粍化石燃料增加了空气中二氧化碳的含量．请你写出天然气中的甲烷完全燃烧的化学方程式：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','元素$###$活性炭$###$煮沸$###$过滤$###$分子在不断的运动$###$清除可燃物$###$石油$###$CH<SUB>4</SUB>+2O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>CO<SUB>2</SUB>+2H<SUB>2</SUB>O','【解答】解：<br />（1）含钙奶粉、补锌口服液等，此处钙，锌是指元素；<br />（2）自来水厂生产自来水的过程中，常加入活性炭，用来吸附水中一些溶解性的杂质，除去臭味．生活中使硬水软化的一种最常用方法是煮沸；<br />（3）通过过滤除去海水中的不溶性杂质，能使浑浊的海水变得澄清；<br />（4）分子在不断的运动，故会闻到菜香味，故填：分子在不断的运动；<br />（5）可燃物燃烧时需：可燃物、氧气、着火点三者缺一不可；森林发生火灾时，将大火蔓延线路前的一片树木砍掉，形成隔离带，其原理是清除可燃物；故答案为：清除可燃物；<br />（6）目前，人们使用的燃料大多来自化石燃料，如煤、石油、天然气等，甲烷燃烧生成了二氧化碳和水，反应的方程式是：CH<SUB>4</SUB>+2O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>CO<SUB>2</SUB>+2H<SUB>2</SUB>O．<br />答案：<br />（1）元素；<br />（2）活性炭；煮沸；<br />（3）过滤；<br />（4）分子在不断的运动；<br />（5）清除可燃物；<br />（6）石油；CH<SUB>4</SUB>+2O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>CO<SUB>2</SUB>+2H<SUB>2</SUB>O．','【分析】（1）根据物质是由元素组成的解答；<br />（2）根据活性炭具有吸附性，生活中使硬水软化的一种最常用方法是煮沸解答；<br />（3）根据净化水的措施过滤、蒸馏的原理分析回答；<br />（4）由于分子在不断运动，菜香分子运动到客厅，所以能闻到菜香味去分析解答；<br />（5）根据灭火的原理解答；<br />（6）根据甲烷燃烧生成了二氧化碳和水，分析写出反应的方程式．','书写',3.00,'32c81817c0abbaac5fc932fb63349c09',9,400,'过滤的原理、方法及其应用,自来水的生产过程与净化方法,利用分子与原子的性质分析和解决问题,元素的概念,书写化学方程式、文字表达式、电离方程式,灭火的原理和方法,化石燃料及其综合利用','',2015,'37','2015秋•曹县校级月考',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839918,'小明同学取少量浑浊的河水倒入烧杯中，先加入少量明矾粉末搅拌溶解，静置一会儿，过滤．请回答：<br />（1）过滤时所需要的玻璃仪器是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）过滤过程中发现过滤速度很慢，原因可能是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）过滤后观察发现，滤液仍浑浊，可能的原因是什么？（写出2条即可）','','','','','','漏斗、烧杯、玻璃棒$###$固体颗粒阻碍了液体通过滤纸孔隙，还可能是滤纸没紧贴漏斗内壁，中间留有气泡或滤纸的规格不对等','【解答】解：（1）过滤是把不溶于液体的固体与液体分离的一种方法，过滤时所需要的仪器和用品是铁架台、漏斗、烧杯、玻璃棒、滤纸．其中玻璃仪器是漏斗、烧杯、玻璃棒．<br />（2）实验过程中发现过滤速度较慢，可能的原因是固体颗粒阻碍了液体通过滤纸孔隙，还可能是滤纸没紧贴漏斗内壁，中间留有气泡或滤纸的规格不对等．<br />（3）过滤后滤液仍浑浊，可能原因是滤纸破损（会使得液体中的不溶物进入下面的烧杯，从而使得滤液浑浊）、液面高于滤纸边缘（会使部分液体未经过滤纸的过滤直接流下，该操作会使滤液仍然浑浊）或盛接滤液的烧杯不干净等．<br />故答案为：（1）漏斗、烧杯、玻璃棒；<br />（2）固体颗粒阻碍了液体通过滤纸孔隙，还可能是滤纸没紧贴漏斗内壁，中间留有气泡或滤纸的规格不对等；<br />（3）滤纸破损、液面高于滤纸边缘．','【分析】（1）过滤是把不溶于液体的固体与液体分离的一种方法，据此判断所需的仪器．<br />（2）根据过滤操作的注意事项分析．<br />（3）凡是不经滤纸的过滤就直接进入滤液的操作，都能造成滤液浑浊；另外接滤液的烧杯不干净也会造成同样的结果．','填空题',3.00,'ef199cc737f75755c1459b209ade9182',9,400,'过滤的原理、方法及其应用','',2015,'35','2015秋•平顶山校级期中',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839919,'根据下列装置图回答问题：<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao69/10d394b0-94d4-11e9-a280-b42e9921e93e_xkb30.png\" style=\"vertical-align:middle\" /><br />（1）图中标示①仪器的名称是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（2）实验室用高锰酸钾制取氧气时，发生装置通常选择图中的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填装置编号），反应的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（3）实验室制取的二氧化碳中常含有氯化氢气体和水蒸气辚获得纯净、干燥的二氧化碳气体，可选用如图所示的F装置和G装置进行除杂和干燥，导管口连接的正确顺序是：气体→<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>→D；<br />（4）实验室通常用加热无水醋酸钠和碱石灰的固体混合物制甲烷，则制取并收集甲烷的装置应选择<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','长颈漏斗$###$A$###$2KMnO<SUB>4</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>K<SUB>2</SUB>MnO<SUB>4</SUB>+MnO<SUB>2</SUB>+O<SUB>2</SUB>↑$###$cdab$###$AC或AE','【解答】解：（1）①是长颈漏斗；<br />（2）高锰酸钾加热生成锰酸钾、二氧化锰和氧气，发生装置选择A；反应的方程式为：2KMnO<SUB>4</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>K<SUB>2</SUB>MnO<SUB>4</SUB>+MnO<SUB>2</SUB>+O<SUB>2</SUB>↑；&nbsp;&nbsp;<br />（3）二氧化碳含有氯化氢气体和水蒸气，要先除杂质再干燥，要长进短出；故答案为：cdab；<br />（4）实验室通常用加热无水醋酸钠和碱石灰的固体混合物制甲烷，甲烷密度比空气小，不易溶于水，故发生装置选择A；收集装置选择C或E．<br />答案：<br />故答案为：（1）长颈漏斗；（2）A、2KMnO<SUB>4</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>K<SUB>2</SUB>MnO<SUB>4</SUB>+MnO<SUB>2</SUB>+O<SUB>2</SUB>↑；&nbsp;&nbsp;（3）cdab；<br />（4）AC或AE．','【分析】（1）要熟悉各种仪器的名称、用途和使用方法； <br />（2）高锰酸钾加热生成锰酸钾、二氧化锰和氧气；<br />（3）二氧化碳含有氯化氢气体和水蒸气，要先除杂质再干燥，要长进短出；若将装置A和D连接制取并收集氧气；<br />（4）实验室通常用加热无水醋酸钠和碱石灰的固体混合物制甲烷，甲烷密度比空气小，不易溶于水．','书写',3.00,'4ee9d3da7da94492443821a7a79e02d7',9,400,'常见气体的检验与除杂方法,氧气的制取装置,二氧化碳的实验室制法,书写化学方程式、文字表达式、电离方程式','',2016,'32','2016•营口模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839920,'下列4个图象能正确反映对应变化关系是（　　）','<img  src=\"/tikuimages/9/2016/400/shoutiniao40/10d98821-94d4-11e9-8ed9-b42e9921e93e_xkb2.png\" style=\"vertical-align:middle\" />向一定量稀H<SUB>2</SUB>SO<SUB>4</SUB>中加入NaOH溶液','<img  src=\"/tikuimages/9/2016/400/shoutiniao12/10dbf921-94d4-11e9-ba33-b42e9921e93e_xkb7.png\" style=\"vertical-align:middle\" />用酒精灯加热一定量KMnO<SUB>4</SUB>固体','<img  src=\"/tikuimages/9/2016/400/shoutiniao10/10dcbc70-94d4-11e9-9b09-b42e9921e93e_xkb76.png\" style=\"vertical-align:middle\" />将水通电电解一段时间','<img  src=\"/tikuimages/9/2016/400/shoutiniao77/10de430f-94d4-11e9-86be-b42e9921e93e_xkb96.png\" style=\"vertical-align:middle\" />向一定量的二氧化锰中加入H<SUB>2</SUB>O<SUB>2</SUB>溶液','','C','【解答】解：A．原来是溶液，其中就含有一定质量的水，所以图象的起点不为零，故错误；<br />B．高锰酸钾加热会生成锰酸钾二氧化锰和氧气，所以剩余的固体中还含有氧元素，因此反应结束后固体中氧元素质量不能为零，该特点与图象不符，故错误；<br />C．通电分解水时产生氢气的体积是氧气的二倍，故正确；<br />D．催化剂的质量和化学性质在反应前后均不发生变化，故错误．<br />故选C．','【分析】A．根据中和反应来分析；<br />B．高锰酸钾加热会生成锰酸钾二氧化锰和氧气；<br />C．根据通电分解水的实验现象及结论进行解答；<br />D．根据催化剂的特征来分析．','选择题',3.00,'fabc348f28f3f1601b73f255fcbaa8ae',9,400,'催化剂的特点与催化作用,电解水实验,中和反应及其应用,盐的化学性质','',2016,'37','2016•顺义区一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839921,'回答下列有关对比实验的相关问题：<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao41/10e2d6f0-94d4-11e9-a560-b42e9921e93e_xkb60.png\" style=\"vertical-align:middle\" /><br />（1）实验甲，K在关闭时白磷不燃烧；打开K，并使A中液体进入B中，白磷燃烧，则B中发生反应的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，此实验得出的燃烧条件是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）实验乙是利用体积相同并充满CO<SUB>2</SUB>的软塑料瓶、等体积的水（瓶①）和NaOH溶液（瓶②）进行实验对比的，证明CO<SUB>2</SUB>&nbsp;与NaOH溶液中的溶质确实发生了反应的实验现象是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','2H<SUB>2</SUB>O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span>&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O+O<SUB>2</SUB>↑$###$与氧气接触$###$瓶②比瓶①变瘪程度大','【解答】解：（1）甲图实验：过氧化氢在二氧化锰做催化剂的条件下生成水和氧气，要注意配平，其化学方程式2H<SUB>2</SUB>O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span>&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O+O<SUB>2</SUB>↑；通过图二实验说明可燃物燃烧需要与氧气接触；<br />（2）二氧化碳和NaOH溶液反应生成碳酸钠和水，二氧化碳被消耗，瓶内压强变小，软塑料瓶变瘪的程度变大；故填瓶②比瓶①变瘪程度大．<br />故答案为：<br />（1）2H<SUB>2</SUB>O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span>&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O+O<SUB>2</SUB>↑，与氧气接触；（2）瓶②比瓶①变瘪程度大．','【分析】（1）通过甲图实验说明燃烧需要：与氧气接触；<br />（2）根据二氧化碳和NaOH溶液反应生成碳酸钠和水进行解答；','简答题',3.00,'7b0d5db009d717546971474e69623cd6',9,400,'碱的化学性质,燃烧与燃烧的条件','',2016,'37','2016•怀柔区一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839922,'课堂上老师演示了如下实验：从盛有煤油的试剂瓶中取出一块金属，用小刀切下一小块，把金属投放在盛有水的烧杯中，发现金属浮在水面上，并跟水剧烈反应，且在水面上急速运动发出“嘶嘶”声，立即熔化成一个闪亮的小球，并逐渐缩小，最后完全消失．根据文中所述推导金属的<br />（1）物理性质<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />（2）化学性质<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','质软（硬度小）、密度比水小、熔点低$###$能与水反应','【解答】解：可以用刀切，说明钠硬度小，能浮在水面，说明密度比水的密度小，能在水中熔化成小球，说明熔点低．<br />故填：（1）物理性质&nbsp;质软（硬度小）、密度比水小、熔点低；（2）化学性质&nbsp;能与水反应．','【分析】物质的化学性质是指在化学变化中表现出来的性质．化学性质主要有：可燃性、毒性、氧化性、还原性、稳定性等，物质的物理性质是指不需要通过化学变化表现出来的性质，物理性质主要有：颜色、状态、气味、密度、硬度、熔点、沸点等，可以根据题干提供的信息进行分析．','填空题',3.00,'5fbad94d01168633844722043fbcf039',9,400,'化学性质与物理性质的差别及应用','',2016,'35','2016春•五华区校级期中',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839925,'某化学实验小组对课本中“硫酸铜溶液等对过氧化氢的分解也具有催化作用”这句话颇感兴趣，便取5%的过氧化氢溶液5mL于试管中，滴入几滴硫酸铜溶液，立即伸入带火星的木条，观察到：“有大量气泡产生，带火星的木条复燃”．他们对此作出猜想：<br />猜想1：可能是水分子加快过氧化氢的分解；<br />猜想2：可能是硫酸根离子加快过氧化氢的分解；<br />猜想3：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（1）大家认为猜想1不成立，理由是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）请你设计一个实验来证明猜想2不成立（写出步骤、可能的现象及结论）<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','可能是铜离子加快过氧化氢的分解$###$过氧化氢溶液中有水，可见水不是催化剂$###$向盛有5mL15%的过氧化氢溶液的试管中加入少量稀H<SUB>2</SUB>SO<SUB>4</SUB>，把带火星的木条伸人试管中，无明显现象，起催化作用的不是SO<SUB>4</SUB><SUP>2-</SUP>','【解答】解：<br />猜想1：可能是水分子加快过氧化氢的分解；<br />猜想2：可能是硫酸根离子加快过氧化氢的分解；<br />猜想3：可能是铜离子加快过氧化氢的分解；<br />（1）过氧化氢溶液中含有水分子，所以水不可能是催化剂；故填：过氧化氢溶液中有水，可见水不是催化剂；<br />（2）稀硫酸中含有硫酸根离子，所以实验方案是：向盛有5mL&nbsp;15%的过氧化氢溶液的试管中加入少量稀H<SUB>2</SUB>SO<SUB>4</SUB>，把带火星的木条伸人试管中，无明显现象，起催化作用的不是SO<SUB>4</SUB><SUP>2-</SUP>；<br />答案：<br />可能是铜离子加快过氧化氢的分解；<br />（1）过氧化氢溶液中有水，可见水不是催化剂；<br />（2）向盛有5mL&nbsp;15%的过氧化氢溶液的试管中加入少量稀H<SUB>2</SUB>SO<SUB>4</SUB>，把带火星的木条伸人试管中，无明显现象，起催化作用的不是SO<SUB>4</SUB><SUP>2-</SUP>；','【分析】根据题意进行猜想；<br />（1）根据过氧化氢溶液中含有水分子进行解答；<br />（2）根据稀硫酸中含有硫酸根离子以及硝酸铜溶液中含有铜离子，然后分别用稀硫酸和硝酸铜溶液进行实验验证即可．','填空题',3.00,'81c1972b29afdf68f672d77c82a39d90',9,400,'实验探究物质的性质或变化规律,催化剂的特点与催化作用','',2016,'32','2016•海口模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839926,'（1）用下列四种元素（O、H、Na、S）各写一个化学式：<br />酸<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；碱<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；盐<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；单质<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />（2）上述酸与碱发生反应的化学方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','H<SUB>2</SUB>SO<SUB>4</SUB>$###$NaOH$###$Na<SUB>2</SUB>SO<SUB>4</SUB>$###$H<SUB>2</SUB>$###$H<SUB>2</SUB>SO<SUB>4</SUB>+2NaOH=Na<SUB>2</SUB>SO<SUB>4</SUB>+2H<SUB>2</SUB>O','【解答】解：（1）酸是指在电离时产生的阳离子全部是氢离子的化合物，根据提供的元素硫酸符合要求，其化学式为：H<SUB>2</SUB>SO<SUB>4</SUB>；<br />碱是指在电离时产生的阴离子全部是氢氧根离子的化合物，根据提供的元素氢氧化钠属于碱，其化学式为：NaOH；<br />盐是由金属离子和酸根离子组成的，根据提供的元素硫酸钠属于盐，其化学式为：Na<SUB>2</SUB>SO<SUB>4</SUB>；<br />单质是由同种元素组成的纯净物，如氧气、氢气、钠、硫．<br />故答案为：H<SUB>2</SUB>SO<SUB>4</SUB>；NaOH；Na<SUB>2</SUB>SO<SUB>4</SUB>；H<SUB>2</SUB>（其他合理均可）；<br />（2）氢氧化钠和硫酸反应生成硫酸钠和水，化学方程式为：H<SUB>2</SUB>SO<SUB>4</SUB>+2NaOH=Na<SUB>2</SUB>SO<SUB>4</SUB>+2H<SUB>2</SUB>O．<br />故填：H<SUB>2</SUB>SO<SUB>4</SUB>+2NaOH=Na<SUB>2</SUB>SO<SUB>4</SUB>+2H<SUB>2</SUB>O．','【分析】首先根据题意确定物质的化学名称，然后根据题目所提供的元素、书写化学式的方法和步骤写出物质的化学式即可．','书写',3.00,'bf3da62d5ccc96bc5d4ff1d4c5d1584a',9,400,'化学式的书写及意义,书写化学方程式、文字表达式、电离方程式','',2016,'37','2016春•仙游县月考',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839927,'实验室用50g质量分数为98%的浓硫酸稀释成质量分数为20%的稀硫酸（水的密度为1g/cm<SUP>3</SUP>）．请回答下列问题：<br />（1）需要水的质量为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>g<br />（2）实验的主要步骤有计算、<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>、溶解、装瓶并贴标签<br />（3）实验需要用到下列仪器中的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填序号）和细口瓶．<br />①烧杯②玻璃棒③托盘天平④胶头滴管⑤量筒．','','','','','','195$###$量取$###$①②④⑤','【解答】解：（1）设要加水的质量为x，根据溶液稀释前后溶质的质量不变，<br />则50g×98%=（50g+x）×20%&nbsp;&nbsp;&nbsp; x=195g．<br />（2）实验室用50g质量分数为98%的浓硫酸稀释成质量分数为20%的稀硫酸，首先计算配制溶液所需浓硫酸和水的质量，再量取所需的浓硫酸和水，最后进行溶解、装瓶并贴标签．<br />（3）实验操作步骤是计算、量取、溶解，量筒和胶头滴管用于量取浓硫酸和水，烧杯、玻璃棒用于进行溶解操作；无需使用托盘天平；细口瓶用于盛装稀硫酸．<br />故答案为：（1）195；（2）量取；（3）①②④⑤．','【分析】（1）根据溶液稀释前后溶质的质量不变，结合题意进行分析解答．<br />（2）利用浓溶液配制稀溶液，采用的加水稀释的方法，由浓溶液配制稀溶液的实验步骤，结合题意进行分析解答．<br />（3）利用浓溶液配制稀溶液，采用的加水稀稀释的方法，其操作步骤是计算、量取、溶解、装瓶，据此判断所需的仪器．','填空题',3.00,'fabbda299aa3b881779ef00fe501f7ab',9,400,'一定溶质质量分数的溶液的配制','',0,'37','',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839929,'<img  src=\"/tikuimages/9/2016/400/shoutiniao59/10f79770-94d4-11e9-8489-b42e9921e93e_xkb72.png\" style=\"vertical-align:middle;FLOAT:right\" />（2016•沈阳模拟）小何同学欲用氯酸钾来制取氧气，并回收提纯氯化钾和二氧化锰．<br />实验一：制取氧气 （1）根据下列装置回答问题．<br />（1）仪器a的名称是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）小红认为装置A，还可以用于高锰酸钾制氧气，其化学反应方程式是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）若用B装置来收集氧气，氧气应从<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填序号）端通入；若用B装置应来除去制得氧气中混有的水蒸气，则B装置中盛放的试剂是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填名称）．<br />（4）若用B装置进行适当变化，采用排水法收集氧气，请简述其操作：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />实验二：分离提纯 （2）根据下面的流程图回答问题．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao77/10fc5261-94d4-11e9-947d-b42e9921e93e_xkb47.png\" style=\"vertical-align:middle\" /><br />（5）在反应过程中MnO<SUB>2</SUB>起到的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>作用．<br />（6）蒸发操作中，当观察到蒸发皿中出现<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>时，停止加热．','','','','','','酒精灯$###$2KMnO<SUB>4</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>K<SUB>2</SUB>MnO<SUB>4</SUB>+MnO<SUB>2</SUB>+O<SUB>2</SUB>↑$###$①$###$浓硫酸$###$在瓶中装满水，从②口入气$###$催化$###$较多固体析出','【解答】解：（1）酒精灯是常用的加热仪器，故答案为：酒精灯；<br />（2）高锰酸钾受热分解生成锰酸钾和二氧化锰和氧气，要注意配平；故答案为：2KMnO<SUB>4</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>K<SUB>2</SUB>MnO<SUB>4</SUB>+MnO<SUB>2</SUB>+O<SUB>2</SUB>↑；<br />（3）若用B装置来收集氧气，氧气应从长管进入，因为氧气的密度比空气大；氧气可以用浓硫酸干燥；故答案为：①；浓硫酸；<br />（4）若用B装置进行适当变化，采用排水法收集氧气，其操作是：在瓶中装满水，从②口入气，因为氧气的密度比水小；故答案为：在瓶中装满水，从②口入气；<br />（5）如果用氯酸钾和二氧化锰制氧气就需要加热，其中二氧化锰起催化作用，故答案为：催化；<br />（6）蒸发操作中，当观察到蒸发皿中出现较多固体析出时，停止加热，故答案为：较多固体析出；','【分析】酒精灯是常用的加热仪器，制取装置包括加热和不需加热两种，如果用双氧水和二氧化锰制氧气就不需要加热，如果用高锰酸钾或氯酸钾制氧气就需要加热．氧气的密度比空气的密度大，不易溶于水，因此能用向上排空气法和排水法收集．若用B装置进行适当变化，采用排水法收集氧气，其操作是：在瓶中装满水，从②口入气，因为氧气的密度比水小；蒸发操作中，当观察到蒸发皿中出现较多固体析出时，停止加热．','书写',3.00,'8f92c34a455840e33681bbd883a9fd5b',9,400,'蒸发与蒸馏操作,气体的干燥（除水）,氧气的制取装置,氧气的收集方法,催化剂的特点与催化作用,书写化学方程式、文字表达式、电离方程式','',2016,'32','2016•沈阳模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839930,'请用化学用语表示：<br />三个铁原子<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>、五个氧离子<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>、金刚石<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>、两个氢氧根离子<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>氯化钠<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>、氮气<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>、氢氧化钙<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>、氯酸钾<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>、高锰酸钾<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','3Fe$###$5O<SUP>2-</SUP>$###$C$###$2OH<SUP>-</SUP>$###$NaCl$###$N<SUB>2</SUB>$###$Ca（OH）<SUB>2</SUB>$###$KClO<SUB>3</SUB>$###$KMnO<SUB>4</SUB>','【解答】解：由原子的表示方法，用元素符号来表示一个原子，表示多个该原子，就在其元素符号前加上相应的数字，故三个铁原子表示为：3Fe．<br />由离子的表示方法：在表示该离子的元素符号右上角，标出该离子所带的正负电荷数，数字在前，正负符号在后，带1个电荷时，1要省略．若表示多个该离子，就在其离子符号前加上相应的数字．五个氧离子、两个氢氧根离子分别可表示为：5O<SUP>2-</SUP>、2OH<SUP>-</SUP>．<br />金刚石属于固态非金属单质，直接用元素符号表示其化学式，其化学式为C．<br />氯化钠中钠元素显+1价，氯元素显-1价，其化学式为：NaCl．<br />氮气属于气态非金属单质，在元素符号的右下角写上表示分子中所含原子数的数字，其化学式为：N<SUB>2</SUB>．<br />氢氧化钙中钙元素显+2价，氢氧根显-1价，其化学式为：Ca（OH）<SUB>2</SUB>．<br />氯酸钾中钾元素显+1价，氯酸根显-1价，其化学式为：KClO<SUB>3</SUB>．<br />高锰酸钾中钾元素显+1价，高锰酸根显-1价，其化学式为：KMnO<SUB>4</SUB>．<br />故答案为：3Fe；5O<SUP>2-</SUP>；C；2OH<SUP>-</SUP>；NaCl；N<SUB>2</SUB>；Ca（OH）<SUB>2</SUB>；KClO<SUB>3</SUB>；KMnO<SUB>4</SUB>．','【分析】原子的表示方法，用元素符号来表示一个原子，表示多个该原子，就在其元素符号前加上相应的数字．<br />离子的表示方法：在表示该离子的元素符号右上角，标出该离子所带的正负电荷数，数字在前，正负符号在后，带1个电荷时，1要省略．若表示多个该离子，就在其离子符号前加上相应的数字．<br />金属单质、固态非金属单质、稀有气体单质，直接用元素符号表示其化学式；气态非金属单质，在元素符号的右下角写上表示分子中所含原子数的数字．<br />化合物化学式的书写一般规律：金属在前，非金属在后；氧化物中氧在后，原子个数不能漏，正负化合价代数和为零．','书写',3.00,'784f0c29bd9caa5b6d5d49408721a4e0',9,400,'化学式的书写及意义,化学符号及其周围数字的意义','',0,'37','',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839932,'<img  src=\"/tikuimages/9/2016/400/shoutiniao45/110775f0-94d4-11e9-a3fa-b42e9921e93e_xkb36.png\" style=\"vertical-align:middle;FLOAT:right\" />（2016•孝义市二模）如图是A、B两种物质的溶解度曲线，请根据图回答下列问题．<br />（1）A的溶解度小于B的溶解度的温度范围是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）50℃时，将45gA物质放入50g水中，充分溶解后，形成的溶液中<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“能”或“不能”）继续溶解B固体．<br />（3）如果要配置相同溶质质量分数的A和B的饱和溶液，需要把温度控制在<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>℃．<br />（4）除去A中混有少量B的方法是先在高温下配置的饱和溶液，然后<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．最后过滤、洗涤、干燥．<br />（5）在50℃时，向等质量的水中加入A和B形成饱和溶液，分别降低温度到10℃时．下列关系正确的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />A．溶质质量：A＞B&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; B．溶质质量：A＜B&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; C．溶剂质量：A＞B<br />D．溶解度：A＜B&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; E．析出晶体的质量：A＞B．','','','','','','0℃-t℃$###$能$###$t$###$降温结晶$###$BDE','【解答】解：（1）通过分析溶解度曲线可知，A的溶解度小于B的溶解度的温度范围是0℃-t℃；<br />（2）50℃时，A物质的溶解度是85.5g，所以将45gA物质放入50g水中，充分溶解后，对于A物质是饱和溶液，对于B物质不是饱和溶液，所以形成的溶液中能继续溶解B固体；<br />（3）t℃时，A、B物质的溶解度相等，所以要配置相同溶质质量分数的A和B的饱和溶液，需要把温度控制在t℃；<br />（4）A物质的溶解度受温度影响较大，所以除去A中混有少量B的方法是先在高温下配置的饱和溶液，然后降温结晶，最后过滤、洗涤、干燥；<br />（5）在50℃时，向等质量的水中加入A和B形成饱和溶液，分别降低温度到10℃时，10℃时，B物质的溶解度大于A物质的溶解度，所以溶质质量：A＜B，50℃时，A物质的溶解度大于B物质，降低温度，析出晶体的质量：A＞B，故选：BDE．<br />故答案为：（1）0℃-t℃；<br />（2）能；<br />（3）t；<br />（4）降温结晶；<br />（5）BDE．','【分析】根据固体的溶解度曲线可以：①查出某物质在一定温度下的溶解度，从而确定物质的溶解性，②比较不同物质在同一温度下的溶解度大小，从而判断饱和溶液中溶质的质量分数的大小，③判断物质的溶解度随温度变化的变化情况，从而判断通过降温结晶还是蒸发结晶的方法达到提纯物质的目的．','填空题',3.00,'890f9413bdf699d4844585619a1be759',9,400,'结晶的原理、方法及其应用,固体溶解度曲线及其作用,溶质的质量分数、溶解性和溶解度的关系','',2016,'37','2016•孝义市二模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839933,'有资料显示，人体含钙质量分数在1.5%左右，过多或过少都会引起病变，这里说的“钙”是指（　　）','钙单质','钙原子','钙离子','钙元素','','D','【解答】解：宏观物质的组成，用宏观概念元素来表示；分子的构成，用微观粒子来表示．元素是具有相同核电荷数（即核内质子数）的一类原子的总称，是宏观概念，只讲种类，不讲个数．这里的钙是宏观概念，指的是元素；<br />故选D．','【分析】宏观物质的组成用宏观概念元素来表示，而这里面说的“钙”为宏观概念，即表示元素．','选择题',3.00,'4526d8a530d35ed7943e20669ed06629',9,400,'元素的概念','',2012,'33','2012秋•潮南区期末',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839934,'小明用质量分数为8%的氯化钠溶液配制质量分数为4%的氯化钠溶液50g．<br />（1）稀释过程中需要加水<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>g．<br />（2）配制过程中除了胶头滴管外，还需要的玻璃仪器主要有哪些？<br />（3）配制过程中，哪些操作可能导致配制的溶液质量分数偏小？（写两条即可）','','','','','','25','【解答】解：（1）设要加水的质量为x，根据溶液稀释前后溶质的质量不变，<br />则50g×4%=（50g-x）×8%&nbsp;&nbsp;&nbsp; x=25g．<br />（2）用浓溶液配制一定溶质质量分数的稀溶液，其操作步骤是计算、量取、溶解，量筒用于量取浓盐酸和水，烧杯、玻璃棒用于进行溶解操作．<br />（3）用量筒量取水时，仰视液面，读数比实际液体体积小，会造成实际量取的水的体积偏大，则使溶质质量分数偏小；用量筒量取8%的氯化钠溶液时，俯视液面，读数比实际液体体积大，会造成实际量取的8%的氯化钠溶液的体积偏小，则使溶质质量分数偏小（合理即可）．<br />故答案为：（1）25；（2）量筒、烧杯、玻璃棒；（3）量取8%的NaCl溶液时俯视读数；量取水时仰视读数（合理即可）．','【分析】（1）根据溶液稀释前后溶质的质量不变，结合题意进行分析解答．<br />（2）利用浓溶液配制稀溶液，采用的加水稀稀释的方法，其操作步骤是计算、量取、溶解，据此判断所需的仪器．<br />（3）溶质质量分数变小，则可能是溶质质量偏小或溶剂质量偏大，可以分析出可能造成这两个方面错误的原因进行分析解答．','填空题',3.00,'5284d559d370694ce28c3521879375cf',9,400,'一定溶质质量分数的溶液的配制','',2016,'32','2016•信阳模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839935,'众所周知，二氧化碳能使澄清的石灰水变浑浊（生成白色沉淀）．但是，在分组实验中，向石灰水中通入二氧化碳（如图1所示），出现了意想不到的现象：有的石灰水未变浑浊；有的出现浑浊后又变澄清（沉淀消失）．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao1/11130eae-94d4-11e9-b055-b42e9921e93e_xkb55.png\" style=\"vertical-align:middle\" /><br />【提出问题】分组实验中，出现意想不到现象的原因是什么呢？<br />【猜想与假设】<br />I．石灰水未变浑浊，是因为二氧化碳中混有少量氯化氢气体．<br />Ⅱ．石灰水未变浑浊，是因为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />Ⅲ．石灰水出现浑浊后又变澄清，与石灰水的浓度、通入二氧化碳的时间有关．<br />【进行实验】实验装置如图2所示．<br />提示：①实验中所用盐酸与分组实验所用盐酸相同；<br />②CO<SUB>2</SUB>在饱和NaHCO<SUB>3</SUB>溶液中的溶解度非常小．<br />步骤1：分别取一定体积的饱和澄清石灰水与一定体积的蒸馏水混合配制成50mL溶液．<br />步骤2：分别向50mL溶液中通入一段时间的二氧化碳，记录现象．<br /><table class=\"edittable\"><TBODY><TR><td width=45 rowSpan=2>实验序号</TD><td width=84 rowSpan=2>V<SUB>饱和石灰水</SUB>/mL</TD><td width=73 rowSpan=2>V<SUB>蒸馏水</SUB><br />/mL</TD><td width=419 colSpan=4>出现现象所需时间/s</TD></TR><TR><td>开始浑浊</TD><td>明显浑浊</TD><td>沉淀减少</TD><td>是否澄清</TD></TR><TR><td>①</TD><td>50</TD><td>0</TD><td>19</TD><td>56</TD><td>366</TD><td rowSpan=3>持续通入CO<SUB>2</SUB>8min以上，沉淀不能完全消失</TD></TR><TR><td>②</TD><td>40</TD><td>10</TD><td>24</TD><td>51</TD><td>245</TD></TR><TR><td>③</TD><td>30</TD><td>20</TD><td>25</TD><td>44</TD><td>128</TD></TR><TR><td>④</TD><td>20</TD><td>30</TD><td>27</TD><td>35</TD><td>67</TD><td>89s后完全澄清</TD></TR><TR><td>⑤</TD><td>10</TD><td>40</TD><td colSpan=4>通3min以上，均无明显现象</TD></TR></TBODY></TABLE>【解释与结论】<br />（1）实验中，所用石灰水为饱和溶液的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填序号）．<br />（2）二氧化碳使澄清石灰水变浑浊原因是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（用化学方程式表示）．<br />（3）饱和NaHCO<SUB>3</SUB>溶液的作用是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（4）通过实验分析，猜想 I<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“成立”或“不成立”），猜想Ⅱ、Ⅲ成立．<br />【反思与评价】<br />（5）向澄清的石灰水中通入二氧化碳，为避免出现“意想不到的现象”，你的建议是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','石灰水的浓度过小（或通入二氧化碳的时间过短）$###$①$###$Ca（OH）<SUB>2</SUB>+CO<SUB>2</SUB>═CaCO<SUB>3</SUB>↓+H<SUB>2</SUB>O$###$除去氯化氢气体$###$不成立$###$所用石灰水的浓度不能太小（或用饱和的石灰水），通入二氧化碳的时间不能过短也不能过长','【解答】解：<br />【猜想与假设】石灰水的浓度过小（或通入二氧化碳的时间过短），也可以造成石灰水未变浑浊<br />【解释与结论】<br />（1）根据表中数据可知：实验中，所用石灰水为饱和溶液的是①；<br />（2）二氧化碳使澄清石灰水变浑浊原因是Ca（OH）<SUB>2</SUB>+CO<SUB>2</SUB>═CaCO<SUB>3</SUB>↓+H<SUB>2</SUB>O；<br />（3）饱和NaHCO<SUB>3</SUB>溶液的作用是除去氯化氢气体；&nbsp;&nbsp;<br />（4）通过实验分析，猜想 I不成立；<br />（5）向澄清的石灰水中通入二氧化碳，为避免出现“意想不到的现象”，建议是所用石灰水的浓度不能太小（或用饱和的石灰水），通入二氧化碳的时间不能过短也不能过长．<br />答案：<br />【猜想与假设】<br />Ⅱ．石灰水的浓度过小（或通入二氧化碳的时间过短）<br />（1）①；<br />（2）Ca（OH）<SUB>2</SUB>+CO<SUB>2</SUB>═CaCO<SUB>3</SUB>↓+H<SUB>2</SUB>O；<br />（3）除去氯化氢气体；&nbsp;&nbsp;&nbsp;<br />（4）不成立；<br />（5）所用石灰水的浓度不能太小（或用饱和的石灰水），通入二氧化碳的时间不能过短也不能过长．','【分析】【猜想与假设】根据石灰水的浓度过小（或通入二氧化碳的时间过短），也可以造成石灰水未变浑浊解答；<br />【解释与结论】<br />（1）根据表中数据进行分析解答；<br />（2）根据二氧化碳和氢氧化钙反应生成碳酸钙沉淀和水解答；<br />（3）根据饱和NaHCO<SUB>3</SUB>溶液的作用是除去氯化氢气体&nbsp;解答；<br />（4）根据实验分析猜想；<br />（5）根据所用石灰水的浓度不能太小（或用饱和的石灰水），通入二氧化碳的时间不能过短也不能过长解答．','书写',3.00,'6a89921661e5581a934129319a38ed5f',9,400,'实验探究物质的性质或变化规律,常见气体的检验与除杂方法,碱的化学性质,书写化学方程式、文字表达式、电离方程式','',2016,'37','2016•房山区一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839936,'从微观角度解释下列事实，错误的是（　　）<br /><table class=\"edittable\"><TBODY><TR><td width=48>选项</TD><td width=171>事实</TD><td width=218>解释</TD></TR><TR><td>A</TD><td>酒香不怕巷子深</TD><td>分子不断运动</TD></TR><TR><td>B</TD><td>氯化钠溶液具有导电性</TD><td>溶液中有自由移动的离子</TD></TR><TR><td>C</TD><td>夏天湿衣服晾干快</TD><td>温度越高，分子运动速度越快</TD></TR><TR><td>D</TD><td>聚乙烯塑料受热会熔化</TD><td>受热情况下，分子的体积变大</TD></TR></TBODY></TABLE>','A','B','C','D','','D','【解答】解：A、酒香不怕巷子深，是因为酒精中含有的分子是在不断运动的，向四周扩散，使人们闻到酒香，故选项解释正确．<br />B、氯化钠溶液具有导电性，是因为溶液中由自由移动的离子，故选项解释正确．<br />C、夏天湿衣服晾干快，是因为温度越高，分子运动速度越快，故选项解释正确．<br />D、聚乙烯塑料受热会熔化，是因为温度升高，分子间的间隔增大，故选项解释错误．<br />故选：D．','【分析】根据分子的基本特征：分子质量和体积都很小；分子之间有间隔；分子是在不断运动的；同种的分子性质相同，不同种的分子性质不同，可以简记为：“两小运间，同同不不”，结合事实进行分析判断即可．','选择题',3.00,'26fadc5e8f648a1a8ed4d5671a6e0409',9,400,'利用分子与原子的性质分析和解决问题','',2016,'37','2016•道外区二模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839937,'配制10%的硝酸钾溶液160g，需溶质和溶剂各多少克水？','','','','','','','【解答】解：溶质质量=溶液质量×溶质的质量分数，配制10%的160克硝酸钾溶液，需硝酸钾的质量=160g×10%=16g；溶剂质量=溶液质量-溶质质量，则所需水的质量=160g-16g=144g．<br />答：需要硝酸钾16g，水144g．','【分析】利用溶质质量=溶液质量×溶质的质量分数，可根据溶液的质量和溶质的质量分数计算配制溶液所需要的溶质的质量；再根据溶剂质量=溶液质量-溶质质量即可求得水的质量．','解答题',3.00,'8c51c5b58ae7b8afdca4c951e0e00a43',9,400,'有关溶质质量分数的简单计算','',2016,'37','2016春•凉州区校级月考',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839940,'<img  src=\"/tikuimages/9/2016/400/shoutiniao42/1121668f-94d4-11e9-8f63-b42e9921e93e_xkb22.png\" style=\"vertical-align:middle;FLOAT:right;\" />（2016•峨眉山市二模）如图是两种原子的原子结构示意图：<br />（1）写出它们能形成单质的化学式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>、<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（2）写出它们能形成化合物的化学式：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />（3）写出它们能形成的阴离子的化学符号．<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','Al$###$Cl<SUB>2</SUB>$###$AlCl<SUB>3</SUB>$###$Cl<SUP>-</SUP>','【解答】解：<br />（1）由原子结构示意图可知，核内质子数为13，为铝元素，形成单质的化学式为Al；<br />元素的核内质子数为17，为氯元素，形成单质的化学式为Cl<SUB>2</SUB>；<br />（2）由原子结构示意图，其核内质子数为13，为铝元素，元素的核内质子数为17，为氯元素；铝原子的最外层电子数为3，在化学反应中易失去3个电子而形成带3个单位正电荷的阳离子，化合价的数值等于离子所带电荷的数值，且符号一致，则铝元素的化合价为+3价；氯原子的最外层电子数为7，在化学反应中易得到1个电子而形成带1个单位正电荷的阴离子，则氯元素的化合价为-1价；形成化合物的化学式为AlCl<SUB>3</SUB>．<br />（3）氯原子的最外层电子数为7，在化学反应中易得到1个电子而形成带1个单位正电荷的阴离子，形成的阴离子的化学符号为：Cl<SUP>-</SUP>．<br />答案：（1）Al、Cl；（2）AlCl<SUB>3</SUB>；（3）Cl<SUP>-</SUP>．','【分析】根据原子结构示意图中，圆圈内数字表示核内质子数，弧线表示电子层，弧线上的数字表示该层上的电子数，离圆圈最远的弧线表示最外层．若最外层电子数≥4，在化学反应中易得电子，若最外层电子数＜4，在化学反应中易失去电子解答．','填空题',3.00,'4acd9968bbae0ddd39ca96621f1084b8',9,400,'原子结构示意图与离子结构示意图','峨眉山市',2016,'37','2016•峨眉山市二模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839941,'Ca表示<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>或<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；2Na表示<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；氦气<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；3个氮分子<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','钙元素$###$一个钙原子$###$2个钠原子$###$He$###$3N<SUB>2</SUB>．','【解答】解：Ca表示钙元素或一个钙原子；2Na表示2个钠原子；氦气的化学式为：He；3个氮分子化学式为：3N<SUB>2</SUB>．<br />故答案为：钙元素；一个钙原子；2个钠原子；He；3N<SUB>2</SUB>．','【分析】原子的表示方法，用元素符号来表示一个原子，表示多个该原子，就在其元素符号前加上相应的数字．<br />氦气属于稀有气体单质，直接用元素符号表示其化学式．<br />分子的表示方法，正确书写物质的化学式，表示多个该分子，就在其化学式前加上相应的数字．','填空题',3.00,'9a6b2452d7ae084828098386000c88d9',9,400,'化学符号及其周围数字的意义','',2016,'32','2016•泰山区模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839942,'下列实验操作符合安全要求的是（　　）','<img  src=\"/tikuimages/9/2016/400/shoutiniao60/1129cb00-94d4-11e9-a320-b42e9921e93e_xkb12.png\" style=\"vertical-align:middle\" /><br />验证氢气的可燃性','<img  src=\"/tikuimages/9/2016/400/shoutiniao53/112cd840-94d4-11e9-88ae-b42e9921e93e_xkb40.png\" style=\"vertical-align:middle\" /><br />闻气体气味','<img  src=\"/tikuimages/9/2016/400/shoutiniao68/112de9b0-94d4-11e9-b8b3-b42e9921e93e_xkb55.png\" style=\"vertical-align:middle\" /><br />移走蒸发皿','<img  src=\"/tikuimages/9/2016/400/shoutiniao3/11300c8f-94d4-11e9-a5d0-b42e9921e93e_xkb82.png\" style=\"vertical-align:middle\" /><br />稀释浓硫酸','','D','【解答】解：A、可燃性气体与空气混合后点燃可能发生爆炸，氢气具有可燃性，为防止氢气与空气的混合气体在点燃时发生爆炸，产生的气体不能立即点燃，图中所示装置错误．<br />B、闻气体的气味时，应用手在瓶口轻轻的扇动，使极少量的气体飘进鼻子中，不能将鼻子凑到集气瓶口去闻气体的气味，图中所示操作错误．<br />C、正在加热的蒸发皿温度较高，为防止烫伤手，不能用手直接拿热的蒸发皿，应用坩埚钳夹取，图中所示操作错误．<br />D、稀释浓硫酸时，要把浓硫酸缓缓地沿器壁注入水中，同时用玻璃棒不断搅拌，以使热量及时地扩散；一定不能把水注入浓硫酸中；图中所示操作正确．<br />故选：D．','【分析】A、根据可燃性气体与空气混合后点燃可能发生爆炸进行分析判断．<br />B、根据闻气体的气味时的方法（招气入鼻法）进行分析判断．<br />C、根据蒸发操作的注意事项进行分析判断．<br />D、根据浓硫酸的稀释方法（酸入水，沿器壁，慢慢倒，不断搅）进行分析判断．','选择题',3.00,'5460938b9a259e39c962ee01a9e7ca67',9,400,'浓硫酸的性质及浓硫酸的稀释,蒸发与蒸馏操作,氢气、一氧化碳、甲烷等可燃气体的验纯','',2016,'37','2016•邻水县一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839943,'下列家庭小实验只发生物理变化的是（　　）','<img  src=\"/tikuimages/9/2016/400/shoutiniao85/113367f0-94d4-11e9-89bf-b42e9921e93e_xkb84.png\" style=\"vertical-align:middle\" /><br />用蜡烛燃烧生成炭黑的实验','<img  src=\"/tikuimages/9/2016/400/shoutiniao17/11345251-94d4-11e9-bab0-b42e9921e93e_xkb45.png\" style=\"vertical-align:middle\" /><br />自制叶脉书签','<img  src=\"/tikuimages/9/2016/400/shoutiniao97/11358acf-94d4-11e9-bfdb-b42e9921e93e_xkb7.png\" style=\"vertical-align:middle\" /><br />保鲜膜保鲜能力测试','<img  src=\"/tikuimages/9/2016/400/shoutiniao96/1137adb0-94d4-11e9-a137-b42e9921e93e_xkb98.png\" style=\"vertical-align:middle\" /><br />自制汽水','','C','【解答】解：A、用蜡烛燃烧生成炭黑的实验，炭黑是新物质，属于化学变化，故选项错误；<br />B、自制叶脉书签是利用碱溶液与树叶的成分反应，发生了化学变化，故选项错误；<br />C、保鲜膜保鲜能力测试没有新物质生成，属于物理变化，故选项正确；<br />D、自制汽水，汽水是新物质，属于化学变化，故选项错误；<br />故选C','【分析】有新物质生成的变化叫化学变化，没有新物质生成的变化叫物理变化．化学变化的特征是：有新物质生成．判断物理变化和化学变化的依据是：是否有新物质生成．','选择题',2.00,'c7f7b2e92b39b9d9e8e0e76d1a766be4',9,400,'化学变化和物理变化的判别','',2016,'37','2016•南京一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839945,'根据如图所示回答问题．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao98/114457de-94d4-11e9-a1b6-b42e9921e93e_xkb66.png\" style=\"vertical-align:middle\" /><br />（1）图A、B是测量&nbsp;空气中氧气体积的实验：<br />①实验小组按A装置进行实验后，发现进入集气瓶中水的体积小于总容积的<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">1</td></tr><tr><td>5</td></tr></table></span>，你认为导致这一结果的原因可能是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填序号）．<br />A．实验前未将橡胶管夹紧；B．实验前导管中未加满水；C．红磷过量；D．氧气未消耗完．<br />②实验后，实验小组又进行反思，设计了B装置，利用气压变化使白磷燃烧，与A装置比较，其优点是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填序号）．<br />A．防止空气污染；&nbsp;B．减少误差；&nbsp;C．节约燃料；&nbsp;D．便于直接读数．<br />（2）两位同学探究物质发生化学变化前后，总质量是否发生改变？小刘设计的实验装置和药品如图C所示；小李设计的实验装置和药品如图D所示；他们在反应前后都进行规范的操作、准确的称量和细致的观察：<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao74/11454240-94d4-11e9-9614-b42e9921e93e_xkb11.png\" style=\"vertical-align:middle\" /><br />①小刘的实验后，天平平衡，从而验证了质量守恒定律，也由此说明，化学反应前后，一定不变的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填序号）；<br />a．原子种类&nbsp;b．原子数目&nbsp;c．分子种类&nbsp;d．分子数目&nbsp;<br />e．元素种类&nbsp;f．物质种类&nbsp;g．原子质量&nbsp;h．元素质量．<br />②若小刘进行实验时，白磷刚引燃，立即将锥形瓶放在天平上称量，天平也不平衡．导致这一结果的原因是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />③小李的实验后，发现天平不平衡，若依然使用图D中的药品，最好选择如图2中<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>仪器（填序号）组装成新的装置，才能达到实验目的．','','','','','','BD$###$ABCD$###$ab$###$白磷刚燃烧时放出热，使气体膨胀，气球胀大，受到的浮力增大$###$②④⑧','【解答】解：（1）①A实验前未将橡胶管夹紧，会使瓶内的部分氧气和其它气体在红磷燃烧时排出，等装置冷却后，使瓶内的气压变得更小，从而使进入瓶内水的体积超过瓶内气体体积的<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">1</td></tr><tr><td>5</td></tr></table></span>；<br />B实验前导管中未加满水，等装置冷却后，只能使流入瓶内的水减少，而不能使流入瓶内水的体积超过瓶内气体体积的<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">1</td></tr><tr><td>5</td></tr></table></span>；<br />C红磷过量，只能使瓶内的氧气耗尽，不会使流入瓶内水的体积超过瓶内气体体积的<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">1</td></tr><tr><td>5</td></tr></table></span>；<br />D氧气未消耗完，只能使流入瓶内水的体积小于瓶内气体体积的<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">1</td></tr><tr><td>5</td></tr></table></span>．<br />②使用A、B装置时会有少量的五氧化二磷释放到空气中，污染空气；C装置属于微型装置，因此能节约燃料；使用的仪器少，因此能减少误差；又有比较精确的刻度，因此便于直接读数．故选①②③④．<br />（2）（2）①天平平衡，说明反应前后物质的总质量相等，从而验证了质量守恒定律；化学反应前后，原子种类、原子数目、原子质量、元素种类、元素质量一定不变，分子种类、物质种类一定改变，分子数目可能改变；<br />故答案为：质量守恒；abegh；<br />②由于白磷刚引燃时燃烧放出大量的热，使装置内气体受热膨胀将气球胀大，装置受到的浮力增大，所以天平不平衡；<br />故答案为：白磷刚引燃时燃烧放出大量的热，使装置内气体受热膨胀将气球胀大，装置受到的浮力增大；<br />③碳酸钠和稀盐酸反应生成二氧化碳气体，对于有气体参加或生成的反应，在验证质量守恒定律时，应在密闭的体系中进行，因此最好选择锥形瓶、橡皮塞和注射器组装成新的装置，才能达到实验目的；<br />故答案为：②④⑧．<br />答案：（1）①BD；&nbsp;②ABCD；&nbsp;&nbsp;<br />（2）①ab；&nbsp;&nbsp;②白磷刚燃烧时放出热，使气体膨胀，气球胀大，受到的浮力增大；&nbsp;③②④⑧．','【分析】（1）只有熟悉用红磷测定空气组成的实验原理、操作、现象、结论和注意事项，才能正确解答本题．<br />（2）①根据实验现象分析结论；根据质量守恒定律的微观解释分析；<br />②白磷刚引燃时燃烧放出大量的热，使装置内气体受热膨胀将气球胀大，装置受到的浮力增大；<br />③对于有气体参加或生成的反应，在验证质量守恒定律时，应在密闭的体系中进行．','填空题',3.00,'e9d85ed7a78f938a0a2e85b0c237a20f',9,400,'测定空气里氧气含量的探究,质量守恒定律的实验探究','',2016,'32','2016•丰都县校级模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839947,'下列化学符号表示一种元素，一个原子，还能表示该元素组成的物质的是（　　）','H','O<SUB>2</SUB>','2N','Fe','','D','【解答】解：元素符号能表示一种元素，还能表示该元素的一个原子；化学式能表示一种物质，当元素符号又是化学式时，就同时具备了上述三层意义．<br />A、H属于可表示氢元素，表示一个氢原子，但不能表示一种物质，故选项不符合题意．<br />B、该符号是氧气的化学式，不是元素符号，故选项不符合题意．<br />C、该符号可表示2个氮原子，不是元素符号，故选项不符合题意．<br />D、Fe属于金属元素，可表示铁元素，表示一个铁原子，还能表示铁这一纯净物，故选项符合题意．<br />故选D．','【分析】根据元素符号的含义进行分析解答，金属、固体非金属、稀有气体都是由原子直接构成的，故它们的元素符号，既能表示一个原子，又能表示一种元素，还能表示一种物质．','选择题',3.00,'ee8a200237e0bdb900f99366ea9d84fe',9,400,'元素的符号及其意义,化学式的书写及意义','',2016,'37','2016•邵阳县二模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839948,'<img  src=\"/tikuimages/9/2016/400/shoutiniao97/114f7b70-94d4-11e9-b274-b42e9921e93e_xkb32.png\" style=\"vertical-align:middle;FLOAT:right;\" />（2016•东兴区一模）已知白磷和红磷的着火点分别为40℃、240℃．某学习小组按照图所示装置去探究燃烧的三个条件，通过多次实验，他们对燃烧的三个条件始终未能探究全面，你认为他们不能探究到的燃烧条件是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','物质具有可燃性','【解答】解：白磷和红磷都属于可燃物，因此不能探究到的燃烧条件是物质具有可燃性．若要探究物质具有可燃性是燃烧的条件，应该设计对比试验，即选择一种可燃物和一种非可燃物．<br />故填：物质具有可燃性．','【分析】燃烧的条件是：物质具有可燃性，与氧气接触，温度达到可燃物的着火点，三者必须同时具备，缺一不可；','填空题',3.00,'4f2c02b217596acb9af0557119f83b29',9,400,'燃烧与燃烧的条件','',2016,'37','2016•东兴区一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839949,'用元素符号或化学式填空．<br />（1）3个硫原子：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；（2）硫酸：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（3）8个水分子：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；（4）氧化铁中铁的化合价为+3价：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','3S$###$H<SUB>2</SUB>SO<SUB>4</SUB>$###$8H<SUB>2</SUB>O$###$<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td style=\"font-size: 90%\">+3</td></tr><tr><td>Fe</td></tr></table></span><SUB>2</SUB>O<SUB>3</SUB>','【解答】解：（1）由原子的表示方法，用元素符号来表示一个原子，表示多个该原子，就在其元素符号前加上相应的数字，故3个硫原子表示为：3S．<br />（2）硫酸中氢元素显+1价，硫酸根显-2价，其化学式为：H<SUB>2</SUB>SO<SUB>4</SUB>．<br />（3）由分子的表示方法，正确书写物质的化学式，表示多个该分子，就在其化学式前加上相应的数字，则8个水分子可表示为：8H<SUB>2</SUB>O．<br />（4）由化合价的表示方法，在该元素的上方用正负号和数字表示，正负号在前，数字在后，故氧化铁中铁的化合价为+3价可表示为：<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td style=\"font-size: 90%\">+3</td></tr><tr><td>Fe</td></tr></table></span><SUB>2</SUB>O<SUB>3</SUB>．<br />故答案为：（1）3S；（2）H<SUB>2</SUB>SO<SUB>4</SUB>；（3）8H<SUB>2</SUB>O；（4）<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td style=\"font-size: 90%\">+3</td></tr><tr><td>Fe</td></tr></table></span><SUB>2</SUB>O<SUB>3</SUB>．','【分析】（1）原子的表示方法，用元素符号来表示一个原子，表示多个该原子，就在其元素符号前加上相应的数字．<br />（2）硫酸中氢元素显+1价，硫酸根显-2价，写出其化学式即可．<br />（3）分子的表示方法，正确书写物质的化学式，表示多个该分子，就在其化学式前加上相应的数字．<br />（4）化合价的表示方法，在该元素的上方用正负号和数字表示，正负号在前，数字在后．','填空题',3.00,'f6af9cc0c3caa5e0fee3bc84da49dc08',9,400,'化学符号及其周围数字的意义','',2016,'37','2016•江门一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839951,'除去食盐水中CaCl<SUB>2</SUB>和Na<SUB>2</SUB>SO<SUB>4</SUB>杂质的操作有：①加过量BaCl<SUB>2</SUB>溶液；②过滤；③加过量Na<SUB>2</SUB>CO<SUB>3</SUB>溶液；④蒸发结晶；⑤加适量盐酸．（提示：BaSO<SUB>4</SUB>、BaCO<SUB>3</SUB>难溶于水）下列操作顺序正确的是（　　）','③①②⑤④','①⑤③②④','①③②⑤④','①③⑤④②','','C','【解答】解：硫酸根离子用钡离子沉淀，加入过量的氯化钡可以将硫酸根离子沉淀，钙离子用碳酸根离子沉淀，除钙离子加入碳酸钠转化为沉淀，但是加入的碳酸钠要放在加入的氯化钡之后，这样碳酸钠会除去反应剩余的氯化钡，离子都沉淀了，再进行过滤，最后再加入盐酸除去反应剩余的氢氧根离子和碳酸根离子，所以正确的顺序为：①③②⑤④．<br />故选：C．','【分析】根据硫酸根离子用钡离子沉淀，钙离子用碳酸根离子沉淀，过滤要放在所有的沉淀操作之后，加碳酸钠要放在加氯化钡之后，可以将过量的钡离子沉淀最后再用盐酸处理溶液中的碳酸根离子和氢氧根离子进行分析．','选择题',3.00,'d978606ec32e92a2ee7602760ae5f874',9,400,'氯化钠与粗盐提纯,盐的化学性质','',2016,'32','2016•锦江区模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839953,'小林在某品牌“雪饼”包装袋内发现一袋生石灰干燥剂，查阅资料发现生石灰干燥剂为常见的食品干燥剂，请回答下列问题：<br />（1）生石灰干燥剂的主要成分为生石灰，其化学式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．生石灰吸水时发生的反应方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）生石灰干在由于具有强腐蚀性，经常发生伤害小孩或老人眼睛的事情，对此，在处理生石灰干燥剂时的应注意<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','CaO$###$CaO+H<SUB>2</SUB>O═Ca（OH）<SUB>2</SUB>$###$远离孩子，防止溅入眼睛（合理即可）','【解答】解：（1）生石灰是氧化钙的俗称，其化学式为：CaO；生石灰与水反应生成氢氧化钙，反应的化学方程式为：CaO+H<SUB>2</SUB>O═Ca（OH）<SUB>2</SUB>．<br />（2）生石灰干在由于具有强腐蚀性，经常发生伤害小孩或老人眼睛的事情，在处理生石灰干燥剂时的应注意远离孩子，防止溅入眼睛（合理即可）．<br />故答案为：（1）CaO；CaO+H<SUB>2</SUB>O═Ca（OH）<SUB>2</SUB>；（2）远离孩子，防止溅入眼睛（合理即可）．','【分析】（1）生石灰是氧化钙的俗称，与水反应生成氢氧化钙，据此进行分析解答．<br />（2）根据题意，生石灰干在由于具有强腐蚀性，经常发生伤害小孩或老人眼睛的事情，进行分析解答．','书写',3.00,'8b423b59d8bdda2b88f630c90f075765',9,400,'生石灰的性质与用途,书写化学方程式、文字表达式、电离方程式','',2016,'32','2016•广东模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839954,'实验室中含有盐酸的废水倒入下水道会造成铸铁管道腐蚀，下列对废水处理措施可行的是（　　）','加水稀释','加入硼酸','加入铜粉','加入熟石灰','','D','【解答】解：<br />A、加水稀释，酸性减弱，不能消除酸，故错误；<br />B、加入硼酸，酸性增强，不能消除酸，故错误；<br />C、加入铜粉，铜在氢的后面，与酸不反应，不能消除酸，故错误；<br />D、加入熟石灰，加碱性物质氢氧化钙处理成中性在排放，该反应的化学方程式为：Ca（OH）<SUB>2</SUB>+2HCl═CaCl<SUB>2</SUB>+2H<SUB>2</SUB>O，可以防止铸铁管道腐蚀，故正确．<br />答案：D','【分析】根据盐酸具有酸性，直接倒入下水道，会造成管道腐蚀，所以需处理成中性后再排放解答．','选择题',3.00,'5cc630b64476f155cc817b396efe1499',9,400,'酸、碱性废水的处理','',2016,'37','2016•官渡区一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839955,'下列有关资源、能源的叙述正确的是（　　）','可利用的淡水资源只占全球水储量的30.4%','煤的综合利用可以得到煤油、煤焦油、柴油等有用的物质','空气的稀有气体的质量分数约为0.94%','宝贵的金属资源主要分布于地壳和海洋中','','B','【解答】解：<br />A、地球上的淡水只约占全球水储量的2.53%，其中可利用的淡水不足1%，故错误；<br />B、煤隔绝空气加强热可以分解成焦炭、煤焦油、焦炉气等物质，故正确；<br />C、空气的稀有气体的体积分数约为0.94%，故错误；<br />D、金属资源不都是以化合物形式存在于地壳和浩瀚的海洋中，例如金是单质，错误．<br />故选B．','【分析】A、根据水资源的分布进行解答；<br />B、根据煤隔绝空气加强热可以分解的产物分析；<br />C、根据空气的成分进行解答；<br />D、根据金属资源可以以单质存在，例如金解答．','选择题',3.00,'15d9cd2acee1356839e4112780702e0b',9,400,'空气的成分及各成分的体积分数,水资源状况,金属元素的存在及常见的金属矿物,化石燃料及其综合利用','',0,'37','',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839956,'图表资料可以为我们提供很多信息，下面是某学生对图表资料的使用，其中不正确的是（　　）','根据金属活动性顺序表，判断金属能否置换出稀硫酸中的氢','根据熔点数据表判断降低温度时，混合在一起的溶液中哪种物质先行变为固体析出','根据溶解度曲线图，判断从溶液中获得晶体的方法','根据酸、碱、盐的溶解性表，判断某些复分解反应能否进行','','B','【解答】解：A、根据金属活动性顺序表，排在氢前面的金属能置换出酸中的氢，使用正确；<br />B、根据物质的溶解度判断降低温度时，混合在一起的溶液中哪种物质先行变为固体析出，故使用不正确；<br />C、根据溶解度曲线图，溶解度受温度影响较大和不大的物质，可以分别采用降温结晶和蒸发结晶的方法结晶，使用正确；<br />D、复分解反应发生的条件是生成物中有沉淀、气体或水生成，故根据溶解度曲线图，可判断有无沉淀生成，故选项说法正确，使用正确．<br />故选B．','【分析】A、根据金属与稀硫酸反应的条件限制来考虑本题；<br />B、根据混合溶液降温固体析出的原因考虑；<br />C、根据从溶液中获得晶体的方法来考虑；<br />D、根据复分解反应的条件分析考虑．','选择题',2.00,'b9912aa84e107e5e213ecfb14e34d6ed',9,400,'固体溶解度曲线及其作用,金属活动性顺序及其应用,酸碱盐的溶解性','',2016,'37','2016•黄冈校级自主招生',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839957,'结构示意图<img  src=\"/tikuimages/9/2016/400/shoutiniao74/11601d40-94d4-11e9-8daf-b42e9921e93e_xkb58.png\" style=\"vertical-align:middle\" />表示的微粒属于（　　）','分子','原子','阳离子','阴离子','','C','【解答】解：A、图中结构示意图中质子数=13，核外电子数=10，质子数＞核外电子数，为阳离子，故选项错误．<br />B、图中结构示意图中质子数=13，核外电子数=10，质子数＞核外电子数，为阳离子，故选项错误．<br />C、图中结构示意图中质子数=13，核外电子数=10，质子数＞核外电子数，为阳离子，故选项正确．<br />D、图中结构示意图中质子数=13，核外电子数=10，质子数＞核外电子数，为阳离子，故选项错误．<br />故选：C．','【分析】根据当质子数=核外电子数，为原子；当质子数＞核外电子数，为阳离子；当质子数＜核外电子数，为阴离子；据此进行分析解答．','选择题',3.00,'f7c258bc3b9b0541bff7361162462257',9,400,'原子结构示意图与离子结构示意图','',2016,'32','2016•苏州模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839958,'现有Mg（OH）<SUB>2</SUB>和MgCO<SUB>3</SUB>的固体混合物6g，经测定镁元素的质量分数为40%，向固体混合物中加入100g一定溶质质量分数的稀盐酸，恰好完全反应，下列说法不正确的是（　　）','反应过程中有气泡产生','生成MgCl<SUB>2</SUB>的质量为9.5g','反应后溶液中的溶质只有MgCl<SUB>2</SUB>','所用稀盐酸中溶质的质量分数为14.6%','','D','【解答】解：A、碳酸镁和盐酸反应生成氯化镁、水和二氧化碳，所以反应过程中有气泡产生，故A正确；<br />B、有Mg（OH）<SUB>2</SUB>和MgCO<SUB>3</SUB>的固体混合物6g，经测定镁元素的质量分数为40%，所以镁元素的质量为：6g×40%=2.4g，<br />设生成氯化镁的质量为x，参加反应的盐酸质量为y，<br />Mg--2HCl---MgCl<SUB>2</SUB>，<br />24&nbsp;&nbsp;&nbsp;73&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 95<br />2.4g&nbsp;&nbsp;y&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; x<br /><span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">24</td></tr><tr><td>2.4g</td></tr></table></span>=<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">73</td></tr><tr><td>y</td></tr></table></span>=<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">95</td></tr><tr><td>x</td></tr></table></span><br />&nbsp;x=9.5g<br />&nbsp;y=7.3g<br />所以生成MgCl<SUB>2</SUB>的质量为9.5g，故B正确；<br />C、Mg（OH）<SUB>2</SUB>和MgCO<SUB>3</SUB>的固体混合物6g，经测定镁元素的质量分数为40%，向固体混合物中加入100g一定溶质质量分数的稀盐酸，恰好完全反应，所以反应后溶液中的溶质只有MgCl<SUB>2</SUB>，故C正确；<br />D、所用稀盐酸中溶质的质量分数为：<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">7.3g</td></tr><tr><td>100g</td></tr></table></span>×100%=7.3%，故D错误．<br />故选：D．','【分析】根据固体混合物中镁元素的质量分数为80%，可以计算混合物镁元素的质量，依据氯化氢与各物质的关系式进行计算．','选择题',3.00,'0ee1930f7e8823b9b42c81541f5cac23',9,400,'有关溶质质量分数的简单计算,酸的化学性质,盐的化学性质,根据化学反应方程式的计算','',0,'37','',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839959,'CO<SUB>2</SUB>在高温下能与木炭反应生成CO．为验证该反应，某课外兴趣小组设计了一套实验装置，装置图如下：<img  src=\"/tikuimages/9/2016/400/shoutiniao46/1167e56e-94d4-11e9-8533-b42e9921e93e_xkb15.png\" style=\"vertical-align:middle\" /><br />（1）根据装置图，填写如表：<br /><table class=\"edittable\"><TBODY><TR><td width=85>仪器标号</TD><td width=242>仪器中所加物质</TD><td width=207>作用</TD></TR><TR><td>A</TD><td>石灰石，稀盐酸</TD><td>石灰石与盐酸作用产生CO<SUB>2</SUB></TD></TR><TR><td>B</TD><td>饱和碳酸氢钠溶液（与CO<SUB>2</SUB>不反应）</TD><td>除去CO<SUB>2</SUB>中的HCl气体</TD></TR><TR><td>C</TD><td><u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD><td><u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD></TR><TR><td>D</TD><td><u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD><td><u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD></TR><TR><td>E</TD><td><u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD><td><u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD></TR></TBODY></TABLE>（2）D处要增加的仪器的名称是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）写出CO<SUB>2</SUB>在高温下与木炭反应生成CO的化学方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（4）验证CO的方法是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','浓硫酸$###$干燥CO<SUB>2</SUB>气体$###$干燥的木炭粉$###$与CO<SUB>2</SUB>反应生成CO$###$Na0H溶液$###$吸收未反应的CO<SUB>2</SUB>$###$酒精喷灯$###$CO<SUB>2</SUB>+C<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2CO$###$点燃气体，火焰呈蓝色，再用内壁附着有澄清石灰水的烧杯罩在火焰上，烧杯内壁的石灰水变浑浊','【解答】解：（1）由题意可知，石灰石和稀盐酸生成的二氧化碳气体中常含有氯化氢气体和水蒸气，应先通过通过饱和碳酸氢钠溶液处于氯化氢气体，再通过中浓硫酸除去水蒸气，在高温条件下，碳与二氧化碳反应生成了一氧化碳，生成一氧化碳中混油二氧化碳，可通过Na0H溶液吸收未反应的CO<SUB>2</SUB>C中装有浓硫酸，浓硫酸有吸水性，故其作用是吸水作干燥剂；D中装有木炭，作用是在高温下和二氧化碳反应生成一氧化碳；E中装有Na0H溶液，作用是吸收未反应的CO<SUB>2</SUB>；<br />（2）CO<SUB>2</SUB>在高温下与木炭反应生成CO，所以，D处要增加的仪器的名称是酒精喷灯；<br />（3）CO<SUB>2</SUB>在高温下与木炭反应生成CO，反应的方程式是：CO<SUB>2</SUB>+C<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2CO；<br />（4）验证CO的方法是：点燃气体，火焰呈蓝色，再用内壁附着有澄清石灰水的烧杯罩在火焰上，烧杯内壁的石灰水变浑．<br />故答案为：<table class=\"edittable\"><TBODY><TR><td width=91>仪器标号</TD><td width=198>仪器中所加物质</TD><td width=181>作用</TD></TR><TR><td>C</TD><td>浓硫酸</TD><td>干燥CO<SUB>2</SUB>气体</TD></TR><TR><td>D</TD><td>干燥的木炭粉</TD><td>与CO<SUB>2</SUB>反应生成CO</TD></TR><TR><td>E</TD><td>Na0H溶液</TD><td>吸收未反应的CO<SUB>2</SUB></TD></TR></TBODY></TABLE>（2）酒精喷灯；<br />（3）CO<SUB>2</SUB>+C<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2CO；<br />（4）点燃气体，火焰呈蓝色，再用内壁附着有澄清石灰水的烧杯罩在火焰上，烧杯内壁的石灰水变浑浊．','【分析】（1）根据实验的目的和过程可知，石灰石和稀盐酸生成的二氧化碳气体中常含有氯化氢气体和水蒸气，应先通过通过饱和碳酸氢钠溶液处于氯化氢气体，再通过中浓硫酸除去水蒸气，在高温条件下，碳与二氧化碳反应生成了一氧化碳，生成一氧化碳中混油二氧化碳，可通过Na0H溶液吸收未反应的CO<SUB>2</SUB>；<br />（2）根据CO<SUB>2</SUB>在高温下与木炭反应分析用的仪器；<br />（3）根据反应写出化学方程式；<br />（4）根据检验一氧化碳的方法分析．','书写',3.00,'d27f617abd8fe938ff0c0b503783e2db',9,400,'常见气体的检验与除杂方法,二氧化碳的实验室制法,二氧化碳的化学性质,书写化学方程式、文字表达式、电离方程式','',2016,'37','2016•番禺区一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839960,'请从如图中选择适当的仪器组装实验装置．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao95/116d1591-94d4-11e9-8c7a-b42e9921e93e_xkb65.png\" style=\"vertical-align:middle\" /><br />（1）若用石灰石和稀盐酸制取二氧化碳，应选择上面仪器中的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填字母代号），反应的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）若用高锰酸钾制取氧气，还需补充的一种仪器是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；反应的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','ABEF$###$CaCO<SUB>3</SUB>+2HCl=CaCl<SUB>2</SUB>+CO<SUB>2</SUB>↑+H<SUB>2</SUB>O$###$试管$###$2KMnO<SUB>4</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>K<SUB>2</SUB>MnO<SUB>4</SUB>+MnO<SUB>2</SUB>+O<SUB>2</SUB>↑．','【解答】解：（1）实验室制取二氧化碳属于固液常温型，所以选择试管、铁架台、单孔塞，或试管长颈漏斗、铁架台、双孔塞；二氧化碳的密度比空气大，所以可用向上排空气法收集，反应的方程式为：CaCO<SUB>3</SUB>+2HCl=CaCl<SUB>2</SUB>+CO<SUB>2</SUB>↑+H<SUB>2</SUB>O；<br />故答案为：ABEF；&nbsp;CaCO<SUB>3</SUB>+2HCl=CaCl<SUB>2</SUB>+CO<SUB>2</SUB>↑+H<SUB>2</SUB>O；<br />（2）实验室用高锰酸钾制氧气需加热，所以需试管，高锰酸钾加热生成锰酸钾、二氧化锰和氧气；反应的方程式为：2KMnO<SUB>4</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>K<SUB>2</SUB>MnO<SUB>4</SUB>+MnO<SUB>2</SUB>+O<SUB>2</SUB>↑．<br />故答案为：试管；2KMnO<SUB>4</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>K<SUB>2</SUB>MnO<SUB>4</SUB>+MnO<SUB>2</SUB>+O<SUB>2</SUB>↑．','【分析】（1）实验室通常用大理石或石灰石和稀盐酸反应制取二氧化碳，反应不需要加热，大理石和石灰石的主要成分是碳酸钙，能和稀盐酸反应生成氯化钙、水和二氧化碳；<br />（2）高锰酸钾受热时能够分解生成锰酸钾、二氧化锰和氧气．','书写',3.00,'22f582aa47e4304f3bd79d307008f99d',9,400,'氧气的制取装置,氧气的收集方法,二氧化碳的实验室制法,书写化学方程式、文字表达式、电离方程式','',2016,'32','2016•安阳模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839961,'<img  src=\"/tikuimages/9/2016/400/shoutiniao40/1170e61e-94d4-11e9-875c-b42e9921e93e_xkb59.png\" style=\"vertical-align:middle;FLOAT:right\" />按照实验主要目的可将实验分为：探究化学反应前后物质的质量关系、探究物质的性质、探究物质的含量、探究物质的组成等．下列实验的主要目的与铁丝在氧气中燃烧实验相同的是（　　）','<img  src=\"/tikuimages/9/2016/400/shoutiniao21/1171a970-94d4-11e9-aedd-b42e9921e93e_xkb61.png\" style=\"vertical-align:middle\" />','<img  src=\"/tikuimages/9/2016/400/shoutiniao71/1177c3f0-94d4-11e9-bdcf-b42e9921e93e_xkb29.png\" style=\"vertical-align:middle\" />','<img  src=\"/tikuimages/9/2016/400/shoutiniao94/117a830f-94d4-11e9-8280-b42e9921e93e_xkb20.png\" style=\"vertical-align:middle\" />','<img  src=\"/tikuimages/9/2016/400/shoutiniao30/117d4230-94d4-11e9-8b1a-b42e9921e93e_xkb50.png\" style=\"vertical-align:middle\" />','','C','【解答】解：铁丝在氧气中燃烧生成四氧化三铁，说明铁能够在氧气中燃烧，主要是探究铁的化学性质；<br />A、利用红磷在空气中燃烧可以测定空气中氧气的含量，该实验的主要目的是探究物质的含量；<br />B、利用氢氧化钠和硫酸铜反应可以验证质量守恒定律，该实验的主要目的是探究化学反应前后物质的质量关系；<br />C、二氧化碳能和水反应生成碳酸，碳酸能使石蕊试液变红色，该实验主要目的是探究物质的性质；<br />D、电解水生成氢气和氧气，说明水是由氢元素和氧元素组成的，该实验主要目的是探究物质的组成．<br />故选：C．','【分析】利用红磷在空气中燃烧可以测定空气中氧气的含量；<br />利用氢氧化钠和硫酸铜反应可以验证质量守恒定律；<br />二氧化碳能和水反应生成碳酸，碳酸能使石蕊试液变红色；<br />电解水生成氢气和氧气．','选择题',3.00,'ed6567de1d0c2199b9133f25edde3a84',9,400,'化学实验方案设计与评价,空气组成的测定,氧气的化学性质','',2016,'32','2016•泰州模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839962,'<img  src=\"/tikuimages/9/2016/400/shoutiniao57/11824b40-94d4-11e9-b41b-b42e9921e93e_xkb80.png\" style=\"vertical-align:middle;FLOAT:right\" />（2016•陕西模拟）根据图中A、B、C三种物质的溶解度曲线，回答下列问题．<br />（1）三种物质中有一种是气体，则<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>是气体（填序号，下同）．<br />（2）t<SUB>1</SUB>℃时，三种物质的溶解度的大小关系是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）现有A、B、C三种物质的浓溶液，适用海水晒盐原理进行结晶提纯的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（4）将t<SUB>2</SUB>℃时A物质的饱和溶液降温至t<SUB>1</SUB>℃，溶液的溶质质量分数<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“减小”“增大”或“不变”）．','','','','','','C$###$B＞A=C$###$B$###$减小','【解答】解：（1）三种物质中C的溶解度随温度升高而减小，是气体；<br />（2）由图可知：t<SUB>1</SUB>℃时，三种物质的溶解度的大小关系是B＞A=C；<br />（3）现有A、B、C三种物质的浓溶液，适用海水晒盐原理进行结晶提纯的是B，因为B的溶解度受温度的影响很小；<br />（4）将t<SUB>2</SUB>℃时A物质的饱和溶液降温至t<SUB>1</SUB>℃，A的溶解度减小析出晶体，溶液中溶质的质量减小，溶剂的质量不变，故溶液的溶质质量分数减小；<br />故答案为：（1）C；&nbsp;&nbsp;（2）B＞A=C；&nbsp;&nbsp;&nbsp;（3）B；&nbsp;&nbsp;（4）减小．','【分析】（1）气体的溶解度随温度升高而减小；<br />（2）据溶解度曲线可比较同一温度下不同物质的溶解度大小；<br />（3）适用海水晒盐原理进行结晶提纯的是B，因为B的溶解度受温度的影响很小；<br />（4）A的溶解度随温度降低而减小，降温析出晶体，据此解答．','填空题',3.00,'dc5912a90e266ab00ae84fd71be18731',9,400,'结晶的原理、方法及其应用,固体溶解度曲线及其作用,溶质的质量分数、溶解性和溶解度的关系','',2016,'32','2016•陕西模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839963,'下列图示的实验操作，正确的是（　　）','<img  src=\"/tikuimages/9/2016/400/shoutiniao2/118a3a80-94d4-11e9-91d7-b42e9921e93e_xkb72.png\" style=\"vertical-align:middle\" /><br />稀释浓硫酸','<img  src=\"/tikuimages/9/2016/400/shoutiniao83/118d20ae-94d4-11e9-af64-b42e9921e93e_xkb78.png\" style=\"vertical-align:middle\" /><br />配制氯化钠溶液','<img  src=\"/tikuimages/9/2016/400/shoutiniao58/118de400-94d4-11e9-884e-b42e9921e93e_xkb96.png\" style=\"vertical-align:middle\" /><br />用滴管取液体','<img  src=\"/tikuimages/9/2016/400/shoutiniao79/118ece61-94d4-11e9-b45b-b42e9921e93e_xkb94.png\" style=\"vertical-align:middle\" /><br />过滤','','D','【解答】解：A、稀释浓硫酸时，要把浓硫酸缓缓地沿器壁注入水中，同时用玻璃棒不断搅拌，以使热量及时地扩散；一定不能把水注入浓硫酸中；图中所示操作错误．<br />B、量筒不能用于配制溶液，图中所示操作错误．<br />C、试剂瓶塞取下应倒放在桌面上，图中所示操作错误．<br />D、过滤液体时，要注意“一贴、二低、三靠”的原则，图中所示操作正确．<br />故选：D．','【分析】A、根据浓硫酸的稀释方法（酸入水，沿器壁，慢慢倒，不断搅）进行分析判断．<br />B、根据量筒不能用于配制溶液进行分析判断．<br />C、根据试剂瓶塞应倒放在桌面上进行分析判断．<br />D、过滤液体时，注意“一贴、二低、三靠”的原则．','选择题',3.00,'7173ebff2865548f30c75fa88704a542',9,400,'液体药品的取用,物质的溶解,浓硫酸的性质及浓硫酸的稀释,过滤的原理、方法及其应用','中山市',2016,'37','2016春•中山市月考',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839964,'下列是某化学小组探究物质组成与性质的实验装置图，根据要求回答：<br /><table class=\"edittable\"><TBODY><TR><td width=140>A</TD><td width=85>B</TD><td width=133>C</TD><td width=193>D</TD></TR><TR><td>测定空气中氧气的含量</TD><td>稀释浓硫酸</TD><td>铁钉在空气中的变化</TD><td>一氧化碳还原氧化铁</TD></TR><TR><td>木炭<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao51/119277de-94d4-11e9-9823-b42e9921e93e_xkb91.png\" style=\"vertical-align:middle\" /></TD><td><img  src=\"/tikuimages/9/2016/400/shoutiniao77/1196bda1-94d4-11e9-b7cf-b42e9921e93e_xkb61.png\" style=\"vertical-align:middle\" /></TD><td>水<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao37/119a8e30-94d4-11e9-9e75-b42e9921e93e_xkb56.png\" style=\"vertical-align:middle\" /></TD><td><img  src=\"/tikuimages/9/2016/400/shoutiniao94/119f220f-94d4-11e9-8a36-b42e9921e93e_xkb68.png\" style=\"vertical-align:middle\" /></TD></TR></TBODY></TABLE>（l）上述实验设计中有错误的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填字母序号）；A中可燃物应取过量的原因是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）B中玻璃棒的作用是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；C中会出现的实验现象为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）D中硬质玻璃管内发生反应的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，酒精灯的作用是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','AB$###$可将装置中的氧气耗尽$###$搅拌，使产生的热量迅速扩散（加快散热）$###$铁钉生锈；U型管液面左高右低$###$3CO+2Fe<SUB>2</SUB>O<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2Fe+3CO<SUB>2</SUB>$###$点燃尾气中的CO，防止污染环境','【解答】解：<br />（1）经分析A、B、C、D四个实验，A实验生成物为二氧化碳气体，不能形成气压差，故不能达到实验目的；而B稀释浓硫酸，应将浓硫酸倒入水中，故不能达到实验目的；A实验测定空气中氧气的含量，故应充分消耗氧气；<br />（2）B中玻璃棒的作用是搅拌、散热；一段时间后，C中的实验现象为铁丝生锈，U型管中液面左高右低；<br />（3）D中硬质玻璃管内发生反应的化学方程式为Fe<SUB>2</SUB>O<SUB>3</SUB>+3CO<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2Fe+3CO<SUB>2</SUB>．酒精灯的作用是点燃未反应的CO，防止其污染空气．<br />答案：<br />（1）AB；可将装置中的氧气耗尽；<br />（2）搅拌，使产生的热量迅速扩散（加快散热）；铁钉生锈；U型管液面左高右低<br />（3）3CO+2Fe<SUB>2</SUB>O<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>&nbsp;2Fe+3CO<SUB>2； </SUB>点燃尾气中的CO，防止污染环境．','【分析】（1）A、B、C、D四个实验中：A实验生成物为二氧化碳气体，不能形成气压差，故不能达到实验目的；而B稀释浓硫酸，应将浓硫酸倒入水中，故不能达到实验目的；A实验测定空气中氧气的含量，故应充分消耗氧气；<br />（2）B实验稀释浓硫酸，玻璃棒搅拌并及时散热；C实验铁丝生锈消耗空气中的氧气，管内外形成气压差，故U型管中液面左高右低；<br />（3）D实验一氧化碳还原氧化铁，已知反应物一氧化碳和氧化铁，生成物为二氧化碳和铁，反应条件为高温；反应完，将有尾气一氧化碳，故需要处理掉．','书写',3.00,'ce490eedd91ba293f528df54d2979e8b',9,400,'浓硫酸的性质及浓硫酸的稀释,空气组成的测定,一氧化碳还原氧化铁,金属锈蚀的条件及其防护,书写化学方程式、文字表达式、电离方程式','',2016,'37','2016•苏州一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839965,'根据如图实验回答问题．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao25/11a6750f-94d4-11e9-9189-b42e9921e93e_xkb37.png\" style=\"vertical-align:middle\" /><br />取甲反应后的溶液，滴加Na<SUB>2</SUB>CO<SUB>3</SUB>溶液，溶液pH的变化如图乙所示，则b点的含义是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，与CaCl<SUB>2</SUB>反应的碳酸钠溶液的质量为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>g．','','','','','','所加的碳酸钠与盐酸恰好反应，溶液呈中性$###$40','【解答】解：溶液pH的变化图乙可知ab段是碳酸钠与盐酸反应，bc段是碳酸钠与氯化钙反应，c点之后碳酸钠过量；所以b点的含义是所加的碳酸钠与盐酸恰好反应，溶液呈中性，与CaCl<SUB>2</SUB>反应的碳酸钠溶液的质量为70.0g-30.0g=40.0g．<br />故答案为：所加的碳酸钠与盐酸恰好反应，溶液呈中性（合理给分）、40．','【分析】根据题目的信息可知，pH变化情况溶液最初pH＜7，所以甲反应后的溶液中一定含CaCl<SUB>2</SUB>和HCl，逐滴加入Na<SUB>2</SUB>CO<SUB>3</SUB>溶液，图乙可以判断是盐酸先反应，因此PH值逐渐增大到7，氯化钙再与碳酸钠反应PH=7，到c点之后碳酸钠过量．','填空题',3.00,'bd3d38c61bb32fca4212428e474106e7',9,400,'酸的化学性质,溶液的酸碱性与pH值的关系,盐的化学性质','',2016,'37','2016•延庆县一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839966,'<img  src=\"/tikuimages/9/2016/400/shoutiniao49/11ad2bcf-94d4-11e9-841f-b42e9921e93e_xkb80.png\" style=\"vertical-align:middle;FLOAT:right;\" />如图是甲、乙、丙三种物质的溶解度曲线，将甲、乙、丙三种物质t<SUB>2</SUB>℃时的饱和溶液降温至t<SUB>1</SUB>℃，所得溶液的溶质质量分数关系正确的是（　　）','甲＞乙＞丙','甲=乙＞丙','丙＞甲=乙','甲=乙=丙','','B','【解答】解：甲、乙物质降低温度，溶解度减小，饱和溶液会析出晶体，丙物质不会析出晶体，t<SUB>1</SUB>℃甲、乙的溶解度大于t<SUB>2</SUB>℃丙物质的溶解度，所以将甲、乙、丙三种物质t<SUB>2</SUB>℃时的饱和溶液降温至t<SUB>1</SUB>℃，所得溶液的溶质质量分数关系是：甲=乙＞丙．<br />故选：B．','【分析】根据固体的溶解度曲线可以：①查出某物质在一定温度下的溶解度，从而确定物质的溶解性，②比较不同物质在同一温度下的溶解度大小，从而判断饱和溶液中溶质的质量分数的大小，③判断物质的溶解度随温度变化的变化情况，从而判断通过降温结晶还是蒸发结晶的方法达到提纯物质的目的．','选择题',3.00,'7ef4a2fb1fd963c07e4c7a15f74dc461',9,400,'固体溶解度曲线及其作用,溶质的质量分数、溶解性和溶解度的关系','',2016,'37','2016•大悟县二模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839968,'浓盐酸敞口放置一段时间后，将会（　　）','质量变小','质量变大','质量不变','质量分数变大','','A','【解答】解：浓盐酸具有挥发性，浓盐酸敞口放置一段时间后，溶质的质量减少，溶剂的质量不变，溶质的质量分数会变小．<br />故选：A．','【分析】根据浓盐酸具有挥发性，进行分析判断．','选择题',3.00,'8b3e9ebfadef127bd76c62396486aaa4',9,400,'酸的物理性质及用途','',2016,'37','2016•石景山区一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839969,'把干燥的蓝色石蕊试纸放在CO<SUB>2</SUB>气体中，石蕊试纸不变色．将CO<SUB>2</SUB>气体通入滴有石蕊指示剂的水中，溶液变红色．再将该溶液敞口加热一段时间，发现溶液的红色褪去，冷却到原温度后，再次通入CO<SUB>2</SUB>，溶液又变红色．上述实验现象不能说明的是（　　）','CO<SUB>2</SUB>具有酸性','CO<SUB>2</SUB>能和水发生化学反应','CO<SUB>2</SUB>的水溶液具有酸性','加热能使CO<SUB>2</SUB>水溶液的酸性减弱','','A','【解答】解：A、把干燥的蓝色石蕊试纸放在CO<SUB>2</SUB>气体中，石蕊试纸不变色，说明二氧化碳不具有酸性，故错误，可选．<br />B、二氧化碳与水反应生成碳酸，所以CO<SUB>2</SUB>能和水发生化学反应是正确的．故不可选；<br />C、二氧化碳的水溶液是碳酸溶液，能使滴有石蕊指示剂的溶液变红色．所以CO<SUB>2</SUB>的水溶液具有酸性正确．故不可选；<br />D、碳酸极易分解，使溶液的酸性减弱，题干中的“将该溶液敞口加热一段时间，发现溶液的红色褪去，”也同样说明了这点，所以加热能使CO<SUB>2</SUB>水溶液的酸性减弱正确，故不可选；<br />故选A．','【分析】首先了解二氧化碳能跟水反应生成碳酸的性质，碳酸显酸性，就是说二氧化碳的水溶液显酸性；碳酸极易分解．我们可以根据这些知识做出判断．另外还要了解紫色石蕊试液遇酸变红．','选择题',3.00,'d95f5ad50fa2dfe26c0abe9c07183468',9,400,'二氧化碳的化学性质','',2016,'37','2016•博白县一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839971,'“珍爱生命，远离毒品”．冰毒是一种毒品，其主要成分是甲基苯丙胺（化学式为C<SUB>10</SUB>H<SUB>15</SUB>N）．有关甲基苯丙胺的说法错误的是（　　）','属于纯净物','燃烧后只生成二氧化碳和水','甲基苯丙胺由碳、氢、氮元素组成','1个甲基苯丙胺分子由10个碳原子、15个氢原子和1个氮原子组成','','B','【解答】解：A．可以写出物质的化学式，由此可见它是一种由单一物质组成的物质，属于纯净物，故正确；<br />B．由物质的化学式C<SUB>10</SUB>H<SUB>15</SUB>N可知，其中含有氮元素，因此燃烧后还有含氮元素的化合物，故错误；<br />C．根据化学式C<SUB>10</SUB>H<SUB>15</SUB>N可知，甲基苯丙胺由C、H、N三种元素组成，故正确；<br />D．由物质的化学式C<SUB>10</SUB>H<SUB>15</SUB>N可知，一个甲基苯丙胺分子中含有10个碳原子、15个氢原子和1个氮原子，故正确．<br />故选B．','【分析】A．根据纯净物的概念来分析；<br />B．根据物质的组成和元素守恒来分析；<br />C．根据化学式的意义来分析；<br />D．根据分子的构成来分析．','选择题',3.00,'e6b6b88c512fa47072b14c5ba1b3b5ad',9,400,'纯净物和混合物的判别,化学式的书写及意义','',2016,'32','2016•梅州模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839972,'区分硬水和软水通常用<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．长期引用硬水有害健康，在生活中，人们通常用<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>办法降低水的硬度，工业上通常用<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>办法降低水的硬度．','','','','','','肥皂水$###$煮沸$###$蒸馏','【解答】解：区分硬水和软水通常用肥皂水．长期引用硬水有害健康，在生活中，人们通常用煮沸办法降低水的硬度，工业上通常用蒸馏办法降低水的硬度．<br />故答案为：肥皂水；煮沸；蒸馏．','【分析】根据硬水和软水的区别方法及转化方法进行分析．','填空题',3.00,'ac07e2dd4c6c1adc23c415db7282a4d8',9,400,'硬水与软水','',0,'37','',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839974,'下列关于甲烷（CH<SUB>4</SUB>）的说法中，错误的是（　　）','甲烷是最简单的有机化合物','甲烷是由碳元素和氢元素组成的','甲烷分子由碳原子和氢原子构成','甲烷中碳、氢元素质量比为1：4','','D','【解答】解：A、甲烷是结构最简单的有机物，故A正确；<br />B、由化学式可知，甲烷是由碳元素和氢元素组成的，故B正确；<br />C、甲烷是由甲烷分子构成的，1个甲烷分子中含有1个碳原子和4个氢原子，故C正确；<br />D、甲烷中碳、氢元素质量比为：12：（1×4）=3：1，故D错误．<br />故选：D．','【分析】A、根据最简单的有机物是甲烷进行分析判断；<br />B、根据物质是由元素组成的进行分析；<br />C、根据甲烷的微观构成进行分析判断；<br />D、根据甲烷中碳、氢元素质量比等于相对原子质量和原子个数乘积的比值进行分析．','选择题',3.00,'ebfbc41c0e05cb4ef326ab5018db597b',9,400,'有机物与无机物的区别,化学式的书写及意义,元素质量比的计算','',2016,'37','2016•南关区一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839975,'如图是某元素的原子结构示意图，下列说法不正确的是（　　）<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao66/11c19e30-94d4-11e9-9658-b42e9921e93e_xkb38.png\" style=\"vertical-align:middle\" />','该原子在化学反应中容易得到2个电子','该原子的质子数为12','该原子的核外有三个电子层','该原子的核外电子数为12','','A','【解答】解：A、由某元素的原子结构示意图，最外层电子数是2，在化学反应中易失去2个电子而形成阳离子，故选项说法错误．<br />B、由某元素的原子结构示意图，圆圈内数字表示核内质子数，该原子的质子数为12，故选项说法正确．<br />C、由某元素的原子结构示意图，弧线表示电子层，该原子的核外有三个电子层，故选项说法正确．<br />D、由某元素的原子结构示意图，该原子的核外电子数为12，故选项说法正确．<br />故选：A．','【分析】原子结构示意图中，圆圈内数字表示核内质子数，弧线表示电子层，弧线上的数字表示该层上的电子数，离圆圈最远的弧线表示最外层．若最外层电子数≥4，在化学反应中易得电子，若最外层电子数＜4，在化学反应中易失去电子．','选择题',3.00,'2bb2e223e675993ae8e1558e1bad7b98',9,400,'原子结构示意图与离子结构示意图','',2016,'37','2016•新化县二模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839976,'下列现象和事实可以用同一原理解释的是（　　）','淀粉遇到碘酒和碘水都会变蓝','生石灰和氢氧化钠放入水中都出现放热现象','浓盐酸和浓硫酸长期露置在空气中浓度均降低','蜡烛剪去灯芯后和罩在玻璃杯中都会熄灭','','A','【解答】解：A、淀粉遇到碘酒和碘水都会变蓝，可以用同一个原理解答，正确；<br />B、生石灰与水反应放出热量，氢氧化钠固体溶于水放出热量，不能用同一个原理解答，错误；<br />C、浓盐酸具有挥发性，浓度减小，浓硫酸具有吸水性，浓度减小，不能用同一个原理解答，错误；<br />D、蜡烛剪去灯芯熄灭后熄灭，是移走可燃物，罩在玻璃杯中会熄灭是隔绝氧气，不能用同一个原理解答，错误；<br />故选A．','【分析】A、根据淀粉的检验方法解答；<br />B、根据生石灰、氢氧化钠与水混合的现象解答；<br />C、根据浓盐酸和浓硫酸的性质解答；<br />D、根据灭火的方法解答．','选择题',3.00,'8742c177d1108639456e1fbd90ba86b5',9,400,'蜡烛燃烧实验,溶解时的吸热或放热现象,生石灰的性质与用途,酸的物理性质及用途,鉴别淀粉、葡萄糖的方法与蛋白质的性质','',2016,'37','2016•苏州一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839977,'<img  src=\"/tikuimages/9/0/400/shoutiniao50/11c7b8b0-94d4-11e9-a005-b42e9921e93e_xkb42.png\" style=\"vertical-align:middle;FLOAT:right\" />某地石灰石资源丰富，某兴趣小组同学取40g样品进行实验：先将样品粉碎，然后取140g稀盐酸分成7等份，逐次加到样品粉末中（假设石灰石中其他成分不与盐酸反应，也不溶于水），得到部分数据与图象如下，请根据有关信息回答问题． <table class=\"edittable\"><TBODY><TR><td width=112></TD><td width=51>　第2次</TD><td width=56>　第5次</TD><td width=70>第6次　</TD></TR><TR><td>加入盐酸质量（g）</TD><td>　20</TD><td>20　</TD><td>20　</TD></TR><TR><td>　剩余固体质量（g）　</TD><td>　30</TD><td>　a</TD><td>10</TD></TR></TBODY></TABLE>（1）表中a=<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />（2）计算该石灰石样品中CaCO<SUB>3</SUB>的质量分数（写出计算过程）<br />（3）求该盐酸中溶质的质量分数为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（结果保留一位小数）','','','','','','15$###$18.25%','【解答】解：（1）第二次加入盐酸后剩余固体质量为30g，所以每次加入盐酸固体质量减少5g，第5次加入盐酸后，剩余固体的质量为：40g-5×5g=15g，所以a是15；<br />（2）由图示可知第6次加入盐酸后气体质量不再增加，所以参加反应的碳酸钙质量为：40g-10g=30g，所以该石灰石样品中CaCO<SUB>3</SUB>的质量分数为：<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">30g</td></tr><tr><td>40g</td></tr></table></span>×100%=75%；<br />（3）设参加反应的盐酸质量为x<br />CaCO<SUB>3</SUB>+2HCl═CaCl<SUB>2</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑<br />&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 73&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;44<br />&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; x&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;13.2g<br />&nbsp;&nbsp;&nbsp;&nbsp; <span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">73</td></tr><tr><td>x</td></tr></table></span>=<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">44</td></tr><tr><td>13.2g</td></tr></table></span><br />&nbsp;&nbsp;&nbsp; x=21.9g<br />所以该盐酸中溶质的质量分数为<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">21.9g</td></tr><tr><td>120g</td></tr></table></span>×100%=18.25%．<br />故答案为：（1）15；<br />（2）75%；<br />（3）该盐酸中溶质的质量分数为18.25%．','【分析】（1）根据第二次加入盐酸后剩余固体质量为30g，所以每次加入盐酸固体质量减少5g，第5次加入盐酸后，剩余固体的质量为：40g-5×5g=15g进行分析；<br />（2）根据图示可知第6次加入盐酸后气体质量不再增加，所以参加反应的碳酸钙质量为：40g-10g=30g进行分析；<br />（3）根据碳酸钙和盐酸反应生成氯化钙、水和二氧化碳，依据题中的数据进行计算．','填空题',3.00,'7ac20f180abb69925b1f8e9eeaad4a8b',9,400,'有关溶质质量分数的简单计算,根据化学反应方程式的计算','',0,'37','',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839979,'石灰岩因其坚硬致密，可用作建筑石材，还可用于生产水泥和生石灰．<br />（1）高温煅烧石灰石可得到生石灰和二氧化碳，生石灰可用作干燥剂的原因是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．在炼铁高炉中，二氧化碳与碳在高温下反应得到一氧化碳，在此反应中作为还原剂的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）石灰石与稀盐酸反应制取二氧化碳，写出反应的化学方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','生石灰能与水发生反应$###$碳（C）$###$CaCO<SUB>3</SUB>+2HCl=CaCl<SUB>2</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑','【解答】解：<br />（1）生石灰可用作干燥剂的原因是生石灰能与水发生反应；在炼铁高炉中，二氧化碳与碳在高温下反应得到一氧化碳，在此反应中作为还原剂的是碳；<br />（2）石灰石与稀盐酸反应制取二氧化碳，同时生成氯化钙、水，反应的化学方程式是CaCO<SUB>3</SUB>+2HCl=CaCl<SUB>2</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑．<br />答案：<br />（1）生石灰能与水发生反应；&nbsp;&nbsp;&nbsp;&nbsp;碳（C）；<br />（2）CaCO<SUB>3</SUB>+2HCl=CaCl<SUB>2</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑','【分析】（1）根据生石灰具有吸水性解答；根据在炼铁高炉中，二氧化碳与碳在高温下反应得到一氧化碳，在此反应中作为还原剂的是碳解答；<br />（2）首先根据反应原理找出反应物、生成物、反应条件，根据化学方程式的书写方法、步骤（写、配、注、等）进行书写即可．','书写',3.00,'f5570fbc3460bd7cebe4703ecb6e702d',9,400,'二氧化碳的化学性质,生石灰的性质与用途,盐的化学性质,书写化学方程式、文字表达式、电离方程式','',2016,'32','2016•井研县模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839980,'如表为元素周期表中某一周期元素的原子结构示意图．<br /><img  src=\"/tikuimages/9/0/400/shoutiniao99/11cf0bb0-94d4-11e9-97c1-b42e9921e93e_xkb20.png\" style=\"vertical-align:middle\" /><br />请回答下列问题：<br />①表中磷原子的核电荷数X=<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />②表中具有相对稳定结构的元素是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>_<br />③铝与硫两种元素组成的化合物的化学式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />④上述元素在周期表中处于第<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>周期，它们属于同一周期的原因是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />⑤磷化钠是剧毒结晶，化学式Na<SUB>3</SUB>P，遇水、潮湿空气反应释出剧毒和自燃的磷化氢 （PH<SUB>3</SUB>）气体和另一物质，试写出该反应的化学反应方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','15$###$氩$###$Al<SUB>2</SUB>S<SUB>3</SUB>$###$三$###$它们原子的核外电子层数相同$###$Na<SUB>3</SUB>P+3H<SUB>2</SUB>O=3NaOH+PH<SUB>3</SUB>↑','【解答】解：①原子中质子数=核外电子数，则磷原子的核电荷数x=2+8+5=15．故填：15；<br />②最外层具有8个电子的原子具有相对稳定结构，表中具有相对稳定结构的元素氩元素．故填：氩；<br />③铝元素显+3价，硫元素显-2价，二者组成的化合物硫化铝的化学式为：Al<SUB>2</SUB>S<SUB>3</SUB>；故填：Al<SUB>2</SUB>S<SUB>3</SUB>；<br />④分析上述元素的特点可知：原子所处的周期数与该原子的电子层数相同，所以在周期表中处于同一周期的原子，核外电子层数相同，因为原子核外有3个电子层，所以位于第三周期．故填：三；它们原子的核外电子层数相同；<br />⑤由质量守恒定律分析可知，磷化钠与水反应生成磷化氢和氢氧化钠，反应的化学方程式为：Na<SUB>3</SUB>P+3H<SUB>2</SUB>O=3NaOH+PH<SUB>3</SUB>↑；故填：Na<SUB>3</SUB>P+3H<SUB>2</SUB>O=3NaOH+PH<SUB>3</SUB>↑．','【分析】①根据原子中质子数=核外电子数，进行分析解答．<br />②最外层具有8个电子的原子具有相对稳定结构．<br />③根据铝元素与硫元素的化合价来分析；<br />④根据元素周期与电子层数的关系考虑；<br />⑤根据化学方程式的书写方法来分析．','书写',3.00,'3d8b1e76ff1a4e8ccaf754ccb6e3cc2c',9,400,'原子结构示意图与离子结构示意图,元素周期表的特点及其应用,书写化学方程式、文字表达式、电离方程式','',0,'37','',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839981,'<img  src=\"/tikuimages/9/2015/400/shoutiniao67/11d462de-94d4-11e9-9224-b42e9921e93e_xkb79.png\" style=\"vertical-align:middle;FLOAT:right\" />（2015秋•周村区校级期中）如图是甲、乙两种固体物质的溶解度曲线．据图回答：<br />（1）P点的含义<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（2）要使接近饱和的甲溶液变成饱和溶液，可采用的方法有<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（写出一种），当甲中含有少量乙时，可采用<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>的方法提纯甲．<br />（3）20℃时，将30g甲物质放入盛有50g水的烧杯中，所得溶液溶质的质量分数为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（计算结果保留到0.1%）．<br />（4）从图中还可获得的信息是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（答出一条）．','','','','','','30℃时甲的溶解度是60g$###$增加甲物质（或蒸发溶剂、降低温度）$###$降温结晶（或冷却热饱和溶液）$###$28.6%$###$t℃时甲乙两物质的溶解度相等（合理均可）','【解答】解：（1）P点表示30℃时甲的溶解度是60g；<br />（2）甲的溶解度随温度升高而增大，所以要使接近饱和的甲溶液变成饱和溶液，可采用增加甲物质或蒸发溶剂、降低温度的方法；甲的溶解度受温度影响较大，乙的溶解度受温度影响较小，所以当甲中含有少量乙时，可采用降温结晶或冷却热饱和溶液的方法提纯甲；<br />（3）20℃时甲的溶解度是40g，即100g水中最多溶解40g的甲，所以将30g甲物质放入盛有50g水的烧杯中，最多溶解20g，所得溶液溶质的质量分数为<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">20g</td></tr><tr><td>70g</td></tr></table></span>×100%≈28.6%；<br />（4）从图中还可知道t℃时甲乙两物质的溶解度相等，20℃时甲的溶解度大于乙 的溶解度等；<br />故答案为：（1）30℃时甲的溶解度是60g；<br />（2）增加甲物质（或蒸发溶剂、降低温度）；降温结晶（或冷却热饱和溶液）；<br />（3）28.6%；<br />（4）t℃时甲乙两物质的溶解度相等（合理均可）．','【分析】（1）据溶解度曲线可知某温度下物质的溶解度；<br />（2）不饱和溶液变为饱和溶液可采取蒸发溶剂、加入溶质或降温大的，并据甲乙的溶解度受温度影响情况分析提纯物质的方法；<br />（3）据该温度下甲的溶解度和溶质的质量分数计算方法解答；<br />（4）据溶解度曲线可知交点表示该温度下物质的溶解度相等，比较同一温度下不同物质的溶解度大小等．','填空题',3.00,'7f2c75c1294f74c39ef1befa075807f2',9,400,'结晶的原理、方法及其应用,饱和溶液和不饱和溶液相互转变的方法,固体溶解度曲线及其作用,有关溶质质量分数的简单计算','',2015,'35','2015秋•周村区校级期中',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839983,'某离子的结构示意图为<img  src=\"/tikuimages/9/2015/400/shoutiniao29/11deea2e-94d4-11e9-9296-b42e9921e93e_xkb8.png\" style=\"vertical-align:middle\" />则与该离子电子数相同的一组粒子是（　　）','Na、Ne','Mg<SUP>2+</SUP>、H<SUB>2</SUB>O','Mg<SUP>2+</SUP>、Ne','O、Al<SUP>3+</SUP>','','B','【解答】解：A、Na的核外电子数为11，Ne核外电子数为10，故A不符合题意；<br />B、OH-、的核外电子数为10，H<SUB>2</SUB>O的核外电子数为10，故B符合题意；<br />C、Mg<SUP>2+</SUP>的核外电子数10，Ne的核外电子数8，故C不符合题意；<br />D、O、的核外电子数为8，Al<SUP>3+</SUP>&nbsp;的核外电子数为10，故D不符合题意．<br />故选B．','【分析】根据离子的结构示意图<img  src=\"/tikuimages/9/2015/400/shoutiniao64/11e8ae30-94d4-11e9-95a9-b42e9921e93e_xkb36.png\" style=\"vertical-align:middle\" />，可知该离子电子数为10，分析选项找出电子数相同的一组即可．','选择题',3.00,'83500f62bbba9608e9e2ccb03bd08bde',9,400,'原子结构示意图与离子结构示意图','',2015,'37','2015秋•烟台校级月考',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839985,'牙膏摩擦剂的类型很多，如CaCO<SUB>3</SUB>，SiO<SUB>2</SUB>或它们的混合物．某兴趣小组对牙膏摩擦剂的成分进行了如下探究．<img  src=\"/tikuimages/9/2015/400/shoutiniao53/11f8dacf-94d4-11e9-983b-b42e9921e93e_xkb39.png\" style=\"vertical-align:middle\" /><br />【提出问题1】某品牌牙膏中是否含有CaCO<SUB>3</SUB>？<br />【查阅资料】二氧化硅不与稀盐酸反应．<br />【实验方案】<br />①取少量牙膏于试管中，加入适量的稀盐酸，出现气泡，将气体通入澄清石灰水，石灰水变浑浊，证明摩擦剂中含有<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填离子符号）<br />②取实验①中的溶液，加入碳酸钾溶液，生成白色沉淀．<br />【提出问题2】该牙膏中CaCO<SUB>3</SUB>的质量分数是多少？<br />【实验装置】<br />【实验步骤】<br />（1）按如图1连接好装置后，发现一处明显的错误，改正为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（2）重新连接好装置，并检查装置的气密性；<br />（3）在B装置中加入牙膏样品8.00g；<br />（4）<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>K<SUB>1</SUB>，<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>K<SUB>2</SUB>（填“打开”或“关闭”），向牙膏样品中滴入10%的盐酸，至B中无气泡产生时，停止滴加盐酸；<br />（5）<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，至C中不再产生沉淀；<br />（6）将C装置中的固液混合物过滤、洗涤、烘干后称量其质量．实验数据如图2（已知此品牌牙膏中的其他成分不和盐酸反应，装置内试剂均足量．）<br />【实验分析及数据处理】<br />（1）若没有A装置，则测定结果将<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．（填“偏大”、“偏小”或“不变”）<br />（2）D装置的作用是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）据上面数据，计算该牙膏样品中碳酸钙的质量分数是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','CO<SUB>3</SUB><SUP>2-</SUP>$###$C中导管应“长进短出”$###$关闭$###$打开$###$打开K<SUB>1</SUB>，关闭K<SUB>2</SUB>，缓缓通入空气$###$偏大$###$吸收空气中的二氧化碳干扰，防止干扰实验测定$###$25%','【解答】解：【实验方案】①取少量牙膏于试管中，加入过量稀盐酸，出现气泡，将气体通入澄清石灰水中，石灰水变浑浊，说明生成二氧化碳，从而证明摩擦剂中含有碳酸根离子；<br />【实验步骤】（1）C的作用是吸收反应生成的二氧化碳，装置中导管应长进短出；<br />（4）关闭K<SUB>1</SUB>，打开K<SUB>2</SUB>，向牙膏样品中滴入10%的盐酸，至不再生成气体，停止滴加盐酸；<br />（5）再次打开K<SUB>1</SUB>，关闭K<SUB>2</SUB>，缓缓通入一段时间空气，使生成的二氧化碳全部排出，充分反应，至C中不再产生沉淀；<br />【实验分析及数据处理】<br />（1）A装置的作用是吸收空气中的二氧化碳，若没有该装置，则进入C的气体还有空气中的二氧化碳，使测量结果偏大；<br />（2）D装置的作用是吸收空气中的二氧化碳，防止空气中二氧化碳进入，使测量结果不准确；<br />（3）依据所测生成沉淀数据，计算反应的二氧化碳的质量，并进而计算牙膏中碳酸钙的质量，计算过程如下；<br />解：根据数据得生成沉淀为3.94g，设参加反应的二氧化碳的质量为x<br />Ba（OH）<SUB>2</SUB>+CO<SUB>2</SUB>═BaCO<SUB>3</SUB>↓+H<SUB>2</SUB>O<br />&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;44&nbsp;&nbsp; 197<br />&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;x&nbsp;&nbsp;&nbsp; 3.94g<br />&nbsp;<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">44</td></tr><tr><td>x</td></tr></table>=<table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">197</td></tr><tr><td>3.94g</td></tr></table></span><br />&nbsp;&nbsp; x=0.88g&nbsp;<br />设碳酸钙的质量为y<br />CaCO<SUB>3</SUB>+2HCl═CaCl<SUB>2</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑<br />100&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp; 44<br />y&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;0.88g<br /><span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">100</td></tr><tr><td>y</td></tr></table>=<table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">44</td></tr><tr><td>0.88g</td></tr></table></span><br />&nbsp;&nbsp; y=2.00g<br />该牙膏样品中碳酸钙的质量分数是<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">2.00g</td></tr><tr><td>8g</td></tr></table></span>×100%=25%．<br />故答案为：【实验方案】①CO<SUB>3</SUB><SUP>2-</SUP>；<br />【实验步骤】（1）C中导管应“长进短出”；<br />（4）关闭；打开；<br />（5）打开K<SUB>1</SUB>，关闭K<SUB>2</SUB>，缓缓通入空气；<br />【实验分析及数据处理】<br />（1）偏大；（2）吸收空气中的二氧化碳干扰，防止干扰实验测定；<br />（3）25%．','【分析】【实验方案】依据碳酸根离子的检验方法分析解答；<br />【实验步骤】（1）洗气装置中导管应长进短出；<br />（4）依据装置设计原理和实验目的分析解答；<br />（5）再次打开K<SUB>1</SUB>，关闭K<SUB>2</SUB>，缓缓通入一段时间空气，使生成的二氧化碳充分反应；<br />【实验分析及数据处理】<br />（1）A装置的作用是吸收空气中的二氧化碳，防止测量结果不准确；<br />（2）D装置的作用是吸收空气中的二氧化碳，防止测量结果不准确；<br />（3）依据所测生成沉淀数据的平均值，计算反应的二氧化碳的质量，并进而计算参加反应的碳酸钙的质量．','填空题',3.00,'914446f8fb702b15f610d181e4c8f0b0',9,400,'实验探究物质的组成成分以及含量,常见气体的检验与除杂方法,盐的化学性质,根据化学反应方程式的计算','',2015,'37','2015秋•绍兴校级月考',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839986,'<img  src=\"/tikuimages/9/2016/400/shoutiniao22/11ffdfb0-94d4-11e9-a494-b42e9921e93e_xkb67.png\" style=\"vertical-align:middle;FLOAT:right;\" />固体物质X在水、乙醇两种溶剂中的溶解度随温度变化的曲线如图所示．下列说法不正确的是（　　）','在水和乙醇中物质X的溶解度都随温度上升高而增大','t<sub>1</sub>℃时，物质X在水中的溶解度为mg','t<sub>2</sub>℃时物质X的水溶液降温至t<sub>1</sub>℃有晶体析出','t<sub>2</sub>℃时，物质X在水中与在乙醇中的溶解度相同','','C','【解答】解：A、由溶解度曲线可知，物质X在水和乙醇中的溶解度均随温度的升高而增大，故A正确；<br />B、根据物质X在水中的溶解度曲线可知，在t<sub>1</sub>℃时，X在水的溶解度是mg，故B正确；<br />C、X的溶解度随温度的升高而增大，所以将t<sub>2</sub>℃时物质X在水中的溶液降温至t<sub>1</sub>℃，X溶液的状态不能确定，所以是否有晶体析出，也不能确定，故C错误；<br />D、由于在t<sub>2</sub>℃时，物质X在两种溶剂中的溶解度曲线交于一点，说明该物质在这两种溶剂中的溶解度相等，故D正确．<br />故选：C．','【分析】根据固体物质w在水、乙醇两种溶剂中的溶解度随温度变化的曲线图，可以判断物质w在水和乙醇中的溶解性；可以查出t<sub>1</sub>℃时，物质w在水中的溶解度；根据w的溶解度随温度变化的变化情况，可以确定升温后是否有溶质从溶液中结晶析出；可以比较同一物质在同一温度下、不同溶剂中的溶解度大小．','选择题',3.00,'9ba3d8866059282adcd29fee22e7756f',9,400,'固体溶解度曲线及其作用,晶体和结晶的概念与现象','',2016,'37','2016春•阜阳校级月考',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839987,'在一定条件下，某反应的微观示意图如下，有关说法正确的是（　　）<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao96/120732b0-94d4-11e9-9d05-b42e9921e93e_xkb25.png\" style=\"vertical-align:middle\" />','属于化合反应','甲在空气中燃烧，产生蓝紫色火焰','乙中H的化合价为+1','该反应的化学方程式为CO+H<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;一定条件&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>CH<SUB>4</SUB>O','','A','【解答】解：由反应的结构示意图和模型表示的原子种类，可判断甲为CO，乙为H<SUB>2</SUB>，丙为CH<SUB>4</SUB>O，反应的化学方程式为CO+2H<SUB>2</SUB>=CH<SUB>4</SUB>O，因此：<br />A、该反应是两种物质生成一种物质的化合反应，故说法正确；<br />B、一氧化碳在空气中燃烧，产生淡蓝色火焰，故说法错误；<br />C、乙是氢气，属于单质，氢元素的化合价为0，故说法错误；<br />D、反应的化学方程式为CO+2H<SUB>2</SUB>=CH<SUB>4</SUB>O，故说法错误；<br />故选项为：A．','【分析】根据反应的结构示意图和模型表示的原子种类，可判断甲为CO，乙为H<SUB>2</SUB>，丙为CH<SUB>4</SUB>O，反应的化学方程式为CO+2H<SUB>2</SUB>=CH<SUB>4</SUB>O，据此分析有关的问题．','选择题',3.00,'49281b1cb7152eb3b330ada44d8ba17f',9,400,'微粒观点及模型图的应用,有关元素化合价的计算,反应类型的判定','',2016,'32','2016•深圳校级模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839988,'下列实验现象与事实不相符的是（　　）','磷在空气中燃烧产生大量棕色的烟','打开盛有浓盐酸的试剂瓶，看到瓶口有大量白雾','氢氧化钠固体曝露在空气中，固体表面逐渐潮湿','碳酸氢铵在蒸发皿中用酒精灯充分加热，最后蒸发皿中无物质剩余','','A','【解答】解：A、磷在空气中燃烧，产生大量的白烟，而不是棕色的烟，故选项说法错误．<br />B、浓盐酸具有挥发性，打开盛有浓盐酸的试剂瓶，看到瓶口有大量白雾，故选项说法正确．<br />C、氢氧化钠固体具有吸水性，氢氧化钠固体曝露在空气中，固体表面逐渐潮湿，故选项说法正确．<br />D、碳酸氢铵在加热条件下生成氨气、水和二氧化碳，碳酸氢铵在蒸发皿中用酒精灯充分加热，最后蒸发皿中无物质剩余，故选项说法正确．<br />故选：A．','【分析】A、根据磷在空气中燃烧的现象进行分析判断．<br />B、根据浓盐酸具有挥发性，进行分析判断．<br />C、根据氢氧化钠固体具有吸水性，进行分析判断．<br />D、根据碳酸氢铵在加热条件下生成氨气、水和二氧化碳，进行分析判断．','选择题',3.00,'e41abf5a622152910d8428c64a06c3fd',9,400,'氧气与碳、磷、硫、铁等物质的反应现象,酸的物理性质及用途,常见碱的特性和用途,盐的化学性质','',2016,'32','2016•南平模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839990,'利用如图装置进行燃烧条件的探究．（资料：S+O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>SO<SUB>2</SUB>）<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao80/121367b0-94d4-11e9-a193-b42e9921e93e_xkb20.png\" style=\"vertical-align:middle\" /><br />（1）打开K<SUB>1</SUB>、K<SUB>2</SUB>、K<SUB>4</SUB>，关闭K<SUB>3</SUB>，向装置中通入二氧化碳气体，当观察到<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，则证明该装置内的空气已经排净．关闭K<SUB>1</SUB>，点燃酒精灯，加热硫粉，观察到硫粉并不燃烧．以上操作都需要打开K<SUB>4</SUB>，原因是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）待冷却后，打开K<SUB>1</SUB>，向装置中通入氧气，证明装置内的二氧化碳已经排净的方法和现象是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）打开<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，关闭<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，点燃酒精灯，加热硫粉，观察到硫粉燃烧，发出明亮的蓝紫色火焰．该实验装置的主要优点是消除了有害气体对环境的污染．','','','','','','石灰水变浑浊$###$平衡装置内外大气压$###$用带火星木条放在K<SUB>4</SUB>处，木条复燃$###$K<SUB>3</SUB>$###$K<SUB>2</SUB>','【解答】解：<br />（1）二氧化碳能使澄清的石灰水变浑浊，打开K<SUB>1</SUB>、K<SUB>2</SUB>、K<SUB>4</SUB>，关闭K<SUB>3</SUB>，向装置中通入二氧化碳气体，当观察到石灰水变浑浊，则证明该装置内的空气已经排净．关闭K<SUB>1</SUB>，点燃酒精灯，加热硫粉，观察到硫粉并不燃烧．以上操作都需要打开K<SUB>4</SUB>，原因是衡装置内外大气压；<br />（2）待冷却后，打开K<SUB>1</SUB>，向装置中通入氧气，证明装置内的二氧化碳已经排净的方法和现象是带火星木条放在K<SUB>4</SUB>处，木条复燃；<br />（3）打开K<SUB>3</SUB>，关闭K<SUB>2</SUB>，点燃酒精灯，加热硫粉，观察到硫粉燃烧．硫燃烧生成二氧化硫，二氧化硫有毒，为防止污染空气，用氢氧化钠溶液吸收．<br />故答案为：<br />（1）石灰水变浑浊；平衡装置内外大气压（合理给分）；<br />（2）用带火星木条放在K<SUB>4</SUB>处，木条复燃（合理给分）；<br />（3）K<SUB>3</SUB>；&nbsp;K<SUB>2</SUB>．','【分析】（1）根据二氧化碳能使澄清的石灰水变浑浊解答；根据压强原理分析解答．<br />（2）根据二氧化碳不能支持燃烧，氧气支持燃烧的性质解答；<br />（3）根据硫粉燃烧的注意事项解答．','填空题',3.00,'c5f1ae4fc0e71c026c2d3158a6d5087f',9,400,'燃烧的条件与灭火原理探究,二氧化碳的化学性质','',2016,'37','2016•通州区一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839992,'铁是一种化学性质比较活泼的金属，在一定条件下能跟多种物质发生化学反应．回答下列问题：<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao83/1221234f-94d4-11e9-bb35-b42e9921e93e_xkb76.png\" style=\"vertical-align:middle\" /><br />（1）以铁为研究对象，依照图中实例在框图处填写物质的化学式（所填写的反应物不属同一类别的物质，生成物必须填写铁的化合物）<br />（2）实验室中要除去细碎铜屑中的少量铁屑，请用化学方法和物理方法将其除去．（只用简单的原理描述即可）<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','化学方法：Fe+2HCl=FeCl<SUB>2</SUB>+H<SUB>2</SUB>↑；或 Fe+CuSO<SUB>4</SUB>=Cu+FeSO<SUB>4</SUB>；物理方法：用磁铁吸','【解答】解：铁可以与盐酸或稀硫酸反应生成亚铁盐，可以与位于其后金属的盐溶液反应生成亚铁盐和金属．<br />（1）本题考查的是金属的化学性质．金属主要有3个化学性质：①金属与O<SUB>2</SUB>的反应；②金属与酸（盐酸，稀硫酸）反应；③金属与金属化合物溶液反应，题目图中给出了Fe与O<SUB>2</SUB>的反应，故余下两个即为②③性质，这里一定要注意题目括号中的内容，即反应物必须填不同类物质，一种为酸，另一种为金属化合物溶液，且生成物必须写铁的化合物，不能写氢气或金属，所以<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao94/1225de40-94d4-11e9-9d72-b42e9921e93e_xkb37.png\" style=\"vertical-align:middle\" /><br />（2）化学方法：①放入到盐酸或稀硫酸中，原理：Fe+2HCl=FeCl<SUB>2</SUB>+H<SUB>2</SUB>↑，Fe+H<SUB>2</SUB>SO<SUB>4</SUB>=FeSO<SUB>4</SUB>+H<SUB>2</SUB>↑，再过滤即可；②放入到CuSO<SUB>4</SUB>溶液中，原理：Fe+CuSO<SUB>4</SUB>=FeSO<SUB>4</SUB>+Cu，此时，不仅可以除去铁还可以变废为宝，再过滤即可；<br />物理方法：即不通过化学反应即可实现：利用铁的磁性，可用磁铁吸出Fe即可．<br />故答案为：（1）<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao94/1225de40-94d4-11e9-9d72-b42e9921e93e_xkb37.png\" style=\"vertical-align:middle\" /><br />（2）化学方法：Fe+2HCl=FeCl<SUB>2</SUB>+H<SUB>2</SUB>↑；或 Fe+CuSO<SUB>4</SUB>=Cu+FeSO<SUB>4</SUB>；物理方法：用磁铁吸．','【分析】根据铁是活动性较强的排在氢前金属，可以与氧气反应生成金属氧化物、与酸反应生成亚铁盐和氢气和某些盐的溶液反应把位于其后金属的盐溶液反应生成亚铁盐和金属进行分析．','书写',3.00,'6c073fc3dfcc23abc27aff9ddee578da',9,400,'常见金属的特性及其应用,金属的化学性质,书写化学方程式、文字表达式、电离方程式','',2016,'32','2016•泰山区模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839994,'砷化镓（GaAs）是一种“LED”绿色节能光源材料，镓元素的相关信息如图．下列有关镓的说法正确的是（　　）<img  src=\"/tikuimages/9/2015/400/shoutiniao97/122f2d0f-94d4-11e9-83d4-b42e9921e93e_xkb7.png\" style=\"vertical-align:middle;FLOAT:right;\" />','属于非金属元素','原子的核内质子数是31','原子的核内中子数是31','1个镓原子的质量为69.72g','','B','【解答】解：A、根据元素周期表中的一格可知，中间的汉字表示元素名称，该元素的名称是镓，属于金属元素，故选项说法错误．<br />B、根据元素周期表中的一格可知，左上角的数字为31，表示原子序数为31；根据原子序数=核电荷数=质子数，则该元素的原子核内质子数为31，故选项说法正确．<br />C、根据元素周期表中的一格可知，左上角的数字为31，表示原子序数为31；根据原子序数=核电荷数=质子数=核外电子数，则该元素的原子核内质子数和核外电子数均为31，而不是中子数为31，故选项说法错误．<br />D、根据元素周期表中的一格可知，汉字下面的数字表示相对原子质量，该元素的相对原子质量为69.72，而不是1个镓原子的质量为69.72g，故选项说法错误．<br />故选：B．','【分析】根据图中元素周期表可以获得的信息：左上角的数字表示原子序数；字母表示该元素的元素符号；中间的汉字表示元素名称；汉字下面的数字表示相对原子质量，进行分析判断即可．','选择题',3.00,'fca7b40045869b1adce8418d09836f97',9,400,'元素周期表的特点及其应用','',2015,'33','2015秋•平和县期末',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839996,'下列A～F是初中化学中的六个实验，请按要求填空：<br /><img  src=\"/tikuimages/9/2014/400/shoutiniao92/123c7380-94d4-11e9-b29f-b42e9921e93e_xkb16.png\" style=\"vertical-align:middle\" /><br />（1）上述实验中能达到实验目的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>&nbsp;（填字母）．<br />（2）D实验横放的玻管内发生反应化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）E实验中，发生反应化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（4）F实验中，发生反应化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />①试管中收集到的气体为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>&nbsp;（填化学式）．','','','','','','D、E、F$###$CuO+CO<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>Cu+CO<SUB>2</SUB>$###$2KMnO<SUB>4</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>K<SUB>2</SUB>MnO<SUB>4</SUB>+MnO<SUB>2</SUB>+O<SUB>2</SUB>↑$###$2H<SUB>2</SUB>O<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;通电&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>↑+O<SUB>2</SUB>↑$###$H<SUB>2</SUB>','【解答】解：（1）A、二氧化碳密度大于空气密度应用向上排空气法收集，因此A不能达到实验目的；<br />B、从所用药品是碳燃烧会生成二氧化碳气体，不能使水进入集气瓶，因此B不能达到实验目的；<br />C、从燃烧的三个条件进行分析，两处的实验现象只能证明燃烧与是否与氧气接触有关，不能证明燃烧与温度是否达到着火点有关，因此不能达到实验目的；<br />D、从实验装置、所用药品、实验结论进行分析，一氧化碳能还原氧化铁，此实验也可达到实验目的；<br />E、从实验操作装置和氧气的收集方法可知，此实验也可达到实验目的；<br />F、电解水时正极生成氧气负极生成氢气，且两者的体积比是1：2，此实验能达到实验的目的；<br />故选：DEF；<br />（2）D实验是一氧化碳和氧化铜在加热的条件下生成铜和二氧化碳，化学方程式为：CuO+CO<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>Cu+CO<SUB>2</SUB>；<br />（3）该装置是固体加热制取气体并用排水法收集的图示，因为试管口部有一团棉花，因此可判断是高锰酸钾制氧气的装置，化学方程式为：2KMnO<SUB>4</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>K<SUB>2</SUB>MnO<SUB>4</SUB>+MnO<SUB>2</SUB>+O<SUB>2</SUB>↑；<br />（4）电解水时正极生成氧气负极生成氢气，且两者的体积比是1：2，①试管中产生的气体多所以是氢气，化学方程式为：2H<SUB>2</SUB>O<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;通电&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>↑+O<SUB>2</SUB>↑．<br />故答案为：（1）D、E、F；<br />（2）CuO+CO<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>Cu+CO<SUB>2</SUB>；<br />（3）2KMnO<SUB>4</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>K<SUB>2</SUB>MnO<SUB>4</SUB>+MnO<SUB>2</SUB>+O<SUB>2</SUB>↑；<br />（4）2H<SUB>2</SUB>O<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;通电&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>↑+O<SUB>2</SUB>↑，H<SUB>2</SUB>．','【分析】本题以图示的形式综合考查了正确判断实验现象及规范进行基本的实验操作的能力的综合题．解题时应充分利用题中给予的信息，并进行剔除筛选分析装置的作用及特点进行解答即可．','书写',3.00,'a9a39b4dd3ea396d3c7576c989792d19',9,400,'空气组成的测定,氧气的制取装置,二氧化碳的实验室制法,一氧化碳的化学性质,电解水实验,书写化学方程式、文字表达式、电离方程式,燃烧与燃烧的条件','',2014,'33','2014秋•顺庆区期末',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839997,'小亮在做“盐酸中和氢氧化钠溶液”的实验时，滴加盐酸前忘了加入指示剂，导致无法判断该中和反应进行的程度．于是他对所得溶液的酸碱性进行探究．<br />[探究目的]探究所得溶液的酸碱性．<br />[提出猜想]<br />所得溶液可能呈碱性，也可能呈<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>性，还可能呈<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>性．<br />[实验验证]<br /><table class=\"edittable\"><TBODY><TR><td width=248>实验操作</TD><td width=165>实验现象</TD><td width=140>结&nbsp;&nbsp;论</TD></TR><TR><td rowSpan=2>用试管取该溶液lmL～2mL，滴入1-2滴无色酚酞试液，振荡</TD><td>无色酚酞试液变<u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD><td>溶液呈碱性</TD></TR><TR><td>无色酚酞试液不变色</TD><td>溶液呈<u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD></TR></TBODY></TABLE>[分析与思考]<br />（1）若溶液呈碱性，则溶液中使其呈碱性的离子是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填写微粒符号），为避免碱性溶液污染环境，小亮对所得溶液进行了如下处理：向溶液中逐滴加入<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>溶液，直到不产生沉淀为止．然后过滤，把滤液倒人蒸发皿中加热，得到氯化钠晶体，该过程涉及的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）若无色酚酞试液不变色，为了进一步确定溶液的酸性、小亮提出了以下方案：<br />方案一：取样，加入碳酸钠溶液，若观察到<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，则所取溶液呈酸性；若无明显现象，则呈中性．<br />方案二：取样，加入金属锌，若有气泡产生，则所取溶液酸性，该反应的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；若无明显现象，则呈中性．','','','','','','中$###$酸$###$红$###$中性或酸性$###$OH<SUP>-</SUP>$###$CuCl<SUB>2</SUB>$###$CuCl<SUB>2</SUB>+2NaOH═Cu（OH）<SUB>2</SUB>↓+2NaCl$###$有气泡产生$###$Zn+2HCl═ZnCl<SUB>2</SUB>+H<SUB>2</SUB>↑','【解答】解：（提出猜想）盐酸与氢氧化钠混合，会发生反应，由于酸碱的量不固定故溶液可能呈碱性、中性或酸性，所以本题答案为：中，酸性；<br />（实验验证）酚酞在碱性溶液中变红，在酸性和中性溶液中为无色，所以本题答案为：<table class=\"edittable\"><TBODY><TR><td>实验操作</TD><td>实验现象</TD><td>结论</TD></TR><TR><td rowSpan=2>&nbsp;</TD><td>红</TD><td>&nbsp;</TD></TR><TR><td>&nbsp;</TD><td>中性或酸性</TD></TR></TBODY></TABLE>（1）溶液呈碱性的实质是氢氧根离子的缘故，氢氧化钠能与氯化铜反应生成氢氧化铜沉淀和氯化钠，所以本题答案为：OH<SUP>-</SUP>，CuCl<SUB>2</SUB>，CuCl<SUB>2</SUB>+2NaOH═Cu（OH）<SUB>2</SUB>↓+2NaCl；<br />（2）所取溶液呈酸性，与碳酸钠接触能产生二氧化碳，所以本题答案为：有气泡产生；<br />（3）所取溶液呈酸性，则能与锌反应生成氢气，所以本题答案为：Zn+2HCl═ZnCl<SUB>2</SUB>+H<SUB>2</SUB>↑．','【分析】氢氧化钠溶液呈碱性，能使酚酞变红，盐酸呈酸性，氢氧化钠与盐酸反应生成氯化钠和水，呈中性，在酸性和中性溶液中酚酞不变色，氢氧化钠能与某些物质反应生成沉淀，碳酸盐遇酸化气，活泼金属能与酸反应生成氢气．','书写',3.00,'4f3d84acaf4b913829635293c8519569',9,400,'酸的化学性质,碱的化学性质,中和反应及其应用,溶液的酸碱性测定,书写化学方程式、文字表达式、电离方程式','',2016,'37','2016•白银校级二模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839998,'铝、铁、铜是人类广泛使用的三种金属，与我们生活息息相关．<br />（1）人们大量使用的是合金而不是纯金属，这是因为合金具有更多优良性能，例如钢比纯铁硬度<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“大”或“小”）．<br />（2）用下列试剂验证这三种金属的活动性顺序，能达到目的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填序号）．<br />A．硫酸铝溶液&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;B．硫酸亚铁溶液&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;C．硫酸铜溶液<br />（3）铁生锈是铁与<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>发生的复杂的化学反应，硫酸和盐酸都能除铁锈，写出硫酸与铁锈的主要成分发生反应的化学方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','大$###$B$###$水和氧气$###$3H<SUB>2</SUB>SO<SUB>4</SUB>+Fe<SUB>2</SUB>O<SUB>3</SUB>═Fe<SUB>2</SUB>（SO<SUB>4</SUB>）<SUB>3</SUB>+3H<SUB>2</SUB>O','【解答】解：（1）一般来说合金的硬度要比组成它的金属大，钢是铁的合金，硬度比纯铁大；<br />（2）A、铁和铜都不与硫酸铝溶液反应，说明铁和铜都排在铝的后面，但是不能说明铁和铜的顺序，故A错误；<br />B、铝能与硫酸亚铁反应，说明铝在铁的前面，铜不与硫酸亚铁反应，说明铁在铜的前面，能证明铝、铁、铜的顺序，故B正确；<br />C、铝和铁都能与硫酸铜溶液反应，说明铝和铁在铜的前面，但是不能说明铝和铁的顺序，故C错误；<br />（3）铁生锈是铁与水和氧气发生的复杂的化学反应；铁锈的主要成分是氧化铁，氧化铁和稀硫酸反应生成硫酸铁和水，反应的化学方程式是：3H<SUB>2</SUB>SO<SUB>4</SUB>+Fe<SUB>2</SUB>O<SUB>3</SUB>═Fe<SUB>2</SUB>（SO<SUB>4</SUB>）<SUB>3</SUB>+3H<SUB>2</SUB>O．<br />故答案为：（1）大；（2）B；（3）水和氧气；3H<SUB>2</SUB>SO<SUB>4</SUB>+Fe<SUB>2</SUB>O<SUB>3</SUB>═Fe<SUB>2</SUB>（SO<SUB>4</SUB>）<SUB>3</SUB>+3H<SUB>2</SUB>O．','【分析】（1）根据合金的硬度比成分金属的硬度大进行分析；<br />（2）根据要验证金属活动性顺序，可以依据金属的性质进行分析；<br />（3）根据铁生锈的条件进行分析；氧化铁和稀硫酸反应生成硫酸铁和水分析解答．','书写',3.00,'d341c5caf2cd6b2373d313ccb61d0309',9,400,'合金与合金的性质,金属活动性顺序及其应用,金属锈蚀的条件及其防护,酸的化学性质,书写化学方程式、文字表达式、电离方程式','',2016,'37','2016•宿迁二模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1839999,'如图是元素周期表的一部分，请按要求填空．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao85/1244d7ee-94d4-11e9-860e-b42e9921e93e_xkb65.png\" style=\"vertical-align:middle\" /><br />（1）由表中空格处元素组成的单质在氧气中燃烧的化学方程式是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）表中不同元素最本质的区别是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填字母）．<br />A．各元素原子的质子数不同&nbsp;&nbsp;B．各元素原子的中子数不同&nbsp;&nbsp;C．各元素相对原子质量不同<br />（3）钠元素原子的结构示意图为<img  src=\"/tikuimages/9/2016/400/shoutiniao93/12488170-94d4-11e9-971f-b42e9921e93e_xkb53.png\" style=\"vertical-align:middle\" />，该原子在化学反应中易<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“得到”或“失去”）电子．<br />（4）元素的化学性质主要由其原子的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>决定．','','','','','','2Mg+O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2MgO$###$A$###$失去$###$最外层电子数','【解答】解：（1）镁在氧气中燃烧生成氧化镁，故填：2Mg+O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2MgO；&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<br />（2）元素是具有相同核电荷数（即核内质子数）的一类原子的总称，不同种元素之间的本质区别是质子数不同，所以元素的种类由质子数（或核电荷数）决定，故填：A；<br />（3）由钠原子结构示意图可知，其最外层有1个电子，在化学变化中易失去一个电子，元素的化学性质主要由其最外层电子数来决定，故填：失去；<br />（4）元素的化学性质主要由其最外层电子数来决定，故填：最外层电子数．','【分析】（1）空格处是镁元素；<br />（2）根据元素的概念来分析；<br />（3）根据原子结构示意图的意义分析；<br />（4）在原子中，最外层的电子数主要决定元素的性质．','填空题',3.00,'a2f28e8c20143a7b987e4ad2d494137a',9,400,'原子结构示意图与离子结构示意图,元素周期表的特点及其应用','',2016,'32','2016•濠江区模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840000,'下列应用和相应的原理用化学方程式表示及基本反应类型都正确的是（　　）','用过氧化氢制取氧气&nbsp;&nbsp;&nbsp;&nbsp;2H<SUB>2</SUB>O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span>．&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O+O<SUB>2</SUB>↑&nbsp;&nbsp;&nbsp;&nbsp;分解反应','用木炭还原氧化铜&nbsp;&nbsp;&nbsp;&nbsp;C+2CuO<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>CO<SUB>2</SUB>↑+Cu&nbsp;&nbsp;&nbsp;&nbsp;氧化反应','用红磷测定空气中氧气的含量&nbsp;&nbsp;P+O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>P<SUB>2</SUB>O<SUB>5</SUB>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;化合反应','用盐酸除去铁锈&nbsp;&nbsp;&nbsp;&nbsp;Fe<SUB>2</SUB>O<SUB>3</SUB>+4HCl═FeCl<SUB>2</SUB>+2H<SUB>2</SUB>0&nbsp;&nbsp;&nbsp;&nbsp;复分解反应','','A','【解答】解：A、该化学方程式书写完全正确，且该反应符合“一变多”的特征，属于分解反应，故选项正确．<br />B、该化学方程式书写完全正确；但氧化反应不是基本的化学反应类型，故选项错误．<br />C、该化学方程式没有配平，正确的化学方程式应为：4P+5O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2P<SUB>2</SUB>O<SUB>5</SUB>，故选项错误．<br />D、铁锈的主要成分是氧化铁，与盐酸反应生成氯化铁和水，反应的化学方程式是：Fe<SUB>2</SUB>O<SUB>3</SUB>+6HCl═2FeCl<SUB>3</SUB>+3H<SUB>2</SUB>O，故选项错误．<br />故选：A．','【分析】根据化学方程式判断正误的方法需考虑：应用的原理是否正确；化学式书写是否正确；是否配平；反应条件是否正确；↑和↓的标注是否正确．若化学方程式书写正确，再根据反应特征确定反应类型．','选择题',3.00,'2ca40c6e296b3f3149462499d032ccb3',9,400,'反应类型的判定,书写化学方程式、文字表达式、电离方程式','',2016,'37','2016•道外区二模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840002,'小昊发现月饼包装袋中有一包“双吸剂”．该“双吸剂”的标签标示其主要成分为铁粉、生石灰．小昊对这包久置的“双吸剂”固体样品很好奇，设计实验进行探究．<br />【提出问题久置“双吸剂”固体的成分是什么？<br />【查阅资料】铁与氯化铁溶液在常温下会发生如下反应：Fe+2FeCl<SUB>3</SUB>═3X，X的化学式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />【作出猜想】久置固体中可能含有Fe、Fe<SUB>2</SUB>O<SUB>3</SUB>、CaO、Ca（OH）<SUB>2</SUB>和CaCO<SUB>3</SUB>．久置固体中可能含有Ca（OH）<SUB>2</SUB>的原因是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>&nbsp;（用化学方程式表示）．<br />【实验探究】下表是小昊设计的实验方案和记录的实验报告，请你补充完整．<br /><table class=\"edittable\"><TBODY><TR><td width=209>实验操作</TD><td width=145>实验现象</TD><td width=165>实验结论</TD></TR><TR><td>①取少量固体放于试管中，加入蒸馏水充分搅拌</TD><td>用手触摸试管，试管外壁<u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD><td>固体中一定含有氧化钙</TD></TR><TR><td>②另取少量固体放于试管中，滴加足量稀盐酸</TD><td>固体逐渐消失，有大量无色气体产生，得到浅绿色溶液</TD><td>固体中一定含有<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，一定不含Fe<SUB>2</SUB>O<SUB>3</SUB></TD></TR><TR><td>③将步骤②中产生的气体通入澄清的石灰水中</TD><td><u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD><td>固体中一定含有CaCO<SUB>3</SUB></TD></TR></TBODY></TABLE>【实验质疑】小月通过讨论认为实验②并不能得出一定不含Fe203的结论，理由是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；若要探究是否含有氧化铁，应取样品先将铁除尽，然后加入稀盐酸观察现象，除去样品中的铁，可采用的物理方法为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','FeCl<SUB>2</SUB>$###$CaO+H<SUB>2</SUB>O═Ca（OH）<SUB>2</SUB>$###$发烫$###$铁$###$澄清石灰水变浑浊$###$少量氧化铁能与稀盐酸反应生成的黄色氯化铁溶液，氯化铁与铁反应生成浅绿色氯化亚铁溶液$###$粉碎样品，用磁铁吸引','【解答】解：【查阅资料】<br />根据质量守恒定律可知，铁与氯化铁溶液在常温下反应的化学方程式为：Fe+2FeCl<SUB>3</SUB>═3FeCl<SUB>2</SUB>．<br />故填：FeCl<SUB>2</SUB>．<br />【作出猜想】<br />久置固体中可能含有Ca（OH）<SUB>2</SUB>的原因是氧化钙和水反应生成了氢氧化钙，反应的化学方程式为：CaO+H<SUB>2</SUB>O═Ca（OH）<SUB>2</SUB>．<br />故填：CaO+H<SUB>2</SUB>O═Ca（OH）<SUB>2</SUB>．<br />【实验探究】<br />实验过程如下表所示：<br /><table class=\"edittable\"><TBODY><TR><td width=209>实验操作</TD><td width=145>实验现象</TD><td width=165>实验结论</TD></TR><TR><td>①取少量固体放于试管中，加入蒸馏水充分搅拌</TD><td>用手触摸试管，试管外壁发烫</TD><td>固体中一定含有氧化钙</TD></TR><TR><td>②另取少量固体放于试管中，滴加足量稀盐酸</TD><td>固体逐渐消失，有大量无色气体产生，得到浅绿色溶液</TD><td>固体中一定含有 铁，一定不含Fe<SUB>2</SUB>O<SUB>3</SUB></TD></TR><TR><td>③将步骤②中产生的气体通入澄清的石灰水中</TD><td>澄清石灰水变浑浊</TD><td>固体中一定含有CaCO<SUB>3</SUB></TD></TR></TBODY></TABLE>【实验质疑】<br />小月通过讨论认为实验②并不能得出一定不含Fe203的结论，理由是：少量氧化铁能与稀盐酸反应生成的黄色氯化铁溶液，氯化铁与铁反应生成浅绿色氯化亚铁溶液；<br />若要探究是否含有氧化铁，可采用的物理方法为粉碎样品，用磁铁吸引．<br />故填：少量氧化铁能与稀盐酸反应生成的黄色氯化铁溶液，氯化铁与铁反应生成浅绿色氯化亚铁溶液；粉碎样品，用磁铁吸引．','【分析】【查阅资料】化学反应前后，元素种类不变，原子种类和总个数不变；<br />【作出猜想】氧化钙能和水反应生成氢氧化钙；<br />【实验探究】氧化钙和水反应时放出大量的热；<br />铁能和稀盐酸反应生成氯化亚铁和氢气；<br />二氧化碳能使澄清石灰水变浑浊；<br />【实验质疑】稀盐酸能和氧化铁反应生成氯化铁和水，氯化铁能和铁反应生成氯化亚铁；<br />铁能被磁铁吸引．','书写',3.00,'628a7144292491892d307893fffcb3ba',9,400,'实验探究物质的组成成分以及含量,金属的化学性质,生石灰的性质与用途,碱的化学性质,盐的化学性质,书写化学方程式、文字表达式、电离方程式','',2016,'32','2016•永丰县模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840003,'下列设计的实验方案中（括号中为方法或试剂），你认为可行的是（　　）','验证Mg、Fe、Cu三种金属的金属活动性顺序（Fe、Cu、MgSO<SUB>4</SUB>溶液）','除去二氧化碳中混有少量的一氧化碳&nbsp;（点燃）','鉴别固体：氢氧化钠、氯化钠、硝酸铵（水）','除去K<SUB>2</SUB>SO<SUB>4</SUB>溶液中的KOH（加入适量稀盐酸）','','C','【解答】解：A、镁比铁活泼，铁比铜活泼，铁和铜都不能和硫酸镁反应，因此不能验证Mg、Fe、Cu三种金属的金属活动性顺序；<br />B、二氧化碳不能燃烧，不支持燃烧，因此不能用点燃的方法除去二氧化碳中混有少量的一氧化碳；<br />C、氢氧化钠溶于水放热，硝酸铵溶于水吸热，氯化钠溶于水无明显现象，因此可以用水鉴别氢氧化钠、氯化钠、硝酸铵；<br />D、氢氧化钾和稀盐酸反应生成氯化钾和水，利用稀盐酸除去氢氧化钾的同时，带入新的杂质氯化钾，因此不能用稀盐酸除去硫酸钾中的氢氧化钾．<br />故选：C．','【分析】镁比铁活泼，铁比铜活泼；<br />二氧化碳不能燃烧，不支持燃烧；<br />氢氧化钠溶于水放热，硝酸铵溶于水吸热，氯化钠溶于水无明显现象；<br />氢氧化钾和稀盐酸反应生成氯化钾和水．','选择题',3.00,'2d29bbe2acfeb86c283df0dc747e43d0',9,400,'化学实验方案设计与评价,常见气体的检验与除杂方法,金属活动性顺序及其应用,碱的化学性质,酸、碱、盐的鉴别','',2016,'32','2016•蓬安县模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840006,'“嫦娥一号”绕月探测卫星于2007年10月24日成功发射，实现了中国人民的飞月梦．担任此次发射卫星任务的是“长征三号甲”运载火箭，所用燃料是液态氮，助燃剂是液态氧，燃烧时产生巨大的推动力将“嫦娥一号”送上了太空．<br />（1）可燃物燃烧需要的条件是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>、<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（2）氢气燃烧的化学方程式是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（3）消防队员用消防直升机在空中向燃烧区域喷洒的是干粉，其主要成分是碳酸氢钠．碳酸<br />氢钠在加热状态下发生分解反应，生成了水、二氧化碳、碳酸钠（Na以色列轰炸加沙地带时使用贫铀弹，敦促国际<SUB>2</SUB>CO<SUB>3</SUB>）．请写出碳酸氢钠（NaHCO<SUB>3</SUB>）受热分解的化学方程式：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','必须与氧气或空气接触$###$温度达到可燃物的着火点$###$2H<SUB>2</SUB>+O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O$###$2NaHCO<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>Na<SUB>2</SUB>CO<SUB>3</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑','【解答】解：（1）根据燃烧的条件可知，物质燃烧必须具有可燃性（可燃物），必须与氧气或空气接触，温度必须达到燃烧所需的最低温度即着火点，这三个条件缺一不可．故答案为：必须与氧气或空气接触；温度达到可燃物的着火点．<br />（2）氢气燃烧是氢气与氧气反应生成水，条件是点燃，所以反应方程式为：2H<SUB>2</SUB>+O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O，故答案为：2H<SUB>2</SUB>+O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O；<br />（3）碳酸氢钠在加热状态下发生分解反应，生成了水、二氧化碳、碳酸钠（Na<SUB>2</SUB>CO<SUB>3</SUB>），反应的化学方程式为：2NaHCO<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>Na<SUB>2</SUB>CO<SUB>3</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑．<br />故填：2NaHCO<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>Na<SUB>2</SUB>CO<SUB>3</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑．','【分析】（1）根据燃烧的条件（同时满足三个条件：①可燃物，②氧气或空气，③达到燃烧所需的最低温度即着火点）进行分析解答即可．<br />（2）根据氢气燃烧生成水结合化学方程式的书写分析；<br />（3）首先根据反应原理找出反应物、生成物、反应条件，根据化学方程式的书写方法、步骤进行书写即可．','书写',3.00,'a62974fcc5f44b4219e804f472223589',9,400,'书写化学方程式、文字表达式、电离方程式,燃烧与燃烧的条件,氢气的化学性质与燃烧实验','',2016,'37','2016春•平凉月考',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840010,'化学教师为了调动同学们探究的积极性，布置了一个探究任务：粉笔的成份是初中化学中常见的盐类，请同学们探究粉笔的化学成份，某兴趣小组进行以下探究实验．<br />【提出猜想】<br />（1）粉笔的成份为氯化物；<br />（2）粉笔的成份为碳酸盐；<br />（3）粉笔的成份为硫酸盐；<br />【实验探究】<br />（1）取一个白色的粉笔（约1g），在研钵中研成粉末，倒入大烧杯中，加入100ml蒸馏水充分搅拌，静置．<br />（2）从烧杯中取少许上层清液注入试管，滴加几滴酚酞，无明显现象，粉笔溶液（酸碱性）不可能呈<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>性；<br />（3）从烧杯中取少许上层清液注入试管，滴加盐酸溶液，无明显现象产生．<br />（4）从烧杯中取少许上层清液注入试管，滴加Na<SUB>2</SUB>CO<SUB>3</SUB>溶液，产生白色沉淀；<br />（5）从烧杯中取少许上层清液注入试管，滴加BaCl<SUB>2</SUB>溶液，产生白色沉淀．<br />【实验判断】根据探究实验中出现的现象，请做出粉笔化学成份可能合理判断．<br />（6）写出粉笔的主要成份化学式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（7）写出【实验探究】（4）中化学反应的化学方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（8）写出【实验探究】（5）中化学反应的化学方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','碱$###$CaSO<SUB>4</SUB>$###$CaSO<SUB>4</SUB>+Na<SUB>2</SUB>CO<SUB>3</SUB>═Na<SUB>2</SUB>SO<SUB>4</SUB>+CaCO<SUB>3</SUB>↓$###$CaSO<SUB>4</SUB>+BaCl<SUB>2</SUB>═CaCl<SUB>2</SUB>+BaSO<SUB>4</SUB>↓','【解答】解：（2）因为所碱能使酚酞试液变红色，从烧杯中取少许上层清液注入试管，滴加几滴酚酞，无明显现象，可判断粉笔溶液（酸碱性）不可能呈碱性，猜想（1）不成立；<br />（3）从烧杯中取少许上层清液注入试管，滴加盐酸溶液，无明显现象产生，说明清液中无CO<SUB>3</SUB><SUP>2-</SUP>；<br />（4）从烧杯中取少许上层清液注入试管，滴加Na<SUB>2</SUB>CO<SUB>3</SUB>溶液，产生白色沉淀，说明清液中有Ca<SUP>2+</SUP>；<br />（5）从烧杯中取少许上层清液注入试管，滴加BaCl<SUB>2</SUB>溶液，产生白色沉淀，说明清液中有SO<SUB>4</SUB><SUP>2</SUP>；<br />【实验判断】（6）根据探究实验中出现的现象，知粉笔的化学成份为CaSO<SUB>4</SUB>，故粉笔的主要成份化学式为：CaSO<SUB>4</SUB>；<br />（7）硫酸钙和碳酸钠反应生成碳酸钙沉淀和硫酸钠，化学方程式为：CaSO<SUB>4</SUB>+Na<SUB>2</SUB>CO<SUB>3</SUB>═Na<SUB>2</SUB>SO<SUB>4</SUB>+CaCO<SUB>3</SUB>↓；<br />（8）硫酸钙和氯化钡反应生成硫酸钡沉淀和氯化钙，化学方程式为：CaSO<SUB>4</SUB>+BaCl<SUB>2</SUB>═CaCl<SUB>2</SUB>+BaSO<SUB>4</SUB>↓．<br />故答案为：（2）碱；<br />（6）CaSO<SUB>4</SUB>；<br />（7）CaSO<SUB>4</SUB>+Na<SUB>2</SUB>CO<SUB>3</SUB>═Na<SUB>2</SUB>SO<SUB>4</SUB>+CaCO<SUB>3</SUB>↓；<br />（8）CaSO<SUB>4</SUB>+BaCl<SUB>2</SUB>═CaCl<SUB>2</SUB>+BaSO<SUB>4</SUB>↓．','【分析】根据粉笔的成份是初中化学中常见的盐类，应为钙盐；<br />根据氧化钙与水反应生成氢氧化钙，溶液显碱性考虑加入酸碱指示剂；<br />根据碳酸根离子的检验方法考虑是否含有碳酸钙，根据硫酸根离子的检验方法检验是否含有硫酸钙．','书写',3.00,'ad393f3916a61b1f7f6b80f42cbb17f5',9,400,'实验探究物质的组成成分以及含量,证明硫酸和可溶性硫酸盐,证明碳酸盐,书写化学方程式、文字表达式、电离方程式','',2016,'37','2016•红桥区一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840013,'<img  src=\"/tikuimages/9/2016/400/shoutiniao41/126e8000-94d4-11e9-ac6e-b42e9921e93e_xkb8.png\" style=\"vertical-align:middle;FLOAT:right\" />（2016•孝义市二模）如图是某校初三（二）班阳光小组的同学们设计的探究铁制品锈蚀条件下的实验装置图，请你一起参与探究，并回答相关问题．<br />（1）在实验室静置一周后，A、B试管中分别观察到的现象是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）智慧小组的同学们认为阳光小组设计的方案不严谨，你认为需要对此实验进行的改造是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．改进实验后，探究得出铁制品锈蚀的条件是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）实验后同学们将实验中铁钉表面的铁锈除去，并对铁钉做了防锈处理，他们的具体做法是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','A试管中的铁钉表面出现红色的铁锈（或红色的物质），B试管中的铁钉表面无明显变化．$###$增加一个干燥的试管，将光亮的铁钉放入其中，用橡胶塞将试管口塞紧，与A、B试管同时放在实验室中，静置同样长的时间$###$铁与氧气、水同时接触$###$在铁制品表面涂油、刷油漆、烧制搪瓷、喷塑','【解答】解：<br />（1）一段时间观察试管A中的铁钉明显锈蚀，A试管中即提供了空气，也提供了水，B试管中植物油隔绝了氧气，在实验室静置一周后，A、B试管中分别观察到的现象是A试管中的铁钉表面出现红色的铁锈&nbsp;（或红色的物质），B试管中的铁钉表面无明显变化；<br />（2）智慧小组的同学们认为阳光小组设计的方案不严谨，你认为需要对此实验进行的改造是增加一个干燥的试管，将光亮的铁钉放入其中，用橡胶塞将试管口塞紧，与A、B试管同时放在实验室中，静置同样长的时间．铁与氧气、水同时接触．<br />（3）除锈方法可考虑用稀盐酸或稀硫酸发生化学反应除去，也可考虑用砂纸打磨除去．防锈方法只要从防止铁制品与水接触或与氧气接触两个角度写即可．如：制成合金；在铁制品表面涂油、刷油漆、烧制搪瓷、喷塑；在铁制品表面电镀、热镀一层其它金属（如锌、锡等）；用化学方法使铁制品表面生成一层致密而稳定的氧化膜等．<br />答案：<br />（1）A试管中的铁钉表面出现红色的铁锈&nbsp;（或红色的物质），B试管中的铁钉表面无明显变化．<br />（2）增加一个干燥的试管，将光亮的铁钉放入其中，用橡胶塞将试管口塞紧，与A、B试管同时放在实验室中，静置同样长的时间．铁与氧气、水同时接触．<br />（3）在铁制品表面涂油、刷油漆、烧制搪瓷、喷塑．','【分析】根据铁生锈的主要条件是铁与水和空气直接接触以及除去铁锈可用盐酸清洗方法进行解答．','填空题',3.00,'67a169a9b4b0c60c54639ff3f82ee605',9,400,'金属锈蚀的条件及其防护','',2016,'37','2016•孝义市二模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840014,'某校化学兴趣小组同学在实验室整理药品时，发现氢氧化钠溶液的试剂瓶未塞瓶塞，且试剂瓶口周围有一些白色固体，他们怀疑氢氧化钠已经变质，于是进行了如下实验探究．<br />【提出问题】氢氧化钠溶液是否变质？<br />【查阅资料】氯化钠溶液、氯化钡溶液呈中性．<br />【实验探究】小王、小李两同学分别设计了不同方案并加以实验．<br /><table class=\"edittable\"><TBODY><TR><td></TD><td>实验步骤</TD><td>实验现象</TD><td>结论</TD></TR><TR><td>小王</TD><td>用pH试纸测溶液的pH</TD><td>溶液的pH＞7</TD><td>说明氢氧化钠溶液没有变质</TD></TR><TR><td>小李</TD><td>取少量溶液于试管中，加入适量氯化钡溶液</TD><td>A</TD><td>说明氢氧化钠溶液已经变质</TD></TR></TBODY></TABLE>（1）小李同学认为小王同学的结论不科学，其理由是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）小李同学的实验中观察到实验现象A为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，该反应的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />【拓展应用】<br />（3）若要除去上述变质的氢氧化钠溶液中的杂质应加入适量的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填化学式）溶液而除去．','','','','','','Na<SUB>2</SUB>CO<SUB>3</SUB>溶液也显碱性，即使NaOH变质，溶液的pH也大于7$###$产生白色沉淀$###$Na<SUB>2</SUB>CO<SUB>3</SUB>+BaCl<SUB>2</SUB>=2NaCl+BaCO<SUB>3</SUB>↓$###$Ba（OH）<SUB>2</SUB>','【解答】解：（1）氢氧化钠能与二氧化碳反应生成碳酸钠，碳酸钠和氢氧化钠的水溶液都呈碱性，pH大于7，故填：Na<SUB>2</SUB>CO<SUB>3</SUB>溶液也显碱性，即使NaOH变质，溶液的pH也大于7；<br />（2）碳酸钠能与氯化钡反应生成碳酸钡沉淀，故填：产生白色沉淀；Na<SUB>2</SUB>CO<SUB>3</SUB>+BaCl<SUB>2</SUB>=2NaCl+BaCO<SUB>3</SUB>↓；<br />（3）要除去碳酸钠，可以加入氢氧化钡溶液，故填：Ba（OH）<SUB>2</SUB>．','【分析】氢氧化钠能与二氧化碳反应生成碳酸钠，根据氯化钡能与碳酸钠反应生成碳酸钡沉淀以及物质的性质进行分析解答即可．','书写',3.00,'4d6370ccb1e7467923bdcb376160ef78',9,400,'药品是否变质的探究,碱的化学性质,盐的化学性质,书写化学方程式、文字表达式、电离方程式','',2016,'37','2016•莒县一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840015,'请从下列物质中，选择适当的物质填空（填序号）．<br />①硝酸钾&nbsp; ②氮气&nbsp; ③氧气&nbsp; ④金刚石&nbsp; ⑤小苏打&nbsp; ⑥纯碱&nbsp; ⑦火碱<br />（1）可用来裁玻璃的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（2）属于复合肥料的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（3）充入食品包装可防腐的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（4）焙制糕点所用的发酵粉的主要成分之一是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（5）广泛应用于肥皂、石油、造纸等工业的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','④$###$①$###$②$###$⑤$###$⑦','【解答】解：（1）金刚石硬度大，可用来裁玻璃；<br />（2）硝酸钾中含有氮元素和钾元素，属于复合肥料；<br />（3）氮气的化学性质稳定，无毒，可充入食品包装可防腐；<br />（4）小苏打是焙制糕点所用的发酵粉的主要成分之一；<br />（5）火碱广泛应用于肥皂、石油、造纸等工业．<br />故填：（1）④；（2）①；（3）②；（4）⑤；（5）⑦．','【分析】物质的性质决定物质的用途，根据已有的物质的性质进行分析解答即可．','填空题',3.00,'e231b095a86bce41cc726b9f93b925a0',9,400,'常见气体的用途,常见碱的特性和用途,常用盐的用途,常见化肥的种类和作用,碳单质的物理性质及用途','',2016,'37','2016•河西区一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840017,'除去括号内杂质选用的试剂正确的是（　　）','CaO（CaCO<SUB>3</SUB>）用稀盐酸','KCl（MgCl<SUB>2</SUB>）用氢氧化钠','Cu（NO<SUB>3</SUB>）<SUB>2</SUB>（AgNO<SUB>3</SUB>）&nbsp;用铜粉','CO<SUB>2</SUB>（CO）用澄清石灰水','','C','【解答】解：A、CaO和CaCO<SUB>3</SUB>均能与稀盐酸反应，不但能把杂质除去，也会把原物质除去，不符合除杂原则，故选项所采取的方法错误．<br />B、MgCl<SUB>2</SUB>能与氢氧化钠溶液反应生成氢氧化镁沉淀和氯化钠，能除去杂质但引入了新的杂质氯化钠，不符合除杂原则，故选项所采取的方法错误．<br />C、铜粉能与AgNO<SUB>3</SUB>溶液反应生成硝酸铜溶液和银，能除去杂质且没有引入新的杂质，符合除杂原则，故选项所采取的方法正确．<br />D、CO<SUB>2</SUB>能与澄清石灰水反应生成碳酸钙沉淀和水，CO不与澄清石灰水反应，反而会把原物质除去，不符合除杂原则，故选项所采取的方法错误．<br />故选：C．','【分析】根据原物质和杂质的性质选择适当的除杂剂和分离方法，所谓除杂（提纯），是指除去杂质，同时被提纯物质不得改变．除杂质题至少要满足两个条件：①加入的试剂只能与杂质反应，不能与原物质反应；②反应后不能引入新的杂质．','选择题',3.00,'ac214d0863da6d6cd6e3fb79f6a24ba5',9,400,'物质除杂或净化的探究,常见气体的检验与除杂方法,盐的化学性质','',2016,'37','2016•邵阳县二模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840019,'下列有关描述中，不正确的是（　　）','清理沼气池前做灯火实验','进入山洞时用火把照明','煮沸能降低水的硬度','天然气泄漏，立即关闭阀门','','A','【解答】解：A、沼气池中含有可燃性气体甲烷，遇明火可能发生爆炸，因此清理沼气池前不应做灯火实验，以防爆炸，故说法不正确；<br />B、山洞中可能含有大量二氧化碳，不支持燃烧，不供给呼吸，故应用火把照明检测二氧化碳的含量，故说法正确；<br />C、煮沸能够降低水的硬度，故说法正确；<br />D、天然气的主要成分是甲烷，甲烷是可燃性气体，遇明火可能发生爆炸，因此天然气泄漏应立即关闭阀门并打开门窗通风，故说法正确．<br />故选：A．','【分析】A、根据沼气池中的气体成分解答；<br />B、根据山洞中的气体成分解答；<br />C、根据煮沸的作用解答；<br />D、根据天然气的成分以及泄漏时的处理办法分析．','选择题',3.00,'093da08eda9c3f85d80f7e077a497d8e',9,400,'二氧化碳的化学性质,硬水与软水,防范爆炸的措施','',2016,'37','2016•桂林二模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840020,'生活中的下列变化不属于化学变化的是（　　）','铁器生锈','葡萄酿酒','剩饭变馊','海水晒盐','','D','【解答】解：A、铁器生锈是生成了新物质铁锈，有新物质生成，发生了化学变化，故A错；<br />B、葡萄酿酒的过程中有新物质酒精生成，属于化学变化，故B错；<br />C、剩饭变馊的过程中有对人体有害的新物质生成，属于化学变化，故C错；<br />D、海水晒盐的过程中没有新物质生成，属于物理变化，故D正确．<br />故选D','【分析】本题考查学生对物理变化和化学变化的确定．判断一个变化是物理变化还是化学变化，要依据在变化过程中有没有生成其他物质，生成其他物质的是化学变化，没有生成其他物质的是物理变化．','选择题',3.00,'e8c7c4ad7cdf353beea148ee1eafbb1b',9,400,'化学变化和物理变化的判别','',2016,'37','2016•海陵区一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840021,'下列“家庭小实验”中，不能达到预期目的是（　　）','用某些植物的花瓣制酸碱指示剂','用碎鸡蛋壳和食盐水制二氧化碳','用柠檬酸、果汁、白糖、水、小苏打等自制汽水','用塑料瓶、小卵石、石英砂、活性炭、膨松棉、纱布等制作简易净水器','','B','【解答】解：A、由于某些植物的花瓣中含有能够遇酸碱会变色的物质，所以用酒精浸泡捣烂的某些植物的花瓣制酸碱指示剂，故A正确；<br />B、鸡蛋壳的成分是碳酸钙，碳酸钙不能和食盐水反应的，所以不能用碎鸡蛋壳和食盐水制二氧化碳，故B错误；<br />C、小苏打是碳酸氢钠，柠檬酸中含有酸，碳酸盐和酸能反应生成二氧化碳，再加入果汁、白糖和水即可制得汽水，故C正确；<br />D、小卵石和石英砂能够起到过滤的作用，而活性炭能吸附可溶性的杂质、异味及色素，而膨松棉可以阻挡活性炭进入水中，可以达到净水的目的，故D正确．<br />故选：B．','【分析】A、根据某些植物的花瓣中含有能够遇酸碱会变色的物质分析解答；<br />B、依据鸡蛋壳的成分是碳酸钙，碳酸钙不能和食盐水反应的情况分析解答；<br />C、小苏打是碳酸氢钠，柠檬酸中含有酸，碳酸盐和酸能反应生成二氧化碳；<br />D、根据过滤难溶性杂质、吸附可溶性杂质等净水的措施来分析解答．','选择题',3.00,'2aabff010045f4bd66844b3868d1f16b',9,400,'化学实验方案设计与评价,水的净化,酸碱指示剂及其性质,酸的化学性质','',2016,'37','2016•海淀区一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840022,'下面是几种实验室制取气体的发生装置和收集装置，请回答下列问题：<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao5/1282f25e-94d4-11e9-af07-b42e9921e93e_xkb99.png\" style=\"vertical-align:middle\" /><br />（1）指出图中标有数字的仪器名称：①<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）实验室用双氧水和二氧化锰制取氧气时应选用的发生装置是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填字母标号，下同），该反应的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；若要收集较纯净的氧气最好选用的收集装置是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）F是一种可用于集气，洗气等的多功能装置．若将F装置内装满水，再连接量筒，可用于收集并测定不溶于水且不与水反应的气体体积，实验时气体应从<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“a”或“b”）进入F中．<br />（4）实验室可用A和E组合的装置制取并收集X气体，据此可推断<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />A．制取X气体的药品状态一定是一种固体和一种液体<br />B．X气体一定不易溶于水，也不与水反应<br />C．X气体的密度一定比空气的小<br />D．X气体一定是可燃性气体．','','','','','','酒精灯$###$B$###$2H<SUB>2</SUB>O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span>&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O+O<SUB>2</SUB>↑$###$D$###$b$###$C','【解答】解：（1）依据实验室常用仪器可知标号仪器是酒精灯；<br />（2）用双氧水和二氧化锰制取氧气属于固液常温型，发生装置为B，反应化学方程式为2H<SUB>2</SUB>O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span>&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O+O<SUB>2</SUB>↑，因为氧气密度比空气大，且不易溶于水，所以收集较为纯净的氧气最好选用D；<br />（3）F装置用排水法收集气体为气体的密度都小于液体的密度，必须短进长出，长管便于排水入量筒，故b进；<br />（4）A装置适用于固体的加热反应，E可收集密度小于空气密度的气体，分析选项只有C是正确的；<br />故答案为：<br />（1）酒精灯；集气瓶．<br />（2）B；2H<SUB>2</SUB>O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span>&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O+O<SUB>2</SUB>↑；D；<br />（3）b；<br />（4）C．','【分析】（1）依据实验室常用仪器的认识解决此题；<br />（2）依据用双氧水和二氧化锰制取氧气属于固液常温型选择发生装置，根据反应原理写化学方程式，因为氧气密度比空气大，且不易溶于水，所以收集较为纯净的氧气最好选用D；<br />（3）用排水法收集气体，因为气体的密度都小于液体的密度，必须短进长出；<br />（4）根据A装置适用于固体的加热反应，E可收集密度小于空气密度的气体解答．','书写',3.00,'0c278195d8722d3de3e60f9f91d68d6a',9,400,'氧气的制取装置,氧气的收集方法,书写化学方程式、文字表达式、电离方程式','',2016,'37','2016•沈河区一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840023,'下列图象能正确反映对应变化关系的是（　　）','<img  src=\"/tikuimages/9/2016/400/shoutiniao78/128a1e51-94d4-11e9-b466-b42e9921e93e_xkb78.png\" style=\"vertical-align:middle\" /><br />加热一定量的高锰酸钾固体','<img  src=\"/tikuimages/9/2016/400/shoutiniao84/128f2761-94d4-11e9-93a7-b42e9921e93e_xkb12.png\" style=\"vertical-align:middle\" /><br />煅烧石灰石','<img  src=\"/tikuimages/9/2016/400/shoutiniao26/12905fe1-94d4-11e9-af4a-b42e9921e93e_xkb73.png\" style=\"vertical-align:middle\" /><br />用水稀释浓硫酸','<img  src=\"/tikuimages/9/2016/400/shoutiniao0/12914a40-94d4-11e9-862b-b42e9921e93e_xkb0.png\" style=\"vertical-align:middle\" /><br />饱和NaCl溶液恒温蒸发水','','B','【解答】解：A、加热一定量的高锰酸钾固体时，高锰酸钾分解生成锰酸钾、二氧化锰和氧气，随着反应的进行，固体中锰元素质量不变，而固体质量不断减小，因此锰元素质量分数不断减小，该选项对应关系不正确；<br />B、煅烧石灰石时，开始时温度较低，碳酸钙不能分解，固体质量不变，一段时间后，碳酸钙分解生成氧化钙和二氧化碳，固体质量不断减小，当碳酸钙完全分解后，固体质量不再减小，该选项对应关系正确；<br />C、用水稀释浓硫酸时，温度升高，浓硫酸完全溶解后，温度不再升高，接着温度降低，当降低至室温时不再降低，该选项对应关系不正确；<br />D、饱和NaCl溶液恒温蒸发水时，仍然是饱和溶液，由于温度不变，氯化钠的溶解度不变，因此氯化钠的质量分数不变，该选项对应关系不正确．<br />故选：B．','【分析】高锰酸钾受热分解生成锰酸钾、二氧化锰和氧气；<br />高温条件下，碳酸钙分解生成氧化钙和二氧化碳；<br />浓硫酸溶于水放热；<br />饱和溶液中溶质质量分数=<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">溶解度</td></tr><tr><td>100g+溶解度</td></tr></table></span>×100%．','选择题',3.00,'9bab6e4f9aee5bba841ccd658ec0106b',9,400,'浓硫酸的性质及浓硫酸的稀释,溶质的质量分数,盐的化学性质','',2016,'32','2016•潮州校级模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840024,'下列化学实验不能实现实验目的是（　　）','<img  src=\"/tikuimages/9/2016/400/shoutiniao90/129541e1-94d4-11e9-a078-b42e9921e93e_xkb75.png\" style=\"vertical-align:middle\" /><br />检查装置的气密性','<img  src=\"/tikuimages/9/2016/400/shoutiniao7/129ae730-94d4-11e9-a384-b42e9921e93e_xkb49.png\" style=\"vertical-align:middle\" /><br />&nbsp;&nbsp; 氧气的制取','<img  src=\"/tikuimages/9/2016/400/shoutiniao84/12a06570-94d4-11e9-a525-b42e9921e93e_xkb38.png\" style=\"vertical-align:middle\" /><br />检验CO<SUB>2</SUB>是否收集满','<img  src=\"/tikuimages/9/2016/400/shoutiniao48/12a19dee-94d4-11e9-8c5b-b42e9921e93e_xkb10.png\" style=\"vertical-align:middle\" /><br />细铁丝在氧气中燃烧','','D','【解答】解：A、用手握住试管，如果导管口有气泡冒出，则证明装置气密性良好，故A正确；<br />B、二氧化锰和过氧化氢溶液混合会放出氧气，且氧气的密度大于空气的密度，可以用向上排空气法收集，故B正确；<br />C、二氧化碳能使燃着的木条熄灭，所以可用燃着的木条放到集气瓶口，如果木条熄灭，则证明二氧化碳已经收集满，故C正确；<br />D、铁丝在氧气中燃烧会生成高温熔融的溅落物，所以应在集气瓶中放少量的水或细沙，故D错误．<br />故选：D．','【分析】A、根据检查装置的气密性的方法进行解答；<br />B、根据二氧化锰和过氧化氢溶液混合会放出氧气，且氧气的密度大于空气的密度进行解答；<br />C、根据二氧化碳能使燃着的木条熄灭进行解答；<br />D、根据铁丝在氧气中燃烧会生成高温熔融的溅落物进行解答．','选择题',3.00,'57123714a31f9d7dd29daba0d119bcd4',9,400,'化学实验方案设计与评价,检查装置的气密性,氧气的化学性质,氧气的制取装置,二氧化碳的检验和验满','',2016,'32','2016•云南模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840025,'甲和乙在一定条件下反应生成丙和丁，微观示意图如下：<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao85/12a8c9e1-94d4-11e9-bfdf-b42e9921e93e_xkb61.png\" style=\"vertical-align:middle\" /><br />请回答下列问题<br />（1）丙的相对分子质量为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）属于氧化物的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>[填序号]．<br />（3）该反应中参加反应的甲和乙的质量比为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','32$###$甲、丁$###$22：3','【解答】解：由反应的微观示意图可知，该反应是二氧化碳和氢气在一定条件下反应生成了甲醇和水，反应的方程式是：CO<SUB>2</SUB>+3H<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;一定条件&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>CH<SUB>3</SUB>OH+H<SUB>2</SUB>O．<br />（1）由丙物质的微观构成示意图可知，丙的化学式是CH<SUB>3</SUB>OH，相对分子子质量是：12+1×4+16=32．<br />（2）由上述方程式可知，甲是二氧化碳，丁是水，都属于氧化物；<br />（3）由反应的方程式可知，发生反应的甲和乙的质量比为44：（3×1×2）=22：3．<br />故答为：（1）32；（2）甲、丁；（3）22：3．','【分析】观察微观示意图，分析反应物、生成物的化学式，根据反应写出反应的化学方程式．根据化学式、方程式的意义计算、分析、判断有关的问题．','填空题',3.00,'7763e6815558c252df8c2402a83c1708',9,400,'从组成上识别氧化物,微粒观点及模型图的应用,相对分子质量的概念及其计算','',2016,'37','2016•平谷区一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840027,'下列有关盐酸的说法中错误的是（　　）','盐酸是混合物，其溶质是一种气体','盐酸和氢氧化钠溶液反应放出大量的热','浓盐酸敞口放置一段时间后，溶质的质量分数会变大','若有很少量的稀盐酸不慎沾到皮肤上，应立即用较多的水冲洗','','C','【解答】解：A、盐酸溶液是氯化氢气体溶于水形成，正确；<br />B、盐酸和氢氧化钠溶液反应属于中和反应，会放出热量，正确；<br />C、浓盐酸具有挥发性，挥发出氯化氢气体，使溶质HCl减少，溶质减少溶剂不变，所以溶质质量分数减少，错误；<br />D、若有很少量的稀盐酸不慎沾到皮肤上，应立即用较多的水冲洗，正确．<br />故选C．','【分析】A、盐酸溶液是氯化氢气体溶于水形成；<br />B、中和反应会放出热量；<br />C、浓盐酸具有挥发性，挥发出氯化氢气体，使溶质HCl减少，溶质减少溶剂不变，所以溶质质量分数减少；<br />D、若有很少量的稀盐酸不慎沾到皮肤上，应立即用较多的水冲洗．','选择题',3.00,'2a94cce308233527585821a4da8a4eab',9,400,'常见的意外事故的处理方法,溶液、溶质和溶剂的相互关系与判断,酸的物理性质及用途,酸的化学性质','昆山市',2016,'37','2016•昆山市二模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840030,'物质变化的奥秘<br />现有下列变化：①铁粉加入到氯化铜溶液中；②双氧水中加入二氧化锰；③给水通直流电；④植物的光合作用；⑤用电热水壶给水加热；⑥用熟石灰处理含硫酸的废水．请回答下列问题：<br />（1）在上述变化中，属于化学变化的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填序号），写出属于置换反应的化学方程式：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（2）在上述变化中，属于其他形式的能量转化为化学能的是（填序号）<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（3）现有下列选项：a．原子的数目；b．元素的种类；c．分子的种类；d．分子的数目；e．元素的化合价．在变化②中，不变的是（填序号）<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（4）物质之所以能够发生化学变化是由于反应物的微粒之间发生了有效地相互作用，变化①和⑥的微观实质分别是（用符号表示）<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>、<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（5）变化①再中国古代就用于湿法冶金，近代又被科学家们用于化学电池的制作，由此可知，研究化学变化的目的就是利用化学反应<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>和<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','①②③④⑥$###$Fe+CuCl<SUB>2</SUB>═FeCl<SUB>2</SUB>+Cu$###$③④$###$ab$###$Fe+Cu<SUP>2+</SUP>=Fe<SUP>2+</SUP>+Cu$###$H<SUP>+</SUP>+OH<SUP>-</SUP>=H<SUB>2</SUB>O$###$制取新物质$###$获取能量','【解答】解：（1）①铁粉加入到氯化铜溶液中；②双氧水中加入二氧化锰；③给水通直流电；④植物的光合作用；⑥用熟石灰处理含硫酸的废水等过程中都有新物质生成，都属于化学变化；<br />铁和氯化铜的反应属于置换反应；双氧水在二氧化锰的催化作用下，分解生成水和氧气，属于分解反应；电解水生成氢气和氧气，属于分解反应；用电热水壶给水加热，不属于化学反应；铁和氯化铜反应能够生成氯化亚铁和铜，反应的化学方程式为：Fe+CuCl<SUB>2</SUB>═FeCl<SUB>2</SUB>+Cu．<br />故填：①②③④⑥；Fe+CuCl<SUB>2</SUB>═FeCl<SUB>2</SUB>+Cu．<br />（2）铁和氯化铜反应、双氧水的分解反应，都是化学能与化学能之间的转化；<br />电解水的反应，是电能转化为化学能；<br />光合作用过程中，是光能转化为化学能；<br />用电热水壶给水加热的过程中，是电能转化为内能．<br />故填：③④．<br />（3）双氧水在二氧化锰的催化作用下，分解生成水和氧气，反应的化学方程式为：2H<SUB>2</SUB>O<SUB>2</SUB> <span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span>&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O+O<SUB>2</SUB>↑，由化学方程式可知，反应前后分子的数目发生了变化；在双氧水中，氧元素的化合价是-1，而在水中，氧元素的化合价是-2，在氧气中，氧元素的化合价是0，所以元素的化合价发生了变化；分子的种类也发生了变化；而原子的数目、元素的种类没有改变．<br />故填：ab；<br />（4）变化①和⑥的微观实质分别是Fe+Cu<SUP>2+</SUP>=Fe<SUP>2+</SUP>+Cu、H<SUP>+</SUP>+OH<SUP>-</SUP>=H<SUB>2</SUB>O．故填：Fe+Cu<SUP>2+</SUP>=Fe<SUP>2+</SUP>+Cu、H<SUP>+</SUP>+OH<SUP>-</SUP>=H<SUB>2</SUB>O．<br />（5）研究化学变化的目的就是利用化学反应制取新物质和获取能量，故填：制取新物质；获取能量．','【分析】化学变化是指有新物质生成的变化，物理变化是指没有新物质生成的变化，化学变化和物理变化的本质区别是否有新物质生成；置换反应是指由一种单质和一种化合物反应，生成另外一种单质和一种化合物的反应；<br />不同形式的能量之间可以相互转化；在化学反应中遵循质量守恒定律，即反应前后元素的种类不变，原子的种类、个数不变，据此分析判断．','书写',3.00,'039595596314c11e9f52073bb758aebd',9,400,'化学变化和物理变化的判别,物质发生化学变化时的能量变化,置换反应及其应用,质量守恒定律及其应用,书写化学方程式、文字表达式、电离方程式','乳山市',2016,'35','2016春•乳山市期中',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840031,'下列实验操作或设计中正确的是（　　）','<img  src=\"/tikuimages/9/2016/400/shoutiniao75/12bf5f21-94d4-11e9-8f19-b42e9921e93e_xkb52.png\" style=\"vertical-align:middle\" /><br />分离碘酒中的碘','<img  src=\"/tikuimages/9/2016/400/shoutiniao58/12c1a90f-94d4-11e9-9b41-b42e9921e93e_xkb75.png\" style=\"vertical-align:middle\" /><br />验证质量守恒定律','<img  src=\"/tikuimages/9/2016/400/shoutiniao13/12c4dd61-94d4-11e9-8f11-b42e9921e93e_xkb43.png\" style=\"vertical-align:middle\" /><br />食盐的结晶','<img  src=\"/tikuimages/9/2016/400/shoutiniao98/12c9e670-94d4-11e9-8ad7-b42e9921e93e_xkb25.png\" style=\"vertical-align:middle\" /><br />稀释浓硫酸','','C','【解答】解：A、过滤是一种把不溶性的物质与液体相分离的一种方法，由于碘溶解在酒精中，故不可以用过滤的方法将碘除去，故错误；<br />B、验证质量守恒定律时，有气体参与的反应一定要在密闭容器中进行，故操作错误；<br />C、食盐的结晶时注意要用玻璃棒不断搅拌，操作正确；<br />D、稀释浓硫酸时要把浓硫酸慢慢注入水中，并不断搅拌，操作错误．<br />故选C．','【分析】A、据分离物质的方法分析；<br />B、据验证质量守恒定律时的注意事项分析；<br />C、根据蒸发实验的注意事项分析；<br />D、根据稀释浓硫酸的正确方法分析．','选择题',3.00,'d3291f71f36c8a0fe802930cd15cd858',9,400,'浓硫酸的性质及浓硫酸的稀释,过滤的原理、方法及其应用,蒸发与蒸馏操作,质量守恒定律及其应用','',2016,'32','2016•滑县模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840032,'下列图示实验操作中，不正确的是（　　）','<img  src=\"/tikuimages/9/2016/400/shoutiniao5/12d07621-94d4-11e9-a676-b42e9921e93e_xkb1.png\" style=\"vertical-align:middle\" /><br />&nbsp;&nbsp;&nbsp; 检查气密性','<img  src=\"/tikuimages/9/2016/400/shoutiniao13/12d2e721-94d4-11e9-87ae-b42e9921e93e_xkb68.png\" style=\"vertical-align:middle\" /><br />&nbsp;&nbsp; 稀释浓硫酸','<img  src=\"/tikuimages/9/2016/400/shoutiniao51/12d3d180-94d4-11e9-a7cb-b42e9921e93e_xkb23.png\" style=\"vertical-align:middle\" /><br />&nbsp;&nbsp; CO<SUB>2</SUB>的验满','<img  src=\"/tikuimages/9/2016/400/shoutiniao96/12d72cde-94d4-11e9-ab4d-b42e9921e93e_xkb39.png\" style=\"vertical-align:middle\" /><br />&nbsp;&nbsp;倾倒液体','','C','【解答】解：A、检查装置气密性的方法：把导管的一端浸没在水里，双手紧贴容器外壁，若导管口有气泡冒出，装置不漏气；图中所示操作正确．<br />B、稀释浓硫酸时，要把浓硫酸缓缓地沿器壁注入水中，同时用玻璃棒不断搅拌，以使热量及时地扩散；一定不能把水注入浓硫酸中；图中所示操作正确．<br />C、检验二氧化碳是否收集满时，应将燃着的木条放在集气瓶口，不能伸入瓶中，图中所示操作错误．<br />D、向试管中倾倒液体药品时，瓶塞要倒放，标签要对准手心，瓶口紧挨，图中所示操作正确．<br />故选：C．','【分析】A、根据检查装置气密性的方法进行分析判断．<br />B、根据浓硫酸的稀释方法（酸入水，沿器壁，慢慢倒，不断搅）进行分析判断．<br />C、根据二氧化碳气体的验满方法进行分析判断．<br />D、根据向试管中倾倒液体药品的方法进行分析判断．','选择题',3.00,'7ec57ea704eefd37bfa0b4624c241172',9,400,'液体药品的取用,浓硫酸的性质及浓硫酸的稀释,检查装置的气密性,二氧化碳的检验和验满','',2016,'37','2016•黑龙江二模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840033,'选择下列适当的物质填空（填字母序号）：<br />A．活性炭&nbsp;&nbsp;&nbsp;B．不锈钢&nbsp;&nbsp;&nbsp;C．纯碱&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;D．硝酸钾<br />E．维生素&nbsp;&nbsp;&nbsp;F．硫酸&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;G．蛋白质&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;H．烧碱<br />（1）复合肥料<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;（2）可用于制造医疗器械的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（3）侯氏制碱法制得的“碱”是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；&nbsp;（4）铅蓄电池中的酸<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（5）蔬菜、水果中富含<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;（6）可用于冰箱除味的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','D$###$B$###$C$###$F$###$E$###$A','【解答】解：（1）硝酸钾中含有钾元素和氮元素，属于复合肥，故填：D；&nbsp;&nbsp;&nbsp;<br />（2）不锈钢可以用于制造医疗器械，故填：B；&nbsp;&nbsp;<br />（3）侯氏制碱法制得的碱是纯碱碳酸钠，故填：C；&nbsp;&nbsp;<br />（4）铅蓄电池中的酸是稀硫酸，故填：F；<br />（5）蔬菜水果中富含维生素，故填：E；<br />（6）活性炭具有吸附性，能用于除去冰箱内的异味，故填：A．','【分析】物质的性质决定物质的用途，根据已有的物质的性质进行分析解答即可．','填空题',3.00,'c65e02d98797611ea9bf8ab8669c2a77',9,400,'酸碱盐的应用,常见化肥的种类和作用,碳单质的物理性质及用途,食品、药品与健康食品中的有机营养素','丹阳市',2016,'32','2016•丹阳市模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840034,'如图是实验室制取某些常见气体的两套装置图，请根据要求回答相关的问题．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao53/12dcab21-94d4-11e9-ad2f-b42e9921e93e_xkb23.png\" style=\"vertical-align:middle\" /><br />（1）甲、乙两套装置均可用于实验室制备氧气．若采用<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（选填“甲”或“乙”）装置，产生氧气的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，所选装置中收集氧气的方法是根据氧气<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>的性质而确定的．<br />（2）实验室制取并收集二氧化碳宜用乙装置，检验二氧化碳是否收集满的方法是：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）甲装置也可用于实验室制取甲烷（CH<SUB>4</SUB>）气体．<br />①下列四组试剂中有一组可用于制取甲烷，该试剂是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填标号）．<br />A．二氧化锰和浓盐酸<br />B．固体电石（CaC<SUB>2</SUB>）和水<br />C．氯化铵（NH<SUB>4</SUB>Cl）和氢氧化钙[Ca（OH）<SUB>2</SUB>]的固体混合物<br />D．无水醋酸钠（CH<SUB>3</SUB>COONa）和碱石灰（NaOH、CaO）的固体混合物<br />②实验开始后，当观察到<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，方可进行气体收集的操作；如果要停止实验，进行的操作是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />③甲烷气体收集满后集气瓶的放置方式如图2所示，据此可推断甲烷一定具备的物理性质是：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','甲或乙$###$2KMnO<SUB>4</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>K<SUB>2</SUB>MnO<SUB>4</SUB>+MnO<SUB>2</SUB>+O<SUB>2</SUB>↑或2H<SUB>2</SUB>O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span>&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O+O<SUB>2</SUB>↑$###$氧气不易溶于水或氧气的密度比空气大$###$将一根燃着的木条平放在集气瓶口，木条熄灭，证明满了$###$D$###$气泡均匀连续的冒出时$###$先移出导管，再熄灭酒精灯$###$甲烷的密度比空气小','【解答】解：（1）如果用双氧水和二氧化锰制氧气就不需要加热，如果用高锰酸钾或氯酸钾制氧气就需要加热；氧气的密度比空气的密度大，不易溶于水，因此能用向上排空气法和排水法收集；高锰酸钾受热分解生成锰酸钾和二氧化锰和氧气，要注意配平；过氧化氢在二氧化锰做催化剂的条件下生成水和氧气，要注意配平；故答案为：甲或乙；2KMnO<SUB>4</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>K<SUB>2</SUB>MnO<SUB>4</SUB>+MnO<SUB>2</SUB>+O<SUB>2</SUB>↑或2H<SUB>2</SUB>O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span>&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O+O<SUB>2</SUB>↑；氧气不易溶于水或氧气的密度比空气大；<br />（2）二氧化碳的验满方法是：将一根燃着的木条平放在集气瓶口，木条熄灭，证明满了；故答案为：将一根燃着的木条平放在集气瓶口，木条熄灭，证明满了；<br />（3）无水醋酸钠（CH<SUB>3</SUB>COONa）和碱石灰（NaOH、CaO）的固体混合物共热可以制取甲烷，因此需要加热；实验开始后，当观察到气泡均匀连续的冒出时，就开始收集；如果要停止实验，进行的操作是：先移出导管，再熄灭酒精灯，防止水倒流而炸裂试管；甲烷的密度比空气小，因此倒放在桌面上；故答案为：①D；②先移出导管，再熄灭酒精灯；③甲烷的密度比空气小；','【分析】制取装置包括加热和不需加热两种，如果用双氧水和二氧化锰制氧气就不需要加热，如果用高锰酸钾或氯酸钾制氧气就需要加热．氧气的密度比空气的密度大，不易溶于水，因此能用向上排空气法和排水法收集．实验室制取CO<SUB>2</SUB>，是在常温下，用大理石或石灰石和稀盐酸制取的，碳酸钙和盐酸互相交换成分生成氯化钙和水和二氧化碳，因此不需要加热．二氧化碳能溶于水，密度比空气的密度大，因此只能用向上排空气法收集．二氧化碳的验满方法是：将一根燃着的木条平放在集气瓶口，木条熄灭，证明满了．氯化铵（NH<SUB>4</SUB>Cl）和氢氧化钙[Ca（OH）<SUB>2</SUB>]的固体混合物共热可以制取氨气，因此需要加热；无水醋酸钠（CH<SUB>3</SUB>COONa）和碱石灰（NaOH、CaO）的固体混合物共热可以制取甲烷，因此需要加热；实验开始后，当观察到气泡均匀连续的冒出时，就开始收集；如果要停止实验，进行的操作是：先移出导管，再熄灭酒精灯，防止水倒流而炸裂试管；甲烷的密度比空气小，因此倒放在桌面上．','书写',3.00,'146a4fc0a97956341f48f68a23bdbbdc',9,400,'常用气体的发生装置和收集装置与选取方法,实验室制取氧气的反应原理,二氧化碳的检验和验满,书写化学方程式、文字表达式、电离方程式','',2016,'37','2016•惠安县一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840035,'下列有关金属的描述不正确的是（　　）','地壳中含量最多的金属元素是铝','车船表面涂油漆可防止生锈','合金的熔点比其组分金属的熔点低','黄铜比纯铜的硬度小','','D','【解答】解：A．地壳中含量较多的元素是氧、硅、铝、铁、钙等，其中含量最多的金属元素是铝，故A正确；<br />B．防锈就是破坏铁生锈的条件，车船表面涂油漆能使铁于水或氧气隔绝，可防止生锈，故B正确；<br />C、合金的熔点一般比各成分金属低，故C正确；<br />D、合金的硬度大于成分金属，故黄铜的硬度比纯铜大，故D错误；<br />故选D．','【分析】A．根据地壳中元素的含量来分析；<br />B．根据金属防锈的措施来分析；<br />C．根据合金的优点分析；<br />D．根据合金的优点来分析．','选择题',3.00,'17206d4163c5aca083c2431cca59fcdd',9,400,'合金与合金的性质,金属锈蚀的条件及其防护,地壳中元素的分布与含量','',2016,'32','2016•潮州校级模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840038,'根据如图所示列举的初中常见的装置图，回答下列问题：<img  src=\"/tikuimages/9/2015/400/shoutiniao13/12e75980-94d4-11e9-b26c-b42e9921e93e_xkb80.png\" style=\"vertical-align:middle\" />&nbsp;<br />（1）标号为①的仪器的名称是：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）实验室用高锰酸钾制取并收集较纯净的氧气，该反应的文字表达式是：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，他应选择上述装置中的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>和<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填字母）进行组合；<br />（3）实验室用过氧化氢溶液和二氧化锰制氧气，应该选择的发生装置是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填字母）．<br />（4）用C装置收集氧气后，验满的方法是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（5）空气中氧气含量测定的实验装置如图F所示，该实验得出的结论是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，下列有关该实验的说法正确的是（填字母）<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />A．实验中的红磷越多，水位上升越高<br />B．燃烧匙中的红磷不可以换成硫和木炭<br />C．选用足量红磷进行实验是为了耗尽集气瓶内氧气<br />D．本实验可以证明空气含有N<SUB>2</SUB>、O<SUB>2</SUB>、CO<SUB>2</SUB>和稀有气体．','','','','','','长颈漏斗$###$高锰酸钾<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td style=\"font-size: 90%\">加热</td></tr><tr><td><div hassize=\"9\"><div style=\"background: url(\'/upimages/images/part/8594L.png\') repeat-x; width:1px; height:5px;float:left;overflow:hidden\" muststretch=\"h\"></div><div style=\"background: url(\'/upimages/images/part/8594.png\') no-repeat; width:9px; height:5px;float:left;overflow:hidden\"></div></div></td></tr></table></span>锰酸钾+二氧化锰+氧气$###$A$###$E$###$B$###$把带火星的木条放在集气瓶口，若木条复燃，证明氧气已满$###$氧气约占空气体积的五分之一$###$BC','【解答】解：（1）①仪器的名称是长颈漏斗；故填：长颈漏斗；<br />（2）高锰酸钾在加热条件下生成锰酸钾、二氧化锰和氧气，反应的文字表达式是高锰酸钾<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td style=\"font-size: 90%\">加热</td></tr><tr><td><div hassize=\"9\"><div style=\"background: url(\'/upimages/images/part/8594L.png\') repeat-x; width:1px; height:5px;float:left;overflow:hidden\" muststretch=\"h\"></div><div style=\"background: url(\'/upimages/images/part/8594.png\') no-repeat; width:9px; height:5px;float:left;overflow:hidden\"></div></div></td></tr></table></span>锰酸钾+二氧化锰+氧气；属于固体加热型，应该选用的发生装置是A；氧气不易溶于水，用排水法收集的气体比较纯净，则收集装置为E．故填：高锰酸钾<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td style=\"font-size: 90%\">加热</td></tr><tr><td><div hassize=\"9\"><div style=\"background: url(\'/upimages/images/part/8594L.png\') repeat-x; width:1px; height:5px;float:left;overflow:hidden\" muststretch=\"h\"></div><div style=\"background: url(\'/upimages/images/part/8594.png\') no-repeat; width:9px; height:5px;float:left;overflow:hidden\"></div></div></td></tr></table></span>锰酸钾+二氧化锰+氧气；A；E；<br />（3）实验室用过氧化氢溶液和二氧化锰制氧气，属于固液常温型，应该选择的发生装置是B．故填：B；<br />（4）用C装置收集氧气后，验满方法为把带火星的木条放在集气瓶口，若木条复燃，证明氧气已满．故填：把带火星的木条放在集气瓶口，若木条复燃，证明氧气已满；<br />（5）红磷燃烧产生大量白烟；测定空气中氧气含量的实验原理利用红磷燃烧消耗空气中的氧气，使左边集气瓶内气体体积变小，压强变小，从而在外部大气压的作用下使烧杯内的水进入集气瓶，通过测量进入集气瓶中水的量来确定空气中氧气的体积分数，所以：<br />A、由于空气中氧气的含量是一定的，当氧气完全反应后，再多的红磷也不和剩余的气体反应，水位就不会再上升，故A错误；<br />B、由于硫和木炭燃烧时，虽然也消耗空气中氧气但是生成了同是气体的二氧化硫和二氧化碳，导致了集气瓶内体积没有减小，烧杯内的水不会进入集气瓶，故B正确；<br />C、选用红磷是因为反应可以耗尽氧气，生成固态的P<SUB>2</SUB>O<SUB>5</SUB>体积减小明显，故C正确；<br />D、本实验无法证明剩余的气体中含有N<SUB>2</SUB>、O<SUB>2</SUB>、CO<SUB>2</SUB>和稀有气体，只能说明剩余的气体的体积及剩余的气体不和红磷反应、不溶于水，故D错误；<br />故填：氧气约占空气体积的五分之一；BC．','【分析】（1）熟记仪器的名称；<br />（2）实验室用高锰酸钾制取氧气，属于固体加热型制取气体，据此确定发生装置；根据氧气的密度及溶解性，进行分析解答；<br />（3）实验室用过氧化氢溶液和二氧化锰制氧气，属于固液常温型，进行分析解答．<br />（4）根据氧气的验满方法进行分析解答；<br />（5）根据红磷燃烧产生大量白烟，以及测定空气中氧气含量的实验原理利用红磷燃烧消耗空气中的氧气，使左边集气瓶内气体体积变小，压强变小，从而在外部大气压的作用下使烧杯内的水进入集气瓶，通过测量进入集气瓶中水的量来确定空气中氧气的体积分数去分析解答．','书写',3.00,'10b6e5d6b1da7a2cbf68d09f8294310a',9,400,'空气组成的测定,氧气的制取装置,氧气的收集方法,氧气的检验和验满,书写化学方程式、文字表达式、电离方程式','',2015,'35','2015秋•淮安校级期中',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840039,'下列物质的用途中，利用其化学性质的是（　　）','干冰用于人工降雨','液氮用作冷冻剂','天然气用作燃料','银用于制作导线','','C','【解答】解：A、干冰用于人工降雨是利用干冰升华吸热，使周围温度降低，水蒸气冷凝成水滴，利用了物理性质，故A错；<br />B、液氮用作冷冻剂是利用液氮气化吸热，没有新物质生成，利用了物理性质，故B错；<br />C、天然气用作燃料是利用天然气的可燃性，天然气燃烧生成二氧化碳和水，属于化学变化，所以利用了化学性质，故C正确；<br />D、银用于制作导线是利用银的导电性，属于物理性质，故D错．<br />故选C．','【分析】物理性质是指物质不需要发生化学变化就表现出来的性质．化学性质是指物质在化学变化中表现出来的性质．而化学变化的本质特征是变化中有新物质生成，因此，判断物理性质还是化学性质的关键就是看表现物质的性质时是否有新物质产生．','选择题',3.00,'2ccca2907f7a3e6144c55589ecd1d37d',9,400,'化学性质与物理性质的差别及应用','',2016,'37','2016•大悟县二模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840040,'下列对相关事实的解释合理的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br /><table class=\"edittable\"><TBODY><TR><td width=190>选项</TD><td width=190>事实</TD><td width=190>解释</TD></TR><TR><td>A</TD><td>CO<SUB>2</SUB>能使紫色石蕊溶液变红</TD><td>CO<SUB>2</SUB>是一种酸</TD></TR><TR><td>B</TD><td>金刚石和石墨性质不同</TD><td>构成它们的原子排列方式不同</TD></TR><TR><td>C</TD><td>水变成水蒸气</TD><td>质量守恒</TD></TR><TR><td>D</TD><td>氧气可用于切割金属</TD><td><u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD></TR></TBODY></TABLE>','','','','','','B$###$氧气能支持燃烧','【解答】解：A、CO<SUB>2</SUB>能使紫色石蕊溶液变红，是因为二氧化碳与水反应生成碳酸，碳酸能使紫色石蕊溶液变红，CO<SUB>2</SUB>不是酸，故选项说法错误．<br />B、金刚石和石墨性质不同，是因为它们的碳原子排列方式不同，故选项说法正确．<br />C、水变成水蒸气属于物理变化，不能用质量守恒定律解释，故选项说法错误．<br />D、氧气可用于切割金属，是利用了氧气能支持燃烧的性质，氧气不具有可燃性．<br />故选：B；氧气能支持燃烧．','【分析】A、根据二氧化碳的化学性质进行分析判断．<br />B、根据金刚石和石墨碳原子排列方式不同，进行分析判断．<br />C、质量守恒定律适用于一切化学变化．<br />D、根据氧气能支持燃烧，不具有可燃性，进行分析判断．','填空题',3.00,'a52b995a77d4e7a4e0b9b9f7ac439c60',9,400,'氧气的用途,二氧化碳的化学性质,碳元素组成的单质,质量守恒定律及其应用','',2016,'37','2016春•赣州校级月考',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840041,'下列关于物质的组成、结构、性质以变化规律的总结中，正确的是（　　）','在CO<SUB>2</SUB>、SO<SUB>2</SUB>、H<SUB>2</SUB>O<SUB>2</SUB>三种物质中，都含有氧气','Ca、Ca<SUP>2+</SUP>两种微粒中质子数相同，化学性质也相同','在高温条件下，C和CO都能使Fe<SUB>2</SUB>O<SUB>3</SUB>失去氧，C和CO都发生了氧化反应','因为蜡烛燃烧生成二氧化碳和水，所以蜡烛的组成中一定含有碳、氢、氧三种元素','','C','【解答】解：A、CO<SUB>2</SUB>、SO<SUB>2</SUB>、H<SUB>2</SUB>O<SUB>2</SUB>三种物质均属于纯净物，均不含氧气，三种物质中都含有氧元素，故选项说法错误．<br />B、元素的化学性质跟它的原子的最外层电子数目关系非常密切，Ca、Ca<SUP>2+</SUP>两种微粒中质子数相同，但它们的最外层电子数不同，化学性质不相似，故选项说法错误．<br />C、在高温条件下，C和CO都能使Fe<SUB>2</SUB>O<SUB>3</SUB>失去氧，C和CO都夺取了氧化铁中的氧，都发生了氧化反应，故选项说法正确．<br />D、因为蜡烛燃烧生成二氧化碳和水，CO<SUB>2</SUB>和H<SUB>2</SUB>O两种物质中含有碳、氢、氧三种元素，根据质量守恒定律，反应前后，元素种类不变，反应物氧气中只含有氧元素，则某化合物中一定含有碳、氢两种元素，可能含有氧元素，故选项说法错误．<br />故选：C．','【分析】A、根据CO<SUB>2</SUB>、SO<SUB>2</SUB>、H<SUB>2</SUB>O<SUB>2</SUB>三种物质均属于纯净物，进行分析判断．<br />B、根据元素的化学性质跟它的原子的最外层电子数目关系非常密切，进行分析判断．<br />C、根据C和CO都能使Fe<SUB>2</SUB>O<SUB>3</SUB>失去氧，进行分析判断．<br />D、根据质量守恒定律，反应前后元素种类不变，进行分析判断．','选择题',3.00,'d7fc63ba6e6d1f7678c2bbb27e49ccee',9,400,'一氧化碳的化学性质,分子、原子、离子、元素与物质之间的关系,原子和离子的相互转化,碳的化学性质,质量守恒定律及其应用','',0,'37','',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840045,'<img  src=\"/tikuimages/9/2016/400/shoutiniao22/12fb56b0-94d4-11e9-95b8-b42e9921e93e_xkb47.png\" style=\"vertical-align:middle;FLOAT:right\" />（2016•罗平县三模）钢铁是我们日常生活中使用最广泛的金属材料之一．<br />（1）生铁与钢都是铁的合金，但钢的新能比生铁优越，这是因为生铁的含碳量比钢的含碳量<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“多”或“少”）．<br />（2）据统计，每年锈蚀的钢铁约占世界钢铁年产量的<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">1</td></tr><tr><td>3</td></tr></table></span>，某同学为探究铁的锈蚀条件，他将一支洁净铁钉放入置于空气中盛有少量水的试管中（如图所示），一段时间后，发现铁钉<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“A”或“B”或“C”）处最先出现生锈，该实验说明，钢铁常温下接触<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>最容易生锈，因此，平时应注意保持铁制品表面的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）工业上常用稀硫酸对生锈的铁制品进行酸洗除锈，请写出稀硫酸与铁锈（主要成分为Fe<SUB>2</SUB>O<SUB>3</SUB>）反应的化学式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（4）将过量铁粉加入到AgNO<SUB>3</SUB>、Cu（NO<SUB>3</SUB>）<SUB>2</SUB>的混合溶液中，充分反应过滤，向滤渣中加入稀盐酸有气泡产生，则滤渣中一定有<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','多$###$B$###$水和氧$###$干燥$###$Fe<SUB>2</SUB>O<SUB>3</SUB>+3H<SUB>2</SUB>SO<SUB>4</SUB>=Fe<SUB>2</SUB>（SO<SUB>4</SUB>）<SUB>3</SUB>+3H<SUB>2</SUB>O$###$Ag、Cu、Fe','【解答】解：（1）生铁和钢都是铁的合金，其中生铁中的含碳量高于钢中额含碳量，故填：多；<br />（2）B处铁钉与水和氧气充分接触，最先生锈；保持铁制品表面的洁净干燥能防止铁生锈，故填：B，水和氧，干燥；<br />（3）氧化铁能与硫酸反应生成硫酸铁和水，故填：Fe<SUB>2</SUB>O<SUB>3</SUB>+3H<SUB>2</SUB>SO<SUB>4</SUB>=Fe<SUB>2</SUB>（SO<SUB>4</SUB>）<SUB>3</SUB>+3H<SUB>2</SUB>O．<br />（4）将过量铁粉加入到AgNO<SUB>3</SUB>、Cu（NO<SUB>3</SUB>）<SUB>2</SUB>的混合溶液中，根据金属活动性．<br />故答案为：（1）多；<br />（2）B，水和氧，干燥；<br />（3）Fe<SUB>2</SUB>O<SUB>3</SUB>+3H<SUB>2</SUB>SO<SUB>4</SUB>=Fe<SUB>2</SUB>（SO<SUB>4</SUB>）<SUB>3</SUB>+3H<SUB>2</SUB>O；<br />（4）Ag、Cu、Fe．','【分析】根据已有的金属的性质进行分析解答即可，生铁和钢都是铁的合金，铁在与水和氧气并存时易生锈，防锈就是破坏铁生锈的条件；氧化铁能与硫酸反应生成硫酸铁和水，据此解答；将过量铁粉加入到AgNO<SUB>3</SUB>、Cu（NO<SUB>3</SUB>）<SUB>2</SUB>的混合溶液中，根据金属活动性解答．','书写',3.00,'53f9ffb568b6e17be710214f90102ec9',9,400,'金属的化学性质,生铁和钢,金属锈蚀的条件及其防护,酸的化学性质,书写化学方程式、文字表达式、电离方程式','',2016,'37','2016•罗平县三模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840047,'家庭生活中处处都有物质变化，下列属物理变化的是（　　）','酵母粉发酵','洁厕精除垢','鲜牛奶变质','西瓜榨成汁','','D','【解答】解：A、酵母粉发酵过程中有新物质生成，属于化学变化．<br />B、洁厕精除垢过程中有新物质生成，属于化学变化．<br />C、鲜牛奶变质过程中有新物质生成，属于化学变化．<br />D、西瓜榨成汁过程中没有新物质生成，属于物理变化．<br />故选D．','【分析】化学变化是指有新物质生成的变化，物理变化是指没有新物质生成的变化，化学变化和物理变化的本质区别是否有新物质生成；据此分析判断．','选择题',3.00,'c65d3390c77a5e50b0b619410c04539d',9,400,'化学变化和物理变化的判别','',2016,'32','2016•郑州模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840051,'下列各组生活现象中，前者属于物理变化，后者属于化学变化的一组是（　　）','蜡烛燃烧，葡萄酒化','汽车爆胎、铁器生锈','水果腐烂、蔗糖溶解','玻璃破碎、鲜肉冷冻','','B','【解答】解：A．蜡烛燃烧时能生成水、二氧化碳等物质，属于化学变化；葡萄酿酒的过程中，葡萄糖转化成酒精，有新物质生成，属于化学变化，故A错误；<br />B．汽车爆胎时，只是轮胎的形状发生了变化，没有新物质生成，属于物理变化；铁器生锈的过程中，生成了新物质--铁锈，属于化学变化，故B正确；<br />C．水果腐烂的过程中，水果发生了缓慢氧化，生成了水、二氧化碳等物质，属于化学变化；蔗糖溶解的过程中，只是蔗糖和水混合，没有新物质生成，属于物理变化，故C错误；<br />D．玻璃破碎的过程中，只是玻璃的形状发生了变化，没有新物质生成，属于物理变化；鲜肉冷冻的过程中，因为降温，鲜肉由软变硬，过程中没有新物质生成，属于物理变化，故D错误．<br />故选B．','【分析】化学变化是指有新物质生成的变化，物理变化是指没有新物质生成的变化，化学变化和物理变化的本质区别是否有新物质生成．','选择题',3.00,'9f06fff395d4a35073f2d17644d09789',9,400,'化学变化和物理变化的判别','普兰店市',2015,'35','2015秋•普兰店市校级期中',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840052,'化学使我们的生活变的更加的绚丽多彩．<br />（1）活性炭净水器主要利用活性炭<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>水中的杂质，净化后水变澄清了，所得到的水是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“纯净物”或“混合物”）．<br />（2）在实验室为了除去水中不溶性固体颗粒，我们常用<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>操作，该操作需要的玻璃仪器有烧杯、玻璃棒、<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）化石燃料包括煤、<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>和天然气，属于不可再生资源，开发利用新能源迫在眉睫，氢能可作为理想的能源，氢化镁（化学式：MgH<SUB>2</SUB>）固体可作为氢动力汽车的能源提供剂，提供能源时氢化镁与水反应生成氢氧化镁和氢气，反应的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','吸附$###$混合物$###$过滤$###$、漏斗$###$石油$###$MgH<SUB>2</SUB>+2H<SUB>2</SUB>O=Mg（OH）<SUB>2</SUB>+2H<SUB>2</SUB>↑','【解答】解：<br />（1）活性炭具有吸附性，能吸附水中的色素和异味，净化得到的水中含有可溶性杂质，属于混合物，故填：吸附，混合物；<br />（2）净化水时，为除去水中不溶性杂质常用过滤的方法，该实验需要的玻璃仪器有：烧杯、漏斗、玻璃棒．<br />（3）化石燃料包括煤、石油和天然气，属于不可再生资源，MgH<SUB>2</SUB>与水反应生成氢氧化镁和氢气，反应的化学方程式为：MgH<SUB>2</SUB>+2H<SUB>2</SUB>O=Mg（OH）<SUB>2</SUB>+2H<SUB>2</SUB>↑．<br />答案：<br />（1）吸附；混合物；<br />（2）过滤；漏斗；<br />（3）石油；MgH<SUB>2</SUB>+2H<SUB>2</SUB>O=Mg（OH）<SUB>2</SUB>+2H<SUB>2</SUB>↑．','【分析】（1）根据已有的水的净化的以及净水中各物质的作用进行分析解答即可；<br />（2）根据过滤的操作分析用的仪器；<br />（3）根据MgH<SUB>2</SUB>与水反应生成氢氧化镁和氢气解答．','书写',3.00,'ea5666a8e21c59b8924c58ac025f3afd',9,400,'过滤的原理、方法及其应用,水的净化,纯净物和混合物的判别,碳单质的物理性质及用途,书写化学方程式、文字表达式、电离方程式,化石燃料及其综合利用','',2016,'32','2016•合肥校级模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840054,'下列实验现象描述正确的是（　　）','木炭燃烧生成灰色固体','红磷燃烧产生大量白烟','铁丝在氧气中剧烈燃烧生成四氧化三铁','镁带在空气中燃烧发出耀眼的白光','','B|D','【解答】解：A、木炭燃烧，生成能使澄清石灰水变浑浊的气体，故选项说法错误．<br />B、红磷燃烧，产生大量的白烟，故选项说法正确．<br />C、铁丝在氧气中剧烈燃烧生成四氧化三铁，是实验结论而不是实验现象，故选项说法错误．<br />D、镁带在空气中燃烧，发出耀眼的白光，故选项说法正确．<br />故选：BD．','【分析】A、根据木炭燃烧的现象进行分析判断．<br />B、根据红磷燃烧的现象进行分析判断．<br />C、根据铁丝在氧气中燃烧的现象进行分析判断．<br />D、根据镁带在空气中燃烧的现象进行分析判断．','多选题',3.00,'119ddae21e8fc5281c0827d14a58d394',9,400,'氧气与碳、磷、硫、铁等物质的反应现象','龙口市',2016,'35','2016春•龙口市期中',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840056,'阅读下面科普短文．<br />绿色植物的光合作用<br />沐浴在阳光下的绿色植物在悄悄地进行着一个重要的化学变化--光合作用．它是绿色植物通过叶绿体，利用光能，把二氧化碳和水转化成储存能量的有机物（淀粉），并且释放氧气的过程．绿色植物的光合作用需要什么条件呢？<br />【实验1】实验装置如图1所示．一段时间后，取玻璃瓶内罩住的叶片及未罩住的一个叶片，分别经过脱色等处<br />理，加入碘溶液．观察到前者无变化，后者变蓝．<br />【实验2】操作步骤：（如图2）①取淡蓝色的BTB溶液倒入试管A中．②另取BTB溶液放入大烧杯中，向溶液中吹气，溶液颜色变黄．③将大烧杯中的BTB溶液倒入B、C、D试管中，在C、D试管中各放入一棵植物．④用橡皮塞将A、B、C、D&nbsp;4个试管塞紧，并用锡纸将试管D包裹．⑤将4个试管置于光照下，观察BTB溶液的变色情况．实验现象：一段时间后，观察到A试管中溶液颜色仍为淡蓝色，B试管中溶液颜色仍为黄色，C试管中溶液颜色变为淡蓝色，D试管中溶液颜色仍为黄色．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao6/131c7340-94d4-11e9-9bb5-b42e9921e93e_xkb85.png\" style=\"vertical-align:middle\" /><br />小资料：1、淀粉遇碘会变为蓝色．<br />2、BTB即溴麝香草酚蓝．溴麝香草酚蓝是一种酸碱指示剂、吸附指示剂．BTB溶液在pH＜6时呈黄色，在pH＞7.6时呈蓝色，在pH&nbsp;为6～7.6时呈过渡颜色绿色或淡蓝．<br />其实，绿色植物的光合作用是一个很复杂的过程．通过光合作用，将无机物转变成了有机物，将光能转变成了化学能，同时维持大气中氧气和二氧化碳的平衡．<br />依据文章内容，回答下列问题：<br />（1）光合作用的反应物是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）实验1的目的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；&nbsp;澄清石灰水的作用是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）实验2中，A、B、C、D&nbsp;4个试管做了3组对照实验，其中能说明光合作用需要在光照条件下进行的对照实验是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填试管序号）．<br />（4）北京市蔬菜需求量庞大，请你为种植大棚蔬菜的农民提出一条增产的建议：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','CO<SUB>2</SUB>、H<SUB>2</SUB>O$###$证明光合作用需要二氧化碳$###$检验空气中的二氧化碳是否除尽$###$CD$###$增加光照（充入气体肥料二氧化碳）','【解答】解：（1）光合作用是二氧化碳与水在一定条件下转化为有机物和氧气，故填：CO<SUB>2</SUB>、H<SUB>2</SUB>O；<br />（2）实验中缺少二氧化碳，经检验没有生成淀粉，澄清的石灰水是用来检验空气中的二氧化碳是否被除尽；故填：证明光合作用需要二氧化碳；检验空气中的二氧化碳是否除尽；<br />（3）A、B、C、D&nbsp;4个试管做了3组对照实验，其中能说明光合作用需要在光照条件下进行的对照实验是CD；故填：CD；<br />（4）适当增大二氧化碳的浓度或延长光照时间都能增强光合作用的强度，增加农作物的产量；故填：增加光照（充入气体肥料二氧化碳）．','【分析】（1）根据光合作用的原理来分析；<br />（2）根据控制变量法来分析；<br />（3）根据四组实验中的相同条件、不同条件来分析；<br />（4）根据光合作用的原料来分析．','填空题',3.00,'77b2495ea0559cfc6e1d65fbcaee69c4',9,400,'光合作用与呼吸作用','',2016,'37','2016•大兴区一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840057,'下列实验操作正确的是（　　）','过滤时，用玻璃棒在过滤器中不断搅拌','排水法收集氧气时，当导管口开始有气泡冒出时立即收集','氢气、甲烷等可燃性气体点燃前应检验纯度','检查装置气密性，应先手握容器外壁，在将导管伸入水中，观察导管口是否有气泡','','C','【解答】解：A、过滤时，用玻璃棒引流，不能在过滤器中不断搅拌，以免损坏滤纸，故错误；<br />B、排水法收集氧气时，不能有气泡冒出立即收集，而是要等到气泡冒出均匀后再收集，故错误；<br />C、可燃性气体燃烧时易发生爆炸，点燃前进行验纯，故正确；<br />D、检查气密性时，先用双手捂住容器外壁后将导管插入水中，这样容器内的空气先从导管冒出，会导致装置气密性好，导管口也没有气泡冒出，故错误．<br />故选C．','【分析】A、根据过滤时的注意事项解答；<br />B、根据实验室制取氧气的步骤进行分析判断；<br />C、根据可燃性气体燃烧时易发生爆炸，点燃前进行验纯解答；<br />D、根据检验装置气密性的正确操作进行分析．','选择题',3.00,'c2068f55b1ec5ca23f15e724bfef5d0e',9,400,'过滤的原理、方法及其应用,检查装置的气密性,制取氧气的操作步骤和注意点,氢气、一氧化碳、甲烷等可燃气体的验纯','胶州市',2016,'35','2016春•胶州市期中',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840059,'<img  src=\"/tikuimages/9/2015/400/shoutiniao94/132b4051-94d4-11e9-a9f2-b42e9921e93e_xkb90.png\" style=\"vertical-align:middle;FLOAT:right;\" />（2015•当涂县模拟）铁及其合金在生产、生活中应用广泛，如图是铁元素的部分信息．下列说法不正确的是（　　）','铁原子的质子数为26','铁元素的相对原子质量为55.85','铁元素是地壳中含量最多的金属元素','铁原子的核外电子数为26','','C','【解答】解：A、根据元素周期表中的一格可知，左上角的数字为26，表示原子序数为26；根据原子序数=核电荷数=质子数=核外电子数，则铁原子的质子数为26，故选项说法正确．<br />B、根据元素周期表中的一格可知，汉字下面的数字表示相对原子质量，该元素的相对原子质量为55.85，故选项说法正确．<br />C、铝元素是地壳中含量最多的金属元素，故选项说法错误．<br />D、根据元素周期表中的一格可知，左上角的数字为26，表示原子序数为26；根据原子序数=核电荷数=质子数=核外电子数，则该元素的原子核外电子数为26，故选项说法正确．<br />故选：C．','【分析】根据图中元素周期表可以获得的信息：左上角的数字表示原子序数；字母表示该元素的元素符号；中间的汉字表示元素名称；汉字下面的数字表示相对原子质量，进行分析判断即可．','选择题',3.00,'96f3b9702112ebb474aac1fff3518e2b',9,400,'元素周期表的特点及其应用','',2015,'32','2015•当涂县模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840060,'某化学兴趣小组探究稀盐酸与以下三种物质是否发生化学反应：<br /><table class=\"edittable\"><TBODY><TR><td width=195>K<SUB>2</SUB>CO<SUB>3</SUB>溶液</TD><td width=195>AgNO<SUB>3</SUB>溶液</TD><td width=195>NaOH溶液</TD></TR></TBODY></TABLE>（1）甲、乙、丙三位同学设计了以下三个实验，请你填写相关内容：<br /><table class=\"edittable\"><TBODY><TR><td width=67></TD><td width=226>实验操作</TD><td width=146>实验现象</TD><td width=146>实验结论</TD></TR><TR><td>实验1</TD><td>在K<SUB>2</SUB>CO<SUB>3</SUB>溶液中加入稀盐酸</TD><td><u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD><td>稀盐酸与K<SUB>2</SUB>CO<SUB>3</SUB>溶液<br />发生了化学反应</TD></TR><TR><td>实验2</TD><td>在AgNO<SUB>3</SUB>溶液加入稀盐酸</TD><td><u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD><td>稀盐酸与在AgNO<SUB>3</SUB>溶液发生了化学反应</TD></TR><TR><td>实验3</TD><td>在NaOH溶液加入稀盐酸</TD><td>没有现象</TD><td><u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD></TR></TBODY></TABLE>（2）写出实验2的化学方程式：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />（3）在实验3中，因为没有明显实验现象，故无法判断该反应能否进行，为了证明该反应能否发生，甲、乙、丙三位同学又设计了如下实验方案，并继续进行实验．<br />【查阅资料】酸与碱发生中和反应会放出热量．<img  src=\"/tikuimages/9/2015/400/shoutiniao64/1332ba5e-94d4-11e9-a6d4-b42e9921e93e_xkb17.png\" style=\"vertical-align:middle;FLOAT:right\" /><br />【实验设计】甲、乙、丙三位同学分别设计了如下实验方案，请你帮他们完成实验报告．<br />方案一：先用pH试纸测定NaOH溶液的pH，再滴加盐酸并不断震荡，同时测定溶液的pH，若测得溶液的pH<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填序号），则证明NaOH溶液与稀盐酸发生了化学反应．<br />a、逐渐增大并≥7&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;b、逐渐减小并≤7&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;c、始终不变并等于7<br />方案二：向NaOH溶液中滴加酚酞试液，溶液显红色，再滴加稀盐酸，若<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，则证明NaOH溶液与稀盐酸发生了化学反应．<br />方案三：丙同学借助酸碱中和反应会放出热量，设计了如图所示的实验证明盐酸与氢氧化钠发生了反应，他观察到红墨水向<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>边（填“左”或“右”）移动．<br />【评价反思】有的同学提出，方案三的现象不足以证明NaOH与稀盐酸发生了化学反应，他的理由是：<br />①<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />②<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />【拓展延伸】有位同学向自己配制的氢氧化钠溶液中滴加酚酞试液时，溶液变红色，一会儿后红色消失了．①甲同学认为可能是酚酞溶液变质的缘故．你认为他的猜想是否合理？<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“合理”或“不合理”）②乙同学认为可能是氢氧化钠溶液与空气中的二氧化碳反应的缘故．丙同学认为乙同学的猜想不正确，他的理由是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','有气泡产生$###$生成白色沉淀$###$无法判断该反应能否进行$###$HCl+AgNO<SUB>3 </SUB>=HNO<SUB>3 </SUB>+AgCl↓$###$b$###$红色褪去$###$右$###$氢氧化钠固体溶于水会放出热量$###$到底热量是氢氧化钠与盐酸反应放出的，还是氢氧化钠固体溶于水放出的，无法判断$###$不合理$###$碳酸钠呈碱性','【解答】解：（1）实验一：稀盐酸与K<SUB>2</SUB>CO<SUB>3</SUB>溶液反应，生成氯化钾、二氧化碳和水，符合发生复分解反应的条件；<br />实验二：稀盐酸与AgNO<SUB>3</SUB>溶液反应，生成氯化银沉淀和硝酸，符合发生复分解反应的条件；<br />实验三：没有明显实验现象，故无法判断该反应能否进行；<br />（2）稀盐酸与AgNO<SUB>3</SUB>溶液反应生成氯化银沉淀和硝酸，反应的方程式为：HCl+AgNO<SUB>3 </SUB>=HNO<SUB>3 </SUB>+AgCl↓；<br />（3）【实验设计】方案一：氢氧化钠与盐酸反应，碱性减弱，pH减小，故选b；<br />方案二：氢氧化钠呈碱性，使酚酞变红，中性溶液或酸性溶液都不能使酚酞试液变色，故红色褪去，则证明NaOH溶液与稀盐酸发生了化学反应；<br />方案三：氢氧化钠固体溶于水会放出大量的热，氢氧化钠与盐酸反应也能放出热量，会导致瓶内温度升高，红墨水右移；<br />【评价反思】方案三的现象不足以证明NaOH与稀盐酸发生了化学反应，理由是：<br />①氢氧化钠固体溶于水会放出热量；<br />②到底热量是氢氧化钠与盐酸反应放出的，还是氢氧化钠固体溶于水放出的，无法判断；<br />【拓展延伸】氢氧化钠浓度太高会破坏酚酞的结构，因此会呈现先变红后红色即可消失的情况；氢氧化钠溶液吸收了空气中二氧化碳生成碳酸钠，其溶液也是显碱性，也能使酚酞试液变红，故氢氧化钠变质，不会影响其红色．<br />故答案为：（1）<table class=\"edittable\"><TBODY><TR><td width=67></TD><td width=226>实验操作</TD><td width=146>实验现象</TD><td width=146>实验结论</TD></TR><TR><td>实验1</TD><td></TD><td>有气泡产生</TD><td></TD></TR><TR><td>实验2</TD><td></TD><td>生成白色沉淀</TD><td></TD></TR><TR><td>实验3</TD><td></TD><td></TD><td>无法判断该反应能否进行</TD></TR></TBODY></TABLE>（2）HCl+AgNO<SUB>3 </SUB>=HNO<SUB>3 </SUB>+AgCl↓；<br />【实验设计】方案一：b；方案二：红色褪去．方案三：右．<br />【评价反思】①氢氧化钠固体溶于水会放出热量<br />②到底热量是氢氧化钠与盐酸反应放出的，还是氢氧化钠固体溶于水放出的，无法判断<br />【拓展延伸】不合理，碳酸钠呈碱性．','【分析】（1）根据反应的实验现象和结论进行分析；<br />（2）根据化学方程式的书写注意事项写出化学方程式；<br />（3）【实验设计】可测定溶液的pH的变化证明NaOH溶液与稀盐酸是否反应；中性溶液或酸性溶液都不能使酚酞试液变色；反应放热，瓶内压强增大，红墨水右移；<br />【评价反思】由于方案三用的是氢氧化钠固体溶于水，由于氢氧化钠固体溶于水会放出热量，到底热量是氢氧化钠与盐酸反应放出的，还是氢氧化钠固体溶于水放出的，无法判断，所以说法错误；<br />【拓展延伸】氢氧化钠浓度太高会破坏酚酞的结构，因此会呈现先变红后红色即可消失的情况．','书写',3.00,'6276d4e29ff9f8ea711dca7bcce0465a',9,400,'探究酸碱的主要性质,酸的化学性质,溶液的酸碱性与pH值的关系,书写化学方程式、文字表达式、电离方程式','普宁市',2015,'37','2015•普宁市一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840061,'空气中氮气的含量最多，氮气在高温、高能量条件下可与某些物质发生反应．<br />如图1是以空气和其它必要的原料合成化肥NH<SUB>4</SUB>NO<SUB>3</SUB>的工业流程．请按要求回答下列问题：<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao12/1337c36e-94d4-11e9-bb15-b42e9921e93e_xkb72.png\" style=\"vertical-align:middle\" /><br />（1）步骤①中发生的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>变化（选填“物理”或“化学”）．<br />（2）写出步骤③中发生反应的化学方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）硝酸铵属于<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>肥（填写化肥类别）．分析图2标签上的信息，可得出硝酸铵具有的性质是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（选填字母）．<br />A．易溶于水&nbsp;&nbsp;&nbsp;&nbsp;B．有挥发性<br />C．受热易分解&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;D．能与熟石灰发生反应<br />（4）图2标签所标注的“含氮量≤30%”是否合理<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（选填“是”或“否”）．','','','','','','物理$###$4NH<SUB>3</SUB>+5O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>催化剂</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>△</td></tr></table></span>4NO+6H<SUB>2</SUB>O$###$氮$###$ACD$###$是','【解答】解：（1）从空气中分离出原有的氮气和氧气是利用二者的沸点不同，没有产生新的物质，发生的是物理变化；<br />（2）步骤③是氨气和氧气在催化剂加热的条件下生成一氧化氮和水，化学方程式为：4NH<SUB>3</SUB>+5O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>催化剂</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>△</td></tr></table></span>4NO+6H<SUB>2</SUB>O；<br />（3）硝酸铵中含有的农作物所需要的营养元素氮元素，属于氮肥；由图2标签上的信息“防潮防晒”、“隔绝热源”可知，NH<SUB>4</SUB>NO<SUB>3</SUB>易溶于水且受热易分解，不能说明硝酸铵具有挥发性；由“不能与草木灰混用”说明了硝酸铵能与熟石灰发生反应．<br />（4）硝酸铵中氮元素的质量分数数：<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">14×2</td></tr><tr><td>80</td></tr></table></span>×100%=35%，所以，图2标签所标注的“含氮量≤30%”合理；<br />故答案为：（1）物理&nbsp;&nbsp;&nbsp;（2）4NH<SUB>3</SUB>+5O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>催化剂</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>△</td></tr></table></span>4NO+6H<SUB>2</SUB>O；（3）氮；&nbsp;A&nbsp;C&nbsp;D；&nbsp;（4）是．','【分析】（1）根据分离液态空气法获得气体的变化特征分析；<br />（2）根据步骤③是氨气和氧气在催化剂加热的条件下生成一氧化氮和水进行分析；<br />（3）根据硝酸铵中含有的农作物所需要的营养元素和保存方法分析判断．<br />（4）根据硝酸铵中氮元素的质量分数分析判断．','书写',3.00,'71f43fabbd7c5e5df9c88bd8f8ed7438',9,400,'常见化肥的种类和作用,物质的相互转化和制备,元素的质量分数计算,化学变化和物理变化的判别,书写化学方程式、文字表达式、电离方程式','',2016,'32','2016•海南模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840064,'阅读下面科普短文（原文作者：袁越）．<br />&nbsp;&nbsp;&nbsp; 二氧化碳是著名的温室气体，它能让太阳光顺利通过，却会阻止地表热量的散失．在地球形成的初期，太阳的辐射强度只有现在的四分之一，为什么那时的地球没有被冻成冰球呢？最新的理论认为，液态的水（比如降雨）能够溶解空气中的二氧化碳，再把它变为碳酸盐，沉积到岩石层中．同时，地球的内部很热，沉积在地壳中的碳经常会随着火山喷发而重新变为二氧化碳释放到大气中，这就形成了一个碳循环．经过几亿年的时间，这个碳循环逐渐达到了某种平衡，使大气中的二氧化碳保持一定的含量．正是由于这些二氧化碳产生的温室效应，使得地球的温度不至于太冷．<br />&nbsp;&nbsp;&nbsp; 生命的诞生促成了另一个碳循环．众所周知，生命的基础是光合作用，就是利用太阳提供的能量，把二氧化碳中的碳元素提取出来，连接成一条长短不一的碳链．这样的碳链被称为“有机碳”，因为它既能作为生命的“建筑材料”，搭建成生命所需的各种有机分子（碳水化合物、蛋白质和氨基酸等），又能“燃烧自己”，产生能量供生命使用．有机碳的燃烧过程又可以称之为“氧化反应”，其产物就是二氧化碳和水．<br />南极冰钻的结果证明，地球大气中的二氧化碳浓度在过去的1万年里基本保持稳定．但自工业化以来，由于化石燃料的大量使用，二氧化碳浓度开始逐年上升，同时段内地球大气层的平均温度也发生了变化．（见表一和表二）<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao36/13404eee-94d4-11e9-a53b-b42e9921e93e_xkb13.png\" style=\"vertical-align:middle\" /><br />目前燃烧化石能源而产生的二氧化碳，大气中的实际含量比理论计算值少一倍．有证据显示，大气中二氧化碳浓度的提高加快了森林的生长速度，促进了土壤对二氧化碳的吸收，这说明大自然正在努力地试图平衡人类带来的影响．但是，大自然的平衡能力是有限的，面对突然多出来的这些“碳”，大自然一时也应付不过来了，所以人类必须自己想办法．下图是当前人类为降低大气中的二氧化碳浓度所采取的措施．（有删改）<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao91/1344e2d1-94d4-11e9-8664-b42e9921e93e_xkb86.png\" style=\"vertical-align:middle\" /><br />依据文章内容，回答下列问题．<br />（1）在地球形成的初期，地球没有被冻成冰球的原因是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）人类使用化石燃料使大气中二氧化碳浓度逐年上升．下列物质中，属于化石燃料的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填字母序号，下同）．<br />A．天然气&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;B．煤&nbsp;&nbsp;&nbsp;C．酒精&nbsp;D．石油<br />（3）观察表一和表二后，小德认为二氧化碳排放量的持续增加导致地球大气层平均温度持续升高，但小威却不同意他的观点．下列不同年份间数据变化的总趋势能支持小威说法的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />A．1900～1910&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;B．1910～1940&nbsp;&nbsp;&nbsp;C．1940～1980&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;D．1980以后<br />（4）人类为降低大气中二氧化碳浓度采取的措施有：开发新能源、植树造林、<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（写出一条即可）等．<br />（5）关于碳循环，下列说法正确的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />A．二氧化碳是温室气体，对人类的生活有害无利<br />B．保持自然界的碳循环处于平衡状态是至关重要的<br />C．人类过多的工业活动产生大量的二氧化碳，打破了碳循环的平衡<br />D．当碳循环的平衡状态被破坏时，大自然完全具有自我修复的能力．','','','','','','由于二氧化碳产生的温室效应，使得地球的温度不至于太冷$###$ABD$###$AC$###$碳封存$###$BC','【解答】解：（1）有题干信息可知，二氧化碳产生的温室效应，使得地球的温度不至于太冷，所以在地球形成的初期，地球没有被冻成冰球；故填：由于二氧化碳产生的温室效应，使得地球的温度不至于太冷；<br />（2）化石燃料包括煤、石油和天然气，酒精不属于化石燃料；故填：ABD；<br />（3）1900～1910年二氧化碳的排放量增加，但是温度呈下降趋势；1940～1980年，二氧化碳的排放量迅速增大，但是温度变化不明显，二者均能支持小威说法；故填：AC；<br />（4）人类为降低大气中二氧化碳浓度采取的措施有：开发新能源、植树造林或将碳封存起来；故填：碳封存（答案合理）；<br />（5）A．二氧化碳是绿色植物光合作用的原料，故错误；<br />B．保持自然界的碳循环处于平衡状态是至关重要的，故正确；<br />C．人类过多的工业活动产生大量的二氧化碳，打破了碳循环的平衡，故正确；<br />D．当碳循环的平衡状态被破坏时，大自然并不能完全自我修复，超过大自然自我修复的能力，就会引发自然灾害，故错误．<br />故填：BC．','【分析】（1）根据题干信息结合二氧化碳是造成温室效应的气体之一来分析；<br />（2）天然气、煤、石油属于化石燃料；<br />（3）根据表格数据来分析；<br />（4）根据降低大气中二氧化碳含量的措施分析判断；<br />（5）根据二氧化碳的性质、用途及碳循环来分析．','填空题',3.00,'276e28719e42e895f04d323cc405ced9',9,400,'二氧化碳对环境的影响,化石燃料及其综合利用','',2016,'37','2016•海淀区一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840065,'化学式一门以实验为基础的科学，正确的实验操作是完成实验认为的保证．下列做法中，不合理的是（　　）','测定空气中氧气的含量时，可以使用白磷','做铁丝在氧气中燃烧实验时，预先在集气瓶中加少量水','稀释浓硫酸时，将浓硫酸注入盛水的量筒中','给烧杯中的液体加热时应使用石棉网','','C','【解答】解：<br />A、白磷燃烧生成五氧化二磷固体，使用白磷测定空气中氧气的含量，故正确；<br />B、做铁丝在氧气中燃烧实验时，预先在集气瓶中加少量水，以防止生成物熔化溅落下来，使瓶底炸裂，故正确；<br />C、稀释浓硫酸时，要把浓硫酸缓缓地沿器壁注入水中，同时用玻璃棒不断搅拌，以使热量及时地扩散；不能在量筒中稀释浓硫酸；故错误；<br />D、给烧杯中的液体加热时为防止烧杯破裂，应使用石棉网，故正确；<br />故选C．','【分析】A、根据使用白磷测定空气中氧气的含量分析；<br />B、根据铁丝在氧气中燃烧实验的注意事项，进行分析判断；<br />C、根据稀释浓硫酸的方法分析；<br />D、根据给烧杯中的液体加热时的注意事项分析．','选择题',3.00,'742acc54c67e80ca787be18dfc7add1e',9,400,'用于加热的仪器,浓硫酸的性质及浓硫酸的稀释,空气组成的测定,氧气的化学性质','',2016,'37','2016•潍坊二模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840066,'2011年3月11日下午，日本宫城县发生里氏9.0级大地震，致使日本福岛第一核电站发生严重的核辐射泄漏，日本政府向核电站附近居民发放碘片（碘-127），以降低放射性碘对人体的伤害．已知放射性碘（碘-131）的核电荷数为53，中子数为78，则下列说法正确的是（　　）','核裂变是一种化学变化','碘原子和放射性碘原子属于同种元素','碘-131和碘-127的化学性质不同','碘-127原子和碘-131原子的相对原子质量相同','','B','【解答】解：A、核能的利用证实原子的可分性；但化学变化不涉及到原子核的变化，所以核裂变不是化学变化．故不正确；<br />B、元素是具有相同核电荷数（即核内质子数）的一类原子的总称，碘I-127原子和放射性碘原子I-133质子数相同，因此属于同种元素；故正确；<br />C、碘-131和碘-127的化学性质相同；故不正确；<br />D、碘127原子和碘131原子的相对原子质量不同；故不正确；<br />故选B．','【分析】A、核能的利用证实原子的可分性；<br />B、根据元素是具有相同核电荷数（即核内质子数）的一类原子的总称，碘I-127原子和放射性碘原子I-133质子数相同，因此属于同种元素，进行解答；<br />C、碘-131和碘-127的化学性质相同；<br />D、根据碘127原子和碘131原子的相对原子质量不同；进行解答．','选择题',3.00,'6d637c9d290c04bd8eb99444bd61dc0d',9,400,'原子的有关数量计算,元素的概念,化学变化和物理变化的判别','',2016,'32','2016•内江模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840068,'如图是实验室常用的一些装置，请根据如图所示回答：<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao5/134fdf4f-94d4-11e9-bb9c-b42e9921e93e_xkb35.png\" style=\"vertical-align:middle\" /><br />①A、B两装置均可供实验室制取氧气，与A装置相比较，用B装置的优点是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．写出用B装置制取氧气的化学方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，若用D装置来干燥氧气，则D内应装的干燥剂是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />②在不做改动的情况下，C装置<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（选填“能”或“不能”）用来做氢气还原氧化铜的实验．<br />③氨气（NH<SUB>3</SUB>）在通常情况下是一种无色、有刺激性气味的气体，密度比空气小，极易溶于水．实验室常用加热氯化铵和熟石灰两种固体混合物来制取氨气，反应的化学方程式为：2NH<SUB>4</SUB>Cl+Ca（OH）<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>CaCl<SUB>2</SUB>+2NH<SUB>3</SUB>↑+2H<SUB>2</SUB>O．实验室制取氨气应选择的发生装置是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填写装置对应的字母），若用D装置来收集氨气，则气体应从<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（选填“a”或“b”）通入．','','','','','','可控制反应速率$###$2H<SUB>2</SUB>O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span>&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O+O<SUB>2</SUB>↑$###$浓硫酸$###$不能$###$C$###$a','【解答】解：<br />①比较B和A装置可以知道它们的主要差别在于B选择了分液漏斗，通过旋转分液漏斗的活塞可以控制反应速率；<br />B装置属于固液常温型，适合过氧化氢溶液和二氧化锰制取氧气，二氧化锰是催化剂，反应的方程式为：2H<SUB>2</SUB>O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span>&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O+O<SUB>2</SUB>↑；<br />装置D适合液体作为干燥剂，因此干燥氧气可以使用具有吸水性的浓硫酸；<br />②因为多余的氢气要溢出试管，所以氢气还原氧化铜的试管口不能有塞子，故该装置不能作为氢气还原氧化铜的装置，应该去掉橡皮塞，将导管伸到试管底部氧化铜的上方；<br />③实验室常用加热氯化铵和熟石灰两种固体混合物来制取氨气，因此发生装置的特点是固体加热型的C装置；<br />因为氨气的密度比空气小，会聚集在集气瓶的上方，故进气口是短管，将空气挤压到集气瓶上部排出；<br />故答案为：①可以控制反应速率；&nbsp;&nbsp;&nbsp;2H<SUB>2</SUB>O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span>&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O+O<SUB>2</SUB>↑；　　浓硫酸；<br />&nbsp;②不能；<br />&nbsp;③C；a．','【分析】①比较装置A和B，根据它们的差别来进行分析的优点；根据B装置的特点选择氧气的制取并写出反应的原理；根据D装置的特点选择干燥剂；<br />②根据氢气还原氧化铜的装置特点分析：试管口略向下倾斜；不使用橡皮塞，将导管伸到试管底部氧化铜的上方；<br />③实验室制取氨气时需要加热，反应物都是固体，应该选择能够加热的实验装置；并根据氨气的密度比空气小，选择进气口．','书写',3.00,'9689374c2350f0fda4381509b2faa70a',9,400,'气体的干燥（除水）,氧气的制取装置,氧气的收集方法,书写化学方程式、文字表达式、电离方程式','',2016,'37','2016•济宁一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840069,'把少量下列物质分别放入水中，充分搅拌，可以得到溶液的是（　　）','泥土','面粉','植物油','高锰酸钾','','D','【解答】解：A、泥土不溶于水，与水混合形成悬浊液，故A错；<br />B、面粉不溶于水，与水混合形成悬浊液，不属于溶液，故B错；<br />C、植物油不溶于水，与水混合形成乳浊液，不属于溶液，故D错；<br />D、高锰酸钾易溶于水，形成均一、稳定的混合物，故D正确．<br />故选D．','【分析】本题考查溶液的概念，在一定条件下溶质分散到溶剂中形成的是均一稳定的混合物．','选择题',3.00,'813860acf5efbd429c31cc5b22bdbce3',9,400,'溶液的概念、组成及其特点','',2016,'37','2016•天津一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840073,'新版《生活饮用水卫生标准》（简称新国标）中水质检测指标从原来的35项增加到106项，对供水各环节的水质量提出了相应的要求．<br />（1）新国标在无机物指标中修订了镉、铅等的限量．这里的镉、铅指的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填序号）．<br />A．原子　　　　　B．分子　　　　　C．元素　　　　　　D．单质<br />（2）新国标中对水的pH的规定为6.5≤pH≤8.5．实验室用pH试纸测得自来水的pH=8，说明该自来水显<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“酸”、“碱”或“中”）性．<br />（3）新国标中消毒剂由1项增至4项，加入了对用臭氧、二氧化氯和氯胺消毒的规定．<br />①臭氧（O<SUB>3</SUB>）在消毒过程中转化为氧气．臭氧转化为氧气属于<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“物理”或者“化学”）变化．<br />②二氧化氯消毒过程中产生的次氯酸根离子（ClO<SUP>-</SUP>）也有消毒作用．ClO<SUP>-</SUP>中氯元素的化合价为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>价．<br />③氯胺（NH<SUB>2</SUB>Cl）由<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填数字）种元素组成．用氯胺消毒时，反应的化学方程式是NH<SUB>2</SUB>Cl+X=NH<SUB>3</SUB>+HClO，其中X的化学式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','C$###$碱$###$化学$###$+1$###$3$###$H<SUB>2</SUB>O','【解答】解：（1）物质是由元素组成，这里的镉、铅指的是元素；<br />（2）溶液的pH＜7，显酸性，pH=7，显中性，pH＞7，显碱性，验室用pH试纸测得自来水的pH=8，说明该自来水显碱性；<br />（3）①氧气和臭氧属于不同的物质，所以氧气转化为臭氧属于化学变化；<br />②设氯元素的化合价为x<br />x+（-2）=-1，解得：x=+1<br />③NH<SUB>2</SUB>Cl含有氮、氢、氯3种元素；根据所给化学方程式可以看出：化学方程式的右边比左边多出了2个氢原子和1个氧原子；根据质量守恒定律中的元素种类不变和原子个数不变的特点，可知多出的原子全部来自1个X分子中，即X的化学式为H<SUB>2</SUB>O．<br />答案：（1）C；（2）碱；（3）①化学；②+1；③3；H<SUB>2</SUB>O．','【分析】（1）根据物质是由元素组成的解答；<br />（2）根据溶液的pH＜7，显酸性，pH=7，显中性，pH＞7，显碱性解答；<br />（3）①根据氧气和臭氧属于不同的物质解答；<br />②根据化合价的求法解答；<br />③根据化学式的意义分析；根据质量守恒定律可知：反应前后元素种类不变，原子个数不变，由以上依据可推出X的化学式．','书写',2.00,'7987b97bb75eebdb2da9deac4484d021',9,400,'自来水的生产过程与净化方法,溶液的酸碱性与pH值的关系,元素的概念,化学式的书写及意义,有关元素化合价的计算,化学变化和物理变化的判别,质量守恒定律及其应用','',2016,'37','2016•萍乡一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840074,'下列是某学生实验现象记录笔记，错误的是（　　）','二氧化碳通入澄清的石灰水中，石灰水会变浑浊','镁在空气中燃烧，发出耀眼白光，放出大量的热，生成一种白色粉末','二氧化碳通入紫色石蕊试剂中，加热，溶液颜色变化为：紫色→红色→无色','将光亮的铁片伸入稀盐酸中，铁片表面有气泡生成，溶液逐渐变成浅绿色','','C','【解答】解：A、二氧化碳能与氢氧化钙生成了碳酸钙沉淀和水，所以通入澄清的石灰水中，石灰水会变浑浊，故A正确；<br />B、镁在空气中燃烧，发出耀眼白光，放出大量的热，生成一种白色粉末，故B正确；<br />C、由于二氧化碳能与水化合生成了碳酸，碳酸不稳定，所以二氧化碳通入紫色石蕊试剂中，加热，溶液颜色变化为：紫色→红色→紫色，故C错误；<br />D、由于铁能与稀盐酸反应生成了氯化亚铁和氢气，所以将光亮的铁片伸入稀盐酸中，铁片表面有气泡生成，溶液逐渐变成浅绿色．故D正确．<br />故选C．','【分析】A、根据二氧化碳与氢氧化钙的反应分析判断；<br />B、根据镁在空气中燃烧的现象分析判断；<br />C、根据二氧化碳能与水化合生成了碳酸，碳酸不稳定分析；<br />D、根据铁与盐酸反应的现象分析判断．','选择题',3.00,'336a1426fc2212683c641d032621f32b',9,400,'氧气与碳、磷、硫、铁等物质的反应现象,二氧化碳的化学性质,金属的化学性质','',2016,'37','2016•延平区一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840078,'纯碱是重要的化工原料<br />（一）制备探究：如图是工业生产纯碱的主要流程示意图．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao21/13603300-94d4-11e9-8a71-b42e9921e93e_xkb81.png\" style=\"vertical-align:middle\" /><br />[查阅资料]<br />①粗盐水中含有杂质MgCl<SUB>2</SUB>、CaCl<SUB>2</SUB>；<br />②常温下，NH<SUB>3</SUB>极易溶于水，CO<SUB>2</SUB>能溶于水，<br />③NaHCO<SUB>3</SUB>加热易分解，Na<SUB>2</SUB>CO<SUB>3</SUB>加热不易分解．<br />（1）写出除去粗盐水中MgCl<SUB>2</SUB>的化学方程式：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）在工业生产纯碱工艺流程中，先“氨化”后“碳酸化”的目的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，“碳酸化”时，NaCl、NH<SUB>3</SUB>、CO<SUB>2</SUB>和H<SUB>2</SUB>O相互作用析出NaHCO<SUB>3</SUB>，写出该反应的化学方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）“碳酸化”后过滤获得的NH<SUB>4</SUB>Cl可用作氮肥，也可先加热NH<SUB>4</SUB>Cl溶液，再加入熟石灰获得循环使用的物质是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（4）煅烧制得纯碱的化学方程式是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（二）成分探究<br />[提出问题]纯碱样品中含有哪些杂质？<br />[猜想]<br />猜想一：可能含有NaHCO<SUB>3</SUB>；&nbsp;猜想二：可能含有NaCl；猜想三：NaHCO<SUB>3</SUB>和NaCl<br />[实验探究]确定纯碱中是否含NaHCO<SUB>3</SUB>．实验装置和主要实验步骤如下：<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao9/1364c6de-94d4-11e9-b4c3-b42e9921e93e_xkb89.png\" style=\"vertical-align:middle\" /><br />①称量D、E装置总质量为200.0g，检查装置的气密性，将10.6&nbsp;g纯碱试样放入锥形瓶中，按上图组装仪器，进行如下操作<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，后缓缓鼓入一段时间空气；<br />②按一定要求正确操作后，加入足量稀硫酸，待锥形瓶中不再产生气泡时，再次打开止水夹K1，从导管a处再次缓缓鼓入空气；<br />③一段时间后再次称量装置D、E的总质量为204.84g．<br />[实验讨论]<br />（5）步骤①中“如下操作”是指<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（6）装置B中一定发生的化学反应方程式为：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>； （7）装置B中生成CO<SUB>2</SUB>的质量为4.84g，通过计算说明纯碱中<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>含NaHCO<SUB>3</SUB>（填字母）．（写出计算过程）<br />A．一定&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;B．一定不&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;C．可能&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;D．无法确定<br />（8）另取10.6样品，加入a&nbsp;g&nbsp;14.6%的盐酸恰好完全反应，再将所得溶液蒸干后得到固体的质量为W，当W的值满足<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>条件时，样品中含有NaCl．','','','','','','MgCl<SUB>2</SUB>+2NaOH=Mg（OH）<SUB>2</SUB>+2NaCl$###$有利于溶液吸收二氧化碳$###$NaCl+NH<SUB>3</SUB>+CO<SUB>2</SUB>+H<SUB>2</SUB>O=NaHCO<SUB>3</SUB>+NH<SUB>4</SUB>Cl$###$NH<SUB>3</SUB>$###$2NaHCO<SUB>3</SUB>=Na<SUB>2</SUB>CO<SUB>3</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑$###$打开K<SUB>1</SUB>、K<SUB>2</SUB>，关闭K<SUB>3</SUB>$###$打开K<SUB>1</SUB>、K<SUB>2</SUB>，关闭K<SUB>3</SUB>$###$Na<SUB>2</SUB>CO<SUB>3</SUB>+H<SUB>2</SUB>SO<SUB>4</SUB>=Na<SUB>2</SUB>SO<SUB>4</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑$###$A$###$W＞0.234a','【解答】解：（1）除去粗盐水中MgCl<SUB>2</SUB>用的试剂是氢氧化钠溶液，故该反应的化学方程式为：MgCl<SUB>2</SUB>+2NaOH=Mg（OH）<SUB>2</SUB>↓+2NaCl；<br />（2）在工业生产纯碱工艺流程中，先“氨化”后“碳酸化”的目的是有利于溶液吸收CO<SUB>2</SUB>气体；氯化钠、氨气、二氧化碳和水反应生成碳酸氢钠和氯化铵，反应的化学方程式为：NaCl+NH<SUB>3</SUB>+CO<SUB>2</SUB>+H<SUB>2</SUB>O=NaHCO<SUB>3</SUB>+NH<SUB>4</SUB>Cl；<br />（3）“碳酸化”后过滤获得的NH<SUB>4</SUB>Cl含有氮元素，所以可用作氮肥；氯化铵不稳定受热易分解或遇到碱放出氨气，也可先加热NH<SUB>4</SUB>Cl溶液，再加入熟石灰获得循环使用的物质是NH<SUB>3</SUB>（或氨气）；<br />（4）因为碳酸氢钠受热分解生成碳酸钠、水和二氧化碳，故反应的化学方程式为：2NaHCO<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>Na<SUB>2</SUB>CO<SUB>3</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑；<br />【猜想三】纯碱样品中还可能含有NaHCO<SUB>3</SUB>和NaCl的混合物；<br />（5）加入样品前还应检查装置气密性；反应结束后打开止水夹K<SUB>1</SUB>，缓缓鼓入空气的目的是将生成的二氧化碳全部送到D中；故步骤①中“如下操作”是指打开K<SUB>1</SUB>、K<SUB>2</SUB>，关闭K<SUB>3</SUB>；<br />（6）装置B中碳酸钠和硫酸反应生成硫酸钠、水和二氧化碳，故一定发生的化学反应方程式为：Na<SUB>2</SUB>CO<SUB>3</SUB>+H<SUB>2</SUB>SO<SUB>4</SUB>=Na<SUB>2</SUB>SO<SUB>4</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑；<br />（7）根据D、E增加的质量为二氧化碳的质量，装置B中生成CO<SUB>2</SUB>的质量=204.84g-200.0g=4.84g；<br />假设10.6g样品全为碳酸钠，生成二氧化碳的质量为x．<br />Na<SUB>2</SUB>CO<SUB>3</SUB>+H<SUB>2</SUB>SO<SUB>4</SUB>=Na<SUB>2</SUB>SO<SUB>4</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑<br />106&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;44<br />10.6g&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; x<br />则<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">106</td></tr><tr><td>44</td></tr></table>=<table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">10.6g</td></tr><tr><td>x</td></tr></table></span>，解得x=4.4g．<br />则4.4g＜4.84g，所以样品中含有碳酸氢钠；<br />（8）根据碳酸钠和盐酸反应生成氯化钠、水和二氧化碳，假设全部是碳酸钠求出增加的质量．<br />Na<SUB>2</SUB>CO<SUB>3</SUB>+2HCl=2NaCl+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑<br />&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 73&nbsp;&nbsp; &nbsp;117<br />&nbsp;&nbsp;&nbsp;ag×14.6%&nbsp;&nbsp;&nbsp; W<br />则<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">73</td></tr><tr><td>117</td></tr></table>=<table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">ag×14.3%</td></tr><tr><td>W</td></tr></table></span>，解得W=0.234ag．<br />所以当W的值满足条件W＞0.234ag时，样品中含有NaCl；<br />故答案为：<br />（1）MgCl<SUB>2</SUB>+2NaOH=Mg（OH）<SUB>2</SUB>+2NaCl．<br />（2）有利于溶液吸收二氧化碳，NaCl+NH<SUB>3</SUB>+CO<SUB>2</SUB>+H<SUB>2</SUB>O=NaHCO<SUB>3</SUB>+NH<SUB>4</SUB>Cl．<br />（3）NH<SUB>3</SUB>，（4）2NaHCO<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>Na<SUB>2</SUB>CO<SUB>3</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑；<br />（5）打开K<SUB>1</SUB>、K<SUB>2</SUB>，关闭K<SUB>3</SUB>；（6）Na<SUB>2</SUB>CO<SUB>3</SUB>+H<SUB>2</SUB>SO<SUB>4</SUB>=Na<SUB>2</SUB>SO<SUB>4</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑<br />（7）A；解题略；（8）W＞0.234a．','【分析】（1）根据除去粗盐水中MgCl<SUB>2</SUB>用的试剂是氢氧化钠溶液进行解答；<br />（2）根据在工业生产纯碱工艺流程中，先“氨化”后“碳酸化”的目的是有利于溶液吸收CO<SUB>2</SUB>气体以及氯化钠、氨气、二氧化碳和水反应生成碳酸氢钠和氯化铵进行解答；<br />（3）根据“碳酸化”后过滤获得的NH<SUB>4</SUB>Cl含有氮元素、氯化铵不稳定受热易分解或遇到碱放出氨气进行解答；<br />（4）根据碳酸氢钠受热分解生成碳酸钠、水和二氧化碳进行解答；<br />【猜想三】根据纯碱样品中还可能含有NaHCO<SUB>3</SUB>和&nbsp;NaCl进行解答；<br />（5）根据加入样品前还应检查装置气密性进行解答；反应结束后打开止水夹K<SUB>1</SUB>，缓缓鼓入空气的目的是将生成的二氧化碳全部送到D中；<br />（6）根据反应结束后打开止水夹K<SUB>1</SUB>，缓缓鼓入空气的目的是将生成的二氧化碳全部送到D中、碳酸钠和硫酸反应生成硫酸钠、水和二氧化碳以及氢氧化钠溶液和二氧化碳反应生成碳酸钠和水进行解答；<br />（7）根据D、E增加的质量为二氧化碳的质量进行解答；<br />（8）根据碳酸钠和盐酸反应生成氯化钠、水和二氧化碳，假设全部是碳酸钠求出增加的质量进行解答．','书写',3.00,'da707faa3126136fb027d30c1295e2a0',9,400,'实验探究物质的组成成分以及含量,常见气体的检验与除杂方法,盐的化学性质,纯碱的制取,书写化学方程式、文字表达式、电离方程式','江阴市',2016,'32','2016•江阴市模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840079,'已知，过氧化钠（化学式为Na<SUB>2</SUB>O<SUB>2</SUB>）是一种浅黄色固体粉末，可用作呼吸面具里的供氧剂，利用人呼出的二氧化碳与Na<SUB>2</SUB>O<SUB>2</SUB>反应放出O<SUB>2</SUB>，供给人的正常生命活动，该反应的化学方程式为：2Na<SUB>2</SUB>O<SUB>2</SUB>+2CO<SUB>2</SUB>═2Na<SUB>2</SUB>CO<SUB>3</SUB>+O<SUB>2</SUB>．某同学利用该反应原理制取氧气，设计了如图一的实验装置．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao54/136c6800-94d4-11e9-a994-b42e9921e93e_xkb61.png\" style=\"vertical-align:middle\" /><br />（1）装置①中CO<SUB>2</SUB>的制取方法有多种：若用加热NaHCO<SUB>3</SUB>固体制取CO<SUB>2</SUB>，应该选用的发生装置为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（选填图二字母）<br />（2）为收集纯净干燥的O<SUB>2</SUB>，装置④中盛放的试剂为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，⑤中收集装置可选用<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填序号）．<br />A．排水集气法　　　　B．向下排空气法　　　　C．向上排空气法　　　　D．真空袋收集法<br />（3）当观察到装置②中<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>时应停止通CO<SUB>2</SUB>．<br />（4）此外，过氧化钠也能与水迅速而剧烈反应，生成碱和氧气，试写出该反应的化学方程式：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．若用此反应制取O<SUB>2</SUB>，可选用的发生装置为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（选填图二中字母）．','','','','','','A$###$浓硫酸$###$D$###$淡黄色粉末都变成了白色$###$2Na<SUB>2</SUB>O<SUB>2</SUB>+2H<SUB>2</SUB>O=4NaOH+O<SUB>2</SUB>$###$C','【解答】解：（1）碳酸氢钠在加热的条件下分解产生二氧化碳，所应该选择固体加热型的反应装置即A装置；故填：A；<br />（2）浓硫酸具有吸水性，可用装置④盛放浓硫酸来对氧气进行干燥；收集纯净干燥的O<SUB>2</SUB>，可用真空袋收集；故填：浓硫酸；D；<br />（3）当看到淡黄色粉末都变成了白色，说明过氧化钠已完全参加了反应；故填：淡黄色粉末都变成了白色；<br />（4）过氧化钠也能与水迅速而剧烈反应，生成碱和氧气，方程式为2Na<SUB>2</SUB>O<SUB>2</SUB>+2H<SUB>2</SUB>O=4NaOH+O<SUB>2</SUB>，属于固液常温下反应制取气体，选择装置C通过分液漏斗来控制反应速率；故填：2Na<SUB>2</SUB>O<SUB>2</SUB>+2H<SUB>2</SUB>O=4NaOH+O<SUB>2</SUB>；C．','【分析】（1）加热NaHCO<SUB>3</SUB>固体制取CO<SUB>2</SUB>，属于固体加热型，故选发生装置A；<br />（2）根据干燥氧气的方法以及收集方法来分析；<br />（3）根据实验现象来判断反应的进行；<br />（4）根据化学方程式的写法以及反应的条件来分析．','书写',3.00,'087e57d6f53604a7d305d0566ec8d818',9,400,'常见气体的检验与除杂方法,气体的干燥（除水）,二氧化碳的实验室制法,书写化学方程式、文字表达式、电离方程式','招远市',2016,'32','2016•招远市模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840081,'请用所学化学知识解释下列原理．<br />（1）洗涤剂能除去油污，是因为它具有<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>功能．<br />（2）用化学方程式表示生石灰做干燥剂原理．<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />（3）用化学方程式表示含Mg（OH）<SUB>2</SUB>的药物治疗胃酸过多症的原理．<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />（4）我国古代将炉甘石&nbsp;（ZnCO<SUB>3</SUB>）、赤铜（Cu<SUB>2</SUB>O）和木炭粉混合后加热到800℃得到一种外观似金子的锌和铜的合金．用化学方程式表示所发生的化学反应（提示：ZnCO<SUB>3</SUB>加热可分解为ZnO）．<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />（5）水垢主要成分是碳酸钙和氢氧化镁，而厨房中的食醋可用于除水垢，食醋主要成分为醋酸（CH<SUB>3</SUB>COOH），醋酸在水溶液中能电离出CH<SUB>3</SUB>COO<SUP>-</SUP>和H<SUP>+</SUP>，请用化学反应方程式表示食醋清洗水垢中的氢氧化镁<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（6）冶炼2000t含杂质3%的生铁，需要含Fe<SUB>3</SUB>O<SUB>4</SUB>90%的磁铁矿石的质量是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','乳化$###$CaO+H<SUB>2</SUB>O═Ca（OH）<SUB>2</SUB>$###$Mg（OH）<SUB>2</SUB>+2HCl═MgCl<SUB>2</SUB>+2H<SUB>2</SUB>O$###$ZnCO<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;800℃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>ZnO+CO<SUB>2</SUB>↑，2ZnO+C<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;800℃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2Zn+CO<SUB>2</SUB>↑，2Cu<SUB>2</SUB>O+C<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;800℃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>4Cu+CO<SUB>2</SUB>↑$###$2CH<SUB>3</SUB>COOH+Mg（OH）<SUB>2</SUB>=（CH<SUB>3</SUB>COO）<SUB>2</SUB>Mg+2H<SUB>2</SUB>O$###$2976.72t','【解答】解：（1）洗涤剂能使植物油在水中分散成无数细小的液滴，而不聚成大的油珠，这种现象叫乳化，乳化后的细小液滴能随着水流动，所以洗涤剂能清洗餐具上的油污；故答案为：乳化；<br />（2）生石灰能和水反应生成氢氧化钙，其反应的化学方程式为：CaO+H<SUB>2</SUB>O═Ca（OH）<SUB>2</SUB>；故答案为：CaO+H<SUB>2</SUB>O═Ca（OH）<SUB>2</SUB>；<br />（3）氢氧化镁和盐酸反应生成氯化镁和水，其反应的化学方程式为：Mg（OH）<SUB>2</SUB>+2HCl═MgCl<SUB>2</SUB>+2H<SUB>2</SUB>O．故填：Mg（OH）<SUB>2</SUB>+2HCl═MgCl<SUB>2</SUB>+2H<SUB>2</SUB>O．<br />（4）根据题干提供的信息可以知道，碳酸锌首先高温分解生成氧化锌和二氧化碳，碳还原氧化锌生成锌和二氧化碳，碳还原氧化亚铜生成铜和二氧化碳，根据化学方程式的书写方法，故答案为：ZnCO<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;800℃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>ZnO+CO<SUB>2</SUB>↑，2ZnO+C<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;800℃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2Zn+CO<SUB>2</SUB>↑，2Cu<SUB>2</SUB>O+C<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;800℃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>4Cu+CO<SUB>2</SUB>↑．<br />（5）根据所给的反应规律可知醋酸与氢氧化镁反应的生成物是醋酸镁和水，反应的化学方程式为：2CH<SUB>3</SUB>COOH+Mg（OH）<SUB>2</SUB>=（CH<SUB>3</SUB>COO）<SUB>2</SUB>Mg+2H<SUB>2</SUB>O，故答案为：2CH<SUB>3</SUB>COOH+Mg（OH）<SUB>2</SUB>=（CH<SUB>3</SUB>COO）<SUB>2</SUB>Mg+2H<SUB>2</SUB>O．<br />（5）解：生铁中铁的质量为2000t×（1-3%）=1940t<br />设需四氧化三铁的质量为x．<br />4CO+Fe<SUB>3</SUB>O<SUB>4</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>3Fe+4CO<SUB>2</SUB><br />&nbsp;&nbsp;&nbsp; 232&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;168<br />&nbsp;&nbsp;&nbsp; x&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;1940t<br />&nbsp;&nbsp; <span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">292</td></tr><tr><td>168</td></tr></table>=<table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">x</td></tr><tr><td>1940t</td></tr></table></span>，<br />&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; x≈2679.05t<br />需磁铁矿的质量为2679.05t÷90%≈2976.72t<br />故答案为：2976.72t．','【分析】（1）从洗涤剂能使植物油在水中分散成无数细小的液滴去分析解答；<br />（2）从生石灰能和水反应生成氢氧化钙去分析解答；<br />（3）根据方程式的写法考虑；<br />（4）根据反应物、生成物以及反应的条件就可以正确写出有关反应的化学方程式；<br />（5）根据所给的反应规律可知醋酸与碳酸镁反应的生成物是醋酸镁和水，写出反应的化学方程式即可；<br />（6）由生铁的质量和杂质的质量分数可以计算出铁的质量；由铁的质量根据一氧化碳与四氧化三铁反应的化学方程式可以计算出四氧化三铁的质量；由四氧化三铁和磁铁矿中四氧化三铁的质量分数可以计算出磁铁矿的质量．','书写',3.00,'723b66c6d1499b3438477d87a98b885b',9,400,'乳化现象与乳化作用,含杂质物质的化学反应的有关计算,书写化学方程式、文字表达式、电离方程式','',2013,'37','2013•广东校级一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840082,'某研究性学习小组的同学认为“保护环境，人人有责”，他们在课外活动时对“二氧化硫能否与水反应生成酸”进行了实验探究．请你参与他们的探究活动，并回答有关问题．<br />【查资料】常温下二氧化硫是一种无色气体，易溶于水．<br />【假设】二氧化硫能与水反应生成酸．<br />【设计实验】<br />（1）小雨同学将二氧化硫气体通入紫色石蕊试液中，观察溶液颜色的变化．他认为，若紫色石蕊试液变<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>色，则证明假设成立．<br />（2）小涛同学认为小雨同学的实验方案不严密．小涛同学用紫色石蕊试液将白色滤纸染成紫色，干燥后做成三朵紫色的小花，然后按下列图示进行实验，在（I）、（Ⅱ）、（Ⅲ）中分别观察到变色和不变色的两种现象．<br /><img  src=\"/tikuimages/9/0/400/shoutiniao44/137bd14f-94d4-11e9-ab00-b42e9921e93e_xkb22.png\" style=\"vertical-align:middle\" /><br />小涛同学认为，若<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填序号）中的紫色小花变色，则证明假设成立．<br />【交流与反思】小明同学对小涛同学的实验方案提出了质疑，他认为上述实验还不足以证明“二氧化硫和水反应生成了酸”，其理由是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，三位同学讨论后补充了一个实验，使探究活动获得了成功，你认为这个实验应该是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．（简述实验步骤和现象）．<br />【结论】通过上述实验得出：二氧化硫能与水反应生成酸．','','','','','','红$###$Ⅲ$###$SO<SUB>2</SUB>也可能使干燥的紫色小花变红$###$干燥的紫色小花投入到盛有SO<SUB>2</SUB>的集气瓶中，紫色小花不变红色（或将SO<SUB>2</SUB>气体通过盛有紫色小花的瓶中，紫色小花不变红色）','【解答】解：（1）根据酸性溶液可使紫色石蕊试液变红；<br />（2）小涛的设计思路是：I喷稀醋酸小花变红可证明酸性溶液可使石蕊变红，Ⅱ喷水小花不变色可证明水不能使石蕊变红，Ⅲ小花喷水后放入二氧化硫，若变红即可证明二氧化硫溶于水后生成了酸；<br />【交流与反思】依据课本上证明二氧化碳是否溶于水生成酸的实验设计，可以发现小涛的实验存在漏洞：未证明二氧化硫本身能否使石蕊变红；<br />（4）为了证明二氧化硫能否使石蕊变红，要排除水的干扰，所以设计为：将干燥的紫色小花投入到盛有SO<SUB>2</SUB>的集气瓶中，紫色小花不变红色．<br />故答案为：（1）红；<br />（2）Ⅲ；<br />交流与反思】SO<SUB>2</SUB>也可能使干燥的紫色小花变红；<br />将干燥的紫色小花投入到盛有SO<SUB>2</SUB>的集气瓶中，紫色小花不变红色（或将SO<SUB>2</SUB>气体通过盛有紫色小花的瓶中，紫色小花不变红色）','【分析】本题设计紧密联系生活实际，视角新颖，这类题通常起点高，落点低，可用类比的方法寻找与该物质性质相似的课本中的物质．观察图示，大家很容易就回忆起二氧化碳曾做过类似的实验，联想验证二氧化碳溶于水的实验设计思路即可解答．','填空题',3.00,'7ea7b19a50900e688b446cf26d8f106c',9,400,'实验探究物质的性质或变化规律','',0,'37','',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840085,'以锶原子为钟摆的“晶格钟”是当今世界最精准的钟．结合如图分析，下列说法错误的是（　　）<img  src=\"/tikuimages/9/2016/400/shoutiniao86/138c2500-94d4-11e9-a28c-b42e9921e93e_xkb90.png\" style=\"vertical-align:middle\" />','Sr元素在化学反应中易得电子','图中X=2','Sr为金属元素','锶原子的质子数是38','','A','【解答】解：A、根据元素周期表中的一格可知，左上角的数字为38，表示原子序数为38；根据原子序数=核电荷数=质子数=核外电子数，则该元素的原子核外电子数为38，2+8+18+8+X=38，X=2，在化学反应中易失去电子，故选项说法错误．<br />B、由A选项的分析，图中X=2，故选项说法正确．<br />C、Sr是锶元素的元素符号，带“钅”字旁，属于金属元素，故选项说法正确．<br />D、由锶原子的结构示意图，圆圈内数字表示核内质子数，则锶原子的质子数是38，故选项说法正确．<br />故选：A．','【分析】根据图中元素周期表可以获得的信息：左上角的数字表示原子序数；字母表示该元素的元素符号；中间的汉字表示元素名称；汉字下面的数字表示相对原子质量；原子结构示意图中，圆圈内数字表示核内质子数，弧线表示电子层，弧线上的数字表示该层上的电子数，离圆圈最远的弧线表示最外层．若原子的最外层电子数≥4，在化学反应中易得电子，若最外层电子数＜4，在化学反应中易失去电子．','选择题',3.00,'b5c6f883a13b7deb1d2bbf07e03d006d',9,400,'原子结构示意图与离子结构示意图,元素周期表的特点及其应用','',2016,'37','2016•滨海县二模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840088,'下列图象分别与选项中的操作相对应，其中不合理的是（　　）','<img  src=\"/tikuimages/9/2016/400/shoutiniao60/139f10c0-94d4-11e9-a80d-b42e9921e93e_xkb66.png\" style=\"vertical-align:middle\" />','<img  src=\"/tikuimages/9/2016/400/shoutiniao49/13a26c21-94d4-11e9-b953-b42e9921e93e_xkb25.png\" style=\"vertical-align:middle\" />向一定量氯化铜溶液中加入铝','<img  src=\"/tikuimages/9/2016/400/shoutiniao79/13a32f6e-94d4-11e9-bb9c-b42e9921e93e_xkb41.png\" style=\"vertical-align:middle\" />向一定量二氧化锰固体中加入过氧化氢溶液','<img  src=\"/tikuimages/9/2016/400/shoutiniao97/13a5c780-94d4-11e9-a39f-b42e9921e93e_xkb57.png\" style=\"vertical-align:middle\" />向等质量、等浓度的稀硫酸中分别加入锌粉和铁粉','','D','【解答】解：A、向一定量氢氧化钠溶液中加入水，碱性减弱，pH减小，不会小于等于7，正确；<br />B、向一定量氯化铜溶液中加入铝，铝与氯化铜反应生成氯化铝和铜<br />2Al+3CuCl<SUB>2</SUB>=2AlCl<SUB>3</SUB>+3Cu<br />溶液质量减小，氯化铜完全反应则溶液质量不变，正确；<br />C、向一定量二氧化锰固体中加入过氧化氢溶液，二氧化锰是该反应的催化剂，质量不变，正确；<br />D、向等质量、等浓度的稀硫酸中分别加入锌粉和铁粉，等质量的金属铁比锌反应生成的氢气质量大，错误；<br />故选D．','【分析】A、根据碱溶液的稀释解答；<br />B、根据铝与氯化铜的反应解答；<br />C、根据二氧化锰是过氧化氢分解的催化剂解答；<br />D、根据锌和铁与稀硫酸的反应解答．','选择题',3.00,'6b7fd28f0bd3f069788eb1b4987d8bdb',9,400,'催化剂的特点与催化作用,金属的化学性质,酸碱溶液的稀释,溶液的酸碱性与pH值的关系','',2016,'37','2016•朝阳区一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840089,'请阅读下面短文，回答有关问题．<br />&nbsp;&nbsp;&nbsp; 人类创造肥皂的历史至少已有2300年了．早期的文明古国将肥皂用作染发剂或药物，直到公园100-199年，人们才逐渐认识到肥皂有去污作用，可用作清洗剂．起初人们利用草木灰（主要成分K<SUB>2</SUB>CO<SUB>3</SUB>）产生的碱液与油脂反应制造肥皂．大约到了1790年，法国科学家Nicolaos Leblanc发现碱可以用普通的食盐和水制得，这一发明使制皂变得更为容易．<br />&nbsp;&nbsp;&nbsp; 肥皂溶液显碱性，它的主要成分是高级脂肪酸纳（如硬脂肪酸纳C<SUB>17</SUB>H<SUB>35</SUB>COONa），它是通过酸（如硬脂酸C<SUB>17</SUB>H<SUB>35</SUB>COOH）和碱反应而得到的产品．<br />&nbsp;&nbsp;&nbsp; 肥皂有许多种类，如：加入Na<SUB>2</SUB>CO<SUB>3</SUB>等物质后的白色洗衣皂去污能力更强；在肥皂中加入硫磺即得到具有消炎、杀菌、控油作用的药皂-硫磺皂；香皂则是在肥皂中加入某些香料制成，其碱性较弱，对皮肤的刺激性较小．<br />（1）若用硬水洗衣服，当肥皂溶于水后产生的现象是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（2）硬脂酸中氢、氧元素的质量比是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（3）有皮肤病的人适宜使用的肥皂是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；白色洗衣皂中与去污能能力增强有关的离子是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','泡沫少，浮渣多$###$9：8$###$硫磺皂$###$CO<SUB>3</SUB><SUP>2-</SUP>','【解答】解：（1）在硬水中加入肥皂水，会形成大量浮渣，没有泡沫，衣物洗不干净；<br />故答案为：泡沫少，浮渣多；<br />（2）硬脂酸中氢、氧元素的质量比为（36×1）：（2×16）=9：8．<br />故答案为：9：8；<br />（3）在肥皂中加入硫磺即得到具有消炎、杀菌、控油作用的药皂-硫磺皂；故有皮肤病的人适宜使用硫磺皂；白色洗衣皂中与去污能能力增强有关的离子是CO<SUB>3</SUB><SUP>2-</SUP>；<br />故答案为：硫磺皂；CO<SUB>3</SUB><SUP>2-</SUP>．','【分析】（1）根据硬水加入肥皂水会产生大量浮渣，无泡沫进行分析；<br />（2）根据元素质量比计算方法进行分析；<br />（3）根据硫磺皂的作用进行分析；根据白色洗衣皂中与去污能能力增强有关的离子进行分析．','填空题',3.00,'34b973b52984f5fa85bff666eec3a164',9,400,'硬水与软水,元素质量比的计算','',2016,'37','2016•太原二模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840093,'如图1，鸡蛋主要由蛋壳、蛋壳膜、蛋白、蛋黄等几个部分组成，其中蛋壳的主要成分是CaCO<SUB>3</SUB>，里面覆盖一层蛋壳膜．某校兴趣小组的同学为此开展了以下探究活动：<br />【提出问题1】<br />（1）鸡蛋白中主要的营养成分是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（2）蛋壳中的CaCO<SUB>3</SUB>含量是多少呢？<br />【查阅资料】<br />a、CaCO<SUB>3</SUB>灼烧分解生成氧化钙和一种无毒的氧化物．<br />b、蛋壳中的其它成分受热不分解，也不与稀盐酸反应．<br />【设计方案】他们分别称取12.0g蛋壳设计了以下不同的实验方案：<br />（1）小华的方案：<br />称取12.0g蛋壳研磨成粉末，灼烧至质量不再减少，再称量剩余固体的质量为7.16g．则：减少的质量是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填化学式）的质量，经计算得蛋壳中的CaCO<SUB>3</SUB>的质量分数为91.7%．<br />（2）小明的方案：小明设计了如图2所示的实验装置．已知仪器a装有10%的稀盐酸，装置内试剂均足量．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao63/13af164f-94d4-11e9-b202-b42e9921e93e_xkb22.png\" style=\"vertical-align:middle\" /><br />【实验步骤】<br />①按如图2连接好装置后，并<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />②向B装置中加入蛋壳样品后，打开K1关闭K2，通入空气一会儿；<br />③接下来的实验操作是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，直到<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>为止（填实验现象）；<br />④打开K1关闭K2，再通入空气一会儿，至装置C中不再产生沉淀为止；<br />⑤将装置C中的固液混合物过滤、洗涤、烘干后称量其质量；<br />⑥重复上述实验．<br />【实验数据】重复实验，3次数据记录如下：<br /><table class=\"edittable\"><TBODY><TR><td width=232>实验次数</TD><td width=97>实验1</TD><td width=97>实验2</TD><td width=97>实验3</TD></TR><TR><td>装置C中沉淀质量（g）</TD><td>19.68</TD><td>19.75</TD><td>19.67</TD></TR></TBODY></TABLE>【实验分析及数据处理】<br />①上述数据是否为托盘天平称量得到的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“是”或“不是”）．<br />②若无A装置，直接通入空气，则测定结果将<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“偏大”、“偏小”或“不变”）．<br />③利用3次实验数据的平均值，计算该蛋壳中的CaCO<SUB>3</SUB>质量分数为83.3%．<br />【实验反思】<br />下列各项措施中，能提高小明测定准确度的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填字母）．<br />A．缓缓滴加稀盐酸<br />B．在装置A、B之间增加盛有浓硫酸的洗气瓶<br />C．在装置B、C之间增加盛有饱和NaHCO<SUB>3</SUB>的洗气瓶．','','','','','','蛋白质$###$CO<SUB>2</SUB>$###$检查装置的气密性$###$关闭K<SUB>1</SUB>，打开K<SUB>2</SUB>，加稀盐酸$###$不产生气泡$###$不是$###$偏大$###$A','【解答】解：【提出问题1】<br />（1）鸡蛋白中主要的营养成分是蛋白质．<br />故填：蛋白质．<br />【设计方案】<br />称取12.0g蛋壳研磨成粉末，灼烧至质量不再减少，再称量剩余固体的质量为7.16g．则：减少的质量是CO<SUB>2</SUB>的质量，经计算得蛋壳中的CaCO<SUB>3</SUB>的质量分数为91.7%．<br />故填：CO<SUB>2</SUB>．<br />【实验步骤】<br />①按如图2连接好装置后，并检查装置的气密性．<br />故填：检查装置的气密性．<br />③接下来的实验操作是关闭K<SUB>1</SUB>，打开K<SUB>2</SUB>，加稀盐酸，直到不产生气泡为止．<br />故填：关闭K<SUB>1</SUB>，打开K<SUB>2</SUB>，加稀盐酸；不产生气泡．<br />【实验分析及数据处理】<br />①上述数据不是托盘天平称量得到的，因为还有其它物质．<br />故填：不是．<br />②若无A装置，直接通入空气，则空气中的二氧化碳能被C装置吸收，则测定结果将偏大．<br />故填：偏大．<br />【实验反思】<br />A．缓缓滴加稀盐酸时，能使产生的二氧化碳被C装置充分吸收，能提高小明测定准确度；<br />B．在装置A、B之间增加盛有浓硫酸的洗气瓶时，对测定准确度不产生影响；<br />C．在装置B、C之间增加盛有饱和NaHCO<SUB>3</SUB>的洗气瓶时，对测定准确度不产生影响．<br />故填：A．','【分析】【提出问题1】<br />鸡蛋白中主要的营养成分是蛋白质；<br />【查阅资料】<br />称取12.0g蛋壳研磨成粉末，灼烧至质量不再减少，再称量剩余固体的质量为7.16g．则：减少的质量是二氧化碳的质量；<br />【实验步骤】<br />按如图2连接好装置后，并检查装置的气密性；<br />【实验分析及数据处理】<br />A装置能够除去空气中的二氧化碳；<br />【实验反思】<br />缓缓滴加稀盐酸时，能使产生的二氧化碳被C装置充分吸收．','填空题',3.00,'57489c5b7e1ba3b1fe43342ae0bd5fe3',9,400,'实验探究物质的组成成分以及含量,常见气体的检验与除杂方法,盐的化学性质,食品、药品与健康食品中的有机营养素','',2016,'37','2016•玄武区一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840094,'化学就在我们身边．请运用所学的化学知识，回答下列问题：<br />（1）白酒的主要成分是乙醇，其化学式为C<SUB>2</SUB>H<SUB>5</SUB>OH，则乙醇由<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>种元素组成，乙醇燃烧的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）铁制品锈蚀实际上是铁与空气中的氧气和<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>等发生了化学反应．用稀盐酸可以除去铁制品表面的铁锈，其反应的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）我们常用洗涤剂清洗餐具上的油污，这是因为洗涤剂具有<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>功能．<br />（4）长期饮用硬水对人体健康不利，生活中可用<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>来区别硬水和软水．','','','','','','三$###$C<SUB>2</SUB>H<SUB>5</SUB>OH+3O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2CO<SUB>2</SUB>+3H<SUB>2</SUB>O$###$水$###$Fe<SUB>2</SUB>O<SUB>3</SUB>+6HCl═2FeCl<SUB>3</SUB>+3H<SUB>2</SUB>O$###$乳化$###$肥皂水','【解答】解：（1）由乙醇的化学式C<SUB>2</SUB>H<SUB>5</SUB>OH可知，它是由三种元素组成的；乙醇燃烧生成水和二氧化碳，其化学方程式为：C<SUB>2</SUB>H<SUB>5</SUB>OH+3O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2CO<SUB>2</SUB>+3H<SUB>2</SUB>O．<br />故填：三；C<SUB>2</SUB>H<SUB>5</SUB>OH+3O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2CO<SUB>2</SUB>+3H<SUB>2</SUB>O；<br />（2）铁在空气中锈蚀，实际上是铁跟空气中的氧气和水共同作用的结果．铁锈的主要成分是氧化铁，与盐酸反应生成氯化铁和水，反应的化学方程式是：Fe<SUB>2</SUB>O<SUB>3</SUB>+6HCl═2FeCl<SUB>3</SUB>+3H<SUB>2</SUB>O；<br />故填：水；Fe<SUB>2</SUB>O<SUB>3</SUB>+6HCl═2FeCl<SUB>3</SUB>+3H<SUB>2</SUB>O；<br />（3）洗洁精具有乳化作用，能使油以小液滴的形式分散到水中，形成较稳定的乳浊液，从而除去油污；<br />故填：乳化；<br />（4）生活中可用肥皂水鉴别硬水和软水，泡沫多的是软水，泡沫少的是硬水；<br />故填：肥皂水．','【分析】（1）根据乙醇的化学式来分析解答；根据乙醇燃烧生成水和二氧化碳写出化学方程式即可；<br />（2）铁在空气中锈蚀，实际上是铁跟空气中的氧气和水共同作用的结果；铁锈的主要成分是氧化铁，与稀盐酸反应生成氯化铁和水，据此进行分析解答；<br />（3）根据洗洁精的乳化作用进行分析；<br />（4）根据生活中区别硬水和软水的方法进行分析．','书写',3.00,'618ee6da665299a73d7f7bbda3bb91d8',9,400,'硬水与软水,乳化现象与乳化作用,金属锈蚀的条件及其防护,甲烷、乙醇等常见有机物的性质和用途,书写化学方程式、文字表达式、电离方程式','',2016,'37','2016•郯城县校级一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840095,'如表，除去物质所含杂质的方法正确的是（　　）<br /><table class=\"edittable\"><TBODY><TR><td width=119>　　　　　　　选项</TD><td width=157>物质</TD><td width=112>所含杂质</TD><td width=228>除去杂质的方法</TD></TR><TR><td>A</TD><td>H<SUB>2</SUB>气体</TD><td>HCl气体</TD><td>通过NaOH溶液，再通过浓硫酸</TD></TR><TR><td>B</TD><td>NaCl溶液</TD><td>Na<SUB>2</SUB>SO<SUB>4</SUB></TD><td>加入过量氯化钡溶液，过滤</TD></TR><TR><td>C</TD><td>CO</TD><td>CO<SUB>2</SUB></TD><td>通过灼热的Fe<SUB>2</SUB>O<SUB>3</SUB></TD></TR><TR><td>D</TD><td>CaCO<SUB>3</SUB>粉末</TD><td>CaCl<SUB>2</SUB></TD><td>加足量水溶解、过滤、洗涤、干燥</TD></TR></TBODY></TABLE>','A','B','C','D','','A|D','【解答】解：A、HCl气体能与氢氧化钠溶液反应生成氯化钠和水，氢气不与氢氧化钠溶液反应，再通过浓硫酸进行干燥，能除去杂质且没有引入新的杂质，符合除杂原则，故选项所采取的方法正确．<br />B、Na<SUB>2</SUB>SO<SUB>4</SUB>能与过量氯化钡溶液反应生成硫酸钡沉淀和氯化钠，能除去杂质但引入了新的杂质氯化钡（过量的），不符合除杂原则，故选项所采取的方法错误．<br />C、CO能与灼热的Fe<SUB>2</SUB>O<SUB>3</SUB>反应铁和二氧化碳，反而会把原物质除去，不符合除杂原则，故选项所采取的方法错误．<br />D、CaCl<SUB>2</SUB>易溶于水，CaCO<SUB>3</SUB>粉末难溶于水，可采取加水溶解、过滤、洗涤、干燥的方法进行分离除杂，故选项所采取的方法正确．<br />故选：AD．','【分析】根据原物质和杂质的性质选择适当的除杂剂和分离方法，所谓除杂（提纯），是指除去杂质，同时被提纯物质不得改变．除杂质题至少要满足两个条件：①加入的试剂只能与杂质反应，不能与原物质反应；②反应后不能引入新的杂质．','多选题',3.00,'10c4cfd6e927ebbd30b35d51ce751208',9,400,'物质除杂或净化的探究,常见气体的检验与除杂方法,酸的化学性质,盐的化学性质','',2016,'37','2016•青岛二模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840096,'富硒大米中的“硒”是指（　　）','原子','分子','元素','单质','','C','【解答】解：富硒大米中的“硒”不是以单质、分子、原子等形式存在，这里所指的“硒”是强调存在的元素，与具体形态无关．<br />故选：C．','【分析】食品、药品、营养品、矿泉水等物质中的“硒”等不是以单质、分子、原子等形式存在，而是指元素，通常用元素及其所占质量（质量分数）来描述．','选择题',3.00,'bba19d451aac566475ff0ea441e80715',9,400,'元素的概念','高邮市',2016,'37','2016•高邮市一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840097,'现有两种没有标签的无色溶液，分别是碳酸钠溶液和澄清石灰水．为了鉴别这两种溶液，小明分别取两种溶液于试管中，然后分别滴加某种试剂，第一支试管产生气泡，第二支试管无明显现象．<br />（1）小明加入的试剂是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（2）第二支试管中的溶液是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（3）写出第一支试管中反应的化学方程式及反应类型．<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','稀盐酸等$###$澄清石灰水$###$Na<SUB>2</SUB>CO<SUB>3</SUB>+2HCl═2NaCl+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑；复分解反应','【解答】解：（1）碳酸钠溶液与稀盐酸反应生成氯化钠、水和二氧化碳，与石灰水反应生成氯化钙和水，但无明显变化，可以鉴别．<br />（2）第二支试管中的溶液是澄清石灰水．<br />（3）碳酸钠溶液与稀盐酸反应生成氯化钠、水和二氧化碳，反应的化学方程式为：Na<SUB>2</SUB>CO<SUB>3</SUB>+2HCl═2NaCl+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑；该反应是两种化合物相互交换成分生成两种新的化合物的反应，属于复分解反应．<br />故答案为：（1）稀盐酸等；（2）澄清石灰水；（3）Na<SUB>2</SUB>CO<SUB>3</SUB>+2HCl═2NaCl+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑；复分解反应．','【分析】鉴别物质时，首先对需要鉴别的物质的性质进行对比分析找出特性，再根据性质的不同，选择适当的试剂，出现不同的现象的才能鉴别．','书写',3.00,'ec51f30f3acaec13373bc4928d1ec840',9,400,'酸、碱、盐的鉴别,反应类型的判定,书写化学方程式、文字表达式、电离方程式','',2016,'32','2016•海口模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840099,'金属在日常生活中应用广泛．<br />（1）下列金属制品中，主要利用其导电性的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填字母序号）．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao16/13bea6b0-94d4-11e9-9480-b42e9921e93e_xkb94.png\" style=\"vertical-align:middle\" /><br />（2）锌能促进人体的生长发育，我国在明朝末年就已开始用氧化锌和硫化锌治病，氧化锌的化学式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','C$###$ZnO','【解答】解：（1）铜制成导线是利用铜的导电性，故填：C；<br />（2）氧化锌的化学式为：ZnO，故填：ZnO．','【分析】根据已有的金属的性质结合物质化学式的书写的知识进行分析解答即可．','书写',3.00,'b3d847c1acfcb260ff54b0455017619d',9,400,'金属的物理性质及用途,化学式的书写及意义','',2016,'37','2016•石景山区一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840100,'化学就在我们身边，它能改善我们的生活．请从下列物质中，选择适当的物质填空（填序号）<br />①干冰 ②肥皂水 ③食盐 ④硝酸钾<br />（1）炒菜调味的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；&nbsp;&nbsp;&nbsp;&nbsp; （2）属于复合肥料的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（3）可用于人工降雨的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；（4）可用于区别硬水和软水的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','③$###$④$###$①$###$②','【解答】解：（1）食盐的主要成分是氯化钠，具有咸味，可用于炒菜调味．<br />（2）硝酸钾中含有钾元素和氮元素，属于复合肥．<br />（3）干冰（固体的二氧化碳）升华时吸收大量的热，可用于人工降雨．<br />（4）硬水和软水的区别在于所含的钙镁离子的多少，可用等量的肥皂水进行鉴别，产生泡沫较多的是软水，较少的是硬水．<br />故答案为：（1）③；（2）④；（3）①；（4）②．','【分析】物质的性质决定物质的用途，根据食盐的主要成分是氯化钠，时含有氮、磷、钾三种元素中的两种或两种以上的肥料称为复合肥，干冰（固体的二氧化碳）升华时吸收大量的热，硬水和软水的区别在于所含的钙镁离子的多少，进行分析解答．','填空题',3.00,'ddc051d91da05425b674e28ee2b73907',9,400,'二氧化碳的用途,硬水与软水,氯化钠与粗盐提纯,常见化肥的种类和作用','',2016,'37','2016春•石城县校级月考',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840101,'下列实验方案能达到预期目的是（　　）<br /><table class=\"edittable\"><TBODY><TR><td width=38>编号</TD><td width=95>A</TD><td width=127>B</TD><td width=165>C</TD><td width=120>D</TD></TR><TR><td>实验<br />方案</TD><td><img  src=\"/tikuimages/9/2016/400/shoutiniao28/13c4c12e-94d4-11e9-90b9-b42e9921e93e_xkb39.png\" style=\"vertical-align:middle\" /></TD><td><img  src=\"/tikuimages/9/2016/400/shoutiniao28/13c55d70-94d4-11e9-8f61-b42e9921e93e_xkb58.png\" style=\"vertical-align:middle\" /></TD><td><img  src=\"/tikuimages/9/2016/400/shoutiniao81/13c66ee1-94d4-11e9-a177-b42e9921e93e_xkb50.png\" style=\"vertical-align:middle\" /></TD><td><img  src=\"/tikuimages/9/2016/400/shoutiniao58/13cd73c0-94d4-11e9-8c70-b42e9921e93e_xkb65.png\" style=\"vertical-align:middle\" /></TD></TR><TR><td>实验<br />目的</TD><td>检验NaOH溶液是否变质</TD><td>证明分子在不断运动</TD><td>测定空气中氧气的含量</TD><td>检验鸡蛋壳中是否含有CO<SUB>3</SUB><SUP>2-</SUP></TD></TR></TBODY></TABLE>','A','B','C','D','','D','【解答】解：A、无论氢氧化钠是否变质，溶液都显碱性，都能够使酚酞试液变红色，该选项不能达到实验目的；<br />B、浓盐酸中的氯化氢分子是不断运动的，当运动到A烧杯中时，能和水结合成盐酸，盐酸显酸性，不能使酚酞试液变红色，因此该实验不能证明微观粒子是不断运动的，该选项不能达到实验目的；<br />C、木炭燃烧生成二氧化碳气体，虽然集气瓶中的氧气被消耗，但是生成的二氧化碳气体会占据氧气的体积，从而不能使水进入集气瓶中，因此不能测定空气中氧气的含量，该选项不能达到实验目的；<br />D、碳酸钙和稀盐酸反应生成二氧化碳，生成的二氧化碳气体能使玻璃片上的澄清石灰水变浑浊，说明鸡蛋壳中含有碳酸根离子，该选项能够达到实验目的．<br />故选：D．','【分析】氢氧化钠能和空气中的二氧化碳反应生成碳酸钠和水，碳酸钠和氢氧化钠的水溶液都显碱性，都能够使酚酞试液变红色；<br />微观粒子是不断运动的；<br />木炭燃烧生成二氧化碳气体；<br />碳酸钙和稀盐酸反应生成氯化钙、水和二氧化碳，二氧化碳能使澄清石灰水变浑浊．','选择题',3.00,'d1dd06fb92ca2e5ab2739c842d09e607',9,400,'化学实验方案设计与评价,证明碳酸盐,空气组成的测定,碱的化学性质,分子的定义与分子的特性','',2016,'37','2016•商水县一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840102,'下列关于碳及其化合物的说法，正确的是（　　）','金刚石和石墨由于碳原子排列方式不同，所以物理性质有很大差异','CO和CO<SUB>2</SUB>都能与NaOH溶液反应','CO<SUB>2</SUB>的排放是导致酸雨形成的原因之一','在室内放一盆澄清石灰水可防止CO中毒','','A','【解答】解：A、金刚石和石墨由于碳原子排列方式不同，所以物理性质有很大差异，故选项说法正确．<br />B、CO<SUB>2</SUB>能与NaOH溶液反应生成碳酸钠和水，CO不能与氢氧化钠溶液反应，故选项说法错误．<br />C、CO<SUB>2</SUB>的排放是导致温室效应的原因之一，故选项说法错误．<br />D、一氧化碳不能与石灰水反应，在室内放一盆澄清石灰水不能防止CO中毒，故选项说法错误．<br />故选：A．','【分析】A、根据金刚石和石墨物理性质差异较大的原因，进行分析判断．<br />B、根据一氧化碳与二氧化碳的化学性质，进行分析判断．<br />C、根据二氧化碳对环境的影响，进行分析判断．<br />D、根据一氧化碳不能与石灰水反应，进行分析判断．','选择题',3.00,'c0642565e103409f6a5d1e987afba4ad',9,400,'二氧化碳的化学性质,二氧化碳对环境的影响,一氧化碳的物理性质,一氧化碳的化学性质,碳单质的物理性质及用途','湘乡市',2016,'35','2016春•湘乡市校级期中',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840104,'下列实验操作正确的是（　　）','把浓硫酸注入盛有水的量筒内稀释','用镊子取用块状固体药品','将pH试纸浸入待测液中测定pH','将实验剩余的药品放回原试剂瓶中','','B','【解答】解：A、稀释浓硫酸时，要把浓硫酸缓缓地沿器壁注入水中，同时用玻璃棒不断搅拌，以使热量及时地扩散；一定不能把水注入浓硫酸中；不能在量筒内稀释浓硫酸，故选项说法错误．<br />B、可用镊子取用块状固体药品，故选项说法正确．<br />C、用pH试纸测定未知溶液的pH时，正确的操作方法为用玻璃棒蘸取少量待测液滴在干燥的pH试纸上，与标准比色卡对比来确定pH．不能将pH试纸伸入待测液中，以免污染待测液，故选项说法错误．<br />D、对化学实验中的剩余药品，既不能放回原瓶，也不可随意丢弃，更不能带出实验室，应放入的指定的容器内，故选项说法错误．<br />故选：B．','【分析】A、根据浓硫酸的稀释方法（酸入水，沿器壁，慢慢倒，不断搅）进行分析判断．<br />B、根据固体药品的取用方法进行分析判断．<br />C、根据用pH试纸测定未知溶液的pH的方法进行分析判断．<br />D、根据实验室剩余药品的处理原则（三不一要），进行分析判断．','选择题',3.00,'8d947f0cb15fb6af28b08a58c8c87d6c',9,400,'固体药品的取用,浓硫酸的性质及浓硫酸的稀释,溶液的酸碱度测定','',2016,'32','2016•泉州模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840105,'能源在人类社会发展过程中，起着重要的作用．<br />（1）我国的原油储量仅占世界的2.43%，天然气储量只占世界的1.20%．天然气的主要成分为（写化学式）<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，它与石油和<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>均属于化石燃料．<br />（2）氢气是理想的清洁能源，水是未来氢气之源．发射“嫦娥一号”火箭上用的燃料是液态氢，以氢燃料电池为动力的汽车已在北京试运行．<br />①写出氢气在空气中燃烧的化学方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />②写出以水为原料制取氢气的化学方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）人们正在利用和开发太阳能、核能及<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>另填一种）等新能源．其中氘的核能开发，被认为是解决未来世界能源、环境等问题的主要途径之一．已知氘和氢（氢的核电荷数为1）是同种元素，则氘原子核内的质子数为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','CH<SUB>4</SUB>$###$煤$###$2H<SUB>2</SUB>+O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O$###$2H<SUB>2</SUB>O<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;通电&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>↑+O<SUB>2</SUB>↑$###$风能$###$1','【解答】解：（1）天然气的主要成分为甲烷．故填：CH<SUB>4</SUB>．天然气、煤、石油属于化石燃料．故填：煤．<br />（2）①氢气在空气中燃烧的化学方程式为：2H<SUB>2</SUB>+O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O．<br />②以水为原料制取氢气的化学方程式为：2H<SUB>2</SUB>O<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;通电&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>↑+O<SUB>2</SUB>↑．<br />（3）人们正在利用和开发太阳能、核能及风能（或地热能、潮汐能等）等新能源．故填：风能．<br />已知氘和氢（氢的核电荷数为1）是同种元素，同种元素的原子的原子核内质子数相等．故填：1．<br />故答案为：（1）CH<SUB>4</SUB>；煤；（2）①2H<SUB>2</SUB>+O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O；②2H<SUB>2</SUB>O<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;通电&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>↑+O<SUB>2</SUB>↑；（3）风能；1．','【分析】（1）三大化石燃料是煤石油天然气，其中天然气的主要成分是甲烷．<br />（2）用元素符号和数字可以表示物质的组成，氢气能燃烧生成水，水电解能生成氢气和氧气；<br />（3）化石燃料主要有石油煤天然气等，是当今世界普遍使用的能源，新能源则是不同于化石燃料的能源，①地热能②潮汐能均属新型能源，元素是具有相同核电荷属或质子数的一类原子的总称，故元素相同质子数一定相同，可以据此答题．','书写',3.00,'752affa2d4b6f65a0f5b9d01cb6cc36a',9,400,'原子的有关数量计算,书写化学方程式、文字表达式、电离方程式,化石燃料及其综合利用,资源综合利用和新能源开发,氢气的化学性质与燃烧实验','',2016,'32','2016•揭西县模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840106,'下列实验现象不正确的是（　　）','磷在空气中燃烧时产生大量白烟','一氧化碳在空气中燃烧时产生淡蓝色火焰','铁与稀盐酸反应时铁表面有气泡，溶液由无色变浅绿色','氢氧化钠溶液与硫酸铜溶液反应产生白色沉淀','','B|D','【解答】解：A、磷在空气中燃烧时产生大量白烟，故选项说法正确．<br />B、一氧化碳在空气中燃烧时，产生蓝色火焰，故选项说法错误．<br />C、铁与稀盐酸反应时生成氯化亚铁溶液和氢气，会观察到铁表面有气泡，溶液由无色变浅绿色，故选项说法正确．<br />D、氢氧化钠溶液与硫酸铜溶液反应，产生蓝色沉淀，故选项说法错误．<br />故选：BD．','【分析】A、根据磷在空气中燃烧的现象，进行分析判断．<br />B、根据一氧化碳在空气中燃烧的现象，进行分析判断．<br />C、根据金属的化学性质，进行分析判断．<br />D、根据碱的化学性质进行分析判断．','多选题',3.00,'75256924f6c7f5d0660d0ab0317bba84',9,400,'氧气与碳、磷、硫、铁等物质的反应现象,金属的化学性质,碱的化学性质','',2016,'37','2016•黑龙江二模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840107,'<img  src=\"/tikuimages/9/2015/400/shoutiniao57/13e54180-94d4-11e9-8675-b42e9921e93e_xkb42.png\" style=\"vertical-align:middle;FLOAT:right;\" />（2015秋•柘城县校级月考）写出如图所示的二氧化碳灭火实验说明了二氧化碳的物理性质与化学性质各一条．<br />物理性质：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；化学性质：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','二氧化碳的密度比空气大$###$不燃烧，也不支持燃烧','【解答】解：因为二氧化碳的密度比空气大，不燃烧，也不支持燃烧．实验现象是：低处的蜡烛先熄灭，高处的蜡烛后熄灭．低处的蜡烛先熄灭，高处的蜡烛后熄灭，说明二氧化碳的密度比空气大，不燃烧，也不支持燃烧．<br />故答案为：二氧化碳的密度比空气大；不燃烧，也不支持燃烧．','【分析】根据量碳酸钠和盐酸反应生成二氧化碳气体，二氧化碳密度比空气密度大，以及不燃烧，也不支持燃烧的特点分析．','填空题',3.00,'d96611c74a3d411de361ac81febd6537',9,400,'二氧化碳的物理性质,二氧化碳的化学性质','',2015,'37','2015秋•柘城县校级月考',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840109,'<img  src=\"/tikuimages/9/2016/400/shoutiniao47/13f3c070-94d4-11e9-8727-b42e9921e93e_xkb26.png\" style=\"vertical-align:middle;FLOAT:right\" />如图是制取氨气的装置图，原料为熟石灰和氯化铵固体．下列说法正确的是（　　）','直接给试管加热，不必进行预热','氨气的密度比空气大','将一湿润的蓝色石蕊试纸放在试管口，若变红，证明已满','铵态氮肥不能与碱性物质混合使用，以防氮肥损失','','D','【解答】A、加热前未给试管进行预热，容易造成试管受热不均而炸裂，故错误．<br />B、从图中看出，氨气用向下排空气法收集，故氨气的密度比空气小，故错误；<br />C、将一湿润的红色石蕊试纸放在试管口，若变蓝，证明已满，故错误；<br />D、因为铵态氮肥与碱性物质反应生成氨气，降低肥效，故正确．<br />故选：D．','【分析】A、根据给试管内固体加热的操作、注意事项等进行分析；<br />B、氨气密度比 空气小，用向下排空气法收集；<br />C、据氨气的检验方法解答；<br />D、根据铵态氮肥与碱性物质反应生成氨气分析','选择题',3.00,'07c1af14d5f95bcab0f20a1ab0e9acd8',9,400,'常用气体的发生装置和收集装置与选取方法,常见气体的检验与除杂方法','',2016,'32','2016•泰州模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840110,'今年夏天，我国西南地区遭遇持续干旱．全国人民团结一致，共同抗旱．<br /><img  src=\"/tikuimages/9/2015/400/shoutiniao94/13f80630-94d4-11e9-a575-b42e9921e93e_xkb80.png\" style=\"vertical-align:middle\" /><br />（1）有些村庄打深井取用地下水．检验地下水是硬水还是软水可用的物质是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）有些村民取浑浊的坑水作生活用水．有同学利用所学的知识将浑浊的坑水用如图所示的简易净水器进行净化，其中小卵石、石英沙的作用是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）如图所示的简易净水器<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>降低水的硬度（填“能”或者“不能”）．如果地下水硬度大，或者坑水中病原微生物过多，都可以采取<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>方法，来降低硬度和杀灭病原微生物．<br />（4）当你身处在地震灾区时，水资源受到严重污染，必须经过净化处理后才能饮用．下列各项与水的净化过程无关的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填标号）．<br />A．吸附&nbsp;B．消毒&nbsp;&nbsp;C．电解&nbsp;D．过滤．','','','','','','肥皂水$###$过滤$###$不能$###$加热煮沸$###$C','【解答】解：（1）硬水与肥皂水混合产生大量的浮渣，软水与肥皂水混合产生较多的泡沫，检验地下水是硬水还是软水，可用的物质是肥皂水，故答案为：肥皂水；<br />（2）在简易净水器中，小卵石和石英砂能阻止不溶性固体颗粒通过，起到了过滤的作用，故答案为：过滤；<br />（3）简易净水器不能降低水的硬度；地下水硬度大，或者坑水中病原微生物过多，通过加热煮沸可使可溶性钙、镁化合物转变为沉淀，使硬水软化，同时还可以杀灭细菌；故答案为：不能；加热煮沸；<br />（4）A、吸附可除去水中有色、有味的杂质，是净化水的一种常用方法；<br />B、消毒，可对水消毒杀菌，是净化水的一种常用方法；<br />C、电解是水分解的条件，不能净化水；<br />D、过滤可除去水中不溶性固体杂质，是净化水的一种常用方法．<br />故答案为：C．','【分析】（1）硬水与肥皂水混合产生大量的浮渣，软水与肥皂水混合产生较多的泡沫；<br />（2）在简易净水器中，小卵石和石英砂能阻止不溶性固体颗粒通过，起到了过滤的作用，活性炭的作用是吸附；<br />（3）根据简易净水器不能降低水的硬度，硬水降低硬度的方法（煮沸、蒸馏）进行分析解答；<br />（4）分析各项操作的过程，判断与自来水的净化过程无关的操作．','填空题',3.00,'52e21a208afee3eae139f1f07628b49a',9,400,'水的净化,硬水与软水','',2015,'33','2015秋•邢台校级期末',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840112,'“见著知微、见微知著”是化学思维方法．．<br />（1）从宏观知微观<br />50mL水与50mL乙醇混合后，溶液体积小于100mL，微观解释为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）从微观知宏观<br />①微粒A在化学反应中容易<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“得”或“失”）电子．<br />②微粒A、B、C、D、E中，对应单质化学性质最稳定的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填编号，下同），属于同一种元素的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>和<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />③甲和乙在高温高压、催化剂条件下反应生成丙、三种物质结构示意图如图2所示，该反应的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao16/13ffce5e-94d4-11e9-8ed1-b42e9921e93e_xkb33.png\" style=\"vertical-align:middle\" />','','','','','','分子间有间隔$###$得$###$B$###$C$###$E$###$CO+2H<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;催化剂&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>高温高压</td></tr></table></span>CH<SUB>3</SUB>OH','【解答】解：（1）由于分子间有间隔，50mL水与50mL乙醇混合后，一部分乙醇分子和水分子相互占据了间隔，所以溶液体积小于100mL．<br />（2）①微粒A的最外层电子数是7，大于4，在化学反应中容易得电子．<br />②微粒A、B、C、D、E中，由于B的原子的最外层达到了稳定结构，对应单质化学性质最稳定；C和E的质子数相同，属于同一种元素．<br />③由微粒的构成的可知，该反应是一氧化碳和氢气在高温高压、催化剂条件下反应生成甲醇，反应的化学方程式为：CO+2H<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;催化剂&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>高温高压</td></tr></table></span>CH<SUB>3</SUB>OH．<br />故答为：（1）分子间有间隔；（2）①得到；&nbsp;②B；&nbsp;C；&nbsp;E；③CO+2H<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;催化剂&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>高温高压</td></tr></table></span>CH<SUB>3</SUB>OH．','【分析】（1）根据分子间有间隔的特征分析回答；<br />（2）①根据微粒A的最外层电子数目的特点分析回答．<br />②根据微粒的最外层电子数目的特点和元素的定义分析回答；<br />③根据微粒的构成分析反应物、生成物，写出反应的化学方程式．','书写',3.00,'8a9a8509702beffc3faa4fed17b2d666',9,400,'微粒观点及模型图的应用,原子结构示意图与离子结构示意图,利用分子与原子的性质分析和解决问题,书写化学方程式、文字表达式、电离方程式','',2016,'32','2016•梅州模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840113,'初中化学学习了化合反应、分解反应、置换反应和复分解反应等常见的基本反应类型，而高中化学我们将重点学习氧化还原反应，有化合价升降的反应就是氧化还原反应，氧化还原反应不是基本反应类型．下列反应是氧化还原反应但不属于基本反应类型的是（　　）','2KmnO<SUB>4</SUB>K<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2MnO<SUB>4</SUB>+MnO<SUB>2</SUB>+O<SUB>2</SUB>↑','2CO+O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2CO<SUB>2</SUB>','Fe<SUB>2</SUB>O<SUB>3</SUB>+3CO<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2Fe+3CO<SUB>2</SUB>','2CuO+C<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2Cu+CO<SUB>2</SUB>↑','','C','【解答】解：<br />A、2KMnO<SUB>4</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>K<SUB>2</SUB>MnO<SUB>4</SUB>+MnO<SUB>2</SUB>+O<SUB>2</SUB>↑属于分解反应，Mn、O元素的化合价变化，也是氧化还原反应，故A错误；<br />B、2CO+O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2CO<SUB>2</SUB>属于化合反应，C、O元素的化合价变化，也是氧化还原反应，故B错误；<br />C、Fe<SUB>2</SUB>O<SUB>3</SUB>+3CO<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2Fe+3CO<SUB>2</SUB>中C、Fe元素的化合价变化，是氧化还原反应，不属于基本反应类型，故C正确；<br />D、2CuO+C<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2Cu+CO<SUB>2</SUB>↑属于置换反应，C、Cu元素的化合价变化，也是氧化还原反应，故D错误．<br />故选：C．','【分析】根据含元素化合价变化的反应一定为氧化还原反应，以此来解答．','选择题',3.00,'725d56dd799c819cbed8ce7e3fdabd94',9,400,'氧化反应,还原反应','',2016,'37','2016•柳江县一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840117,'如图是初中化学常见的基本实验，回答下列问题．<br />（1）图1中，正极和负极产生的气体体积比约为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；质量比为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）图2中，观察到的现象是：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；该实验说明<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）若图3装置中大试管内黑色粉末逐渐变成了红色，该试管中的固体药品是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．酒精灯上的金属网作用是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />&nbsp;<img  src=\"/tikuimages/9/2016/400/shoutiniao23/1418ad91-94d4-11e9-892e-b42e9921e93e_xkb23.png\" style=\"vertical-align:middle\" />','','','','','','1：2$###$8：1$###$酚酞试液都变红，B比A变红的快$###$说明分子是不断运动的，温度越高分子运动速率越快$###$碳$###$氧化铜$###$集中火焰，提高温度','【解答】解：（1）图1是电解水的实验，正极和负极产生的气体分别是氧气和氢气，体积比约为 1：2；质量比为8：1．<br />（2）图2中，观察到的现象是：酚酞试液都变红，B比A变红的快；该实验说明分子是不断运动的，温度越高分子运动速率越快．<br />（3）由于在高温条件下，碳能还原氧化铜生成了铜和二氧化碳，若图3装置中大试管内黑色粉末逐渐变成了红色，该试管中的固体药品是碳和氧化铜．酒精灯上的金属网作用是 集中火焰，提高温度．<br />故答案为：（1）1：2，8：1；<br />（2）酚酞试液都变红，B比A变红的快；说明分子是不断运动的，温度越高分子运动速率越快；<br />（3）碳和氧化铜；集中火焰，提高温度．','【分析】（1）根据电解水试验的现象和结论分析回答；<br />（2）根据分子是不断运动的，运动的速率与温度有关进行分析回答；<br />（3）根据装置的特点和实验的现象分析药品，酒精灯上的金属网能集中火焰，提高温度．','简答题',3.00,'c4d7648ff081aa6f8e511111ef9897c2',9,400,'电解水实验,分子的定义与分子的特性,碳的化学性质','',2016,'37','2016•黑龙江一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840118,'联合国政府间气候变化专业委员会发布的一份报告指出，“全球气温本世纪可能上升1.1℃～6.4℃，海平面上升18cm～59cm，如果气温上升超过3.5℃，40%～70%的物种面临灭绝．”回答下列问题：<br />（1）导致气候变暖的主要气体是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，它所产生的这种效应称为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）写出上述气体的两种用途：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>、<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','二氧化碳$###$温室效应$###$参与植物的光合作用$###$固体能用于人工降雨','【解答】解：（1）导致气候变暖的主要气体是二氧化碳，它所产生的这种效应称为温室效应，故填：二氧化碳；温室效应；<br />（2）二氧化碳能参与植物的光合作用，固体干冰能用于人工降雨，还是重要的化工原料，可以用于制取碳酸饮料等，故填：参与植物的光合作用，固体能用于人工降雨．','【分析】根据已有的知识进行分析，二氧化碳是造成温室效应的主要气体，二氧化碳能参与植物的光合作用，固体干冰能用于人工降雨，还是重要的化工原料．','填空题',3.00,'d86e30e0d4531c76438f13e2a8120df8',9,400,'二氧化碳的用途,二氧化碳对环境的影响','',2016,'32','2016•内江模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840119,'下列实验操作中，正确的是（　　）','<img  src=\"/tikuimages/9/2016/400/shoutiniao92/1421fc5e-94d4-11e9-bb2b-b42e9921e93e_xkb93.png\" style=\"vertical-align:middle\" /><br />取用少量液体','<img  src=\"/tikuimages/9/2016/400/shoutiniao50/142530ae-94d4-11e9-897a-b42e9921e93e_xkb86.png\" style=\"vertical-align:middle\" /><br />过滤','<img  src=\"/tikuimages/9/2016/400/shoutiniao43/142816de-94d4-11e9-aef3-b42e9921e93e_xkb27.png\" style=\"vertical-align:middle\" /><br />闻气味','<img  src=\"/tikuimages/9/2016/400/shoutiniao91/1428da30-94d4-11e9-a44d-b42e9921e93e_xkb87.png\" style=\"vertical-align:middle\" /><br />存放酒精灯','','C','【解答】解：A、使用胶头滴管滴加少量液体时，注意胶头滴管不能伸入到试管内或接触试管内壁．应垂直悬空在试管口上方滴加液体，防止污染胶头滴管，图中所示操作错误；<br />B、过滤液体时，要注意“一贴、二低、三靠”的原则，倾倒液体时应用玻璃棒引流，图中所示操作错误；<br />C、闻气气味时，防止气体有毒，应用手扇动，不能直接用鼻子闻；图中所示操作正确；<br />D、存放酒精灯时，应盖上灯帽，以防酒精挥发，图中所示操作错误．<br />故选：C．','【分析】A、根据胶头滴管的使用方法进行分析判断；<br />B、根据过滤的注意事项进行分析判断；<br />C、根据闻气味的方法进行分析判断；<br />D、根据酒精灯的使用方法进行分析判断．','选择题',3.00,'5be3b93a7a03994dbfa2a5772d3fa602',9,400,'加热器皿-酒精灯,液体药品的取用,过滤的原理、方法及其应用','',2016,'37','2016•柳州二模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840120,'活性炭常用于防毒面具，利用了活性炭的（　　）','吸附性','可燃性','不溶性','还原性','','A','【解答】解：活性炭具有吸附性，能吸附色素和异味，防毒面具中使用活性炭，这是利用了活性炭的吸附作用．<br />答案：A','【分析】根据已有的物质的性质进行分析解答即可活性炭具有吸附性．','选择题',3.00,'50ceadd7bc93b1cb14070e40c6d312db',9,400,'碳单质的物理性质及用途','',0,'37','',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840125,'下列实验现象或实验原理正确的是（　　）','<img  src=\"/tikuimages/9/2016/400/shoutiniao58/143cfe70-94d4-11e9-9634-b42e9921e93e_xkb36.png\" style=\"vertical-align:middle\" /><br />白磷始终不燃烧','<img  src=\"/tikuimages/9/2016/400/shoutiniao88/143cfe71-94d4-11e9-8247-b42e9921e93e_xkb6.png\" style=\"vertical-align:middle\" /><br />验证甲烷燃烧生成二氧化碳和水','<img  src=\"/tikuimages/9/2016/400/shoutiniao29/14420780-94d4-11e9-900c-b42e9921e93e_xkb48.png\" style=\"vertical-align:middle\" /><br />U形管内液面左高右低','<img  src=\"/tikuimages/9/2016/400/shoutiniao26/144318f0-94d4-11e9-a33f-b42e9921e93e_xkb29.png\" style=\"vertical-align:middle\" /><br />验证Fe、Cu、Ag的金属或定性强弱','','C','【解答】解：A、热水的温度是80℃已经达到白磷的着火点，又与氧气接触，所以可以燃烧，故A错误；<br />B、澄清石灰水变浑浊，只能证明验证甲烷燃烧生成二氧化碳，但不能证明生成水水，故B错误；<br />C、硝酸铵溶于水吸热，使瓶内的压强减小，使U型管两端的液面左高右低，故C正确；<br />D、铜和银都不能与硫酸亚铁溶液反应，只能说明它们排在了铁的后面，但不能排铜和银的顺序，故D错误．<br />故选C．','【分析】A、根据燃烧的条件分析；<br />B、根据澄清石灰水变浑浊分析；<br />C、根据硝酸铵溶于水吸热，使瓶内的压强减小进行解答；<br />D、根据金属与盐反应的条件考虑本题，能反应说明该金属排在盐中金属的前面，不反应说明该金属排在盐中金属的后面进行解答．','选择题',3.00,'12ba793863a0af0721a399a1e5ddb577',9,400,'化学实验方案设计与评价,溶解时的吸热或放热现象,金属活动性顺序及其应用,甲烷、乙醇等常见有机物的性质和用途,燃烧与燃烧的条件','靖江市',2016,'37','2016•靖江市一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840126,'原子序数为16的元素符号是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，它的原子的最外层电子为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，它的常见化合价是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','S$###$6$###$-2，+4，+6','【解答】解：核电荷数为16的元素是硫元素，元素符号为S；其最外层有6个电子，在化学变化中易得到电子；它的常见化合价是-2，+4，+6．<br />答案：S；6；-2，+4，+6','【分析】根据核电荷数确定元素名称进行分析；','填空题',3.00,'b91e9e9cea23cedde1109e2e0626d2cc',9,400,'原子的有关数量计算,元素的符号及其意义,常见元素与常见原子团的化合价','',2012,'35','2012秋•阜南县校级期中',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840128,'<img  src=\"/tikuimages/9/2016/400/shoutiniao53/144b564f-94d4-11e9-a7dc-b42e9921e93e_xkb91.png\" style=\"vertical-align:middle;FLOAT:right\" />（2016春•连云港期中）空气、水、金属等是重要的资源<br />（1）工业上常把煤块粉碎后使其充分燃烧，其原理是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />家中堆放杂物的纸箱着火时，可用水浇灭，其原理是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）铜制品在空气中会发生锈蚀[铜锈的主要成分是Cu<SUB>2</SUB>（OH）<SUB>2</SUB>CO<SUB>3</SUB>]．<br />①根据铜锈的主要成分中含有氢元素，可以判断出铜生锈需要空气中的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；根据铜锈的主要成分中含有碳元素，可以判断出铜生锈需要空气中的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />②某化学兴趣小组欲通过实验探究“铜生锈是否需要空气中的氧气？”如图所示实验中（试管内的“<img  src=\"/tikuimages/9/2016/400/shoutiniao6/144d7930-94d4-11e9-8048-b42e9921e93e_xkb81.png\" style=\"vertical-align:middle\" />”均为铜片），只需完成实验<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>和<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>即可达到探究目的（填字母）．<br />（3）铝是世界上年产量第二的金属．<br />铝和氢氧化钠溶液反应的化学方程式为2Al+2NaOH+2X═2NaAlO<SUB>2</SUB>+3H<SUB>2</SUB>↑，其中X的化学式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．足量的稀盐酸与氢氧化钠溶液分别和等质量的铝粉充分反应，产生氢气的质量分别为m<SUB>1</SUB>、m<SUB>2</SUB>，下列有关m<SUB>1</SUB>、m<SUB>2</SUB>大小关系的说法正确的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填字母）．<br />A．m<SUB>1</SUB>＞m<SUB>2</SUB>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;B．m<SUB>1</SUB>＜m<SUB>2</SUB>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;C．m<SUB>1</SUB>=m<SUB>2</SUB>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;D．无法比较．','','','','','','增大燃料与空气的接触面$###$水可以吸收可燃物燃烧释放出的大量热，使可燃物的温度降到其着火点以下$###$H<SUB>2</SUB>O$###$CO<SUB>2</SUB>$###$B$###$D$###$H<SUB>2</SUB>O$###$C','【解答】解：（1）工业上常把煤块粉碎，增大了煤与氧气的接触面积，能使煤充分燃烧，减少煤的浪费，堆放杂物的纸箱着火可用水浇灭，是利用了使温度降到可燃物的着火点以下的灭火原理．故填：增大燃料与空气的接触面；水可以吸收可燃物燃烧释放出的大量热，使可燃物的温度降到其着火点以下；<br />（2）①根据质量守恒定律反应前后各元素种类不变，所以铜锈的主要成分中含有氢元素，可以判断出铜生锈需要空气中的含有氢元素的物质，即水蒸气；铜锈的主要成分中含有碳元素，可以判断出铜生锈需要空气中含碳的物质，即二氧化碳参加反应；故填：H<SUB>2</SUB>O；CO<SUB>2</SUB><br />②铜生锈是否需要空气中的氧气，变量是否含有氧气，其余条件必须相同，所以B和D形成对照实验，可以探究是否需要有氧气参加反应．故填：B；D；<br />（3）2Al+2NaOH+2X═2NaAlO<SUB>2</SUB>+3H<SUB>2</SUB>↑，反应前含有2个A铝原子，2个Na原子，2个H原子，2个O原子，2X，反应后含有2个Na原子，2个Al原子，4个H原子，6个H原子，故2X中含有4个H原子和2个O原子，故其化学式为H<SUB>2</SUB>O，足量的稀盐酸与氢氧化钠溶液分别和等质量的铝粉充分反应，产生氢气，<br />2Al+2NaOH+2H<SUB>2</SUB>O═2NaAlO<SUB>2</SUB>+3H<SUB>2</SUB>↑，<br />54&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 6<br />2Al+6HCl═2AlCl<SUB>3</SUB>+3H<SUB>2</SUB>↑<br />54&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 6<br />故产生的氢气的质量相等；<br />故填：H<SUB>2</SUB>O，C．','【分析】（1）促进可燃物燃烧的方法有：增大可燃物与氧气的接触面积或增大氧气的浓度，据此进行分析判断．根据灭火原理：①清除或隔离可燃物，②隔绝氧气或空气，③使温度降到可燃物的着火点以下，据此结合灭火方法进行分析解答；<br />（2）①根据质量守恒定律考虑；②根据对照实验的设计方法考虑；<br />（3）根据质量守恒定律的知识进行分析解答即可．','填空题',3.00,'bbc43cd7b401ad8896f04a9770a5f482',9,400,'金属锈蚀的条件及其防护,质量守恒定律及其应用,燃烧与燃烧的条件,灭火的原理和方法','',2016,'35','2016春•连云港期中',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840131,'我们日常生活中的衣食住行与化学联系密切，现举几例：<br />（1）衣：衣服上的油渍可以用洗衣液清洗．该过程中洗衣液的作用是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）食：重庆本土的缙云毛峰全国有名．好茶需要好水泡，自然界中的水通常以硬水居多，实验室可以通过<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>操作，把硬水制成几乎纯净的水．<br />（3）住：钢铁在建筑中应用广泛，钢铁的锈蚀也给人类带来了巨大的损失．铁在空气中锈蚀的原因是：铁与空气中的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>反应（填化学式）．<br />（4）行：①重庆本土汽车品牌--长安汽车自主研发设计的跑车“氢程”，以氢能源为直接燃料的原因是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（用化学方程式表示）．<br />②重庆轻轨穿楼而过，震撼场面全国罕见．下列轻轨组件中，由有机合成材料制成的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填序号）<br />A．钢铁车架&nbsp;&nbsp;B．玻璃车窗&nbsp;&nbsp;C．塑料座椅．','','','','','','乳化作用$###$蒸馏$###$H<SUB>2</SUB>O、O<SUB>2</SUB>$###$2H<SUB>2</SUB>+O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O$###$C','【解答】解；（1）洗衣液洗去衣服上的油渍是利用的洗涤剂对油污的乳化作用；<br />（2）实验室可以通过蒸馏操作，把硬水制成几乎纯净的水；<br />（3）铁在空气中锈蚀的原因是：铁与空气中的水和氧气反应；<br />（4）①氢气在氧气中燃烧生成水，其化学方程式为：2H<SUB>2</SUB>+O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O；<br />②轻轨组件中，由有机合成材料制成的是塑料座椅；<br />故答案为：（1）乳化作用；（2）蒸馏；（3）H<SUB>2</SUB>O、O<SUB>2</SUB>；（4）①2H<SUB>2</SUB>+O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O；②C．','【分析】（1）根据洗衣液的乳化作用进行分析；<br />（2）根据实验室把硬水制成几乎纯净的水的方法进行分析；<br />（3）根据铁生锈的条件进行分析；<br />（4）①根据氢气在氧气中燃烧生成水写出化学方程式即可；②根据材料的分类进行分析．','书写',3.00,'f9fd41bc06aa39ba99cb8eb7255d46c2',9,400,'硬水与软水,乳化现象与乳化作用,金属锈蚀的条件及其防护,书写化学方程式、文字表达式、电离方程式,氢气的化学性质与燃烧实验,合成材料的使用及其对人和环境的影响','',2016,'32','2016•重庆校级模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840134,'溶液在生产、生活中起着十分重要的作用．<br />（1）高锰酸钾溶液常用于消毒，高锰酸钾溶液中的溶质是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）硝酸钾溶液是常用的无土栽培的营养液之一，硝酸钾属于<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />A．氮肥&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;B．磷肥&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;C．钾肥&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;D．复合肥<br />（3）硝酸钾在不同温度下得溶解度如下表所示：<br /><table class=\"edittable\"><TBODY><TR><td width=138 colSpan=2>温度/℃</TD><td width=69>10</TD><td width=69>20</TD><td width=69>30</TD><td width=69>40</TD><td width=69>50</TD><td width=69>60</TD><td width=69>70</TD></TR><TR><td>溶解度</TD><td>KNO<SUB>3</SUB></TD><td>20.9</TD><td>31.6</TD><td>45.8</TD><td>63.9</TD><td>85.5</TD><td>110</TD><td>138</TD></TR></TBODY></TABLE>在不改变溶液浓度的情况下，将硝酸钾的不饱和溶液变为饱和溶液的方法是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','高锰酸钾$###$D$###$降温','【解答】解：<br />（1）高锰酸钾溶液常用于消毒，高锰酸钾溶液中的溶质是高锰酸钾；<br />（2）硝酸钾中含有钾元素和氮元素，属于复合肥．<br />（3）因为硝酸钾的溶解度随温度变化影响较大；则：降低温度，KNO<SUB>3</SUB>溶解度降低，而没有晶体析出，从而达到饱和，这是其浓度保持不变．<br />答案：<br />（1）高锰酸钾；&nbsp;&nbsp;&nbsp;<br />（2）D；&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<br />（3）降温．','【分析】（1）根据医用高锰酸钾溶液中的溶质是高锰酸钾解答；<br />（2）根据有氮元素的肥料称为氮肥．含有磷元素的肥料称为磷肥．含有钾元素的肥料称为钾肥．同时含有氮、磷、钾三种元素中的两种或两种以上的肥料称为复合肥解答；<br />（3）因为硝酸钾的溶解度随温度升高而变化较大，则降低温度，KNO<SUB>3</SUB>溶解度降低，从而达到饱和；可知一定温度下，将一瓶接近饱和的KNO<SUB>3</SUB>溶液变为饱和溶液，且不改变溶液的浓度的方法．','填空题',3.00,'96bf09fff9b9b059fd5c001d450772e6',9,400,'溶液、溶质和溶剂的相互关系与判断,饱和溶液和不饱和溶液相互转变的方法,常见化肥的种类和作用','',2016,'37','2016•顺义区一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840137,'碳酸钡广泛应用于显像管（CRT）、陶瓷、光学玻璃等行业．以碳酸盐矿石（主要成分为BaCO<SUB>3</SUB>&nbsp;和CaCO<SUB>3</SUB>）为原料可生产碳酸钡．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao38/14612840-94d4-11e9-be8e-b42e9921e93e_xkb15.png\" style=\"vertical-align:middle\" /><br />已知碳酸钡和碳酸钙具有相似的化学性质；氧化钡和氧化钙也具有相似的化学性质．氢氧化钙和氢氧化钡在不同温度下的溶解度如下表．<br /><table class=\"edittable\"><TBODY><TR><td width=144>温度/℃</TD><td width=70>0</TD><td width=86>40</TD><td width=86>80</TD></TR><TR><td>Ca（OH）<SUB>2</SUB>溶解度/g</TD><td>0.187</TD><td>0.141</TD><td>0.094</TD></TR><TR><td>Ba（OH）<SUB>2</SUB>溶解度/g</TD><td>1.67</TD><td>8.22</TD><td>101.4</TD></TR></TBODY></TABLE>（1）焙烧炉中加入焦炭粉和热空气，焦炭和空气发生反应的作用是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）生产过程中需要进行多次过滤．实验室进行过滤时，除铁架台（带铁圈）、烧杯和玻璃棒外，还需要的实验用品有<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）“操作1”是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；通过“操作1”再过滤，得到的白色固体的化学式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（4）“吸收”中发生的化学反应方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','提供热量$###$漏斗，滤纸$###$加热$###$Ca（OH）<SUB>2</SUB>$###$Ba（OH）<SUB>2</SUB>+CO<SUB>2</SUB>=BaCO<SUB>3</SUB>↓+H<SUB>2</SUB>O','【解答】解：由题目给出的信息可知：焙烧炉中碳酸钡在高温的条件下生成氧化钡和二氧化碳，碳酸钙在高温的条件下生成氧化钙和二氧化碳；氧化钡和水反应生成氢氧化钡，氧化钙和水反应生成氢氧化钙；因此除了水之外，进入分离池中的物质是氢氧化钡和氢氧化钙；分离池中操作1的方法是：升温、过滤，因为氢氧化钙的溶解度随温度的升高而减小；由溶液制取产品是二氧化碳与氢氧化钡反应生成碳酸钡白色沉淀和水；对矿石进行预处理“研磨成粉状”有利于充分反应；焙烧炉中添加焦炭粉和热空气是为了维持炉内的高温状态；废渣需要经过洗涤才能弃渣，是为了保护环境，同时充分利用原料是．则：<br />（1）焙烧炉中加入焦炭粉和热空气，焦炭和空气发生反应的作用是提供热量．<br />（2）生产过程中需要进行多次过滤，实验室进行过滤时，除铁架台（带铁圈）、烧杯和玻璃棒外，还需要的实验用品有漏斗，滤纸．<br />（3）“操作1”是加热；通过“操作1”再过滤，因为氢氧化钙的溶解度随温度的升高而减小，故得到的白色固体的化学式为Ca（OH）<SUB>2</SUB>．<br />（4）“吸收”中发生的化学反应方程式为Ba（OH）<SUB>2</SUB>+CO<SUB>2</SUB>=BaCO<SUB>3</SUB>↓+H<SUB>2</SUB>O．<br />故答案为：<br />（1）提供热量；（2）漏斗、滤纸；（3）加热；Ca（OH）<SUB>2</SUB>；<br />（4）Ba（OH）<SUB>2</SUB>+CO<SUB>2</SUB>=BaCO<SUB>3</SUB>↓+H<SUB>2</SUB>O．','【分析】本题是酸碱盐的应用中的物质的制备，由题目给出的信息可知：焙烧炉中碳酸钡在高温的条件下生成氧化钡和二氧化碳，碳酸钙在高温的条件下生成氧化钙和二氧化碳；氧化钡和水反应生成氢氧化钡，氧化钙和水反应生成氢氧化钙；因此除了水之外，进入分离池中的物质是氢氧化钡和氢氧化钙；分离池中操作1的方法是：升温、过滤，因为氢氧化钙的溶解度随温度的升高而减小；由溶液制取产品是二氧化碳与氢氧化钡反应生成碳酸钡白色沉淀和水；对矿石进行预处理“研磨成粉状”有利于充分反应；焙烧炉中添加焦炭粉和热空气是为了维持炉内的高温状态；废渣需要经过洗涤才能弃渣，是为了保护环境，同时充分利用原料是．','书写',3.00,'68df00e0abb55596eb383f8d717b2f40',9,400,'过滤的原理、方法及其应用,盐的化学性质,物质的相互转化和制备,书写化学方程式、文字表达式、电离方程式','',2016,'37','2016•南通一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840139,'（1）现有以下四种物质：A．硝酸钾&nbsp;&nbsp;B．熟石灰&nbsp;&nbsp;C．干冰&nbsp;&nbsp;D．稀硫酸根据下列要求用字母填空：<br />①能用于配制波尔多液的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />②可用作复合肥的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />③可用人工降雨的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />④能用于除去金属表面锈迹的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）根据题意用下列选项的字母填空：A．熔点&nbsp;&nbsp;&nbsp;B．沸点&nbsp;&nbsp;&nbsp;&nbsp;C．硬度<br />①工业制氧气和工业石油分馏都是利用混合物中各成分的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>不同．<br />②武德合金用作保险丝比用纯铜丝更安全，是因为前者的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>低．','','','','','','B$###$A$###$C$###$D$###$B$###$A','【解答】解：（1）①熟石灰可用于配制波尔多液；②硝酸钾含有氮元素和钾元素，可用作复合肥；③干冰升华吸热，可用于人工降雨；④稀硫酸能用于除去金属表面锈迹．<br />（2）①工业制氧气和工业石油分馏都是利用混合物中各成分的沸点不同．<br />②武德合金用作保险丝比用纯铜丝更安全，是因为前者的熔点低．<br />故填：（1）①B；②A；③C；④D；（2）①B；②A．','【分析】物质的性质决定物质的用途，根据已有的物质的性质进行分析解答即可．','填空题',3.00,'c427d4c10c1bf9bfc6370c65c2a88b46',9,400,'氧气的工业制法,二氧化碳的用途,合金与合金的性质,酸的物理性质及用途,常用盐的用途,常见化肥的种类和作用,石油加工的产物','',2016,'32','2016•常州模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840140,'下列有关溶液的说法不正确的是（　　）','溶液一定是混合物','饱和溶液的质量分数一定比不饱和溶液的质量分数大','凡是溶液一定具有均一性和稳定性','物形成溶液的溶质不一定是固态物质','','B','【解答】解：A．溶液是由溶质和溶剂两部分组成的混合物，故正确；<br />B．若低温下某物质的饱和溶液和高温下该物质的不饱和溶液相比，溶质质量分数有可能大、可能小、也可能相等，因此必须强调同温度下，故错误；<br />C．溶液都具有均一性和稳定性的特征，故正确；<br />D．溶质的状态可能是固体、液体或气体，故正确．<br />故选B．','【分析】A．根据溶液的组成来分析；<br />B．根据溶质质量分数的比较来分析；<br />C．根据溶液的特征来分析；<br />D．根据溶质的状态来分析．','选择题',3.00,'3c34967b44ba41fdbca4fe5a98498f8a',9,400,'溶液的概念、组成及其特点,溶质的质量分数','',2015,'37','2015秋•深圳月考',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840141,'实验室部分仪器或装置如图所示，请回答下列问题．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao22/14712dcf-94d4-11e9-abaa-b42e9921e93e_xkb54.png\" style=\"vertical-align:middle\" /><br />（1）写出仪器B的名称：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）请你用上面仪器中的 ①<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填仪器标号）组装一套用过氧化氢溶液制取氧气<br />发生装置，收集装置可选用图中的 ②<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填仪器标号）．写出对应的化学方程式为③<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）实验开放日，小江抽到了有关金属性质的实验．<br />①他将铝片投入稀硫酸中，没有明显现象，则在投入之前铝片已发生<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填化学方程式）反应．<br />②将一定量的铝粉和铁粉加入到硫酸铜溶液中，充分反应后溶液为无色，则一定发生的反应是：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填化学方程式）．<br />（4）化学学习兴趣小组同学设计如下实验装置（铁架台等仪器省略）制备CO<SUB>2</SUB>和验证CO<SUB>2</SUB>与NaOH反应．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao89/14759aa1-94d4-11e9-be66-b42e9921e93e_xkb20.png\" style=\"vertical-align:middle\" /><br />&nbsp;①写出装置A&nbsp;中发生反应的化学方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，小组同学分析实验装置后，发现用E装置收集CO<SUB>2</SUB>有明显不足，需将装置E中的导管<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“a”或“b”）加长至接近集气瓶底部方可．<br />②改进装置，打开止水夹K<SUB>1</SUB>，检查气密性，加入药品开始实验，当观察到装置F中的现象是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，可确定装置E收集满CO<SUB>2</SUB>，同时证明CO<SUB>2</SUB>能H<SUB>2</SUB>O发生反应．<br />③当装置E中收集满CO<SUB>2</SUB>时，关闭止水夹K<SUB>1</SUB>，打开止水夹K<SUB>2</SUB>，把注射器D中50ml浓NaOH溶液压入装置E中，观察到装置F导管内液柱逐渐上升的现象时，即能证明CO<SUB>2</SUB>与NaOH反应．但有同学认为此现象是CO<SUB>2</SUB>溶于水的结果，为了进一步证明CO<SUB>2</SUB>与NaOH确实反应了，你在上述现象结束后采取的方法是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','长颈漏斗$###$ACD$###$G$###$2H<SUB>2</SUB>O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span>&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O+O<SUB>2</SUB>↑$###$4Al+3O<SUB>2</SUB>=2Al<SUB>2</SUB>O<SUB>3</SUB>$###$2Al+3CuSO<SUB>4</SUB>=3Cu+Al<SUB>2</SUB>（SO<SUB>4</SUB>）<SUB>3</SUB>$###$CaCO<SUB>3</SUB>+2HCl=CaCl<SUB>2</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑$###$b$###$溶液变红$###$取反应后的溶液，滴加稀盐酸有气泡产生','【解答】解：（1）通过分析题中所指仪器的作用可知，B是长颈漏斗；<br />（2）实验室用过氧化氢制取氧气的反应物是固体和液体，反应条件是常温，所以用上面仪器中的ACD组装一套用过氧化氢溶液制取氧气发生装置，氧气的密度比空气大，收集装置可选用图中的G，过氧化氢在二氧化锰的催化作用下生成水和氧气，化学方程式为：2H<SUB>2</SUB>O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span>&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O+O<SUB>2</SUB>↑；<br />（3）铝的化学性质活泼，会与空气中的氧气反应生成氧化铝，化学方程式为：4Al+3O<SUB>2</SUB>=2Al<SUB>2</SUB>O<SUB>3</SUB>；<br />铝和硫酸铜反应生成硫酸铝和铜，化学方程式为：2Al+3CuSO<SUB>4</SUB>=3Cu+Al<SUB>2</SUB>（SO<SUB>4</SUB>）<SUB>3</SUB>；<br />（4）①碳酸钙和盐酸反应生成氯化钙、水和二氧化碳，化学方程式为：CaCO<SUB>3</SUB>+2HCl=CaCl<SUB>2</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑，二氧化碳密度比空气大，所以需将装置E中的导管b加长至接近集气瓶底部方可；<br />②二氧化碳和水反应生成碳酸，碳酸能使紫色石蕊变红色，所以当观察到装置F中的现象是溶液变红，可确定装置E收：取反应后的溶液，滴加稀盐酸有气泡产生．<br />故答案为：（1）长颈漏斗；<br />（2）ACD或ABCE，G，2H<SUB>2</SUB>O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span>&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O+O<SUB>2</SUB>↑；<br />（3）①4Al+3O<SUB>2</SUB>=2Al<SUB>2</SUB>O<SUB>3</SUB>；<br />②2Al+3CuSO<SUB>4</SUB>=3Cu+Al<SUB>2</SUB>（SO<SUB>4</SUB>）<SUB>3</SUB>；<br />（4）①CaCO<SUB>3</SUB>+2HCl=CaCl<SUB>2</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑，b；&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<br />②溶液变红；&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<br />③取反应后的溶液，滴加稀盐酸有气泡产生．','【分析】（1）根据实验室常用仪器的名称和题中所指仪器的作用进行分析；<br />（2）根据实验室用过氧化氢制取氧气的反应物是固体和液体，反应条件是常温，氧气的密度比空气大，过氧化氢在二氧化锰的催化作用下生成水和氧气进行分析；<br />（3）根据铝的化学性质活泼，会与空气中的氧气反应生成氧化铝进行分析；<br />根据铝和硫酸铜反应生成硫酸铝和铜进行分析；<br />（4）①根据碳酸钙和盐酸反应生成氯化钙、水和二氧化碳，二氧化碳密度比空气大进行分析；<br />②根据二氧化碳和水反应生成碳酸，碳酸能使紫色石蕊变红色进行分析；<br />③根据碳酸钠和盐酸反应会生成二氧化碳进行分析．','书写',3.00,'5c6d92c65f343a6b7752d71d62ef3919',9,400,'氧气的制取装置,氧气的收集方法,金属的化学性质,碱的化学性质,书写化学方程式、文字表达式、电离方程式','江阴市',2016,'35','2016春•江阴市期中',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840148,'物质的构成与变化<br />现有以下6个变化：<br />①在点燃的条件下，氢气在氧气中燃烧；<br />②给水通直流电；<br />③锌加入稀硫酸溶液中；<br />④氢氧化钠溶液跟稀盐酸混合；<br />⑤冷却硝酸钾热饱和溶液析出硝酸钾晶体；<br />⑥日本福岛第一核电站发生核反应，铀235裂变产生碘131．<br />请回答下列问题：<br />（1）探究化学变化的规律始终是化学科学的核心任务．上述变化中，属于化学变化的是（填变化的代号，下同）<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，属于物理变化的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）化学变化也叫化学反应，在上述化学反应中，属于分解反应的有<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，属于复分解反应的有<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．将化学反应分为分解反应、化合反应、复分解反应和置换反应四种基本反应类型的依据是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）我们知道，能直接构成物质的微粒有分子、原子、离子．物质之所以发生化学反应，从微观的角度看，是因为反应物的微粒之间在一定条件下发生有效的相互作用，使微粒的结构发生改变，或微粒重新排列组合，宏观上表现为生成了新的物质．<br />在反应①中，发生有效的相互作用的微粒是（写出微粒的符号）<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />反应③的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，发生有效的相互作用而促使反应发生的微粒是（写出微粒的符号）<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，微粒的转化过程是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />在反应④中，发生有效的相互作用而促使反应发生的微粒是（写出微粒的符号）<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />③和⑥两个变化的实质是否相同？<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．为什么？<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />（4）根据化学反应是放出热量还是吸收热量，将化学反应分为放热反应和吸热反应．在上述化学变化中，属于吸热反应的有<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','①②③④$###$⑤$###$②$###$④$###$根据反应物和生成物的种类和类别$###$H<SUB>2</SUB>和O<SUB>2</SUB>$###$Zn+H<SUB>2</SUB>SO<SUB>4</SUB>=ZnSO<SUB>4</SUB>+H<SUB>2</SUB>↑$###$H<SUP>+</SUP>和Zn$###$锌原子失去电子，变成锌离子，氢离子得到电子变成氢原子，两个氢原子结合生成一个氢分子$###$H<SUP>+</SUP>、OH<SUP>-</SUP>③$###$不相同$###$③是化学变化，原子核（元素和原子的种类）不变；⑥是核变化，原子核（元素和原子的种类）发生改变$###$②','【解答】解：（1）根据是否生成了新物质，上述变化中，属于化学变化的是 ①②③④，属于物理变化的是 ⑤．<br />（2）在上述化学反应中，属于分解反应的有 ②，属于复分解反应的有 ④．将化学反应分为分解反应、化合反应、复分解反应和置换反应四种基本反应类型的依据是 根据反应物和生成物的种类和类别（或：根据反应物和生成物的种类和组成）．<br />（3）氢气在氧气中燃烧，发生有效的相互作用的微粒是氢分子和氧分子，通过反应结合成水分子． <br />锌加入稀硫酸溶液中的化学方程式为 Zn+H<SUB>2</SUB>SO<SUB>4</SUB>═ZnSO<SUB>4</SUB>+H<SUB>2</SUB>↑，<br />发生有效的相互作用而促使反应发生的微粒是 H<SUP>+</SUP> 和Zn，微粒的转化过程是 锌原子失去电子，变成锌离子，氢离子得到电子变成氢原子，两个氢原子结合生成一个氢分子．<br />氢氧化钠溶液跟稀盐酸混合发生中和反应，发生有效的相互作用而促使反应发生的微粒是氢离子和氢氧根离子；③和⑥两个变化的实质不相同．因为 ③是化学变化，原子核（元素和原子的种类）不变；⑥是核变化，原子核（元素和原子的种类）发生改变．．<br />（4）在上述化学变化中，属于吸热反应的有 ②，因为它需要不断通电来提供能量．<br />故答案：<br />（1）①②③④；⑤；（2）②；④；根据反应物和生成物的种类和类别；（3）H<SUB>2</SUB>和O<SUB>2</SUB>； Zn+H<SUB>2</SUB>SO<SUB>4</SUB>═ZnSO<SUB>4</SUB>+H<SUB>2</SUB>↑；H<SUP>+</SUP> 和Zn；锌原子失去电子，变成锌离子，氢离子得到电子变成氢原子，两个氢原子结合生成一个氢分子；&nbsp; H<SUP>+</SUP>、OH<SUP>-</SUP>；不相同； ③是化学变化，原子核（元素和原子的种类）不变；⑥是核变化，原子核（元素和原子的种类）发生改变；（4）②．','【分析】（1）根据是否生成新物质来判断一个变化是物理变化还是化学变化；<br />（2）分解反应的特点是“一变多”，复分解反应的特点是两种化合物相互交换成分，生成两种新的化合物；区分四种基本反应类型的依据要从反应物和生成物的种类和物质的分类上考虑；<br />（3）判断哪些反应物的微粒发生有效的相互作用，要从反应的特点和生成物的组成入手考虑，氢气在氧气中燃烧，氢分子和氧分子之间发生了有效的反应，金属和酸反应实质就是金属原子和氢离子之间的反应，通过电子得失完成离子和原子的相互转化；中和反应的实质是氢离子和氢氧根离子的结合；③和⑥两个变化的实质要从原子核是否发生变化来考虑．<br />（4）吸热反应的表现是发生反应时需要吸收能量．','书写',3.00,'5c7f544680ebd704898e1a7f6c2911b7',9,400,'化学变化和物理变化的判别,物质发生化学变化时的能量变化,分解反应及其应用,复分解反应及其应用,书写化学方程式、文字表达式、电离方程式','荣成市',2016,'35','2016春•荣成市期中',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840149,'请结合下列实验常用装置，回答有关问题．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao5/148aa940-94d4-11e9-b185-b42e9921e93e_xkb16.png\" style=\"vertical-align:middle\" /><br />（1）写出装置图中标号仪器a的名称<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）实验室用氯酸钾和二氧化锰制取较为纯净的氧气，可采用的装置组合为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，发生反应的化学方程式是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．若要回收其中的二氧化锰，可采取的操作步骤是：溶解、<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>、洗涤、烘干．<br />（3）实验室用C装置制取二氧化碳的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，若要使反应停止可采取的操作是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（4）草酸晶体（H<SUB>2</SUB>C<SUB>2</SUB>O<SUB>4</SUB>•2H<SUB>2</SUB>O）是一种无色晶体，100℃开始失水，101.5℃熔化，150℃左右分解产生H<SUB>2</SUB>O、CO和CO<SUB>2</SUB>，发生反应的化学方程式为：H<SUB>2</SUB>C<SUB>2</SUB>O<SUB>4</SUB>•2H<SUB>2</SUB>O<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>&nbsp;CO↑+CO<SUB>2</SUB>↑+3H<SUB>2</SUB>O，若要用加热草酸晶体的方法制取一氧化碳气体，可以选择的气体发生装置是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，选择该发生装置的原因<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，为了除去二氧化碳并收集一氧化碳气体，下列装置中最佳的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，在洗气瓶中装的试剂是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao88/148e049e-94d4-11e9-b730-b42e9921e93e_xkb11.png\" style=\"vertical-align:middle\" />','','','','','','铁架台$###$AD$###$2KClO<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span></td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>△</td></tr></table></span>2KCl+3O<SUB>2</SUB>↑$###$过滤$###$CaCO<SUB>3</SUB>+2HCl=CaCl<SUB>2</SUB>+CO<SUB>2</SUB>↑+H<SUB>2</SUB>O$###$将活塞关闭$###$B$###$①反应条件需要加热②草酸晶体在101.5℃会熔化因此试管口朝上$###$丙$###$氢氧化钠','【解答】解：（1）通过分析题中所指仪器的作用可知，a是铁架台；<br />（2）实验室用氯酸钾制取的反应物是固体，反应条件是加热，氧气密度比空气大，不易溶于水，所以可采用的装置组合为AD，氯酸钾在二氧化锰的催化作用下生成氯化钾和氧气，化学方程式为：2KClO<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span></td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>△</td></tr></table></span>2KCl+3O<SUB>2</SUB>↑，二氧化锰难溶于水，所以要回收其中的二氧化锰，可采取的操作步骤是：溶解、过滤、洗涤、烘干；<br />（3）碳酸钙和盐酸反应生成氯化钙、水和二氧化碳，化学方程式为：CaCO<SUB>3</SUB>+2HCl=CaCl<SUB>2</SUB>+CO<SUB>2</SUB>↑+H<SUB>2</SUB>O，固体和液体分离可以使反应停止，所以要使反应停止可采取的操作是：将活塞关闭；<br />（4）加热草酸晶体的方法制取一氧化碳气体的反应物是固体，反应条件是加热，所以可以选择的气体发生装置是B，选择该发生装置的原因是：①反应条件需要加热②草酸晶体在101.5℃会熔化因此试管口朝上，氢氧化钠会与二氧化碳反应，一氧化碳难溶于水，所以为了除去二氧化碳并收集一氧化碳气体，下列装置中最佳的是丙，在洗气瓶中装的试剂是氢氧化钠．<br />故答案为：（1）铁架台；<br />（2）AD，2KClO<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span></td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>△</td></tr></table></span>2KCl+3O<SUB>2</SUB>↑，过滤；<br />（3）CaCO<SUB>3</SUB>+2HCl=CaCl<SUB>2</SUB>+CO<SUB>2</SUB>↑+H<SUB>2</SUB>O，将活塞关闭；<br />（4）B，①反应条件需要加热②草酸晶体在101.5℃会熔化因此试管口朝上，丙，氢氧化钠．','【分析】（1）根据实验室常用仪器的名称和题中所指仪器的作用进行分析；<br />（2）根据实验室用氯酸钾制取的反应物是固体，反应条件是加热，氧气密度比空气大，不易溶于水，氯酸钾在二氧化锰的催化作用下生成氯化钾和氧气，二氧化锰难溶于水进行分析；<br />（3）根据碳酸钙和盐酸反应生成氯化钙、水和二氧化碳，固体和液体分离可以使反应停止进行分析；<br />（4）根据加热草酸晶体的方法制取一氧化碳气体的反应物是固体，反应条件是加热，加热固体的正确操作，氢氧化钠会与二氧化碳反应，一氧化碳难溶于水进行分析．','书写',3.00,'dfce26c8bf788e6f42813055fd7f677b',9,400,'混合物的分离方法,常用气体的发生装置和收集装置与选取方法,实验室制取氧气的反应原理,二氧化碳的实验室制法,书写化学方程式、文字表达式、电离方程式','',2016,'37','2016•滨湖区一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840152,'“珍惜地球资源，改变发展方式”是当今世界的共识．针对这一主题下列表达正确的是（　　）','地球上水含量非常丰富，无需节约，可以任意使用','煤、石油、天然气属于可再生资源，取之不尽、用之不竭','世界上没有真正的垃圾，只有放错地方的资源','大力提倡使用一次性筷子、纸杯等','','C','【解答】解：<br />A、虽然地球表面积的75%是水，但是淡水资源却比较少，并且分布不均匀，所以要节约用水、保护水资源，故错；<br />B、煤、石油、天然气属于不可再生资源，不是可再生资源，故错；<br />C、垃圾分类放置可以节约资源，保护环境，从而说世界上没有真正的垃圾，只有放错地方的资源，故对；<br />D、提倡使用一次性纸杯，增大了森林的砍伐、消耗的能量、二氧化碳的排放；故选项不符合“低碳生活”理念，故错．<br />答案：C','【分析】A、根据水资源的现状解答；<br />B、根据煤、石油、天然气属于不可再生资源解答；<br />C、根据垃圾分类放置可以节约资源，保护环境解答；<br />D、根据提倡使用一次性纸杯，增大了森林的砍伐、消耗的能量、二氧化碳的排放解答．','选择题',3.00,'9b516fa4fd410d7412b6ba7da7ceaf83',9,400,'水资源状况,常见能源的种类、能源的分类','',2016,'37','2016•滨海县二模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840153,'水是生命之源，水的化学式为H<SUB>2</SUB>O，下列对水的认识正确的是（　　）','水是由氢原子和氧原子构成','水中氢、氧的质量比为2：1','水中含有氧分子','水由氢元素和氧元素组成','','D','【解答】解：A、水分子由氢原子和氧原子构成，故A错误；<br />B、由化学式H<SUB>2</SUB>O，可以计算水中氢、氧元素的质量比为：（1×2）：16=1：8，故B错误；<br />C、水是由水分子构成的，水是由水分子构成的，故C错误；<br />D、水由氢元素和氧元素组成，故D正确．<br />故选D．','【分析】A、根据水分子由氢原子和氧原子构成进行分析；<br />B、由化学式H<SUB>2</SUB>O，可以计算水中氢、氧元素的质量比；<br />C、根据水是由水分子构成的进行分析；<br />D、由化学式H<SUB>2</SUB>O，知水由氢元素和氧元素组成．','选择题',3.00,'d35e69ec4cb6c327d39e873dc10ba0c5',9,400,'水的组成,元素质量比的计算','',2012,'33','2012秋•潮南区期末',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840154,'下列图象能正确表示实验过程中相应量的变化关系的是（　　） <table class=\"edittable\"><TBODY><TR><td width=176>A</TD><td width=178>B</TD><td width=175>C</TD><td width=183>D</TD></TR><TR><td><img  src=\"/tikuimages/9/0/400/shoutiniao48/1494e270-94d4-11e9-bb64-b42e9921e93e_xkb84.png\" style=\"vertical-align:middle\" /></TD><td><img  src=\"/tikuimages/9/0/400/shoutiniao50/1496901e-94d4-11e9-b111-b42e9921e93e_xkb39.png\" style=\"vertical-align:middle\" /></TD><td><img  src=\"/tikuimages/9/0/400/shoutiniao23/14994f40-94d4-11e9-8473-b42e9921e93e_xkb94.png\" style=\"vertical-align:middle\" /></TD><td><img  src=\"/tikuimages/9/0/400/shoutiniao14/149a128f-94d4-11e9-9fba-b42e9921e93e_xkb37.png\" style=\"vertical-align:middle\" /></TD></TR><TR><td>电解水</TD><td>向pH=2的稀盐酸中加入pH=12的氢氧化钠溶液</TD><td>20℃时，向一定量氯化钠饱和溶液中加入氯化钠固体</TD><td>用等质量氯酸钾分别制取氧气</TD></TR></TBODY></TABLE>','A','B','C','D','','A','【解答】解：A．电解水时生成的氢气和氧气的体积比是2：1，故正确；<br />B．向稀盐酸中加入氢氧化钠溶液时，稀盐酸与氢氧化钠发生中和反应，pH值不断增大，但是不会等于加入的碱的pH值，而应该略小，故错误；<br />C．饱和溶液在温度不变的条件下加入该溶质已经不再继续溶解，饱和溶液的质量不再增加，故错误；<br />D．等质量的氯酸钾受热分解生成氧气的质量与是由使用催化剂无关，催化剂只能改变反应速率，不影响生成氧气的量，故错误．<br />故选A．','【分析】A．根据电解水时生成的氢气和氧气的体积关系分析；<br />B．盐酸与氢氧化钠发生中和反应；<br />C．根据饱和溶液的概念来分析；<br />D．根据氯酸钾放出氧气的质量与有无催化剂无关分析．','选择题',3.00,'c340a760ecb3a26bd7bb6efb0982968f',9,400,'催化剂的特点与催化作用,电解水实验,饱和溶液和不饱和溶液,中和反应及其应用,溶液的酸碱性与pH值的关系','',0,'37','',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840156,'A～H是初中化学常见的物质，他们之间的相互关系如图所示（“→”指向生成物），其中B是红棕色粉末，A、C常温下均为气体，F是一种可溶性碱，反应③是复分解反应，请回答下列问题：<img  src=\"/tikuimages/9/2016/400/shoutiniao33/14a50f0f-94d4-11e9-aecc-b42e9921e93e_xkb11.png\" style=\"vertical-align:middle\" /><br />（1）物质E的化学式：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />（2）反应①<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“是”或“不是”）置换反应．<br />（3）反应③的化学反应方程式是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（4）少量气体C通入F溶液中有白色沉淀产生，其反应方程式是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（5）一定量的物质D能与少量G在水溶液中能发生化合反应，且反应后溶液颜色呈浅绿色，则反应的化学方程式是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','H<sub>2</sub>SO<sub>4</sub>$###$不是$###$Fe<sub>2</sub>（SO<sub>4</sub>）<sub>3</sub>+3Ba（OH）<sub>2</sub>=3BaSO<sub>4</sub>↓+2Fe（OH）<sub>3</sub>↓$###$CO<sub>2</sub>+Ba（OH）<sub>2</sub>=BaCO<sub>3</sub>↓+H<sub>2</sub>O$###$Fe+Fe<sub>2</sub>（SO<sub>4</sub>）<sub>3</sub>=3FeSO<sub>4</sub>','【解答】解：（1）B是红棕色粉末，A、C常温下均为气体，常见的红色的粉末有铜、氧化铁、红磷等，常见的还原剂有一氧化碳、木炭、氢气，通过分析可知A是一氧化碳，B是氧化铁，C是二氧化碳，D就是铁，反应③是复分解反应，F是一种可溶性碱，所以硫酸钡沉淀中的硫酸根离子来自G，钡离子来自F，所以E是硫酸，G就是硫酸铁，F是氢氧化钡，氢氧化钡和硫酸亚铁反应生成硫酸钡沉淀和氢氧化铁沉淀，所以H是氢氧化铁，氧化铁和一氧化碳在高温的条件下生成铁和二氧化碳，氧化铁和硫酸反应生成硫酸铁和水，硫酸铁和氢氧化钡反应生成硫酸钡沉淀和氢氧化铁沉淀，推出的各种物质均满足题意，推导合理，所以物质E的化学式：H<sub>2</sub>SO<sub>4</sub>；<br />（2）反应①是两种化合物反应生成一种单质和一种化合物的反应，不满足置换反应的定义，所以该反应不是置换反应；<br />（3）反应③是氢氧化钡和硫酸铁反应生成硫酸钡沉淀和氢氧化铁沉淀，化学方程式为：Fe<sub>2</sub>（SO<sub>4</sub>）<sub>3</sub>+3Ba（OH）<sub>2</sub>=3BaSO<sub>4</sub>↓+2Fe（OH）<sub>3</sub>↓；<br />（4）通过推导可知C是二氧化碳，F是氢氧化钡，二氧化碳和氢氧化钡反应生成碳酸钡沉淀和水，化学方程式为：CO<sub>2</sub>+Ba（OH）<sub>2</sub>=BaCO<sub>3</sub>↓+H<sub>2</sub>O；<br />（5）反应后溶液颜色呈浅绿色，所以溶液中一定含有亚铁离子，通过推导可知，D是铁，G是硫酸铁，根据质量守恒定律可知，铁和硫酸铁会发生化合反应生成硫酸亚铁，反应的方程式为：Fe+Fe<sub>2</sub>（SO<sub>4</sub>）<sub>3</sub>=3FeSO<sub>4</sub>．<br />故答案为：（1）H<sub>2</sub>SO<sub>4</sub>；<br />（2）不是；<br />（3）Fe<sub>2</sub>（SO<sub>4</sub>）<sub>3</sub>+3Ba（OH）<sub>2</sub>=3BaSO<sub>4</sub>↓+2Fe（OH）<sub>3</sub>↓；<br />（4）CO<sub>2</sub>+Ba（OH）<sub>2</sub>=BaCO<sub>3</sub>↓+H<sub>2</sub>O；<br />（5）Fe+Fe<sub>2</sub>（SO<sub>4</sub>）<sub>3</sub>=3FeSO<sub>4</sub>．','【分析】根据B是红棕色粉末，A、C常温下均为气体，常见的红色的粉末有铜、氧化铁、红磷等，常见的还原剂有一氧化碳、木炭、氢气，通过分析可知A是一氧化碳，B是氧化铁，C是二氧化碳，D就是铁，反应③是复分解反应，F是一种可溶性碱，所以硫酸钡沉淀中的硫酸根离子来自G，钡离子来自F，所以E是硫酸，G就是硫酸铁，F是氢氧化钡，氢氧化钡和硫酸铁反应生成硫酸钡沉淀和氢氧化铁沉淀，所以H是氢氧化铁，然后将推出的各种物质代入转化关系中验证即可．','书写',3.00,'f4ec83a106c5eef86c2efde00825234d',9,400,'物质的鉴别、推断,置换反应及其应用,书写化学方程式、文字表达式、电离方程式','',2016,'35','2016春•定陶县期中',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840157,'<img  src=\"/tikuimages/9/2016/400/shoutiniao57/14aadb70-94d4-11e9-b150-b42e9921e93e_xkb54.png\" style=\"vertical-align:middle;FLOAT:right\" />（2016•永新县一模）如图所示为A、B、C三种固体物质（不含结晶水）的溶解度曲线．下列有关说法正确的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />A．t<SUB>1</SUB>℃三种物质的饱和溶液中，溶质质量最大的是B<br />B．t<SUB>2</SUB>℃三种物质的饱和溶液中，溶质的质量分数最大的是A<br />C．可用降低温度的方法使t<SUB>2</SUB>℃时C的不饱和溶液变为饱和溶液<br />D．将t<SUB>2</SUB>℃时的三种物质的饱和溶液降温至t<SUB>1</SUB>℃时，溶质的质量分数由大到小的顺序是：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','B$###$B＞A＞C','【解答】解：A、t<SUB>1</SUB>℃三种物质的饱和溶液质量不能确定，所以溶质质量也不能判断，故A错误；<br />B、t<SUB>2</SUB>℃时，A物质的溶解度最大，所以三种物质的饱和溶液中，溶质的质量分数最大的是A，故B正确；<br />C、C物质的溶解度随温度的降低而增大，所以不能用降低温度的方法使t<SUB>2</SUB>℃时C的不饱和溶液变为饱和溶液，故C错误；<br />D、t<SUB>1</SUB>℃时，B物质的溶解度最大，A物质次之，C物质降低温度不会析出晶体，应该按照t<SUB>2</SUB>℃时的溶解度计算，所以将t<SUB>2</SUB>℃时的三种物质的饱和溶液降温至t<SUB>1</SUB>℃时，溶质的质量分数由大到小的顺序是B＞A＞C，故D正确．<br />故答案为：B，B＞A＞C．','【分析】根据固体的溶解度曲线可以：①查出某物质在一定温度下的溶解度，从而确定物质的溶解性，②比较不同物质在同一温度下的溶解度大小，从而判断饱和溶液中溶质的质量分数的大小，③判断物质的溶解度随温度变化的变化情况，从而判断通过降温结晶还是蒸发结晶的方法达到提纯物质的目的．','填空题',3.00,'b55382aaba64ae48be878e17aaa79af6',9,400,'饱和溶液和不饱和溶液相互转变的方法,固体溶解度曲线及其作用,溶质的质量分数,溶质的质量分数、溶解性和溶解度的关系','',2016,'37','2016•永新县一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840162,'现有X、Y、Z&nbsp;三种金属，根据下列方程式可知，三种金属活动性顺序是（　　）<br />①Z+YSO<SUB>4</SUB>=ZSO<SUB>4</SUB>+Y&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<br />②X+YSO<SUB>4</SUB>=XSO<SUB>4</SUB>+Y<br />③Z+H<SUB>2</SUB>SO<SUB>4</SUB>=ZSO<SUB>4</SUB>+H<SUB>2</SUB>↑&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<br />④X+H<SUB>2</SUB>SO<SUB>4</SUB>不发生反应．','Z＞X＞Y','Z＞Y＞X','X＞Y＞X','X＞Z＞Y','','A','【解答】解：由方程式可知<br />①Z+YSO<SUB>4</SUB>═ZSO<SUB>4</SUB>+Y，说明了Z比Y活泼；<br />②X+YSO<SUB>4</SUB>═XSO<SUB>4</SUB>+Y；说明了X比Y活泼；<br />③Z+H<SUB>2</SUB>SO4═ZSO<SUB>4</SUB>+H<SUB>2</SUB>；说明了Z位于氢的前边；<br />④X+H<SUB>2</SUB>SO<SUB>4</SUB>-不反应，说明了X位于氢的后边；<br />综合以上分析可知，Z＞X＞Y，所以A正确．<br />故选A．','【分析】检验金属活动性时，能与酸反应生成氢气的金属的活动性大于不能与酸反应的金属，或根据金属间的置换进行活动性强弱的比较：能将金属从其盐溶液中置换出来的金属活动性大于盐中金属的活动性．','选择题',3.00,'41edf726ae26c800f63f5f2e3d915990',9,400,'金属活动性顺序及其应用','',0,'37','',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840164,'<img  src=\"/tikuimages/9/2016/400/shoutiniao19/14bc408f-94d4-11e9-9e07-b42e9921e93e_xkb94.png\" style=\"vertical-align:middle;FLOAT:right\" />（2016•合肥模拟）探究氯酸钾和二氧化锰制氧气实验中反应物与催化剂的最佳质量比．实验中采用右图实验装置，取20g氯酸钾，分成10等份，每份2g，分别装入10支试管中．在10支试管中分别加入不同质量的二氧化锰，分别测量二氧化锰与氯酸钾在不同质量比时，生成200ml氧气所消耗的时间．实验记录的数据如下：<br /><table class=\"edittable\"><TBODY><TR><td width=140>&nbsp;实验编号</TD><td width=47>&nbsp;1</TD><td width=47>&nbsp;2</TD><td width=47>&nbsp;3</TD><td width=57>&nbsp;4</TD><td width=47>&nbsp;5</TD><td width=57>&nbsp;6</TD><td width=47>&nbsp;7</TD><td width=42>8&nbsp;</TD><td width=30>9&nbsp;</TD><td width=30>&nbsp;10</TD></TR><TR><td>&nbsp;二氧化锰的质量（g）</TD><td>&nbsp;0.1</TD><td>&nbsp;0.2</TD><td>0.33&nbsp;</TD><td>&nbsp;0.4</TD><td>&nbsp;0.5</TD><td>0.66&nbsp;</TD><td>&nbsp;1</TD><td>&nbsp;1.3</TD><td>&nbsp;2</TD><td>&nbsp;3</TD></TR><TR><td>二氧化锰与氯酸钾的质量比&nbsp;</TD><td>&nbsp;&nbsp;&nbsp;&nbsp; 1：20</TD><td>&nbsp;&nbsp;&nbsp;&nbsp; 1：10</TD><td>&nbsp;&nbsp;&nbsp; 1：6</TD><td>&nbsp;&nbsp;&nbsp;&nbsp;1：5</TD><td>&nbsp;1：4</TD><td>&nbsp;1：3</TD><td>&nbsp;1：2</TD><td>&nbsp;2：3</TD><td>&nbsp;&nbsp;&nbsp; 1：1</TD><td>&nbsp;&nbsp; 3：2&nbsp;</TD></TR><TR><td>&nbsp;生成200mL氧气时间（秒）</TD><td>&nbsp;235</TD><td>&nbsp;186</TD><td>&nbsp;162</TD><td>&nbsp;147</TD><td>&nbsp;133</TD><td>117&nbsp;</TD><td>&nbsp;143</TD><td>&nbsp;160</TD><td>&nbsp;211</TD><td>&nbsp;244</TD></TR></TBODY></TABLE>请回答下列问题：<br />（1）实验中收集一定体积的氧气采用的实验方法利用了氧气<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>的性质．收集气体前集气瓶内未装满水，对实验结果<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>&nbsp;（填“有”或“没有”）影响．<br />（2）分析上述数据，你认为利用此法制氧气时，二氧化锰与氯酸钾的最佳质量比为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）利用双氧水和二氧化锰的混合物制氧气比加热氯酸钾和二氧化锰混合物制氧气更好．请写出前者的两个优点①<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>②<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（4）反思：催化剂能影响化学反应速度，温度、反应物的浓度、接触面积等也会对化学反应速度带来影响．分析上表数据，结合影响化学反应速度的分析：<br />①当二氧化锰与氯酸钾的质量比高于最佳质量比时，反应速度变低的原因：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />②当二氧化锰与氯酸钾的质量比低于最佳质量比时，反应速度变低的原因是：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','不易溶于水$###$没有$###$1：3$###$操作简便$###$节能$###$二氧化锰用量多，导致反应物氯酸钾的质量分数变小$###$二氧化锰用量少，导致氯酸钾与二氧化锰的接触面积变小','【解答】解：（1）利用排水装置收集的气体应具备的性质是难溶于水；空气不易溶于水，那么排出水的气体即为所收集的气体体积，．<br />故答案为：不易溶于水；没有；<br />（2）通过观察图表中的数据可知，二氧化锰与氯酸钾的质量比是 1：3时，所用的时间最少，反应速率最快．<br />故答案为：1：3；<br />（3）双氧水和二氧化锰混合制氧气不需加热，可以节能，同时更安全；副产物是水无污染，更环保；价格低廉，操作简便等；<br />故答案为：操作简便；节能；<br />（4）①二氧化锰用量过多氯酸钾的质量分数就降低了，反应物的接触面积就少了，就影响了反应速度；<br />②反应速度的影响因素主要是反应物的接触面积，接触面积越大反应速度就越快，二氧化锰的用量过少与氯酸钾的接触面积就越小，所以反应速度慢；<br />故答案为：①二氧化锰用量多，导致反应物氯酸钾的质量分数变小；<br />②二氧化锰用量少，导致氯酸钾与二氧化锰的接触面积变小．','【分析】（1）根据氧气的溶解性和排出水的体积即为气体体积来分析<br />（2）根据图表中的数据分析，所用时间少的就是反应速率快的；<br />（3）双氧水和二氧化锰混合制氧气不需加热，可以节能等分析；<br />（4）根据反应速度的影响因素主要是反应物的接触面积；混合物的混合情况考虑，催化剂多了相对来说反应物就少了，反应速度就收到了影响．','填空题',3.00,'1f445338da8719dd902a1bf6d722fa2f',9,400,'影响化学反应速率的因素探究,催化剂的特点与催化作用','',2016,'32','2016•合肥模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840165,'水是一种重要的资源．<br />（1）自来水厂净水过程中用到活性炭，其作用是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）硬水的危害很大，生活中将硬水软化的方法是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）下列做法中，有利于保护水资源的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />A．使用节水型马桶　　　　　&nbsp;B．生活污水任意排放<br />C．合理使用农药和化肥&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;D．工业废水处理达标后排放<br />（4）下列各组白色固体能用水区分的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />A．纯碱、烧碱&nbsp;&nbsp;&nbsp;B．生石灰、熟石灰&nbsp;&nbsp;&nbsp;C．淀粉、白糖&nbsp;&nbsp;&nbsp;D．氯化钠、硝酸铵．','','','','','','吸附作用$###$煮沸$###$ACD$###$ABCD','【解答】解：<br />（1）自来水厂净水过程中用到活性炭，其作用是吸附作用；<br />（2）生活中将硬水软化的方法是加热煮沸；<br />（3）A、使用节水型马桶，可以节约用水，有利于保护水资源，故选项正确．<br />B、合理使用化肥农药，不会造成水体的污染，有利于保护水资源，故选项正确．<br />C、生活污水任意排放，会造成水体的污染，不利于保护水资源，故选项错误．<br />D、工业废水处理达标后排放，不会造成水体的污染，有利于保护水资源，故选项正确．<br />（4）A、碳酸钠溶于水没有明显的现象，而氢氧化钠溶于水会放出大量的热，现象不同，可以鉴别；<br />B、碳酸钙不溶于水，且遇到水没有明显的变化，而生石灰遇到水会与水反应放出大量的热，现象不同，可以鉴别；<br />C、淀粉不易溶于水、白糖易溶于水，可以鉴别；<br />D、氯化钠溶于水没有明显的变化，而硝酸铵溶于水时会吸热，导致溶液的温度降低，现象不同，可以鉴别；<br />答案：<br />（1）吸附作用；&nbsp;&nbsp;&nbsp;（2）煮沸；&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;（3）ACD；&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;（4）ABCD．','【分析】（1）根据活性炭具有吸附性解答；<br />（2）根据生活中将硬水软化的方法是加热煮沸解答；<br />（3）根据保护水资源要从两个方面做起：一方面要节约用水，另一方面要防止水体污染，据此结合题意进行分析判断；<br />（4）根据已有的知识进行分析，能用水区分的物质，需要遇到水时产生明显的不同现象，如溶液的颜色或物质是否溶于水等．','填空题',3.00,'1aec86b4ee8c21aedc7038307f8c367b',9,400,'自来水的生产过程与净化方法,硬水与软水,物质的鉴别、推断,保护水资源和节约用水','',2016,'37','2016•朝阳区一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840166,'20℃时，将等质量的甲、乙两种固体物质，分别加入到盛有100g水的烧杯中，充分搅拌后现象如图1，加热到50℃时现象如图2，甲、乙两种物质的溶解度曲线如图3．请结合图示回答下列问题：<br /><img  src=\"/tikuimages/9/2014/400/shoutiniao70/14c3456e-94d4-11e9-810d-b42e9921e93e_xkb85.png\" style=\"vertical-align:middle\" /><br />（1）图1中一定为饱和溶液的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）图2中甲、乙两溶液中溶质质量分数的大小关系为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）图3中表示乙的溶解度曲线是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；P点表示的含义是：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','甲$###$甲=乙$###$N$###$甲、乙两种固体物质的溶解度相等','【解答】解：（1）从图1中可以知道甲没有完全溶解，则说明甲中形成的溶液为饱和溶液；<br />（2）图2中相同质量的溶剂溶解了相同质量的溶质，所以它们形成溶液的质量也相等，结合溶质的质量分数的计算公式可以知道，它们形成的溶液中溶质的质量分数也相等；<br />（3）根据图1中信息可以知道20℃时甲的溶解度小于乙的溶解度，而图3中在20℃时，N曲线的溶解度大于M曲线的溶解度，故N表示的是乙物质的溶解度曲线；根据溶解度曲线的意义可以知道P点表示30℃时，甲、乙两种固体物质的溶解度相等．<br />故答案为：（1）甲；&nbsp;&nbsp;（2）相等或甲=乙；&nbsp;&nbsp;（3）N、30℃时，甲、乙两种固体物质的溶解度相等．','【分析】（1）根据饱和溶液的定义进行解答；<br />（2）根据题意可以知道图2中相同质量的溶剂溶解了相同质量的溶质，结合溶质的质量分数的计算公式可以完成解答；<br />（3）根据图中信息可以知道20℃时甲的溶解度小于乙的溶解度，结合图3可以完成解答，两条溶解度曲线的交点表示该点所示的温度下，两物质的溶解度是相同的，可以据此判断P点的意义．','填空题',3.00,'07d4bbed3b28d85d1efe542dce74b79c',9,400,'饱和溶液和不饱和溶液,固体溶解度曲线及其作用,溶质的质量分数','',2014,'32','2014•番禺区模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840167,'请结合图示实验装置，回答下列问题．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao89/14c89c9e-94d4-11e9-b66d-b42e9921e93e_xkb98.png\" style=\"vertical-align:middle\" /><br />（1）写出有标号的仪器名称：①<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，②<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）若B装置中固体为锌粒，②中加入<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>试剂可制H<SUB>2</SUB>．要获得干燥的H<SUB>2</SUB>，应先将气体通过盛有<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，（填名称）的D装置，再用E装置收集，气体应从<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>端通入（选填“a”或“b”）．<br />（3）实验室可用B或C装置制CO<SUB>2</SUB>，C相对于B装置的优点有<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（4）若用F装置收集NH<SUB>3</SUB>，油层的作用是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','酒精灯$###$分液漏斗$###$稀硫酸$###$浓硫酸$###$a$###$能够随时使反应进行或停止$###$防止氨气与水反应','【解答】解：（1）①是酒精灯，常用作热源；②是分液漏斗，通过分液漏斗可以向反应容器中注入液体药品．<br />故填：酒精灯；分液漏斗．<br />（2）实验室制取氢气是用锌和稀硫酸反应制的，利用洗气瓶干燥氢气可用浓硫酸，氢气的密度比空气小，可用向下排空气法；故填：稀硫酸；浓硫酸；a；<br />（3）与B装置相比其优点是能够随时使反应进行或停止，如用F装置收集H<SUB>2</SUB>，由于氢气的密度比空气小，则气体应从d端通入．故填：能够随时使反应进行或停止；<br />（4）氨气的密度比空气小，能溶于水，要想用排水法收集，必须使氨气和水分离，故用油层隔离水；故填：防止氨气与水反应．','【分析】（1）要熟悉各种仪器的名称、用途和使用方法；&nbsp;&nbsp; <br />（2）实验室制取氢气是用锌和稀硫酸反应制的，利用洗气瓶干燥氢气可用浓硫酸，氢气的密度比空气小，可用向下排空气法；&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<br />（3）C装置能够随时使反应进行或停止，原理是：关闭开关时，试管中的气体增多，压强增大，把液体压入长颈漏斗，固体和液体分离，反应停止；打开开关时，气体导出，试管中的气体减少，压强减小，液体和固体混合，反应进行；<br />（4）氨气的密度比空气小，能溶于水，要想用排水法收集，必须使氨气和水分离；','填空题',3.00,'d24ad3638b8aa823d28c04c3a3d2c2d6',9,400,'气体的干燥（除水）,二氧化碳的实验室制法,氢气的制取和检验','',2016,'32','2016•相城区模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840171,'<img  src=\"/tikuimages/9/2016/400/shoutiniao11/14d1010f-94d4-11e9-98ef-b42e9921e93e_xkb63.png\" style=\"vertical-align:middle;FLOAT:right\" />（2016•丰台区一模）小东在科学课上自制了一瓶饮料．<br />（1）蔗糖（C<SUB>12</SUB>H<SUB>22</SUB>O<SUB>11</SUB>）属于<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“有机物”或“无机物”）．<br />（2）汽水中的气体是柠檬酸与配方中的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>反应生成的，人体胃液中的酸也可以与该成分反应，该反应的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','有机物$###$小苏打$###$NaHCO<SUB>3</SUB>+HCl═NaCl+CO<SUB>2</SUB>↑+H<SUB>2</SUB>O','【解答】解：（1）蔗糖是含有碳元素的化合物，属于有机物．<br />故填：有机物．<br />（2）汽水中的气体是柠檬酸与配方中的小苏打反应生成的，人体胃液中的盐酸能和碳酸氢钠反应，该反应的化学方程式为：NaHCO<SUB>3</SUB>+HCl═NaCl+CO<SUB>2</SUB>↑+H<SUB>2</SUB>O．<br />故填：有机物；小苏打；NaHCO<SUB>3</SUB>+HCl═NaCl+CO<SUB>2</SUB>↑+H<SUB>2</SUB>O．','【分析】含有碳元素的化合物一般属于有机物；<br />小苏打是碳酸氢钠的俗称，能和柠檬酸反应生成二氧化碳气体，能和稀盐酸反应生成氯化钠、水和二氧化碳．','书写',3.00,'b08247c4c40f1ea23d292b543c7e26a5',9,400,'酸的化学性质,有机物与无机物的区别,书写化学方程式、文字表达式、电离方程式','',2016,'37','2016•丰台区一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840175,'下列做法可以达到目的是（　　）','寻找新的催化剂使水变成油','测定空气中氧气含量的实验，用白磷代替红磷','用燃着的木条检验二氧化碳','通过加高烟囱排放工业废气，防止空气污染','','B','【解答】解：A、根据质量守恒定律，反应前后元素的种类不变，水是由氢、氧两种元素组成，油中含有碳元素，水永远不会变成油，所以错误．<br />B、白磷、红磷燃烧的产物相同，故测定空气中氧气含量的实验，用白磷代替红磷，所以正确．<br />C、二氧化碳能灭火，说明二氧化碳不能燃烧、不能支持燃烧，除了二氧化碳外，氮气等一些气体也不能燃烧、不能支持燃烧，也可用于灭火，所以检验二氧化碳应用澄清的石灰水，不能用燃着的木条检验．<br />D、加高烟囱排放工业废气，只是把工业废气排放到高空中，还是会造成空气污染，所以错误．<br />故选B．','【分析】A、根据质量守恒定律判断．<br />B、根据白磷、红磷燃烧的产物相同判断．<br />C、检验是利用物质典型的特性或特征现象来证明物质，除了二氧化碳，还有其它不能支持燃烧的气体，据此进行分析解答．<br />D、从加高烟囱排放工业废气，只是把工业废气排放到高空中，还是会造成空气污染去分析解答．','选择题',3.00,'365628e23150d578fb223084de50b02c',9,400,'空气组成的测定,防治空气污染的措施,二氧化碳的检验和验满,质量守恒定律及其应用','',2015,'33','2015秋•平和县期末',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840177,'下列实验操作正确的是（　　）','<img  src=\"/tikuimages/9/2016/400/shoutiniao53/14d8a230-94d4-11e9-8db1-b42e9921e93e_xkb98.png\" style=\"vertical-align:middle\" /><br />测溶液pH','<img  src=\"/tikuimages/9/2016/400/shoutiniao36/14d96580-94d4-11e9-8b25-b42e9921e93e_xkb24.png\" style=\"vertical-align:middle\" /><br />过滤','<img  src=\"/tikuimages/9/2016/400/shoutiniao44/14daec21-94d4-11e9-a95e-b42e9921e93e_xkb39.png\" style=\"vertical-align:middle\" /><br />稀释浓硫酸','<img  src=\"/tikuimages/9/2016/400/shoutiniao65/14dbaf70-94d4-11e9-be00-b42e9921e93e_xkb39.png\" style=\"vertical-align:middle\" />检查气密性','','D','【解答】解：A、用pH试纸测定未知溶液的pH时，正确的操作方法为用玻璃棒蘸取少量待测液滴在干燥的pH试纸上，与标准比色卡对比来确定pH．不能将pH试纸伸入待测液中，以免污染待测液，图中所示操作错误．<br />B、过滤液体时，要注意“一贴、二低、三靠”的原则，图中缺少玻璃棒引流、漏斗下端没有紧靠在烧杯内壁上，图中所示操作错误．<br />C、稀释浓硫酸时，要把浓硫酸缓缓地沿器壁注入水中，同时用玻璃棒不断搅拌，以使热量及时地扩散；一定不能把水注入浓硫酸中；图中所示操作错误．<br />D、检查装置气密性的方法：把导管的一端浸没在水里，双手紧贴容器外壁，若导管口有气泡冒出，装置不漏气；图中所示操作正确．<br />故选：D．','【分析】A、根据用pH试纸测定未知溶液的pH的方法进行分析判断．<br />B、过滤液体时，注意“一贴、二低、三靠”的原则．<br />C、根据浓硫酸的稀释方法（酸入水，沿器壁，慢慢倒，不断搅）进行分析判断．<br />D、根据检查装置气密性的方法进行分析判断．','选择题',3.00,'e9cca641275865d658826404a3a13ab7',9,400,'浓硫酸的性质及浓硫酸的稀释,过滤的原理、方法及其应用,检查装置的气密性,溶液的酸碱度测定','',2016,'37','2016•龙岗区二模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840178,'有关水和溶液的知识是初中化学的重要内容，根据你所学的知识，回答下列问题．<br />（1）自来水厂净水的主要过程是沉淀、<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>、吸附、消毒．<br />（2）家庭中常用<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>来区别硬水和软水．<br />（3）水电解的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，由此说明水是由<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>组成的．','','','','','','过滤$###$肥皂水$###$2H<SUB>2</SUB>O<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;通电&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>↑+O<SUB>2</SUB>↑$###$氢元素和氧元素','【解答】解：<br />（1）生产自来水的净水方法有沉淀、吸附、过滤、消毒等，其中沉淀、吸附、过滤，没有新物质生成，属于物理变化；<br />（2）家庭中常用肥皂水来区别硬水和软水，泡沫多的是软水，泡沫少的是硬水；<br />（3）水电解生成的是氢气和氧气，化学方程式为：2H<SUB>2</SUB>O<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;通电&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>↑+O<SUB>2</SUB>↑，说明水是由氢元素和氧元素组成的．<br />答案：（1）过滤；（2）煮沸；蒸馏；（3）2H<SUB>2</SUB>O<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;通电&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>↑+O<SUB>2</SUB>↑，氢元素和氧元素．','【分析】（1）根据已有的知识进行分析，生产自来水的净水方法有沉淀、吸附、过滤、消毒等解答；<br />（2）根据家庭中常用肥皂水来区别硬水和软水解答；<br />（3）水电解生成的是氢气和氧气，据此解答．','书写',3.00,'c84a3408a27b3d0c3c2c03bbcd980942',9,400,'电解水实验,自来水的生产过程与净化方法,硬水与软水,书写化学方程式、文字表达式、电离方程式','',2016,'32','2016•洛阳模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840180,'分类、类比是初中化学常用的学习方法．<br />请根据某种方式对下列说法进行分类．<br />a．通常状况下，水是无色、无臭的液体；b．在通电条件下，水分解生成氢气和氧气；e．水与二氧化碳反应生成碳酸；f．在101kPa，水加热至100℃水变成水蒸气；g．通常状况下，水能与金属钠反应生成氢氧化钠和氢气；h．用水和氯化钠配制生理盐水．<br />选出的一类的序号是①<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，它们归为一类的依据是②<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','afh或beg$###$属于物理性质或属于化学性质','【解答】解：属于物理性质的有：a．通常状况下，水是无色、无臭的液体；f．在101kPa，水加热至100℃水变成水蒸气；h．用水和氯化钠配制生理盐水；属于化学性质的有：b．在通电条件下，水分解生成氢气和氧气；e．水与二氧化碳反应生成碳酸；g．通常状况下，水能与金属钠反应生成氢氧化钠和氢气；故答案为：afh或beg；属于物理性质或属于化学性质；','【分析】物质在化学变化中表现出来的性质叫化学性质，物质不需要发生化学变化就表现出来的性质，叫物理性质；物理性质经常表现为：颜色、状态、气味、密度、硬度、熔点、沸点、导电性、导热性、溶解性、挥发性等．属于物理性质的有：a．通常状况下，水是无色、无臭的液体；f．在101kPa，水加热至100℃水变成水蒸气；h．用水和氯化钠配制生理盐水；属于化学性质的有：b．在通电条件下，水分解生成氢气和氧气；e．水与二氧化碳反应生成碳酸；g．通常状况下，水能与金属钠反应生成氢氧化钠和氢气．','填空题',2.00,'b574cb49eb75e9e76fbd7ae240496f5d',9,400,'化学性质与物理性质的差别及应用','',0,'37','',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840184,'如图所示为实验室中常见的气体制备和收集装置．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao62/14f48ea1-94d4-11e9-b47c-b42e9921e93e_xkb64.png\" style=\"vertical-align:middle\" /><br />请回答下列问题：<br />（1）标有序号&nbsp;m的仪器的名称为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）实验室用高锰酸钾制取氧气，应选用的发生装置<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填字母序号），反应的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．欲使用装置E用排空气法收集一瓶氧气，则气体应从<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“a”或“b”）端通入．<br />（3）实验室用碳酸钙和稀盐酸制取二氧化碳，若选用C做发生装置，你认为选用C的优点是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（4）已知实验室制取一瓶一氧化氮气体时选用F作为收集装置，而不能选用D装置来收集，由此可以推断一氧化氮气体可能具有的性质是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．（填字母序号）<br />①能溶于水&nbsp;&nbsp;&nbsp;②难溶于水&nbsp;&nbsp;&nbsp;&nbsp;③密度比空气大&nbsp;&nbsp;&nbsp;&nbsp;④能和空气中某种气体发生反应．','','','','','','铁架台$###$A$###$2KMnO<SUB>4</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>K<SUB>2</SUB>MnO<SUB>4</SUB>+MnO<SUB>2</SUB>+O<SUB>2</SUB>↑$###$a$###$可以准确的控制药品的添加量，便于控制反应速率，节约药品$###$②④','【解答】解：（1）m是铁架台；<br />（2）用高锰酸钾制氧气时，反应物是高锰酸钾，生成物是锰酸钾、二氧化锰、氧气，反应条件是加热，因此发生装置的特点属于固体加热型的A装置；<br />反应的化学方程式为：2KMnO<SUB>4</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>K<SUB>2</SUB>MnO<SUB>4</SUB>+MnO<SUB>2</SUB>+O<SUB>2</SUB>↑；因为氧气的密度比空气大，故氧气应该从长管a进入，把空气从短管排出；<br />（3）实验室用碳酸钙和稀盐酸制取二氧化碳，若选用C做发生装置，因为通过注射器可以控制液体的流速，所以选用C的优点是：可以准确的控制药品的添加量，便于控制反应速率，节约药品；<br />（4）根据信息一氧化氮气体时选用F-排水法作为收集装置，而不能选用D-向上排空气法装置来收集，说明一氧化氮难溶于水，能和空气中某种气体发生反应，故选择②④．<br />故答案为：（1）铁架台<br />（2）A，2KMnO<SUB>4</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>K<SUB>2</SUB>MnO<SUB>4</SUB>+MnO<SUB>2</SUB>+O<SUB>2</SUB>↑，a；<br />（3）可以准确的控制药品的添加量，便于控制反应速率，节约药品（合理即可）<br />（4）②④．','【分析】（1）熟练掌握常见化学仪器的名称及其用途；<br />（2）根据用高锰酸钾制取氧气的反应物的状态和反应条件选择发生装置，并写出反应的方程式；根据氧气的密度选择进气管；<br />（3）根据发生装置C中注射器的作用分析优点；<br />（4）根据信息中的收集方法推测一氧化氮的性质：溶解性和密度等','书写',3.00,'7cee3c2b112362dc68b8ef939bb11112',9,400,'实验室制取氧气的反应原理,二氧化碳的实验室制法,书写化学方程式、文字表达式、电离方程式','',2016,'37','2016•娄星区一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840188,'豆科植物的根瘤菌，能直接固定空气中某气体成为农作物养料，这种气体是（　　）','氨气','氮气','碳酸气','氧气','','B','【解答】解：豆科植物的根瘤菌有固氮作用，它能将空气中不能被植物吸收的氮气转化为可被植物吸收的氮．而氮元素是植物生长的必须元素．<br />故选：B','【分析】根据各种气体的性质和用途分析解答．','选择题',3.00,'aaa75b1d6de80857fc83315125c766c0',9,400,'常见气体的用途','',0,'37','',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840190,'实验室提供了下列仪器，请选用合适的仪器完成指定的实验．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao73/150e7f40-94d4-11e9-b8e1-b42e9921e93e_xkb65.png\" style=\"vertical-align:middle\" /><br />（1）写出图中仪器a的名称：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）实验室制取氧气已经选用f仪器，则组装气体的发生装置还需要选用的三种仪器是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填序号），该反应的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，实验室制取二氧化碳已经选用a仪器，为了便于添加液体药品，组装气体的发生装置还需要<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填需要），该反应的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，收集以上两种气体都可以采用<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>法．<br />（3）提纯粗盐，溶解时需要选用上述一种仪器是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填序号），除此之外还需添加的仪器是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填名称），该仪器溶解过程中的作用是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，若再添加天平、量筒可进行一定溶质质量分数溶液的配制，配制过程中取水时若仰视量筒刻度，其他步骤都正确，则配制的溶液的溶质质量分数会<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“偏大”、“偏小”或“不变”）．','','','','','','锥形瓶$###$bdh$###$2KMnO<SUB>4</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>K<SUB>2</SUB>MnO<SUB>4</SUB>+MnO<SUB>2</SUB>+O<SUB>2</SUB>↑$###$cg$###$CaCO<SUB>3</SUB>+2HCl═CaCl<SUB>2</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑$###$向上排空气$###$e$###$玻璃棒$###$加快溶解速率$###$偏小','【解答】解：（1）图中仪器a的名称为：锥形瓶；<br />（2）实验室制取氧气已经选用f仪器，需要加热，则组装气体的发生装置还需要选用的三种仪器是bdh，该反应的化学方程式为2KMnO<SUB>4</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>K<SUB>2</SUB>MnO<SUB>4</SUB>+MnO<SUB>2</SUB>+O<SUB>2</SUB>↑（或2KClO<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span></td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>△</td></tr></table></span>2KCl+3O<SUB>2</SUB>↑）；实验室制取二氧化碳已经选用a仪器，为了便于添加液体药品，需要长颈漏斗；组装气体的发生装置还需要&nbsp;cg，该反应的化学方程式为CaCO<SUB>3</SUB>+2HCl═CaCl<SUB>2</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑，因为氧气、二氧化碳的密度大于空气，故收集以上两种气体都可以采用向上排空气法．<br />（3）过滤要用到的仪器有：铁架台、烧杯、漏斗、玻璃棒等．所以除图中的铁架台、烧杯之外，还需要玻璃棒和漏斗，该仪器字溶解过程中的作用是加快溶解速率，若再添加天平、量筒可进行一定溶质质量分数溶液的配制，配制过程中取水时若仰视量筒刻度，则量取的水偏多，其他步骤都正确，则配制的溶液的溶质质量分数会偏小．<br />故答案为：<br />（1）锥形瓶；（2）bdh；2KMnO<SUB>4</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>K<SUB>2</SUB>MnO<SUB>4</SUB>+MnO<SUB>2</SUB>+O<SUB>2</SUB>↑（或2KClO<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span></td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>△</td></tr></table></span>2KCl+3O<SUB>2</SUB>↑）；&nbsp;cg；<br />CaCO<SUB>3</SUB>+2HCl═CaCl<SUB>2</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑；&nbsp;向上排空气；<br />（3）e；玻璃棒；加快溶解速率；偏小．','【分析】（1）根据实验室常用仪器的名称和题中所指仪器的作用进行分析；<br />（2）实验室制取氧气已经选用f仪器，考虑需要加热；实验室制取二氧化碳已经选用a仪器，为了便于添加液体药品，考虑长颈漏斗，进行分析．<br />（3）过滤要用到的仪器有：铁架台、烧杯、漏斗、玻璃棒等；考虑取水时若仰视量筒刻度，引起的误差．','书写',3.00,'fdf567e4f723f29b8beb3ba84ff018ee',9,400,'一定溶质质量分数的溶液的配制,实验室制取氧气的反应原理,二氧化碳的实验室制法,氯化钠与粗盐提纯,书写化学方程式、文字表达式、电离方程式','东台市',2016,'37','2016•东台市一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840191,'下列物质由离子构成的是（　　）','铁','金刚石','氧气','硫酸铜','','D','【解答】解：A、铁属于金属单质，是由铁原子直接构成的，故选项错误．<br />B、金刚石属于固态非金属单质，是由碳原子直接构成的，故选项错误．<br />C、氧气属于气态非金属单质，是由氧分子构成的，故选项错误．<br />D、硫酸铜是由铜离子和硫酸根离子构成的，故选项正确．<br />故选：D．','【分析】根据金属、大多数固态非金属单质、稀有气体等由原子构成；有些物质是由分子构成的，气态的非金属单质和由非金属元素组成的化合物，如氢气、水等；有些物质是由离子构成的，一般是含有金属元素和非金属元素的化合物，如氯化钠，进行分析判断即可．','选择题',3.00,'73ca5fc6cc62e8b44c48f511ae770f42',9,400,'物质的构成和含量分析','',2016,'35','2016春•扶沟县期中',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840196,'下列实验操作错误的是（　　）','<img  src=\"/tikuimages/9/2015/400/shoutiniao6/15247840-94d4-11e9-9996-b42e9921e93e_xkb83.png\" style=\"vertical-align:middle\" /><br />&nbsp;&nbsp;&nbsp;给液体加热','<img  src=\"/tikuimages/9/2015/400/shoutiniao23/15305f21-94d4-11e9-b06d-b42e9921e93e_xkb62.png\" style=\"vertical-align:middle\" /><br />&nbsp;&nbsp;&nbsp; 倾倒液体','<img  src=\"/tikuimages/9/2015/400/shoutiniao56/1534f300-94d4-11e9-ae2f-b42e9921e93e_xkb42.png\" style=\"vertical-align:middle\" /><br />&nbsp;&nbsp;蒸发食盐水','<img  src=\"/tikuimages/9/2015/400/shoutiniao78/153715de-94d4-11e9-bd77-b42e9921e93e_xkb75.png\" style=\"vertical-align:middle\" /><br />&nbsp;&nbsp; 溶解','','B','【解答】解：A、给试管中的液体加热，要用酒精灯的外焰加热，试管内液体不能超过其体积的<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">1</td></tr><tr><td>3</td></tr></table></span>，试管夹夹在试管的中上部，图中所示操作正确；<br />B、取用液体药品时，瓶塞要倒放，标签要对准手心，瓶口紧挨；图中所示操作错误；<br />C、蒸发时，应用玻璃棒不断搅拌，以防液体受热不均匀，造成液体飞溅，图中所示操作正确；<br />D、溶解时可用玻璃棒搅拌，可加快溶解的速度，图中所示操作正确；<br />故选B．','【分析】A、根据给试管中的液体加热的方法分析．<br />B、根据倾倒液体时的注意事项进行分析．<br />C、根据蒸发食盐水的方法分析．<br />D、根据溶解的注意事项进行分析．','选择题',3.00,'e472ce154a1fb0b3036dba78249e0447',9,400,'液体药品的取用,给试管里的液体加热,物质的溶解,蒸发与蒸馏操作','',2015,'37','2015秋•高邑县校级月考',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840198,'金属的应用广泛，回答生活中下列金属应用的有关问题．<br />（1）实验室的废酸液不能直接倒入下水道（铸铁制）是因为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）工人师傅在切割钢板时，常用硫酸铜溶液画线是因为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（用化学方程式表示）．<br />（3）铜常用来做精密仪器的电子元件，利用的是其<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．钛及钛合金被认为是21世纪最重要的金属材料，其主要原因是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','酸液会腐蚀铸铁管道$###$Fe+CuSO<SUB>4</SUB>═Cu+FeSO<SUB>4</SUB>$###$导电性$###$抗腐蚀性强','【解答】解：（1）实验室的废酸液不能直接倒入下水道（铸铁制）是因为酸液会腐蚀铸铁管道．<br />故填：酸液会腐蚀铸铁管道．<br />（2）工人师傅在切割钢板时，常用硫酸铜溶液画线是因为铁能和硫酸铜反应生成硫酸亚铁和铜，反应的化学方程式为：Fe+CuSO<SUB>4</SUB>═Cu+FeSO<SUB>4</SUB>．<br />故填：Fe+CuSO<SUB>4</SUB>═Cu+FeSO<SUB>4</SUB>．<br />（3）铜常用来做精密仪器的电子元件，利用的是其导电性；<br />钛及钛合金被认为是21世纪最重要的金属材料，其主要原因是抗腐蚀性强．<br />故填：导电性；抗腐蚀性强．','【分析】铁能和酸反应生成盐和氢气；<br />铁能和硫酸天反应生成硫酸亚铁和铜；<br />铜具有良好的导电性；<br />钛合金的抗腐蚀性强．','填空题',3.00,'b52b166bc3d6e8c1316b85cf1f0d1d12',9,400,'金属的物理性质及用途,金属的化学性质','',0,'37','',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840200,'<img  src=\"/tikuimages/9/2016/400/shoutiniao96/154d5d00-94d4-11e9-88cc-b42e9921e93e_xkb28.png\" style=\"vertical-align:middle;FLOAT:right;\" />（2016•谷城县模拟）手机在人们的生活中使用广泛，如图是充电或连接用的数据线．<br />（1）所标物质中属于金属单质的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，属于有机合成材料的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）数据线的铁制插头在日常潮湿的空气中会生锈，铁生锈是一个复杂的过程，铁锈的主要成分是Fe<SUB>2</SUB>O<SUB>3</SUB>•xH<SUB>2</SUB>O，由此可知参加反应的是铁、氧气和<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）写出只有一步反应表示铁比铜活泼的化学方程式：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','铁或铜$###$塑料$###$水$###$Fe+CuSO<SUB>4</SUB>=Cu+FeSO<SUB>4</SUB>','【解答】解：<br />（1）据图可以看出，铁和铜是金属单质，塑料属于有机合成材料，故填：铁或铜，塑料；<br />（2）铁与水和氧气并存时易生锈，故填：水；<br />（3）铁与硫酸铜反应生成硫酸亚铁和铜，故填：Fe+CuSO<SUB>4</SUB>=Cu+FeSO<SUB>4</SUB>；<br />答案：<br />（1）铁或铜；塑料；<br />（2）水；<br />（3）Fe+CuSO<SUB>4</SUB>=Cu+FeSO<SUB>4</SUB>；','【分析】（1）根据已有的物质的类别解答；<br />（2）根据铁生锈的条件解答；<br />（3）根据铁与硫酸铜反应的知识进行分析解答即可．','书写',3.00,'2b4aed87942acf299ef3c2d3dc150292',9,400,'金属的化学性质,金属材料及其应用,金属锈蚀的条件及其防护,书写化学方程式、文字表达式、电离方程式,合成材料的使用及其对人和环境的影响','',2016,'32','2016•谷城县模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840201,'化学与我们的生活息息相关，请从下列物质中选择相应的序号填空：<br />①氧气&nbsp;&nbsp;②不锈钢&nbsp;③生石灰&nbsp;&nbsp;④食醋&nbsp;&nbsp;⑤氢氧化铝<br />（1）可用于抢救危重病人的物质是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（2）常用作食品干燥剂的物质是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（3）常用作制造刀具的物质是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（4）可用于治疗胃酸过多的物质是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（5）厨房调味品pH＜7的物质是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','①$###$③$###$②$###$⑤$###$④','【解答】解：（1）氧气能供给呼吸，可用于抢救危重病人；<br />（2）生石灰能与水反应生成了氢氧化钙，常用作食品干燥剂；<br />（3）不锈钢的硬度大，常用作制造刀具；<br />（4）氢氧化铝能与盐酸发生中和反应，无毒，可用于治疗胃酸过多；<br />（5）食醋是厨房调味品，主要成分是醋酸，pH＜7．<br />故答为：（1）①；（2）③；（3）②；（4）⑤；（5）④．','【分析】物质的性质决定物质的用途，根据常见物质的性质和用途分析回答．','填空题',3.00,'947d3ecc7e6f40dda3add5c8437be0f9',9,400,'氧气的用途,生铁和钢,生石灰的性质与用途,中和反应及其应用,溶液的酸碱性与pH值的关系','',2016,'37','2016•玉林一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840204,'鉴别下列各组物质所用的方法或试剂错误的是（　　）','氧气和氮气--带火星的木条','硬水和软水--静置观察有无沉淀','碳酸钠溶液和氯化钠溶液--无色的酚酞溶液','熟石灰和生石灰--水','','B','【解答】解：A、氧气可使木条复燃，而氮气无明显变化，故方法正确；<br />B、硬水和软水含有的矿物质不同，静置都不会产生沉淀，应该取样品，加入肥皂水，振荡，产生泡沫较多的是软水，产生泡沫较少的是硬水进行鉴别，故方法错误；<br />C、氯化钠溶液呈中性，不能使酚酞试液变色，而碳酸钠溶液呈碱性，能使酚酞试液变红，现象不同，可以鉴别，故方法正确；<br />D、氧化钙和水反应放热，而氢氧化钙无明显变化，可以鉴别，故方法正确；<br />故选项为：B．','【分析】A、带火星的木条在氧气中会复燃，可以据此解答该题；<br />B、根据硬水和软水的成分不同进行分析；<br />C、根据氯化钠溶液呈中性，不能使酚酞试液变色，而碳酸钠溶液呈碱性，能使酚酞试液变红进行分析；<br />D、根据氧化钙和水反应放热分析．','选择题',3.00,'c69b55faf272f75e0880895f07d02d12',9,400,'常见气体的检验与除杂方法,硬水与软水,酸、碱、盐的鉴别,物质的鉴别、推断','',2016,'32','2016•宜昌模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840205,'化学实验既要操作规范，更要保障安全．下列实验基本操作不符合这一要求的是（　　）','不能直接把鼻孔凑到试剂瓶口闻浓盐酸的气味','给试管里的液体加热时，试管口不能对着自己或他人','不慎将浓硫酸洒在皮肤上，先用大量水冲，再涂抹氢氧化钠溶液','用一氧化碳与氧化铁反应时，先要通入气体将装置内的空气排净后再加热','','C','【解答】解：A、闻药品气味时，要用手在瓶口扇动，让少量气味进入鼻孔，不能直接用鼻子去闻，故A操作正确．<br />B、给试管里的液体加热时，试管口不能对着自己或他人，以防试管中的液体喷出伤人；故B操作正确．<br />C、浓硫酸溶于水放出大量的热，并有强烈的腐蚀性，浓硫酸沾到皮肤上，应先用布擦去，再用大量水冲洗，然后涂上3%～5%的碳酸氢钠溶液；氢氧化钠溶液具有腐蚀性，不能涂氢氧化钠溶液，故C操作错误．<br />D、由于一氧化碳具有可燃性，与空气的混合气体在点燃时已发生爆炸．所以，在一氧化碳还原氧化铁时，要先通入一氧化碳气体后再加热，其目的是赶走试管里的空气，以免试管里的氧气与一氧化碳反应发生爆炸，故D操作正确．<br />故选：C．','【分析】A、根据闻药品气味的方法进行分析判断．<br />B、根据给试管中的液体加热的方法进行分析判断．<br />C、根据浓硫酸的性质、浓硫酸沾到皮肤上的处理方法进行分析判断．<br />D、根据一氧化碳具有可燃性，与空气的混合气体在点燃时易发生爆炸分析．','选择题',3.00,'ad22adc954c2f8481d414b8b191f3f2f',9,400,'给试管里的液体加热,常见的意外事故的处理方法,一氧化碳还原氧化铁','',2016,'37','2016•市中区二模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840206,'海洋中蕴含丰富的资源．如图为从海水中制备纯碱和金属镁的工艺流程：<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao43/155ec21e-94d4-11e9-88f9-b42e9921e93e_xkb21.png\" style=\"vertical-align:middle\" /><br />根据上述资料，请回答下列问题：<br />（1）在氨碱法基础上，著名科学家<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填序号）创立了更为先进的联合制碱法．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao86/15637d11-94d4-11e9-905f-b42e9921e93e_xkb52.png\" style=\"vertical-align:middle\" /><br />（2）反应③的化学方程式是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）上述工艺流程中，没有涉及到的化学反应类型是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>反应．（填“分解”、“化合”、“置换”、“复分解”中之一）<br />（4）纯碱和小苏打在生活、生产中用途非常广泛，请你至少写出一条应用：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','B$###$2NaHCO<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>Na<SUB>2</SUB>CO<SUB>3</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑$###$置换$###$食品','【解答】解：（1）侯德榜创立了更为先进的侯氏制碱法，故填：B；<br />（2）反应③是碳酸氢钠受热分解生成碳酸钠、水和二氧化碳，其化学方程式是2NaHCO<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>Na<SUB>2</SUB>CO<SUB>3</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑，故填：2NaHCO<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>Na<SUB>2</SUB>CO<SUB>3</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑．<br />（3）上述工艺流程中，反应③是分解反应，反应④⑦是复分解反应，反应⑥是分解反应，没有涉及到的化学反应类型是置换反应，故填：置换．<br />（4）纯碱和小苏打在生活中可以用于食品工业，故填：食品．','【分析】根据海水制碱的知识、碳酸氢钠受热分解的化学方程式以及用途进行分析解答即可．','书写',3.00,'1243c7a6d9d47973a03db506a25ca9b1',9,400,'常用盐的用途,盐的化学性质,反应类型的判定,书写化学方程式、文字表达式、电离方程式,对海洋资源的合理开发与利用','',2016,'37','2016•历下区二模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840207,'下列图象正确的是（　　）','<img  src=\"/tikuimages/9/2016/400/shoutiniao52/156bba70-94d4-11e9-b38d-b42e9921e93e_xkb7.png\" style=\"vertical-align:middle\" /><br />向一定量的硝酸钾溶液中不断加水','<img  src=\"/tikuimages/9/2016/400/shoutiniao38/156ec7b0-94d4-11e9-89c2-b42e9921e93e_xkb48.png\" style=\"vertical-align:middle\" /><br />一定质量的KClO<SUB>3</SUB>固体受热','<img  src=\"/tikuimages/9/2016/400/shoutiniao58/157111a1-94d4-11e9-9804-b42e9921e93e_xkb56.png\" style=\"vertical-align:middle\" /><br />向一定量的稀盐酸中不断加氢氧化钠溶液','<img  src=\"/tikuimages/9/2016/400/shoutiniao85/15735b8f-94d4-11e9-bcf3-b42e9921e93e_xkb62.png\" style=\"vertical-align:middle\" /><br />向氢氧化钾和硝酸钡的混合溶液中，逐滴滴加稀硫酸','','B','【解答】解：A．向硝酸钾溶液中不断加水时，溶质质量不变，故图象错误；<br />B．反应后生成氧气跑掉，故固体中氧元素的质量从开反应就不断减少，反应停止后不变，由于二氧化锰中也有氧元素，故最后不能为0，故符合该图象正确．<br />C．向一定量稀盐酸中滴加氢氧化钠溶液时，开始的pH应小于7，因此该选项不能正确反映对应变化关系．<br />D．硝酸钡和硫酸反应生成不溶于水的硫酸钡沉淀，一开始就会出现沉淀，而图象显示是过一段时间才出现沉淀，故错误；<br />答案：B．','【分析】A．根据向硝酸钾溶液中不断加水时，溶质质量不变来分析；<br />B．根据KClO<SUB>3</SUB>固体受热生成氧气跑掉，但二氧化锰中也有氧元素，故最后不能为0，<br />C．氢氧化钠能和稀盐酸反应生成氯化钠和水；<br />D．硝酸钡和硫酸反应生成不溶于水的硫酸钡沉淀来分析．','选择题',3.00,'9300d92e522fc6855c71361c93f890a6',9,400,'用水稀释改变浓度的方法,酸的化学性质,中和反应及其应用,溶液的酸碱性与pH值的关系,盐的化学性质','',2016,'37','2016•大悟县二模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840211,'如图是小梅配制50g溶质质量分数为12%的NaCl溶液的实验操作示意图：<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao19/158276c0-94d4-11e9-b9f2-b42e9921e93e_xkb16.png\" style=\"vertical-align:middle\" /><br />（1）指出图中的一处错误操作<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）配制时应选择容量为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>mL（10mL或50mL或100mL）的量筒量取所需要的水．<br />（3）用上述图示的序号表示配制溶液的操作正确顺序为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','把药品和砝码直接放在天平的托盘上$###$50$###$④②①⑤③','【解答】解：（1）天平称量药品质量时，不能将药品直接放在托盘上；<br />（2）量取水的质量是：50g-6g=44g，合44mL，量筒称量液体体积时，应该采用就近原则，所以配制时应选择容量为50mL的量筒量取所需要的水；<br />（3）正确配置溶液的比值是计算、称量、溶解、装瓶，所以配制溶液的操作正确顺序为④②①⑤③．<br />故答案为：（1）把药品和砝码直接放在天平的托盘上；<br />（2）50；<br />（3）④②①⑤③．','【分析】（1）根据天平的正确使用方法进行分析；<br />（2）根据量筒的正确使用方法进行分析；<br />（3）根据正确配置溶液的比值是计算、称量、溶解、装瓶进行分析．','填空题',3.00,'0e32a71748e941ed8253430fd0abb1f0',9,400,'一定溶质质量分数的溶液的配制','宜城市',2016,'32','2016•宜城市模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840212,'<img  src=\"/tikuimages/9/2016/400/shoutiniao96/1589a2b0-94d4-11e9-b948-b42e9921e93e_xkb10.png\" style=\"vertical-align:middle;FLOAT:right;\" />（2016•黄冈模拟）由&nbsp;C、H、O、Fe四种元素组成的初中常见物质间有如图所示的关系．其中甲、乙、丙是单质，X、Y是化合物．且Y有毒．图中“-”表示相连的物质两两之间可以发生反应，“→”表示由某一物质可制得另一物质．请回答问题：<br />（1）写出物质X和Y的化学式：X<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，Y<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）写出“丙+乙→”的化学反应方程式：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）五种物质中属于氧化物的有<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>种．','','','','','','H<SUB>2</SUB>O$###$CO$###$3Fe+2O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>Fe<SUB>3</SUB>O<SUB>4</SUB>$###$3','【解答】解：（1）由&nbsp;C、H、O、Fe四种元素组成的初中常见物质，甲、乙、丙是单质，X、Y是化合物，且Y有毒，所以Y是一氧化碳，单质丙会与一氧化碳反应，所以丙是氧气，化合物X会转化成单质甲和氧气，所以X是水，甲是氢气，氧气会转化成乙，所以乙是四氧化三铁，经过验证，推导正确，所以X是H<SUB>2</SUB>O，Y是CO；<br />（2）丙+乙→”的反应是铁和氧气在点燃的条件下生成四氧化三铁，化学反应方程式为：3Fe+2O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>Fe<SUB>3</SUB>O<SUB>4</SUB>；<br />（3）五种物质中属于氧化物的有一氧化碳、四氧化三铁、水三种．<br />故答案为：（1）H<SUB>2</SUB>O，CO；<br />（2）3Fe+2O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>Fe<SUB>3</SUB>O<SUB>4</SUB>；<br />（3）3．','【分析】根据由&nbsp;C、H、O、Fe四种元素组成的初中常见物质，甲、乙、丙是单质，X、Y是化合物，且Y有毒，所以Y是一氧化碳，单质丙会与一氧化碳反应，所以丙是氧气，化合物X会转化成单质甲和氧气，所以X是水，甲是氢气，氧气会转化成乙，所以乙是四氧化三铁，然后将推出的物质进行验证即可．','书写',3.00,'400405d8496d13b052bb88050e62cda3',9,400,'从组成上识别氧化物,物质的鉴别、推断,书写化学方程式、文字表达式、电离方程式','',2016,'32','2016•黄冈模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840215,'<img  src=\"/tikuimages/9/2016/400/shoutiniao59/158f961e-94d4-11e9-b0a9-b42e9921e93e_xkb7.png\" style=\"vertical-align:middle;FLOAT:right\" />如图是物质的分类及部分转化关系图，有关说法不正确的是（　　）','转化a一定是化学变化','转化b一定是化学变化','转化b中一定有元素的存在形态发生改变','分离液态空气制取氧气属于转化c','','A','【解答】解：A、转化a为混合物转化为甲，但不一定是化学变化，故说法错误；<br />B、转化b为化合物生成乙，则一定是化学变化，故说法正确；<br />C、转化b中一定有元素的存在形态发生改变，故说法正确；<br />D、根据图可知：分离液态空气制取氧气属于转化c，故说法正确；<br />故选A．','【分析】判断一个变化是物理变化还是化学变化，要依据在变化过程中有没有生成其他物质，生成其他物质的是化学变化，没有生成其他物质的是物理变化．','选择题',3.00,'3473f77dae9586d23bffe4526cb84c48',9,400,'氧气的工业制法,物质的相互转化和制备,化学变化和物理变化的判别','',2016,'37','2016•浦东新区二模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840217,'（1）生活用水常利用活性炭去除异味，活性炭主要起<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>作用．<br />（2）水是最常用的溶剂，汽车中常用的电池是铅酸蓄电池，其中酸用的是稀硫酸．如果用98%的浓硫酸配制980g的20%稀硫酸，需要水的质量为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>&nbsp;g．','','','','','','吸附$###$780','【解答】解：<br />（1）活性炭具有吸附性，可以吸附色素和异味，生活用水常利用活性炭去除异味，活性炭主要起吸附作用．<br />（2）设需加水的质量为x，则<br />（980g-x）×98%=980g×20%<br />解得：x=780g<br />答案：（1）吸附；（2）780','【分析】（1）根据活性炭具有吸附性，可以吸附色素和异味解答；<br />（2）根据溶液稀释前后，溶质的质量不变进行分析解答．','填空题',3.00,'c5a830279b9998bfc38f59519c6a0ab4',9,400,'用水稀释改变浓度的方法,碳单质的物理性质及用途','',2016,'37','2016•延庆县一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840218,'苯甲酸有抑制真菌、细菌、霉菌生长的作用，常用作药物或防腐剂．苯甲酸常温下为片状或针状结晶，在100℃时会迅速升华，苯甲酸在水中的溶解度见表．<br /><table class=\"edittable\"><TBODY><TR><td width=86>温度</TD><td width=76>20℃</TD><td width=76>25℃</TD><td width=76>50℃</TD><td width=76>75℃</TD><td width=67>95℃</TD></TR><TR><td>溶解度</TD><td>0.17g</TD><td>0.35g</TD><td>0.95g</TD><td>2.2g</TD><td>6.8g</TD></TR></TBODY></TABLE>为了提纯某苯甲酸样品（其中含有难溶于水的杂质），某化学小组进行了如下实验：<br />（l）取约1g样品放入烧杯中，加入50mL蒸馏水充分搅拌，发现样品几乎没溶解．原因是苯甲酸属于<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（选填“易溶”、“可溶”、“微溶”、“难溶”）物．<br />（2）将烧杯放在石棉网上加热至样品充分溶解，再加少量蒸馏水，然后趁热过滤．趁热过滤的目的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）将所得滤液<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，使苯甲酸以晶体的形式析出，然后过滤得到较纯净的苯甲酸晶体．<br />（4）过滤后的滤液是苯甲酸的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（选填“饱和”或“不饱和”）溶液．','','','','','','微溶$###$避免苯甲酸析出而损耗$###$降温结晶$###$饱和','【解答】解：（1）苯甲酸的溶解度随温度升高而增大，20℃时其溶解度只有0.17g，取约1g样品放入烧杯中，加入50mL蒸馏水充分搅拌，发现样品几乎没溶解．原因是苯甲酸属于微溶物．<br />（2）温度低时苯甲酸溶解度很低，易析出过滤会使其与泥沙混合滤出导致其损失，<br />（3）缓慢降温使其析出会得到颗粒较大的晶体，便于得到苯甲酸晶体；<br />（4）过滤后的滤液是苯甲酸的饱和溶液．<br />答案：（1）微溶；&nbsp;（2）避免苯甲酸析出而损耗；（3）降温结晶；&nbsp;（4）饱和','【分析】（1）利用20℃甲酸的溶解度解答．<br />（2）温度低时苯甲酸溶解度很低，易析出过滤会使其损失．<br />（3）为使其形成颗粒较大的晶体，应缓慢降温；<br />（4）根据有固体剩余的溶液是饱和溶液解答解答；','填空题',3.00,'ca00820caca6aea0282e023969d2f3a3',9,400,'结晶的原理、方法及其应用,饱和溶液和不饱和溶液,物质的溶解性及影响溶解性的因素','',2016,'37','2016•苏州一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840219,'化学与我们的生产、生活、健康密不可分．请在下列物质中作出正确选择，并用序号填空．<br />①大米②干冰③鸡蛋清④稀有气体⑤氯化钠 ⑥甲烷⑦氢氧化钙<br />（1）可用于人工降雨的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）富含糖类的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）能中和酸性土壤的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（4）可作为焊接金属的保护气的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（5）医疗上用来配制生理盐水的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（6）富含蛋白质的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（7）属于沼气的主要成分的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','②$###$①$###$⑦$###$④$###$⑤$###$③$###$⑥','【解答】解：（1）由于干冰在升华时能吸收大量的热，可用于人工降雨；<br />（2）大米富含糖类．<br />（3）熟石灰能与土壤中的酸反应，所以农业生产中常用熟石灰来中和土壤中的酸．<br />（4）稀有气体化学性质稳定，可作为焊接金属的保护气．<br />（5）医疗上用来配制生理盐水的是氯化钠．<br />（6）鸡蛋清富含蛋白质．<br />（7）沼气的主要成分是甲烷．<br />故答案为：（1）②；（2）①；（3）⑦；（4）④；（5）⑤；（6）③；（7）⑥．','【分析】物质的性质决定物质的用途，根据常见的物质的性质和用途分析回答．','填空题',3.00,'58d82a3df7bcf9eaecf073ec3917ebef',9,400,'常见气体的用途,二氧化碳的用途,常见碱的特性和用途,常用盐的用途,常用燃料的使用与其对环境的影响,食品、药品与健康食品中的有机营养素','',2016,'37','2016•柳江县一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840220,'<img  src=\"/tikuimages/9/2016/400/shoutiniao1/159e6330-94d4-11e9-baef-b42e9921e93e_xkb55.png\" style=\"vertical-align:middle;FLOAT:right\" />如图所示是甲、乙两种物质的溶解度曲线，下列说法正确的是（　　）','t℃时，甲的溶解度大于乙','t℃时，甲的饱和溶液中溶质质量分数为30%','当甲中含有少量乙时，可以用蒸发结晶法提纯甲','在t℃时，100g水中加入30g乙，形成不饱和溶液','','A','【解答】解：A、由溶解度曲线可知：t℃时，甲的溶解度大于乙正确；故选项正确；<br />B、t℃时，甲的饱和溶液中溶质质量分数=<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">30g</td></tr><tr><td>130g</td></tr></table>×100%≈23.1%</span>，30%错误，故选项错误；<br />C、当甲中含有少量乙时，可以用降温结晶法提纯甲，不是蒸发结晶，故选项错误；<br />D、在t℃时，100g水中加入30g乙，形成饱和溶液，并且固体有剩余，形成不饱和溶液错误，故选项错误；<br />故选A','【分析】根据题目信息和溶解度曲线可知：甲、乙两种固体物质的溶解度，都是随温度升高而增大，而甲的溶解度随温度的升高变化比乙大；t℃时，甲的溶解度大于乙正确；t℃时，甲的饱和溶液中溶质质量分数=<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">30g</td></tr><tr><td>130g</td></tr></table>×100%≈23.1%</span>，30%错误；当甲中含有少量乙时，可以用降温结晶法提纯甲，不是蒸发结晶；在t℃时，100g水中加入30g乙，形成饱和溶液，并且固体有剩余．','选择题',3.00,'d11be3c3dabf4a60c3e3aa2cf4b4643d',9,400,'结晶的原理、方法及其应用,固体溶解度曲线及其作用,溶质的质量分数、溶解性和溶解度的关系','',2016,'37','2016•官渡区一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840222,'下列能源属于不可再生的是（　　）','煤','风能','氢能','水能','','A','【解答】解：A、煤属于化石燃料，不能短时期内从自然界得到补充，属于不可再生能源，故选项正确．<br />B、风能属于可再生能源，故选项错误．<br />C、氢能可用水来制取，属于可再生能源，故选项错误．<br />D、水能能属于可再生能源，故选项错误．<br />故选：A．','【分析】从能源是否可再利用的角度可把能源分为可再生能源和不可再生能源．人类开发利用后，在现阶段不可能再生的能源，属于不可再生能源；在自然界中可以不断再生的能源，属于可再生能源．','选择题',3.00,'4527cce307c1bb26325087f9ac35d86d',9,400,'常见能源的种类、能源的分类','',2016,'32','2016•双牌县模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840229,'用金属铁制备自来水新型消毒、净水剂Na<SUB>2</SUB>FeO<SUB>4</SUB>的流程如图，请回答下列问题：<img  src=\"/tikuimages/9/2016/400/shoutiniao74/15c0430f-94d4-11e9-bb2b-b42e9921e93e_xkb8.png\" style=\"vertical-align:middle\" /><br />（1）溶液A的溶质是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（写名称）；写出Na<SUB>2</SUB>FeO<SUB>4</SUB>中铁元素的化合价<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）操作Ⅰ用到玻璃棒的作用是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）请写出反应2的化学方程式：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，反应类型是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','硫酸亚铁$###$+6$###$引流$###$FeSO<SUB>4</SUB>+2NaOH=Fe（OH）<SUB>2</SUB>↓+Na<SUB>2</SUB>SO<SUB>4</SUB>$###$复分解反应','【解答】解：（1）由反应的流程可知，过量的铁粉与硫酸铜反应，生成了硫酸亚铁和铜，故溶液A的溶质是硫酸亚铁；在K<SUB>2</SUB>FeO<SUB>4</SUB>中，由于钾元素显+1价、氧显-2价，设铁的化合价为x，由化合价原则可知：（+1）×2+x+（-2×4）=0，解得：x=+6．<br />（2）由上述反应①得到是固液的混合物，要将固液分开，应进行的操作是过滤，过滤时玻璃棒的作用是引流；<br />（3）由反应的流程可知：反应②是硫酸亚铁与氢氧化钠反应，方程式是：FeSO<SUB>4</SUB>+2NaOH=Fe（OH）<SUB>2</SUB>↓+Na<SUB>2</SUB>SO<SUB>4</SUB>；属于复分解反应；<br />故答案为：<br />（1）硫酸亚铁；+6．（2）引流．（3）FeSO<SUB>4</SUB>+2NaOH=Fe（OH）<SUB>2</SUB>↓+Na<SUB>2</SUB>SO<SUB>4</SUB>，复分解反应．','【分析】（1）根据反应①过量铁粉与硫酸铜的反应分析固体B的成分，根据化合价原则“化合物中各元素的化合价的代数和为0”，由化学式求出元素的化合价；<br />（2）根据过滤可用于固体和液体的分离以及过滤的所需的仪器进行解答；<br />（3）根据发生的反应写出反应的化学方程式．','书写',3.00,'5622bd793521d644c83bf53ba0380971',9,400,'过滤的原理、方法及其应用,物质的相互转化和制备,有关元素化合价的计算,反应类型的判定,书写化学方程式、文字表达式、电离方程式','',2016,'37','2016•阜阳校级二模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840231,'从香料八角中提取的莽草酸（C<SUB>7</SUB>H<SUB>10</SUB>O<SUB>5</SUB>）是合成治疗禽流感药物-达菲的原料，莽草酸易溶于水，能使紫色石蕊试液变红，下列关于莽草酸的说法正确的是（　　）','属于无机物','由7个碳原子、10个氢原子和5个氧原子组成','莽草酸水溶液呈酸性','莽草酸完全燃烧，产物不仅有CO<SUB>2</SUB>和H<SUB>2</SUB>O，还可能有SO<SUB>2</SUB>','','C','【解答】解：A．由莽草酸的化学式C<SUB>7</SUB>H<SUB>10</SUB>O<SUB>5</SUB>可知，它是一种含有碳元素的化合物，属于有机化合物，故错误；<br />B．由莽草酸的化学式C<SUB>7</SUB>H<SUB>10</SUB>O<SUB>5</SUB>可知，它是由碳、氢、氧三种元素组成的，故错误；<br />C．莽草酸易溶于水，能使紫色石蕊试液变红，说明其水溶液显酸性，故正确；<br />D．由质量守恒定律，反应前后元素种类不变，莽草酸中不含氧元素，莽草酸完全燃烧，产物有CO<SUB>2</SUB>和H<SUB>2</SUB>O，不可能含有二氧化硫，故错误．<br />故选：C．','【分析】A．根据无机物与有机物的概念来分析；<br />B．根据物质的组成来分析；<br />C．根据题干信息来分析；<br />D．根据质量守恒定律，反应前后元素种类不变，进行分析判断．','选择题',3.00,'f85f11293c1b3e02750ac3b90b9b917f',9,400,'酸碱指示剂及其性质,有机物与无机物的区别,化学式的书写及意义','',2016,'32','2016•云南模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840232,'地球是一颗美丽的蓝色星球，海水占了全球水资源的97%．海水资源的开发和利用具有非常广阔的前景．<br />（1）海水中含有大量的NaCl，可用于制备钠及其化合物，其流程如图：<br /><img  src=\"/tikuimages/9/0/400/shoutiniao38/15cb66a1-94d4-11e9-904c-b42e9921e93e_xkb58.png\" style=\"vertical-align:middle\" /><br />①上述过程中使用了过滤和蒸发操作，实验室中这两个操作均会使用到的玻璃仪器是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />②加入过量Na<SUB>2</SUB>CO<SUB>3</SUB>溶液目的是除去CaCl<SUB>2</SUB>和<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，流程中不能用过量稀硫酸代替过量稀盐酸的原因是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）根据以上信息，写出制取金属钠的化学方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）工业上利用电解饱和食盐水制备NaOH的方程式为：<br />2NaCl+2H<SUB>2</SUB>O<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;通电&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2NaOH+H<SUB>2</SUB>↑+Cl<SUB>2</SUB>↑<br />该电解原理与实验室电解水类似，由此推测，电解食盐水时电源正极一端产生的气体是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','玻璃棒$###$氯化钡$###$会生成新的杂质硫酸钠$###$2NaCl<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;通电&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2Na+Cl<SUB>2</SUB>↑$###$Cl<SUB>2</SUB>','【解答】解：（1）①碳酸根离子和钙离子、钡离子会生成碳酸钙沉淀、碳酸钡沉淀，所以加入过量Na<SUB>2</SUB>CO<SUB>3</SUB>溶液目的是除去CaCl<SUB>2</SUB>和氯化钡，硫酸和氢氧化钠、碳酸钠会生成硫酸钠，所以不能用过量稀硫酸代替过量稀盐酸的原因是会生成新的杂质硫酸钠；<br />②蒸发过程中用到的玻璃仪器有玻璃棒、酒精灯，过滤过程中用到的玻璃仪器有烧杯、漏斗、玻璃棒，所以两步操作中需要的玻璃仪器有玻璃棒；<br />③氯化钠在通电的条件下生成钠和氯气，化学方程式为：2NaCl<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;通电&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2Na+Cl<SUB>2</SUB>↑；<br />（2）氯化钠溶液在通电的条件下生成氢氧化钠、氯气和氢气，化学方程式为：2NaCl+2H<SUB>2</SUB>O<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;通电&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2NaOH+H<SUB>2</SUB>↑+Cl<SUB>2</SUB>↑，电荷同性相斥，异性相吸，所以电解食盐水时电源正极一端产生的气体是Cl<SUB>2</SUB>．<br />故答案为：（1）①BaCl<SUB>2</SUB>，会生成新的杂质硫酸钠；<br />②玻璃棒；<br />③2NaCl<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;通电&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2Na+Cl<SUB>2</SUB>↑；<br />（2）Cl<SUB>2</SUB>．','【分析】（1）①根据碳酸根离子和钙离子、钡离子会生成碳酸钙沉淀、碳酸钡沉淀，硫酸和氢氧化钠、碳酸钠会生成硫酸钠进行分析；<br />②根据蒸发、过滤过程中需要的玻璃仪器进行分析；<br />③根据氯化钠在通电的条件下生成钠和氯气进行分析；<br />（2）根据氯化钠溶液在通电的条件下生成氢氧化钠、氯气和氢气，依据电荷同性相斥，异性相吸的原则进行分析．','书写',3.00,'3de4e788ae6b9c7af4fc89b0a3324fd7',9,400,'过滤的原理、方法及其应用,蒸发与蒸馏操作,电解水实验,碱的化学性质,盐的化学性质,书写化学方程式、文字表达式、电离方程式,对海洋资源的合理开发与利用','',0,'37','',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840233,'下列有关物质的性质与应用叙述错误的是（　　）','氮气的化学性质稳定，可用作保护气','铜具有良好的导电性，可用来制导线','氢氧化钠固体能吸水，可用来干燥SO<SUB>2</SUB>气体','氧气能支持燃烧，可用作火箭燃料的助燃剂','','C','【解答】解：A、氮气的性质稳定用作保护气，是利用稀有气体不易和其它物质发生反应，常用作保护气，故A说法正确；<br />B、铜具有良好的导电性，可用来制导线，正确；<br />C、氢氧化钠可以与二氧化硫反应，不能用来干燥二氧化硫气体，故C不正确；<br />D、氧气能支持燃烧，可用作火箭燃料的助燃剂，正确；<br />故选C','【分析】物质的性质决定用途，物质的用途反映物质的性质．','选择题',3.00,'5fb41b9f70650daf54baf7dded6e8059',9,400,'氧气的用途,常见气体的用途,金属的物理性质及用途,根据浓硫酸或烧碱的性质确定所能干燥的气体','昆山市',2016,'37','2016•昆山市二模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840235,'关于化学方程式2H<SUB>2</SUB>+O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O的读法错误的是（　　）','每2个氢分子和1个氧分子结合生成2个水分子','氢气和氧气在点燃的条件下化合生成水','氢元素和氧元素在点燃的条件下生成水','每4份质量的氢气与32份质量的氧气完全反应生成36份质量的水','','C','【解答】解：A、由2H<SUB>2</SUB>+O<SUB>2</SUB> <span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O可知，每2个氢分子和1个氧分子在点燃条件下发生反应，生成2个水分子．故选项正确；<br />B、应该是氢气和氧气在点燃条件下化合生成水．故选项正确；<br />C、应该是氢气和氧气在点燃条件下化合生成水．故选项错误；<br />D、由2H<SUB>2</SUB>+O<SUB>2</SUB> <span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O可知，每4份质量的氢气跟32份质量的氧气在点燃的条件下完全反应生成36份质量的水．故选项正确．<br />故选C．','【分析】化学方程式的含义有：反应物和生成物的种类；反应的条件；反应物和生成物的微观粒子个数比；反应物和生成物的质量比等．','选择题',3.00,'998b156379633b9183df1c25437ca5ca',9,400,'化学方程式的概念、读法和含义','普兰店市',2015,'35','2015秋•普兰店市校级期中',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840236,'<img  src=\"/tikuimages/9/2016/400/shoutiniao28/15d4b570-94d4-11e9-8f6e-b42e9921e93e_xkb40.png\" style=\"vertical-align:middle;FLOAT:right\" />（2016•武进区模拟）电视节目《荒野求生》中的生命吸管（如图所示）是一种将污水净化为饮用水的吸管装置，可以除去&nbsp;99.3%&nbsp;的细菌和病毒，但难以除去汞等金属离子．回答下列问题：<br />（1）汞等金属离子危害人体健康的原因是：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）生命吸管<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（“能”或“不能”）使硬水软化．家庭中使硬水软化一般采取<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>方法．<br />（3）吸管中活性炭主要用来除去<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．碘主要起到杀菌消毒作用，生活中消毒还可采用84消毒液，有效成分是次氯酸钠（NaClO），杀菌原理是与空气中的二氧化碳和水反应生成碳酸氢钠和次氯酸（HClO），试写出反应方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','重金属离子能使人体蛋白质变性$###$不能$###$加热煮沸$###$色素、异味$###$CO<SUB>2</SUB>+H<SUB>2</SUB>O+NaClO═NaHCO<SUB>3</SUB>+HClO','【解答】解：（1）汞等金属离子危害人体健康的原因是重金属离子能使人体蛋白质变性；<br />（2）生命吸管不能使硬水软化．家庭中使硬水软化一般采取加热煮沸方法；<br />（3）吸管中活性炭主要用来除去色素、异味；次氯酸钠（NaClO），杀菌原理是与空气中的二氧化碳和水反应生成碳酸氢钠和次氯酸（HClO），反应的化学方程式为：CO<SUB>2</SUB>+H<SUB>2</SUB>O+NaClO═NaHCO<SUB>3</SUB>+HClO．<br />故答案为：（1）重金属离子能使人体蛋白质变性（失去生理活性）；（2）不能；加热煮沸；（3）色素、异味；CO<SUB>2</SUB>+H<SUB>2</SUB>O+NaClO═NaHCO<SUB>3</SUB>+HClO．','【分析】（1）根据重金属离子使人体蛋白质变性进行分析；<br />（2）根据生命吸管净化污水的原理进行分析；根据家庭中使硬水软化的方法进行分析；<br />（3）根据对反应的说明，判断反应的发生，写出反应的化学方程即可．','书写',3.00,'8bd223c112652df1a379c87a76e06d65',9,400,'硬水与软水,碳单质的物理性质及用途,书写化学方程式、文字表达式、电离方程式,鉴别淀粉、葡萄糖的方法与蛋白质的性质','',2016,'32','2016•武进区模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840238,'《西游记》75回中，作者借孙悟空之口描述金箍棒“棒是九转镔铁炼，老君亲手炉中煅．禹王求得号神珍，四海八河为定验．中间星斗暗铺陈，两头箝裹黄金片．花纹密布神鬼惊，上造龙纹与凤篆”．<br />（1）金箍棒的原材料“镔铁”，是中国乃至世界古代中最重要的钢材料之一．“镔铁”<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“属于”或“不属于”）合金．<br />（2）大禹治水时，利用此棒能变大变小的性质测定海水深浅．“能变大变小”体现了金属的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填序号）．<br />A．导电性&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;B．导热性&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;C．延展性&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;D．硬度较大的性质<br />从题干的描述文字找到线索，解释此棒能常年放入海水中而不被腐蚀的原因是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）金箍棒表面使“鬼神”惊怕的花纹是利用硫酸铁来处理的．若实验室提供原料铁、稀硫酸、过氧化氢溶液制取硫酸铁．反应分两步，请写出第一步生成硫酸亚铁的化学反应方程式：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．第二步是向有酸剩余的硫酸亚铁溶液中加入过氧化氢溶液，生成物只有硫酸铁和水，该反应的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','属于$###$C$###$由不活泼的金属包裹$###$Fe+H<SUB>2</SUB>SO<SUB>4</SUB>═FeSO<SUB>4</SUB>+H<SUB>2</SUB>↑$###$2FeSO<SUB>4</SUB>+H<SUB>2</SUB>SO<SUB>4 </SUB>+H<SUB>2</SUB>O<SUB>2</SUB>=Fe<SUB>2</SUB>（SO<SUB>4</SUB>）<SUB>3</SUB>+2H<SUB>2</SUB>O','【解答】解：（1）镔铁又名宾铁，是中国乃至世界古代中最重要的钢材料之一，其制作工艺是把一般易腐蚀的钢表面磨光，再用腐蚀剂进行处理，故“镔铁”属于合金；<br />（2）“能变大变小”体现了金属的延展性；由题干中的信息可知，金箍棒能常年放入海水中而不被腐蚀的原因是因为此棒由不活泼的金属包裹；<br />（3）铁和稀硫酸反应生成硫酸亚铁和氢气，反应的化学方程式为：Fe+H<SUB>2</SUB>SO<SUB>4</SUB>═FeSO<SUB>4</SUB>+H<SUB>2</SUB>↑；<br />（4）向有酸剩余的硫酸亚铁溶液中加入过氧化氢溶液，生成物只有硫酸铁和水，该反应的化学方程式为；2<br />FeSO<SUB>4</SUB>+H<SUB>2</SUB>SO<SUB>4 </SUB>+H<SUB>2</SUB>O<SUB>2</SUB>=Fe<SUB>2</SUB>（SO<SUB>4</SUB>）<SUB>3</SUB>+2H<SUB>2</SUB>O；<br />故答案为：（1）属于；（2）C；由不活泼的金属包裹；（3）Fe+H<SUB>2</SUB>SO<SUB>4</SUB>═FeSO<SUB>4</SUB>+H<SUB>2</SUB>↑；（4）2FeSO<SUB>4</SUB>+H<SUB>2</SUB>SO<SUB>4 </SUB>+H<SUB>2</SUB>O<SUB>2</SUB>=Fe<SUB>2</SUB>（SO<SUB>4</SUB>）<SUB>3</SUB>+2H<SUB>2</SUB>O．','【分析】（1）根据合金的特性进行分析；<br />（2）根据金属的延展性进行分析；根据题干中的信息进行分析；<br />（3）根据反应物、生成物、质量守恒定律以及化学方程式的书写方法、步骤进行书写；<br />（4）根据反应物、生成物、质量守恒定律以及化学方程式的书写方法、步骤进行书写．','书写',3.00,'f49496812334e5c1a17f98374bd169c3',9,400,'金属的物理性质及用途,合金与合金的性质,金属的化学性质,书写化学方程式、文字表达式、电离方程式','',2016,'37','2016春•重庆校级月考',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840244,'<img  src=\"/tikuimages/9/2016/400/shoutiniao27/15e6419e-94d4-11e9-85a9-b42e9921e93e_xkb76.png\" style=\"vertical-align:middle;FLOAT:right;\" />（2016春•苏州校级期中）图为NaCl、KNO<SUB>3</SUB>的溶解度曲线．<br />（1）20℃时，KNO<SUB>3</SUB>的溶解度是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>g．<br />（2）除去KNO<SUB>3</SUB>固体中混有的少量NaCl，提纯的步骤是：加水溶解、蒸发浓缩、<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，然后过滤、洗涤、干燥．<br />（3）根据溶解度曲线判断，能否在某一条件下，配制得到相同溶质质量分数的饱和氯化钠、硝酸钾溶液？答：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“能”或“不能”）．<br />（4）以下实验中（水的密度约为1g/cm<SUP>3</SUP>）：<br />10mL水<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"font-size:90%;\">加2.5gKN<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">3</span></span></td></tr><tr><td><div hassize=\"9\"><div style=\"background: url(\'/upimages/images/part/8594L.png\') repeat-x; width:1px; height:5px;float:left;overflow:hidden\" muststretch=\"h\"></div><div style=\"background: url(\'/upimages/images/part/8594.png\') no-repeat; width:9px; height:5px;float:left;overflow:hidden\"></div></div></td></tr><tr><td style=\"font-size:90%\">搅拌(20℃)</td></tr></table></span>x<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"font-size:90%;\">再加2.5gKN<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">3</span></span></td></tr><tr><td><div hassize=\"9\"><div style=\"background: url(\'/upimages/images/part/8594L.png\') repeat-x; width:1px; height:5px;float:left;overflow:hidden\" muststretch=\"h\"></div><div style=\"background: url(\'/upimages/images/part/8594.png\') no-repeat; width:9px; height:5px;float:left;overflow:hidden\"></div></div></td></tr><tr><td style=\"font-size:90%\">搅拌(20℃)</td></tr></table></span>y<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"font-size:90%;\">升温</td></tr><tr><td><div hassize=\"9\"><div style=\"background: url(\'/upimages/images/part/8594L.png\') repeat-x; width:1px; height:5px;float:left;overflow:hidden\" muststretch=\"h\"></div><div style=\"background: url(\'/upimages/images/part/8594.png\') no-repeat; width:9px; height:5px;float:left;overflow:hidden\"></div></div></td></tr><tr><td style=\"font-size:90%\">至50℃</td></tr></table></span>z<br />①z溶液是否为饱和溶液？答：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．（填“是”或“不是”）<br />②y溶液的溶质质量分数为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．（列出计算式即可）','','','','','','31.6$###$冷却热饱和溶液$###$能$###$不是$###$24%','【解答】解：（1）通过分析溶解度曲线可知，20℃时，KNO<SUB>3</SUB>的溶解度是31.6g；<br />（2）硝酸钾的溶解度受温度影响较大，所以除去KNO<SUB>3</SUB>固体中混有的少量NaCl，提纯的步骤是：加水溶解、蒸发浓缩、冷却热饱和溶液，然后过滤、洗涤、干燥；<br />（3）根据溶解度曲线判断，硝酸钾和氯化钠的溶解度曲线交于一点，所以能在某一条件下，配制得到相同溶质质量分数的饱和氯化钠、硝酸钾溶液；<br />（4）20℃时，硝酸钾的溶解度是31.6g，50℃时，硝酸钾的溶解度是85.5g，所以z溶液不是饱和溶液，<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">3.16g</td></tr><tr><td>13.16g</td></tr></table></span>×100%=24%．<br />故答案为：（1）31.6；<br />（2）冷却热饱和溶液；<br />（3）能；<br />（4）不是，24%．','【分析】根据固体的溶解度曲线可以：①查出某物质在一定温度下的溶解度，从而确定物质的溶解性，②比较不同物质在同一温度下的溶解度大小，从而判断饱和溶液中溶质的质量分数的大小，③判断物质的溶解度随温度变化的变化情况，从而判断通过降温结晶还是蒸发结晶的方法达到提纯物质的目的．','填空题',3.00,'4203d29c03e07dec43d78c9e05146eed',9,400,'结晶的原理、方法及其应用,饱和溶液和不饱和溶液,固体溶解度曲线及其作用,溶质的质量分数、溶解性和溶解度的关系','',2016,'35','2016春•苏州校级期中',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840246,'对空气造成污染&nbsp;的一组气体是（　　）','氮气、二氧化碳','水蒸气、二氧化碳','稀有气体、二氧化碳','二氧化硫、一氧化碳、二氧化氮','','D','【解答】解：A、氮气、二氧化碳不会造成污染；故选项错误；<br />B、水蒸气、二氧化碳不会造成污染，故选项错误；<br />C、稀有气体、二氧化碳不会造成污染，故选项错误；<br />D、有害气体主要有一氧化碳、二氧化硫、二氧化氮等气体，故选项正确；<br />故选D．','【分析】本题主要是空气的污染及其危害，空气污染的途径主要有两个：有害气体和粉尘．有害气体主要有一氧化碳、二氧化硫、二氧化氮、臭氧等气体；粉尘主要指一些固体小颗粒．','选择题',3.00,'c4c70d8ba75277ab40760d380f867db8',9,400,'空气的污染及其危害','',2015,'37','2015秋•海南校级月考',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840254,'<img  src=\"/tikuimages/9/2016/400/shoutiniao6/16051440-94d4-11e9-b268-b42e9921e93e_xkb36.png\" style=\"vertical-align:middle;FLOAT:right\" />（2016•高安市二模）学习完“酸、喊、盐的性质”后，老师将两瓶未贴标签的饱和石灰水、饱和碳酸钠溶液摆放在实验桌上，让同学们区分．下面是“雄鹰小组”同学们的探究过程：<br />【讨论交流】<br />李壮（组长）：区分两种物质的原理是：依据物质的性质，选择一种试剂与两种物质混合，产生两种不同的现象．大家椐此来说一说，区分这两种溶液应选用哪种试剂，并说明理由；<br />王志：我认为可以选择稀盐酸，理由是稀盐酸与碳酸钠反应有气泡产生，稀盐酸与氢氧化钙反应无明显现象．<br />马凌：我选择的是碳酸钠溶液．理由是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />【实验探究】<br />大家按照王志的方案进行了如图所示的实验：<br />（1）实验中同学们根据现察到的现象，判断出甲试管中原溶液为碳酸钠溶液，乙试管中原溶液为石灰水．则乙试管中反应的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />（2）实验后王志将甲、乙两试管中的废液倒入烧杯中，发现烧杯中液体澄清，大家对烧杯中溶液的溶质成分产生了探究兴趣，请你和他们一起探究：<br />①取少量烧杯中的溶液于试管中，滴入几滴无色酚酞溶液，发现溶液仍为无色，由此可知溶液中一定含有<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填化学式）可能含有HCl．<br />②另取少量溶液于试管中，加入<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，观察到有气泡产生，说明该废液中还含有HCl．<br />【反思评价】<br />欲从烧杯废液中得到纯净的NaCl，可先向烧杯中加入适量的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>溶液，经过滤后进行蒸发即可得到纯净的NaCl．<br />探究活动结束了，同学们充分体验到合作探究的乐趣和学习成功的喜悦！','','','','','','碳酸钠与氢氧化钙反应生成沉淀$###$Ca（OH）<SUB>2</SUB>+2HCl=CaCl<SUB>2</SUB>+2H<SUB>2</SUB>O$###$NaCl、CaCl<SUB>2</SUB>$###$锌粒等$###$碳酸钠','【解答】解：【讨论交流】碳酸钠和氢氧化钙反应生成白色的碳酸钙沉淀，所以选择碳酸钠溶液的理由是：碳酸钠和氢氧化钙反应会生成白色沉淀；故填：碳酸钠与氢氧化钙反应生成沉淀；<br />【实验探究】（1）氢氧化钙和盐酸反应生成氯化钙和水，故乙试管中反应的化学方程式为：Ca（OH）<SUB>2</SUB>+2HCl=CaCl<SUB>2</SUB>+2H<SUB>2</SUB>O；<br />（2）①氢氧化钙和盐酸反应生成氯化钙和水，盐酸和碳酸钠会生成氯化钠、水和二氧化碳，取少量烧杯中的溶液于试管中，滴入几滴无色酚酞溶液，发现溶液仍为无色，说明溶液不显碱性，因此没有碳酸钠和氢氧化钙的剩余，故溶质一定有：NaCl、CaCl<SUB>2</SUB>；<br />②盐酸能够和活泼金属反应产生氢气，故填：锌粒等；<br />【反思评价】要除去氯化钠中的氯化钙，可以加入适量的碳酸钠，碳酸钠和氯化钙反应产生碳酸钙沉淀和氯化钠；故填：碳酸钠．','【分析】【讨论交流】根据碳酸钠和氢氧化钙反应生成白色的碳酸钙沉淀进行分析；<br />【实验探究】（2）①根据氢氧化钙和盐酸反应生成氯化钙和水写出反应的方程式；盐酸和碳酸钠会生成氯化钠、水和二氧化碳进行分析；<br />②根据稀盐酸能够和活泼金属反应产生氢气进行分析；<br />【反思评价】根据除杂的原则及碳酸钠的性质进行分析．','书写',3.00,'a1e82eed40e6577ff6674cdd63e3c952',9,400,'缺失标签的药品成分的探究,酸的化学性质,碱的化学性质,盐的化学性质,书写化学方程式、文字表达式、电离方程式','高安市',2016,'37','2016•高安市二模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840257,'下列有关分子、原子、离子的叙述中，正确的是（　　）','分子是保持物质性质的最小粒子','原子是最小的粒子，不能再分','分子和原子都是构成物质的基本粒子，离子不是','二氧化碳气体和干冰化学性质相同，因为它们是由同种分子构成的','','D','【解答】解：A．分子是保持物质化学性质的最小粒子，不能保持物质的物理性质，故错误；<br />B．原子是化学变化中的最小粒子，在化学变化中不能再分，但是用其他方法还可以分成原子核和电子，故错误；<br />C．分子、原子和离子是构成物质的三种基本粒子，故错误；<br />D．因为气体二氧化碳和干冰（固态二氧化碳）中都含有相同的分子即二氧化碳分子，所以它们的化学性质相同，故正确．<br />故选D．','【分析】A．根据分子的概念来分析；<br />B．根据原子的概念来分析；<br />C．根据构成物质的基本粒子来分析；<br />D．同中分子性质相同．','选择题',3.00,'5d3fae4b38955091727d041be17dc57f',9,400,'分子、原子、离子、元素与物质之间的关系,原子的定义与构成,分子的定义与分子的特性','',2013,'37','2013•泾川县校级二模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840259,'实验室要配制50g&nbsp;10%的氯化钠溶液，其操作如图所示．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao47/1614085e-94d4-11e9-be0b-b42e9921e93e_xkb82.png\" style=\"vertical-align:middle\" /><br />（1）①中称量氯化钠固体的质量为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>g．<br />（2）④中玻璃棒的作用是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）下列操作会导致所配溶液的溶质质量分数偏低的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填字母序号）．<br />A．②中将固体倒入烧杯时洒出<br />B．③中将水倒入烧杯时洒出<br />C．溶解完全后，转移时溶液洒出．','','','','','','5$###$搅拌加速氯化钠的溶解$###$A','【解答】解：（1）配制50g质量分数为10%的氯化钠溶液，需要氯化钠的质量为50g×10%=5g；<br />（2）④中玻璃棒的作用是搅拌，加速溶解；故搅拌加速氯化钠的溶解；<br />（3）A、称好的氯化钠倒入烧杯时有少量洒出，会使氯化钠的质量减小，从而导致溶质质量分数偏小；<br />B、将水倒入烧杯时洒出，会使水的质量减小，从而导致溶质质量分数偏大；<br />C、由于溶液的均一性，配好装瓶时，有少量溶液洒出，对溶质质量分数无影响．<br />答案：（1）5；（2）搅拌加速氯化钠的溶解；（3）A．','【分析】（1）利用溶质质量=溶液质量×溶质的质量分数，可根据溶液的质量和溶质的质量分数计算配制溶液所需要的溶质的质量；<br />（2）根据题中信息可知解答，根据玻璃棒的作用解答；<br />（3）配制溶液的质量分数偏小，可以是溶剂多，也可能是溶质少所导致．','填空题',3.00,'f139890b61ba8f83176b2ffb956d2d63',9,400,'一定溶质质量分数的溶液的配制','',2016,'37','2016•海淀区一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840264,'<img  src=\"/tikuimages/9/0/400/shoutiniao93/161f2bf0-94d4-11e9-a189-b42e9921e93e_xkb58.png\" style=\"vertical-align:middle;FLOAT:right;\" />小刚为了净化收集到的河水，自制了一个如图所示的简易净水器．<br />①其中小卵石，石英砂和膨松棉的作用是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />②活性炭的主要作用是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />③小刚在净化后的水中加入<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，确定是否是硬水．为了降低该水的硬度，小刚采取的常用方法是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','过滤$###$吸附$###$肥皂水$###$煮沸','【解答】解：<br />（1）小卵石、石英沙和膨松棉可以滤去不溶于水的物质，所以其中小卵石、石英沙和膨松棉的作用是过滤；<br />（2）活性炭具有吸附作用，可以吸附水中的色素和异味；<br />（3）可用肥皂水判断所得的水是硬水还是软水，其中产生泡沫较多的是软水，产生泡沫较少的是硬水；为了降低该水的硬度，小刚采取的常用方法是加热煮沸．．<br />故答案为：（1）过滤；（2）吸附；（3）肥皂水；煮沸．','【分析】（1）根据小卵石、石英沙和膨松棉可以滤去不溶于水的物质进行解答；<br />（2）根据活性炭具有吸附作用，可以吸附水中的色素和异味进行解答；<br />（3）根据可用肥皂水判断所得的水是硬水还是软水进行解答．','填空题',3.00,'a25fb736e098d50947e8692f62946f93',9,400,'水的净化,硬水与软水','',0,'37','',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840267,'2016年3月25日上海浦东一物流仓库发生火灾，造成严重经济损失．当火灾发生时，下列措施不恰当的是（　　）','隔离可燃物','有烟雾的地方要蹲下或匍匐前进','室内失火时，立即打开所有门窗，尽量让浓烟从门窗排出','把棉被用水湿透，覆盖在可燃物上，防止火势迅速蔓延','','C','【解答】解：A、灭火的原理有三种：清除可燃物、隔绝氧气或空气、降低可燃物的温度到着火点以下，所以正确．<br />B、气体或烟雾受热密度变小，燃烧产生的有毒的气体或烟雾会随热气流上升，所以正确．<br />C、打开所有门窗，会使室内空气对流，为燃烧反而提供了更多的氧气，燃烧会更旺，所以错误．<br />D、棉被用水湿透，覆盖在可燃物上，一来可以隔绝氧气或空气，二来通过水蒸发吸热能降低可燃物的温度到着火点以下，所以正确．<br />故选C．','【分析】A、根据灭火的原理判断．<br />B、根据气体或烟雾受热密度变小判断．<br />C、根据燃烧的条件和灭火的原理判断．<br />D、根据灭火的原理判断．','选择题',3.00,'5b37d80d0c1aea7bb191a30fa4b1fbb8',9,400,'灭火的原理和方法','',2016,'37','2016•黄石二模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840271,'金属是一类重要的材料，其应用十分广泛．<br />（1）铝的利用比铜和铁晚，金属大规摸开发和利用的先后顺序与哪些因素有关？（写出一条即可）<br />（2）写出用盐酸清洗铁锈（主要成分是Fe<SUB>2</SUB>O<SUB>3</SUB>）的化学方程式．','','','','','','','【解答】解：<br />（1）铝的活动性大于铜与铁，从人类大规模开发、利用金属限可以看出，越活泼的金属，开发利用的越晚，说明主要和金属的活动性有关．<br />（2）盐酸与铁锈反应会生成氯化铁与水，化学方程式为：Fe<SUB>2</SUB>O<SUB>3</SUB>+6HCl=2FeCl<SUB>3</SUB>+3H<SUB>2</SUB>O．<br />答案：<br />&nbsp;（1）与金属的活泼性（或与金属的冶炼难易程度）；<br />（2）Fe<SUB>2</SUB>O<SUB>3</SUB>+6HCl═2FeCl<SUB>3</SUB>+3H<SUB>2</SUB>O','【分析】（1）从人类大规模开发、利用金属的年限可以看出，越活泼的金属，开发利用的越晚，说明主要和金属的活动性有关．<br />（2）根据化学方程式的书写要求结合盐酸与铁锈反应会生成氯化铁与水书写．','书写',3.00,'df6b27628762a75d17f35d78b381d413',9,400,'金属材料及其应用,酸的化学性质,书写化学方程式、文字表达式、电离方程式','',2016,'35','2016春•太康县期中',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840272,'已知T℃时四种化合物在水中和液氨中的溶解度如下表：<br /><table class=\"edittable\"><TBODY><TR><td width=115></TD><td width=115>AgNO<SUB>3</SUB></TD><td width=115>Ba（NO<SUB>3</SUB>）<SUB>2</SUB></TD><td width=115>AgCl</TD><td width=115>BaCl<SUB>2</SUB></TD></TR><TR><td>H<SUB>2</SUB>O</TD><td>170g</TD><td>92.0g</TD><td>1.50×10<SUP>-4</SUP>g</TD><td>33.3g</TD></TR><TR><td>NH<SUB>3</SUB></TD><td>86.0g</TD><td>97.2g</TD><td>0.80g</TD><td>0.00g</TD></TR></TBODY></TABLE>（1）T℃时，把100g硝酸钡放入100g水中，充分溶解后，所得溶液的质量为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）上述四种物质能在液氨中发生复分解反应的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','190g$###$Ba（NO<SUB>3</SUB>）<SUB>2</SUB>+2AgCl=2AgNO<SUB>3</SUB>+BaCl<SUB>2</SUB>↓','【解答】解：（1）T℃时Ba（NO<SUB>3</SUB>）<SUB>2</SUB>的溶解度为92.0g，故100g水中只能溶解92.0gBa（NO<SUB>3</SUB>）<SUB>2</SUB>，故所得溶液的质量为：100g+92g=192g；<br />（2）由溶解性表可知，在液氨中氯化钡不溶，液氨中硝酸钡与氯化银发生复分解反应，反应为Ba（NO<SUB>3</SUB>）<SUB>2</SUB>+2AgCl=2AgNO<SUB>3</SUB>+BaCl<SUB>2</SUB>↓；<br />故答案为：（1）190g&nbsp;&nbsp;&nbsp; （2）Ba（NO<SUB>3</SUB>）<SUB>2</SUB>+2AgCl=2AgNO<SUB>3</SUB>+BaCl<SUB>2</SUB>↓','【分析】（1）根据溶解度表和溶解度的定义进行解答．<br />（2）由溶解度表可知，在水中AgCl不溶于水，在液氨中氯化钡不溶，则在液氨中硝酸钡与氯化银发生复分解反应；','书写',3.00,'b1a40656c117a7463dc31fc59e9dea68',9,400,'复分解反应及其发生的条件,书写化学方程式、文字表达式、电离方程式','',2016,'32','2016•新昌县模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840278,'根据下列实验装置图，按要求回答下列问题：<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao77/16452a80-94d4-11e9-ac9e-b42e9921e93e_xkb72.png\" style=\"vertical-align:middle\" /><br />（1）写出标有序号的仪器名称：①<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；②<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）实验室制取CO<SUB>2</SUB>的化学反应方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；制取并收集该气体应选择的装置是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填序号）；若想控制反应速率，可以将仪器②用<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>代替；检验该气体集满的方法是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）如果选择与制取CO<SUB>2</SUB>相同的发生装置来制取O<SUB>2</SUB>，则发生反应的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，其基本反应类型是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','试管$###$长颈漏斗$###$CaCO<SUB>3</SUB>+2HCl=CaCl<SUB>2</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑$###$BD$###$分液漏斗$###$将燃着的木条放在集气瓶口，若熄灭，则表明集气瓶已集满二氧化碳气体$###$2H<SUB>2</SUB>O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span>&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O+O<SUB>2</SUB>↑$###$分解反应','【解答】解：（1）①是试管；②是长颈漏斗；故填：试管；长颈漏斗；<br />（2）碳酸钙和稀盐酸反应生成氯化钙、水和二氧化碳，化学方程式为：CaCO<SUB>3</SUB>+2HCl═CaCl<SUB>2</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑；为固液常温型装置，二氧化碳能够溶于水，密度比空气大，用瓶口向上排空气法收集，故用BE；分液漏斗可控制反应速率；二氧化碳气体用澄清石灰水进行检验，将气体通入澄清的石灰水中，若变浑浊，则为二氧化碳；&nbsp;&nbsp;故填：CaCO<SUB>3</SUB>+2HCl=CaCl<SUB>2</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑；BD；分液漏斗；将燃着的木条放在集气瓶口，若熄灭，则表明集气瓶已集满二氧化碳气体．<br />（3）实验室如果选择与制取CO<SUB>2</SUB>相同的发生装置来制取O<SUB>2</SUB>，应该是利用过氧化氢制取，发生反应的化学方程式为：2H<SUB>2</SUB>O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span>&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O+O<SUB>2</SUB>↑，一变多，为分解反应．<br />故填：2H<SUB>2</SUB>O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span>&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O+O<SUB>2</SUB>↑；分解反应．','【分析】（1）熟记仪器的名称；<br />（2）实验室通常用大理石或石灰石和稀盐酸反应制取二氧化碳；为固液常温型装置，二氧化碳能够溶于水，密度比空气大，用瓶口向上排空气法收集；分液漏斗可控制反应速率；根据二氧化碳的鉴定方法来回答；<br />（3）通常情况下，过氧化氢在二氧化锰的催化作用下，分解生成水和氧气．','书写',3.00,'02554560263644c7b6442c954498a088',9,400,'实验室制取氧气的反应原理,二氧化碳的实验室制法,二氧化碳的检验和验满,书写化学方程式、文字表达式、电离方程式','',2016,'32','2016•双牌县模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840280,'人类的生产生活都离不开水．<br />（1）下列物质放入水中，能形成溶液的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />A．蔗糖&nbsp;&nbsp;&nbsp;&nbsp;B．花生油&nbsp;&nbsp;&nbsp;C．食盐&nbsp;&nbsp;&nbsp;D．碳酸钙<br />（2）水常用于灭火，其主要灭火原理是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）测定地下水的酸碱度可用<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“紫色石蕊溶液”或“无色酚酞溶液”或“pH试纸”）','','','','','','AC$###$降温$###$pH试纸','【解答】解：（1）整糖、食盐易溶于水形成均一、稳定的混合物，都属于溶液；花生油不溶于水，与水混合形成乳浊液；碳酸钙不溶于水，与水混合形成悬浊液；<br />（2）水灭火是利用水汽化吸热降低温度到可燃物着火点以下；<br />（3）测定地下水的酸碱度可用pH试纸．<br />故答案为：（1）AC（2）降温（3）pH试纸','【分析】（1）本题考查溶液的概念，在一定条件下溶质分散到溶剂中形成的是均一稳定的混合物．（2）根据灭火的方法考虑；（3）测酸碱度用pH试纸．','填空题',3.00,'dfa61b0c0376cfd4bfd8cb582bc93ab7',9,400,'溶液的酸碱度测定,溶液的概念、组成及其特点,灭火的原理和方法','',2016,'37','2016•怀柔区一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840281,'<img  src=\"/tikuimages/9/0/400/shoutiniao57/164a0c80-94d4-11e9-9001-b42e9921e93e_xkb97.png\" style=\"vertical-align:middle\" /><br />（1）水是人类最宝贵的自然资源．小明利用如图1所示的装置探究水的组成．通电一段时间后，试管a与试管b所收集到的气体体积之比约为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，电解水的文字反应式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，该反应属于<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填化合或分解）反应．<br />（2）保护水资源，防止水污染是每个公民和全社会的责任．下列做法有利于防止水资源污染的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />①农业生产中要合理使用农药和化肥&nbsp;&nbsp;②工业废水和生活污水处理达标后再排放&nbsp;&nbsp;&nbsp;<br />③不用含磷洗衣粉&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;④将海水淡化<br />A．①②④B．②③④C．①②③D．①②③④<br />（3）长期使用硬水，会给生活和生产带来许多麻烦．区别硬水和软水的常用试剂是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．生活中将硬水软化常用的方法是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（4）如图2是水分子分解的示意图，请回答下列问题：保持水的化学性质的微粒是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，该变化中发<br />生变化的微粒是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；没有发生变化的微粒是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>、<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，产生的新微粒是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>、<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','2：1$###$水<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td style=\"font-size: 90%\">通电</td></tr><tr><td><div hassize=\"9\"><div style=\"background: url(\'/upimages/images/part/8594L.png\') repeat-x; width:1px; height:5px;float:left;overflow:hidden\" muststretch=\"h\"></div><div style=\"background: url(\'/upimages/images/part/8594.png\') no-repeat; width:9px; height:5px;float:left;overflow:hidden\"></div></div></td></tr></table></span>氢气+氧气$###$分解反应$###$C$###$肥皂水$###$煮沸$###$水分子$###$水分子$###$氢原子$###$氧原子$###$氢分子$###$氧分子','【解答】解：（1）由电解水的实验装置可知，通电一段时间后，试管a与试管b所收集到的气体分别为电源的负极、正极产生的气体，体积之比约为2：1，电解水生成了氢气和氧气，反应的文字反应式为：水<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td style=\"font-size: 90%\">通电</td></tr><tr><td><div hassize=\"9\"><div style=\"background: url(\'/upimages/images/part/8594L.png\') repeat-x; width:1px; height:5px;float:left;overflow:hidden\" muststretch=\"h\"></div><div style=\"background: url(\'/upimages/images/part/8594.png\') no-repeat; width:9px; height:5px;float:left;overflow:hidden\"></div></div></td></tr></table></span>氢气+氧气，该反应属于分解反应．<br />（2）①农业生产中要合理使用农药和化肥，能防止水的污染；<br />&nbsp;&nbsp;②工业废水和生活污水处理达标后再排放，能防止水的污染；<br />③不用含磷洗衣粉，能防止水的污染；<br />④将海水淡化，与防止水的污染无关．<br />由上述分析可知C正确；<br />（3）长期使用硬水，会给生活和生产带来许多麻烦．区别硬水和软水的常用试剂是肥皂水，遇肥皂水产生的泡沫少的是硬水，遇肥皂水产生的泡沫多的是软水．生活中将硬水软化常用的方法是煮沸．<br />（4）由微粒的构成及变化可知：保持水的化学性质的微粒是水分子，该变化中发生变化的微粒是水分子；没有发生变化的微粒是氢原子、氧原子，产生的新微粒是氢分子、氧分子．<br />故答为：（1）2：1，水<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td style=\"font-size: 90%\">通电</td></tr><tr><td><div hassize=\"9\"><div style=\"background: url(\'/upimages/images/part/8594L.png\') repeat-x; width:1px; height:5px;float:left;overflow:hidden\" muststretch=\"h\"></div><div style=\"background: url(\'/upimages/images/part/8594.png\') no-repeat; width:9px; height:5px;float:left;overflow:hidden\"></div></div></td></tr></table></span>氢气+氧气，分解反应；（2）C；（3）肥皂水，煮沸；（4）水分子；水分子；氢原子，氧原子；氢分子、氧分子．','【分析】（1）根据电解水试验的现象结论即发生的反应分析回答有关的问题；<br />（2）根据防止水资源污染的措施分析回答；<br />（3）根据硬水和软水的区别和硬水的软化方法分析回答；<br />（4）根据微粒的构成及变化分析回油管的问题．','书写',3.00,'b8c524478e14243ee34355c783bbe96e',9,400,'电解水实验,硬水与软水,水资源的污染与防治,微粒观点及模型图的应用,反应类型的判定,书写化学方程式、文字表达式、电离方程式','',0,'37','',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840282,'如图是1～18号元素原子最外层电子数与原子核电荷数的关系图．试回答：<br /><img  src=\"/tikuimages/9/2015/400/shoutiniao94/164e0421-94d4-11e9-b080-b42e9921e93e_xkb41.png\" style=\"vertical-align:middle\" /><br />（1）发现元素周期律的科学家是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（选名字）．<br />A．道尔顿&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;B．门捷列夫&nbsp;&nbsp;&nbsp;&nbsp;C．拉瓦锡&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;D．卢瑟福<br />（2）一个水分子共有<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>个原子核和<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>个质子．<br />（3）一个Mg<SUP>2+</SUP>核外共有<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>个电子；Cl<SUP>-</SUP>的最外层电子数和<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>原子的最外层电子数相同；Mg<SUP>2+</SUP>和Cl<SUP>-</SUP>形成的化合物为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','门捷列夫$###$3$###$10$###$10$###$Ar$###$MgCl<SUB>2</SUB>','【解答】解：（1）发现元素周期律的科学家是门捷列夫；故填：门捷列夫；<br />（2）一个水分子是由一个氧原子和两个氢原子构成的，由元素原子最外层电子数与原子核电荷数的关系图可知：氧原子的核电荷数是8，核内的质子数是8；氢原子的核电荷数是1，核内的质子数是1，所以，一个水分子共有质子数是：8+1×2=10；故填：3；10；<br />（3）一个镁原子的核电荷数是12，核外有12个电子，核外电子排布是2、8、2，易失去最外层的两个电子，使次外层变为最外层达到8个电子的稳定结构，变成带两个正电荷的镁离子．所以，一个镁离子核外共有10个电子；氯离子的最外层电子数是8，和Ar原子的最外层电子数相同，Mg<SUP>2+</SUP>和Cl<SUP>-</SUP>形成的化合物为MgCl<SUB>2</SUB>．故填：10；Ar；MgCl<SUB>2</SUB>．','【分析】（1）根据化学史来分析；<br />（2）根据水分子的构成以及核电荷数与核内质子数的关系计算；<br />（3）根据原子与离子的之间的关系分析镁离子核外的电子数，根据核外电子排布找出与氯离子的最外层电子数相同相同的原子．','填空题',3.00,'3a912f1db52115d2222fd9ca06615d4c',9,400,'原子和离子的相互转化,元素周期表的特点及其应用','',2015,'37','2015秋•石林县校级月考',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840283,'碳和碳的化合物是自然界中重要的物质，请回答下列问题：<br />（1）金刚石和石墨的物理性质有很大的差异，其原因是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）葡萄糖（C<SUB>6</SUB>H<SUB>12</SUB>O<SUB>6</SUB>）中碳、氢、氧三种元素的原子个数比为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','碳原子排列方式不同$###$1：2：1','【解答】解：（1）由于金刚石的碳原子排列是空间网状结构，石墨的碳原子排列是层状结构，金刚石、石墨的内部碳原子的排列方式不同，导致二种物质物理性质差异很大；故填：碳原子排列方式不同．<br />（2）葡萄糖（C<SUB>6</SUB>H<SUB>12</SUB>O<SUB>6</SUB>）中碳、氢、氧三种元素的原子个数比为：6：12：6=1：2：1，故填：1：2：1．','【分析】（1）根据物质结构决定物质的性质分析，金刚石的碳原子排列是空间网状结构，石墨的碳原子排列是层状结构；<br />（2）根据物质的化学式的计算分析回答．','书写',3.00,'841b15b6ca8fcf2c30e84742c2d1f6ed',9,400,'碳元素组成的单质,化学式的书写及意义','',2016,'35','2016春•嘉峪关校级期中',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840286,'如图是某化学反应的微观示意图，其中不同的圆球代表不同原子．下列说法中正确的是（　　）<img  src=\"/tikuimages/9/2015/400/shoutiniao39/165c34f0-94d4-11e9-896b-b42e9921e93e_xkb53.png\" style=\"vertical-align:middle\" />','反应前后，原子的种类、分子数目都不变','该反应前后，所有元素的化合价都没有变化','参加反应的两种分子的个数比为1：1','参加反应的两种物质中既有单质，又有化合物','','C','【解答】解：由化学反应的微观示意图可知，各物质反应的微粒数目关系是：<img  src=\"/tikuimages/9/2015/400/shoutiniao39/165fde6e-94d4-11e9-a455-b42e9921e93e_xkb43.png\" style=\"vertical-align:middle\" /><br />A、由微粒的变化可知，反应前后，原子的种类没有变化，分子数目发生了变化，故A错误；<br />B、由微粒的构成可知，该反应有单质生成，一定有化合价的变化，故B错误；<br />C、由上图可知，参加反应的两种分子的个数比为1：1，故C正确；<br />D、由微粒的构成可知，参加反应的两种物质都属于有化合物，故D错误．<br />故选C．','【分析】观察化学反应的微观示意图，根据微粒的变化分析分子、原子的变化，参加反应的两种分子的个数比等；根据微粒的构成分析物质的类别及化合价的变化等．','选择题',3.00,'0363dfd74257e71b25a376e0d945139c',9,400,'单质和化合物的判别,微粒观点及模型图的应用','',2015,'35','2015秋•江津区校级期中',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840288,'<img  src=\"/tikuimages/9/2016/400/shoutiniao44/166d4bf0-94d4-11e9-878b-b42e9921e93e_xkb17.png\" style=\"vertical-align:middle;FLOAT:right\" />（2016•潮阳区模拟）在世界范围内，据统计化肥对粮食增产的贡献率已超过40%，如图是两种氦肥标签的部分，请根据如图所示的信息回答下列问题．<br />（1）尿素中碳元素与氮元素的质量比<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（2）碳酸氢铵的相对分子质量<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（3）纯尿素中氮元素的质量分数为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，这种尿素肥料的纯度至少为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．（计算结果精确到0.1%）<br />（4）从标签上看，碳酸氢铵不同于尿素的化学性质是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填字母序号）；<br />A．易溶于水&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;B．受热易分解&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;C．有挥发性．','','','','','','3：7$###$79$###$46.7%$###$90.6%$###$B','【解答】解：（1）尿素中碳、氮元素的质量比12：（14×2）=3：7．故填：3：7．<br />（2）碳酸氢铵（NH<SUB>4</SUB>HCO<SUB>3</SUB>）的相对分子质量=14×1+1×5+12×1+16×3=79；故填：79；<br />（3）尿素中氮元素的质量分数为<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">14×2</td></tr><tr><td style=\"padding-top:1px;font-size:90%\">12+16+(14+1×2)×2</td></tr></table></span>≈46.7%；这种尿素肥料的纯度至少是42.3%÷46.7%≈90.6%．故填：46.7%；90.6%．<br />（4）根据标签可以看出，这两种物质保存时都需要防潮，因此都是易溶于水的物质，碳酸氢铵要避免高温施用，说明碳酸氢铵受热易分解；故填：B．','【分析】（1）根据化合物中各元素质量比=各原子的相对原子质量×原子个数之比，进行分析解答．<br />（2）相对分子质量为构成该分子的各原子的相对质量之和；<br />（3）根据元素的质量分数公式计算出尿素中氮元素的质量分数，然后根据“46%÷尿素中氮元素的质量分数”计算即可；<br />（4）根据标签的内容即可获得物质的性质．','填空题',3.00,'cb0d7fdea630323aba782e097394d393',9,400,'相对分子质量的概念及其计算,元素质量比的计算,元素的质量分数计算,标签上标示的物质成分及其含量','',2016,'32','2016•潮阳区模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840289,'下列实验操作正确的是（　　）','<img  src=\"/tikuimages/9/2016/400/shoutiniao44/1674c600-94d4-11e9-84fe-b42e9921e93e_xkb49.png\" style=\"vertical-align:middle\" /><br />&nbsp;检查装置气密性','<img  src=\"/tikuimages/9/2016/400/shoutiniao14/16784870-94d4-11e9-a6df-b42e9921e93e_xkb26.png\" style=\"vertical-align:middle\" /><br />&nbsp;&nbsp; 测定溶液pH','<img  src=\"/tikuimages/9/2016/400/shoutiniao13/167a9261-94d4-11e9-9651-b42e9921e93e_xkb94.png\" style=\"vertical-align:middle\" />&nbsp;&nbsp; 稀释浓硫酸','<img  src=\"/tikuimages/9/2016/400/shoutiniao30/167d9fa1-94d4-11e9-bc5f-b42e9921e93e_xkb97.png\" style=\"vertical-align:middle\" /><br />&nbsp;&nbsp;&nbsp; 加热液体','','C','【解答】解：A、该装置未构成封闭体系，即左边的长颈漏斗与大气相通；无论该装置气密性是否良好，导管口都不会有气泡产生，不能判断气密性是否良好，图中所示操作错误．<br />B、用pH试纸测定未知溶液的pH时，正确的操作方法为用玻璃棒蘸取少量待测液滴在干燥的pH试纸上，与标准比色卡对比来确定pH．不能将pH试纸伸入待测液中，以免污染待测液，图中所示操作错误．<br />C、稀释浓硫酸时，要把浓硫酸缓缓地沿器壁注入水中，同时用玻璃棒不断搅拌，以使热量及时地扩散；一定不能把水注入浓硫酸中；图中所示操作正确．<br />D、给试管中的液体加热时，用酒精灯的外焰加热试管里的液体，且液体体积不能超过试管容积的<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">1</td></tr><tr><td>3</td></tr></table></span>，图中液体超过试管容积的<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">1</td></tr><tr><td>3</td></tr></table></span>，图中所示操作错误．<br />故选：C．','【分析】A、根据检查装置气密性的方法进行分析判断．<br />B、根据用pH试纸测定未知溶液的pH的方法进行分析判断．<br />C、根据浓硫酸的稀释方法（酸入水，沿器壁，慢慢倒，不断搅）进行分析判断．<br />D、根据给试管中的液体加热的方法进行分析判断．','选择题',3.00,'ec4191546f3c8dfe7d1e5d5a8dcf7d75',9,400,'给试管里的液体加热,浓硫酸的性质及浓硫酸的稀释,检查装置的气密性,溶液的酸碱度测定','宁国市',2016,'37','2016•宁国市一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840290,'<img  src=\"/tikuimages/9/2016/400/shoutiniao26/1686040f-94d4-11e9-b41e-b42e9921e93e_xkb52.png\" style=\"vertical-align:middle;FLOAT:right;\" />（2016•吉安一模）下列制取和收集气体的方案，可以直接采用如图所示装置进行的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />A．用大理石和稀盐酸制取二氧化碳&nbsp; B．用高锰酸钾制取氧气<br />C．用锌粒和稀硫酸制取氢气&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;D．<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','CD$###$过氧化氢制取氧气','【解答】解：此发生装置适用于“固-液常温”型．共同特点是：固体与液体反应；反应在常温即可发生，不需加热．用排水法收集．<br />A、二氧化碳溶于水不能用排水法计算，故错误；<br />B、高锰酸钾制取氧气需加热，故错误；<br />C、锌粒和稀硫酸制取氢气，固液常温反应，生成的氢气难溶于水，故正确<br />D、此装置还可用于过氧化氢制取氧气．<br />故选：CD．','【分析】本题综合考查气体发生装置和收集装置的选取原理等知识． 此发生装置适用于“固-液常温”型．共同特点是：固体与液体反应；反应在常温即可发生，不需加热．用排水法收集．','填空题',3.00,'4b295b037e2fc88174fde0afecb32aa2',9,400,'常用气体的发生装置和收集装置与选取方法','',2016,'37','2016•吉安一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840292,'下列实验方案合理的是（　　）','用碳酸钾和氢氧化镁制备氢氧化钾','用水鉴别氢氧化钠和硝酸铵两种白色固体','在pH为4的某溶液中，加入水，使其pH提高到8','用观察和组内物质相互反应的方法，可鉴别FeCl<SUB>3</SUB>、NaCl、H<SUB>2</SUB>SO<SUB>4</SUB>三种溶液','','B','【解答】解：A、氢氧化镁为难溶性碱，不能和碳酸钾反应，故A错误；<br />B、氢氧化钠溶于水放热，而硝酸铵溶于水吸热，所以可用水鉴别硝酸铵固体和氢氧化钠固体，故B正确；<br />C、在pH为4的某溶液中，加入水稀释只能接近中性，而不能变成碱性，故C错误；<br />D、FeCl<SUB>3</SUB>、为黄色溶液，但三种物质间不能反应，无法鉴别NaCl、H<SUB>2</SUB>SO<SUB>4</SUB>，故D错误；<br />故选：B．','【分析】A、根据氢氧化镁为难溶性碱，不能和碳酸钾反应进行解答；<br />B、根据氢氧化钠溶于水放热，而硝酸铵溶于水吸热进行解答；<br />C、根据溶液加水稀释不能改变溶液的酸碱性分析；<br />D、根据三种溶液的颜色及反应现象分析．','选择题',3.00,'50935b4d2c3b5c1dcd97f8885559f8b0',9,400,'化学实验方案设计与评价,溶液的酸碱性与pH值的关系,盐的化学性质,酸、碱、盐的鉴别','',2016,'32','2016•海珠区模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840293,'下列物质或用品均为生活中常见物质：<br />①碳酸钙&nbsp;&nbsp;②金刚石&nbsp;&nbsp;③苏打&nbsp;&nbsp;④浓硫酸&nbsp;&nbsp;⑤食盐&nbsp;&nbsp;⑥水&nbsp;&nbsp;⑦熟石灰<br />请用序号回答下列问题：（填序号）<br />（1）为防止骨质疏松，人体可以摄入的物质是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，<br />（2）常用于干燥CO<SUB>2</SUB>气体的干燥剂是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>&nbsp;&nbsp;&nbsp;&nbsp;<br />（3）常用作调味品的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，<br />（4）能中和酸性土壤的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．&nbsp;&nbsp;<br />（5）相对分子质量最小的氧化物是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','①$###$④$###$⑤$###$⑦$###$⑥','【解答】解：（1）碳酸钙含有钙元素，食用可以防止骨质疏松；<br />（2）浓硫酸具有吸水性且不和二氧化碳反应，常用于干燥CO<SUB>2</SUB>气体的干燥剂；<br />（3）食盐具有咸味，是常用的调味品；<br />（4）熟石灰具有碱性，常用于改良酸性土壤；<br />（5）水是相对分子质量最小的氧化物．<br />故答案为：（1）①；（2）④；（3）⑤；（4）⑦；（5）⑥．','【分析】物质的性质决定物质的用途，根据已有的物质的性质进行分析解答即可．','填空题',3.00,'66e3768de418c67491ac52b1898944b5',9,400,'常见碱的特性和用途,常用盐的用途,根据浓硫酸或烧碱的性质确定所能干燥的气体,人体的元素组成与元素对人体健康的重要作用','',2016,'32','2016•清远模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840297,'如图所示是实验室常用的几种仪器，请回答有关问题．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao76/16940dcf-94d4-11e9-873c-b42e9921e93e_xkb7.png\" style=\"vertical-align:middle\" /><br />粗盐中含少量Ca<SUP>2+</SUP>、Mg<SUP>2+</SUP>、SO<SUB>4</SUB><SUP>2-</SUP>和泥沙等杂质，粗盐精制过程中涉及常规操作步骤：<br />（1）①加水溶解；②依次加入过量的氯化钡溶液、氢氧化钠溶液、碳酸钠溶液；③<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；④加入适量的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；⑤<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填试剂或操作名称）．<br />（2）过滤用到的玻璃仪器有：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）配置1000g5.85%的NaCl溶液，需<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>gNaCl；在准确称取固体后，用量筒量取水时，仰视读数，则所配置溶液溶质质量分数<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“偏高”或“偏低”）','','','','','','过滤$###$稀盐酸$###$蒸发浓缩$###$A、B、G$###$58.5$###$偏低','【解答】解：（1）粗盐中含少量Ca<SUP>2+</SUP>、Mg<SUP>2+</SUP>、SO<SUB>4</SUB><SUP>2-</SUP>和泥沙等杂质，粗盐精制的过程中涉及常规操作步骤有：<br />①加水溶解；②依次加入过量的BaCl<SUB>2</SUB>、NaOH、Na<SUB>2</SUB>CO<SUB>3</SUB>溶液；③过滤除去沉淀；④加入适量的稀盐酸除去过量的氢氧化钠和碳酸钠；⑤蒸发浓缩、冷却结晶得到精盐；<br />（2）过滤用到的玻璃仪器有：烧杯、漏斗、玻璃棒；<br />（3）蒸发时不能将溶液直接蒸干；<br />（4）配置1000g5.85%的NaCl溶液，需NaCl的质量为：1000g×5.85%=58.5g；在准确称取固体后，用量筒量取水时，仰视读数，使量取的水的体积偏多，则所配置溶液溶质质量分时偏低．<br />故答案为：（1）过滤，稀盐酸，蒸发浓缩；（2）A、B、G．&nbsp;（3）58.5&nbsp;&nbsp;&nbsp;偏低','【分析】（1）根据粗盐中含有的杂质、提纯的过程分析回答；<br />（2）根据过滤的操作分析回答；<br />（3）根据蒸发的注意事项分析回答；<br />（4）根据溶质量分数的计算公式分析回答．','填空题',3.00,'952535bb6c1bae492382a90109a48e81',9,400,'一定溶质质量分数的溶液的配制,氯化钠与粗盐提纯,盐的化学性质','',2016,'35','2016春•淮安校级期中',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840298,'下列是有关物质的俗名，不含钙元素的是（　　）','生石灰','石膏','大理石','火碱','','D','【解答】解：A．生石灰为氧化钙，含钙元素，故A不符合题意； <br />B．石膏是CaSO<SUB>4</SUB>•2H<SUB>2</SUB>O，含钙元素，故B不符合题意；<br />C．大理石为碳酸钙，含钙元素，故C不符合题意；<br />D．火碱是氢氧化钠，不含钙元素，故D符合题意．<br />故选D','【分析】根据物质的成分解题，生石灰为氧化钙，石膏是CaSO<SUB>4</SUB>•2H<SUB>2</SUB>O，大理石为碳酸钙，火碱是氢氧化钠的俗称进行解答．','选择题',3.00,'db67b3c249dc5f7483bede654b05dec2',9,400,'物质的元素组成','',2016,'32','2016•南平模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840301,'自来水厂通过沉降、过滤、吸附、消毒将水净化．<br />（1）为提高沉降效果，通常要加入絮凝剂明矾，其化学式为KAl（SO<SUB>4</SUB>）<SUB>2</SUB>･12H<SUB>2</SUB>O，明矾由<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>种元素组成，其中属于金属元素的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（写元素名称）．<br />（2）在吸附过程中，常用的吸附剂为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）二氧化氯（ClO<SUB>2</SUB>）是新一代自来水消毒剂．它属于<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填字母序号）．<br />A．化合物&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;B．氧化物&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;C．混合物．','','','','','','5$###$钾元素、铝元素$###$活性炭$###$AB','【解答】解：（1）为提高沉降效果，通常要加入絮凝剂明矾，其化学式为KAl（SO<SUB>4</SUB>）<SUB>2</SUB>･12H<SUB>2</SUB>O，明矾由5种元素组成，其中属于金属元素的是钾元素、铝元素．<br />（2）由于活性炭具有吸附性，在吸附过程中，常用的吸附剂为活性炭．<br />（3）二氧化氯（ClO<SUB>2</SUB>）是由氯和氧两种元素组成的化合物，属于氧化物．<br />故答为：（1）5，钾元素、铝元素；（2）活性炭；（3）AB．','【分析】（1）根据化学式为KAl（SO<SUB>4</SUB>）<SUB>2</SUB>･12H<SUB>2</SUB>O的意义分析判断回答；<br />（2）根据活性炭具有吸附性分析回答；<br />（3）根据二氧化氯（ClO<SUB>2</SUB>）的组成分析．','书写',3.00,'0bc97af715bc481770fc780e3fe28b8b',9,400,'自来水的生产过程与净化方法,从组成上识别氧化物,纯净物和混合物的判别,单质和化合物的判别,元素的简单分类,化学式的书写及意义','',2016,'37','2016•大兴区一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840303,'如表所示，要除去下列物质中的少量杂质，所选试剂及操作都正确的是（　　）<br /><table class=\"edittable\"><TBODY><TR><td width=55>序号</TD><td width=85>物质</TD><td width=94>杂质</TD><td width=105>试剂</TD><td width=84>操作</TD></TR><TR><td>A</TD><td>Fe</TD><td>Fe<SUB>2</SUB>O<SUB>3</SUB></TD><td>盐酸</TD><td>结晶</TD></TR><TR><td>B</TD><td>NH<SUB>4</SUB>HCO<SUB>3</SUB></TD><td>NaCl</TD><td>-</TD><td>加热</TD></TR><TR><td>C</TD><td>Cu（NO<SUB>3</SUB>）<SUB>2</SUB></TD><td>BaCl<SUB>2</SUB></TD><td>AgNO<SUB>3</SUB>溶液</TD><td>过滤</TD></TR><TR><td>D</TD><td>O<SUB>2</SUB></TD><td>水蒸气</TD><td>浓硫酸</TD><td>洗气</TD></TR></TBODY></TABLE>','A','B','C','D','','D','【解答】解：A、Fe<SUB>2</SUB>O<SUB>3</SUB>和Fe均能与盐酸反应，不但能把杂质除去，也会把原物质除去，不符合除杂原则，故选项所采取的方法错误．<br />B、NH<SUB>4</SUB>HCO<SUB>3</SUB>在加热条件下生成氨气、水和二氧化碳，反而会把原物质除去，不符合除杂原则，故选项所采取的方法错误．<br />C、BaCl<SUB>2</SUB>能与AgNO<SUB>3</SUB>溶液反应生成氯化银沉淀和硝酸钡，能除去杂质但引入了新的杂质硝酸钡，不符合除杂原则，故选项所采取的方法错误．<br />D、浓硫酸具有吸水性，且不与氧气反应，能除去杂质且没有引入新的杂质，符合除杂原则，故选项所采取的方法正确．<br />故选：D．','【分析】根据原物质和杂质的性质选择适当的除杂剂和分离方法，所谓除杂（提纯），是指除去杂质，同时被提纯物质不得改变．除杂质题至少要满足两个条件：①加入的试剂只能与杂质反应，不能与原物质反应；②反应后不能引入新的杂质．','选择题',3.00,'cdbe1991e4187271ab6a1a1fcb14afee',9,400,'物质除杂或净化的探究,气体的干燥（除水）,盐的化学性质','',2016,'32','2016•洛阳模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840305,'用“物理变化”、“化学变化”、“物理性质”、“化学性质”四个概念填空：<br />①白纸是白色的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />②白纸燃烧了<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />③把白纸撕碎<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />④碳酸钙在高温下能分解<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','物理性质$###$化学变化$###$物理变化$###$化学性质','【解答】解：①白纸是白色的属于物理性质；<br />②物质燃烧生成二氧化碳和水，属于化学变化；<br />③白纸撕碎只是形状的改变，属于物理变化；<br />④碳酸钙分解属于化学变化，能分解属于化学性质；<br />故答案为：①物理性质；②化学变化；③物理变化；④化学性质．','【分析】化学变化是指有新物质生成的变化，物理变化是指没有新物质生成的变化．物质的化学性质是指在化学变化中表现出来的性质，如可燃性、稳定性；物质的物理性质是指不需要通过化学变化表现出来的性质，如颜色、状态、味道、溶解性等；性质和变化，性质是特性，变化是过程，描述性质一般用“能、易、难、会、可以”等．','填空题',3.00,'c01f38ecc4619a9794e4615c85aa2f2c',9,400,'化学变化和物理变化的判别,化学性质与物理性质的差别及应用','',2015,'35','2015秋•淮安校级期中',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840307,'如图所示为实验室中常见的气体制备、干燥、收集和性质实验的部分装置（组装实验装置时，可重复选择），试根据题目要求，回答下列问题：<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao45/16b5c6a1-94d4-11e9-b871-b42e9921e93e_xkb94.png\" style=\"vertical-align:middle\" /><br />Ⅰ．氯酸钾固体和二氧化锰混合加热，能较快产生氧气．小刚同学欲在实验室中以此方法制备并收集干燥的氧气．<br />（1）所选装置的连接顺序为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填写仪器序号字母）．<br />（2）用氯酸钾制备氧气所发生的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />Ⅱ．小强同学欲用锌和稀硫酸反应制取氢气，然后利用氢气测定某氧化铁样品中含Fe<SUB>2</SUB>O<SUB>3</SUB>的质量分数（假设杂质不挥发，也不发生反应）．选择制取氢气的发生装置是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，接下来连接装置的顺序为：D<SUB>1</SUB>→C→D<SUB>2</SUB>→D<SUB>3</SUB>．（已知：Fe<SUB>2</SUB>O<SUB>3</SUB>+3H<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2Fe+3H<SUB>2</SUB>O；其中D<SUB>1</SUB>、D<SUB>2</SUB>、D<SUB>3</SUB>为三个盛浓硫酸洗气瓶）<br />（1）写出仪器a的名称<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，在装药品之前，检查装置A气密性的具体操作是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />（2）小强欲通过测量反应前后D<SUB>2</SUB>浓硫酸洗气瓶的质量变化，测算氧化铁样品中含有Fe<SUB>2</SUB>O<SUB>3</SUB>的质量分数．实验中，在其他操作正常的条件下，若装置中不连接D<SUB>1</SUB>，则测算结果将<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．（填写“偏小”“偏大”“不受影响”）','','','','','','BDF$###$2KClO<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span>&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>△</td></tr></table></span>2KCl+3O<SUB>2</SUB>↑$###$A$###$分液漏斗$###$关闭分液漏斗的开关，先把导管的一端浸入水中，然后两手紧握锥形瓶的外壁，如果导管口有气泡冒出，证明严密$###$偏大','【解答】解：Ⅰ．（1）如果用氯酸钾制氧气就需要加热，氧气可以用浓硫酸干燥，氧气的密度比空气的密度大，不易溶于水，因此能用向上排空气法和排水法收集，用向上排空气法收集的氧气比较纯净；故答案为：BDF；<br />（2）氯酸钾在二氧化锰做催化剂和加热的条件下生成氯化钾和氧气，配平即可，故答案为：2KClO<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span>&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>△</td></tr></table></span>2KCl+3O<SUB>2</SUB>↑；<br />Ⅱ．实验室是用锌粒和稀硫酸在常温下反应制氢气的，氢气难溶于水，密度比空气的密度小，因此可以用排水法和向下排空气法收集；故答案为：A；<br />（1）分液漏斗可以控制反应的发生和停止，检查仪器A气密性的具体操作步骤是：关闭分液漏斗的开关，先把导管的一端浸入水中，然后两手紧握锥形瓶的外壁，如果导管口有气泡冒出，证明严密；故答案为：分液漏斗；关闭分液漏斗的开关，先把导管的一端浸入水中，然后两手紧握锥形瓶的外壁，如果导管口有气泡冒出，证明严密；<br />（2）实验中，在其他操作正常的条件下，若装置中不连接D<SUB>1</SUB>，则测算结果将偏大，因为没有D<SUB>1</SUB>时，氢气中混有的水蒸气被D<SUB>2</SUB>吸收，结果偏大；故答案为：偏大．','【分析】分液漏斗可以控制反应的发生和停止，检查仪器A气密性的具体操作步骤是：关闭分液漏斗的开关，先把导管的一端浸入水中，然后两手紧握锥形瓶的外壁，如果导管口有气泡冒出，证明严密；制取装置包括加热和不需加热两种，如果用双氧水和二氧化锰制氧气就不需要加热，如果用高锰酸钾或氯酸钾制氧气就需要加热．氧气的密度比空气的密度大，不易溶于水，因此能用向上排空气法和排水法收集，用向上排空气法收集的氧气比较纯净．实验室制取CO<SUB>2</SUB>，是在常温下，用大理石或石灰石和稀盐酸制取的，碳酸钙和盐酸互相交换成分生成氯化钙和水和二氧化碳，因此不需要加热．二氧化碳能溶于水，密度比空气的密度大，因此只能用向上排空气法收集．实验室是用锌粒和稀硫酸在常温下反应制氢气的，氢气难溶于水，密度比空气的密度小，因此可以用排水法和向下排空气法收集；实验中，在其他操作正常的条件下，若装置中不连接D<SUB>1</SUB>，则测算结果将偏大，因为没有D<SUB>1</SUB>时，氢气中混有的水蒸气被D<SUB>2</SUB>吸收，结果偏大．','书写',3.00,'0b1512f68c0bc29376226df5ca90db39',9,400,'常用气体的发生装置和收集装置与选取方法,实验室制取氧气的反应原理,书写化学方程式、文字表达式、电离方程式,氢气的化学性质与燃烧实验','',2016,'37','2016•繁昌县二模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840309,'下列微观解释不正确的是（　　）','“春色满园，花香四溢”---------分子不断在运动','将二氧化碳气体制成干冰，体积变小-----------分子体积变小','甘蔗甜，柠檬酸---------不同分子性质不同','50ml酒精和50ml水混合后总体积小于100ml--------分子间有间隔','','B','【解答】解：A、“春色满园，花香四溢”，是因为花香中含有的分子是在不断运动的，向四周扩散，使人们闻到花香，故选项解释正确．<br />B、将二氧化碳气体制成干冰，体积变小，是因为分子间有间隔，气体受压后，分子间隔变小，故选项解释错误．<br />C、甘蔗甜，柠檬酸，是因为它们分子的构成不同，不同种的分子性质不同，故选项解释正确．<br />D、50ml酒精和50ml水混合后总体积小于100ml，是因为分子之间有间隔，一部分水分子和酒精分子会互相占据分子之间的间隔，故选项解释正确．<br />故选：B．','【分析】根据分子的基本特征：分子质量和体积都很小；分子之间有间隔；分子是在不断运动的；同种的分子性质相同，不同种的分子性质不同，可以简记为：“两小运间，同同不不”，结合事实进行分析判断即可．','选择题',3.00,'eeb8d4da9bfc8bfb5e852aff1da7a76c',9,400,'利用分子与原子的性质分析和解决问题','普宁市',2015,'37','2015•普宁市一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840312,'<img  src=\"/tikuimages/9/2016/400/shoutiniao34/16c66870-94d4-11e9-8ae1-b42e9921e93e_xkb28.png\" style=\"vertical-align:middle;FLOAT:right\" />（2016春•召陵区期中）如图是反应的微观示意图．<br />（1）图中单质的化学式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；写出该反应的化学方程式：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）已知HClO是一种酸，能与碱反应生成盐和水．将氯气通入过量NaOH溶液，所得溶液中的溶质有哪些？','','','','','','Cl<SUB>2</SUB>$###$Cl<SUB>2</SUB>+H<SUB>2</SUB>O═HCl+HClO','【解答】解：（1）观察图示可得，该反应是氯气和水生成了氯化氢和次氯酸；反应的化学方程式：Cl<SUB>2</SUB>+H<SUB>2</SUB>O═HCl+HClO；因此图中的单质为氯气；故填：Cl<SUB>2</SUB>；Cl<SUB>2</SUB>+H<SUB>2</SUB>O═HCl+HClO；<br />（2）氯气和水生成了氯化氢和次氯酸，HClO是一种酸，能与碱反应生成盐和水，因此氯气与NaOH溶液反应的化学方程式为：Cl<SUB>2</SUB>+2NaOH=NaCl+NaClO+H<SUB>2</SUB>O；因此溶液中的溶质为NaCl、NaClO和过量的NaOH；故填：NaCl、NaClO、NaOH．','【分析】（1）观察微观示意图，根据微粒的构成分析物质的类别及名称，分析反应物、生成物，写出反应的化学方程式；<br />（2）根据题目信息结合方程式的书写方法写出反应的方程式．','书写',3.00,'f8d856adedb47a325eead4189e461cda',9,400,'中和反应及其应用,单质和化合物的判别,微粒观点及模型图的应用,书写化学方程式、文字表达式、电离方程式','',2016,'35','2016春•召陵区期中',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840316,'2015年世界环境日宣传标语，当我们发展经济的时候，我们忘记了环境，当我们发现我们的天空不再那么湛蓝，空气不再那么新鲜的时候，我们才知道环境的重要性，下列做法合理的是（　　）','大力发展火力发电','严禁使用塑料制品，以免白色污染','重污染企业搬迁到偏远地区','建设垃圾无害化处理设施','','D','【解答】解：A、大力发展火力发电，虽然解决电力紧张问题，但也造成了大气污染，故错误；<br />B、禁止使用任何塑料制品是不现实的，故B错误；<br />C、重污染企业搬迁到偏远地区，同样会造成污染，故C错误；<br />D、将垃圾进行无害化处理，会减少环境污染，做法合理，故D正确；<br />故选D．','【分析】A、根据火力发电的环境污染分析；<br />B、根据要解决白色污染，要尽可能减少使用塑料品，回收废旧塑料，研制开发可降解塑料，不能禁止使用解答；<br />C、重污染企业搬迁到偏远地区，同样会造成污染；<br />D、将垃圾进行无害化处理，会减少环境污染．','选择题',3.00,'d6801f27366c414f79cafd01a9335635',9,400,'防治空气污染的措施','',2015,'32','2015•当涂县模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840319,'下图所示实验操作中，正确的是（　　）','<img  src=\"/tikuimages/9/2016/400/shoutiniao55/16d9f070-94d4-11e9-bc05-b42e9921e93e_xkb91.png\" style=\"vertical-align:middle\" />加热液体','<img  src=\"/tikuimages/9/2016/400/shoutiniao93/16dc134f-94d4-11e9-8553-b42e9921e93e_xkb21.png\" style=\"vertical-align:middle\" />连接仪器','<img  src=\"/tikuimages/9/2016/400/shoutiniao87/16de844f-94d4-11e9-b248-b42e9921e93e_xkb44.png\" style=\"vertical-align:middle\" />读取液体体积','<img  src=\"/tikuimages/9/2016/400/shoutiniao83/16e0ce40-94d4-11e9-a8e1-b42e9921e93e_xkb74.png\" style=\"vertical-align:middle\" />液体的倾倒','','B','【解答】解：A、给试管中的液体加热时，用酒精灯的外焰加热试管里的液体，且液体体积不能超过试管容积的<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">1</td></tr><tr><td>3</td></tr></table></span>，图中没有用外焰加热，图中所示操作错误．<br />B、导管连接胶皮管时，先把导管一端湿润，然后稍用力转动使之插入胶皮管内，图中所示装置正确．<br />C、量取液体时，视线与液体的凹液面最低处保持水平，图中俯视刻度，操作错误．<br />D、向试管中倾倒液体药品时，瓶塞要倒放，标签要对准手心，瓶口紧挨；图中瓶塞没有倒放，所示操作错误．<br />故选：B．','【分析】A、根据给试管中的液体加热的方法进行分析判断．<br />B、根据导管连接胶皮管的方法进行分析判断．<br />C、根据量筒读数时视线要与凹液面的最低处保持水平进行分析判断．<br />D、根据向试管中倾倒液体药品的方法进行分析判断．','选择题',3.00,'789e46c3ac403ab0063dc4007020bc88',9,400,'测量容器-量筒,液体药品的取用,给试管里的液体加热,仪器的装配或连接','',2016,'37','2016春•吉林校级月考',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840322,'<img  src=\"/tikuimages/9/2016/400/shoutiniao90/16eed800-94d4-11e9-a46d-b42e9921e93e_xkb46.png\" style=\"vertical-align:middle;FLOAT:right\" />（2016•保定模拟）某钙片的标签如图所示<br />（1）阅读上图，你认为该该片的主要功能是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>食用方法中“咀嚼后咽下”，“咀嚼”的目的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）你认为该该片除了它的主要功能外，还可以治疗<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．有人认为这种钙片食用量越多越好，你是否赞同，原因是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）通过标签的内容，这种该片的含钙量是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','补钙$###$增大钙片与胃液的接触面积$###$胃酸过多$###$食用过多碳酸钙也会消弱胃的消化功能$###$93.57%','【解答】解：（1）该药物中含有碳酸钙，可用于补钙；通过嚼食可将较大片状的食品变成细小的颗粒，增大反应物的接触面积，促进钙的吸收．故填：补钙；增大钙片与胃液的接触面积；<br />（2）碳酸钙与胃液中的盐酸发生反应，生成了氯化钙、水、二氧化碳，服用过多，会消耗大量的胃酸，造成消化功能下降，故填：胃酸过多；食用过多碳酸钙也会消弱胃的消化功能；<br />（3）这种该片的含钙量为：<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">0.75g</td></tr><tr><td style=\"padding-top:1px;font-size:90%\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">40g</td></tr><tr><td>50</td></tr></table></td></tr></table>×100%</span>=93.57%．故填：93.57%．','【分析】（1）根据药品的成分以及咀嚼后的作用来分析；<br />（2）根据碳酸钙与胃液中的盐酸反应来分析；<br />（3）根据标签信息来分析．','填空题',3.00,'de6ef399d27e80d7a6fef524ad640157',9,400,'盐的化学性质,化合物中某元素的质量计算,标签上标示的物质成分及其含量','',2016,'32','2016•保定模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840329,'利用如图所示的装置可做CO还原Fe<SUB>2</SUB>O<SUB>3</SUB>的实验，并检验该反应的气体产物，已知由A装置制取的CO气体中混有少量的CO<SUB>2</SUB>．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao31/170df8c0-94d4-11e9-8192-b42e9921e93e_xkb74.png\" style=\"vertical-align:middle\" /><br />（1）CO与Fe<SUB>2</SUB>O<SUB>3</SUB>反应的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）气体通过装置的顺序是A→<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．（装置不能重复使用）．B装置的作用是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．C装置的作用是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．（注：氢氧化钠能吸收二氧化碳）<br />（3）从环保角度考虑，对以上装置的改进措施是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','Fe<SUB>2</SUB>O<SUB>3</SUB>+3CO<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2Fe+3CO<SUB>2</SUB>$###$C→D→B$###$检验生成物中有二氧化碳$###$除去混合气的二氧化碳，防止对一氧化碳的生成物检验造成干扰$###$在B装置后增加尾气处理装置','【解答】解：（1）CO与Fe<SUB>2</SUB>O<SUB>3</SUB>在高温的条件下反应生成铁和二氧化碳，化学方程式为：Fe<SUB>2</SUB>O<SUB>3</SUB>+3C0<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2Fe+3CO<SUB>2</SUB>；<br />（2）混合气体通过C装置，浓氢氧化钠溶液吸收由A装置制取的CO气体中混有少量的CO<SUB>2</SUB>，然后CO进入D装置，发生CO还原Fe<SUB>2</SUB>O<SUB>3</SUB>反应，生成二氧化碳和铁，检验二氧化碳应用澄清石灰水，所以正确的顺序为：C→D→B，B装置的作用是检验生成物中有二氧化碳，C装置的作用是：除去混合气的二氧化碳，防止对一氧化碳的生成物检验造成干扰；<br />（3）一氧化碳有毒，排放到空气中会污染空气，装置的改进措施是：将尾气点燃或进行收集处理．<br />故答案为：（1）Fe<SUB>2</SUB>O<SUB>3</SUB>+3C0<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2Fe+3CO<SUB>2</SUB>；<br />（2）C→D→B，检验生成物中有二氧化碳，除去混合气的二氧化碳，防止对一氧化碳的生成物检验造成干扰；<br />（3）将尾气点燃或进行收集处理．','【分析】根据一氧化碳的还原性、检验二氧化碳用澄清石灰水以及一氧化碳有毒污染空气进行解答．','书写',3.00,'e81ab60e2edd1bf9835631aaddd0f033',9,400,'常见气体的检验与除杂方法,一氧化碳还原氧化铁,书写化学方程式、文字表达式、电离方程式','',2016,'32','2016•昆明模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840331,'<img  src=\"/tikuimages/9/2016/400/shoutiniao92/171599e1-94d4-11e9-be83-b42e9921e93e_xkb84.png\" style=\"vertical-align:middle;FLOAT:right\" />（2016•武进区模拟）小兵同学利用棉签设计了如图实验，a处滴浓氨水，b处滴酚酞试液，过一会他观察到b处的棉花变红，a处的棉花不变红．下列说法不正确的是（　　）','氨分子在不断地运动','所有分子的运动速率相等','氨水的pH大于7','该实验药品用量少并能有效防止氨气逸出','','B','【解答】解：A、说明氨分子在不断的运动，正确；<br />B、说明所有分子的运动速率不相等，错误；<br />C、酚酞变成红色，说明氨水的pH大于7，正确；<br />D、该实验药品用量少并能有效防止氨气逸出，正确；<br />故选B．','【分析】根据浓氨水易挥发和分子在不断的运动分析判断即可．','选择题',3.00,'9699ba59dc83641e4fb2bc6a519c92bb',9,400,'碱的化学性质,溶液的酸碱性与pH值的关系,分子的定义与分子的特性','',2016,'32','2016•武进区模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840333,'甲酸（HCOOH）通常是一种无色易挥发的液体，它在浓硫酸作用下易分解，反应方程式为：HCOOH<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;浓硫酸&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>CO↑+H<SUB>2</SUB>O，某课外活动小组的同学欲用该反应来制取CO，并还原红棕色的氧化铁粉末．现有下列仪器或装置供选择：<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao98/171c9ec0-94d4-11e9-878a-b42e9921e93e_xkb75.png\" style=\"vertical-align:middle\" /><br />（1）用甲酸滴入浓硫酸的方法制取CO，应选<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填序号）装置；如果要除去CO中混有少量甲酸气体，最好选择图中<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填序号）进行洗气．<br />（2）用上述方法制取的CO还原氧化铁，并检验气体产物，则各仪器的接口连接顺序为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填字母）．<br />（3）在该制取CO的反应中，浓硫酸所起的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>作用．<br />（4）从开始加热到实验结束，氧化铁粉末的颜色变化为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','②$###$③$###$BDCGHF$###$脱水$###$红色变为黑色','【解答】解：（1）向甲酸中滴加浓硫酸液体，应通过分液漏斗加入．所以应该选择②装置；氢氧化钠溶液、氢氧化钙溶液都能够吸收CO中混有的少量甲酸气体，但是氢氧化钠溶液的吸收能力比氢氧化钙溶液强．所以应选装置③；<br />（2）生成的一氧化碳气体从B中排出，从D进入氢氧化钠溶液除去甲酸气体，从C中排出；再从G进入，在加热的条件下．一氧化碳还原氧化铁，气体再从H排出，最后从F进入氢氧化钙溶液中，检验有二氧化碳生成．所以各仪器的接口连接顺序为：B、D、C、G、H、F．<br />（3）由题意可知，在该制取CO的反应中，浓硫酸使甲酸脱水生成了一氧化碳，起到脱水的作用；<br />（4）在加热的条件下，氧化铁被还原成铁，所以从开始加热到实验结束，氧化铁粉末的颜色变化为：红色变为黑色．<br />故答案为：（1）②；③；（2）BDCGHF；（3）脱水；（4）红色变为黑色．','【分析】（1）根据分液漏斗的用途分析选择用甲酸滴入浓硫酸的装置；根据氢氧化钠和氢氧化钙的溶解性分析除去CO中混有少量甲酸气体应用的装置；<br />（2）用一氧化碳CO还原氧化铁．应先除去出甲酸气体，再还原氧化铁，最后用澄清的石灰水检验生成的二氧化碳，据此分析各仪器的接口连接顺序；<br />（3）根据浓硫酸具有脱水性分析；<br />（4）根据反应物、生成物的颜色分析现象．','填空题',3.00,'ac1a2efe52349bd85dd82ff92c9a8758',9,400,'常见气体的检验与除杂方法,一氧化碳还原氧化铁','',2016,'32','2016•鱼台县模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840336,'下列实验的先后顺序正确的是（　　）','称量物质的质量时，先加质量大的砝码，后加质量小的砝码','向试管中滴加少量液体药品时，先把滴管伸入试管内，后捏橡胶头向试管中滴入液体药品','测定溶液的pH时，先用水润湿pH试纸，后蘸取溶液滴在pH试纸上','稀释浓硫酸时，先在烧杯倒入浓硫酸，后向烧杯中缓慢注入水','','A','【解答】解：A、用托盘天平称量一定质量的药品时，应先放质量大的砝码，后放质量小的砝码，所以本实验操作正确；故本选项符合题意；<br />B、用胶头滴管吸取液体先按住胶头，挤压出空气，再插入被吸液体，故选项说法错误；<br />C、用pH试纸测定溶液的pH时，不能用水润湿pH试纸，润湿pH试纸，等于稀释了溶液，所以本实验操作错误；故本选项不符合题意；<br />D、稀释浓硫酸时，要把浓硫酸沿着玻璃棒缓缓注入水中，并不断搅拌．所以本实验操作错误；故本选项不符合题意；<br />故选A．','【分析】A、根据用托盘天平称量一定量的药品的方法进行分析判断．<br />B、根据胶头滴管的使用方法进行分析判断；<br />C、根据pH试纸的使用方法分析判断；<br />D、根据稀释浓硫酸时，浓硫酸溶解于水要放出大量的热，造成液滴飞溅的注意事项回答．','选择题',3.00,'1a709b4cf3d0dc45e5eb11256d56f806',9,400,'称量器-托盘天平,液体药品的取用,溶液的酸碱度测定','',2016,'32','2016•澄海区模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840338,'金属M与AgNO<SUB>3</SUB>溶液发生反应：M+2AgNO<SUB>3</SUB>═M（NO<SUB>3</SUB>）<SUB>2</SUB>+2Ag，下列说法不正确的是（　　）','M不可能是铝','反应前后M的化合价改变','M的金属活动性比Ag强','在金属活动性顺序表中，M一定排在氢前','','D','【解答】解：A、由M与硝酸银溶液反应的生成了M（NO<SUB>3</SUB>）<SUB>2</SUB>，则M在化合物中的化合价为+2价，铝与硝酸银溶液反应生成了硝酸铝，铝为+3价，故该金属不可能是铝，故说法正确；<br />B、反应前M为单质，元素的化合价为0，反应后表现为+2，故说法正确；<br />C、由于金属M能与硝酸银反应置换出银，说明M的金属活动性比银强，故说法正确；<br />D、由于金属M能与硝酸银反应置换出银，说明M的金属活动性比银强，但M不一定排在氢前，故说法错误．<br />故选：D．','【分析】根据金属活动性顺序的应用分析推断．根据题意，金属M能与硝酸银反应置换出银，说明M的活动性比银强；由生成了M（NO<SUB>3</SUB>）<SUB>2</SUB>可知，M在化合物中的化合价为+2价；结合选项进行分析．','选择题',3.00,'0388d5cce5d480b7c311f70a046a1a61',9,400,'金属的化学性质,化合价规律和原则','',2016,'37','2016•宝丰县二模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840340,'在装有水的四个烧杯中，分别加入少量食盐、菜油、牛奶和面粉，充分搅拌后可形成溶液的是（　　）','食盐','菜油','牛奶','面粉','','A','【解答】解：A．食盐易溶于水，形成均一、稳定的混合物，属于溶液，故选项正确．<br />B．菜油不溶于水，不能和水形成均一、稳定的混合物，即不能够形成溶液，故选项错误．<br />C．牛奶是不溶性的小液滴等分散到液体中形成的，不能和水形成均一、稳定的混合物，即不能够形成溶液，故选项错误．<br />D．面粉不溶于水，不能和水形成均一、稳定的混合物，即不能够形成溶液，故选项错误．<br />故选：A．','【分析】一种或几种物质分散到另一种物质中，形成均一的、稳定的混合物叫做溶液，它的基本特征是均一性和稳定性；只有被分散的物质在水中是可溶的，二者混合后才会形成溶液．','选择题',3.00,'f0ebfb4d730b9fc7d83687d0bcebb9a7',9,400,'溶液的概念、组成及其特点','',2015,'37','2015秋•深圳月考',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840341,'实验是学习化学的重要途径．如图化学实验基本操作不正确的是（　　）','<img  src=\"/tikuimages/9/2016/400/shoutiniao25/173d1f0f-94d4-11e9-a5ae-b42e9921e93e_xkb3.png\" style=\"vertical-align:middle\" /><br />熄灭酒精灯','<img  src=\"/tikuimages/9/2016/400/shoutiniao79/173e0970-94d4-11e9-8886-b42e9921e93e_xkb7.png\" style=\"vertical-align:middle\" /><br />读出液体的体积','<img  src=\"/tikuimages/9/2016/400/shoutiniao73/173fb71e-94d4-11e9-9352-b42e9921e93e_xkb54.png\" style=\"vertical-align:middle\" /><br />加热液体','<img  src=\"/tikuimages/9/2016/400/shoutiniao96/1742c45e-94d4-11e9-a7da-b42e9921e93e_xkb57.png\" style=\"vertical-align:middle\" /><br />稀释浓硫酸','','B','【解答】解：A、熄灭酒精灯，不能用嘴吹灭，要用灯帽盖灭，故A正确；<br />B、读出量筒中液体的体积时，视线要与凹液面的最低处保持相平，故B错误；<br />C、加热试管中的液体，液体取用量不能超过试管容积的三分之一，故C正确；<br />D、稀释浓硫酸要将浓硫酸加入水中，不能将水加入浓硫酸中，水的密度比浓硫酸小并且浓硫酸溶于水放出大量的热，把水加入浓硫酸中水会浮在浓硫酸上面而造成液滴飞溅，故D正确．<br />故选B．','【分析】根据化学实验的基本操作，主要从操作要领和注意事项上来分析．','选择题',3.00,'f1bf29d7dc7c9046867da5560221f436',9,400,'测量容器-量筒,加热器皿-酒精灯,给试管里的液体加热,浓硫酸的性质及浓硫酸的稀释','',2016,'32','2016•潮阳区模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840343,'用“<img  src=\"/tikuimages/9/2016/400/shoutiniao84/174d72c0-94d4-11e9-9fb7-b42e9921e93e_xkb90.png\" style=\"vertical-align:middle\" />”和“<img  src=\"/tikuimages/9/2016/400/shoutiniao93/174e5d21-94d4-11e9-b4b5-b42e9921e93e_xkb33.png\" style=\"vertical-align:middle\" />”分别表示氢原子和氧原子，如图是氢气与氧气在点燃条件下发生反应的微观模拟图．请回答下列问题：<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao68/174e5d22-94d4-11e9-a26f-b42e9921e93e_xkb47.png\" style=\"vertical-align:middle\" /><br />（1）在B图中将相关粒子图形补充完整：<br />（2）此变化中参加反应的氢分子与氧分子的个数比是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（3）结合该图示从微观角度解释由A到C变化的实质是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','2：1$###$在点燃的条件下，氢分子分解为氢原子，氧分子分解为氧原子，每两个氢原子和一个氧原子结合成一个水分子','【解答】解：（1）由A→B是分子的分开过程，故B中应该是原子数目与A中的相同，所以补充两个氢原子，相关粒子图形为：<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao99/175513e1-94d4-11e9-969f-b42e9921e93e_xkb67.png\" style=\"vertical-align:middle\" />；<br />（2）由微粒的变化可知，此变化中参加反应的氢分子与氧分子的个数比是2：1；<br />（3）由反应的微观模拟图可知，由A到B变化的实质是：在点燃的条件下氢分子分解成氢原子，氧分子分解成氧原子．<br />故答为：（1）补充2个氢原子；（2）2：1；（3）在点燃的条件下，氢分子分解为氢原子，氧分子分解为氧原子，每两个氢原子和一个氧原子结合成一个水分子．','【分析】（1）依据化学反应的实质和质量守恒定律分析解答即可；<br />（2）根据图示中微粒变化情况分析解答；<br />（3）根据化学变化中分子会分开而后原子再进行重新组合分析解答．','填空题',3.00,'8d0855e3e02a5d281c3bf74142c6b167',9,400,'微粒观点及模型图的应用,化学反应的实质','',2016,'37','2016•松北区二模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840344,'小强在厨房里发现一瓶没有标签的无色液体．<br />（1）他闻了闻，初步判断为白醋，小强是利用白醋的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“物理”或“化学”）性质作出的判断．<br />（2）他取少量此液体放入玻璃杯中，加入纯碱，产生气体，说明该液体是显<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“酸性”“碱性”或“中性”）物质，进一步判断为白醋．<br />（3）他另取少量此液体滴入石蕊试液，溶液变为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>色，要使其变为蓝色，可向其中加入<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填编号）．<br />a．食盐b．熟石灰c．稀硫酸d．水．','','','','','','物理$###$酸性$###$红$###$b','【解答】解：（1）小华在厨房里发现一瓶无色液体，他闻了闻，初步判断为白醋，是利用白醋具有刺激性气味，不需要发生化学变化就能表现出来，是利用了其物理性质．<br />（2）纯碱能与酸反应生成二氧化碳气体，他取少量此液体放入玻璃杯中，加入纯碱，产生气体，说明该液体是显酸性物质，进一步判断为白醋．<br />（3）他另取少量此液体滴入石蕊试液，溶液变为红色；要使其变为蓝色，应加入碱性物质，食盐、熟石灰、稀硫酸、水分别显中性、碱性、酸性、中性，故可向其中加入熟石灰．<br />故答案为：（1）物理；（2）酸性；（3）红；b．','【分析】（1）根据醋酸的化学性质，醋酸具有刺激性气味，据此进行分析解答．<br />（2）根据纯碱能与酸反应生成二氧化碳气体，进行分析解答．<br />（3）根据紫色石蕊溶液遇酸性溶液变红，遇碱性溶液变蓝，进行分析解答．','填空题',3.00,'66df23efb8e0a17954c84483508dec0d',9,400,'醋酸的性质及醋酸的含量测定,化学性质与物理性质的差别及应用','宁夏',2016,'37','2016•宁夏校级一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840346,'从分子的角度分析，下列解释错误的是（　　）','吹胀气球-气体分子间的间隔变大','浓氨水需要密封保存-分子在不断地运动','干冰升华-分子间间隔变大','过氧化氢分解成水和氧气-分子破裂、原子重新结合成新的分子','','A','【解答】解：A、吹胀气球，是因为吹入气体后，气体增多，体积变大，不是因为分子间的间隔变大，故选项解释错误．<br />B、浓氨水需要密封保存，是为了防止因氨分子在不断的运动，运动到空气中造成浪费，故选项解释正确．<br />C、干冰（固体的二氧化碳）升华，固体变为气体，二氧化碳分子间的间隔增大，故选项金属正确．<br />D、过氧化氢分解成水和氧气，是因为过氧化氢分子分裂成了氢原子和氧原子，然后氢原子、氧原子分别重新组合形成水分子、氧分子，大量的水分子、氧分子分别聚集成水、氧气，该事实说明分子是可以再分的，故选项解释正确．<br />故选：A．','【分析】根据分子的基本特征：分子质量和体积都很小；分子之间有间隔；分子是在不断运动的；同种的分子性质相同，不同种的分子性质不同，可以简记为：“两小运间，同同不不”，结合事实进行分析判断即可．','选择题',3.00,'b139af7120a72ad7528e79c9f4d18354',9,400,'利用分子与原子的性质分析和解决问题','',2016,'32','2016•合肥校级模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840348,'为探究燃烧的条件，小刚同学查阅资料得知：白磷为白色蜡状固体，有剧毒，不溶于水，着火点为40℃，红磷着火点为240℃．然后他进行了如下实验：<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao76/1769381e-94d4-11e9-ac8f-b42e9921e93e_xkb97.png\" style=\"vertical-align:middle\" /><br />（1）按A装置进行实验，可观察到的现象为：<br />①铜片上红磷<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；②铜片上白磷<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；③热水中的白磷<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（2）B装置是用以制取氧气．B和C装置连接，C装置中水下的白磷<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）从以上实验分析得出，燃烧需要的条件为：可燃物、<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>和<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','不燃烧$###$燃烧$###$不燃烧$###$燃烧$###$与氧接触$###$温度达到着火点','【解答】解：（1）铜片上的白磷能够燃烧，是因为具备了物质燃烧的条件：物质具有可燃性、可燃物与氧气接触、温度达到着火点；铜片上的红磷与水中的白磷都不燃烧；<br />（2）由于水中的白磷满足了燃烧的条件，所以水中的白磷会燃烧；<br />（3）A实验现象①铜片上的红磷由于其温度没有达到其着火点，所以不燃烧；②铜片上的白磷满足了燃烧的条件，所以燃烧，可得出物质要燃烧其温度要达到着火点；，③水中的白磷由于没有与氧气接触，所以不燃烧，对照②③和B实验现象得出可得出物质要燃烧必须与氧气接触都有是化合反应，都是氧化反应，都放热&nbsp;有大量的白烟．<br />故答案为：（1）①不燃烧；②燃烧；③不燃烧；<br />（2）燃烧；<br />（3）与氧接触、温度达到着火点．','【分析】（1）（2）了解物质燃烧的三个条件：即物质具有可燃性、与空气接触、温度达到着火点，可以据此解答该题；<br />（3）利用对比对照的思路分析；','填空题',3.00,'1c71a64ab7b0e14a8fd9ff0e349564f9',9,400,'燃烧的条件与灭火原理探究','',2016,'32','2016•呼伦贝尔校级模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840350,'用学过的化学知识填空<br />（1）吸取和滴加少量的液体需要用<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（2）少量溶液相互反应时，需要用<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（3）蒸发溶液时，需要用<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','胶头滴管$###$试管$###$蒸发皿','【解答】解：（1）用于吸取和滴加少量液体的是胶头滴管；故填：胶头滴管；<br />（2）少量溶液相互反应时，需要用试管；故填：试管；<br />（3）蒸发溶液时，需要用蒸发皿．故填：蒸发皿．','【分析】熟悉化学实验中的常用仪器和它们的作用．','填空题',3.00,'5f2144d3415831e7d520ebb8c87a6fa7',9,400,'常用仪器的名称和选用','',2014,'37','2014春•尧都区校级月考',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840353,'某实验小组为探究酸的化学性质，由甲、乙同学设计并完成了如图所示的实验．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao45/178fd2f0-94d4-11e9-91aa-b42e9921e93e_xkb52.png\" style=\"vertical-align:middle\" /><br />（1）乙同学实验中所发生的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />【查阅资料】CaCl<SUB>2</SUB>、NaCl的溶液呈中性，Na<SUB>2</SUB>CO<SUB>3</SUB>溶液呈<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“酸”或“碱”）性．<br />【实验探究一】确定甲、乙同学上述实验后所得溶液中的溶质．<br />（2）取甲所得溶液少许，加入石蕊试液，溶液显红色，甲所得溶液中一定含有的溶质是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（写化学式）．<br />（3）取乙所得溶液少许，加入酚酞试液，溶液显红色，乙所得溶液中一定含有的溶质是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（写化学式）．<br />【实验探究二】<br />（4）甲、乙同学把上述实验所得溶液倒入同一废液缸中，观察到废液缸中产生少量气泡，最终有白色沉淀生成，该白色沉淀是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（写化学式）．<br />【讨论交流】丙同学提出：若将上述废缸中物质取出，过滤，所得滤液中含有哪些物质呢？经过讨论大家一致认为：一定含有NaCl，可能含有CaCl<SUB>2</SUB>、盐酸和Na<SUB>2</SUB>CO<SUB>3</SUB>中的部分物质．<br />【实验探究三】取少量滤液于试管中，滴加过量的稀硝酸，可观察到有大量气泡产生．<br />【归纳总结】该滤液中含有的溶质是氯化钠和<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（写化学式）．','','','','','','Na<SUB>2</SUB>CO<SUB>3</SUB>+2HCl═2NaCl+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑$###$碱$###$CaCl<SUB>2</SUB>、HCl$###$NaCl、Na<SUB>2</SUB>CO<SUB>3</SUB>$###$CaCO<SUB>3</SUB>$###$Na<SUB>2</SUB>CO<SUB>3</SUB>','【解答】解：（1）乙同学实验是碳酸钠与稀盐酸的反应，化学方程式为Na<SUB>2</SUB>CO<SUB>3</SUB>+2HCl═2NaCl+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑；Na<SUB>2</SUB>CO<SUB>3</SUB>溶液显碱性；<br />（2）取甲所得溶液少许，加入石蕊试液，溶液显红色，说明溶液显酸性，所以甲所得溶液中一定含有的溶质CaCl<SUB>2</SUB>和HCl；<br />（3）取乙所得溶液少许，加入酚酞试液，溶液显红色，说明溶液显碱性，所以乙所得溶液中一定含有的溶质NaCl和Na<SUB>2</SUB>CO<SUB>3</SUB>；<br />（4）甲、乙同学把上述实验所得溶液倒入同一废液缸中，发生的化学反应是Na<SUB>2</SUB>CO<SUB>3</SUB>+2HCl=2NaCl+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑，Na<SUB>2</SUB>CO<SUB>3</SUB>+CaCl<SUB>2</SUB>=CaCO<SUB>3</SUB>↓+2NaCl，所以白色沉淀是CaCO<SUB>3</SUB>；取少量滤液于试管中，滴加过量的稀硝酸，可观察到有大量气泡产生，说明溶液中一定有碳酸钠．<br />故答案为：（1）Na<SUB>2</SUB>CO<SUB>3</SUB>+2HCl═2NaCl+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑；碱；<br />（2）CaCl<SUB>2</SUB>、HCl；<br />（3）NaCl、Na<SUB>2</SUB>CO<SUB>3</SUB>；<br />（4）CaCO<SUB>3</SUB>；&nbsp;&nbsp;Na<SUB>2</SUB>CO<SUB>3</SUB>．','【分析】（1）根据乙同学的实验和实验现象解答；<br />（2）根据甲同学的实验和实验现象以及石蕊试液的性质解答；<br />（3）根据乙同学的实验和实验现象以及酚酞试液的性质解答；<br />（4）根据实验和实验现象以及碳酸钠、氯化钙的性质进行分析．','书写',3.00,'3e04e2a326c4cab13915084605d50f3a',9,400,'探究酸碱的主要性质,酸的化学性质,书写化学方程式、文字表达式、电离方程式','',2016,'32','2016•潮南区模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840354,'“远离毒品，珍爱生命”！可卡因就是一种少量能使人昏睡，大量会使人死亡的毒品，其化学式为C<SUB>17</SUB>H<SUB>31</SUB>NO<SUB>4</SUB>．下列有关可卡因的叙述正确的是（　　）','可卡因中氮氧元素的个数比为1：4','每个可卡因分子中含有43个原子核','可卡因是由碳、氢、氧原子构成的有机化合物','可卡因分子中含有17个碳原子、21个氢原子、1个氮原子和2个氧分子','','B','【解答】解：A．元素是个宏观概念，只讲种类、不讲个数，故错误；<br />B．由可卡因的化学式C<SUB>17</SUB>H<SUB>21</SUB>NO<SUB>4</SUB>可知，1个可卡因分子中含有17个碳原子、21个氢原子、1个氮原子和4个氢原子，共43个原子，1个原子有1个原子核，所以每个可卡因分子中含有43个原子核，故正确；<br />C．可卡因是由元素组成的，分子构成的，而不是由原子直接构成的，故错误；<br />D．分子是由原子构成的，而不存在分子，故错误．<br />故选B．','【分析】A．根据元素的规定来分析；<br />B．根据物质的构成情况来分析；<br />C．根据物质的组成来分析；<br />D．根据分子构成来分析．','选择题',3.00,'421bd3811ac2f90d89a1f5de80d8dd02',9,400,'有机物与无机物的区别,化学式的书写及意义','',2016,'37','2016•道里区二模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840359,'河水净化的主要步骤如图所示．有关说法错误的是（　　）<br /><img  src=\"/tikuimages/9/2015/400/shoutiniao21/17acf7de-94d4-11e9-b922-b42e9921e93e_xkb40.png\" style=\"vertical-align:middle\" />','步骤Ⅰ可以除去难溶性杂质','X试剂可以是明矾','步骤Ⅲ可杀菌、消毒','净化后的水仍是混合物','','B','【解答】解：<br />A、过滤能除去不溶性固体杂质，故步骤Ⅰ可出去难溶性杂质，正确；<br />B、明矾可以吸附较大颗粒，加速沉降，活性炭具有吸附性，能除去水中的色素和异味，X试剂可以是活性炭，错误；<br />C、液氯能杀菌，故步骤Ⅲ可杀菌、消毒，正确；<br />D、纯净水是不含有杂质的水，经过这几步净化的水中含有可溶性杂质，不是纯净物，仍是混合物，正确；<br />故选B．','【分析】根据已有的知识进行分析，过滤能除去不溶性固体杂质；活性炭具有吸附性；液氯能杀菌；纯净水是不含有杂质的水，据此解答．','选择题',3.00,'073d4f91056bd73239ca24af4e52054b',9,400,'水的净化','',2015,'35','2015秋•周村区校级期中',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840361,'某兴趣小组在探究金属与酸的反应时发现了意外情况：铜能与稀硫酸反应--将一小块铜片放入试管中，加入适量稀硫酸，无现象；加热该试管，较长时间后铜片上有气泡，溶液逐渐变成了蓝色．<br />【讨论猜想】（1）小王提了两点想法，但遭到同学们的反驳，请写出同学们的反驳意见．<br />小王的想法一：铜片可能不纯，含有的活动性强的金属杂质与稀硫酸反应生成了H<SUB>2</SUB>．同学们的反驳意见是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />小王的想法二：在金属活动性顺序中Cu排在H后面，所以Cu一定未参加反应．同学们的反驳意见是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）小张也提出了两点想法，得到了同学们的拥护．<br />小张的想法一：加热后稀硫酸浓缩成浓硫酸，而浓硫酸具有稀硫酸所不具有的吸水性和脱水性等特性，也许能与Cu反应．<br />小张的想法二：生成的气体可能是SO<SUB>2</SUB>、O<SUB>2</SUB>、H<SUB>2</SUB>中的一种或几种．你认为他猜想的依据是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />【探究实验一】根据小张的想法，兴趣小组的同学设计了下图1、图2所示装置进行实验来探究浓硫酸能否与Cu反应．结果实验现象相同--加热后铜片上有气泡，溶液逐渐变成了蓝色，石蕊试液变红，说明浓硫酸能与Cu反应．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao37/17b61f9e-94d4-11e9-8719-b42e9921e93e_xkb44.png\" style=\"vertical-align:middle\" /><br />请回答：<br />（1）石蕊试液变红说明生成的气体中有<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（2）在图1装置中，“蘸有浓碱液的棉花”的作用是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）比较两实验装置，图2装置的优点是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />【探究实验二】兴趣小组的同学为判断生成的气体中是否含有O<SUB>2</SUB>&nbsp;和H<SUB>2</SUB>，通过查阅资料得知：O<SUB>2</SUB>+4KI+4HCl═2I<SUB>2</SUB>+4KCl+2H<SUB>2</SUB>O，I<SUB>2</SUB>为碘单质．于是他们设计了如下的实验装置：<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao23/17b8909e-94d4-11e9-86e0-b42e9921e93e_xkb17.png\" style=\"vertical-align:middle\" /><br />请回答：A中观察到<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；B中观察到<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，证明无O<SUB>2</SUB>；C中观察到<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，证明无H<SUB>2</SUB>．<br />【实验结论】上述探究说明，所谓铜能与稀硫酸反应实际上是加热时水分蒸发、稀硫酸逐渐变成浓硫酸才与铜发生了反应．请根据以上信息写出浓硫酸与Cu反应的化学反应方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','如果含有活动性强的金属，加热前就应该冒气泡$###$溶液变成蓝色，说明铜参加了反应$###$化学反应前后，元素的种类保持不变$###$二氧化硫$###$吸收多余的二氧化硫，防止污染空气$###$能控制反应的发生与停止、能防止倒吸$###$溶液褪色$###$无明显现象$###$无明显现象$###$Cu+2H<SUB>2</SUB>SO<SUB>4</SUB>=CuSO<SUB>4</SUB>+SO<SUB>2</SUB>+2H<SUB>2</SUB>O','【解答】解：【讨论猜想】（1）小王的想法一：实验中观察到放出稀硫酸的铜片无现象，说明铜片中不含活泼金属；<br />小王的想法二：实验中观察到溶液呈蓝色，说明有硫酸铜生成，因此铜一定参加了反应；<br />（2）小张的想法二：报猜想的气体的组成元素都是反应前物质中所含有的元素，因此可推测其猜想依据为：化学变化前后元素不变；<br />【探究实验一】（1）石蕊遇酸变红，观察到石蕊变红，说明气体遇水可形成酸性溶液，因此，判断气体中含有二氧化硫；<br />（2）利用二氧化硫能与碱溶液发生反应，放置于管口蘸有浓碱液的棉花吸收二氧化硫，防止造成空气污染；<br />【探究实验二】A、二氧化硫能使紫红色高锰酸钾溶液褪色，因此可观察到紫红色溶液逐渐褪色成无色；<br />B、不含有氧气，混合溶液就不能产生碘，因此淀粉溶液也就不会出现颜色变化；<br />C、不含氢气，灼热的氧化铜不能被还原，观察不到红色出现；<br />【实验反思】铜能与浓硫酸反应而与稀硫酸不能反应，随着反应的不断进行，溶液中硫酸的浓度逐渐减小，因此，停止反应后溶液中还会含有少量硫酸；<br />【实验结论】上述探究说明，所谓铜能与稀硫酸反应实际上是加热时水分蒸发、稀硫酸逐渐变成浓硫酸才与铜发生了反应．根据以上信息，则可以写出浓硫酸与Cu反应的化学反应方程式为Cu+2H<SUB>2</SUB>SO<SUB>4</SUB>=CuSO<SUB>4</SUB>+SO<SUB>2</SUB>+2H<SUB>2</SUB>O．<br />故答案为：<br />【讨论猜想】（1）如果含有活动性强的金属，加热前就应该冒气泡；<br />溶液变成蓝色，说明铜参加了反应；<br />（2）化学反应前后，元素的种类保持不变；<br />【探究实验一】（1）二氧化硫；&nbsp;（2）吸收多余的二氧化硫，防止污染空气；<br />（3）能控制反应的发生与停止、能防止倒吸；<br />【探究实验二】溶液褪色；&nbsp;无明显现象；&nbsp;无明显现象；<br />【实验结论】Cu+2H<SUB>2</SUB>SO<SUB>4</SUB>=CuSO<SUB>4</SUB>+SO<SUB>2</SUB>+2H<SUB>2</SUB>O．','【分析】【讨论猜想】（1）根据实验中所出现的现象，利用所掌握的基础知进行分析，评价小王的想法；<br />（2）小张的想法二：根据对其猜想中元素与反应前元素的对比，推断其猜想的理论依据；<br />【探究实验一】（1）结合猜想中的气体，对实验中现象进行分析，对气体组成做出判断；<br />（2）观察装置中蘸有浓碱液的棉花的放置位置，利用碱液的性质，对其作用进行推断；<br />【探究实验二】<br />A、根据二氧化硫能使高锰酸钾溶液褪色的性质，预计实验现象，以说明气体中含有二氧化硫；<br />B、根据所提供的资料，判断若有氧气时会出现的实验现象，不含氧气时此现象不会出现；<br />C、根据氢气还原氧化铜的实验现象，推测不含氢气时所出现的现象；<br />【实验反思】根据探究的结论，铜可与浓硫酸发生反应，分析剩余溶液中硫酸的浓度变化，做出判断．<br />【实验结论】根据题目中信息，正确书写浓硫酸与Cu反应的化学反应方程式．','书写',3.00,'40284d495afd61825ef22fbb9cf5c0a5',9,400,'实验探究物质的性质或变化规律,常见气体的检验与除杂方法,金属的化学性质,书写化学方程式、文字表达式、电离方程式','丹阳市',2016,'32','2016•丹阳市模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840370,'<img  src=\"/tikuimages/9/2016/400/shoutiniao82/17dd088f-94d4-11e9-b30a-b42e9921e93e_xkb61.png\" style=\"vertical-align:middle;FLOAT:right\" />取一滤纸条，等间距离滴入10滴紫色石蕊溶液，装入玻璃瓶中，然后从玻璃管的两端同时放入蘸有浓氨水和浓盐酸的棉球，并迅速封闭两端管口，很快可观察到管内有白烟产生，待白烟消失后还可观察到如图所示的现象．则下列说法中不正确的是（　　）','酸、碱能使酸碱指示剂显示不同颜色','图中黑点颜色为红色','氨水和盐酸都有属于易挥发性物质','该反应属于化合反应','','B','【解答】解：A、据图可以看出，氨水这边变蓝色，浓盐酸这边变红色，说明酸、碱能使酸碱指示剂显示不同颜色，故说法正确；<br />B、氨水呈碱性，能使石蕊试液变蓝，不是红色，故说法错误；<br />C、管内有白烟产生，是因为氨水挥发出的氨气与盐酸挥发出的氯化氢反应生成了氯化铵的缘故，说明氨水和盐酸都属于易挥发性物质，故说法正确；<br />D、该过程中氨气和氯化氢反应生成了氯化铵，发生的是化合反应，故说法正确；<br />故选：B．','【分析】根据已有的分子运动以及酸碱指示剂在酸碱溶液中的变色的知识进行分析解答即可．','选择题',3.00,'47ec4b7c767b6b7ecae2c3d8b08b4e0a',9,400,'酸碱指示剂及其性质,反应类型的判定','',2016,'32','2016•瑶海区模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840371,'<img  src=\"/tikuimages/9/2016/400/shoutiniao27/17e40d70-94d4-11e9-85aa-b42e9921e93e_xkb22.png\" style=\"vertical-align:middle;FLOAT:right\" />（2016春•乳山市期中）如图是小明同学利用测定白磷燃烧的实验来探究化学反应前后质量有无变化，请你回答下列问题：<br />（1）小明同学的实验方案是：（友情提示：取下瓶中的玻璃管在酒精灯上加热，迅速接触瓶中的白磷，使白磷燃烧）<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />（2）白磷燃烧时观察到的现象是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />（3）左盘锥形瓶口在白磷点燃反应并冷却后，为什么不能打开瓶塞称量？<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>由此实验，小明同学获得的结论是：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','①先调节天平平衡，将整套装置放在左盘上进行称量，记录数据为m<SUB>1</SUB>；<br />②将装置取下，取下瓶中的玻璃管在酒精灯上加热，迅速接触瓶中的白磷，使白磷燃烧，；<br />③待反应结束，温度冷却至室温后，再将整套装置放回左盘进行称量，记录数据为m<SUB>2</SUB>；<br />④比较为m<SUB>1</SUB>与m<SUB>2</SUB>的大小．$###$产生大量白烟$###$4P+5O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2P<SUB>2</SUB>O<SUB>5</SUB>$###$打开瓶塞后外界的空气会进入瓶中，导致质量增加$###$化学反应前后，物质的总重量不变','【解答】解：（1）小明同学的实验方案是：<br />①先调节天平平衡，将整套装置放在左盘上进行称量，记录数据为m<SUB>1</SUB>；<br />②将装置取下，取下瓶中的玻璃管在酒精灯上加热，迅速接触瓶中的白磷，使白磷燃烧，；<br />③待反应结束，温度冷却至室温后，再将整套装置放回左盘进行称量，记录数据为m<SUB>2</SUB>；<br />④比较为m<SUB>1</SUB>与m<SUB>2</SUB>&nbsp;的大小．<br />（2）白磷燃烧的现象是产生大量白烟；白磷燃烧的方程式为：4P+5O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2P<SUB>2</SUB>O<SUB>5</SUB>；<br />（3）左盘锥形瓶口在白磷点燃反应并冷却后，不能打开瓶塞称量，原因是：打开瓶塞后外界的空气会进入瓶中，导致质量增加；由此实验，小明同学获得的结论是：化学反应前后，物质的总重量不变．<br />答案：<br />（1）①先调节天平平衡，将整套装置放在左盘上进行称量，记录数据为m<SUB>1</SUB>；<br />②将装置取下，取下瓶中的玻璃管在酒精灯上加热，迅速接触瓶中的白磷，使白磷燃烧，；<br />③待反应结束，温度冷却至室温后，再将整套装置放回左盘进行称量，记录数据为m<SUB>2</SUB>；<br />④比较为m<SUB>1</SUB>与m<SUB>2</SUB>&nbsp;的大小．<br />（2）产生大量白烟；&nbsp;4P+5O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2P<SUB>2</SUB>O<SUB>5</SUB>；防止炸裂锥形瓶．<br />（3）打开瓶塞后外界的空气会进入瓶中，导致质量增加；化学反应前后，物质的总重量不变．','【分析】（1）根据实验目的涉设计实验方案；<br />（2）点燃条件下，白磷能够在空气中剧烈地燃烧，产生大量白烟，生成白色固体；<br />（3）化学反应遵循质量守恒定律，即参加反应的物质的质量之和，等于反应后生成的物质的质量之和；','书写',3.00,'59b3c7a4d8855e90991db739e6cd1aa8',9,400,'质量守恒定律的实验探究,书写化学方程式、文字表达式、电离方程式','乳山市',2016,'35','2016春•乳山市期中',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840374,'除去下列物质中含有的杂质所选用试剂或操作方法不正确的一组是（　　） <table class=\"edittable\"><TBODY><TR><td width=50></TD><td width=124>物　　　质</TD><td width=112>所含杂质</TD><td width=198>除去杂质的试剂或方法</TD></TR><TR><td>A</TD><td>NaCl固体</TD><td>Na<SUB>2</SUB>CO<SUB>3</SUB></TD><td>过量稀盐酸，蒸干</TD></TR><TR><td>B</TD><td>CO</TD><td>CO<SUB>2</SUB></TD><td>氢氧化钠浓溶液，干燥</TD></TR><TR><td>C</TD><td>KCl</TD><td>KClO<SUB>3</SUB></TD><td>MnO<SUB>2</SUB>，加热</TD></TR><TR><td>D</TD><td>FeSO<SUB>4</SUB>溶液</TD><td>CuSO<SUB>4</SUB></TD><td>足量铁粉，过滤</TD></TR></TBODY></TABLE>','A','B','C','D','','C','【解答】解：A、Na<SUB>2</SUB>CO<SUB>3</SUB>能与过量稀盐酸反应生成氯化钠、水和二氧化碳，再蒸干，能除去杂质且没有引入新的杂质，符合除杂原则，故选项所采取的方法正确．<br />B、CO<SUB>2</SUB>能与氢氧化钠溶液反应生成碳酸钠和水，CO不与氢氧化钠溶液反应，再干燥，能除去杂质且没有引入新的杂质，符合除杂原则，故选项所采取的方法正确．<br />C、KClO<SUB>3</SUB>在二氧化锰的催化作用下生成氯化钾和氧气，二氧化锰作催化剂，反应前后质量不变，能除去杂质但引入了新的杂质二氧化锰，不符合除杂原则，故选项所采取的方法错误．<br />D、足量铁粉能与CuSO<SUB>4</SUB>溶液反应生成硫酸亚铁溶液和铜，再过滤，能除去杂质且没有引入新的杂质，符合除杂原则，故选项所采取的方法正确．<br />故选：C．','【分析】根据原物质和杂质的性质选择适当的除杂剂和分离方法，所谓除杂（提纯），是指除去杂质，同时被提纯物质不得改变．除杂质题至少要满足两个条件：①加入的试剂只能与杂质反应，不能与原物质反应；②反应后不能引入新的杂质．','选择题',3.00,'4a392ebebb2602f12b4f084b95684ad0',9,400,'物质除杂或净化的探究,常见气体的检验与除杂方法,盐的化学性质','',2016,'32','2016•淮安模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840375,'<img  src=\"/tikuimages/9/0/400/shoutiniao16/17ee94c0-94d4-11e9-b668-b42e9921e93e_xkb31.png\" style=\"vertical-align:middle;FLOAT:right;\" />A、B、C、D是初中化学中四种常见的不同类别的物质，其中B是红色固体，固体C在空气中易潮解，A、B、C、D和H<SUB>2</SUB>SO<SUB>4</SUB>之间的相互关系如图所示（“→”表示物质间的转化关系，“-”表示物质间可以相互反应），请回答：<br />（1）物质C的化学式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）物质D的类别为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“酸”“碱”“盐”或“氧化物”）．<br />（3）反应①的化学方程式是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，反应②的基本反应类型属于<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（4）请总结出一条酸与其他物质反应的规律：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','NaOH$###$盐$###$Fe+H<SUB>2</SUB>SO<SUB>4</SUB>=FeSO<SUB>4</SUB>+H<SUB>2</SUB>↑$###$复分解反应$###$酸与碱反应产生盐和水','【解答】解：（1）C固体在空气中易潮解，所以C就是氢氧化钠，B是红色固体，和硫酸会发生反应，所以B就是氧化铁，B会转化成A，A和硫酸会发生反应，所以A就是铁，D会转化成氢氧化钠，氢氧化钠和D都会与硫酸反应，A～D是我们已经学过的四种不同类别的常见物质，所以D就是盐，经过验证，推导正确，所以C是NaOH；<br />（2）通过推导可知，D属于盐；<br />（3）反应①是铁和硫酸反应生成硫酸亚铁和氢气，化学方程式是：Fe+H<SUB>2</SUB>SO<SUB>4</SUB>=FeSO<SUB>4</SUB>+H<SUB>2</SUB>↑；<br />反应②是氢氧化钙和碳酸钠反应生成碳酸钙沉淀和氢氧化钠，所以基本反应类型属于复分解反应；<br />（4）依据复分解反应的原理进行总结，酸与其他物质反应的规律为：酸与碱反应产生盐和水．<br />故答案为：（1）NaOH；<br />（2）盐；<br />（3）Fe+H<SUB>2</SUB>SO<SUB>4</SUB>=FeSO<SUB>4</SUB>+H<SUB>2</SUB>↑；复分解反应；<br />（4）酸与碱反应产生盐和水．','【分析】根据C固体在空气中易潮解，所以C就是氢氧化钠，B是红色固体，和硫酸会发生反应，所以B就是氧化铁，B会转化成A，A和硫酸会发生反应，所以A就是铁，D会转化成氢氧化钠，氢氧化钠和D都会与硫酸反应，A～D是我们已经学过的四种不同类别的常见物质，所以D就是盐，然后将推出的物质进行验证即可．','书写',3.00,'d5e2aef2c959eb4e1c1c72a7b60a1649',9,400,'常见的氧化物、酸、碱和盐的判别,物质的鉴别、推断,反应类型的判定,书写化学方程式、文字表达式、电离方程式','',0,'37','',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840384,'如图1是实验室制取气体的常用装置．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao61/18127070-94d4-11e9-b164-b42e9921e93e_xkb97.png\" style=\"vertical-align:middle\" /><br />（1）写出一个用B装置或C装置制取气体的化学方程式．<br />（2）用B装置制取气体时，试管口为什么要稍向下倾斜？<br />（3）在实验室里，可用氯化钠固体和浓硫酸在加热条件下起反应制取氯化氢（HCl）气体，该气体极易溶于水，則制取氯化氢应选择的发生装置和收集装置分別是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填字母代号）<br />（4）如图2装置可用用来收集或干燥气体，现在若用该装置收集一瓶氯化氢气体应从玻璃管的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“a”或“b”）端通入．','','','','','','BD$###$a','【解答】解（1）B装置制取氧气为固体加热型反应，试管口无棉花团，为氯酸钾制取氧气，反应的方程式为：2KClO<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span></td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>△</td></tr></table></span>2KCl+3O<SUB>2</SUB>↑；<br />（2）用B装置制取气体时，加热时试管口要略向下倾斜，防止受热时药品中的水分蒸发出来，在试管口部凝聚成小水珠倒流引起试管破裂；<br />（3）在实验室里，可用氯化钠固体和浓硫酸在加热条件下起反应制取氯化氢（HCl）气体，需加热，发生装置选择B，该气体极易溶于水，密度比空气大，收集装置选择D；<br />（4）如图2装置可用用来收集或干燥气体，现在若用该装置收集一瓶氯化氢气体，因该气体的密度比空气大，故应长进短出．<br />故答案为：（1）B装置制取氧气为固体加热型反应，试管口无棉花团，为氯酸钾制取氧气，反应的方程式为：2KClO<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span></td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>△</td></tr></table></span>2KCl+3O<SUB>2</SUB>↑；<br />（2）为防止受热时药品中的水分蒸发出来，在试管口部凝聚成小水珠倒流引起试管破裂；<br />（3）BD；（4）a．','【分析】（1）常用气体的发生装置和收集装置分别又分为两种和三种．前者有“固体加热型”和“固液常温型”两种装置；后者有“排水法”、“向下排空气法”和“向上排空气法”三种装置；<br />（2）据固体加热反应的实验注意事项解答；<br />（3）发生装置的选择要根据反应物的状态和反应条件，收集装置要根据气体的密度和溶水性；<br />（4）据氯化氢气体的密度比空气大解答．','书写',3.00,'84bdaafa0ad2bdeef03bb39729389e14',9,400,'常用气体的发生装置和收集装置与选取方法,气体的干燥（除水）,书写化学方程式、文字表达式、电离方程式','',2016,'37','2016•宛城区一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840386,'工业上用氢氧化钠溶液吸收工厂排出的二氧化硫气体．请计算：<br />（1）要配制100Kg20%的氢氧化钠溶液，需要<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>Kg氢氧化钠固体；<br />（2）若用60Kg30%的氢氧化钠溶液加<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>Kg水稀释，也可配制20%的氢氧化钠溶液；<br />（3）用含氢氧化钠8Kg的氢氧化钠溶液吸收二氧化硫气体，理论上可处理二氧化硫气体的质量是多少？（有关反应的化学方程式为：2NaOH+SO<SUB>2</SUB>═Na<SUB>2</SUB>SO<SUB>3</SUB>+H<SUB>2</SUB>O）','','','','','','20$###$30','【解答】解：（1）配制100Kg20%的氢氧化钠溶液，需要氢氧化钠固体的质量为：100kg×20%=20kg；故填：20；<br />（2）设需加水的质量为x<br />60kg×30%=（60+x）kg×20%，解得：x=30kg；故填：30；<br />（3）设可吸收SO<SUB>2</SUB>的质量为x<br />2NaOH+SO<SUB>2</SUB>═Na<SUB>2</SUB>SO<SUB>3</SUB>+H<SUB>2</SUB>O<br />80&nbsp;&nbsp;&nbsp;&nbsp;64<br />8kg&nbsp;&nbsp;&nbsp; x<br /><span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">80</td></tr><tr><td>64</td></tr></table>=<table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">8kg</td></tr><tr><td>x</td></tr></table></span><br />x=6.4kg<br />答：理论上可处理二氧化硫气体的质量是6.4kg．','【分析】（1）根据溶质质量等于溶液质量和溶质质量分数的乘积进行分析；<br />（2）根据稀释前后溶质的质量不变分析解答；<br />（3）根据反应的化学方程式，结合氢氧化钠的质量计算出可吸收的二氧化硫的质量．','填空题',3.00,'db43b99d9b8ea9731694a16ce4515235',9,400,'用水稀释改变浓度的方法,有关溶质质量分数的简单计算,根据化学反应方程式的计算','',2016,'37','2016•惠安县一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840387,'图小雨同学画的氢气和氯气在点燃条件下发生变化的微观模拟过程．仔细观察图示．回答问题．<br /><img  src=\"/tikuimages/9/0/400/shoutiniao12/18175270-94d4-11e9-94e7-b42e9921e93e_xkb58.png\" style=\"vertical-align:middle\" /><br />（1）请将该微观图补充完整；<br />（2）此化学变化的微观实质是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（3）此变化中发生变化和没有发生变化的粒子个数比为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','每一个氢分子分解成两个氢原子，每一个氯分子分解成两个氯原子，每一个氢原子和一个氯原子结合成了一个氯化氢分子$###$1：2','【解答】解：（1）化学反应前后元素的种类和原子的个数不变，故乙中应该补充2个氢原子，丙中应该补充1个氯化氢分子，填图如下：<br /><img  src=\"/tikuimages/9/0/400/shoutiniao39/181b4a0f-94d4-11e9-9b0f-b42e9921e93e_xkb22.png\" style=\"vertical-align:middle\" /><br />（2）由微粒的变化可知，每一个氢分子分解成两个氢原子，每一个氯分子分解成两个氯原子，每一个氢原子和一个氯原子结合成了一个氯化氢分子．<br />（3）由微粒的变化可知，此变化中发生变化微粒两个分子，没有发生变化是两个氢原子和两个氯原子，发生变化和没有发生变化的粒子个数比为1：2．<br />故答为：（1）见上图；（2）每一个氢分子分解成两个氢原子，每一个氯分子分解成两个氯原子，每一个氢原子和一个氯原子结合成了一个氯化氢分子；（3）1：2．','【分析】根据图示结合已有的知识进行分析解答．根据质量守恒定律化学反应前后原子的种类和个数不变，将微观图补充完整；根据微粒的变化分析化学变化的微观实质和此变化中发生变化和没有发生变化的粒子个数比．','填空题',3.00,'f89f7b00a318d7325c4422e54e5a0825',9,400,'微粒观点及模型图的应用,化学反应的实质','',0,'37','',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840389,'将10g碳酸钙高温煅烧一段时间后，冷却测得剩余固体质量为8.9g，再向剩余固体中加入足量稀盐酸，充分反应后，蒸发结晶，理论上可得到固体氯化钙的质量为（　　）','11.1g','8.9g','8.4g','无法计算','','A','【解答】解：碳酸钙高温煅烧生成氧化钙和二氧化碳，碳酸钙与稀盐酸反应生成氯化钙、水和二氧化碳，氧化钙与稀盐酸反应生成氯化钙和水，由质量守恒定律，生成的氯化钙的钙元素全部来自于10g碳酸钙，则理论上可得到固体氯化钙的质量为X，则<br />CaCO<SUB>3</SUB>-Ca---CaCl<SUB>2</SUB><br />100&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;111<br />10g&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; X<br /><span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">100</td></tr><tr><td>111</td></tr></table>=<table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">10g</td></tr><tr><td>X</td></tr></table></span>&nbsp;&nbsp; X=11.1g<br />故选：A','【分析】根据题意，碳酸钙高温煅烧生成氧化钙和二氧化碳，碳酸钙与稀盐酸反应生成氯化钙、水和二氧化碳，氧化钙与稀盐酸反应生成氯化钙和水；由质量守恒定律，生成的氯化钙的钙元素全部来自于10g碳酸钙，据此结合元素守恒进行分析解答．','选择题',3.00,'e6f9f5b5e831fb83819765752c0379cf',9,400,'化合物中某元素的质量计算,质量守恒定律及其应用','东台市',2016,'37','2016•东台市一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840391,'同学们一起探究铝、铁、铜三种金属的活动性，小刚同学设计了用铜丝、铁丝、铝丝和稀盐酸，只用一只试管，取一次盐酸的探究方案．请你和他们一起完善下表的探究方案并回答有关问题．<br />（1）填表<br /><table class=\"edittable\"><tbody><tr><td width=\"309\">实验步骤</td><td width=\"193\">观察到的现象</td></tr><tr><td>①在试管中取少量盐酸，插入铁丝，充分作用．</td><td><u>&nbsp;&nbsp;&nbsp;&nbsp;</u></td></tr><tr><td>②在①所得的溶液中，插入<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，充分作用．</td><td>无明显现象</td></tr><tr><td>③在②所得的溶液中，插入<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，充分作用．</td><td><u>&nbsp;&nbsp;&nbsp;&nbsp;</u></td></tr></tbody></table>结论：金属活动性Al＞Fe＞Cu<br />（2）将铝丝插入前应进行的操作是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（3）小华同学认为在小刚设计的方案中，只要补充一个实验，就可得出Al＞Fe＞H＞Cu的结论．小华要补充的实验是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（4）小强同学认为要得到Al＞Fe＞H＞Cu的结论，不必做补充实验，中需将小明同学方案中插入金属的顺序调整即可，你认为调整后插入金属的顺序是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','有气泡冒出，溶液由无色变为浅绿色$###$铜丝$###$铝丝$###$铝丝表面有一层黑色物质，溶液由浅绿色变无色$###$用砂纸将铝表面的氧化物打磨掉$###$将铜丝插入稀盐酸中$###$铜、铁、铝','【解答】解：（1）铁丝插入盐酸中能观察到有气泡冒出，说明铁排在氢前面，得到氯化亚铁溶液，在氯化亚铁溶液中插入铜丝无明显现象，说明铜的活动性比铁弱．而在氯化亚铁溶液中插入铝丝，铝的活动性比铁强，能置换氯化亚铁中铁，表面有黑色物质析出，溶液由浅绿色变无色．同时得出金属活动性顺序为：Al＞Fe＞Cu（Al＞Fe＞H）．<br />（2）铝的表面易生成致密的氧化铝保护膜，影响反应的进行和反应速度，使用前应把保护膜除去．<br />（3）要得出Al＞Fe＞H＞Cu的结论，上述实验没有比较Cu与H的关系，所以补充铜丝与盐酸的反应就可以了．<br />（4）证明这三种金属的活动性方法较多，应当选用简便易行，现象明显的方案．刚开始加进去的金属应该是不能和盐酸发生反应的，这样实验现象比较明显，因此先插入铜丝，无现象，证明活泼性H＞Cu，再插入铁丝，有气泡，证明Fe＞H，再插入铝丝，铝丝表面有一层黑色物质，溶液由浅绿色变无色，证明Al＞Fe，则可以证明Al＞Fe＞H＞Cu的结论．<br />故答案为：<br />（1）<br /><table class=\"edittable\"><tbody><tr><td width=\"309\">实验步骤</td><td width=\"193\">观察到的现象</td></tr><tr><td>①在试管中取少量盐酸，插入铁丝，充分作用．</td><td>有气泡冒出，溶液由无色变为浅绿色</td></tr><tr><td>②在①所得的溶液中，插入 铜丝，充分作用．</td><td>无明显现象</td></tr><tr><td>③在②所得的溶液中，插入 铝丝，充分作用．</td><td>铝丝表面有一层黑色物质，溶液由浅绿色变无色</td></tr></tbody></table>（2）用砂纸将铝表面的氧化物打磨掉．<br />（3）将铜丝插入稀盐酸中；<br />（4）铜、铁、铝．','【分析】根据金属活动性顺序表可知三种金属的活动性铝＞铁＞氢＞铜，其中铁和盐酸反应得到浅绿色硫酸亚铁和氢气，铜不能硫酸亚铁反应，铝能置换出硫酸亚铁中的铁并生成无色的氯化铝溶液．小刚同学设计的实验能得出铝＞铁＞铜以及铝＞铁＞氢的活动性，不能证明铜和氢的活动性，要想证明，只要把铜插入盐酸中即可．证明这三种金属的活动性方法较多，应当选用简便易行，现象明显的方案．','书写',3.00,'0718894be0adb72e9575f146f959cbd0',9,400,'金属活动性的探究,书写化学方程式、文字表达式、电离方程式','',0,'37','',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840393,'<img  src=\"/tikuimages/9/0/400/shoutiniao79/182e35d1-94d4-11e9-94c7-b42e9921e93e_xkb28.png\" style=\"vertical-align:middle;FLOAT:right;\" />水是人及一切生物生存所必需的，为了人类和社会经济的可持续发展，我们应该了解一些有关水的知识．请你回答下列问题：<br />（1）如图是电解水的简易装置，则：<br />①电解水时应通<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>电（填“直流”或“交流”）．<br />②a试管中收集到的气体是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />b试管的气体检验方法是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />电解水的文字表达式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />③电解水时发生变化的物质是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，发生变化的粒子是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，不变的粒子是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />④由以上的实验事实可得出结论：水是由<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>组成的；分子与原子的区别是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />⑤在某次电解水的实验中加入了少量的NaOH溶液，分别测得a、b中产生的气体的实验数据如下：<br /><table class=\"edittable\"><TBODY><TR><td width=158>时间/分</TD><td width=38>1</TD><td width=38>2</TD><td width=38>3</TD><td width=38>4</TD><td width=38>5</TD><td width=38>6</TD><td width=38>7</TD><td width=38>8</TD><td width=38>9</TD><td width=38>10</TD></TR><TR><td>a中气体体积（cm<SUP>3</SUP>）</TD><td>6</TD><td>12</TD><td>20</TD><td>29</TD><td>39</TD><td>49</TD><td>55</TD><td>65</TD><td>75</TD><td>85</TD></TR><TR><td>b中气体体积（cm<SUP>3</SUP>）</TD><td>2</TD><td>4</TD><td>7</TD><td>11</TD><td>16</TD><td>21</TD><td>26</TD><td>31</TD><td>36</TD><td>41</TD></TR></TBODY></TABLE>仔细分析以上实验数据，1～5分钟内a、b中生成的气体体积之比大于2：l，可能的原因是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）天然水中含有许多杂质，可利用吸附、沉淀、过滤和蒸馏等方法净化，其中净化程度最高的方法是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．该方法发生的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>变化（填“化学”或“物理”）．<br />（3）实验室常用过滤的方法除去水中悬浮的杂质，过滤需要用到的玻璃仪器有④玻璃棒，还需要<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填序号）．<br />①带铁圈的铁架台&nbsp;&nbsp;②酒精灯&nbsp;&nbsp;③漏斗&nbsp;&nbsp;④玻璃棒&nbsp;&nbsp;⑤量筒&nbsp;&nbsp;⑥烧杯&nbsp;&nbsp;⑦滤纸<br />其中玻璃棒的作用为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．在过滤时，若过滤后的滤液仍然浑浊，其原因可能是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（写出一点即可）<br />（4）上海世博园内有许多饮水台，可取水直接饮用．其中的饮用水处理步骤如图所示；<br /><img  src=\"/tikuimages/9/0/400/shoutiniao62/1830f4f0-94d4-11e9-93c7-b42e9921e93e_xkb60.png\" style=\"vertical-align:middle\" /><br />步骤①的作用是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，步骤③的作用是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（5）地球上的总储水量虽然很大，但淡水很少，爱护水资源是每个公民的责任和义务．下列行为属于节约用水的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填序号）．<br />A．公共场所随手关水龙头<br />B．洗完菜的水用来浇花<br />C．不间断放水刷牙．','','','','','','直流$###$氢气$###$将带火星的木条伸入试管中，带火星的木条复燃，证明产生的气体是氧气$###$水<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td style=\"font-size: 90%\">通电</td></tr><tr><td><div hassize=\"9\"><div style=\"background: url(\'/upimages/images/part/8594L.png\') repeat-x; width:1px; height:5px;float:left;overflow:hidden\" muststretch=\"h\"></div><div style=\"background: url(\'/upimages/images/part/8594.png\') no-repeat; width:9px; height:5px;float:left;overflow:hidden\"></div></div></td></tr></table></span>氢气+氧气$###$水$###$水分子$###$氢原子和氧原子$###$氢元素和氧元素$###$在化学反应中能否再分$###$氧气的溶解性大于氢气的溶解性$###$蒸馏$###$物理$###$③⑥$###$引流$###$液面高于滤纸边缘$###$吸附$###$杀菌消毒$###$AB','【解答】解：（1）<br />①电解水时应通直流电；<br />②由电解水实验装置图可知，电解生成的气体气体a较多为氢气；<br />由电解水实验装置图可知，电解生成的气体B较少是氧气，氧气有助燃性，所以，实验室检验气体B的方法是将带火星的木条伸入试管中，带火星的木条复燃，证明产生的气体是氧气；电解水生成了氢气和氧气，该反应的文字表达式：水<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td style=\"font-size: 90%\">通电</td></tr><tr><td><div hassize=\"9\"><div style=\"background: url(\'/upimages/images/part/8594L.png\') repeat-x; width:1px; height:5px;float:left;overflow:hidden\" muststretch=\"h\"></div><div style=\"background: url(\'/upimages/images/part/8594.png\') no-repeat; width:9px; height:5px;float:left;overflow:hidden\"></div></div></td></tr></table></span>氢气+氧气；<br />③电解水时发生变化的物质是水，发生变化的粒子是水分子，不变的粒子是氢原子和氧原子；<br />④由以上的实验事实可得出结论：水是由氢元素和氧元素组成的；分子与原子的区别是在化学反应中能否再分；<br />⑤由于实验数据，1～6分钟内阴、阳两极生成的气体体积之比大于2：l，可能的原因是氧气的溶解性大于氢气的溶解性．<br />（2）天然水里含有许多杂质，可利用吸附、沉淀、过滤和蒸馏等方法进行净化，其中净化程度最高的是蒸馏．该方法发生的是物理变化；<br />（3）实验室常用过滤的方法除去水中悬浮的杂质，过滤需要用到的玻璃仪器有④玻璃棒，还需要漏斗和烧杯；<br />实验过程中玻璃棒的作用是引流．<br />过滤后滤液仍浑浊，可能原因是滤纸破损（会使得液体中的不溶物进入下面的烧杯，从而使得滤液浑浊）、液面高于滤纸边缘（会使部分液体未经过滤纸的过滤直接流下，该操作会使滤液仍然浑浊）或盛接滤液的烧杯不干净等．<br />（4）在步骤①中，由于炭罐中有活性炭，活性炭有吸附性可以吸附色素和异味；在步骤③中用紫外灯管照射，紫外线能杀菌消毒；<br />（5）<br />A、公共场所不用随手关水龙头，这是浪费水的做法，不可取，<br />B、洗完菜的水用来浇花，这种做法能提高水的利用率，可节约用水，<br />C、不间断放水刷牙，这会浪费水；<br />故答案为：<br />（1）<br />①直流；<br />②氢气；将带火星的木条伸入试管中，带火星的木条复燃，证明产生的气体是氧气；水<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td style=\"font-size: 90%\">通电</td></tr><tr><td><div hassize=\"9\"><div style=\"background: url(\'/upimages/images/part/8594L.png\') repeat-x; width:1px; height:5px;float:left;overflow:hidden\" muststretch=\"h\"></div><div style=\"background: url(\'/upimages/images/part/8594.png\') no-repeat; width:9px; height:5px;float:left;overflow:hidden\"></div></div></td></tr></table></span>氢气+氧气，分解反应；<br />③水；水分子；氢原子和氧原子；<br />④氢元素和氧元素；在化学反应中能否再分；<br />⑤氧气的溶解性大于氢气的溶解性；<br />（2）蒸馏；物理；<br />（3）③⑥；引流；液面高于滤纸边缘；<br />（4）吸附；杀菌消毒；<br />（5）AB．','【分析】（1）根据电解水时，生成的气体较多的是氢气；根据氧气由助燃性分析；根据电解水的反应写出反应的文字表达式，根据氧气和氢气的溶解性分析．<br />（2）根据蒸馏可以得到纯水解答；<br />（3）根据过滤需要用到的仪器及注意事项分析解答；<br />（4）可以根据物质的性质和图中提供的信息方面进行分析、判断：活性炭有吸附性，可以吸附色素和异味；紫外线能杀菌消毒；<br />（5）根据节约用水的方法解答．','书写',3.00,'4d6da7781ef310f12b68765acb07be20',9,400,'过滤的原理、方法及其应用,常见气体的检验与除杂方法,电解水实验,水的净化,书写化学方程式、文字表达式、电离方程式,保护水资源和节约用水','',0,'37','',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840396,'图1是常见固体物质的溶解度曲线，根据图示回答：<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao28/18398070-94d4-11e9-9bb3-b42e9921e93e_xkb3.png\" style=\"vertical-align:middle\" /><br />（1）t<SUB>1</SUB>℃时，60g水中最多溶解A物质<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>g．<br />（2）将t<SUB>1</SUB>℃一定量A的饱和溶液升温至t<SUB>3</SUB>℃时，可用图的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>点表示．<br />（3）若将t<SUB>3</SUB>℃等质量的三种物质的饱和溶液分别降温到t<SUB>1</SUB>℃，则所得的溶液[的溶质质量分数由大到小的顺序是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．（析出晶体都不含结晶水）<br />（4）气体的溶解度也有一定的变化规律．<br />①图2中a、b代表不同压强下气体的溶解度曲线．<br />若曲线a对应的气压为2个大气压，则曲线b对应的气压<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（选填编号，下同）2个大气压．<br />A．小于&nbsp;&nbsp;&nbsp;&nbsp;B．大于&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;C．等于&nbsp;&nbsp;D．无法确定<br />②不同温度下，氧气的溶解度随压强变化如图3所示．若t<SUB>1</SUB>对应的温度为40℃，则t<SUB>2</SUB>对应的温度<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />A．小于40℃B．等于40℃<br />C．大于40℃D．无法确定．','','','','','','24$###$b$###$B＞C＞A$###$B$###$C','【解答】解：（1）通过分析溶解度曲线可知，t<SUB>1</SUB>℃时，60g水中最多溶解A物质24g；<br />（2）将t<SUB>1</SUB>℃一定量A的饱和溶液升温至t<SUB>3</SUB>℃时，溶液中的固体质量不变，温度升高，所以可用图的b点表示；<br />（3）t<SUB>1</SUB>℃时，B物质的溶解度最大，A物质次之，C物质的溶解度降低温度不会析出晶体，A、B物质会析出晶体，所以将t<SUB>3</SUB>℃等质量的三种物质的饱和溶液分别降温到t<SUB>1</SUB>℃，则所得的溶液[的溶质质量分数由大到小的顺序是B＞C＞A；<br />（4）①图2中a、b代表不同压强下气体的溶解度曲线，温度相同时，压强增大，气体的溶解度增大，所以曲线a对应的气压为2个大气压，则曲线b对应的气压大于2个大气压，故选：B；<br />②不同温度下，氧气的溶解度随压强变化如图3所示，<br />若t<SUB>1</SUB>对应的温度为40℃，压强不变，温度升高，气体的溶解度减小，所以t<SUB>2</SUB>对应的温度大于40℃，故选：C．<br />故答案为：（1）24；<br />（2）b；<br />（3）B＞C＞A；<br />（4）①B；<br />②C．','【分析】根据固体的溶解度曲线可以：①查出某物质在一定温度下的溶解度，从而确定物质的溶解性，②比较不同物质在同一温度下的溶解度大小，从而判断饱和溶液中溶质的质量分数的大小，③判断物质的溶解度随温度变化的变化情况，从而判断通过降温结晶还是蒸发结晶的方法达到提纯物质的目的．','填空题',3.00,'0e7bae732af5cf163422c72d9188b06e',9,400,'固体溶解度曲线及其作用,气体溶解度的影响因素,溶质的质量分数、溶解性和溶解度的关系','',2016,'32','2016•广东模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840397,'下列实验装置或操作正确的是（　　）','<img  src=\"/tikuimages/9/2016/400/shoutiniao13/183b2e1e-94d4-11e9-9df2-b42e9921e93e_xkb88.png\" style=\"vertical-align:middle\" /><br />处理尾气','<img  src=\"/tikuimages/9/2016/400/shoutiniao87/183c3f91-94d4-11e9-885c-b42e9921e93e_xkb54.png\" style=\"vertical-align:middle\" /><br />盛放氧气','<img  src=\"/tikuimages/9/2016/400/shoutiniao52/1846038f-94d4-11e9-b8f7-b42e9921e93e_xkb55.png\" style=\"vertical-align:middle\" /><br />收集一定体积氧气','<img  src=\"/tikuimages/9/2016/400/shoutiniao43/1847b140-94d4-11e9-a8ab-b42e9921e93e_xkb39.png\" style=\"vertical-align:middle\" /><br />蒸发食盐水','','C','【解答】解：A、尾气处理的操作，根据一氧化碳的性质，一氧化碳难溶于水，所以用水不能吸收一氧化碳，故A操作错误；<br />B、根据氧气的性质，氧气的密度比空气大，所以收集满的集气瓶应正放在桌面上，故B操作错误；<br />C、根据氧气的性质，氧气不易溶于水，可以用排水法收集，利用洗气瓶进行收集时，因为水的密度比气体的密度大，所以短进长出；故C操作正确；<br />D、蒸发食盐要用玻璃棒不断搅拌，要外焰加热，但加热蒸发皿不需加石棉网，故D错误；<br />故选：C．','【分析】A、运用尾气处理的操作，一氧化碳的性质解答；<br />B、运用氧气的性质，氧气的密度解答；<br />C、运用氧气的收集方法解答；<br />D、运用蒸发操作要领解答．','选择题',3.00,'34bc275fa099dbe3ee19e3f971e0a46b',9,400,'量气装置,蒸发与蒸馏操作,氧气的物理性质,一氧化碳的物理性质','',2016,'32','2016•瑶海区模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840400,'下列有关能源和资源的说法错误的是（　　）','稀土是储量较少的一类金属的统称，有“工业的维生素”的美誉','煤分解可以制取焦炉煤气、粗氨水、焦炭和煤焦油','金属的回收利用，可以节约金属资源，目前，世界上有90%以上的铁得到回收利用','海水和陆地咸水约占全球总储水量的97.47%','','C','【解答】解：A、稀土有“工业的维生素”的美誉，是不可再生的重要战略资源；故正确；<br />B、煤的干馏是指在隔绝空气条件下加热、分解，生成焦炭（或半焦）、煤焦油、煤气等产物的过程，故不正确；<br />C、金属的回收再利用不仅可以节约资源以及因冶炼金属而消耗的能源，世界上有90%以上的铁得到回收利用；故正确；<br />D、海水和陆地咸水约占全球总储水量的97.47%，正确<br />故答案为：B．','【分析】A、根据稀土金属是不可再生的重要战略资源分析．<br />B、煤的干馏是将煤隔绝空气加强热使煤发生复杂的分解反应的过程；<br />C、金属的回收再利用不仅可以节约资源以及因冶炼金属而消耗的能源，还对环保有一定好处；<br />D、海水和陆地咸水约占全球总储水量的97.47%．','选择题',3.00,'dbdbd4e8cc55fc82f08d5fe0906625c6',9,400,'水资源状况,金属元素的存在及常见的金属矿物,金属的回收利用及其重要性,化石燃料及其综合利用','',2016,'37','2016•南岗区三模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840401,'除去下列物质中的少量杂质，所选用的试剂及操作方法均正确的是（　　）<br /><table class=\"edittable\"><TBODY><TR><td width=67>选项</TD><td width=171>物质（括号内为杂质）</TD><td width=104>试剂</TD><td width=95>操作方法</TD></TR><TR><td>A</TD><td>CO<SUB>2</SUB>气体（CO）</TD><td>过量的氧气</TD><td>点燃</TD></TR><TR><td>B</TD><td>Fe<SUB>2</SUB>O<SUB>3</SUB>（Fe）</TD><td>足量的盐酸</TD><td>过滤</TD></TR><TR><td>C</TD><td>FeSO<SUB>4</SUB>溶液（CuSO<SUB>4</SUB>）</TD><td>足量的锌</TD><td>过滤</TD></TR><TR><td>D</TD><td>O<SUB>2</SUB>（H<SUB>2</SUB>O）</TD><td>浓硫酸</TD><td>洗气</TD></TR></TBODY></TABLE>','A','B','C','D','','D','【解答】解：A、除去二氧化碳中的一氧化碳不能够通氧气点燃，这是因为除去气体中的气体杂质不能使用气体，否则会引入新的气体杂质，故选项所采取的方法错误．<br />B、Fe和Fe<SUB>2</SUB>O<SUB>3</SUB>均能与足量的盐酸反应，不但能把杂质除去，也会把原物质除去，不符合除杂原则，故选项所采取的方法错误．<br />C、FeSO<SUB>4</SUB>溶液和CuSO<SUB>4</SUB>均能与足量的锌反应，不但能把杂质除去，也会把原物质除去，不符合除杂原则，故选项所采取的方法错误．<br />D、浓硫酸具有吸水性，且不与氧气反应，能除去杂质且没有引入新的杂质，符合除杂原则，故选项所采取的方法正确．<br />故选：D．','【分析】根据原物质和杂质的性质选择适当的除杂剂和分离方法，所谓除杂（提纯），是指除去杂质，同时被提纯物质不得改变．除杂质题至少要满足两个条件：①加入的试剂只能与杂质反应，不能与原物质反应；②反应后不能引入新的杂质．','选择题',3.00,'aff9f77ab23f954ec8b9d5b34d8e8d7c',9,400,'物质除杂或净化的探究,常见气体的检验与除杂方法,金属的化学性质,盐的化学性质','',2016,'37','2016•大兴区一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840403,'2015年10月，我国科学家屠呦呦获得了诺贝尔生理学或医学奖，这是中国人的骄傲．她在治疗疟疾方面作出了卓越贡献．青蒿素化学式为C<SUB>15</SUB>H<SUB>22</SUB>O<SUB>5</SUB>，关于青蒿素的说法正确的是（　　）','属于氧化物','每个青蒿素分子共有42个原子','碳、氢、氧三种元素质量之比15：22：5','氢元素的质量分数最大','','B','【解答】解：A．氧化物是只含有两种元素且其中一种元素是氧元素的化合物，青蒿素中含有碳、氢、氧三种元素，不属于氧化物，故错误；<br />B.1个青蒿素分子是由15个碳原子、22个氢原子和5个氧原子构成的，则一个青蒿素分子含有42个原子．故正确；<br />C．青蒿素中碳、氢、氧三种元素的质量比为（12×15）：（1×22）：（16×5）=90：11：40，故错误；<br />D．青蒿素中碳、氢、氧三种元素的质量比为（12×15）：（1×22）：（16×5）=90：11：40，可见其中碳元素的质量分数最大，故错误．<br />故选B．','【分析】A．根据氧化物的概念来分析；<br />B．根据分子结构来分析；<br />C．根据化合物中元素的质量比来分析；<br />D．根据化合物中元素的质量比来分析．','选择题',3.00,'2cc3069e20c0d81ae13bec6e78404220',9,400,'从组成上识别氧化物,化学式的书写及意义,元素质量比的计算,元素的质量分数计算','',2016,'37','2016•芜湖二模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840405,'如图所示是自然界中常见的两个变化过程．请据图并结合你学过的知识，回答下列问题：<img  src=\"/tikuimages/9/2016/400/shoutiniao90/1865c08f-94d4-11e9-8876-b42e9921e93e_xkb40.png\" style=\"vertical-align:middle\" /><br />（1）从物质组成的角度看，葡萄糖和纤维素均属于<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“无机物”或“有机物”）．<br />（2）图中方框所示变化过程为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，从物质转化的角度看，该变化过程属于<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“物理”或“化学”）变化．<br />（3）从能量转化的角度看，上述过程中能量转化的方式是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />（4）在图中两个过程中实现了自然界中的重要循环，参与循环的元素有<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','有机物$###$光合作用$###$化学$###$光能转化为化学能$###$C、H、O','【解答】解：<br />（1）葡萄糖和纤维素是含碳元素的化合物，属于有机物；故填：有机物；<br />（2）光合作用是二氧化碳和水在光照的作用下，在叶绿体中合成葡萄糖，并释放出氧气；图中方框所示变化过程为光合作用，光合作用及葡萄糖转化为纤维素都是有新物质生成，属于化学变化；<br />（3）植物进行光合作用时，消耗光能，产生化学能，故是将光能转化为化学能；<br />（4）在图中光合作用与燃烧反应过程中参与循环的元素有碳、氢和氧元素，故填：C、H、O．<br />答案：<br />（1）有机物；<br />（2）光合作用；化学；<br />（3）光能转化为化学能；<br />（4）C、H、O．','【分析】（1）含碳元素的化合物叫有机化合物，一氧化碳、二氧化碳和碳酸盐除外；<br />（2）根据物质变化的过程来分析判断并书写化学方程式；<br />（3）断出每个过程中消耗了哪种形式的能，进而产生了哪种形式的能是解决该题的关键；<br />（4）根据图示进行分析解答．','填空题',3.00,'23ef92149089cf526bcdb8bf57eb9e01',9,400,'自然界中的氧循环,自然界中的碳循环,有机物与无机物的区别,化学变化和物理变化的判别,物质发生化学变化时的能量变化','',2016,'32','2016•泰安模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840411,'<img  src=\"/tikuimages/9/2016/400/shoutiniao93/187e78ae-94d4-11e9-a284-b42e9921e93e_xkb40.png\" style=\"vertical-align:middle;FLOAT:right\" />（2016•东城区一模）晚上，静静和妈妈一起做晚饭．<br />（1）静静发现煮饭所用的大米为“富硒大米”，查阅资料后发现硒元素具有抗衰老、抑制癌细胞的功能．硒元素在元素周期表中有如图所示信息．下列有关说法不正确的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填字母序号）．<br />A．硒属于金属元素<br />B．硒的原子核内有34个质子<br />C．硒的相对原子质量为78.96<br />D．富硒食品可以预防癌症<br />（2）炒菜时静静看到铁锅底部有炭黑．静静家所用的燃料是天然气，天然气的主要成分是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；产生炭黑的原因是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','A$###$甲烷$###$燃料燃烧不充分','【解答】解：（1）A．根据元素周期表中的一格可知，中间的汉字表示元素名称，该元素的名称是硒，属于固态非金属元素，故选项说法错误．<br />B．根据元素周期表中的一格可知，左上角的数字为34，表示原子序数为34；根据原子序数=核电荷数=质子数，则该元素的原子核内有34个质子，故选项说法正确．<br />C．根据元素周期表中的一格可知，汉字下面的数字表示相对原子质量，元素的相对原子质量为78.96，故选项说法正确．<br />D．硒有防癌、抗癌作用，缺硒可能引起表皮角质化和癌症，故选项说法正确．<br />故填：A；<br />（2）天然气的主要成分是甲烷，甲烷不完全燃烧就会产生炭黑，故填：甲烷；燃料燃烧不充分．','【分析】（1）根据图中元素周期表可以获得的信息：左上角的数字表示原子序数；字母表示该元素的元素符号；中间的汉字表示元素名称；汉字下面的数字表示相对原子质量；硒有防癌、抗癌作用，据此进行分析判断即可．<br />（2）根据天然气的主要成分以及不完全燃烧来分析．','填空题',3.00,'d54963d0b1a2119942889c56e3d26e63',9,400,'元素周期表的特点及其应用,完全燃烧与不完全燃烧,常用燃料的使用与其对环境的影响','',2016,'37','2016•东城区一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840412,'利用化学知识判断下列事故处理恰当的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />A．发现煤气泄漏应立即打开排风扇以利通风&nbsp; <br />B．炒菜油锅着火立即泼冷水以降低温度<br />C．碱液沾到皮肤上应立即用大量水冲洗，再涂上适量的硼酸溶液<br />D．<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','C$###$菜时油锅着火，立即盖上锅盖','【解答】解：A、可燃性的气体与空气或氧气的混合气体遇明火、静电、电火花或加热可能发生爆炸，煤气是可燃性气体，煤气泄露打开排风扇会产生电火花，故A错误；<br />B、油的密度小于水的密度，所以炒菜油锅着火立即泼冷水会使得油和空气接触更充分，燃烧更旺，故B错误；<br />C、碱液沾到皮肤上应立即用大量水冲洗，再涂上适量的硼酸溶液，故C正确；<br />D、炒菜时油锅着火，立即盖上锅盖，盖上锅盖可以隔绝空气或氧气，以达到灭火的目的．<br />故答案为：C；菜时油锅着火，立即盖上锅盖．','【分析】A、根据可燃性的气体与空气或氧气的混合气体遇明火、静电、电火花或加热能发生爆炸判断．<br />B、根据油的密度小于水的密度判断．<br />C、根据碱液沾到皮肤的处理方法判断．<br />D、根据灭火的原理判断．','填空题',3.00,'4bf1cc2119d69fe759d18b69812dff29',9,400,'常见的意外事故的处理方法,灭火的原理和方法,防范爆炸的措施','',2016,'37','2016春•石城县校级月考',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840413,'用5000t含Fe<SUB>2</SUB>O<SUB>3</SUB>80%的赤铁矿石冶炼生铁．<br />（1）原料中含Fe<SUB>2</SUB>O<SUB>3</SUB>的质量为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>t；<br />（2）理论上，冶炼出含铁96%的生铁的质量是多少？（要求写出计算过程．）','','','','','','4000','【解答】解：<br />（1）原料中含Fe<SUB>2</SUB>O<SUB>3</SUB>的质量为5000t×80%=4000t<br />（2）设理论上可炼取纯铁质量为x；&nbsp;&nbsp;&nbsp; <br />Fe<SUB>2</SUB>O<SUB>3</SUB>+3CO<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2Fe+3CO<SUB>2</SUB><br />&nbsp;160&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 112<br />4000t&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; x<br /><span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">160</td></tr><tr><td>4000t</td></tr></table>=<table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">112</td></tr><tr><td>x</td></tr></table></span><br />x=2800t<br />理论上可炼取含铁96%的生铁的质量为：<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">2800t</td></tr><tr><td>95%</td></tr></table></span>≈2917t<br />答案：<br />（1）4000<br />（2）理论上，冶炼出含铁96%的生铁的质量是2917t．','【分析】（1）根据题意，5000吨含Fe<SUB>2</SUB>O<SUB>3 </SUB>80%的赤铁，据此可计算出含纯铁的质量．<br />（2）由生铁的质量分数即可解答．','填空题',3.00,'bb7258706c9234eddd96f546873d0b9d',9,400,'含杂质物质的化学反应的有关计算','',2016,'32','2016•深圳校级模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840416,'氢氧化钠固体及溶液在空气中很容易发生变质生成碳酸钠．<br />（1）甲同学在实验室里发现一瓶敞口放置的氢氧化钠溶液，对其变质情况进行了如下探究．<br />【猜想与假设】该氢氧化钠溶液已部分变质．<br />【实验与结论】<br /><table class=\"edittable\"><TBODY><TR><td width=313>实验步骤</TD><td width=168>实验现象</TD><td width=171>实验结论</TD></TR><TR><td>a．取少量该溶液于一支试管中，加入足量<u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD><td><u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD><td>该氢氧化钠溶液已变质</TD></TR><TR><td>b．将步骤a所得的液体过滤，取滤液于另一支试管中，滴加<u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD><td><u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD><td>该氢氧化钠溶液已部分变质</TD></TR></TBODY></TABLE>【拓展探究】甲同学又尝试通过实验除去该氢氧化钠溶液中含有的碳酸钠，方案如下：<br />向该溶液中加入适量<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，反应后，经过滤操作，即可除去该氢氧化钠溶液中含有的杂质．<br />（2）乙同学在实验室里发现少量部分变质的氢氧化钠固体，尝试通过实验测定其中碳酸钠的质量分数．<br />【设计和进行实验】乙同学设计了如图所示的实验装置（铁架台、铁夹等固定用装置已略去，碱石灰是固体氢氧化钠和生石灰的混合物），取10.0g待测样品，进行实验．<br />实验步骤如图：<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao33/18855680-94d4-11e9-9c86-b42e9921e93e_xkb38.png\" style=\"vertical-align:middle\" /><br />a．连接好装置，并检查气密性；<br />b．打开弹簧夹，缓缓通入一段时间空气；<br />c．称量装置D的质量为83.4g；<br />d．关闭弹簧夹，慢慢滴加Y溶液，至不再产生气泡为止；<br />e．打开弹簧夹，再次缓缓通入一段时间空气；<br />f．再次称量装置D的质量为84.5g．<br />【交流讨论】<br />①试剂X、Y、Z依次最适宜选用<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（选填字母）．<br />A．氢氧化钠溶液&nbsp;&nbsp;&nbsp;浓盐酸&nbsp;&nbsp;&nbsp;浓硫酸&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;B．澄清石灰水&nbsp;&nbsp;&nbsp;稀盐酸&nbsp;&nbsp;&nbsp;稀硫酸<br />C．氢氧化钠溶液&nbsp;&nbsp;&nbsp;稀硫酸&nbsp;&nbsp;&nbsp;浓硫酸&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;D．浓硫酸&nbsp;&nbsp;稀硫酸&nbsp;&nbsp;&nbsp;氢氧化钠溶液<br />②步骤e中通入空气的作用是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />③若没有装置C或E，则会导致测定结果<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（选填“偏大”、“偏小”或“无影响”）<br />【数据处理】根据实验中测得的有关数据，计算部分变质的氢氧化钠固体中碳酸钠的质量分数是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','氯化钙溶液$###$有白色沉淀生成$###$无色酚酞试液$###$酚酞试液变红$###$氢氧化钙$###$C$###$将B中产生的二氧化碳全部赶到D装置中$###$偏大$###$26.5%','【解答】解：（1）【实验与结论】根据实验结论：该氢氧化钠溶液已变质，说明其中含有碳酸钠，所以要检验是否含有碳酸根即可，又因为第二步中需要检验是否含有氢氧根，而碳酸钠也显碱性，所以第一步加入的试剂不仅要验证出有碳酸根离子，还需要除去碳酸根离子，所以加入试剂氯化钙即可，因为氯化钙显中性，与碳酸钠反应生成碳酸钙沉淀和氯化钠，氯化钠也显中性，所以加入氯化钙，如果有白色沉淀生成，说明有碳酸钠存在，即已经变质；将步骤a所得的液体静置，取上层清液于另一支试管中，滴加酚酞试液，如果变红，说明溶液显碱性，即含有氢氧化钠，说明部分变质；<br />【拓展探究】除去碳酸钠主要是除去其中的碳酸根离子，可以用钙离子把碳酸根离子沉淀出来，又因为不能引入新的杂质，所以加入氢氧化钙就行，然后通过过滤将碳酸钙沉淀过滤出来；<br />（2）①A中试剂的作用主要是除去空气中的二氧化碳，所以用氢氧化钠溶液即可，B中加入的试剂主要是与碳酸钠反应生成二氧化碳，所以加入酸，不能是浓盐酸，可以是稀硫酸或稀盐酸；C中作用是除去混在二氧化碳中的水蒸气，所以用浓硫酸即可；<br />②由于反应结束后，B装置中会含有很多二氧化碳，所以通过通入空气将B装置中的二氧化碳排到D装置中；<br />③若没有装置C，会把水蒸气误认为是二氧化碳，导致二氧化碳质量偏多，通过计算得到的碳酸钠质量偏大，所以结果偏大；<br />【数据处理】解：生成二氧化碳的质量=84.5g-83.4g=1.1g<br />设部分变质的氢氧化钠固体中碳酸钠的质量为x<br />Na<SUB>2</SUB>CO<SUB>3</SUB>+H<SUB>2</SUB>SO<SUB>4</SUB>═Na<SUB>2</SUB>SO<SUB>4</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑<br />106&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;44<br />x&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;1.1g<br /><span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">106</td></tr><tr><td>x</td></tr></table></span>=<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">44</td></tr><tr><td>1.1g</td></tr></table></span>&nbsp;&nbsp; x=2.65g<br />部分变质的氢氧化钠固体中碳酸钠的质量分数：<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">2.65g</td></tr><tr><td>10g</td></tr></table></span>×100%=26.5%<br />答：部分变质的氢氧化钠固体中碳酸钠的质量分数为26.5%<br />故答案为：（1）【实验与结论】<br /><table class=\"edittable\"><TBODY><TR><td>实验步骤</TD><td>实验现象</TD><td>实验结论</TD></TR><TR><td>a．取少量该溶液于一支试管中，加入 足量的氯化钙溶液</TD><td>有白色沉淀生成</TD><td>该氢氧化钠溶液已变质</TD></TR><TR><td>b．将步骤a所得的液体静置，取上层清液于另一支试管中，滴加 无色酚酞试液</TD><td>酚酞试液变红</TD><td>该氢氧化钠溶液已部分变质</TD></TR></TBODY></TABLE>【拓展探究】氢氧化钙；<br />（2）①C；②将B中产生的二氧化碳全部赶到D装置中；③偏大；<br />【数据处理】26.5%．','【分析】（1）【实验与结论】主要看物质中是否含有碳酸钠，即测定碳酸根的存在，所以用钙离子来验证碳酸根的存在，并除去碳酸根离子，再利用酚酞试液测定氢氧根离子的存在即可；【拓展与探究】除杂质要根据加入的试剂只能与杂质反应，不能引入新的杂质考虑．<br />（2）①本题主要是根据D装置中质量的增加量，就是生成的二氧化碳的质量，所以一开始要除去空气中二氧化碳，通过加入的试剂与碳酸钠反应生成二氧化碳，反应后还需除去二氧化碳中的水蒸气，最后将二氧化碳吸收；②主要考虑反应完成后B中会存有很多的二氧化碳；③没有C就会把水蒸气误认为是二氧化碳进行分析．<br />【数据处理】根据化学方程式结合质量分数进行计算．','填空题',3.00,'31bedfe26276195e4a0097f80d38dab1',9,400,'药品是否变质的探究,常见气体的检验与除杂方法,碱的化学性质,盐的化学性质','',2016,'37','2016•上饶二模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840419,'<img  src=\"/tikuimages/9/2016/400/shoutiniao49/18931221-94d4-11e9-9562-b42e9921e93e_xkb93.png\" style=\"vertical-align:middle;FLOAT:right\" />（2016•丹东模拟）如图中：A、B、C、D、E、F是初中化学中的常见物质，其中B是一种气体单质，D是大理石的主要成分，“---”表示两端的物质能发生反应，“→”表示一种物质可转化为另一种物质，请回答：<br />（1）A和F反应的现象为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（2）写出下列反应的化学方程式：<br />①A→B的化学方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />②C→D的化学方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）物质E的类型应属于<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“酸”、“碱”、“盐”或“氧化物”）；<br />（4）上述反应中未涉及到的基本反应类型是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','放热$###$2H<SUB>2</SUB>O<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;通电&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>↑+O<SUB>2</SUB>↑$###$Ca（OH）<SUB>2</SUB>+CO<SUB>2</SUB>═CaCO<SUB>3</SUB>↓+H<SUB>2</SUB>O$###$酸$###$置换反应','【解答】解：（1）A、B、C、D、E、F是初中化学中的常见物质，D是大理石的主要成分，所以D是碳酸钙，B是一种气体单质，B转化成的C和碳酸钙可以相互转化，所以B是氧气，C是二氧化碳，A和二氧化碳会反应，会生成氧气，所以A是水，碳酸钙生成的F会与水反应，所以F是氧化钙，E会与氧化钙、碳酸钙反应，所以E可以是稀盐酸，经过验证，推导正确，所以A和F的反应是氧化钙和水反应生成氢氧化钙，实验现象为放热；<br />（2）①A→B的反应是水在通电的条件下生成氢气和氧气，化学方程式为：2H<SUB>2</SUB>O<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;通电&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>↑+O<SUB>2</SUB>↑；<br />②C→D的反应是氢氧化钙和二氧化碳反应生成碳酸钙沉淀和水，化学方程式为：Ca（OH）<SUB>2</SUB>+CO<SUB>2</SUB>═CaCO<SUB>3</SUB>↓+H<SUB>2</SUB>O；<br />（3）通过推导可知，物质E的类型应属于酸；<br />（4）水通电属于分解反应，二氧化碳和水生成碳酸属于化合反应，盐酸和碳酸钙的反应属于复分解反应，所以上述反应中未涉及到的基本反应类型是置换反应．<br />故答案为：（1）放热；<br />（2）①2H<SUB>2</SUB>O<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;通电&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>↑+O<SUB>2</SUB>↑；<br />②Ca（OH）<SUB>2</SUB>+CO<SUB>2</SUB>═CaCO<SUB>3</SUB>↓+H<SUB>2</SUB>O；<br />（3）酸；<br />（4）置换反应．','【分析】根据A、B、C、D、E、F是初中化学中的常见物质，D是大理石的主要成分，所以D是碳酸钙，B是一种气体单质，B转化成的C和碳酸钙可以相互转化，所以B是氧气，C是二氧化碳，A和二氧化碳会反应，会生成氧气，所以A是水，碳酸钙生成的F会与水反应，所以F是氧化钙，E会与氧化钙、碳酸钙反应，所以E可以是稀盐酸，然后将推出的物质进行验证即可．','书写',3.00,'be0b672de5765a997f4673d6736934a3',9,400,'常见的氧化物、酸、碱和盐的判别,物质的鉴别、推断,反应类型的判定,书写化学方程式、文字表达式、电离方程式','',2016,'32','2016•丹东模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840422,'请你运用分子的性质判断，以下事实的解释错误的是（　　）','空气是混合物----空气由不同种分子构成','乒乓球瘪了，用热水泡会重新鼓起来----分子变大了','固体酒精和液体酒精都能燃烧----相同物质的分子，其化学性质相同','柚子花开，香满园----分子在不断运动','','B','【解答】解：A、空气是混合物，是因为空气是由不同种分子构成的，故选项解释正确．<br />B、乒乓球瘪了，用热水泡会重新鼓起来，是因为温度升高，分子间的间隔变大，故选项解释错误．<br />C、固体酒精和液体酒精都能燃烧，是因为它们是由酒精分子构成的，同种的分子化学性质相同，故选项解释正确．<br />D、柚子花开，香满园，是因为柚子花香中含有的分子是在不断运动的，向四周扩散，使人们闻到柚子花香，故选项解释正确．<br />故选：B．','【分析】根据分子的基本特征：分子质量和体积都很小；分子之间有间隔；分子是在不断运动的；同种的分子性质相同，不同种的分子性质不同，可以简记为：“两小运间，同同不不”，结合事实进行分析判断即可．','选择题',3.00,'1b32a21c2ec369b91b8ebfd5ef65545f',9,400,'利用分子与原子的性质分析和解决问题','',2015,'33','2015秋•平和县期末',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840423,'下列有关实验现象的描述，错误的是（　　）','铁丝在氧气中剧烈燃烧时，火星四射，生成黑色固体','分别点燃一小团棉花和羊毛，都产生了烧焦羽毛的气味','红磷在空气中燃烧时，产生大量白烟，放出大量热','打磨后的铝丝放入硫酸铜溶液中，铝丝表面有红色物质析出','','B','【解答】解：A、铁丝在氧气中剧烈燃烧，火星四射，生成一种黑色固体，故选项说法正确．<br />B、分别点燃一小团棉花和羊毛，羊毛的主要成分是蛋白质，燃烧会产生烧焦羽毛的气味，棉花燃烧有烧纸的气味，故选项说法错误．<br />C、红磷在空气中燃烧，产生大量的白烟，放出大量热，故选项说法正确．<br />D、打磨后的铝丝放入硫酸铜溶液中，铝与硫酸铜溶液反应生成硫酸铝和铜，会观察到铝丝表面有红色物质析出，故选项说法正确．<br />故选：B．','【分析】A、根据铁丝在氧气中燃烧的现象进行分析判断．<br />B、根据羊毛、棉花燃烧的现象进行分析判断．<br />C、根据红磷在空气中燃烧的现象进行分析判断．<br />D、根据金属的化学性质进行分析判断．','选择题',3.00,'17bdcd3db8390c162718cd9a5be45c24',9,400,'氧气与碳、磷、硫、铁等物质的反应现象,金属的化学性质,棉纤维、羊毛纤维和合成纤维的鉴别','江阴市',2016,'35','2016春•江阴市校级期中',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840425,'<img  src=\"/tikuimages/9/2016/400/shoutiniao50/18ad77f0-94d4-11e9-bec5-b42e9921e93e_xkb75.png\" style=\"vertical-align:middle;FLOAT:right\" />（2016•定州市校级模拟）化学复习课上，某实验小组同学利用盐酸与大理石反应制取二氧化碳时，发现产生的气体经反复实验也不能使澄清石灰水变浑浊．<br />【做出猜想】<br />甲：澄清石灰水可能保存不当已完全变质．<br />乙：用浓盐酸进行实验，制得的二氧化碳中混有<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>气体，写出二氧化碳中的气体杂质与澄清石灰水反应的化学方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />【进行实验】<br />1．通过实验证明原瓶中的澄清石灰水没有变质，猜想甲是错误的，请你写出实验方法：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />2．取原瓶中的盐酸少量，加适量水稀释后再与大理石反应，将产生的气体通入从原瓶取出的少量澄清石灰水中，结果石灰水边浑浊，由此证明猜想乙是正确的．<br />【拓展与交流】<br />为了除去上述实验中二氧化碳气体内的杂质，得到纯净干燥的二氧化碳气体，设计了如图所示装置，a瓶内的液体是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填编号，下同），b瓶内的液体是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />A．浓NaOH溶液　 B．浓H<SUB>2</SUB>SO<SUB>4</SUB>　 C．NaHCO<SUB>3</SUB>溶液<br />老师说：“化学实验中难免失败或者有意外现象发生，通过查阅资料或请教他人、再进行实验后认真分析，常常会有意想不到的收货．”<br />例如：稀硫酸和块状大理石不能用于实验室制取二氧化碳，因为稀硫酸与块状大理石中的碳酸钙反应生成硫酸钙较多时不再溶解，会附着在大理石表面，阻止硫酸和碳酸钙进一步反应，而大理石粉末与稀硫酸可以持续剧烈反应，有时也可以用于实验室制取二氧化碳，大理石粉末与稀硫酸持续剧烈反应的原因是：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />再如：足量CO<SUB>2</SUB>通入澄清石灰水时，石灰水先变浑浊，然后逐渐澄清，这是因为还发生了反应：CaCO<SUB>3</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>═Ca（HCO<SUB>3</SUB>）<SUB>x</SUB>，Ca（HCO<SUB>3</SUB>）<SUB>x</SUB>易溶于水，x=<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','氯化氢$###$Ca（OH）<SUB>2</SUB>+2HCl═CaCl<SUB>2</SUB>+2H<SUB>2</SUB>O$###$取适量澄清石灰水于试管中，向试管中滴加几滴酚酞试液，酚酞试液变红色$###$C$###$B$###$大理石粉末和稀硫酸充分接触$###$2','【解答】解：【做出猜想】<br />乙：用浓盐酸进行实验，制得的二氧化碳中混有氯化氢气体；<br />二氧化碳中的气体杂质氯化氢与澄清石灰水反应的化学方程式为：Ca（OH）<SUB>2</SUB>+2HCl═CaCl<SUB>2</SUB>+2H<SUB>2</SUB>O．<br />故填：氯化氢；Ca（OH）<SUB>2</SUB>+2HCl═CaCl<SUB>2</SUB>+2H<SUB>2</SUB>O．<br />【进行实验】<br />1．实验方法是：取适量澄清石灰水于试管中，向试管中滴加几滴酚酞试液，酚酞试液变红色，说明澄清石灰水没有变质．<br />故填：取适量澄清石灰水于试管中，向试管中滴加几滴酚酞试液，酚酞试液变红色．<br />【拓展与交流】<br />为了除去上述实验中二氧化碳气体内的杂质，得到纯净干燥的二氧化碳气体，设计了如图所示装置，a瓶内的液体是碳酸氢钠溶液，用来吸收氯化氢气体，b瓶内的液体是浓硫酸，用来吸收水蒸气．<br />故填：C；B．<br />大理石粉末与稀硫酸可以持续剧烈反应，有时也可以用于实验室制取二氧化碳，大理石粉末与稀硫酸持续剧烈反应的原因是大理石粉末和稀硫酸充分接触．<br />故填：大理石粉末和稀硫酸充分接触．<br />由CaCO<SUB>3</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>=Ca（HCO<SUB>3</SUB>）<SUB>x</SUB>可知，反应前钙原子是1个，碳原子是2个，氢原子是2个，氧原子是6个，反应后钙原子是1个，碳原子是2个，氢原子是2个，氧原子是6个，因此x=2．<br />故填：2．<br />答案：<br />【做出猜想】氯化氢；Ca（OH）<SUB>2</SUB>+2HCl═CaCl<SUB>2</SUB>+2H<SUB>2</SUB>O．<br />【进行实验】取适量澄清石灰水于试管中，向试管中滴加几滴酚酞试液，酚酞试液变红色．<br />【拓展与交流】C；B；大理石粉末和稀硫酸充分接触；2．','【分析】【做出猜想】<br />浓盐酸易挥发；<br />氢氧化钙和稀盐酸反应生成氯化钙和水；<br />【进行实验】<br />澄清石灰水显碱性，能使酚酞试液变红色；<br />【拓展与交流】<br />饱和碳酸氢钠溶液能够吸收氯化氢气体，不能吸收二氧化碳气体；<br />浓硫酸具有吸水性，可以用来干燥二氧化碳气体；<br />反应物检查越充分，反应速率越快；<br />化学反应遵循质量守恒定律，即反应前后元素种类不变，原子种类和总个数不变．','书写',3.00,'84845f3e7eec5c0659c06160b537aa17',9,400,'实验探究物质变化的条件和影响物质变化的因素,常见气体的检验与除杂方法,碱的化学性质,质量守恒定律及其应用,书写化学方程式、文字表达式、电离方程式','定州市',2016,'32','2016•定州市校级模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840429,'某课外兴趣小组的同学设计了一系列实验方案．其中能达到实验目的是（　　）','用50&nbsp;ml的量筒量取8.0ml水','用向上排空气法收集氢气','用燃着的木条伸入集气瓶中检验瓶中的气体是否为CO<SUB>2</SUB>','用带火星的木条放在集气瓶口检验氧气是否集满','','D','【解答】解：A、在量取水时，应选择稍大于所要量取水体积的量筒，一次性量取，所以用1OmL的量筒量取9．OmL水，故A错误；<br />B、氢气的密度小于空气的密度，所以用向下排空气法收集氢气，故B错误；<br />C、能使燃着的木条熄灭的不一定是二氧化碳，例如氮气也能使燃着的木条熄灭，故C错误；<br />D、氧气具有助燃性，能使带火星的木条复燃，所以可用带火星的木条放在集气瓶口检验氧气是否集满，故D正确；<br />故选D．','【分析】A、根据量筒量程的选取方法考虑；<br />B、根据氢气的密度小于空气的密度考虑；<br />C、根据能使燃着的木条熄灭的不一定是二氧化碳考虑；<br />D、根据氧气具有助燃性，能使带火星的木条复燃考虑．','选择题',3.00,'820a4025052450bd52bfa2971cb53451',9,400,'化学实验方案设计与评价,测量容器-量筒,常用气体的收集方法,氧气的检验和验满,二氧化碳的检验和验满','',2016,'37','2016•邵阳县二模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840431,'下列不属于化学研究范围的是（　　）','水的电解','水的浮力','水的分子结构','水的组成','','B','【解答】解：A、水的电解，属于研究物质的变化，属于化学研究的范畴，故选项错误．<br />B、水的浮力，属于物理学领域研究的范畴，故选项正确．<br />C、水的分子结构，属于物质的微观构成，属于化学研究的范畴，故选项错误．<br />D、水的组成，属于研究物质的宏观组成，属于化学研究的范畴，故选项错误．<br />故选：B．','【分析】根据化学的定义和研究内容进行分析判断，化学是一门在分子、原子的层次上研究物质的性质、组成、结构及其变化规律的科学，研究对象是物质，研究内容有组成、结构、性质、变化、用途等．','选择题',3.00,'597ed91288f23b622526d3ecfca458df',9,400,'化学的研究领域','',2015,'32','2015•武侯区模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840433,'化学与我们的生产、生活息息相关．现有以下物质<br />①食醋&nbsp;&nbsp;&nbsp;②硝酸铵&nbsp;&nbsp;&nbsp;③维生素A&nbsp;&nbsp;&nbsp;④活性炭&nbsp;&nbsp;&nbsp;⑤熟石灰．<br />请选择合适物质的序号填空：<br />（1）常用于改良酸性土地壤的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（2）导致夜盲症是因为人体缺乏：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（3）常用作厨房调味品的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（4）可制成“冰袋”用于降温的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（5）常用于冰箱除异味的物质是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','⑤$###$③$###$①$###$②$###$④','【解答】解：（1）熟石灰显碱性，常用于改良酸性土地壤；<br />（2）人体缺乏后能导致盲症的是维生素A；<br />（3）食盐具有咸味，常用作厨房调味品；<br />（4）硝酸铵溶于水吸热，可制成“冰袋”用于降温；<br />（5）活性炭具有吸附性，常用于冰箱除异味的物质．<br />故答案为：（1）⑤；（2）③；（3）①；（4）②；（5）④．','【分析】分析生活应用中所需要物质具有的性质，结合所给物质的性质和用途，判断与所列出用途相符合的物质．','填空题',3.00,'9a06871c562aec3c499d6aa314f71d21',9,400,'溶解时的吸热或放热现象,常见碱的特性和用途,常用盐的用途,碳单质的物理性质及用途,微量元素、维生素与健康的关系及摄取方法','张家港市',2016,'32','2016•张家港市模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840434,'金属材料与人类的生产和生活密切相关．请回答：<br />（1）下列用品中，主要利用金属导电性的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填字母）<br />A．铂金饰品B．铁锅&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;C．铝导线<br />（2）电池中含有铅、镍、镉、汞等金属，随意丢弃废旧电池，会造成水体和土壤污染，威胁人类健康．回收废旧电池的重要意义是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />A．节约金属资料B．增加金属产量C．减少环境污染<br />（3）为了验证铝、铜、银三种金属的活动性顺序，设计了下列四种方案，其中可行的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填序号）<br />①将铝、银非别浸入到硫酸铜溶液中②将银分别浸入到硫酸铝、硫酸铜溶液中<br />③将铝、铜、银分别浸入到稀硫酸溶液中④将铜分别浸入到硫酸铝、硝酸银溶液中<br />（4）在氯化铜和氯化亚铁的混合溶液中加入一定质量的镁粉，充分反应后过滤，得到滤渣和滤液．向滤渣中滴加稀盐酸，有气泡产生，则滤渣中可能含有的物质是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填化学式）','','','','','','C$###$AC$###$①④$###$Cu、Fe','【解答】解：（1）在常见的金属材料用品中，铝导线主要利用金属导电性．<br />（2）回收废旧电池能节约金属资源、减少环境污染．故填：AC．<br />（3）要验证铝、铜、银三种金属的活动性顺序，可采用“三取中”的方法将铝、银非别浸入到硫酸铜溶液中；将铜分别浸入到硫酸铝、硝酸银溶液中；<br />（4）镁的金属活动性比铜、铁强，镁会先于氯化铜反应，置换出铜，所以在氯化铜和氯化亚铁的混合溶液中加入一定质量的镁粉，充分反应后过滤，得到滤渣和滤液．向滤渣中滴加稀盐酸，有气泡产生，则滤渣中一定含有的物质是Cu、Fe．<br />故答案为：<br />（1）C；（2）AC；（3）①④；（4）Cu、Fe．','【分析】（1）根据金属材料的性质和用途分析回答；<br />（2）根据回收废旧金属的意义进行分析解答即可；<br />（3）在金属活动性顺序中，氢前的金属能与酸反应生成氢气，位置在前的金属能将位于其后的金属从其盐溶液中置换出来；<br />（4）根据镁的金属活动性比铜、铁强，镁会先于氯化铜反应，置换出铜进行分析．','简答题',3.00,'e32f57a62484b418dd7a839e0e2adb4e',9,400,'金属的物理性质及用途,金属的化学性质,金属活动性顺序及其应用,金属的回收利用及其重要性','',2016,'37','2016•重庆校级一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840439,'镁是一种用途很广的金属，目前世界上60%的镁是从海水中提取的．主要步骤如下：<br /><img  src=\"/tikuimages/9/2015/400/shoutiniao55/18d79530-94d4-11e9-b76b-b42e9921e93e_xkb3.png\" style=\"vertical-align:middle\" /><br />（1）上述步骤中试剂①应过量，理由是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）请写出加入试剂②后，发生反应的化学方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）无水MgCl<SUB>2</SUB>在熔融状态下通电得到金属镁和一种气态单质，其化学式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','确保海水中的氯化镁完全反应$###$Mg（OH）<SUB>2</SUB>+2HCl=MgCl<SUB>2</SUB>+2H<SUB>2</SUB>O$###$Cl<SUB>2</SUB>','【解答】解：（1）步骤中试剂①应过量，目的是确保海水中的氯化镁完全反应；<br />（2）氢氧化镁和盐酸反应生成氯化镁和水，化学方程式为：Mg（OH）<SUB>2</SUB>+2HCl=MgCl<SUB>2</SUB>+2H<SUB>2</SUB>O；<br />（3）无水氯化镁在通电的条件下生成镁和氯气，化学式为：Cl<SUB>2</SUB>．<br />故答案为：（1）确保海水中的氯化镁完全反应；<br />（2）Mg（OH）<SUB>2</SUB>+2HCl=MgCl<SUB>2</SUB>+2H<SUB>2</SUB>O；<br />（3）Cl<SUB>2</SUB>．','【分析】（1）根据步骤中试剂①应过量，目的是确保海水中的氯化镁完全反应进行分析；<br />（2）根据氢氧化镁和盐酸反应生成氯化镁和水进行分析；<br />（3）根据无水氯化镁在通电的条件下生成镁和氯气进行分析．','书写',3.00,'7ed05500ad0e259f86fa02c297890a34',9,400,'盐的化学性质,书写化学方程式、文字表达式、电离方程式,对海洋资源的合理开发与利用','',2015,'35','2015秋•周村区校级期中',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840444,'化学反应有的需要吸收热，有的放出热，下列反应属于吸热反应的是（　　）','C+O<SUB>2</SUB>═CO<SUB>2</SUB>&nbsp;（充分燃烧）','2C+O<SUB>2</SUB>═2CO&nbsp;&nbsp;（不充分燃烧）','C+CO<SUB>2</SUB>═2CO','2H<SUB>2</SUB>+0<SUB>2</SUB>═H<SUB>2</SUB>O','','C','【解答】解：<br />A．木炭燃烧是放热反应，故A错误； <br />B．木炭不充分燃烧是放热反应，故B错误；<br />C．C和CO<SUB>2</SUB>在高温下反应，需要吸收热量，是吸热反应，故C正确；<br />D．氢气在氧气中燃烧，是放热反应，故D错误．<br />故选C．','【分析】根据常见的放热反应有：所有的物质燃烧、所有金属与酸反应、金属与水反应，所有中和反应；绝大多数化合反应和铝热反应；<br />常见的吸热反应有：绝大数分解反应，个别的化合反应（如C和CO<SUB>2</SUB>），少数分解置换以及某些复分解（如铵盐和强碱）．','选择题',3.00,'61b72d0def1bd1bdb7038964748c0570',9,400,'物质发生化学变化时的能量变化','',0,'37','',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840445,'<img  src=\"/tikuimages/9/0/400/shoutiniao46/18e63b30-94d4-11e9-b04b-b42e9921e93e_xkb53.png\" style=\"vertical-align:middle;FLOAT:right;\" />某兴趣小组的同学对氢氧化钠进行了以下探究活动．<br />（1）氢氧化钠与二氧化碳反应没有明显规象，某同学设计了如图1装置．<br />①挤压胶头滴管后能观察到的现象是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />②有小组成员认为该实验不足以证明二氧化碳与氢氧化钠发生了化学反应，请你在上述实验的基础上进一步作出补充．<br />③有小组成员想探究该实验后集气瓶内溶液是否还含有氢氧化钠，请你帮助他们设计实验方案．<br />（2）氢氧化钠溶液与稀盐酸反应也没有明显现象，小组同学利用数字实验室的设备绘制了反应过程中溶液pH变化的图象2．<br />①A点表示的溶液中的溶质为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />②小组成员为证明氢氧化钠与盐酸是否发生反应，在两者混合后的溶液中加入硝酸银溶液，看是否有白色沉淀生成．你认为此方案是否正确，请简述理由．<br />③实验用去50g溶质质量分数7.3%的稀盐酸，则恰好完全反应时，所得溶液的溶质质量分数是多少？','','','','','','U形管内出现左高右低的现象$###$盐酸，氯化钠','【解答】解：（1）①若胶头滴管中的物质是浓NaOH溶液，锥形瓶中充满CO<SUB>2</SUB>，则挤压胶头滴管后氢氧化钠与二氧化碳反应生成碳酸钠和水，由于气体被消耗，U形管内出现左高右低的现象；故答案为：U形管内出现左高右低的现象；<br />②要证明氢氧化钠和二氧化碳确实发生了化学反应，只需要证明反应生成了碳酸钠即可，取反应后的溶液加入试管中，滴加稀盐酸，有气泡产生，说明氢氧化钠和二氧化碳确实发生了化学反应．故答案为：取反应后的溶液加入试管中，滴加稀盐酸，有气泡产生；<br />③紫色石蕊试液遇碱变蓝，无色酚酞试液遇碱变红，取反应后的溶液加入试管中，滴加无色酚酞试液，如果变红，则说明集气瓶内溶液还含有氢氧化钠，否则不含氢氧化钠．故答案为：取反应后的溶液加入试管中，滴加无色酚酞试液．<br />（2）氢氧化钠溶液与稀盐酸发生反应过程中溶液pH变化如图所示．<br />①A点溶液成酸性，溶质含有盐酸和氯化钠．故答案为：盐酸、氯化钠；<br />②氢氧化钠与盐酸反应生成氯化钠，氯化钠能与硝酸银反应生成白色沉淀，而盐酸也能与硝酸银反应生成白色沉淀．故答案为：滴加硝酸银有白色沉淀，只能说明溶液中含有氯离子，不能说明生成了新物质氯化钠，因为盐酸与硝酸银溶液反应有白色沉淀．<br />③<br />（4）解：设所用氢氧化钠的质量为x，反应生成氯化钠的质量为y<br />&nbsp;&nbsp; HCl+NaOH═H<SUB>2</SUB>O+NaCl<br />36.5&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 40&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;58.5<br />50g×7.3%&nbsp;&nbsp; x&nbsp;&nbsp; &nbsp;&nbsp; &nbsp;y<br /><span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">36.5</td></tr><tr><td>50g×7.3%</td></tr></table></span>=<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">40</td></tr><tr><td>x</td></tr></table></span>=<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">58.5</td></tr><tr><td>y</td></tr></table></span><br />x=4g，y=5.85g<br />反应后所得溶液中溶质的质量分数=<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">5.85g</td></tr><tr><td>50g+4g</td></tr></table></span>×100%=10.8%<br />答反应后所得溶液中溶质的质量分数为10.8%','【分析】根据复分解反应的条件以及物质间反应的实验现象进行分析，紫色石蕊试液遇碱变蓝，无色酚酞试液遇碱变红；根据图象结合盐酸和氢氧化钠的反应解答；根据化学方程式结合题干提供的数据解答．','书写',3.00,'6416c92d918f3761d476e741412c596f',9,400,'探究酸碱的主要性质,有关溶质质量分数的简单计算,碱的化学性质,书写化学方程式、文字表达式、电离方程式,根据化学反应方程式的计算','',0,'37','',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840446,'规范的操作是实验成功的保证，下列实验操作正确的是（　　）','<img  src=\"/tikuimages/9/2016/400/shoutiniao39/18ebb970-94d4-11e9-acaa-b42e9921e93e_xkb94.png\" style=\"vertical-align:middle\" /><br />收集O<SUB>2</SUB>','<img  src=\"/tikuimages/9/2016/400/shoutiniao76/18ef8a00-94d4-11e9-9b96-b42e9921e93e_xkb19.png\" style=\"vertical-align:middle\" /><br />称取食盐','<img  src=\"/tikuimages/9/2016/400/shoutiniao63/18f29740-94d4-11e9-9e91-b42e9921e93e_xkb64.png\" style=\"vertical-align:middle\" /><br />稀释浓硫酸','<img  src=\"/tikuimages/9/2016/400/shoutiniao48/18f5f2a1-94d4-11e9-9aa2-b42e9921e93e_xkb78.png\" style=\"vertical-align:middle\" /><br />加入锌粒','','A','【解答】解：A、在用排水法收集氧气时，气体要从短管进入，故A正确；<br />B、天平称量方法：左物右码，要将药品放在纸上称量，故B错；<br />C、稀释浓硫酸的方法：将浓硫酸沿着容器壁慢慢倒入水中，并用玻璃棒不断搅拌，故C错；<br />D、固体的放法：将试管横放，用镊子将药品放在试管口，将试管慢慢竖起，故D错．<br />故选A．','【分析】A、在用排水法收集氧气时，气体要从短管进入；B、天平称量方法：左物右码，要将药品放在纸上称量；C、稀释浓硫酸的方法考虑；D、根据固体的放法考虑．','选择题',3.00,'1511693883c6a73ff51a0e8bde9b4efa',9,400,'称量器-托盘天平,固体药品的取用,浓硫酸的性质及浓硫酸的稀释,氧气的收集方法','',2016,'32','2016•合肥模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840447,'水下作业时，潜水员需要呼吸“富氧空气”．<br />（1）潜水员水下作业时携带“富氧空气”，是因为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）“富氧空气”中氧气含量高于普通空气，其中氧气与气体气体体积比可能是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />A、1：2&nbsp;&nbsp;&nbsp;&nbsp;B、1：5&nbsp;&nbsp;&nbsp;&nbsp;C、1：4&nbsp;&nbsp;&nbsp;&nbsp;D、21：79．','','','','','','氧气能供给呼吸；$###$A','【解答】解：（1）潜水员水下作业时携带“富氧空气”，是因为氧气能供给呼吸．<br />（2）氧气的体积分数大于空气中的平均氧气体积分数的空气称为富氧空气，即氧气的体积分数大于21%，<br />A、1：2大约是33.3%，属于富氧空气，故A正确；<br />B、1：5不可以，比空气中的还低，故B错误；<br />C、1：4，属于普通空气的组成，故C错误；<br />D、21：79，与空气的成分相差不大，故D错误．<br />故D答案为：（1）氧气能供给呼吸；&nbsp;（2）A．','【分析】（1）根据氧气能供给呼吸分析回答；<br />（2）根据空气的成分和“富氧空气”中含有较多的氧气分析判断．','填空题',3.00,'f80c43a4ab86e41f4c3394d12c676d92',9,400,'空气的成分及各成分的体积分数,氧气的用途','',2015,'35','2015秋•朝阳区校级期中',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840448,'A～G是目前初中化学常见的七种物质，在一定条件下能发生如图所示的化学反应，其中F是自然界中最常见的液体，C、E都是气体，D在反应③中作催化剂使用，B可溶于水，D不溶于水．<br /><img  src=\"/tikuimages/9/0/400/shoutiniao69/18fbbf00-94d4-11e9-a188-b42e9921e93e_xkb45.png\" style=\"vertical-align:middle\" /><br />（1）推断有关物质：E<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>、F<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>、G<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）写出反应①的表达式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；写出反应②的表达式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（3）以上4个化学反应中，分解反应有<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>个．','','','','','','H<SUB>2</SUB>$###$H<SUB>2</SUB>O$###$H<SUB>2</SUB>O<SUB>2</SUB>$###$2KMnO<SUB>4</SUB> <span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>K<SUB>2</SUB>MnO<SUB>4</SUB>+MnO<SUB>2</SUB>+O<SUB>2</SUB>↑$###$2H<SUB>2</SUB>+O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O$###$3','【解答】解：F是自然界中最常见的液体，能够通电分解产生C和E，C、E都是气体，D在反应③中作催化剂使用，可以产生C，因此F是水，C是氧气，E是氢气；A分解产生B、氧气和D，B可溶于水，D不溶于水，因此A是高锰酸钾，B是锰酸钾，D是二氧化锰，则G是过氧化氢，带入验证符合转化关系，因此：<br />（1）E是氢气，F是水，G是过氧化氢；故填：H<SUB>2</SUB>；H<SUB>2</SUB>O；H<SUB>2</SUB>O<SUB>2</SUB>；<br />（2）反应①是高锰酸钾分解产生锰酸钾、二氧化锰和氧气，反应②是氢气和氧气点燃产生水；故填：2KMnO<SUB>4</SUB> <span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>K<SUB>2</SUB>MnO<SUB>4</SUB>+MnO<SUB>2</SUB>+O<SUB>2</SUB>↑；2H<SUB>2</SUB>+O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O；<br />（3）反应①是高锰酸钾分解产生锰酸钾、二氧化锰和氧气，属于分解反应；反应②是氢气和氧气点燃产生水，属于化合反应；反应③是过氧化氢分解产生水喝氧气，反应④是谁通电产生氢气和氧气，属于分解反应；故填：3．','【分析】根据F是自然界中最常见的液体，能够通电分解产生C和E，C、E都是气体，D在反应③中作催化剂使用，可以产生C，因此F是水，C是氧气，E是氢气；A分解产生B、氧气和D，B可溶于水，D不溶于水，因此A是高锰酸钾，B是锰酸钾，D是二氧化锰，则G是过氧化氢，带入验证解答该题．','书写',3.00,'d523db5f6d5553b2f395bc3adbd7df4f',9,400,'物质的鉴别、推断,分解反应及其应用,书写化学方程式、文字表达式、电离方程式','',0,'37','',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840449,'氢氧化钠（NaOH）是一种腐蚀性很强的碱，下列有关说法正确的是（　　）','由钠、氢、氧三个元素组成','氢元素的化合价为-1','氢、氧元素质量比为1：16','农业上可用于改良酸性土壤','','C','【解答】解：A．元素是个宏观概念，只讲种类、不讲个数，故错误；<br />B．在化合物中，氢元素通常显+1价，故错误；<br />C．氢氧化钠（NaOH）中，氢元素与氧元素的质量比为：1：16，故正确；<br />D．氢氧化钠具有强烈的腐蚀性，不能用来改良酸性土壤，故错误．<br />故选C．','【分析】A．根据元素的规定来分析；<br />B．根据元素的化合价来分析；<br />C．根据化合物元素的质量比来分析；<br />D．根据物质的性质来分析．','选择题',3.00,'31759d1ce1bbded5d3e6a7790673a1b3',9,400,'化学式的书写及意义,有关元素化合价的计算,元素质量比的计算','',2016,'32','2016•长春模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840452,'某学习小组围绕“气体的实验室制取”进行研讨，请你参与完成下面的问题．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao24/190d2421-94d4-11e9-9ee1-b42e9921e93e_xkb1.png\" style=\"vertical-align:middle\" /><br />（1）图①装置制取气体的适用条件是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）图③装置相比图②装置的主要优点是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）实验室不宜用镁来代替锌与稀H<SUB>2</SUB>SO<SUB>4</SUB>反应制取H<SUB>2</SUB>，主要原因是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（4）实验室若用图④装置收集O<SUB>2</SUB>时，气体应从<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>端通入．（填字母）','','','','','','反应物是固体，反应需要加热$###$能随时控制反应的开始和结束$###$镁与稀盐酸反应速率太快，不利于实验操作（或“镁与稀盐酸反应速率太快，不利于气体的收集”）$###$a','【解答】解：<br />（1）图①装置属于故固加热型，制取气体的适用条件是反应物是固体，反应需要加热；<br />（2）图③装置相比图②装置的主要优点是能随时控制反应的开始和结束；<br />（3）实验室不宜用镁来代替锌与稀H<SUB>2</SUB>SO<SUB>4</SUB>反应制取H<SUB>2</SUB>，主要原因是镁与稀盐酸反应速率太快，不利于实验操作&nbsp;（或“镁与稀盐酸反应速率太快，不利于气体的收集”）；<br />（4）氧气的密度比空气的大，实验室若用图④装置收集O<SUB>2</SUB>时，气体应从a进．<br />答案：<br />（1）反应物是固体，反应需要加热；<br />（2）能随时控制反应的开始和结束<br />（3）镁与稀盐酸反应速率太快，不利于实验操作&nbsp;（或“镁与稀盐酸反应速率太快，不利于气体的收集”）；&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<br />（4）a．','【分析】（1）根据图①装置属于故固加热型，制取气体的适用条件是反应物是固体，反应需要加热解答；<br />（2）根据装置特点分析解答；<br />（3）根据镁与稀盐酸反应速率太快解答；<br />（4）根据氧气的密度比空气的大解答．','填空题',3.00,'6be08c8a12bf3d9397cfd3ec59516878',9,400,'实验室制取气体的思路,常用气体的发生装置和收集装置与选取方法','',2016,'32','2016•临沂模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840455,'善于归纳知识，利于培养素质．下面是小青同学对初中所学部分化学知识的归纳，其中有错误的是<br /><table class=\"edittable\"><TBODY><TR><td width=32 rowSpan=2>A</TD><td width=344>安全常识</TD><td width=41 rowSpan=2>B</TD><td width=253>物质鉴别</TD></TR><TR><td>①使用煤炉子取暖-注意室内通风，防止一氧化碳中毒<br />②浓硫酸沾到皮肤上-立即用水冲洗</TD><td>①食盐与纯碱-加入食醋<br />②氧气与二氧化碳-用燃着的木条</TD></TR><TR><td rowSpan=2>C</TD><td>实验误差分析</TD><td rowSpan=2>D</TD><td>实验操作顺序</TD></TR><TR><td>①实验室配制一定溶质质量分数溶液，量取水的体积时仰视读数，使实验结果偏大<br />②测定盐酸溶液pH值时先用水将pH试纸湿润后再滴入试液，所得pH值偏大</TD><td>①先检验装置气密性，再加药品<br />②加热固体时先均匀加热，再固定加热</TD></TR></TBODY></TABLE>（　　）','A','B','C','D','','C','【解答】解；A．一氧化碳俗称煤气，通风可防止一氧化碳中毒，浓硫酸沾到皮肤上，要先用大量的水冲洗然后涂上3%-5%的硼酸溶液，故正确；<br />B．加入食醋后产生气泡的是纯碱，无明显变化的是食盐；将燃着的木条伸入集气瓶内，能使木条燃烧更旺盛的是氧气，木条立即熄灭的则是二氧化碳，故正确；<br />C．实验室配制一定溶质质量分数溶液，量取水的体积时仰视读数，量取的水的体积偏大，造成所配制的溶液的浓度偏小，使实验结果偏小；故错误；<br />D．先检验装置气密性，再加药品，以防装置漏气；加热固体时先均匀加热，再固定加热，以防仪器受热不均破裂，故正确．<br />故选C．','【分析】A．根据安全知识来分析；<br />B．根据物质的鉴别方法来分析；<br />C．根据实验的误差分析来解答；<br />D．根据实验的步骤来分析．','选择题',3.00,'24595ce023d6033b48346fd6fbd62647',9,400,'给试管里的固体加热,一定溶质质量分数的溶液的配制,检查装置的气密性,常见的意外事故的处理方法,溶液的酸碱度测定,一氧化碳的毒性,物质的鉴别、推断','',2015,'33','2015秋•滨州期末',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840456,'（1）生活中蕴藏科学，留心处皆为学问．请用序号填空．<br />①NaCl、②CaO、③NH<SUB>4</SUB>H<SUB>2</SUB>PO<SUB>4</SUB>、④C<SUB>6</SUB>H<SUB>12</SUB>O<SUB>6 </SUB>⑤CH<SUB>4</SUB><br /><u>&nbsp;&nbsp;&nbsp;&nbsp;</u>与水反应放热可以煮鸡蛋；<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>为最简单的有机物；<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>可作复合肥料；<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>是人体血糖中的“糖”；<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>可用作生活中可用作调味品；．<br />（2）氢氧化钾是我国古代纺织业常用做漂洗的洗涤剂，古人将贝壳（主要成分是碳酸钙）燃烧后的固体（主要成分是氧化钙）与草木灰（主要成分是碳酸钾）在水中相互作用，就生成了氢氧化钾．请按要求化学方程式表达上述反应．<br />①分解反应：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />②化合反应：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />③复分解反应：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','②$###$⑤$###$③$###$④$###$①$###$CaCO<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>CaO+CO<SUB>2</SUB>↑$###$CaO+H<SUB>2</SUB>O=Ca（OH）<SUB>2</SUB>$###$K<SUB>2</SUB>CO<SUB>3</SUB>+Ca（OH）<SUB>2</SUB>=CaCO<SUB>3</SUB>↓+2KOH','【解答】解：（1）生石灰和水反应生成氢氧化钙，该反应放出大量的热，与水反应放热可以煮鸡蛋；<br />最简单的有机物是甲烷；复合肥是指含有氮、磷、钾三种元素中两种或两种以上的化肥，NH<SUB>4</SUB>H<SUB>2</SUB>PO<SUB>4</SUB>中含有氮和磷两种元素；食物中的糖类，主要是淀粉，在体内酶的催化作用下，最终水解成葡萄糖，葡萄糖通过毛细血管进入血液后即称血糖；食盐有咸味，生活中可用作调味品；<br />（2）①碳酸钙灼烧时可以分解生成氧化钙和二氧化碳，根据分解反应的特点“一变多”，可以知道，该反应属于分解反应，故答案为：CaCO<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>CaO+CO<SUB>2</SUB>↑<br />②氧化钙遇水会化合生成氢氧化钙，根据化合反应的特点“多变一”可知，该反应属于化合反应，故答案为：CaO+H<SUB>2</SUB>O=Ca（OH）<SUB>2</SUB><br />③两种化合物相互交换成分生成两种新的化合物：氢氧化钙与碳酸钾发生复分解反应生成碳酸钙沉淀和氢氧化钾，故答案为：K<SUB>2</SUB>CO<SUB>3</SUB>+Ca（OH）<SUB>2</SUB>=CaCO<SUB>3</SUB>↓+2KOH故答案为：<br />（1）②⑤③④①④；<br />（2）①CaCO<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>CaO+CO<SUB>2</SUB>↑；②CaO+H<SUB>2</SUB>O=Ca（OH）<SUB>2 </SUB>；③K<SUB>2</SUB>CO<SUB>3</SUB>+Ca（OH）<SUB>2</SUB>=CaCO<SUB>3</SUB>↓+2KOH','【分析】（1）根据血糖的定义判断．根据氧化钙和水反应是放热反应判断．根据复合肥的定义判断．根据食盐的性质判断；最简单的有机物是甲烷解答；<br />（2）根据教材中学习的知识，碳酸钙灼烧时可以分解生成氧化钙和二氧化碳，而氧化钙遇水会化合生成氢氧化钙，氢氧化钙与碳酸钾发生复分解反应生成碳酸钙沉淀和氢氧化钾．','书写',3.00,'868b538a6df09b4c1bf513b3bcdc7548',9,400,'生石灰的性质与用途,常用盐的用途,盐的化学性质,常见化肥的种类和作用,甲烷、乙醇等常见有机物的性质和用途,书写化学方程式、文字表达式、电离方程式','启东市',2016,'37','2016•启东市一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840457,'过氧化氢又称双氧水，化学式（H<SUB>2</SUB>O<SUB>2</SUB>），下列关于该物质说法中错误的是（　　）','它属于纯净物、化合物、氧化物','它是由过氧化氢分子构成','它由氢氧两种元素组成','它由氢分子和氧分子构成','','D','【解答】解：A．由过氧化氢的化学式可知，H<SUB>2</SUB>O<SUB>2</SUB>是由氢、氧两种元素组成的纯净物，属于化合物，属于氧化物，故正确；<br />B．过氧化氢由过氧化氢分子构成的，故正确；<br />C．根据过氧化氢的分子式H<SUB>2</SUB>O<SUB>2</SUB>，可知过氧化氢由氢元素和氧元素组成；故正确；<br />D．过氧化氢由过氧化氢分子构成的，里面没有氢分子和氧分子，故错误．<br />故选D．','【分析】根据化学式的意义进行分析：①宏观意义：a．表示一种物质； b．表示该物质的元素组成； ②微观意义：a．表示该物质的一个分子； b．表示该物质的分子构成；据此进行分析判断．','选择题',3.00,'ad79da5688f59763f900a56f3f8e8aaf',9,400,'从组成上识别氧化物,纯净物和混合物的判别,单质和化合物的判别,化学式的书写及意义','',2016,'35','2016春•渠县校级期中',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840458,'下列物质中不存在分子这种微粒的是（　　）','氯化钠','氮气','二氧化碳','氢氧化钠溶液','','A','【解答】解：A、氯化钠是由钠离子和氯离子构成的，故选项正确．<br />B、氮气属于气态非金属单质，是由氮分子构成的，故选项错误．<br />C、二氧化碳是由二氧化碳分子构成的，故选项错误．<br />D、氢氧化钠溶液是氢氧化钠的水溶液，氢氧化钠是由钠离子和氢氧根离子构成的，水是由水分子构成的，故选项错误．<br />故选：A．','【分析】根据金属、大多数固态非金属单质、稀有气体等由原子构成；有些物质是由分子构成的，气态的非金属单质和由非金属元素组成的化合物，如氢气、水等；有些物质是由离子构成的，一般是含有金属元素和非金属元素的化合物，如氯化钠，进行分析判断即可．','选择题',3.00,'1ea6ae72b24ac44902d4c9bbbc29a4fd',9,400,'物质的构成和含量分析','',2016,'32','2016•濮阳校级模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840459,'<img  src=\"/tikuimages/9/2014/400/shoutiniao6/191b06cf-94d4-11e9-a83d-b42e9921e93e_xkb88.png\" style=\"vertical-align:middle;FLOAT:right;\" />（2014秋•宁县月考）&nbsp;根据如图写出下列仪器名称：A<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>B<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>C<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>D<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','试管$###$铁架台$###$酒精灯$###$集气瓶','【解答】解：根据图中所标，可知：A是试管，B是铁架台，C是酒精灯，D是集气瓶．<br />故答案为：试管；铁架台；酒精灯；集气瓶．','【分析】熟悉常用实验仪器，了解名称．','填空题',3.00,'1a3df74a750cf21995f721b37d6431f8',9,400,'常用仪器的名称和选用','',2014,'37','2014秋•宁县月考',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840465,'<img  src=\"/tikuimages/9/2016/400/shoutiniao34/1946839e-94d4-11e9-a853-b42e9921e93e_xkb32.png\" style=\"vertical-align:middle;FLOAT:right\" />（2016•昆明一模）水是最重要的自然资源，也是生命之源．<br />（1）为了人类和社会经济的吋持续发展，我们必须爱护水资源，节约用水是每个公民<br />的责任和义务．下列做法属于节约用水的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（选填序号）<br />A．城市生活废水直接排入河流&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; B．农业上合理使用化肥和农药<br />C．工业上冷却水循环使用&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; D．用水后及时关闭水龙头<br />（2）饮用酸碱度过大或硬度过大的水都不利于人体健康，在实验室里可用<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>测定水的酸碱度；在生活中可用<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>区分硬水和软水，可通过<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>的方法来降低水的硬度．<br />（3）用如图装置进行电解水的实验，发生反应的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，该实验说明水是由<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />组成的．<br />（4）高铁酸钾（K<SUB>2</SUB>FeO<SUB>4</SUB>）是-种新型、高效的多功能水处理剂高铁酸钾受热时发生的反应为4K<SUB>2</SUB>FeO<SUB>4</SUB>═2R+4K<SUB>2</SUB>O+3O<SUB>2</SUB>↑，则R是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填化学式）．<br />（5）请举一例生活中节约用水的具体措施<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','CD$###$pH试纸$###$肥皂水$###$加热煮沸$###$2H<SUB>2</SUB>O<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;通电&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>↑+O<SUB>2</SUB>↑$###$氢元素和氧元素组成的$###$Fe<SUB>2</SUB>O<SUB>3</SUB>$###$洗菜水浇花','【解答】解：（1）A．城市生活废水直接排入河流会污染水，不属于节约用水；&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<br />B．农业上合理使用化肥和农药可以防止水污染，不属于节约用水；<br />C．工业上冷却水循环使用属于节约用水；&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<br />D．用水后及时关闭水龙头属于节约用水．<br />故填：CD．<br />（2）用pH试纸测定溶液的酸碱度，硬水中含有较多的可溶性的Ca<SUP>2+</SUP>和Mg<SUP>2+</SUP>，向水中加入肥皂水时，如果产生的泡沫较多，是软水，如果产生大量浮渣，是硬水；&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 加热煮沸时，水中的钙离子、镁离子能够以碳酸钙沉淀、氢氧化镁沉淀的形式从水中析出，从而降低水的硬度．<br />故填：pH试纸；肥皂水；加热煮沸．<br />（3）电解水时，正极产生的是氧气，负极产生的是氢气，该实验说明水是由氢元素和氧元素组成的，电解水的化学方程式为：2H<SUB>2</SUB>O<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;通电&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>↑+O<SUB>2</SUB>↑；故填：2H<SUB>2</SUB>O<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;通电&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>↑+O<SUB>2</SUB>↑；氢元素和氧元素组成的；<br />（4）由质量守恒定律：反应前后，原子种类、数目均不变，由反应的化学方程式4K<SUB>2</SUB>FeO<SUB>4</SUB>=2R+4K<SUB>2</SUB>O+3O<SUB>2</SUB>↑，反应前钾、铁、氧原子个数分别为8、4、16，反应后的生成物中钾、铁、氧原子个数分别为8、0、10，根据反应前后原子种类、数目不变，则2R分子中含有4个铁原子和6个氧原子，则每个R分子由2个铁原子和3个氧原子构成，则物质R的化学式为Fe<SUB>2</SUB>O<SUB>3</SUB>；故答案为：Fe<SUB>2</SUB>O<SUB>3</SUB>；<br />（5）生活中节约用水的具体措施很多，例如洗菜水浇花、洗衣水冲厕等．故填：洗菜水浇花．','【分析】（1）我国水资源匮乏，节约用水、防止水污染具有重要意义；<br />（2）根据用pH试纸测定溶液的酸碱度，利用肥皂水可以区分硬水和软水，加热煮沸可以降低水的硬度；<br />（3）根据电解水时，正极产生的是氧气，负极产生的是氢气进行解答；<br />（4）根据由质量守恒定律：反应前后，原子种类、数目均不变，据此由反应的化学方程式推断生成物R的化学式进行解答；<br />（5）所谓节约用水，就是要提高水的利用效益．使用新技术、改革工艺和改变习惯可以减少大量农业和生活用水．','填空题',3.00,'27159fdf4d968f6cfdf26eed482fdf6c',9,400,'电解水实验,硬水与软水,质量守恒定律及其应用,保护水资源和节约用水','',2016,'37','2016•昆明一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840466,'根据金属活动性顺序Al＞Fe＞Cu＞Ag判断，下列说法错误的是（　　）','反应Cu+FeSO<SUB>4</SUB>═Fe+CuSO<SUB>4</SUB>可以发生，属于置换反应','反应Fe+2AgNO<SUB>3</SUB>═2Ag+Fe（NO<SUB>3</SUB>）<SUB>2</SUB>可以发生，属于置换反应','反应Cu+2AgNO<SUB>3</SUB>═2Ag+Cu（NO<SUB>3</SUB>）<SUB>2</SUB>可以发生，溶液颜色由无色变成蓝色','反应2Al+3CuSO<SUB>4</SUB>═3Cu+Al<SUB>2</SUB>（SO<SUB>4</SUB>）<SUB>3</SUB>可以表面有红色物质析出，溶液由蓝色变无色','','A','【解答】解：A、由于金属活动性Fe＞Cu，所以反应Cu+FeSO<SUB>4</SUB>═Fe+CuSO<SUB>4</SUB>不可以发生．<br />B、由于金属活动性Fe＞Ag，反应Fe+2AgNO<SUB>3</SUB>═2Ag+Fe（NO<SUB>3</SUB>）<SUB>2</SUB>可以发生，反应物是一种单质和一种化合物．生成物是一种单质和一种化合物．属于置换反应．<br />C、由于金属活动性Cu＞Ag，反应Cu+2AgNO<SUB>3</SUB>═2Ag+Cu（NO<SUB>3</SUB>）<SUB>2</SUB>可以发生，由于硝酸银溶液为无色，硝酸铜溶液为蓝色，所以溶液颜色由无色变成蓝色．<br />D、由于金属活动性Al＞Cu，反应2Al+3CuSO<SUB>4</SUB>═3Cu+Al<SUB>2</SUB>（SO<SUB>4</SUB>）<SUB>3</SUB>可以发生，由于硫酸铜溶液为蓝色，硫酸铝溶液为无色，铜单质为红色，所以表面有红色物质析出，溶液由蓝色变无色．<br />故选：A．','【分析】在金属活动性顺序中，只有排在氢前的金属可以和稀酸溶液反应生成氢气，只有前面的金属可以把排在它后面的金属从它的盐溶液中置换出来，而且越靠前的金属活动性越强．','选择题',3.00,'6e6bed3ed907358588550a1fc595401d',9,400,'金属活动性顺序及其应用,置换反应及其应用','',2015,'32','2015•成都模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840467,'<img  src=\"/tikuimages/9/2016/400/shoutiniao60/194e72e1-94d4-11e9-bf06-b42e9921e93e_xkb20.png\" style=\"vertical-align:middle;FLOAT:right;\" />（2016•丰台区一模）同学们查阅资料，镁是一种化学性质非常活泼的金属，可以在多种气体中燃烧．同学们用下图所示装置分别进行镁条在二氧化碳和氮气中燃烧的实验．A中观察到镁冒出淡黄色的烟，并闻到刺激性气味．同学们对两个反应进行了探究．<br /><br />[查阅资料]<br />①氧化镁、碳酸镁均可以和盐酸反应．<br />②氮化镁（Mg<SUB>3</SUB>N<SUB>2</SUB>），淡黄色固体，无味，遇水产生白色沉淀氢氧化镁，并产生有刺激<br />性气味的氨气（NH<SUB>3</SUB>）．<br />实验1&nbsp;&nbsp;&nbsp;探究A中生成黑色固体和白烟的成分．<br />[猜想与假设]1黑色固体是炭．<br />2白烟可能是MgO、MgCO<SUB>3</SUB>或<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />[进行实验]<br /><table class=\"edittable\"><TBODY><TR><td width=197>实验操作</TD><td width=142>实验现象</TD><td width=231>实验结论</TD></TR><TR><td>①取A中生成的固体于试管中，加入适量稀盐酸</TD><td>白色固体消失，无色气泡生成．</TD><td><u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．</TD></TR><TR><td>②将①反应后的剩余物过滤，将滤渣洗涤干燥后在盛有氧气的集气瓶中点燃，熄灭后，向瓶中倒入澄清石灰水</TD><td>黑色固体燃烧，发白光，放热，澄清石灰水变浑浊．</TD><td>黑色固体是炭．<br />写出该实验中澄清石灰水发生反应的化学方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．</TD></TR></TBODY></TABLE>实验2&nbsp;&nbsp;&nbsp;探究B中闻到刺激性气味的原因．<br />同学们根据查阅的资料分析，镁条在氮气中燃烧闻到刺激性气味的原因与集气瓶中的水有关．请你在已有实验的基础上，设计一个实验方案证明同学们的猜想正确．<br /><u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />[反思与评价]<br />通过探究，同学们对燃烧和灭火有了新的认识<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','MgO和MgCO<SUB>3</SUB>$###$白色固体是碳酸镁$###$Ca（OH）<SUB>2</SUB>+CO<SUB>2</SUB>═CaCO<SUB>3</SUB>↓+H<SUB>2</SUB>O$###$将点燃镁条伸入盛有氮气的集气瓶中，没有闻到刺激性气味$###$燃烧不一定需要氧气','【解答】解：实验1【猜想与假设】依据质量守恒定律进行猜想，白烟可能是氧化镁、碳酸镁或氧化镁、碳酸镁的混合物；<br />【进行实验】碳酸镁和盐酸反应会生成二氧化碳气体，二氧化碳和氢氧化钙反应会生成碳酸钙沉淀，所以<br /><table class=\"edittable\"><TBODY><TR><td width=197>实验操作</TD><td width=142>实验现象</TD><td width=231>实验结论</TD></TR><TR><td>①取A中生成的固体于试管中，加入适量稀盐酸</TD><td>白色固体消失，无色气泡生成．</TD><td>白色固体是碳酸镁</TD></TR><TR><td>②将①反应后的剩余物过滤，将滤渣洗涤干燥后在盛有氧气的集气瓶中点燃，熄灭后，向瓶中倒入澄清石灰水</TD><td>黑色固体燃烧，发白光，放热，澄清石灰水变浑浊．</TD><td>黑色固体是炭．<br />写出该实验中澄清石灰水发生反应的化学方程式<br />Ca（OH）<SUB>2</SUB>+CO<SUB>2</SUB>═CaCO<SUB>3</SUB>↓+H<SUB>2</SUB>O</TD></TR></TBODY></TABLE>实验2&nbsp;为了验证镁条在氮气中燃烧闻到刺激性气味的原因与集气瓶中的水有关，所以可以设计实验：将点燃镁条伸入盛有氮气的集气瓶中，没有闻到刺激性气味；<br />【反思与评价】金属镁可以在二氧化碳或氮气中燃烧，所以对燃烧产生的新的认识是：燃烧不一定需要氧气．<br />故答案为：实验1【猜想与假设】MgO和MgCO<SUB>3</SUB>；<br />【进行实验】白色固体是碳酸镁；<br />Ca（OH）<SUB>2</SUB>+CO<SUB>2</SUB>═CaCO<SUB>3</SUB>↓+H<SUB>2</SUB>O；<br />实验2将点燃镁条伸入盛有氮气的集气瓶中，没有闻到刺激性气味；<br />【反思与评价】燃烧不一定需要氧气．','【分析】实验1【猜想与假设】根据质量守恒定律进行猜想，白烟可能是氧化镁、碳酸镁或氧化镁、碳酸镁的混合物进行分析；<br />【进行实验】根据碳酸镁和盐酸反应会生成二氧化碳气体，二氧化碳和氢氧化钙反应会生成碳酸钙沉淀进行分析；<br />实验2&nbsp;根据控制变量法的具体操作进行分析；<br />【反思与评价】根据金属镁可以在二氧化碳或氮气中燃烧进行分析．','书写',3.00,'ac883a98a386f48ee9817fc2a1d301a8',9,400,'实验探究物质的组成成分以及含量,酸的化学性质,碳的化学性质,书写化学方程式、文字表达式、电离方程式,燃烧与燃烧的条件','',2016,'37','2016•丰台区一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840469,'如图所示写出仪器名称：<br /><img  src=\"/tikuimages/9/2014/400/shoutiniao35/19532dcf-94d4-11e9-8b0e-b42e9921e93e_xkb11.png\" style=\"vertical-align:middle\" /><br />A．<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>&nbsp;B．<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>&nbsp;C．<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>&nbsp;D．<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','酒精灯$###$烧杯$###$滴瓶$###$试管刷','【解答】解：图中A是酒精灯，B是烧杯，C是滴瓶，D是试管刷．<br />故答案为：酒精灯；烧杯；滴瓶；试管刷．','【分析】熟悉常见仪器，了解它们的名称．','填空题',3.00,'2f568229ddc21687883d7f94cc6ba8ce',9,400,'常用仪器的名称和选用','',2014,'37','2014秋•宁县月考',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840471,'化学符号有的可以①表示一种元素&nbsp; ②表示一个原子&nbsp; ③表示一种单质．下列符号不具有上述三种意义的是<br />A．Cu&nbsp;&nbsp;&nbsp; B．N&nbsp;&nbsp;&nbsp; C．S&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; D．<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','H','【解答】解：A、因Cu是由原子直接构成的，则可表示一种元素、一个原子、一种单质，故A不选；<br />B、因元素符号N可表示一种元素、一个氮原子，但氮气是由分子构成的，且1个分子中含有2个原子，则符号N不表示单质，故B选；<br />C、因元素符号S可表示一种元素、一个原子，单质可用化学式“S”来表示，故C不选；<br />D、因元素符号H可表示一种元素、一个氢原子，但氢气是由分子构成的，且1个分子中含有2个原子，则符号H不表示单质．<br />故选B，补充：H．','【分析】根据元素符号可表示1个原子和一种元素，由原子直接构成的物质，其元素符号为物质的化学式，则可表示一种物质来分析解答．','书写',3.00,'a40347708ab59cd4a9c3dfc7b17281ee',9,400,'元素的符号及其意义,化学式的书写及意义','',2016,'32','2016•宜春模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840473,'下列说法不正确的是（　　）','白磷在空气中燃烧会产生大量的白烟','打开盛有浓盐酸的试剂瓶盖，瓶口会出现白雾','一氧化碳通过灼热氧化铁，固体由红色变为黑色','水通电后，与正极相连的电极上产生的气体是氢气','','D','【解答】解：A、白磷在空气中燃烧，产生大量的白烟，故A正确；<br />B、浓盐酸具有挥发性，与空气中的水蒸气结合成盐酸小液滴，所以会看到白雾，故B正确；<br />C、一氧化碳和氧化铁在高温的条件下生成铁和二氧化碳，固体由红色变为黑色，故C正确；<br />D、水通电后，带正电的氢离子向负极运动，带负电的氧离子向正极运动，所以与正极相连的电极上产生的气体是氧气，故D错误．<br />故选：D．','【分析】A、根据白磷在空气中燃烧的现象进行分析判断；<br />B、根据浓盐酸具有挥发性进行分析；<br />C、根据一氧化碳和氧化铁在高温的条件下生成铁和二氧化碳进行分析；<br />D、根据水通电后，带正电的氢离子向负极运动，带负电的氧离子向正极运动进行分析．','选择题',3.00,'c3c56a1c4d4937f571879e0d0fb27d6e',9,400,'氧气与碳、磷、硫、铁等物质的反应现象,电解水实验,一氧化碳还原氧化铁,酸的物理性质及用途','',2016,'37','2016•玄武区一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840475,'下列实验操作先后顺序正确的是（　　）','蒸发食盐水时，先在铁圈上放置蒸发皿，后放酒精灯','如果不慎将碱液沾到皮肤上，先涂上硼酸液，后用大量水冲洗','测定溶液pH，先将试纸湿润，后用干净的玻璃棒蘸取待测液点在试纸上','氯酸钾制氧气结束时，先停止加热，再把导管从盛有水的烧杯中取出','','D','【解答】解：A、蒸发食盐水时，先在放酒精灯，后铁圈上放置蒸发皿，故选项说法错误．<br />B、碱溶液沾到皮肤上，立即用大量水冲洗，再涂上硼酸溶液，故选项说法错误．<br />C、用pH试纸测定未知溶液的pH时，正确的操作方法为用玻璃棒蘸取少量待测液滴在干燥的pH试纸上，与标准比色卡对比来确定pH．不能将pH试纸伸入待测液中，以免污染待测液，故选项说法错误．<br />D、氯酸钾制氧气结束时，先停止加热，再把导管从盛有水的烧杯中取出，以防止水倒吸入试管，使试管因骤然冷却而炸裂，故选项说法正确．<br />故选：D．','【分析】A、根据蒸发操作的注意事项进行分析判断．<br />B、根据碱溶液不慎沾到皮肤上的处理方法，进行分析判断．<br />C、根据用pH试纸测定未知溶液的pH的方法进行分析判断．<br />D、根据实验室加热氯酸钾制取氧气的实验步骤、注意事项进行分析判断．','选择题',3.00,'e061b6a7768630e6d0b9d8bace7f20f8',9,400,'蒸发与蒸馏操作,常见的意外事故的处理方法,溶液的酸碱度测定,制取氧气的操作步骤和注意点','',2016,'37','2016•邵阳二模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840477,'A～F是实验室制取、收集、验证气体的常见装置图，请根据装置图回答．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao11/1965a45e-94d4-11e9-9f3c-b42e9921e93e_xkb47.png\" style=\"vertical-align:middle\" /><br />（1）写出图中仪器①的名称<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）若用A、D组合制取实验室的一种气体，发生装置中需装入的药品是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填化学式）．<br />（3）实验室用B、E、C装置制取并收集干燥的二氧化碳，则E中应盛放的试剂是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填名称）．<br />（4）图F中若X是石蕊试液，能观察到的现象是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．若X是澄清石灰水，写出发生反应的化学方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','长颈漏斗$###$KMnO<SUB>4</SUB>$###$浓硫酸$###$紫色石蕊试液变红$###$Ca（OH）<SUB>2</SUB>+CO<SUB>2</SUB>=CaCO<SUB>3</SUB>↓+H<SUB>2</SUB>O','【解答】解：（1）标号仪器是可随时添加液体的长颈漏斗；<br />（2）A装置是加热固体制取气体，且试管口有棉花，说明是用加热高锰酸钾的方法制取氧气，高锰酸钾的化学式是KMnO<SUB>4</SUB>；<br />（3）浓硫酸具有吸水性，可用来干燥二氧化碳，是一种常用的液体干燥剂，所以E中试剂是浓硫酸；<br />（4）二氧化碳可与水反应生成碳酸，碳酸可使紫色石蕊试液变红，二氧化碳与石灰水主要成分氢氧化钙反应生成碳酸钙沉淀和水，反应方程式是：Ca（OH）<SUB>2</SUB>+CO<SUB>2</SUB>=CaCO<SUB>3</SUB>↓+H<SUB>2</SUB>O；<br />故答案为：（1）长颈漏斗；&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;（2）KMnO<SUB>4</SUB>；&nbsp;&nbsp;&nbsp;&nbsp;（3）浓硫酸；（4）紫色石蕊试液变红；&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Ca（OH）<SUB>2</SUB>+CO<SUB>2</SUB>=CaCO<SUB>3</SUB>↓+H<SUB>2</SUB>O．','【分析】（1）据常用仪器回答；<br />（2）A装置是加热固体制取气体，且试管口有棉花，说明是用加热高锰酸钾的方法制取氧气，据此解答；<br />（3）浓硫酸具有吸水性，可用来干燥二氧化碳；<br />（4）二氧化碳可与水反应生成碳酸，碳酸可使紫色石蕊试液变红，二氧化碳与石灰水主要成分氢氧化钙反应生成碳酸钙沉淀和水，据此书写方程式．','书写',3.00,'42b54a1e4dbe4489d2f8b4486cc5f8b9',9,400,'常用气体的发生装置和收集装置与选取方法,实验室制取氧气的反应原理,二氧化碳的实验室制法,制取二氧化碳的操作步骤和注意点,二氧化碳的化学性质,书写化学方程式、文字表达式、电离方程式','',2016,'35','2016春•宜宾期中',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840479,'味精是常见调味料之一，主要成分是谷氨酸钠[化学式为C<SUB>5</SUB>H<SUB>8</SUB>NO<SUB>4</SUB>Na]，熔点232℃，20℃时溶解度71.7g．下列有关叙述错误的是（　　）','谷氨酸钠相对分子质量为169','谷氨酸钠是由5个碳原子、8个氢原子、4个氧原子、1个氮原子和1个钠原子构成','20℃时，100g水最多可配制成171.7g谷氨酸钠的饱和溶液','谷氨酸钠中钠元素的质量分数约为13.6%','','B','【解答】解：A、谷氨酸钠相对分子质量为12×5+1×8+14+16×4+23=169，故选项说法正确．<br />B、一个谷氨酸钠分子是由5个碳原子、8个氢原子、4个氧原子、1个氮原子和1个钠原子构成，故选项说法错误．<br />C、20℃溶解度71.7g，则20℃时100g水最多溶解71.7g谷氨酸钠，20℃时100g水最多可配制成171.7g谷氨酸钠的饱和溶液，故选项说法正确．<br />D、谷氨酸钠中钠元素的质量分数为<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">23</td></tr><tr><td>12×5+1×8+14+16×4+23</td></tr></table></span>×100%≈13.6%，故选项说法正确．<br />故选：B．','【分析】A、根据相对分子质量的计算方法进行分析判断．<br />B、根据物质化学式的意义进行分析．<br />C、根据题意，20℃溶解度71.7g，进行分析判断．<br />D、根据化合物中元素的质量分数公式进行分析判断．','选择题',3.00,'94aea610e462129b1909eeea42548cd1',9,400,'固体溶解度的概念,化学式的书写及意义,相对分子质量的概念及其计算,元素的质量分数计算','江阴市',2016,'35','2016春•江阴市校级期中',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840480,'锌对人体有很重要的作用，能促进人体的生长发育、增进人体正常食欲、增强人体免疫力、促进伤口和创伤的愈合．硫酸锌常用作补锌药，可作为食品锌强化剂的原料．工业上常用菱锌矿生产硫酸锌，菱锌矿的主要成分是ZnCO<SUB>3</SUB>，并含少量的Fe<SUB>2</SUB>O<SUB>3</SUB>、FeCO<SUB>3</SUB>、MgO等，工艺流程简化示意图如下：<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao56/1972eacf-94d4-11e9-92c6-b42e9921e93e_xkb9.png\" style=\"vertical-align:middle\" /><br />根据如图回答下列问题：<br />（1）将菱锌矿研磨成粉的目的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）请写出氢氧化锌和稀硫酸反应的化学方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）“滤液3”之前加入锌粉的目的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（4）操作a的名称是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，实验室进行该操作时用到玻璃棒的作用是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（5）儿童、青少年正处于生长发育期，需要摄入适量的锌．写出一种合理的补锌方法<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','菱锌矿粉与盐酸充分反应$###$Zn（OH）<SUB>2</SUB>+H<SUB>2</SUB>SO<SUB>4</SUB>=ZnSO<SUB>4</SUB>+2H<SUB>2</SUB>O$###$除去剩余的硫酸$###$蒸发或蒸发结晶$###$防止局部温度过高造成液体飞溅$###$多吃含锌的食物','【解答】解：（1）将菱锌矿研磨成粉可以增加反应时的接触面积，使得反应更加充分；故填：菱锌矿粉与盐酸充分反应；<br />（2）氢氧化锌与硫酸反应生成硫酸锌和水，化学方程式为Zn（OH）<SUB>2</SUB>+H<SUB>2</SUB>SO<SUB>4</SUB>=ZnSO<SUB>4</SUB>+2H<SUB>2</SUB>O，故填：Zn（OH）<SUB>2</SUB>+H<SUB>2</SUB>SO<SUB>4</SUB>=ZnSO<SUB>4</SUB>+2H<SUB>2</SUB>O；<br />（3）由于加入的硫酸是过量的，要得到纯净的硫酸锌溶液须除去过量硫酸，因此加入锌粉的目的是除去剩余的硫酸；故填：除去剩余的硫酸；<br />（4）滤液3是硫酸锌溶液，从溶液中得到溶质可采用蒸发结晶，然后干燥的方法；在蒸发结晶中，玻璃棒搅拌的目的是防止局部温度过高造成液体飞溅；故填：蒸发或蒸发结晶；防止局部温度过高造成液体飞溅；<br />（5）多吃含锌的食物如虾皮等可以补充锌元素，故填：多吃含锌的食物（其他合理答案均可）．','【分析】（1）根据影响化学反应速率的因素回答，固体表面积越大，反应速率越大；<br />（2）根据书写方程式原则写出方程式；<br />（3）根据生成物中只有硫酸锌分析；<br />（4）根据从溶液中制得干燥纯净的固体一般有蒸发、结晶、过滤后洗涤并干燥等操作来回答；<br />（5）根据获取锌元素的方法来分析．','书写',3.00,'8aa99ee94f3f19f83d5ea2c29c3acae9',9,400,'过滤的原理、方法及其应用,金属的化学性质,酸的化学性质,物质的相互转化和制备,书写化学方程式、文字表达式、电离方程式,微量元素、维生素与健康的关系及摄取方法','',2016,'37','2016•马鞍山二模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840481,'下列不属于氧气的用途是（　　）','气割','炼钢','致冷剂','供给呼吸','','C','【解答】解：A、利用氧气的助燃性，可以用于气割，正确；<br />B、利用氧气的助燃性，可以用于炼钢，正确；<br />C、氧气不能用于制冷剂，错误；<br />D、氧气能帮助呼吸，正确；<br />故选C．','【分析】物质的性质决定物质的用途，根据已有的氧气的助燃性和帮助呼吸的性质进行分析解答即可．','选择题',3.00,'fe95b63311881ca2193380169e00f764',9,400,'氧气的用途','',0,'37','',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840482,'育才中学化学社的同学欲用大理石（杂质不参与反应且难溶于水）和稀盐酸来制取二氧化碳，并回收提纯氯化钙．<br />实验一：制取二氧化碳&nbsp;&nbsp;&nbsp;<br />（1）根据下列装置回答问题．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao93/1978de40-94d4-11e9-95bf-b42e9921e93e_xkb49.png\" style=\"vertical-align:middle\" /><br />①装置A中发生的化学反应方程式是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />②若用C装置来收集二氧化碳，二氧化碳应从<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填序号）端通入；若用B装置应来制取二氧化碳，优点为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />实验二：分离提纯&nbsp;&nbsp;<br />（2）根据下面的流程图回答问题．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao12/197c87c0-94d4-11e9-95d7-b42e9921e93e_xkb23.png\" style=\"vertical-align:middle\" /><br />①两个步骤中都用到的玻璃仪器是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />②蒸发操作中，当观察到蒸发皿中出现<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>时，停止加热．','','','','','','CaCO<SUB>3</SUB>+2HCl=CaCl<SUB>2</SUB>+CO<SUB>2</SUB>↑+H<SUB>2</SUB>O$###$①$###$能控制反应的发生和停止$###$玻璃棒$###$较多固体时','【解答】解：（1）实验室制取CO<SUB>2</SUB>，是在常温下，用大理石或石灰石和稀盐酸制取的，碳酸钙和盐酸互相交换成分生成氯化钙和水和二氧化碳，因此不需要加热；二氧化碳能溶于水，密度比空气的密度大，因此只能用向上排空气法收集，用C装置来收集二氧化碳，应长进短出；装置B的优点是：可以控制反应的速度；<br />（2）过滤用玻璃棒引流，蒸发用玻璃棒不断搅拌，以防止局部温度过高而使液体飞溅；当加热至较多固体出现时，应停止加热利用余热蒸干．<br />故答案为：①CaCO<SUB>3</SUB>+2HCl=CaCl<SUB>2</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑；②①、可以控制反应的速；（2）①玻璃棒；②较多固体出现时．','【分析】（1）实验室制取CO<SUB>2</SUB>，是在常温下，用大理石或石灰石和稀盐酸制取的，碳酸钙和盐酸互相交换成分生成氯化钙和水和二氧化碳，因此不需要加热．二氧化碳能溶于水，密度比空气的密度大，因此只能用向上排空气法收集．装置C的优点是：可以控制反应的速度；<br />（2）过滤用玻璃棒引流，蒸发用玻璃棒不断搅拌，以防止局部温度过高而使液体飞溅；当加热至较多固体出现时，应停止加热利用余热蒸干．','书写',3.00,'72fbb5dd305b9afac1e1c130bf738c64',9,400,'混合物的分离方法,过滤的原理、方法及其应用,蒸发与蒸馏操作,二氧化碳的实验室制法,书写化学方程式、文字表达式、电离方程式','',2016,'32','2016•重庆校级模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840484,'下列关于物质的用途与性质，对应关系不正确的是（　　）','氧气用于火箭发射--可燃性','氩气能制霓虹灯--通电时能发出不同颜色的光','不锈钢制成炊具--导热耐腐蚀','二氧化碳可制碳酸饮料--能溶于水且与水反应','','A','【解答】解：A、氧气用于发射火箭，利用了氧气的助燃性，而不是可燃性．故A说法不正确．<br />B、稀有气体用于制霓虹灯，是由于稀有气体在通电时能发出不同颜色的光．故B说法正确．<br />C、不锈钢制成炊具--导热耐腐蚀．故C说法正确．<br />D、二氧化碳可制碳酸饮料--能溶于水且与水反应．故D说法正确．<br />故选：A．','【分析】主要运用氧气、二氧化碳、氮气、稀有气体的物理性质、化学性质和用途分析来解决．','选择题',3.00,'d9902b8d9883646a4fefbc8a8fdbb814',9,400,'常见气体的用途,生铁和钢','',2016,'37','2016•嘉祥县一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840485,'某净水丸分内外两层，对饮用水有杀菌消毒作用．使用时直接将它投入适量水中，外层的优氯净（C<SUB>3</SUB>O<SUB>3</SUB>N<SUB>3</SUB>Cl<SUB>2</SUB>Na）先与水反应，生成的次氯酸（HClO）具有杀菌消毒作用，几分钟后内层的亚硫酸钠（Na<SUB>2</SUB>SO<SUB>3</SUB>）会将水中的余氯（HClO等）除去．下列关于该净水丸的叙述不正确的是（　　）','优氯净由5种元素组成','Na<SUB>2</SUB>SO<SUB>3</SUB>中S元素的化合价为+4','经过净水丸处理过的水为纯净物','除去余氯的反应之一：Na<SUB>2</SUB>SO<SUB>3</SUB>+HClO=Na<SUB>2</SUB>SO<SUB>4</SUB>+HCl，该反应不属于复分解反应','','C','【解答】解：A．由优氯净（C<SUB>3</SUB>O<SUB>3</SUB>N<SUB>3</SUB>Cl<SUB>2</SUB>Na）的化学式可知，它是由碳、氧、氮、氯和钠五种元素组成的，故正确；<br />B．Na<SUB>2</SUB>SO<SUB>3</SUB>中钠元素显+1价，氧元素显-2价，根据化合物的代数和为零，设硫元素的化合价为x，则（+1）×2+x+（-2）×3=0，x=+4，故正确；<br />C．经过净水丸处理过的水中还含有可溶性杂质，属于混合物，故错误；<br />D．该反应不符合两种化合物相互交换成分生成两种新的化合物，不属于复分解反应，故正确．<br />故选C．','【分析】A．根据化学式的意义来分析；<br />B．根据化合物中元素的化合价计算方法来分析；<br />C．根据纯净物与混合物的概念来分析；<br />D．根据复分解反应的概念来分析．','选择题',3.00,'29ba8154b3e71e2759e226a796a080ba',9,400,'纯净物和混合物的判别,化学式的书写及意义,有关元素化合价的计算,反应类型的判定','',2016,'37','2016•房山区一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840488,'氢能是最清的能源是因为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；含碳物质的不充分燃烧会产生<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>污染空气；常言道“真金不怕火炼”其中蕴含的科学道理是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','无污染、燃烧值高、资源广泛$###$一氧化碳$###$金的化学性质稳定，高温下金不与氧气反应','【解答】解：氢气燃烧产物是水，不污染环境；氢气的燃烧值高；工业上制取氢气是通过电解水得到的，而地球上水资源丰富，可以从水中提取氢气，说明资源广泛；含碳物质不充分燃烧会产生一氧化碳等污染气体，会浪费燃料、污染空气；金的化学性质稳定，不易和其它物质发生反应，即便是在高温下金也不会与氧气反应，所以真金不怕火炼．<br />故答案为：无污染、燃烧值高、资源广泛；一氧化碳；金的化学性质稳定，高温下金不与氧气反应．','【分析】根据氢能源的优点，碳燃烧的产物，金的化学性质稳定即可作答．','填空题',3.00,'aaab3774dac2d77ce291cc647f6ee790',9,400,'金属的化学性质,完全燃烧与不完全燃烧,氢气的用途和氢能的优缺点','',2016,'32','2016•河南模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840495,'下列说法正确的是（　　）','使用催化剂只能使化学反应速率加快','化学反应前后，催化剂的质量和性质改变','硬水是化合物，软水是单质','氧化反应不一定就是化合反应','','D','【解答】解：A、催化剂能够改变化学反应速率，既可加快反应速率，也可以减慢反应速率，故选项说法错误．<br />B、催化剂在化学反应前后其质量和化学性质一定不变，故选项说法错误．<br />C、硬水和天然软水中都含有钙镁化合物和其它杂质，都是混合物，故说法错误．<br />D、氧化反应不一定是化合反应，如CH<SUB>4</SUB>+2O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>CO<SUB>2</SUB>+2H<SUB>2</SUB>O，属于氧化反应，但不是化合反应，故选项说法正确．<br />故选D．','【分析】在化学反应里能改变其他物质的化学反应速率，而本身的质量和化学性质在反应前后都没有发生变化的物质叫做催化剂（又叫触媒）．催化剂的特点可以概括为“一变二不变”，一变是能够改变化学反应速率，二不变是指质量和化学性质在化学反应前后保持不变．硬水和天然软水中都含有钙镁化合物和其它杂质．化合反应：两种或两种以上物质反应后生成一种物质的反应，其特点可总结为“多变一”；是物质与氧气发生的化学反应，属于氧化反应；据此进行分析判断即可．','选择题',3.00,'742940e4471a2dc826a964f6c7e417e9',9,400,'催化剂的特点与催化作用,硬水与软水,化合反应及其应用,氧化反应','',2011,'37','2011秋•蓝山县校级月考',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840499,'下列实验操作中，不正确的是（　　）','<img  src=\"/tikuimages/9/2016/400/shoutiniao24/19b216ae-94d4-11e9-8681-b42e9921e93e_xkb29.png\" style=\"vertical-align:middle\" /><br />稀释浓硫酸','<img  src=\"/tikuimages/9/2016/400/shoutiniao14/19b4d5cf-94d4-11e9-98cc-b42e9921e93e_xkb67.png\" style=\"vertical-align:middle\" /><br />蒸发食盐水','<img  src=\"/tikuimages/9/2016/400/shoutiniao11/19b8cd70-94d4-11e9-9c9a-b42e9921e93e_xkb87.png\" style=\"vertical-align:middle\" /><br />称量粗盐质量','<img  src=\"/tikuimages/9/2016/400/shoutiniao72/19b91b8f-94d4-11e9-a88c-b42e9921e93e_xkb2.png\" style=\"vertical-align:middle\" /><br />测雨水pH','','D','【解答】解：A、稀释浓硫酸时，要把浓硫酸缓缓地沿器壁注入水中，同时用玻璃棒不断搅拌，以使热量及时地扩散；一定不能把水注入浓硫酸中；图中所示操作正确．<br />B、蒸发时，应用玻璃棒不断搅拌，以防液体受热不均匀，造成液体飞溅，图中所示操作正确．<br />C、托盘天平的使用要遵循“左物右码”的原则，图中所示操作正确．<br />D、用pH试纸测定未知溶液的pH时，正确的操作方法为用玻璃棒蘸取少量待测液滴在干燥的pH试纸上，与标准比色卡对比来确定pH．不能将pH试纸伸入待测液中，以免污染待测液，图中所示操作错误．<br />故选：D．','【分析】A、根据浓硫酸的稀释方法（酸入水，沿器壁，慢慢倒，不断搅）进行分析判断．<br />B、根据蒸发操作的注意事项进行分析判断．<br />C、根据托盘天平的使用要遵循“左物右码”的原则进行分析判断．<br />D、根据用pH试纸测定未知溶液的pH的方法进行分析判断．','选择题',3.00,'d562a831cc98a59c08f646922449d7da',9,400,'称量器-托盘天平,浓硫酸的性质及浓硫酸的稀释,蒸发与蒸馏操作,溶液的酸碱度测定','',2016,'37','2016•玄武区一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840501,'下列家庭小实验不能成功的是（　　）','用燃烧的方法区别羊毛和涤纶','用铅笔芯代替石墨试验导电性','用食盐水浸泡菜刀除去表面的铁锈','用肥皂水区别软水和硬水','','C','【解答】解：A、羊毛的主要成分是蛋白质，燃烧时有烧焦羽毛的气味，而涤纶的主要成分是纤维素，燃烧时无此气味，所以能够成功；<br />B、铅笔芯的主要成分是石墨，所以可以用铅笔芯代替石墨试验导电性；<br />C、食盐的主要成分是氯化钠，氯化钠不能和水垢发生反应，所以不能用食盐除水垢；<br />D、生活中可用肥皂水来区分硬水和软水，产生泡沫较多的是软水，较少的硬水，所以可以鉴别；<br />故选项为：C．','【分析】A、根据羊毛和涤纶的成分和该成分的化学性质判断；<br />B、铅笔芯的主要成分是石墨；<br />C、食盐和水垢不能反应；<br />D、硬水含有较多的钙、镁离子等．','选择题',3.00,'a9e76ecee2fdc72949b02afd81de01e8',9,400,'化学实验方案设计与评价,硬水与软水,盐的化学性质,碳单质的物理性质及用途,棉纤维、羊毛纤维和合成纤维的鉴别','',2016,'37','2016•莒县一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840504,'某兴趣小组同学在探究碱溶液与酚酞作用的实验时，发现了一个意外现象：氢氧化钠溶液滴入酚酞试液中，溶液变成了红色，一会儿红色就消失了．<br />【猜想】可能与氢氧化钠溶液质量分数大小有关．<br />【实验设计一】<br /><table class=\"edittable\"><TBODY><TR><td width=193>实验方法</TD><td width=320>观察到的现象和结论</TD></TR><TR><td>分别配制不同质量分数的氢氧化钠溶液，然后各滴加2滴酚酞试液．</TD><td>质量分数小的溶液中红色不消失，质量分数大的溶液中红色会消失，则证明<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．</TD></TR></TBODY></TABLE>【反思与评价】<br />通过大家分析讨论，认为可能还有以下猜想：<br />猜想一、可能是酚酞变质的缘故；<br />猜想二、可能是氢氧化钠溶液与空气中二氧化碳反应的缘故；<br />猜想三、可能是酚酞与空气中氧气反应，使红色消失的缘故．&nbsp;…<br />【实验设计二】为证实猜想三，做如下实验，请完成下表．<br /><table class=\"edittable\"><TBODY><TR><td width=280>实验步骤</TD><td width=237>设计这一步骤的目的</TD></TR><TR><td>1．用煮沸过的蒸馏水配制氢氧化钠溶液．</TD><td><u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD></TR><TR><td>2．在氢氧化钠溶液中滴入酚酞，并在上<br />方滴一些植物油．</TD><td><u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD></TR></TBODY></TABLE>【理论分析】猜想一不对，理由是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />猜想二也不正确，理由是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />猜想三也不正确，理由是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','与NaOH溶液质量分数大小有关$###$排除水中溶解的O<SUB>2</SUB>$###$隔绝O<SUB>2 </SUB>$###$若酚酞试液变质，就不可能出现遇氢氧化钠溶液变红的现象$###$碳酸钠溶液显碱性，也可使酚酞试液变红$###$实验过程中试液一直与氧气接触','【解答】解：<br />【实验设计一】<br />分别配制不同质量分数的氢氧化钠溶液，然后各滴加2滴酚酞试液．质量分数小的溶液中红色不消失，质量分数大的溶液中红色会消失，则证明与NaOH溶液质量分数大小有关（是因为NaOH溶液质量分数过大）；<br />【实验设计二】据气体的溶解度随温度的升高而降低，那么将氢氧化钠溶液加热可以使溶解在氢氧化钠溶液中的氧气排除，同样的道理，在第二步中加入植物油的目的是利用植物油密度小于水，且空气几乎不溶于植物油，会浮在液面上方起到隔绝空气的作用，防止氧气再次溶于氢氧化钠溶液中；<br />【理论分析】<br />猜想一的猜想错误，因为若酚酞试剂变质，就不可能出现遇氢氧化钠溶液变红的现象；<br />猜想二提出：可能是氢氧化钠溶液与空气中的二氧化碳反应的缘故，这一说法明显不对，因为氢氧化钠与二氧化碳反应要生成碳酸钠即：2NaOH+CO<SUB>2</SUB>═Na<SUB>2</SUB>CO<SUB>3 </SUB>+H<SUB>2</SUB>O，而碳酸钠溶液也呈碱性，可以使酚酞变红；<br />猜想三也不正确，理由是实验过程中试液一直与氧气接触．<br />答案：<br />【实验设计一】与NaOH溶液质量分数大小有关（是因为NaOH溶液质量分数过大）<br />【实验设计二】排除水中溶解的O<SUB>2</SUB>；隔绝O<SUB>2</SUB>；<br />【理论分析】若酚酞试液变质，就不可能出现遇氢氧化钠溶液变红的现象；<br />碳酸钠溶液显碱性，也可使酚酞试液变红；<br />实验过程中试液一直与氧气接触．','【分析】【实验设计一】根据题中信息分析解答；<br />【实验设计二】根据气体的溶解度与温度有关，温度越高，溶解度越小；在氢氧化钠溶液上方滴一些植物油能将氢氧化钠溶液与空气隔绝解答；<br />【理论分析】<br />根据若酚酞试剂变质，遇氢氧化钠溶液不会变红解答；<br />根据氢氧化钠能与二氧化碳反应生成碳酸钠，生成的碳酸钠呈碱性，能使酚酞试液变红解答；<br />根据实验过程中试液一直与氧气接触解答．','填空题',3.00,'c62392812cbe58e47825a3cf2817162b',9,400,'实验探究物质变化的条件和影响物质变化的因素,碱的化学性质','',2016,'37','2016•抚州校级三模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840505,'已知1个C-12原子的质量为a千克，则实际质量为b千克的另一种原子的相对原子质量为（　　）','<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">12b</td></tr><tr><td>a</td></tr></table></span>&nbsp;千克','<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">12b</td></tr><tr><td>a</td></tr></table></span>','<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">12a</td></tr><tr><td>b</td></tr></table></span>','<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">12a</td></tr><tr><td>b</td></tr></table></span>&nbsp;千克','','B','【解答】解：由于相对原子质量=<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">原子的质量</td></tr><tr><td style=\"padding-top:1px;font-size:90%\">碳原子的质量×<table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">1</td></tr><tr><td>12</td></tr></table></td></tr></table></span>，故该原子的相对原子质量为：<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">bkg</td></tr><tr><td style=\"padding-top:1px;font-size:90%\">akg×<table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">1</td></tr><tr><td>12</td></tr></table></td></tr></table></span>=<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">12b</td></tr><tr><td>a</td></tr></table></span>．<br />故选B．','【分析】弄清相对原子质量的概念，然后根据相对原子质量的计算公式计算即可．','选择题',3.00,'902dc8b85c60102e5cccf1f93f8cb682',9,400,'相对原子质量的概念及其计算方法','',0,'37','',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840517,'<img  src=\"/tikuimages/9/2016/400/shoutiniao13/19f49df0-94d4-11e9-9c99-b42e9921e93e_xkb39.png\" style=\"vertical-align:middle;FLOAT:right\" />焦亚硫酸钠（Na<SUB>2</SUB>S<SUB>2</SUB>O<SUB>5</SUB>）是常用的食品抗氧化剂之一．利用下图装置（实验前已除尽装置内空气）可制取Na<SUB>2</SUB>S<SUB>2</SUB>O<SUB>5</SUB>． II中发生的反应为：Na<SUB>2</SUB>SO<SUB>3</SUB>+SO<SUB>2</SUB>═Na<SUB>2</SUB>S<SUB>2</SUB>O<SUB>5</SUB>．下列说法不正确的是（　　）','I中反应生成SO<SUB>2</SUB>','II中发生的反应为化合反应','III中x为NaOH溶液','实验中含硫元素的物质，硫元素化合价均相同','','D','【解答】解：A、根据图示可以看出I中反应生成二氧化硫，与II中亚硫酸钠反应制备焦亚硫酸钠（Na<SUB>2</SUB>S<SUB>2</SUB>O<SUB>5</SUB>），故A正确；B、化合反应是“多变一”，二氧化硫与亚硫酸钠反应制备焦亚硫酸钠（Na<SUB>2</SUB>S<SUB>2</SUB>O<SUB>5</SUB>），属于化合反应，故B正确；<br />C、二氧化硫是有毒气体，排放到空气中，会污染大气，需要进行吸收，吸收二氧化硫用氢氧化钠溶液，故C正确；<br />D、在化合物中正负化合价的代数和为零，所以二氧化硫中硫元素显+4价，硫酸中硫元素显+6价，故硫元素化合价不同，故D错误；<br />故选D．','【分析】A、根据二氧化硫的制备方法进行分析；<br />B、根据反应类型的判断，化合反应是“多变一”进行分析；<br />C、常见气体的制备与除杂，吸收二氧化硫用氢氧化钠溶液进行分析；<br />D、根据化合价的计算进行分析．','选择题',3.00,'df11a7b8e60474f34380fd8be3d5a820',9,400,'常见气体的检验与除杂方法,物质的相互转化和制备,有关元素化合价的计算,反应类型的判定','',2016,'37','2016•大兴区一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840524,'图书馆中的重要书籍着火，可以采用以下哪种灭火器（　　）','干粉灭火器','二氧化碳灭火器','泡沫灭火器','以上三种都可以','','B','【解答】解：用来扑灭图书资料的火灾，不能有水，否则容易受损．<br />A、干粉灭火器是利用压缩气体吹出干粉来灭火，可以用来扑灭一般火灾，还可用于扑灭油，气等引起的火灾，不适合用来扑灭图书资料等处的火灾，故选项错误．<br />B、二氧化碳灭火器后不留有痕迹，不会造成图书资料的损坏，可以用来扑灭图书资料等火灾，故选项正确．<br />C、泡沫灭火器可用于扑灭木材棉布等一般火灾，喷出的溶液会损坏图书资料，不适合用来扑灭图书资料等处的火灾，故选项错误．<br />D根据上述选项的分析，不能使用干粉灭火器、泡沫灭火器，故选项错误，<br />故选：B．','【分析】用来扑灭图书资料、精密仪器等处的火灾不能有水，否则容易受损；根据灭火器的原理和适用范围进行分析判断即可．','选择题',3.00,'9a51056b9a1c82346bba4439d449a400',9,400,'几种常用的灭火器','',2015,'33','2015秋•甘肃校级期末',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840530,'下列关于水的说法不正确的是（　　）','水是人类宝贵的资源','长期饮用蒸馏水有益健康','用洗过衣服的水冲马桶','自然界的水都是混合物','','B','【解答】解：A、水是人类宝贵的资源，人和一切动植物都离不开水，在工农业生产中起着重要作用，故选项说法正确．<br />B、人喝水的目的之一是补充人体所需要的矿物质，太纯净的水中矿物质含量太少，长期饮用蒸馏水不利于人体健康，故选项说法错误．<br />C、用洗过衣服的水冲马桶，能节约用水，故选项说法正确．<br />D、自然界的水中含有水、可溶性钙镁化合物等，都是混合物，故选项说法正确．<br />故选：B．','【分析】A、根据水的重要用途，进行分析判断．<br />B、根据人喝水的目的之一是补充人体所需要的矿物质，进行分析判断．<br />C、根据节约用水的措施，进行分析判断．<br />D、自然界的水中含有水、可溶性钙镁化合物等，进行分析判断．','选择题',3.00,'862dd6010ea004b0ce2173242e69bddc',9,400,'纯净物和混合物的判别,保护水资源和节约用水','',0,'37','',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840531,'下列图示实验操作中，正确的是（　　）','<img  src=\"/tikuimages/9/2016/400/shoutiniao37/1a35295e-94d4-11e9-b525-b42e9921e93e_xkb91.png\" style=\"vertical-align:middle\" /><br />测定溶液的pH','<img  src=\"/tikuimages/9/2016/400/shoutiniao35/1a37252e-94d4-11e9-94a6-b42e9921e93e_xkb76.png\" style=\"vertical-align:middle\" /><br />稀释浓硫酸','<img  src=\"/tikuimages/9/2016/400/shoutiniao44/1a37e880-94d4-11e9-89bc-b42e9921e93e_xkb60.png\" style=\"vertical-align:middle\" /><br />氧气的验满','<img  src=\"/tikuimages/9/2016/400/shoutiniao10/1a39962e-94d4-11e9-b51f-b42e9921e93e_xkb74.png\" style=\"vertical-align:middle\" /><br />检查装置的气密性','','D','【解答】解：A、用pH试纸测定未知溶液的pH时，正确的操作方法为用玻璃棒蘸取少量待测液滴在干燥的pH试纸上，与标准比色卡对比来确定pH．不能将pH试纸伸入待测液中，以免污染待测液，图中所示操作错误．<br />B、稀释浓硫酸时，要把浓硫酸缓缓地沿器壁注入水中，同时用玻璃棒不断搅拌，以使热量及时地扩散；一定不能把水注入浓硫酸中；图中所示操作错误．<br />C、检验氧气是否收集满时，应将带火星的木条放在集气瓶口，不能伸入瓶中，图中所示操作错误．<br />D、检查装置气密性的方法：把导管的一端浸没在水里，双手紧贴容器外壁，若导管口有气泡冒出，装置不漏气；图中所示操作正确．<br />故选：D．','【分析】A、根据用pH试纸测定未知溶液的pH的方法进行分析判断．<br />B、根据浓硫酸的稀释方法（酸入水，沿器壁，慢慢倒，不断搅）进行分析判断．<br />C、根据氧气的验满方法进行分析判断．<br />D、根据检查装置气密性的方法进行分析判断．','选择题',3.00,'6f5446811c2d2f4e186153a586fd8b37',9,400,'浓硫酸的性质及浓硫酸的稀释,检查装置的气密性,溶液的酸碱度测定,氧气的检验和验满','',2016,'32','2016•清远模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840539,'物质的结构决定性质，下列有关说法错误的是（　　）','过氧化氢与水化学性质不同的原因是由于分子构成不同','盐酸与硫酸化学性质相似的原因是在水溶液中都含有H<SUP>+</SUP>','氢氧化钠与氢氧化钙化学性质相似的原因是在水溶液中都含有OH<SUP>-</SUP>','金刚石与石墨物理性质差异较大的原因是由于碳原子不同','','D','【解答】解：A、过氧化氢与水化学性质不同，是因为它们分子的构成不同，不同种的分子性质不同，故选项解释正确．<br />B、盐酸与硫酸化学性质相似，是因为在水溶液中都含有H<SUP>+</SUP>，故选项解释正确．<br />C、氢氧化钠与氢氧化钙化学性质相似是，因为在水溶液中都含有OH<SUP>-</SUP>，故选项解释正确．<br />D、金刚石与石墨物理性质差异较大，是因为碳原子的排列方式不同，故选项解释错误．<br />故选：D．','【分析】根据分子的基本特征：分子质量和体积都很小；分子之间有间隔；分子是在不断运动的；同种的分子性质相同，不同种的分子性质不同，可以简记为：“两小运间，同同不不”，结合酸与碱具有相似化学性质的原因，进行分析判断即可．','选择题',3.00,'94fbb4b1b02e05f19bc360d63663050b',9,400,'酸的化学性质,碱的化学性质,利用分子与原子的性质分析和解决问题','',2016,'37','2016春•姜堰区校级月考',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840540,'为了除去物质中的杂质（括号内物质为杂质），所选用试剂和操作方法都正确的是（　　）<br /><table class=\"edittable\"><TBODY><TR><td width=36></TD><td width=144>物&nbsp;&nbsp;质</TD><td width=109>选用试剂（过量）</TD><td width=231>操作方法</TD></TR><TR><td>A</TD><td>CO<SUB>2</SUB>（HCl）</TD><td>氢氧化钠溶液</TD><td>通入装有氢氧化钠溶液的洗气瓶</TD></TR><TR><td>B</TD><td>Na<SUB>2</SUB>CO<SUB>3</SUB>（NH<SUB>4</SUB>HCO<SUB>3</SUB>）</TD><td></TD><td>加热至质量不再减少</TD></TR><TR><td>C</TD><td>Cu（NO<SUB>3</SUB>）<SUB>2</SUB>溶液（AgNO<SUB>3</SUB>）</TD><td>铜粉</TD><td>过滤</TD></TR><TR><td>D</TD><td>NaCl溶液（CaCl<SUB>2</SUB>）</TD><td>K<SUB>2</SUB>CO<SUB>3</SUB></TD><td>过滤</TD></TR></TBODY></TABLE>','A','B','C','D','','B|C','【解答】解：A、CO<SUB>2</SUB>（HCl）通入装有氢氧化钠溶液的洗气瓶不可行，因为二氧化碳和氯化氢气体都能与氢氧化钠反应；故选项错误；<br />B、Na<SUB>2</SUB>CO<SUB>3</SUB>（NH<SUB>4</SUB>HCO<SUB>3</SUB>）加热至质量不再减少可行，因为碳酸钠不易分解，碳酸氢铵在加热的条件下生成氨气、水和二氧化碳；故选项正确；<br />C、Cu（NO<SUB>3</SUB>）<SUB>2</SUB>溶液（AgNO<SUB>3</SUB>）加入过量的铜粉可行，铜和硝酸银反应生成银和硝酸铜，过滤即可；故选项正确；<br />D、在NaCl溶液（CaCl<SUB>2</SUB>）中加入过量的碳酸钾不可行，因为氯化钙与碳酸钾反应生成碳酸钙白色沉淀和氯化钾，氯化钾是新杂质，故选项错误；<br />故选B、C','【分析】本题属于除杂质题，一般的除杂质题必须同时满足两个条件：①加入的试剂只能与杂质反应，不能与原物质反应②反应时不能加入新的杂质．对于本题A中的CO<SUB>2</SUB>（HCl），通入装有氢氧化钠溶液的洗气瓶不可行，因为二氧化碳和氯化氢气体都能与氢氧化钠反应；B中Na<SUB>2</SUB>CO<SUB>3</SUB>（NH<SUB>4</SUB>HCO<SUB>3</SUB>）加热至质量不再减少可行，因为碳酸钠不易分解，碳酸氢铵在加热的条件下生成氨气、水和二氧化碳；C中Cu（NO<SUB>3</SUB>）<SUB>2</SUB>溶液（AgNO<SUB>3</SUB>）加入过量的铜粉可行，铜和硝酸银反应生成银和硝酸铜，过滤即可；D中，在NaCl溶液（CaCl<SUB>2</SUB>）中加入过量的碳酸钾不可行，因为氯化钙与碳酸钾反应生成碳酸钙白色沉淀和氯化钾，氯化钾是新杂质．','多选题',4.00,'8876e3899cdfc8d4348400ba93c44dd4',9,400,'物质除杂或净化的探究,常见气体的检验与除杂方法,盐的化学性质','句容市',2016,'32','2016•句容市模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840541,'下图是初中化学常见气体的发生与收集装置，据图回答下列问题．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao39/1a64c4de-94d4-11e9-8289-b42e9921e93e_xkb61.png\" style=\"vertical-align:middle\" /><br />（1）写出图中仪器a的名称<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（2）实验室用B装置可以制取<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>气体（填2种气体的化学式）；<br />（3）实验室采用A装置制CH<SUB>4</SUB>，推测其反应物是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填序号）；<br />A．H<SUB>2</SUB>和C&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; B．Al<SUB>4</SUB>C<SUB>3</SUB>固体和水&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; C．无水醋酸钠固体和碱石灰固体<br />（4）若用E装置收集一瓶O<SUB>2</SUB>，则装置明显的错误是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','长颈漏斗$###$CO<SUB>2</SUB>、O<SUB>2</SUB>$###$C$###$集气瓶中有空气，收集的氧气不纯','【解答】解：（1）图中所给仪器的名称为长颈漏斗；<br />（2）B装置为固液常温型的反应，所以可以用来制取二氧化碳，过氧化氢与二氧化锰混合制取氧气，实验室制取氢气；<br />（3）分析题中所给的制取气体的方法，只有D中的是固体且是在加热的条件下进行反应，故选D；<br />（4）排水法收集气体时，要先在集气瓶中装满水，倒立于水槽中，瓶底不能有气泡，否则收集的气体不纯．<br />故答案为：<br />（1）长颈漏斗；（2）CO<SUB>2</SUB>、H<SUB>2</SUB>、O<SUB>2</SUB>（填其中的2种给分）<br />（3）C；&nbsp;（4）集气瓶中有空气，收集的氧气不纯．','【分析】（1）直接写出仪器的名称；<br />（2）B装置为固液常温型的反应，所以可以用来制取二氧化碳，可以据此完成解答；<br />（3）A装置为固固加热型的发生装置，可以据此结合题给的选项进行解答；<br />（4）据排水法收集气体的注意事项解答．','填空题',3.00,'bfdfd145ff0f3765cf6394c77539cc63',9,400,'实验室制取气体的思路,常用气体的发生装置和收集装置与选取方法','安陆市',2016,'32','2016•安陆市模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840542,'加油站应张贴的警示标志是（　　）','<img  src=\"/tikuimages/9/2016/400/shoutiniao54/1a6a431e-94d4-11e9-a3dd-b42e9921e93e_xkb56.png\" style=\"vertical-align:middle\" />','<img  src=\"/tikuimages/9/2016/400/shoutiniao53/1a6deca1-94d4-11e9-82a3-b42e9921e93e_xkb16.png\" style=\"vertical-align:middle\" />','<img  src=\"/tikuimages/9/2016/400/shoutiniao2/1a6f4c2e-94d4-11e9-92f1-b42e9921e93e_xkb12.png\" style=\"vertical-align:middle\" />','<img  src=\"/tikuimages/9/2016/400/shoutiniao82/1a714800-94d4-11e9-8b07-b42e9921e93e_xkb64.png\" style=\"vertical-align:middle\" />','','D','【解答】解：可燃性的气体与氧气或空气混合后点燃易发生爆炸，加油站内的空气中混有可燃的汽油蒸气，要防止引燃混合气体发生爆炸．<br />A、图中所示标志是塑料制品回收标志，不适合在加油站使用，故选项错误．<br />B、图中所示标志是腐蚀品标志，不适合在加油站使用，故选项错误．<br />C、图中所示标志是有毒品标志，不适合在加油站使用，故选项错误．<br />D、图中所示标志是禁止吸烟标志，适合在加油站使用，故选项正确．<br />故选：D．','【分析】根据图标所表示的含义来考虑，并结合加油站应注意的事项进行分析判断．','选择题',3.00,'7b765e0546a8a13446a0bd520d27c91a',9,400,'几种常见的与化学有关的图标','',2016,'32','2016•澄海区模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840545,'写出下列仪器的作用<br />①量筒：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />②烧杯：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />③集气瓶：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','量取液体$###$溶解固体、配制溶液、加热较多量液体试剂$###$收集或储存少量气体','【解答】解：①量筒：用于量取液体；故填：量取固体．<br />②烧杯：用来溶解固体、配制溶液、加热较多量液体试剂．故填：溶解固体、配制溶液、加热较多量液体试剂．<br />③集气瓶：用于收集或储存少量气体．故填：收集或储存少量气体．','【分析】量筒用于量取液体；烧杯用来溶解固体、配制溶液、加热较多量液体试剂；集气瓶用于收集或储存少量气体．','填空题',3.00,'7bdf5bd07b5b28c6e5a6bb3fe6a3da8f',9,400,'常用仪器的名称和选用','',2016,'37','2016春•武汉校级月考',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840548,'磷酸二氢铵（NH<SUB>4</SUB>H<SUB>2</SUB>PO<SUB>4</SUB>）是农业上一种常用的化肥．下列叙述正确的是（　　）','磷酸二氢铵属于氮肥','磷酸二氢铵中，氮、氢、磷、氧元素的质量比是1：6：1：4','磷酸二氢铵的相对分子质量是115','磷酸二氢铵是一种氧化物','','C','【解答】解：A、磷酸二氢铵中含有氮元素和磷元素，属于复合肥，故说法错误；<br />B、磷酸二氢铵（NH<SUB>4</SUB>H<SUB>2</SUB>PO<SUB>4</SUB>）中，氮、氢、磷、氧元素的质量比是14：（1×6）：31：（16×4）=14：6：31：64，故说法错误；<br />C、磷酸二氢铵的相对分子质量=14+1×6+31+16×4=115，故说法正确；<br />D、磷酸二氢铵含有氮、氢、磷、氧四种元素，不属于氧化物，故说法错误．<br />故选C．','【分析】A、含有氮元素的肥料称为氮肥，含有磷元素的肥料称为磷肥，含有钾元素的肥料称为钾肥，同时含有氮、磷、钾三种元素中的两种或两种以上的肥料称为复合肥．<br />B、元素的质量比等于各元素原子量与原子个数乘积之比．<br />C、相对分子质量等于各元素的相对原子质量之和．<br />D、氧化物是只含有两种元素，且其中一种是氧的化合物','选择题',3.00,'7ada3ec7e74cf0267ebc81a1c1444949',9,400,'常见化肥的种类和作用,从组成上识别氧化物,相对分子质量的概念及其计算,元素质量比的计算','',2016,'35','2016春•无锡期中',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840549,'汽车在受到猛烈撞击时安全气囊内的物质瞬间分解，产生的是空气中体积分数最大的气体．该气体是（　　）','氮气','氧气','氦气','水蒸气','','A','【解答】解：A、汽车在受到猛烈撞击时安全气囊内的物质瞬间分解，产生的是空气中体积分数最大的气体，该气体是氮气，故选项正确；<br />B、氧气大约占空气体积的21%，不是最多，故选项错误；<br />C、氦气是稀有气体，含量比较少，故选项错误；<br />D、水蒸气、其它气体和杂质大约占0.03%，含量比较少，故选项错误；<br />故选A','【分析】空气中各成分的体积分数分别是：氮气大约占78%、氧气大约占21%、稀有气体大约占0.94%、二氧化碳大约占0.03%、水蒸气和其它气体和杂质大约占0.03%；汽车在受到猛烈撞击时安全气囊内的物质瞬间分解，产生的是空气中体积分数最大的气体，该气体是氮气．','选择题',2.00,'152b9323922b6b67cd01c2df7f3d6b08',9,400,'空气的成分及各成分的体积分数','',2016,'37','2016•通州区一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840554,'<img  src=\"/tikuimages/9/2016/400/shoutiniao7/1a8e6cee-94d4-11e9-a07c-b42e9921e93e_xkb51.png\" style=\"vertical-align:middle;FLOAT:right\" />如图是甲、乙两种物质的溶解度曲线．下列说法正确的是（　　）','甲的溶解度大于乙的溶解度','t<SUB>1</SUB>℃时甲、乙两物质的饱和溶液中溶质质量相等','t<SUB>2</SUB>℃时等质量甲、乙两物质的饱和溶液降温至t<SUB>1</SUB>℃时，两溶液中溶质质量分数相等','氧气溶解度随温度变化的规律与图中的乙相似','','D','【解答】解：A、甲的溶解度大于乙的溶解度错误，因为没有指明温度，故选项错误；<br />B、t<SUB>1</SUB>℃时，甲、乙两物质的饱和溶液中溶质质量相等错误，因为没有指明是等质量的饱和溶液，故选项错误；<br />C、t<SUB>2</SUB>℃时等质量甲、乙两物质的饱和溶液降温至t<SUB>1</SUB>℃时，两溶液中溶质质量分数是甲大于乙，因为乙降温后有饱和变为不饱和溶液，质量分数不变，还是比甲小，相等是错误的，故选项错误；<br />D、氧气溶解度随温度变化的规律与图中的乙相似正确，因为气体的溶解度随温度的升高而减小，故选项正确；<br />故选D','【分析】根据题目信息和溶解度曲线可知：甲固体物质的溶解度，都是随温度升高而增大，而乙的溶解度随温度的升高而减少；甲的溶解度大于乙的溶解度错误，因为没有指明温度；t<SUB>1</SUB>℃时，甲、乙两物质的饱和溶液中溶质质量相等错误，因为没有指明是等质量的饱和溶液；t<SUB>2</SUB>℃时等质量甲、乙两物质的饱和溶液降温至t<SUB>1</SUB>℃时，两溶液中溶质质量分数是甲大于乙，因为乙降温后有饱和变为不饱和溶液，质量分数不变，还是比甲小；氧气溶解度随温度变化的规律与图中的乙相似正确，因为气体的溶解度随温度的升高而减小．','选择题',3.00,'a184b056a93d4647716cea5173164a76',9,400,'固体溶解度曲线及其作用,气体溶解度的影响因素,溶质的质量分数、溶解性和溶解度的关系','',2016,'37','2016•河北区一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840560,'从2Mg+O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2MgO中，可以获取的信息：<br />（1）<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；（2）<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；（3）<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','该反应的反应物是镁和氧气，生成物是氧化镁$###$反应的条件是点燃$###$属于化合反应','【解答】解：（1）可知道：该反应的反应物是镁和氧气，生成物是氧化镁；<br />（2）等号上方的是反应条件，反应的条件是点燃；<br />（3）该反应符合“多变一”的特征，属于化合反应；<br />故答案为：该反应的反应物是镁和氧气，生成物是氧化镁；反应的条件是点燃；属于化合反应．','【分析】从化学方程式获得的信息主要有：反应物、生成物、反应条件、各物质间质量的关系，据此结合题意进行分析解答．','填空题',3.00,'e6da64abb9906ef97e6bdef023db861d',9,400,'化学方程式的概念、读法和含义','',2015,'37','2015春•临汾月考',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840562,'已知在一定条件下发生反应：甲十乙→丙+丁．下列判段正确的是（　　）','若丙为Cu（NO<SUB>3</SUB>）<SUB>2</SUB>，则甲可能为Cu，也可能为CuO','若丙为盐、丁为水．则该反应一定为复分解反应','若甲、丙为单质，乙、丁为化合物，则该反应可能为置换反应','甲、乙、丙、丁不可能含有同一种元素','','A','【解答】解：A、丙为Cu（NO<SUB>3</SUB>）<SUB>2</SUB>，则甲可能为铜或氧化铜，例如铜和硝酸银溶液反应能生成硝酸铜和银，氧化铜和硝酸反应能生成硝酸铜和水．故判断正确．<br />B、氢氧化钙和二氧化碳反应生成碳酸钙和水，有盐和水生成，但不是复分解反应．故判断错误；<br />C、若甲、丙为单质，乙、丁为化合物，则该反应一定为置换反应．故判断错误；<br />D、甲、乙、丙、丁有可能含有同一种元素，例如二氧化碳和氢氧化钠反应生成盐碳酸钠和水，反应物和生成物中均含有氧元素．故判断错误；<br />故选A．','【分析】A、举例说明丙为Cu（NO<SUB>3</SUB>）<SUB>2</SUB>，则甲可能为铜或氧化铜．<br />B、氢氧化钙和二氧化碳反应生成碳酸钙和水，有盐和水生成，但不是复分解反应；<br />C、根据置换反应的概念进行分析；<br />D、二氧化碳和氢氧化钠反应生成盐碳酸钠和水．','选择题',3.00,'4d6a997cedf4a48a9f87a9eba3f57a76',9,400,'复分解反应及其发生的条件,置换反应及其应用,质量守恒定律及其应用','',0,'37','',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840570,'化学是一门以实验为基础的自然学科，而正确的选择和使用化学仪器是顺利完成化学实验的前提．列举了常见的化学仪器，请根据要求完成以下各题．<br /><img  src=\"/tikuimages/9/2015/400/shoutiniao89/1abbe58f-94d4-11e9-9447-b42e9921e93e_xkb45.png\" style=\"vertical-align:middle\" /><br />首先，全面了解一种仪器，可以从名称、用途和使用注意事项=个方面．<br />例如：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．用作<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，注意<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（1）<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（写名称），用作配制溶液的容器，加热时应放在<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>上，使受热均匀；<br />（2）<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（写序号），用作量度液体的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，注意不能加热，不能<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（3）酒精灯，用于<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，使用时不能用嘴吹灭酒精灯火焰，应用<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>盖灭；<br />请仿照如上格式，从图中再选择两种仪器，进行全面的说明：<br />（4）<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（5）<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />其次，要完成一项基本的实验操作，需要选择几种仪器进行组合，例如：<br />（6）要想彻底的除去黄泥水中的不溶性固体，需要进行的操作是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，此时，可以从上图中选择的仪器为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（写序号），还需要补充的仪器和用品为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>和<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','试管$###$少量试剂的反应容器$###$加热后不能冷却，以防炸裂$###$B$###$石棉网$###$C$###$体积$###$作反应的容器（或用来配制溶液）$###$加热$###$灯帽$###$集气瓶的作用是收集或贮存少量气体，且瓶口要用毛玻璃片遮盖$###$胶头滴管的作用是：吸取或滴加少量液体，用完之后，滴管不能水平放在桌面上，以防药品倒流进入滴管，腐蚀胶头．用完的滴管必须涮洗干净放回试管架上$###$过滤$###$BGI$###$铁架台$###$滤纸','【解答】解：（1）给烧杯内液体加热，不能直接加热，要放置在石棉网上，防止受热不均炸裂；<br />（2）量筒可用于量取液体体积，不能加热或用作反应的容器（或用来配制溶液）；<br />（3）酒精灯是热源，用于加热，点燃酒精灯用火柴，熄灭酒精灯应用灯帽盖灭；<br />（4）集气瓶的作用是收集或贮存少量气体，且瓶口要用毛玻璃片遮盖；<br />（5）胶头滴管的作用是：吸取或滴加少量液体，用完之后，滴管不能水平放在桌面上，以防药品倒流进入滴管，腐蚀胶头．用完的滴管必须涮洗干净放回试管架上．<br />（6）过滤操作可除去水中不溶性固体，过滤操作需要用到的仪器有：烧杯、漏斗、玻璃棒、铁架台、滤纸，所以可以从上图中选择的仪器为BGI，所以还需要铁架台和滤纸．<br />故答案为：（1）烧杯&nbsp;石棉网&nbsp;<br />（2）C&nbsp;体积&nbsp;作反应的容器（或用来配制溶液）&nbsp;<br />（3）加热&nbsp;灯帽&nbsp;<br />（4）集气瓶的作用是收集或贮存少量气体，且瓶口要用毛玻璃片遮盖；<br />（5）胶头滴管的作用是：吸取或滴加少量液体，用完之后，滴管不能水平放在桌面上，以防药品倒流进入滴管，腐蚀胶头．用完的滴管必须涮洗干净放回试管架上．<br />（6）过滤&nbsp;BGI&nbsp;铁架台和滤纸','【分析】根据常用仪器的用途和使用注意事项分析回答，要注意与前边的仪器相对应．','填空题',3.00,'f4d6083d24cd822f8110de6a0adafb6c',9,400,'测量容器-量筒,加热器皿-酒精灯,过滤的原理、方法及其应用,常用仪器的名称和选用','',2015,'35','2015秋•周村区校级期中',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840573,'如表是氯化钠和碳酸钠在不同温度时的溶解度，根据此表回答：<br /><table class=\"edittable\"><TBODY><TR><td width=175 colSpan=2>温度/℃</TD><td width=95>10</TD><td width=95>20</TD><td width=95>30</TD><td width=95>40</TD></TR><TR><td rowSpan=2>溶解度<br />g/100gH2O</TD><td>氯化钠</TD><td>35.8</TD><td>36.0</TD><td>36.3</TD><td>36.6</TD></TR><TR><td>碳酸钠</TD><td>12.2</TD><td>21.8</TD><td>39.7</TD><td>53.2</TD></TR></TBODY></TABLE>①40℃时，氯化钠的溶解度为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>g/100g水；<br />②碳酸钠溶液中混有少量的氯化钠，可通过<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>的方法提纯；<br />③碳酸钠的溶解度随温度的升高而<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“增大”或“减小”）．在20℃时，将100g水加入30g碳酸钠中，充分搅拌后得到的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“饱和”或“不饱和”）溶液，将上述溶液升温到30℃，该溶液的溶质质量分数为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','36.6$###$降温结晶（或冷却热的饱和溶液）$###$增大$###$饱和$###$23.1%','【解答】解：①由图表可知40℃时，氯化钠的溶解度为36.6g．<br />②碳酸钠的溶解度受温度的影响变化大，所以碳酸钠溶液中混有少量的氯化钠，可通过降温结晶（或冷却热的饱和溶液）的方法提纯；<br />③由碳酸钠在不同温度时的溶解度表，碳酸钠的溶解度随温度的升高而增大．<br />20℃时，碳酸钠的溶解度为21.8g，在20℃时，将30g碳酸钠加入到100g水中，充分搅拌后，最多只能溶解21.8g，得到的是饱和溶液；<br />由表格知，30℃碳酸钠的溶解度为39.7g，将上述溶液升温到30℃，将100g的水加入30g碳酸钠中，充分搅拌后得到的是不饱和溶液，碳酸钠完全溶解，该溶液的溶质质量分数=<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">30g</td></tr><tr><td>100g+30g</td></tr></table></span>×100%=23.1%．<br />故答案为：①36.6；②降温结晶（或冷却热的饱和溶液）；③增大；饱和；23.1%．','【分析】①由氯化钠和碳酸钠在不同温度时的溶解度表，查出40℃时氯化钠的溶解度即可；<br />②根据碳酸钠的溶解度受温度的影响变化大进行解答；<br />③由碳酸钠在不同温度时的溶解度表，确定碳酸钠的溶解度受温度影响的变化趋势即可．20℃时，碳酸钠的溶解度为21.8g，进行分析解答．','填空题',3.00,'61181d604707a7413ad5088a420768fa',9,400,'结晶的原理、方法及其应用,固体溶解度的影响因素,溶质的质量分数','',2016,'37','2016•浦东新区二模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840574,'根据如图所示实验装置图，按要求回答有关问题．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao16/1ac6e20f-94d4-11e9-babb-b42e9921e93e_xkb82.png\" style=\"vertical-align:middle\" /><br />（1）写出图中带有标号仪器的名称：b<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）用高锰酸钾制取并收集氧气时，应选用的装置是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填序号，下同），在试管口处放一团棉花的作用是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；高锰酸钾分解制氧气的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）实验室用锌粒和稀硫酸反应制取并收集氢气选用的装置是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（4）实验室制取并收集二氧化碳常选用的装置是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．所选用的收集方法的依据是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','铁架台$###$A$###$可以防止加热时高锰酸钾粉末进入导管$###$2KMnO<SUB>4</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>K<SUB>2</SUB>MnO<SUB>4</SUB>+MnO<SUB>2</SUB>+O<SUB>2</SUB>↑$###$C$###$B$###$二氧化碳密度比空气大','【解答】解：（1）熟记常用仪器的名称可知a为长颈漏斗，b为铁架台．<br />（2）反应物的状态是固态，反应条件是加热，应选固-固加热型的发生装置，氧气的密度比空气大，不易溶于水，可用排水法或向上排空气法收集；高锰酸钾加热时会飞溅，因而用高锰酸钾制氧气时应在试管口塞一团棉花；高锰酸钾加热生成锰酸钾、二氧化锰和氧气．故答案为：A；&nbsp;&nbsp;防止加热时高锰酸钾粉末进入导管；2KMnO<SUB>4</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>K<SUB>2</SUB>MnO<SUB>4</SUB>+MnO<SUB>2</SUB>+O<SUB>2</SUB>↑&nbsp;<br />（3）反应物的状态是固态和液态，反应条件是常温，应选固-液不加热型的发生装置，氢气的密度比空气小，难溶于水，可用向下排空气法或排水法收集．故答案为：C<br />（4）反应物的状态是固态和液态，反应条件是常温，应选固-液不加热型的发生装置，二氧化碳的密度比空气大，可用向上排空气法收集．故答案为：B；&nbsp;&nbsp; 二氧化碳的密度比空气大<br />答案：<br />（1）铁架台；<br />（2）A；可以防止加热时高锰酸钾粉末进入导管；2KMnO<SUB>4</SUB>&nbsp;<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>&nbsp;K<SUB>2</SUB>MnO<SUB>4</SUB>+MnO<SUB>2</SUB>+O<SUB>2</SUB>↑；<br />（3）C；<br />&nbsp;（4）B；&nbsp;二氧化碳密度比空气大．','【分析】（1）a、b均为实验室常用的仪器．<br />（2）根据反应物的状态和反应条件确定发生装置，根据气体的密度和溶水性确定收集装置；高锰酸钾加热时会飞溅；高锰酸钾加热生成锰酸钾、二氧化锰和氧气．<br />（3）根据反应物的状态和反应条件确定发生装置，根据气体的密度和溶水性确定收集装置．<br />（4）根据反应物的状态和反应条件确定发生装置，根据气体的密度和溶水性确定收集装置．','书写',3.00,'9e798d755e19724e55aa2c3633bc3764',9,400,'常用气体的发生装置和收集装置与选取方法,制取氧气的操作步骤和注意点,书写化学方程式、文字表达式、电离方程式','',2016,'32','2016•峨边县模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840578,'<img  src=\"/tikuimages/9/2016/400/shoutiniao30/1ad512de-94d4-11e9-9373-b42e9921e93e_xkb66.png\" style=\"vertical-align:middle;FLOAT:right\" />（2016•安庆一模）实验室常用二氧化锰催化分解过氧化氢的方法制取氧气，请写出反应的化学方程式：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．某同学从课外读物中得知新鲜的动物肝脏中有较多的过氧化氢酶，过氧化氢酶与无机催化剂Fe<SUP>3+</SUP>都可以催化过氧化氢分解．<br />现有实验材料如下：<br />新鲜的含过氧化氢酶3.5%的猪肝研磨液、3.5%的FeCl<SUB>3</SUB>溶液、新配制的3%的过氧化氢溶液．<br />量筒、滴管、试管、试管架、试管夹、小木条、火柴、酒精灯等．<br />某同学为比较过氧化氢酶和Fe<SUP>3+</SUP>催化效率，利用上述材料进行探究：<br />【提出问题】比较过氧化氢酶和Fe<SUP>3+</SUP>对过氧化氢分解的催化效率的高低．<br />【提出猜想】<br />猜想1：过氧化氢酶的催化效率比Fe<SUP>3+</SUP>高；<br />猜想2：Fe<SUP>3+</SUP>的催化效率比过氧化氢酶高；<br />【实验步骤】<br />（1）取两支洁净的试管，分别注入2ml3%的过氧化氢溶液．<br />（2）分别向两支试管内<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）将<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>分别放入两支试管内液面的上方，观察现象．<br />【实验现象与结论】<br />（4）观察到<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，证明猜想2正确．<br />【继续探究】<br />（5）该同学继续探究温度对猪肝研磨液和FeCl<SUB>3</SUB>溶液催化效率的影响，实验结果绘成如图，图中代表猪肝研磨液催化效果的曲线是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，你能得到的实验结论是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，（写一条即可）．<br />（6）你认为猪肝研磨液的催化效率还可能与<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>有关．（填一种因素即可）','','','','','','2H<SUB>2</SUB>O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span>&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O+O<SUB>2</SUB>↑$###$滴入等量的猪肝研磨液和氯化铁溶液$###$带火星的小木条$###$滴入氯化铁溶液的试管内产生的气泡更迅速，带火星的木条易复燃$###$乙$###$30～40℃时，猪肝研磨液催化效果最佳（或40℃以内，温度对FeCl<SUB>3</SUB>溶液催化效率几乎无影响）$###$猪肝研磨液中过氧化氢酶的浓度','【解答】解：<br />（1）实验室常用二氧化锰催化分解过氧化氢的方法制取氧气，反应的化学方程式为：2H<SUB>2</SUB>O<SUB>2</SUB>&nbsp;&nbsp;<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span>&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O+O<SUB>2</SUB>↑；<br />（2）取两支洁净的试管，分别注入2ml3%的过氧化氢溶液．分别向两支试管内滴入等量的猪肝研磨液和氯化铁溶液；<br />（3）氧气支持燃烧，将带火星的小木条分别放入两支试管内液面的上方，观察现象．<br />（4）观察到滴入氯化铁溶液的试管内产生的气泡更迅速，带火星的木条易复燃，证明猜想2正确．<br />（5）该同学继续探究温度对猪肝研磨液和FeCl<SUB>3</SUB>溶液催化效率的影响，实验结果绘成如图，图中代表猪肝研磨液催化效果的曲线是乙；得到的实验结论是：30～40℃时，猪肝研磨液催化效果最佳（或40℃以内，温度对FeCl<SUB>3</SUB>溶液催化效率几乎无影响）；<br />（6）猪肝研磨液的催化效率还可能与猪肝研磨液中过氧化氢酶的浓度有关．<br />答案：<br />（1）2H<SUB>2</SUB>O<SUB>2</SUB>&nbsp;&nbsp;<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span>&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O+O<SUB>2</SUB>↑；<br />（2）滴入等量的猪肝研磨液和氯化铁溶液；<br />（3）带火星的小木条（合理即可）；<br />（4）滴入氯化铁溶液的试管内产生的气泡更迅速，带火星的木条易复燃（合理即可）．<br />（5）乙；&nbsp;30～40℃时，猪肝研磨液催化效果最佳（或40℃以内，温度对FeCl<SUB>3</SUB>溶液催化效率几乎无影响）（合理即得分）<br />（6）猪肝研磨液中过氧化氢酶的浓度（合理即可）','【分析】（1）根据实验室常用二氧化锰催化分解过氧化氢的方法制取氧气解答；<br />（2）根据对比新鲜的含过氧化氢酶3.5%的猪肝研磨液、3.5%的FeCl<SUB>3</SUB>溶液解答．<br />（3）根据氧气支持燃烧的性质解答；<br />（4）根据实验现象分析猜想解答；<br />（5）根据高温会破坏酶的空间结构，使酶变性失活解答；<br />（6）根据猪肝研磨液的催化效率还可能与猪肝研磨液中过氧化氢酶的浓度有关解答．','书写',3.00,'2083db863d983cac9d03a18396a3528f',9,400,'实验探究物质变化的条件和影响物质变化的因素,催化剂的特点与催化作用,书写化学方程式、文字表达式、电离方程式','',2016,'37','2016•安庆一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840582,'甲、乙两原子的相对原子质量分别A、B，它们的质量分别为akg，bkg．一个碳原子的质量为W&nbsp;Kg请完成下面的计算．<br />（1）A=<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />（2）B=<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />（3）试求两原子的质量之比与相对原子质量之比的关系？','','','','','','<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">12a</td></tr><tr><td>W</td></tr></table></span>$###$<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">12b</td></tr><tr><td>W</td></tr></table></span>','【解答】解：（1）甲原子的相对原子质量为A，它们的质量为akg，一个碳原子的质量为WKg，则A=<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">akg</td></tr><tr><td style=\"padding-top:1px;font-size:90%\">Wkg×<table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">1</td></tr><tr><td>12</td></tr></table></td></tr></table></span>=<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">12a</td></tr><tr><td>W</td></tr></table></span>．<br />（2）乙原子的相对原子质量为B，它们的质量为bkg，一个碳原子的质量为WKg，则b=<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">bkg</td></tr><tr><td style=\"padding-top:1px;font-size:90%\">Wkg×<table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">1</td></tr><tr><td>12</td></tr></table></td></tr></table></span>=<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">12b</td></tr><tr><td>W</td></tr></table></span>．<br />（3）两原子的质量之比为akg：bkg=a：b；相对原子质量之比为<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">12a</td></tr><tr><td>W</td></tr></table></span>：<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">12b</td></tr><tr><td>W</td></tr></table></span>=a：b；则两原子的质量之比与相对原子质量之比相等．<br />故答案为：（1）<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">12a</td></tr><tr><td>W</td></tr></table></span>；（2）<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">12b</td></tr><tr><td>W</td></tr></table></span>；（3）两原子的质量之比与相对原子质量之比相等．','【分析】根据某原子的相对原子质量=<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">该元素的一个原子的质量</td></tr><tr><td style=\"padding-top:1px;font-size:90%\">一种碳原子质量×<table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">1</td></tr><tr><td>12</td></tr></table></td></tr></table></span>，结合题意进行分析解答即可．','填空题',3.00,'187b3af758c47fa6ee63d7119529dadc',9,400,'相对原子质量的概念及其计算方法','',2016,'35','2016春•白银期中',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840587,'回答下列关于水的问题：<br />（1）配制生理盐水时水作<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（2）实验室区别软水和硬水可用<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（3）消防队员用水灭火的主要原理是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','溶剂$###$肥皂水$###$降低温度至着火点以下','【解答】解：<br />（1）水作为常见溶剂，配制生理盐水时，把氯化钠溶于水，故水为溶剂；<br />（2）检验水样是硬水还是软水，可用的物质是肥皂水，其中产生泡沫较少的是硬水，泡沫较多的是软水；（3）根据灭火的原理，消防队员用水灭火，是利用水降低温度至着火点以下；<br />答案：<br />（1）溶剂&nbsp;&nbsp;&nbsp;&nbsp;（2）肥皂水&nbsp;（3）降低温度至着火点以下','【分析】（1）根据水作为常见溶剂进行分析；<br />（2）根据检验水样是硬水还是软水，可用的物质是肥皂水进行分析；<br />（3）根据灭火的原理进行分析．','填空题',3.00,'4a1def422f61440945fcdc683783535b',9,400,'硬水与软水,溶液、溶质和溶剂的相互关系与判断,灭火的原理和方法','',2016,'37','2016•长春二模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840590,'二甲醚（C<SUB>2</SUB>H<SUB>6</SUB>O）是一种重要的化工原料，下列关于二甲醚的说法错误的是（　　）','由3种元素组成','相对分子质量为9','碳、氢、氧元素质量比为12：3：8','碳元素的质量分数约为52.2%','','B','【解答】解：A、由二甲醚的化学式：C<SUB>2</SUB>H<SUB>6</SUB>O可知由C、H、O三种元素组成，故A说法正确；<br />B、二甲醚的相对分子质量：12×2+1×6+16=46，故B说法错误；<br />C、碳、氢、氧三种元素的质量比为：12×2：1×6：16=12：3：8，故C说法正确；<br />D、二甲醚（C<SUB>2</SUB>H<SUB>6</SUB>O）中碳元素的质量分数为：<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">12×2</td></tr><tr><td>46</td></tr></table></span>×100%═52.2%，故D说法正确．<br />故选B','【分析】A、根据二甲醚的化学式考虑元素组成；B、根据相对分子质量的计算方法考虑；C、根据元素质量的计算方法考虑；D、根据元素质量分数的计算方法考虑','选择题',3.00,'d1956bb98481e3ba2c57d644ac1c89b9',9,400,'化学式的书写及意义,相对分子质量的概念及其计算,元素质量比的计算,元素的质量分数计算','高邮市',2016,'37','2016•高邮市一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840591,'如图表示家用燃料使用的发展历程（括号内表示主要成分），下列说法错误的是（　　）<img  src=\"/tikuimages/9/2016/400/shoutiniao79/1b06aa30-94d4-11e9-9101-b42e9921e93e_xkb70.png\" style=\"vertical-align:middle\" />','煤中主要含有碳元素，还&nbsp;含有氢、硫等元素','液化石油气是石油化工的一种产品','家用燃料燃烧时均放出热量','煤、石油、天然气和氢气都是化石燃料','','D','【解答】解：A、煤中主要含有碳元素，还&nbsp;含有氢、硫等元素，故A正确；<br />B、液化石油气是石油化工的一种产品，故B正确；<br />C、家用燃料燃烧时均放出热量，故C正确；<br />D、煤、石油、天然气都是化石燃料，氢气不属于化石燃料，故D错误．<br />故选D．','【分析】A、根据煤的组成分析；<br />B、根据石油化工的产品分析；<br />C、根据燃料燃烧时均放出热量分析；<br />D、根据化石燃料的种类分析．','选择题',2.00,'44c6e774358c52652d04147e620b8799',9,400,'常用燃料的使用与其对环境的影响,化石燃料及其综合利用,石油加工的产物','',2016,'32','2016•洛阳模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840595,'下列实验方案设计合理的是（　　）','用燃着的小木条区分N<SUB>2</SUB>与CO<SUB>2</SUB>','用酚酞溶液区分NaOH溶液和Na<SUB>2</SUB>CO<SUB>3</SUB>溶液','用氢氧化钠固体干燥SO<SUB>2</SUB>','用Fe、Ag、CuSO<SUB>4</SUB>溶液验证Fe、Cu、Ag三种金属的活动性','','D','【解答】解：A、氮气和二氧化碳都能使燃着的木条熄灭，所以不能用用燃着的小木条区分N<SUB>2</SUB>与CO<SUB>2</SUB>，故A错误；<br />B、NaOH溶液和Na<SUB>2</SUB>CO<SUB>3</SUB>溶液都呈碱性，都能使酚酞试液变红，所以不能用酚酞溶液区分NaOH溶液和Na<SUB>2</SUB>CO<SUB>3</SUB>溶液，故B错误；<br />C、氢氧化钠能和二氧化硫反应生成亚硫酸钠和水，所以不能用氢氧化钠固体干燥SO<SUB>2</SUB>，故C错误；<br />D、Fe和CuSO<SUB>4</SUB>溶液能反应，说明铁排在了铜的前面，Ag和CuSO<SUB>4</SUB>溶液不反应，说明银排在铜的后面，所以铁大于铜大于银，故D正确．<br />故选：D．','【分析】A、根据氮气和二氧化碳都能使燃着的木条熄灭进行解答；<br />B、根据NaOH溶液和Na<SUB>2</SUB>CO<SUB>3</SUB>溶液都呈碱性，都能使酚酞试液变红进行解答；<br />C、根据氢氧化钠能和二氧化硫反应生成亚硫酸钠和水进行解答；<br />D、根据金属活动性顺序结合实验现象分析．','选择题',3.00,'40b6c41508e903ed48765bc5d3fbdc78',9,400,'化学实验方案设计与评价,常见气体的检验与除杂方法,金属活动性顺序及其应用,根据浓硫酸或烧碱的性质确定所能干燥的气体,酸、碱、盐的鉴别','',2016,'32','2016•荔湾区模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840597,'下列化学用语中，既能表示一种物质，还能表示一种元素的是（　　）','Cu','H','Cl<SUB>2</SUB>','2N','','A','【解答】解：元素符号能表示一种元素，还能表示该元素的一个原子；化学式能表示一种物质，当元素符号又是化学式时，就同时具备了上述三层意义．<br />A、Cu属于金属元素，可表示铜元素，表示一个铜原子，还能表示铜这一纯净物，故选项符合题意．<br />B、H属于气态非金属元素，可表示氢元素，表示一个氢原子，但不能表示一种物质，故选项不符合题意．<br />C、该符号是氮气的化学式，不是元素符号，故选项不符合题意．<br />D、2N可表示2个氮原子，不能表示一种物质，故选项不符合题意．<br />故选：A','【分析】根据化学式与元素符号的含义进行分析解答，金属、大多数固体非金属等都是由原子直接构成的，故它们的元素符号，既能表示一个原子，又能表示一种元素，还能表示一种物质．','选择题',3.00,'1e4fbee005b3857dd212fcbf91f4d5fd',9,400,'元素的符号及其意义,化学式的书写及意义','',2016,'37','2016•工业园区一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840599,'某同学发现，上个月做实验用的氢氧化钠溶液忘记了盖瓶盖．对于溶液是否变质，同学们提出了如下假设：<br />【猜想】猜想一：没有变质；　　　猜想二：部分变质；　　猜想三：（1）<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />【查阅资料】Na<SUB>2</SUB>CO<SUB>3</SUB>溶液呈碱性．<br />【进行实验】对猜想二进行验证．<br /><table class=\"edittable\"><TBODY><TR><td width=212>实验步骤</TD><td width=181>实验现象</TD><td width=110>结论</TD></TR><TR><td>（2）取少许该溶液于试管中，滴<br />入适量的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>溶液．</TD><td>（3）<u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD><td rowSpan=2>猜想二正确</TD></TR><TR><td>（4）静置上述溶液，取上层清液少许于另一支试管中，滴入几滴<br /><u>&nbsp;&nbsp;&nbsp;&nbsp;</u>溶液</TD><td>（5）<u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD></TR></TBODY></TABLE>写出现象（3）的化学方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />【反思】氢氧化钠溶液必须密封保存．<br />（7）氢氧化钠溶液变质的原因<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（用化学方程表示）．','','','','','','全部变质$###$BaCl<SUB>2</SUB>（CaCl<SUB>2</SUB>、Ba（NO<SUB>3</SUB>）<SUB>2</SUB>）$###$有白色沉淀$###$酚酞$###$溶液变红$###$Na<SUB>2</SUB>CO<SUB>3</SUB>+BaCl<SUB>2</SUB>═2NaCl+BaCO<SUB>3</SUB>↓$###$CO<SUB>2</SUB>+2NaOH═Na<SUB>2</SUB>CO<SUB>3</SUB>+H<SUB>2</SUB>O','【解答】解：【猜想】氢氧化钠露置在空气的情况是不变质、部分变质或是不变质，故填：全部变质；<br />【进行实验】向溶液的样品中滴加氯化钙溶液，利用氯化钙与碳酸钠反应生成碳酸钙沉淀和氯化钠，若观察到有沉淀出现可判断变质生成了碳酸钠；在反应后的溶液中滴入酚酞，由于氢氧化钠能使酚酞变红，若出现变红则可判断有氢氧化钠；因此既出现沉淀又有溶液变红，则溶液部分变质；（合理即可）；故答案为：<br /><table class=\"edittable\"><TBODY><TR><td width=212>实验步骤</TD><td width=181>实验现象</TD><td width=110>结论</TD></TR><TR><td>1、取少许该溶液于试管中，滴<br />入适量的BaCl<SUB>2</SUB>（CaCl<SUB>2</SUB>、Ba（NO<SUB>3</SUB>）<SUB>2</SUB>）溶液．</TD><td>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;有白色沉淀</TD><td rowSpan=2>猜想二正确</TD></TR><TR><td>2、静置上述溶液，取上层清液少许于另一支试管中，滴入几滴<br />酚酞溶液</TD><td>溶液变红</TD></TR></TBODY></TABLE>现象③是碳酸钠和氯化钡的反应生产了氯化钠和碳酸钡沉淀，故填：Na<SUB>2</SUB>CO<SUB>3</SUB>+BaCl<SUB>2</SUB>═2NaCl+BaCO<SUB>3</SUB>↓；<br />【反思】因为氢氧化钠吸收二氧化碳生成碳酸钠和水而变质，故填：CO<SUB>2</SUB>+2NaOH═Na<SUB>2</SUB>CO<SUB>3</SUB>+H<SUB>2</SUB>O．','【分析】【猜想】根据已有的知识进行分析，物质露置在空气的情况是不变质、部分变质或是不变质分析；<br />【进行实验】验证溶液部分变质，既要证明有氢氧化钠，又要证明有碳酸钠进行分析：碳酸钠能够和氯化钡等溶液产生沉淀既具有氢氧化钠变质成碳酸钠，又能够除去碳酸钠碱性对氢氧化钠的干扰，再滴加酚酞试液观察变色，确定溶液的组成；并写出反应的方程式；<br />【反思】根据氢氧化钠吸收二氧化碳生成碳酸钠和水，写出反应的方程式．','书写',3.00,'16acc3c78996b01c9e38eea1dccd690a',9,400,'药品是否变质的探究,碱的化学性质,盐的化学性质,书写化学方程式、文字表达式、电离方程式','',2016,'37','2016•黑龙江二模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840600,'类推是学习中常用的思维方法．现有以下类推结果，其中错误的是（　　）<br />①酸碱中和反应生成盐和水，所以生成盐和水的反应一定是中和反应；②氯化钠的水溶液显中性，所以盐溶液一定显中性；③分子可以构成物质，物质一定是由分子构成的；④碳酸盐与盐酸反应放出气体，所以与盐酸反应放出气体的一定是碳酸盐．','只有②③④','只有①②③','①②③④','只有①','','C','【解答】解：①酸碱中和反应生成盐和水，所以生成盐和水的反应一定是中和反应错误，如金属氧化物与酸反应生成盐和水，不是中和反应；②氯化钠的水溶液显中性，所以盐溶液一定显中性错误，碳酸钠的水溶液显碱性；③分子可以构成物质，物质一定是由分子构成的错误，分子、原子、离子都能构成物质；④碳酸盐与盐酸反应放出气体，所以与盐酸反应放出气体的一定是碳酸盐错误，还可能是比较活泼的金属与盐酸反应放出氢气；故选C','【分析】由题目的信息可知：①酸碱中和反应生成盐和水，所以生成盐和水的反应一定是中和反应错误，如金属氧化物与酸反应生成盐和水，不是中和反应；②氯化钠的水溶液显中性，所以盐溶液一定显中性错误，碳酸钠的水溶液显碱性；③分子可以构成物质，物质一定是由分子构成的错误，分子、原子、离子都能构成物质；④碳酸盐与盐酸反应放出气体，所以与盐酸反应放出气体的一定是碳酸盐错误，还可能是比较活泼的金属与盐酸反应放出氢气．','选择题',3.00,'f585c1ec7e7b4ae7ab5e7b0255f175b1',9,400,'酸的化学性质,中和反应及其应用,盐的化学性质,分子、原子、离子、元素与物质之间的关系','',2016,'37','2016•遂川县一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840601,'<img  src=\"/tikuimages/9/2016/400/shoutiniao62/1b2359f0-94d4-11e9-b68d-b42e9921e93e_xkb0.png\" style=\"vertical-align:middle;FLOAT:right\" />如图所示实验中，①、④为用紫色石蕊溶液润湿的棉球，②、③为用石蕊溶液染成紫色的干燥棉球．下列能说明密度大于空气且能与水反应的现象是（　　）','①变红，③不变红','④变红，③不变红','①、④变红，②、③不变红','④比①先变红，②、③不变红','','D','【解答】解：通二氧化碳，用石蕊溶液染成紫色的干燥棉球②③都不变色，说明二氧化碳不能使石蕊变色，通二氧化碳，用紫色石蕊溶液润湿的棉球都变红，说明二氧化碳和水发生了化学反应．并且④比①先变红，说明二氧化碳的密度比空气的大．<br />故选：D','【分析】根据二氧化碳能和水反应生成碳酸，碳酸显酸性，能使石蕊试液变红色解答．','选择题',3.00,'ef0d38785f343c86329f5d690da81285',9,400,'二氧化碳的物理性质,二氧化碳的化学性质','',2016,'37','2016•雁江区一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840603,'今天是化学兴趣小组活动日，同学们展开了一系列探究，也请你参与．<br />（1）某实验测出人呼吸中各种气体的体积分数，如表所示：<br /><table class=\"edittable\"><TBODY><TR><td width=88>气体</TD><td width=139>吸入气体</TD><td width=139>呼出气体</TD></TR><TR><td>x</TD><td>78%</TD><td>75%</TD></TR><TR><td>y</TD><td>21%</TD><td>15%</TD></TR><TR><td>CO<SUB>2</SUB></TD><td>0.03%</TD><td>3.68%</TD></TR><TR><td>水蒸气</TD><td>0.06%</TD><td>5.44%</TD></TR><TR><td>其他</TD><td>0.91%</TD><td>0.88%</TD></TR></TBODY></TABLE>①请你判断：x是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，因参与人体新陈代谢而消耗的气体是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />②请你分析：x气体在呼吸过程中没有参与化学反应，但在呼出气体中体积分数却减少了，原因是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）老师拿出两瓶无标签的试剂，分别是固体和液体．他取少量试剂在试管中混合，立即产生一种无色气体．此气体是什么物质？请你完成验证它的实验方案：<br /><table class=\"edittable\"><TBODY><TR><td width=180>猜&nbsp;&nbsp;想</TD><td width=180>实验步骤</TD><td width=180>现象及结论</TD></TR><TR><td>该气体可能是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD><td></TD><td></TD></TR></TBODY></TABLE>','','','','','','氮气$###$氧气$###$由于呼出的气体中二氧化碳与水蒸气含量增加而引起的$###$氧气','【解答】解：（1）①由于空气中氮气的体积分数是78%，所以x是氮气，因参与人体新陈代谢而消耗的气体是氧气．<br />②由表中的数据可知，氮气体在呼吸过程中没有参与化学反应，但在呼出气体中体积分数却减少了，原因是由于呼出的气体中二氧化碳与水蒸气的含量增加而引起的．<br />（2）由于过氧化氢在二氧化锰的作用下能放出氧气，氧气具有助燃性，所以填表如下：<br /><table class=\"edittable\"><TBODY><TR><td width=180>猜&nbsp;&nbsp;想</TD><td width=180>实验步骤</TD><td width=180>现象及结论</TD></TR><TR><td>该气体可能是氧气</TD><td>用带火星的木条检验该气体</TD><td>带火星的木条复燃，<br />该气体是氧气</TD></TR></TBODY></TABLE>故答为：：（1）①氮气，氧气；②由于呼出的气体中二氧化碳与水蒸气含量增加而引起的．<br />（2）<br /><table class=\"edittable\"><TBODY><TR><td width=180>猜&nbsp;&nbsp;想</TD><td width=180>实验步骤</TD><td width=180>现象及结论</TD></TR><TR><td>氧气</TD><td>用带火星的木条检验该气体</TD><td>带火星的木条复燃，<br />该气体是氧气</TD></TR></TBODY></TABLE>','【分析】（1）根据吸入的空气和呼出气体成分的变化分析回答；<br />（2）根据过氧化氢在二氧化锰的作用下能放出氧气及氧气的性质分析回答．','填空题',3.00,'56be8fbbfb72ba42d5228ec3c3c45b38',9,400,'常见气体的检验与除杂方法,吸入空气与呼出气体的比较','',2015,'35','2015秋•合肥校级期中',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840606,'下列关于分子、原子的说法中，错误的是（　　）','分子是不断运动','化学变化中原子的种类不变','分子间有一定间隔','原子是不可再分的最小粒子','','D','【解答】解：A、分子始终在不断运动，故A说法正确；<br />B、化学反应前后原子种类、数目、质量不变，故B说法正确；<br />C、分子间一定有间隔，故C说法正确；<br />D、原子是化学变化中的最小粒子，原子是用化学方法不能再分，原子可以再分为原子核和核外电子，故D说法错误．<br />故选D．','【分析】A、根据分子始终在不断运动考虑；B、根据反应前后原子种类、数目、质量不变；C、根据分子的特点考虑；D、原子是化学变化中的最小粒子．','选择题',3.00,'5789316c9152f6bf3a1d614776d313d8',9,400,'原子的定义与构成,分子和原子的区别和联系','',2015,'37','2015•长春二模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840622,'如图是配制溶质质量分数为10%的NaCl溶液的实验操作示意图如图<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao16/1b5cb96e-94d4-11e9-bee6-b42e9921e93e_xkb41.png\" style=\"vertical-align:middle\" /><br />（1）用图1中序号表示配制溶液的制取操作顺序<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />（2）称量NaCl时，天平 平衡后的状态如图⑤所示，游码标尺位置见图，则称取的NaCl称量为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>g<br />（3）NaCl称量完毕放回砝码时，发现10g的砝码有缺损，若其他操作步骤正确，则所配制溶液的质量分数<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“大于”“小于”或“等于”）10%<br />（4）量取蒸馏水操作如下：将蒸馏水注入量筒，待液面接近量取体积对应刻度线时，改用<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>填仪器名称），滴加蒸馏水至刻度线，如图2观察方式正确的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填字母）<br />（5）若用C观察方式量取水的体积，所配制溶液的溶质质量分数会<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“偏大”“偏小”或“无影响”）','','','','','','②⑤①④③$###$18.2$###$小于$###$胶头滴管$###$A$###$偏大','【解答】解：（1）根据用固体溶质氯化钠配制溶液的步骤：计算-称量-溶解，先取氯化钠进行称取，倒入烧杯；然后量取水，倒入盛氯化钠的烧杯中；最后进行溶解；故答案为：②⑤①④③；<br />（2）砝码的质量为10g+5g=15g，游码的质量为3.2g，所称取氯化钠的质量=15g+3.2g=18.2g；<br />（3）NaCl称量完毕放回砝码时，发现10g的砝码有缺损，则会造成实际称取的氯化钠的质量偏少，则所配制溶液的质量分数小于10%．<br />（4）量取蒸馏水操作如下：将蒸馏水注入量筒，待液面接近量取体积对应刻度线时，改用胶头滴管，量筒读数时视线要与量筒内液体的凹液面的最低处保持水平，观察方式正确的是A．<br />（5）俯视读数偏大，实际量取的水体积偏少，会造成配制溶液的溶质质量分数会偏大．<br />故答案为：<br />（1）②⑤①④③；（2）18.2；（3）小于；（4）胶头滴管，A；&nbsp;（5）偏大．','【分析】（1）图①操作为把称取的氯化钠倒入烧杯，图②的操作为用药匙取氯化钠，图③操作为用玻璃棒搅拌溶解氯化钠，图④操作为把量取的水倒入盛有氯化钠的烧杯中，图⑤操作为称取氯化钠；根据利用固体溶质氯化钠配制溶液的步骤，对操作进行排序；<br />（2）砝码与游码的质量和为所称氯化钠的质量，游码读数时应以游码左侧数值为准；<br />（3）NaCl称量完毕放回砝码时，发现10g的砝码有缺损，则会造成实际称取的氯化钠的质量偏少，据此进行分析解答；<br />（4）根据量筒的使用注意事项：量筒读数时视线要与凹液面的最低处保持水平，进行分析解答；<br />（5）考虑俯视读数偏大，实际量取的水体积较少，会造成配制溶液的溶质质量分数会偏大','简答题',3.00,'37290d779f0f6e2e751edbfaec00ec34',9,400,'测量容器-量筒,称量器-托盘天平,一定溶质质量分数的溶液的配制','',2016,'37','2016春•凉山州校级月考',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840623,'推理是化学学习的重要方法，下列由事实得出的结论正确的是（　　）<br /><table class=\"edittable\"><TBODY><TR><td width=46>编号</TD><td width=296>事实</TD><td width=228>结论</TD></TR><TR><td>A</TD><td>某气体不能使带火星木条复燃</TD><td>该气体一定不含氧气</TD></TR><TR><td>B</TD><td>某物质可以导电</TD><td>该物质一定是金属</TD></TR><TR><td>C</TD><td>某元素原子的质子数为8</TD><td>该原子核外电子数也一定为8</TD></TR><TR><td>D</TD><td>某物质在氧气中完全燃烧，生成物只有CO<SUB>2</SUB>和H<SUB>2</SUB>O</TD><td>该物质一定由C、H、O三种元素组成</TD></TR></TBODY></TABLE>','A','B','C','D','','C','【解答】解：A、空气中含有氧气，但不能使带火星的木条复燃；故错误；<br />B、石墨具有导电性，但它是非金属；故错误；<br />C、在原子里质子数=电子数，可知该原子核外电子数一定为8；故正确；<br />D、某物质完全燃烧，生成物只有CO<SUB>2</SUB>和H<SUB>2</SUB>O，由质量守恒定律可知，该物质一定含有C、H元素，可能含有O元素，故错误；<br />故选项为：C．','【分析】A、从空气中含有氧气，但空气不能使带火星的木条复燃去分析；<br />B、从石墨具有导电性，但它是非金属去分析；<br />C、根据在原子里质子数=电子数去分析；<br />D、根据元素守恒分析．','选择题',3.00,'37b8c9d2fc4fdba35878694644343493',9,400,'常见气体的检验与除杂方法,溶液的导电性及其原理分析,核外电子在化学反应中的作用,质量守恒定律及其应用','',2015,'37','2015秋•颍州区校级月考',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840626,'<img  src=\"/tikuimages/9/2016/400/shoutiniao43/1b6df780-94d4-11e9-8d16-b42e9921e93e_xkb17.png\" style=\"vertical-align:middle;FLOAT:right;\" />（2016•句容市校级模拟）A、D、E、F为氧化物，B、C为单质，C不能与稀硫酸反应．除F外都是初中化学中常见的物质．F与A的组成元素相同，且各元素均呈现常见的化合价，F的相对分子质量为144．在一定条件下，它们的转化关系如图所示．<br />（1）物质F的化学式是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）反应①属于<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>反应（填基本反应类型）．<br />（3）反应④<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“放出”或“吸收”）热量．<br />（4）写出反应②的化学方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；写出反应③的化学方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','Cu<SUB>2</SUB>O$###$置换$###$吸收$###$CuO+CO<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>Cu+CO<SUB>2</SUB>$###$C+O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>CO<SUB>2</SUB>','【解答】解：（1）单质C不能与稀硫酸反应，因此猜想C是铜；A+B═C+D，A+E═C+D，且C可以转化成A，可知A是CuO，B是C，C是Cu，D是CO<SUB>2</SUB>，E是CO，（B可以生成D，D可以生成E）C可以生成F，又知F的分子量为144，可知F是Cu<SUB>2</SUB>O（氧化亚铜）．<br />（2）反应①的化学方程式为；2CuO+C<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2Cu+CO<SUB>2</SUB>↑，产物与生成物均是一种单质和一种化合物，所以属于置换反应；<br />（3）反应④的化学方程式为：C+CO<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>CO，由于该反应需要提供热量，所以CO<SUB>2</SUB>生成CO的反应为吸热反应．<br />（4）反应②的化学方程式为：CuO+CO<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>Cu+CO<SUB>2</SUB>；反应③的化学方程式为；C+O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>CO<SUB>2</SUB>．<br />故答案为；（1）Cu<SUB>2</SUB>O；<br />（2）置换吸热；<br />（3）吸收；<br />（4）CuO+CO<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>Cu+CO<SUB>2</SUB>；C+O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>CO<SUB>2</SUB>．','【分析】单质C不能与稀硫酸反应，因此猜测判断C是铜，又知F的分子量为144，可知F是Cu<SUB>2</SUB>O（氧化亚铜）．以此为突破口，逐一分析，即可正确解答．','书写',3.00,'a04755e046b534860174293117152891',9,400,'物质的鉴别、推断,物质发生化学变化时的能量变化,反应类型的判定,书写化学方程式、文字表达式、电离方程式','句容市',2016,'32','2016•句容市校级模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840628,'当前，人类以化石燃料为主要能源<br />（1）化石燃料有煤、<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>和天然气，化石燃料属于<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“可再生”或“不可再生”）能源<br />（2）甲烷完全燃烧的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />（3）从表中数据分析，与煤相比，用天然气作燃料的优点是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（打1点即可）<br /><table class=\"edittable\"><TBODY><TR><td width=142 rowSpan=2>&nbsp;</TD><td width=426 colSpan=3>&nbsp;&nbsp;&nbsp;&nbsp; 1g物质完全燃烧产生</TD></TR><TR><td>&nbsp;放出的热量/kJ</TD><td>&nbsp;CO<SUB>2</SUB>的质量/g</TD><td>&nbsp;产生SO<SUB>2</SUB>的质量/mg</TD></TR><TR><td>&nbsp;甲烷</TD><td>&nbsp;56</TD><td>&nbsp;2.75</TD><td>&nbsp;0.3</TD></TR><TR><td>&nbsp;煤</TD><td>&nbsp;32</TD><td>&nbsp;3.67</TD><td>&nbsp;11</TD></TR></TBODY></TABLE>','','','','','','石油$###$不可再生$###$CH<SUB>4</SUB>+2O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>CO<SUB>2</SUB>+2H<SUB>2</SUB>O$###$产生的二氧化碳少，放出的热量多','【解答】解：（1）化石燃料包括煤、天然气和石油；化石燃料属于不可再生能源；<br />（2）甲烷燃烧生成二氧化碳和水，该反应的化学方程式为：CH<SUB>4</SUB>+2O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>CO<SUB>2</SUB>+2H<SUB>2</SUB>O；<br />（3）由题目提供的信息可知，相同质量的甲烷和煤相比，甲烷燃烧产生的二氧化碳少，放出的热量多；<br />故答案为：（1）石油；不可再生；（2）CH<SUB>4</SUB>+2O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>CO<SUB>2</SUB>+2H<SUB>2</SUB>O；（3）产生的二氧化碳少，放出的热量多．','【分析】（1）根据化石燃料的种类分析；<br />（2）根据甲烷燃烧生成二氧化碳和水，写出反应的化学方程式即可；<br />（3）根据题目提供的信息分析解答．','书写',3.00,'11b07eb1e0d7e1748987f508e0d5a7a3',9,400,'书写化学方程式、文字表达式、电离方程式,化石燃料及其综合利用,常见能源的种类、能源的分类','',2016,'37','2016•鞍山一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840629,'请结合如图回答问题：<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao90/1b7598a1-94d4-11e9-86db-b42e9921e93e_xkb28.png\" style=\"vertical-align:middle\" /><br />（1）写出图中标号的仪器名称：a<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>&nbsp;&nbsp;b<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />（2）给A装置试管里的固体物质加热时，应先<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，然后集中加热盛有药品的部位<br />（3）氢气是一种无色无味气体、难溶于水、密度比空气小的气体．实验室常用稀硫酸和锌粒制取氢气，因为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>和<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，所以发生装置应选用<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．收集氢气时，使用<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>的方法收集到得气体比较纯净．','','','','','','分液漏斗$###$试管$###$预热$###$反应物是固体和液体$###$反应条件是常温$###$A$###$C','【解答】解：（1）通过分析题中所指仪器的作用可知，a是分液漏斗，b是试管；<br />（2）给试管里的固体物质加热时，应先预热，然后集中加热盛有药品部位；<br />（3）实验室制取氢气的反应物是固体和液体，反应条件是常温，所以发生装置应选用A，氢气密度比空气小，难溶于水，所以收集氢气时，使用C所示的方法收集到集气瓶口有气泡冒出．<br />故答案为：<br />（1）分液漏斗，试管；（2）预热；<br />（3）反应物是固体和液体，反应条件是常温，A，C．','【分析】（1）根据实验室常用仪器的名称和题中所指仪器的作用进行分析；<br />（2）根据给试管里的固体物质加热时，应先预热，然后集中加热盛有药品部位进行分析；<br />（3）根据实验室制取氢气的反应物是固体和液体，反应条件是常温，氢气密度比空气小，难溶于水进行分析．','填空题',3.00,'aa6f366825bcaced0133471197caef7c',9,400,'给试管里的固体加热,氢气的制取和检验','',2016,'37','2016•市南区二模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840631,'化学与我们的日常生活紧密相关，现有①石墨&nbsp;②氮气&nbsp;③盐酸&nbsp;④钛合金&nbsp;⑤3%-5%的碳酸氢钠溶液&nbsp;⑥葡萄糖&nbsp;⑦硫酸铜&nbsp;⑧醋酸&nbsp;⑨氢气等物质，请按要求用序号填空：<br />（1）可作为人体重要能量来源的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />（2）与人体具有很好“相容性”，可用来制造人造骨的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（3）人体胃液中主要的酸是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（4）食品包装袋内填充的保护气是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','⑥$###$④$###$③$###$②','【解答】解：（1）可作为人体重要能量来源的是葡萄糖，<br />（2）与人体具有很好“相容性”，可用来制造人造骨的是钛合金；<br />（3）人体胃液中主要的酸是盐酸；<br />（4）氮气的化学性质不活泼，可用作食品包装袋内填充的保护气，<br />故答案为：⑥；④；③；②．','【分析】物质的性质决定物质的用途，根据人体胃液中主要的酸是盐酸，氮气的化学性质不活泼；钛合金与人体具有很好“相容性”进行分析解答．','填空题',3.00,'d8a7a5881d910803924f61d8e5bf00e3',9,400,'常见气体的用途,合金与合金的性质,酸的物理性质及用途,生命活动与六大营养素','',2016,'32','2016•清远模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840632,'下列实验操作正确的是（　　）','<img  src=\"/tikuimages/9/2014/400/shoutiniao81/1b8242cf-94d4-11e9-91b2-b42e9921e93e_xkb81.png\" style=\"vertical-align:middle\" /><br />&nbsp;&nbsp;点燃酒精灯','<img  src=\"/tikuimages/9/2014/400/shoutiniao75/1b848cc0-94d4-11e9-a1a6-b42e9921e93e_xkb36.png\" style=\"vertical-align:middle\" /><br />&nbsp;&nbsp;&nbsp; 称量10.05g固体','<img  src=\"/tikuimages/9/2014/400/shoutiniao26/1b874bde-94d4-11e9-a213-b42e9921e93e_xkb94.png\" style=\"vertical-align:middle\" /><br />&nbsp;&nbsp;&nbsp;&nbsp; 液体加热','<img  src=\"/tikuimages/9/2014/400/shoutiniao64/1b885d4f-94d4-11e9-95eb-b42e9921e93e_xkb68.png\" style=\"vertical-align:middle\" /><br />&nbsp;&nbsp;&nbsp;&nbsp; 量取9.5mL液体','','C','【解答】解：A、使用酒精灯时要注意“两查、两禁、一不可”，禁止用一酒精灯去引燃另一酒精灯，图中所示操作错误．<br />B、托盘天平用于粗略称量药品的质量，准确到0.1g，不能精确到0.01g，故不能用托盘天平称取10.05g固体，图中所示操作错误．<br />C、给试管中的液体加热时，用酒精灯的外焰加热试管里的液体，且液体体积不能超过试管容积的<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">1</td></tr><tr><td>3</td></tr></table></span>，图中所示操作正确．<br />D、选取量筒时，尽量选用能一次量取的最小规格的量筒．用100mL量筒量取9.5mL液体，误差太大，量筒选择不合理，图中所示操作错误．<br />故选C．','【分析】A、根据酒精灯的使用方法进行分析判断．<br />B、托盘天平用于粗略称量药品的质量，准确到0.1g．<br />C、根据给试管中的液体加热的方法进行分析判断．<br />D、从减小实验误差的角度去选择量筒的量程．','选择题',3.00,'09cdb08df08c381ffc230f8ea198c301',9,400,'测量容器-量筒,称量器-托盘天平,加热器皿-酒精灯,给试管里的液体加热','',2014,'35','2014秋•开福区期中',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840637,'制作“叶脉书签”需要100g溶质质量分数为10%的氢氧化钠溶液．请问：若用20%的氢氧化钠溶液和水配制，需要20%的氢氧化钠溶液质量是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>g．','','','','','','50','【解答】解：设需要20%的氢氧化钠溶液质量为x，根据溶液稀释前后溶质的质量不变，<br />则100g×10%=x×20%&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; x=50g．<br />故答案为：50．','【分析】根据溶液稀释前后溶质的质量不变，结合题意进行分析解答．','填空题',3.00,'67d67e82cf450cd46e0514314ef7c582',9,400,'用水稀释改变浓度的方法','枣阳市',2015,'32','2015•枣阳市校级模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840640,'现代社会，酸雨是主要环境问题之一．<br />（1）有研究发现，正常雨水的pH并不是7，主要是因为（用化学方程式表示）：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（2）一般认为，二氧化硫和氮氧化物（NO、NO<SUB>2</SUB>等）是造成酸雨的主要物质，大气中氮氧化物的主要来源是汽车尾气排放，SO<SUB>2</SUB>的主要来源是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（3）汽油的主要成分是碳氢化合物（如C<SUB>8</SUB>H<SUB>18</SUB>等），其中并不含有氮元素，你推测汽车尾气中氮元素的来源是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（4）汽油在汽车发动机里燃烧时，可能生成CO<SUB>2</SUB>，也可能生成CO．以C<SUB>8</SUB>H<SUB>18</SUB>为例，1分子C<SUB>8</SUB>H<SUB>18</SUB>燃烧完全生成CO<SUB>2</SUB>时，需要消耗12.5分子O<SUB>2</SUB>，1分子C<SUB>8</SUB>H<SUB>18</SUB>燃烧C原子完全转化为CO时，需要消耗<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>分子O<SUB>2</SUB>，为提高汽油燃烧效率，可以采取的措施是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','CO<SUB>2</SUB>+H<SUB>2</SUB>O═H<SUB>2</SUB>CO<SUB>3</SUB>$###$煤炭直接燃烧$###$空气$###$8.5$###$增加氧气的供应量','【解答】解：<br />（1）正常雨水的pH≈5.6，小于7，说明正常雨水显弱酸性；正常雨水显弱酸性的原因是二氧化碳与水反应生成碳酸，碳酸显酸性，反应的化学方程式为：CO<SUB>2</SUB>+H<SUB>2</SUB>O═H<SUB>2</SUB>CO<SUB>3</SUB>．<br />（2）氧化硫和氮氧化物（NO、NO<SUB>2</SUB>等）是造成酸雨的主要物质，大气中氮氧化物的主要来源是汽车尾气排放，SO<SUB>2</SUB>的主要来源是煤炭直接燃烧；<br />（3）汽油的主要成分是碳氢化合物（如C<SUB>8</SUB>H<SUB>18</SUB>等），其中并不含有氮元素，空气中含有大量氮气，所以汽车尾气中氮元素的来源是氮气．<br />（4）1体积C<SUB>8</SUB>H<SUB>18</SUB>燃烧完全生成CO<SUB>2</SUB>时的化学方程式是：C<SUB>8</SUB>H<SUB>18</SUB>+12.5O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>8CO<SUB>2</SUB>+9H<SUB>2</SUB>O，1体积C<SUB>8</SUB>H<SUB>18</SUB>燃烧完全生成CO时的化学方程式是C<SUB>8</SUB>H<SUB>18</SUB>+8.5O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>8CO+9H<SUB>2</SUB>O，所以1体积C<SUB>8</SUB>H<SUB>18</SUB>燃烧完全生成CO<SUB>2</SUB>时，需要消耗12.5体积O<SUB>2</SUB>，1体积C<SUB>8</SUB>H<SUB>18</SUB>燃烧完全生成CO时，需要消耗8.5体积O<SUB>2</SUB>；要使燃料燃烧释放的能量得到充分利用，则应使其充分燃烧，故尾气中一氧化碳的含量越低越好．<br />答案：（1）H<SUB>2</SUB>O+CO<SUB>2</SUB>=H<SUB>2</SUB>CO<SUB>3</SUB>&nbsp;&nbsp;&nbsp;（2）煤炭直接燃烧；（3）空气；（4）8.5；&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;增加氧气的供应量．','【分析】（1）根据空气中含有二氧化碳，二氧化碳可与水反应生成碳酸，碳酸显酸性，进行分析解答．<br />（2）根据空气中二氧化硫的来源主要是矿物燃料的燃烧进行分析解答．<br />（3）根据空气中含有大量氮气进行分析解答．<br />（4）依据气体的体积比等于方程式中化学式前的化学计量数之比及C<SUB>8</SUB>H<SUB>18</SUB>燃烧完全生成CO<SUB>2</SUB>，和生成CO的方程式分析回答；要使燃料燃烧释放的能量得到充分利用，则应使其充分燃烧，故尾气中一氧化碳的含量越低越好．','书写',3.00,'36561d0a47a332e43d4e3ccbafd13458',9,400,'酸雨的产生、危害及防治,书写化学方程式、文字表达式、电离方程式,完全燃烧与不完全燃烧','',2016,'32','2016•常州模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840645,'下列关于金属的说法正确的是（　　）','人类使用金属铝的年代早于铜、铁','“真金不怕火炼”表明金的熔点高','通过高炉炼得的铁为生铁','武德合金的熔点高，可用于制造保险丝','','C','【解答】解：<br />A、人类使用最早的金属材料是铜，较晚的是铝，故A错误；<br />B、“真金不怕火炼”表明金（Au）在高温条件下也很难与氧气反应，不是熔点高，故B错误；<br />C、通过高炉炼得的铁为生铁而不是纯铁，故C正确；<br />D、合金的熔点比各成分的熔点低，因此应该是武德合金的熔点低，可用于制造保险丝，故D错误．<br />故选：C．','【分析】A、根据人类对金属的利用情况分析；<br />B、根据金的化学性质分析；<br />C、根据高炉炼得的铁为生铁分析；<br />D、根据合金的性质与保险丝的特点进行分析．','选择题',3.00,'8492166624a772e2806324e8076011be',9,400,'合金与合金的性质,金属的化学性质,金属材料及其应用,铁的冶炼','',2016,'37','2016•淮安校级二模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840646,'在生产、生活中下列做法，你认为不合理的是（　　）','胃酸过少的病人可服用食醋缓解','园林工人将含有硫磺粉的石灰浆涂刷在树上用于防止虫蛀','碳酸钡悬浊液用于医疗上作为检查胃病的钡餐','二氧化碳可用作气体肥料','','B','【解答】解：A、胃酸过少的病人可服用食醋缓解，故选项说法正确．<br />B、含有硫磺粉的石灰浆具有一定的杀菌、杀虫能力，园林工人将含有硫磺粉的石灰浆涂刷在树上用于防止害虫生卵，防止冻伤，故选项说法错误．<br />C、硫酸钡难溶于水、不与酸反应，碳酸钡悬浊液用于医疗上作为检查胃病的钡餐，故选项说法正确．<br />D、二氧化碳是植物光合作用的原料，可用作气体肥料，故选项说法正确．<br />故选：B．','【分析】A、胃酸过少的病人可补充酸性物质，进行分析判断．<br />B、根据含有硫磺粉的石灰浆具有一定的杀菌、杀虫能力，进行分析判断．<br />C、根据硫酸钡难溶于水、不与酸反应，进行分析判断．<br />D、根据二氧化碳是植物光合作用的原料，进行分析判断．','选择题',3.00,'0c34a39bae51e785df1164f20723a8b6',9,400,'二氧化碳的用途,酸碱盐的应用','',0,'37','',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840647,'下列物质的转化关系均能一步实现的是（　　）<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao80/1ba72ff0-94d4-11e9-9ab3-b42e9921e93e_xkb35.png\" style=\"vertical-align:middle\" />','①③','②③','③','①②③','','C','【解答】解：①氯化钠和硝酸银会生成氯化银和硝酸钠，硫酸钠和氯化钡会生成硫酸钡沉淀和氯化钡，硝酸钠不会转化成硫酸钠，故错误；<br />②氢氧化铁和盐酸会生成氯化铁，氯化铁和不会转化成硫酸亚铁，硫酸亚铁不会转化成氢氧化铁，故错误；<br />③二氧化碳通过光合作用会生成氧气，氧气和碳会生成一氧化碳，一氧化碳和氧气会生成二氧化碳，故正确．<br />故选：C．','【分析】根据题目给出的信息和流程图，回顾物质的性质，根据物质之间的转变关系，找出各物质的变化中，每一转化在一定条件下均能一步实现的选择项．','选择题',3.00,'06651273c51345d4110ca4d61e838758',9,400,'二氧化碳的化学性质,一氧化碳的化学性质,碱的化学性质,盐的化学性质,物质的相互转化和制备','',2016,'32','2016•镇江模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840655,'下列对有关事实的解释中，正确的是（　　）','石墨和金刚石物理性质不同--碳原子结构不同','一氧化碳和二氧化碳的化学性质不同--分子构成不同','用CO<SUB>2</SUB>鉴别NaOH和Ca（OH）<SUB>2</SUB>溶液--CO<SUB>2</SUB>只与Ca（OH）<SUB>2</SUB>反应','用酚酞溶液不能鉴别稀硫酸和稀盐酸--酚酞溶液遇两种酸都变红色','','B','【解答】解：A、石墨和金刚石物理性质不同，是因为碳原子的排列方式不同，故选项说法错误．<br />B、一氧化碳和二氧化碳的化学性质不同，是因为它们分子的构成不同，不同种的分子性质不同，故选项解释正确．<br />C、用CO<SUB>2</SUB>鉴别NaOH和Ca（OH）<SUB>2</SUB>溶液，不是因为CO<SUB>2</SUB>只与Ca（OH）<SUB>2</SUB>反应，二氧化碳也能与氢氧化钠溶液反应，但无明显变化，故选项说法错误．<br />D、用酚酞溶液不能鉴别稀硫酸和稀盐酸，是因为酚酞溶液遇两种酸都不变色，故选项说法错误．<br />故选：B．','【分析】A、根据石墨和金刚石物理性质不同的原因，进行分析判断．<br />B、同种的分子性质相同，不同种的分子性质不同，进行分析判断．<br />C、根据碱溶液能与二氧化碳反应进行分析判断．<br />D、无色酚酞溶液遇酸性溶液不变色，遇碱性溶液变红．','选择题',3.00,'fab44aea97eb2471b56986a89f0a4838',9,400,'酸、碱、盐的鉴别,分子的定义与分子的特性,碳元素组成的单质','',2016,'37','2016•南关区一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840657,'在探究二氧化碳性质的实验课上，同学们把一瓶CO<SUB>2</SUB>气体（排空气法收集，并已验满）倒扣在盛水的水槽中，有的小组没有观察到水压入瓶中，有的小组观察到少量水压入瓶中．针对课堂中出现的现象，化学小组的同学在课外进行了探究．<br />【提出问题】按教材的说法“1体积水大约能溶解1体积CO<SUB>2</SUB>”，水槽中的水足够把<br />集气瓶中的CO<SUB>2</SUB>气体全部溶解，为什么集气瓶中的水面不上升或上升很少呢？<br />【作出猜想】1、排空气法收集到的CO<SUB>2</SUB>气体可能不纯．<br />2、<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />3、…<br />【进行实验】小组同学分别用排空气法和排水法各收集2瓶CO<SUB>2</SUB>，重复图B实验，在不同时间测得压入的水量占集气瓶容积的百分数，数据见表1．（已知：CO<SUB>2</SUB>在饱和碳酸氢钠溶液中溶解度很小）<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao9/1bd2acc0-94d4-11e9-9535-b42e9921e93e_xkb93.png\" style=\"vertical-align:middle\" /><br />表1&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;24小时内CO<SUB>2</SUB>溶解情况<br /><table class=\"edittable\"><TBODY><TR><td width=97>气体样本</TD><td width=114>1</TD><td width=114>2</TD><td width=114>3</TD><td width=92>4</TD></TR><TR><td>收集方法</TD><td>向上排空气法</TD><td>向上排空气法</TD><td>排水法</TD><td>排水法</TD></TR><TR><td>2小时</TD><td>13.7%</TD><td>15.0%</TD><td>15.3%</TD><td>15.1%</TD></TR><TR><td>24小时</TD><td>35.7%</TD><td>56.3%</TD><td>94.2%</TD><td>93.6%</TD></TR></TBODY></TABLE>实验室制CO<SUB>2</SUB>的化学反应方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />【得出结论】<br />课堂中同学们没有观察到预期实验现象的原因是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />【实验反思】<br />该实验引发了同学们对向上排空气法收集CO<SUB>2</SUB>的纯度及CO<SUB>2</SUB>收集方法的思考．<br />（1）向上排空气法收集CO<SUB>2</SUB>的纯度．<br />小组同学用容积为&nbsp;150mL&nbsp;的集气瓶，装入部分水，留一定量的空气，用CO<SUB>2</SUB>排出这些水．用此法收集到占空气体积分数不同的CO<SUB>2</SUB>气体，再将燃着的木条伸入集气瓶中，观察是否熄灭．试验不同浓度的CO<SUB>2</SUB>使燃着的木条熄灭的情况，实验结果见表&nbsp;2．<br />表2&nbsp;&nbsp;占空气体积分数不同的CO<SUB>2</SUB>使燃着的木条熄灭情况<br /><table class=\"edittable\"><TBODY><TR><td width=180></TD><td width=95>实验1</TD><td width=95>实验2</TD><td width=95>实验3</TD><td width=76>实验4</TD></TR><TR><td>集气瓶中预留水的体积</TD><td>60mL</TD><td>45mL</TD><td>42mL</TD><td>39mL</TD></TR><TR><td>CO<SUB>2</SUB>的体积分数</TD><td>40%</TD><td>30%</TD><td>X</TD><td>26%</TD></TR><TR><td>燃着的木条状况</TD><td>灭</TD><td>灭</TD><td>灭</TD><td>不灭</TD></TR></TBODY></TABLE>上表中X的值为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）排水法和向上排空气法收集气体所需时间对照<br />小组同学分4组用排水法和排空气法分别收集1瓶（集气瓶容积相同）CO<SUB>2</SUB>气体，记录所需时间，如表3所示．<br />表3&nbsp;排水法和向上排空气法收集1瓶气体所需时间对照<br /><table class=\"edittable\"><TBODY><TR><td width=106></TD><td width=108>1</TD><td width=108>2</TD><td width=107>3</TD><td width=95>4</TD></TR><TR><td>排水法</TD><td>35“</TD><td>43“</TD><td>43“</TD><td>43“</TD></TR><TR><td>排空气法</TD><td>1\'57“</TD><td>1\'15“</TD><td>1\'4“</TD><td>1\'12“</TD></TR></TBODY></TABLE>①每次收集气体时，需要反应开始一段时间后再收集，其原因是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />②综合上述实验，请你对CO<SUB>2</SUB>的两种收集方法做出评价：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','CO<SUB>2</SUB>溶于水需要时间较长$###$CaCO<SUB>3</SUB>+2HCl═CaCl<SUB>2 </SUB>+CO<SUB>2</SUB>↑+H<SUB>2</SUB>O$###$时间短，排空气法收集的气体不纯$###$28%$###$排出装置内的空气$###$排水法比排空气法收集的气体纯度高且收集时间短','【解答】解：【作出猜想】根据探究过程可以看出，要探究在不同时间测得压入的水量占集气瓶容积的百分数所以可以推知猜想，故猜想2：CO<SUB>2</SUB>溶于水需要时间较长；<br />【进行实验】实验室用碳酸钙和稀盐酸制二氧化碳，化学方程式为：CaCO<SUB>3</SUB>+2HCl═CaCl<SUB>2 </SUB>+CO<SUB>2</SUB>↑+H<SUB>2</SUB>O；<br />【得出结论】从实验数据可以看出，时间越长溶解越充分，时间越短溶解越少，收集的气体不纯，所以没有观察到预期实验现象；<br />【实验反思】从实验数据可以看出，42mL/150mL×100%=28%；收集气体时，需要反应开始一段时间后再收集，目的是排出装置内的空气，防止气体不纯；从实验数据可以看出排水法比排空气法收集的气体纯度高且收集时间短；<br />故答案为：【作出猜想】CO<SUB>2</SUB>溶于水需要时间较长；<br />【进行实验】CaCO<SUB>3</SUB>+2HCl═CaCl<SUB>2 </SUB>+CO<SUB>2</SUB>↑+H<SUB>2</SUB>O；<br />【得出结论】时间短，排空气法收集的气体不纯；<br />【实验反思】（ 1）28%；<br />&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; （2）①排出装置内的空气；<br />&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; ②排水法比排空气法收集的气体纯度高且收集时间短．','【分析】【作出猜想】根据探究过程分析要探究的问题从而推出猜想，进行分析；<br />【进行实验】根据实验室制二氧化碳的反应，写出化学方程式；<br />【得出结论】根据实验现象和数据进行分析；<br />【实验反思】根据实验数据，以及制取和收集气体的注意事项进行分析．','书写',3.00,'d4eea778d47bdc18493ec0a4e43d6a2d',9,400,'二氧化碳的实验室制法,探究二氧化碳的性质,书写化学方程式、文字表达式、电离方程式','',2016,'37','2016•大兴区一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840662,'造纸是我国古代四大发明之一，它有效地推动了人类文明的发展．<br />（1）回收农田产生的秸秆作为造纸原料，可以减少焚烧秸秆带来的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>污染．秸秆的主要成分是纤维素[（C<SUB>6</SUB>H<SUB>10</SUB>O<SUB>5</SUB>）<SUB>n</SUB>]，纤维素中各元素的质量比为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（用最简整数比表示）．<br />（2）造纸会产生大量含NaOH的碱性废水，需经处理呈中性后再排放．若某造纸厂废水中NaOH的质量分数为1.6%，现有废硫酸19.6t（H<SUB>2</SUB>SO<SUB>4</SUB>的质量分数为20%），可以处理的废水质量是多少？','','','','','','大气$###$36：5：40','【解答】解：<br />（1）焚烧秸秆会产生大量烟尘等，污染空气；根据纤维素的化学式（C<SUB>6</SUB>H<SUB>10</SUB>0<SUB>5</SUB>）n知，纤维素中C、H、O三种元素的质量比=（12×6）：（1×10）：（16×5）=36：5：40．<br />（2）19.6t废硫酸溶液中含硫酸的质量为19.6t×20%=3.92t<br />设19.6t废硫酸溶液可以反应氢氧化钠的质量为x<br />H<SUB>2</SUB>SO<SUB>4 </SUB>+2NaOH=Na<SUB>2</SUB>SO<SUB>4 </SUB>+2H<SUB>2</SUB>O<br />&nbsp;98&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 80<br />3.92t&nbsp;&nbsp;&nbsp; x<br /><span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">98</td></tr><tr><td>3.92t</td></tr></table>=<table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">80</td></tr><tr><td>x</td></tr></table></span><br />解之得：x=3.2t<br />含1.6t氢氧化钠的废水的质量=<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">3.2t</td></tr><tr><td>1.6%</td></tr></table></span>=200t<br />故答案为：<br />（1）大气，36：5：40；（2）可以处理的废水质量是200t．','【分析】（1）焚烧秸秆会带来大气污染，化学式中各元素的质量比为各元素的相对原子质量×原子个数的之比，据此分析计算回答．<br />（2）可先根据化学方程式求出氢氧化钠的质量，在根据废水中氢氧化钠的质量分数求出废水的质量．','填空题',3.00,'d4940bf08b6d197138cc531fac2f8281',9,400,'空气的污染及其危害,元素质量比的计算,根据化学反应方程式的计算','昆山市',2016,'37','2016•昆山市二模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840666,'铝、铁、铜是我们生活中常见的金属．<br />（1）铝的利用比铜、铁晚是因为金属大规模开发和利用的先后顺序与<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>有关（填字母）<br />A．金属的活动性&nbsp;&nbsp;B．金属的导电性&nbsp;&nbsp;C．金属在地壳中的含量<br />（2）铝具有很好的抗腐蚀性能，原因是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）每年全世界钢铁的产量很高，但钢铁的锈蚀也给人类带来了巨大的损失，铁在空气中锈蚀，实际上是铁跟<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>和<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>共同作用的结果；防止铁生锈的方法是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（写一种）；写出一种你熟悉的金属与酸反应的化学方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（4）印发铜制电路板的“腐蚀性”为FeCl<SUB>3</SUB>溶液反应，化学方程式分别为①Cu+2FeCl<SUB>3</SUB>═2FeCl<SUB>2</SUB>+CuCl<SUB>2</SUB>；②Fe+2FeCl<SUB>3</SUB>═3X．则②中X的化学式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','A$###$4Al+3O<SUB>2</SUB>=2Al<SUB>2</SUB>O<SUB>3</SUB>$###$水$###$氧气$###$刷漆$###$Fe+2HCl═FeCl<SUB>2</SUB>+H<SUB>2</SUB>↑$###$FeCl<SUB>2</SUB>','【解答】解：（1）铝比铜、铁活泼，开发比较困难，所以开发利用较晚，所以金属开发利用的先后顺序与金属的活动性有关，故选：A；<br />（2）铝能与氧气发生化合反应生成氧化铝，化学方程式为：4Al+3O<SUB>2</SUB>=2Al<SUB>2</SUB>O<SUB>3</SUB>；<br />（3）铁金属生锈的条件是水与氧气并存，要想防止钢铁生锈，就需要控制导致其生锈的条件，常用的防止钢铁生锈的方法主要有：涂油漆或保持铁的表面干燥等；铁能与盐酸反应生成氯化亚铁和氢气；<br />（4）根据，反应前含有3个Fe原子，6个Cl原子，故反应后3X中含有Fe原子，6个Cl原子，故X中含有1个Fe原子，2个Cl原子．<br />故答案为：（1）A；<br />（2）4Al+3O<SUB>2</SUB>=2Al<SUB>2</SUB>O<SUB>3</SUB>；<br />（3）水，氧气，刷漆，Fe+2HCl═FeCl<SUB>2</SUB>+H<SUB>2</SUB>↑；<br />（4）FeCl<SUB>2</SUB>．','【分析】（1）根据金属越活泼，越不容易冶炼进行分析；<br />（2）根据铝和氧气反应会生成氧化铝进行分析；<br />（3）根据铁与水和氧气同时接触时容易生锈，根据生锈的条件可以判断防止生锈的方法，根据金属与酸的反应进行分析解答；<br />（4）根据质量守恒定律进行分析．','填空题',3.00,'327bc7e5863699064a49af973334e206',9,400,'金属的化学性质,金属材料及其应用,金属锈蚀的条件及其防护,质量守恒定律及其应用','灵武市',2016,'37','2016•灵武市校级一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840667,'实验桌上摆放有下列仪器，请根据要求填写下列空白．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao82/1bf772d1-94d4-11e9-87dd-b42e9921e93e_xkb14.png\" style=\"vertical-align:middle\" /><br />（1）取用10mL盐酸，要用到的仪器是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填名称）．<br />（2）加热时必须垫上石棉网的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填字母）．<br />（3）在做粗盐的提纯实验中还必须用到的一种玻璃仪器名称叫<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（4）若配制25g质量分数为10%的氯化钠溶液，仪器C的主要作用是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，下列关于配制该溶液的说法不正确的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填字母）．<br />A．上述仪器不能完成该实验<br />B．计算得氯化钠的质量为2.5g，水的体积为22.5mL<br />C．量取水时，若俯视读数，会使配得的溶液浓度偏小<br />D．使用天平时将砝码和物品放反，会导致配得的溶液浓度偏小（1g以下用游码）','','','','','','量筒和胶头滴管$###$E$###$漏斗$###$搅拌，加速溶解的速度$###$C','【解答】解：（1）取用10mL盐酸时，必须要用到的仪器有量筒和胶头滴管，故填：量筒和胶头滴管；<br />（2）加热时必须垫上石棉网的玻璃仪器是烧杯；故选：E；<br />（3）粗盐提纯实验中，需要进行过滤，所以必须用漏斗，故填：漏斗；<br />（4）玻璃棒在粗盐提纯实验中的溶解操作中用于搅拌，加速溶解的速度；<br />A、配制溶液包括计算、称量、量取、溶解，所以需要的仪器还有胶头滴管等，故A正确；<br />B、配制25g质量分数为10%的氯化钠溶液，计算得氯化钠的质量为2.5g，水的体积为22.5mL，故B正确；<br />C、量取水时，若俯视读数，则量取的水的实际体积偏少，会使配得的溶液浓度偏大，故C错误；<br />D、使用天平时将砝码和物品放反，则称取的氯化钠质量偏少，会导致配得的溶液浓度偏小，故D正确；<br />故选C．','【分析】（1）根据取用10mL盐酸时，必须要用到的仪器有量筒和胶头滴管解答；<br />（2）根据常用加热仪器进行分析；<br />（3）根据粗盐提纯实验中，需要进行过滤，进行分析；<br />（4）根据玻璃棒在粗盐提纯实验中的溶解操作中用于搅拌，加速溶解的速度，配制溶液的过程，注意事项进行分析．','填空题',3.00,'090944545bd7a49128a2842175f75929',9,400,'用于加热的仪器,测量容器-量筒,一定溶质质量分数的溶液的配制,氯化钠与粗盐提纯','',2016,'37','2016•重庆校级二模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840668,'区分下列各组物质的两种方法都正确的是（　　）<br /><table class=\"edittable\"><TBODY><TR><td width=68>选项</TD><td width=75>A</TD><td width=102>B</TD><td width=115>C</TD><td width=90>D</TD></TR><TR><td>需区分的物质</TD><td>矿泉水和蒸馏水</TD><td>人体吸入空气和呼出气体</TD><td>氯化钠和硝酸铵</TD><td>纯棉制品和涤纶制品</TD></TR><TR><td>方案一</TD><td>加肥皂水搅拌</TD><td>带火星的木条</TD><td>加适量水溶解后测温度变化</TD><td>观察颜色</TD></TR><TR><td>方案二</TD><td>观察颜色</TD><td>澄清石灰水</TD><td>加熟石灰研磨，闻气味</TD><td>灼烧闻气味</TD></TR></TBODY></TABLE>','A','B','C','D','','C','【解答】解：A、矿泉水和蒸馏水都是无色无味的液体，观察颜色的方法不能鉴别，不符合题意，故A错误；<br />B、吸入的空气和呼出的气体都不能使带火星的木条复燃，不符合题意，故B错误；<br />C、取样品，加水溶解，溶液温度明显降低的是硝酸铵，温度基本不变的是氯化钠，现象不同，可以鉴别，取样品，加入熟石灰研磨，有刺激性气味的气体产生的是硝酸铵，没有明显现象的是氯化钠，现象不同，可以鉴别，符合题意，故C正确；<br />D、纯棉制品和涤纶制品的颜色相同，观察颜色的方法不能鉴别，不符合题意，故D错误．<br />故选：C．','【分析】A、根据矿泉水和蒸馏水都是无色无味的液体进行分析；<br />B、根据吸入的空气和呼出的气体都不能使带火星的木条复燃进行分析；<br />C、根据硝酸铵溶于水溶液温度降低，硝酸铵和熟石灰研磨会产生刺激性气味的气体进行分析；<br />D、根据纯棉制品和涤纶制品的颜色相同进行分析．','选择题',3.00,'9fb0c035125cfa4d21a468d5ae54cd03',9,400,'吸入空气与呼出气体的比较,硬水与软水,酸、碱、盐的鉴别,物质的鉴别、推断,棉纤维、羊毛纤维和合成纤维的鉴别','',2016,'37','2016•乐至县一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840673,'碘盐是指在食盐（NaCl）中添加碘酸钾（KClO<SUB>3</SUB>）的食用盐，碘盐中不含碘单质（I<SUB>2</SUB>）、小明看新闻得知，当地超市出售的某品牌某批次碘盐涉嫌造假（不含碘酸钾），为了检验家里使用的该品牌碘盐是否含有碘酸钾，他决定就进行以下探究：<br />[查阅资料]①碘酸钾可溶于水，碘酸钾、酸和碘化钾（KI）三种物质共同作用可生成碘单质；②碘单质能使白色淀粉溶液变成蓝色．<br />[设计实验]小明首先准备了该品牌碘盐溶液，实验室能提供以下试剂：碘化钾溶液、稀硫酸、无色酚酞试液、氢氧化钠溶液、白色淀粉溶液．请你从中选择适当的几种，并协助完成以下实验：<table class=\"edittable\"><TBODY><TR><td width=189>实验操作</TD><td width=189>&nbsp;实验想象</TD><td width=189>&nbsp;实验推测</TD></TR><TR><td>&nbsp;取少量该碘盐溶液于试管中，然后<br /><u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD><td>&nbsp;<br /><u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD><td>反应生成碘单质，从而推测出该品牌碘盐<br /><u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“含”或“不含”碘酸钾）</TD></TR></TBODY></TABLE>','','','','','','碘化钾溶液、稀硫酸后，振荡，再向溶液中加入淀粉溶液，$###$溶液变蓝色$###$含','【解答】解：根据题中的信息链接可知，先将食盐配成溶液，如果食盐溶液中含有碘酸钾，那么加入碘化钾溶液、稀硫酸后就会有碘单质生成；再向溶液中加入淀粉溶液，溶液就会变蓝色．<br />故答案为：<br /><table class=\"edittable\"><TBODY><TR><td width=189>实验操作</TD><td width=189>&nbsp;实验想象</TD><td width=189>&nbsp;实验推测</TD></TR><TR><td>&nbsp;取少量该碘盐溶液于试管中，然后<br />碘化钾溶液、稀硫酸后，振荡，再向溶液中加入淀粉溶液，</TD><td>&nbsp;<br />溶液变蓝色</TD><td>反应生成碘单质，从而推测出该品牌碘盐<br />含（填“含”或“不含”碘酸钾）</TD></TR></TBODY></TABLE>','【分析】根据题中信息链接结合自己的知识进行分析．','填空题',3.00,'8ad444e298c6e4804ecac90894d98be7',9,400,'实验探究物质的组成成分以及含量,加碘盐的检验','',0,'37','',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840689,'工业上用电解饱和NaCl溶液的方法制烧碱，其反应的化学方程式如下：2NaCl+2H<SUB>2</SUB>O<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;通电&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2NaOH+H<SUB>2</SUB>↑+Cl<SUB>2</SUB>↑<br />（1）NaOH中钠、氧、氢元素的质量比为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）已知20℃时，NaCl的溶解度为36g，其饱和溶液中溶质和溶剂的质量比为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）20℃时，取136g饱和NaCl溶液电解，当NaCl转化了32.5%时，理论上剩余溶液的质量为多少？','','','','','','23：16：1$###$9：25','【解答】解：（1）NaOH中钠、氧、氢元素的质量比为：23：16：1；<br />（2）20℃时，NaCl的溶解度为36g，所以饱和溶液中溶质和溶剂的质量比为：36g：100g=9：25；<br />（3）设生成H<SUB>2</SUB>的质量为x，生成Cl<SUB>2</SUB>的质量为y<br />&nbsp;&nbsp; 2NaCl+2H<SUB>2</SUB>O═2NaOH+H<SUB>2</SUB>↑+Cl<SUB>2</SUB>↑<br />2×58.5&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 2&nbsp;&nbsp;&nbsp;&nbsp;71<br />36g×32.5%&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; x&nbsp;&nbsp;&nbsp; y<br /><span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">117</td></tr><tr><td>36g×32.5%</td></tr></table></span>=<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">2</td></tr><tr><td>x</td></tr></table></span>=<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">71</td></tr><tr><td>y</td></tr></table></span><br />x=0.2g<br />y=7.1g<br />剩余溶液的质量=136g˗0.2g˗7.1g=128.7g，<br />答：剩余溶液的质量为128.7g．<br />故答案为：（1）23：16：1；<br />（2）9：25；<br />（3）剩余溶液的质量为128.7g．','【分析】（1）根据氢氧化钠的化学式和相对原子质量计算元素质量比；<br />（2）根据20℃时，NaCl的溶解度为36g进行分析；<br />（3）根据化学方程式和题中的数据进行计算．','填空题',3.00,'49c0fb685e977370fa5851a2c2b3e542',9,400,'固体溶解度的概念,元素质量比的计算,根据化学反应方程式的计算','',2016,'37','2016•海淀区一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840690,'关于空气有下列表述：①空气中<U>&nbsp;</U>氧气的体积分数；②首要污染物；③空气中氮气的体积分数；④空气污染指数；⑤空气质量级别．空气质量日报的主要内容是下列中的（　　）','①③④','②④⑤','①②③④','①③⑤','','B','【解答】解：空气质量日报的主要内容是：②首要污染物、④空气污染指数、⑤空气质量级别，以及空气质量状况．<br />故选B．','【分析】根据空气质量日报的主要内容考虑本题．','选择题',3.00,'82c919fb7bd972d32ea3075c04483600',9,400,'空气的污染及其危害','',2016,'35','2016春•呼和浩特期中',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840693,'过氧化钙（CaO<SUB>2</SUB>）广泛应用于水产养殖、污水处理等领域，是优良的供氧剂．<br />资料：过氧化钙（CaO<SUB>2</SUB>）常温下能与水反应生成氢氧化钙和氧气．<br />探究一：证明某种供氧剂的主要成分为过氧化钙<br />（1）取一定量该供氧剂放置于试管中，向其中加水，再将带火星的木条伸入试管口中，观察到<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，证明该供氧剂的主要成分为过氧化钙．<br />（2）此供氧剂的保存方法是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）此供氧剂长期暴露在空气中，会变质生成CaCO<SUB>3</SUB>．为检验此供氧剂是否变质，可向其加入稀盐酸，若观察到有气泡生成，则判断此供氧剂已变质．你是否认同上述方案，理由是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />探究二：此类供氧剂中过氧化钙（CaO<SUB>2</SUB>）的含量测定<br />方案一：通过测定产生氧气的体积，最终计算出供氧剂中过氧化钙的含量．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao81/1c537580-94d4-11e9-b2ed-b42e9921e93e_xkb46.png\" style=\"vertical-align:middle\" /><br />（1）取一定质量的供氧剂溶于水，按照图一装置进行装配．充分反应后，待量筒中液面不再变化，还要<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，再读出体积．<br />（2）为保证测量结果的准确性，收集气体的时机是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />a．导管口还没有气泡产生时&nbsp;&nbsp;b．导管口连续均匀冒出气泡时&nbsp;&nbsp;c．导管口冒出气泡很多时<br />（3）用此装置测得的含量会偏大，如果将发生装置换成图二中的装置<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填序号）可以避免此误差．<br />方案二：把样品中的过氧化钙转化为碳酸钙，通过测定碳酸钙沉淀的质量，进而求得过氧化钙的质量．具体流程如图三．<br />（1）写出供氧剂与水反应的化学方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（2）滴加的Na<SUB>2</SUB>CO<SUB>3</SUB>溶液要过量的原因是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（3）洗涤沉淀的目的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（4）检验沉淀已经洗涤干净的方法是：取最后一次洗涤液，滴加<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>溶液，若无明显现象，则表示已经洗净．<br />（5）若供氧剂的质量m=5g，碳酸钙质量n=5g，请通过计算得出供氧剂中的过氧化钙的含量．<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />（6）若CaCO<SUB>3</SUB>沉淀过滤后不洗涤，则过氧化钙的含量将<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>&nbsp;（填“偏大”、“偏小”或“无影响”）．<br />（7）此方案排除所加试剂和操作的原因，测定的过氧化钙含量也可能偏大，原因是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','带火星的木条复燃$###$密封、干燥保存$###$不认同，因为盐酸中的水与CaO<SUB>2</SUB>反应会生成O<SUB>2</SUB>，即使没有CaCO<SUB>3</SUB>与盐酸反应生成CO<SUB>2</SUB>，也会有气泡生成$###$使量筒内外液面相平$###$a$###$BCD$###$CaO<SUB>2</SUB>+H<SUB>2</SUB>O=Ca（OH）<SUB>2</SUB>+O<SUB>2</SUB>↑$###$使氯化钙完全转化为沉淀$###$除去沉淀表面的可溶性杂质，测量结果更准确$###$硝酸银$###$72.0%$###$偏大$###$所测样品中Ca元素还可能还来源于变质生成的Ca（OH）<SUB>2</SUB>、CaCO<SUB>3</SUB>，从而导致所测过氧化钙含量偏大','【解答】解：探究一：（1）取一定量该供氧剂放置于试管中，向其中加水，再将带火星的木条伸入试管口中，观察到带火星的木条复燃，证明该供氧剂的主要成分为过氧化钙．<br />故填：带火星的木条复燃．<br />（2）此供氧剂能和水反应，因此保存方法是密封、干燥保存．<br />故填：密封、干燥保存．<br />（3）因为盐酸中的水与CaO<SUB>2</SUB>反应会生成O<SUB>2</SUB>，即使没有CaCO<SUB>3</SUB>与盐酸反应生成CO<SUB>2</SUB>，也会有气泡生成，因此不能通过观察到有气泡生成来判断此供氧剂已变质．<br />故填：不认同，因为盐酸中的水与CaO<SUB>2</SUB>反应会生成O<SUB>2</SUB>，即使没有CaCO<SUB>3</SUB>与盐酸反应生成CO<SUB>2</SUB>，也会有气泡生成．<br />探究二：方案一：（1）取一定质量的供氧剂溶于水，按照图一装置进行装配．充分反应后，待量筒中液面不再变化，还要使量筒内外液面相平，再读出体积．<br />故填：使量筒内外液面相平．<br />（2）为保证测量结果的准确性，收集气体的时机是导管口还没有气泡产生时．<br />故填：a．<br />（3）用此装置测得的含量会偏大，是因为分液漏斗中滴下的水会占据一部分体积，导致测定的氧气体积偏大，从而导致测定的结构偏大；如果将发生装置换成图二中的装置BCD装置可以避免此误差．<br />故填：BCD．<br />方案二：（1）供氧剂与水反应的化学方程式为：CaO<SUB>2</SUB>+H<SUB>2</SUB>O=Ca（OH）<SUB>2</SUB>+O<SUB>2</SUB>↑．<br />故填：CaO<SUB>2</SUB>+H<SUB>2</SUB>O=Ca（OH）<SUB>2</SUB>+O<SUB>2</SUB>↑．<br />（2）滴加的Na<SUB>2</SUB>CO<SUB>3</SUB>溶液要过量的原因是使氯化钙完全转化为沉淀．<br />故填：使氯化钙完全转化为沉淀．<br />（3）洗涤沉淀的目的是除去沉淀表面的可溶性杂质，测量结果更准确．<br />故填：除去沉淀表面的可溶性杂质，测量结果更准确．<br />（4）检验沉淀已经洗涤干净的方法是：取最后一次洗涤液，滴加硝酸银溶液，若无明显现象，则表示已经洗净．<br />故填：硝酸银．<br />（5）设过氧化钙质量为x，<br />由CaO<SUB>2</SUB>+H<SUB>2</SUB>O=Ca（OH）<SUB>2</SUB>+O<SUB>2</SUB>↑，Ca（OH）<SUB>2</SUB>+CO<SUB>2</SUB>═CaCO<SUB>3</SUB>↓+H<SUB>2</SUB>O可知，CaO<SUB>2</SUB>～CaCO<SUB>3</SUB>，<br />CaO<SUB>2</SUB>～CaCO<SUB>3</SUB>，<br />72&nbsp;&nbsp;&nbsp;&nbsp; 100<br />x&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;5g<br /><span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">72</td></tr><tr><td>x</td></tr></table></span>=<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">100</td></tr><tr><td>5g</td></tr></table></span>，<br />x=3.6g，<br />供氧剂中的过氧化钙的含量为：<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">3.6g</td></tr><tr><td>5g</td></tr></table></span>×100%=72.0%，<br />故填：72.0%．<br />（6）若CaCO<SUB>3</SUB>沉淀过滤后不洗涤，则会导致碳酸钙质量偏大，从而导致计算的过氧化钙质量偏大，进一步导致过氧化钙的含量将偏大．<br />故填：偏大．<br />（7）此方案排除所加试剂和操作的原因，测定的过氧化钙含量也可能偏大，原因是所测样品中Ca元素还可能还来源于变质生成的Ca（OH）<SUB>2</SUB>、CaCO<SUB>3</SUB>，从而导致所测过氧化钙含量偏大．<br />故填：所测样品中Ca元素还可能还来源于变质生成的Ca（OH）<SUB>2</SUB>、CaCO<SUB>3</SUB>，从而导致所测过氧化钙含量偏大．','【分析】探究一：氧气能使带火星的木条复燃；<br />根据物质的性质可以判断保存方法；<br />探究二：为保证测量结果的准确性，收集气体的时机是导管口还没有气泡产生时，因为这样不会导致生成的气体损失，全面进入量筒中；<br />根据反应物、生成物、反应条件及其质量守恒定律可以书写反应的化学方程式；<br />滴加的Na<SUB>2</SUB>CO<SUB>3</SUB>溶液要过量的原因是使氯化钙完全反应；<br />氯化钠能和硝酸银反应生成白色沉淀氯化银和硝酸钠；<br />根据提供的数据和反应的化学方程式可以进行相关方面的计算．','书写',3.00,'0cba43bdf62c787ce10fbeb6f173ba36',9,400,'实验探究物质的组成成分以及含量,证明碳酸盐,氧气的检验和验满,书写化学方程式、文字表达式、电离方程式,根据化学反应方程式的计算','高邮市',2016,'37','2016•高邮市一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840695,'<img  src=\"/tikuimages/9/2016/400/shoutiniao87/1c5ee730-94d4-11e9-ba4e-b42e9921e93e_xkb31.png\" style=\"vertical-align:middle;FLOAT:right\" />如图是空气成分示意图（按体积计算），其中“b”代表的是（　　）','氧气','氮气','二氧化碳','稀有气体','','A','【解答】解：A、氧气大约占21%，由图示可知，b表示氧气，故选项正确；<br />B、氮气大约占78%，因此b一定不是氮气，故选项错误；<br />C、二氧化碳大约占0.03%，因此b一定不是二氧化碳，故选项错误；<br />D、稀有气体大约占0.94%，因此b一定不是稀有气体，故选项错误；<br />故选A．','【分析】空气中各成分的体积分数分别是：氮气大约占78%、氧气大约占21%、稀有气体大约占0.94%、二氧化碳大约占0.03%、水蒸气和其它气体和杂质大约占0.03%．','选择题',3.00,'bac241f2b2cdffeecb9916ada5e98a1d',9,400,'空气的成分及各成分的体积分数','',2016,'37','2016•官渡区一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840706,'我国新修订的《环境空气质量标准》颁布后，各地采取多种措施提高空气质量．下列措施中，不利于提高空气质量的是（　　）','禁止燃放烟花爆竹','将盗版光盘泼上汽油燃烧处理','控制PM2.5的排放，以减少雾霾天气','向煤炭中加入石灰石或生石灰作固硫剂，减少二氧化硫的排放','','B','【解答】解：A、燃放烟花爆竹会形成大量的空气污染物，禁止燃放烟花爆竹，有利于提高空气质量，故不符合题意．<br />B、焚烧盗版光盘会形成大量的空气污染物，不利于提高空气质量，故符合题意；<br />C、PM2.5是形成雾霾天气的主要原因之一，PM2.5属于空气污染物，控制PM2.5的排放可以减少雾霾天气，有利于提高空气质量，故不符合题意；<br />D、二氧化硫是空气污染物，向煤炭中加入石灰石或生石灰作固硫剂，减少二氧化硫的排放，有利于提高空气质量，故不符合题意；<br />故选B．','【分析】A、根据二氧化硫是空气污染物判断；<br />B、PM2.5是形成雾霾天气的主要原因之一；<br />C、焚烧盗版光盘会形成大量的空气污染物；<br />D、燃放烟花爆竹会形成大量的空气污染物','选择题',3.00,'1ed82ff85193602c3bcc6b431f7f0287',9,400,'防治空气污染的措施','',2016,'37','2016•邵阳二模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840709,'判断化学反应发生的根本依据应该是有新物质生成．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao6/1c88b64f-94d4-11e9-b590-b42e9921e93e_xkb71.png\" style=\"vertical-align:middle\" /><br />（1）两种物质的溶液混合后，若有沉淀、气体产生，可以帮助我们判断化学反应是否发生．请写出一有沉淀产生，且属于复分解反应类型的化学方程式．<br />&nbsp;（2）氢氧化钠与CO<SUB>2</SUB>反应时无明显现象．如图1是某同学设计的实验装置，其气密良好．<br />①若胶头滴管中的物质是浓NaOH溶液，锥形瓶中充满CO<SUB>2</SUB>，则挤压胶头滴管后观察到什么现象？并写出发生反应的化学方程式．<br />②小明同学认为原因是二氧化碳溶于了水中所造成的．请你在上述实验的基础，进一步做补充实验，证明氢氧化钠和二氧化碳确实发生了化学反应．（要求写出实验操作及判断依据）<br />（3）氢氧化钠溶液与稀盐酸发生反应过程中溶液pH变化如图所示．<br />①B点表示的溶液中溶质为：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />②该实验过程中，后用滴管加入的溶液为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />③为探究盐酸与氢氧化钠溶液是否发生反应，某同学在两者混合后溶液样品中加硝酸银溶液，看是否产生白色沉淀．同学们认为此方案不正确，请简述其中的理由．<br />（4）称取12.5g石灰石（杂质不参加反应）放入烧杯中，向其中加入50g稀盐酸，二者恰好完全反应．应结束后称量烧杯中剩余物质的总质量为58．lg（不包括烧杯的质量，且气体的溶解忽略不计）．请算：石灰石中杂质的质量分数是多少？','','','','','','氢氧化钠、氯化钠$###$氢氧化钠溶液','【解答】解：（1）氢氧化钠与硫酸铜反应生成硫酸钠和氢氧化铜沉淀，是有沉淀生成的复分解反应，故答案为：2NaOH+CuSO<SUB>4</SUB>═Cu（OH）<SUB>2</SUB>↓+Na<SUB>2</SUB>SO<SUB>4</SUB>；<br />（2）①若胶头滴管中的物质是浓NaOH溶液，锥形瓶中充满CO<SUB>2</SUB>，则挤压胶头滴管后氢氧化钠与二氧化碳反应生成碳酸钠和水，由于气体被消耗，U形管内出现左高右低的现象，故答案为：U形管内出现左高右低的现象；2NaOH+CO<SUB>2</SUB>═Na<SUB>2</SUB>CO<SUB>3</SUB>+H<SUB>2</SUB>O．<br />②要证明氢氧化钠和二氧化碳确实发生了化学反应，只需要证明反应生成了碳酸钠即可，取反应后的溶液加入试管中，滴加稀盐酸，有气泡产生，说明氢氧化钠和二氧化碳确实发生了化学反应，故答案为：取反应后的溶液加入试管中，滴加稀盐酸，有气泡产生；<br />（3）氢氧化钠溶液与稀盐酸发生反应过程中溶液pH变化如图所示．<br />①B点溶液成碱性，溶质含有氢氧化钠和氯化钠，故填：氢氧化钠、氯化钠；<br />②溶液开始呈酸性，后来呈碱性，说明是将氢氧化钠溶液加入盐酸中，故填：氢氧化钠溶液；<br />③氢氧化钠与盐酸反应生成氯化钠，氯化钠能与硝酸银反应生成白色沉淀，而盐酸也能与硝酸银反应生成白色沉淀，故答案为：滴加硝酸银有白色沉淀，只能说明溶液中含有氯离子，不能说明生成了新物质氯化钠，因为盐酸与硝酸银溶液反应有白色沉淀．<br />（4）生成的二氧化碳的质量为：12.5g+50g-58.1g=4.4g<br />设碳酸钙的质量为x<br />CaCO<SUB>3</SUB>+2HCl═CaCl<SUB>2</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑<br />100&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;44<br />x&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 4.4g<br /><span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">100</td></tr><tr><td>44</td></tr></table>=<table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">x</td></tr><tr><td>4.4g</td></tr></table></span>&nbsp; x=10g<br />杂质的质量分数为：<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">12.5g-10g</td></tr><tr><td>12.5g</td></tr></table>×100%</span>=20%<br />答：石灰石中杂质的质量分数是20%．','【分析】根据复分解反应的条件以及物质间反应的实验现象进行分析，根据图象结合盐酸和氢氧化钠的反应解答；根据化学方程式结合题干提供的数据解答．','书写',3.00,'40f4780cfca4f02b4286643b8cbbdafe',9,400,'碱的化学性质,中和反应及其应用,溶液的酸碱性与pH值的关系,复分解反应及其发生的条件,书写化学方程式、文字表达式、电离方程式,根据化学反应方程式的计算','',2016,'35','2016春•召陵区期中',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840715,'某兴趣小组对实验室制备氧气的反应条件进行如下探究实验．<br />（1）为探究不同催化剂对氯酸钾分解速度的影响，甲设计以下对比实验探究，在相同温度下，比较两组实验产生O<SUB>2</SUB>的快慢．<br />Ⅰ．将3.0g&nbsp;KClO<SUB>3</SUB>与1.0g&nbsp;MnO<SUB>2</SUB>均匀混合加热<br />Ⅱ．将x&nbsp;g&nbsp;KClO<SUB>3</SUB>与1.0g&nbsp;CuO均匀混合加热<br />实验中Ⅱ中x的值应为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；Ⅰ中反应的化学方程式是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>_．<br />实验结论：在相同条件下，MnO<SUB>2</SUB>的催化能力比CuO强．<br />（2）乙探究了影响双氧水分解速度的某种因素．实验数据记录和可选装置如下：<br /><table class=\"edittable\"><TBODY><TR><td width=26></TD><td width=101>双氧水的质量</TD><td width=119>双氧水的浓度</TD><td width=111>MnO<SUB>2</SUB>的质量</TD><td width=184>相同时间内产生O<SUB>2</SUB>体积</TD></TR><TR><td>Ⅰ</TD><td>50.0g</TD><td>1%</TD><td>0.1g</TD><td>9mL</TD></TR><TR><td>Ⅱ</TD><td>50.0g</TD><td>2%</TD><td>0.1g</TD><td>16mL</TD></TR><TR><td>Ⅲ</TD><td>50.0g</TD><td>4%</TD><td>0.1g</TD><td>31mL</TD></TR></TBODY></TABLE><img  src=\"/tikuimages/9/2016/400/shoutiniao27/1ca27fde-94d4-11e9-ba02-b42e9921e93e_xkb23.png\" style=\"vertical-align:middle\" /><br />本实验中，测量O<SUB>2</SUB>体积的装置应该是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填编号）．<br />实验结论：在相同条件下，<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，双氧水分解得越快．','','','','','','3.0$###$2KClO<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span></td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>△</td></tr></table></span>2KCl+3O<SUB>2</SUB>↑$###$c$###$双氧水的浓度越大','【解答】解：（1）控制氯酸钾的质量相等，才能比较出二氧化锰和氧化铜的催化效果，所以Ⅱ中x的数值为3.0g；&nbsp;氯酸钾在二氧化锰的作用下加热分解产生氯化钾和氧气，反应的方程式为：2KClO<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span></td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>△</td></tr></table></span>2KCl+3O<SUB>2</SUB>↑；<br />（2）测量氧气的体积采用的是排水法，将排出的水的用量筒收集起来，排出的水的体积就是氧气的体积；气体短管进长管出才能排出试剂瓶中的水，要将瓶内的水排出来，A和B都不可以，应该选择C装置作为实验装置．<br />由表格数据分析可知，在相同条件下，双氧水的浓度越大，双氧水分解得越快、<br />故答案为：（1）3.0；&nbsp;2KClO<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span></td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>△</td></tr></table></span>2KCl+3&nbsp;O<SUB>2</SUB>↑；<br />（2）C；双氧水的浓度越大．','【分析】（1）可采用控制变量法，要控制氯酸钾的质量相等才能比较出二氧化锰和氧化铜的催化效果，据此进行分析解答．<br />（2）测量氧气的体积采用的是排水法，将排出的水的用量筒收集起来，排出的水的体积就是氧气的体积；气体短管进长管出才能排出试剂瓶中的水，据此进行分析解答．','书写',3.00,'790380b2124cd1bf7fcb7a9ad2612fa4',9,400,'影响化学反应速率的因素探究,量气装置,催化剂的特点与催化作用,书写化学方程式、文字表达式、电离方程式','',2016,'37','2016•陕西校级四模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840717,'小明在做铁与硫酸溶液反应的实验时，发现生成的气体有刺激性气味，于是进行了探究．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao31/1cb0fecf-94d4-11e9-8976-b42e9921e93e_xkb44.png\" style=\"vertical-align:middle\" /><br />【提出问题】铁与硫酸溶液反应生成的气体为什么有刺激性气味？<br />【查阅资料】<br />（1）6H<SUB>2</SUB>SO<SUB>4</SUB>（浓）+2Fe═Fe<SUB>2</SUB>（SO<SUB>4</SUB>）<SUB>3</SUB>+6H<SUB>2</SUB>O+3SO<SUB>2</SUB>↑<br />（2）SO<SUB>2</SUB>可使品红溶液的红色褪去<br />【进行猜想】铁与不同浓度的硫酸溶液反应，生成的气体产生可能有二氧化硫<br />【实验探究】小明用图甲所示装置进行实验，并将E中收集到的气体进行如图乙所示的爆鸣实验．<br />请帮助小明完成如表：<br /><table class=\"edittable\"><TBODY><TR><td width=\"12%\">&nbsp;</TD><td width=\"15%\">&nbsp;</TD><td width=\"17%\">实验现象</TD><td width=\"17%\">&nbsp;</TD><td width=\"12%\">&nbsp;</TD><td width=\"22%\">&nbsp;</TD></TR><TR><td>&nbsp;</TD><td>硫酸溶液浓度</TD><td>B中品红溶液</TD><td>D中品红溶液</TD><td>爆鸣实验</TD><td>A中生成气体成分</TD></TR><TR><td>实验一</TD><td>98%</TD><td>&nbsp;<br /><u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD><td>不褪色</TD><td>无爆鸣声</TD><td>只有SO<SUB>2</SUB></TD></TR><TR><td>实验二</TD><td>45%</TD><td>稍有褪色</TD><td>不褪色</TD><td>有爆鸣声</TD><td>&nbsp;<br /><u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD></TR><TR><td>实验三</TD><td>25%</TD><td>&nbsp;<br /><u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD><td>不褪色</TD><td><u>&nbsp;&nbsp;&nbsp;&nbsp;</u>&nbsp;</TD><td>只有H<SUB>2</SUB></TD></TR></TBODY></TABLE>【实验结论】铁与不同浓度的硫酸反应，生成的气体产物可能不同，当硫酸溶液浓度达到足够大时，生成的气体产物中有二氧化硫<br />【交流反思】<br />（1）写出实验三中铁与硫酸溶液反应的化学方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，写出实验二中发生爆鸣反应的化学方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）实验一中C装置的作用是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．【拓展延伸】小明又将A装置进行了如图丙所示的改进，试分析：增加的导管下端伸入液<br />面以下原因<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，该导管所起的作用是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','褪色$###$SO<SUB>2</SUB>和H<SUB>2</SUB>$###$不褪色$###$有爆鸣声$###$Fe+H<SUB>2</SUB>SO<SUB>4</SUB>═FeSO<SUB>4</SUB>+H<SUB>2</SUB>↑$###$2H<SUB>2</SUB>+O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O$###$吸收二氧化硫$###$形成液封，防止生成的气体逸出$###$平衡试管内外的压强，防止橡皮塞飞出','【解答】解：【实验探究】<br />填表如下所示：<br /><table class=\"edittable\"><TBODY><TR><td width=67 rowSpan=2></TD><td width=74 rowSpan=2>A中硫酸浓度</TD><td width=306 colSpan=3>实验现象</TD><td width=123 rowSpan=2>A中生成气体成分</TD></TR><TR><td>B中品红溶液</TD><td>D中品红溶液</TD><td>爆鸣实验</TD></TR><TR><td>实验一</TD><td>98%</TD><td>褪色 </TD><td>不褪色</TD><td>无爆鸣声</TD><td>只有SO<SUB>2</SUB></TD></TR><TR><td>实验二</TD><td>45%</TD><td>稍有褪色</TD><td>不褪色</TD><td>有爆鸣声</TD><td>SO<SUB>2</SUB>和H<SUB>2</SUB> </TD></TR><TR><td>实验三</TD><td>25%</TD><td>不褪色</TD><td>不褪色</TD><td>有爆鸣声 </TD><td>只有H<SUB>2</SUB></TD></TR></TBODY></TABLE>【交流反思】<br />（1）实验二中，铁与硫酸反应生成硫酸亚铁和氢气，反应的化学方程式为：Fe+H<SUB>2</SUB>SO<SUB>4</SUB>═FeSO<SUB>4</SUB>+H<SUB>2</SUB>↑；<br />实验二中，发生爆鸣是因为氢气和氧气反应生成了水，反应的化学方程式为：2H<SUB>2</SUB>+O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O．<br />故填：Fe+H<SUB>2</SUB>SO<SUB>4</SUB>═FeSO<SUB>4</SUB>+H<SUB>2</SUB>↑；2H<SUB>2</SUB>+O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O．<br />（2）实验一中C装置的作用是吸收二氧化硫．<br />故填：吸收二氧化硫．<br />【拓展延伸】<br />增加的导管伸入液面以下的原因是防止产生的气体逸出；<br />该导管所起的作用是平衡试管内外的压强，防止橡皮塞飞出．<br />故填：形成液封，防止生成的气体逸出；平衡试管内外的压强，防止橡皮塞飞出．','【分析】【实验探究】<br />根据实验结论可以判断实验现象，根据实验现象可以判断实验结论；<br />【交流反思】<br />（1）根据铁和稀硫酸反应生成硫酸亚铁和氢气以及氢气燃烧生成水进行解答；<br />（2）根据氢氧化钠溶液能够吸收二氧化硫气体进行解答；<br />【拓展延伸】根据二氧化硫气体有毒，扩散到空气中会污染环境进行解答．','书写',3.00,'9d084a6d5a0df07f78350ef19cece866',9,400,'实验探究物质的性质或变化规律,常见气体的检验与除杂方法,书写化学方程式、文字表达式、电离方程式','寿光市',2016,'32','2016•寿光市校级模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840723,'“铁、铜、铝、锌”是生活中常见的金属<br />（1）人体缺<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>元素会引起食欲不振，生长迟缓，发育不良．<br />（2）铁锅作炊具，主要是利用铁的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>性，工业上利用赤铁矿炼铁，写出该反应的化学方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）只用一种试剂就能一次性判断金属铜、铝、锌的活动性顺序，这种试剂是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填字母）．<br />A．CuSO<SUB>4</SUB>溶液&nbsp;&nbsp;&nbsp; B．Al<SUB>2</SUB>（SO<SUB>4</SUB>）<SUB>3</SUB>溶液&nbsp;&nbsp;&nbsp;&nbsp; C．ZnSO<SUB>4</SUB>溶液．','','','','','','锌$###$导热$###$3CO+Fe<SUB>2</SUB>O<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2Fe+3CO<SUB>2</SUB>$###$C','【解答】解：（1）人体缺锌元素会引起食欲不振，生长迟缓，发育不良．<br />（2）铁锅作炊具，主要是利用铁的导热性，工业上利用赤铁矿炼铁，赤铁矿的主要成分是氧化铁，该反应的化学方程式是：3CO+Fe<SUB>2</SUB>O<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2Fe+3CO<SUB>2</SUB>．<br />（3）三种金属的金属活动性由强到弱的顺序为：铝＞锌＞铜，可使用锌分别放入铝和铜的可溶性盐溶液，也可以采取把另两种金属单质与锌的盐溶液反应来验证三种金属的活动性，由题意可知，本题的试剂可选用硫酸锌溶液．<br />故答为：（1）锌；（2）导热，3CO+Fe<SUB>2</SUB>O<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2Fe+3CO<SUB>2</SUB>；（3）C．','【分析】（1）根据元素与人体健康的关系分析回答；<br />（2）根据铁的性质和应用、炼铁的原理分析回答；<br />（3）根据在验证三种金属的活动性的“三取中”的方法分析判断．','书写',3.00,'e58ae18442616f250f46850c28503fdc',9,400,'金属的物理性质及用途,金属活动性顺序及其应用,铁的冶炼,书写化学方程式、文字表达式、电离方程式,人体的元素组成与元素对人体健康的重要作用','',2016,'32','2016•莆田模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840724,'通过一年的化学学习，你已经掌握了实验室制取气体的有关规律，请结合图回答问题：<img  src=\"/tikuimages/9/0/400/shoutiniao20/1cce71e1-94d4-11e9-ab54-b42e9921e93e_xkb5.png\" style=\"vertical-align:middle\" /><br />（1）写出图中标号仪器的名称①<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；②<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）实验室常用锌粒和稀硫酸反应制取氢气，该选用的发生装置是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填序号，下同）．收集装置是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，反应的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．该B装置还可用来制取的气体是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）C装置与B装置相比，C装置的优点是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','酒精灯$###$长颈漏斗$###$B或C$###$E$###$Zn+H<SUB>2</SUB>SO<SUB>4</SUB>=ZnSO<SUB>4</SUB>+H<SUB>2</SUB>↑$###$氧气、二氧化碳$###$可以控制反应的发生和停止','【解答】解：（1）熟记常用仪器的名称可知①为酒精灯；②为分液漏斗．<br />（2）实验室常用锌粒和稀硫酸反应制取氢气，不需要加热，该选用的发生装置是B或C．收集装置是E，反应的化学方程式为Zn+H<SUB>2</SUB>SO<SUB>4</SUB>=ZnSO<SUB>4</SUB>+H<SUB>2</SUB>↑．该B装置还可用来制取的气体是氧气、二氧化碳．<br />（3）用装置C的优点是可以通过活塞的关闭随时控制反应的发生和停止，打开活塞，固体和液体接触生成氢气，关闭活塞，试管内压强增大，将液体压入长颈漏斗，固液分离，反应停止．<br />故答案为：<br />（1）酒精灯；分液漏斗<br />（2）B或C，E，Zn+H<SUB>2</SUB>SO<SUB>4</SUB>=ZnSO<SUB>4</SUB>+H<SUB>2</SUB>↑． 氧气、二氧化碳．<br />（3）可以控制反应的发生和停止．．','【分析】（1）①②均为实验室常用的仪器，熟记常用仪器的名称即可轻松作答．<br />（2）根据反应物的状态和反应条件确定实验室制取气体的发生装置；根据气体的密度和溶水性确定气体的收集装置；熟记常用的化学方程式．<br />（3）从气体压强的知识进行分析；不同的实验装置，功能不同，据装置特点分析其优越性．','书写',3.00,'de85136efacdea2bc7d89c2a5085879c',9,400,'氢气的制取和检验,书写化学方程式、文字表达式、电离方程式','',0,'37','',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840733,'区分下列物质的两个实验方案都合理的一组是（　　）<br /><table class=\"edittable\"><TBODY><TR><td width=176>需区分的物质</TD><td width=100>摘一</TD><td width=144>方法二</TD></TR><TR><td>A．氢气和二氧化碳</TD><td>带火星的木条</TD><td>紫色石蕊溶液</TD></TR><TR><td>B．羊毛和腈纶</TD><td>点燃、闻气味</TD><td>观察颜色</TD></TR><TR><td>C．硝酸铵和磷酸钙</TD><td>加氢氧化钙研磨</TD><td>加水</TD></TR><TR><td>D．18K黄金和黄铜</TD><td>加热，观察颜色</TD><td>加入硝酸银溶液</TD></TR></TBODY></TABLE>','A','B','C','D','','C','【解答】解：A、带火星的木条不能检验氢气和二氧化碳，不符合题意，故A错误；<br />B、羊毛和腈纶的颜色相同，观察颜色的不能鉴别，不符合题意，故B错误；<br />C、取样品，加入熟石灰研磨，有刺激性气味的气体产生的是硝酸铵，加水溶解，不能溶于水的是磷酸钙，现象不同，可以鉴别，故C正确；<br />D、18K黄金和黄铜的颜色都是金黄色，观察颜色的方法不能鉴别，故D错误．<br />故选：C．','【分析】A、根据带火星的木条不能检验氢气和二氧化碳进行分析；<br />B、根据羊毛和腈纶的颜色相同进行分析；<br />C、根据铵态氮肥和碱混合会生成氨气，磷肥不溶于水进行分析；<br />D、根据18K黄金和黄铜的颜色都是金黄色进行分析．','选择题',3.00,'7ba43f005a2fff61a7a73d2341035651',9,400,'常见气体的检验与除杂方法,金属的化学性质,铵态氮肥的检验,物质的鉴别、推断,棉纤维、羊毛纤维和合成纤维的鉴别','',2016,'37','2016•香坊区二模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840734,'如图四个图象能正确反映对应实验操作的是（　　）<br /><table class=\"edittable\"><TBODY><TR><td width=128><img  src=\"/tikuimages/9/2016/400/shoutiniao17/1ce925cf-94d4-11e9-bde0-b42e9921e93e_xkb13.png\" style=\"vertical-align:middle\" /></TD><td width=116><img  src=\"/tikuimages/9/2016/400/shoutiniao96/1cea1030-94d4-11e9-8c86-b42e9921e93e_xkb13.png\" style=\"vertical-align:middle\" /></TD><td width=133><img  src=\"/tikuimages/9/2016/400/shoutiniao8/1cebe4f0-94d4-11e9-8e96-b42e9921e93e_xkb61.png\" style=\"vertical-align:middle\" /></TD><td width=133><img  src=\"/tikuimages/9/2016/400/shoutiniao9/1ceccf4f-94d4-11e9-a36a-b42e9921e93e_xkb21.png\" style=\"vertical-align:middle\" /></TD></TR><TR><td>A．向一定量的氢氧化钠和氯化钡混合溶液中不断滴入稀硫酸</TD><td>B．加有Na<SUB>2</SUB>SO<SUB>4</SUB>的蒸馏水通电</TD><td>C．向等质量的Mg和MgO中分别加入足量等浓度的稀盐酸</TD><td>D．一定压强下，氧气在水中的溶解度</TD></TR></TBODY></TABLE>','A','B','C','D','','C','【解答】解：A、向一定量的氢氧化钠和氯化钡混合溶液中不断滴入稀硫酸，则硫酸根离子与钡离子接触就反应生成硫酸钡沉淀，故沉淀从0开始，错误；<br />B、电解水时正极产生的是氧气，负极产生的是氢气，且其体积比是1：2，错误；<br />C、向等质量的Mg和MgO中分别加入足量等浓度的稀盐酸，等质量的盐酸生成的氯化镁的质量相等，等质量的镁和氧化镁，镁生成的氯化镁的质量大，正确；<br />D、一定压强下，氧气在水中的溶解度随温度的升高而减小，错误；<br />故选C．','【分析】A、根据硫酸与氢氧化钠和氯化钡的反应解答；<br />B、根据电解水正氧负氢、氢二氧一解答；<br />C、根据盐酸与金属镁和氧化镁的反应解答；<br />D、根据氧气的溶解度与温度和压强的关系解答．','选择题',3.00,'4ba780f50862808026ad650c8a782e73',9,400,'电解水实验,气体溶解度的影响因素,酸的化学性质','',2016,'37','2016•滨海县二模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840739,'下列物质由分子构成的是（　　）','锌','氮气','金刚石','氯化钠','','B','【解答】解：A、锌属于金属单质，是由锌原子直接构成的，故选项错误．<br />B、氮气属于气态非金属单质，是由氮分子构成的，故选项正确．<br />C、金刚石属于固态非金属单质，是由碳原子直接构成的，故选项错误．<br />D、氯化钠是由钠离子和氯离子构成的，故选项错误．<br />故选：B．','【分析】根据金属、大多数固态非金属单质、稀有气体等由原子构成；有些物质是由分子构成的，气态的非金属单质和由非金属元素组成的化合物，如氢气、水等；有些物质是由离子构成的，一般是含有金属元素和非金属元素的化合物，如氯化钠，进行分析判断即可．','选择题',3.00,'f86d8aacf9de764e7649c65dfc496114',9,400,'物质的构成和含量分析','',0,'37','',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840740,'某学生针对下列四组物质的鉴别，设计了两种方案，其中都合理的是（　　）<br /><table class=\"edittable\"><TBODY><TR><td width=75></TD><td width=120>A</TD><td width=120>B</TD><td width=120>C</TD><td width=120>D</TD></TR><TR><td>需区分的物质</TD><td>黄铜和黄金</TD><td>聚乙烯（PE）和聚氯乙烯（PVC）</TD><td>锦纶布料和羊毛布料</TD><td>NH<SUB>4</SUB>NO<SUB>3</SUB>和NaOH固体</TD></TR><TR><td>第一方案</TD><td>分别取样，观察颜色</TD><td>分别取样，观察颜色</TD><td>分别取样，用手触摸</TD><td>分别取样，加水充分溶解，加氯化铵溶液，加热，观察现象</TD></TR><TR><td>第二方案</TD><td>分别取样，在石棉网上高温灼烧，观察现象</TD><td>分别取样，分别点燃，闻燃烧时产生的气味</TD><td>分别取样，点燃，闻气味</TD><td>分别取样，加水，滴加无色酚酞溶液，观察现象</TD></TR></TBODY></TABLE>','A','B','C','D','','D','【解答】解：A、取样品，观察颜色，都是金黄色的固体，不能鉴别，取样灼烧，颜色变黑的是黄铜，颜色不变的是黄金，现象不同，可以鉴别，不符合题意，故A错误；<br />B、取样，观察颜色，颜色相同，不能鉴别，取样点燃，有刺激性气味的气体产生的是聚氯乙烯，没有刺激性气味产生的是聚乙烯，现象不同，可以鉴别，不符合题意，故B错误；<br />C、取样品，用手触摸，感觉相同，不能鉴别，取样品点燃，有烧焦羽毛气味的是羊毛，没有此气味的是锦纶，现象不同，可以鉴别，不符合题意，故C错误；<br />D、取样品，加水溶解，溶液温度升高的是氢氧化钠，溶液温度降低的是硝酸铵，现象不同，可以鉴别，分别取样，加水，滴加无色酚酞溶液，溶液变红的是氢氧化钠，溶液颜色不变的是硝酸铵，现象不同，可以鉴别，符合题意，故D正确．<br />故选：D．','【分析】A、根据黄金和黄铜都是金黄色的固体进行分析；<br />B、根据聚乙烯和聚氯乙烯都是白色的固体进行分析；<br />C、根据锦纶布料和羊毛布料的触感是相同的进行分析；<br />D、根据氢氧化钠溶于水，溶液温度升高，硝酸铵溶于水，溶液温度降低，氢氧化钠的水溶液显碱性进行分析．','选择题',3.00,'6450b5891cc88a101f26c5cefd88e6b3',9,400,'金属的化学性质,酸、碱、盐的鉴别,物质的鉴别、推断,塑料及其应用,棉纤维、羊毛纤维和合成纤维的鉴别','',2016,'37','2016•南岗区二模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840741,'<img  src=\"/tikuimages/9/2016/400/shoutiniao58/1cf7a4c0-94d4-11e9-a9ee-b42e9921e93e_xkb26.png\" style=\"vertical-align:middle;FLOAT:right\" />科学家在低温下合成了一种化合物，其分子模型如图所示，其中●代表碳原子，O代表氢原子，则该物质在氧气中完全燃烧生成水和二氧化碳时，下列关系式正确的是（　　）','m（CO<SUB>2</SUB>）：m（H<SUB>2</SUB>O）=44：18','v（C<SUB>5</SUB>H<SUB>4</SUB>）：v（CO<SUB>2</SUB>）=1：5','m（CO<SUB>2</SUB>）：m（H<SUB>2</SUB>O）=55：18','v（C<SUB>5</SUB>H<SUB>4</SUB>）：v（CO<SUB>2</SUB>）=1：2','','B','【解答】解：化合物的分子模型图，●代表碳原子，O代表氢原子，1个该化合物的分子是由5个碳原子和4个氢原子构成的，其化学式为C<SUB>5</SUB>H<SUB>4</SUB>，该物质在氧气中完全燃烧生成水和二氧化碳时，反应的化学方程式为：C<SUB>5</SUB>H<SUB>4</SUB>+6O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>5CO<SUB>2</SUB>+2H<SUB>2</SUB>O．<br />C<SUB>5</SUB>H<SUB>4</SUB>+6O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>5CO<SUB>2</SUB>+2H<SUB>2</SUB>O<br />64&nbsp;&nbsp;&nbsp;192&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 220&nbsp; 36<br />A、m（CO<SUB>2</SUB>）：m（H<SUB>2</SUB>O）=220：36=55：9，故选项说法错误．<br />B、由化学方程式的微观意义，v（C<SUB>5</SUB>H<SUB>4</SUB>）：v（CO<SUB>2</SUB>）=1：5，故选项说法正确．<br />C、m（CO<SUB>2</SUB>）：m（H<SUB>2</SUB>O）=220：36=55：9，故选项说法错误．<br />D、v（C<SUB>5</SUB>H<SUB>4</SUB>）：v（CO<SUB>2</SUB>）=1：5，故选项说法错误．<br />故选：B．','【分析】根据化合物的分子模型图，●代表碳原子，O代表氢原子，1个该化合物的分子是由5个碳原子和4个氢原子构成的，其化学式为C<SUB>5</SUB>H<SUB>4</SUB>，写出完全燃烧时的化学方程式，利用各物质之间的质量比等于相对分子质量和的比，进行分析解答即可．','选择题',3.00,'77149f2c3f32f124061e5df226e50130',9,400,'常见化学反应中的质量关系','',2016,'32','2016•苏州模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840760,'甲烷和水反应可以制备水煤气（混合气体），其反应的微观示意图如下所示：<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao61/1d3019de-94d4-11e9-8801-b42e9921e93e_xkb99.png\" style=\"vertical-align:middle\" /><br />根据如图所示微观示意图得出的结论中，正确的是（　　）','反应中含氢元素的化合物有三种','反应中甲和丙的质量之比为4：7','反应前后碳元素的化合价没有发生变化','水煤气的成分是一氧化碳和氧气','','B','【解答】解：由图示可知，该反应的反应物是甲烷和水，生成物是一氧化碳和氢气，反应的化学方程式为：CH<SUB>4</SUB>+H<SUB>2</SUB>O<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;一定条件&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>CO+3H<SUB>2</SUB>．<br />A、由反应的方程式可知，反应中含氢元素的化合物有CH<SUB>4</SUB>、H<SUB>2</SUB>O两种，故不正确；<br />B、由上述方程式可知，反应中甲和丙的质量之比为（12+1×4）：（12+16）=4：7，故正确；<br />C、反应前在甲烷中碳元素的化合价是-4，反应后在一氧化碳中碳元素的化合价是+2，发生了改变，故错误．<br />D、由图示可知，水煤气的成分是一氧化碳和氢气，故不正确．<br />故选B．','【分析】根据反应的微观示意图分析，反应物是甲烷和水，生成物是一氧化碳和氢气，反应的化学方程式为CH<SUB>4</SUB>+H<SUB>2</SUB>O<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;一定条件&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>CO+3H<SUB>2</SUB>，据此分析解答即可．','选择题',3.00,'7d63ba950e9ef830e9d066001852d086',9,400,'单质和化合物的判别,微粒观点及模型图的应用,化合价规律和原则','',2016,'37','2016•泰安一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840764,'“从生活走进化学，从化学走向社会”．请你用化学知识回答以下问题：<br />燃料和食品在人类社会的发展中起着相当重要的作用．<br />（1）下面是人类大规模使用燃料的大致顺序：木柴→木炭→煤→石油、天然气．<br />①上述燃料中属于化石燃料的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；这些化石燃料属于<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>能源（填“可再生”或“不可再生”）；上述燃料中通常被称为“清洁能源”的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />②天然气的主要成分是（填化学式）<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，其完全燃烧的化学方程式为：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）现代社会对能量的需求量越来越大，化学反应提供的能量已不能满足人类的需求．人们正在利用和开发新能源，如<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．（举一例）<br />（3）节假日或双休日，去郊游野炊实为一大快事．<br />①对野炊有经验的人会告诉你，餐具最好是导热性好，不易破碎的材料制的，你准备带的餐具是用：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>制的；（填字母）<br />A．金属&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;B．陶瓷&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;C．塑料<br />②在“灶”上悬挂野炊锅时，应调节野炊锅到合适的高度．你觉得原因可能是：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />③野餐食谱如下：主食：馒头&nbsp;&nbsp;&nbsp;&nbsp;配菜：烤火腿、鲫鱼汤、五香豆干．<br />主食和配菜中主要含蛋白质、<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>、油脂、无机盐和水等营养素，从营养角度来看，你准备再添-样<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；（填字母）<br />A．炒鸡蛋&nbsp;&nbsp;&nbsp;B．牛奶&nbsp;&nbsp;&nbsp;C．黄瓜&nbsp;&nbsp;&nbsp;D．烤野兔肉<br />④在引燃细枯枝后，如果迅速往“灶”里塞满枯枝，结果反而燃烧不旺，并产生很多浓烟，说明物质充分燃烧需要的条件是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','煤、石油、天然气$###$不可再生$###$天然气$###$CH<SUB>4</SUB>$###$CH<SUB>4</SUB>+2O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>CO<SUB>2</SUB>+2H<SUB>2</SUB>O$###$太阳能$###$A$###$能使用外焰加热$###$糖类$###$C$###$使可燃物与氧气充分接触','【解答】解：（1）①煤、石油、天然气属于化石燃料，属于不可再生能源；天然气燃烧产生的污染物较少，称为“清洁能源”；故答案为：煤、石油、天然气；不可再生；天然气；<br />②天然气的主要成分是甲烷，化学式是CH<SUB>4</SUB>，甲烷和氧气在点燃的条件下生成二氧化碳和水；故填：CH<SUB>4</SUB>；CH<SUB>4</SUB>+2O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>CO<SUB>2</SUB>+2H<SUB>2</SUB>O；<br />（2）人们正在利用和开发的新能源有太阳能、地热能、核能、潮汐能等；故答案为：太阳能；<br />（3）①金属具有导热性且不易破碎的性质，故填A；<br />②调节锅和火的距离，可以想到外焰的温度最高，故答案为：能使用外焰加热．<br />③所给的食物中主要含蛋白质、糖、油脂、无机盐和水等营养素，还需要有富含维生素的物质，蔬菜富含维生素，故填：糖类；C；<br />④“灶”里塞满枯枝，结果反而燃烧不旺，并产生很多浓烟，说明燃烧不充分，氧气不足．故答案为：使可燃物与氧气充分接触．','【分析】（1）煤、石油、天然气属于化石燃料，属于不可再生能源；②根据甲烷的主要成分及反应原理书写方程式；<br />（2）根据人们正在利用和开发的新能源回答；<br />（3）根据已有的知识进行分析，根据燃烧的条件、金属具有导热性且不易破碎的性质，人体所需的营养素有水、无机盐、糖类、油脂、维生素和蛋白质解答．','填空题',3.00,'b515ce5bee8fd10bdbc610ae3ebb01bc',9,400,'金属材料及其应用,燃烧与燃烧的条件,化石燃料及其综合利用,资源综合利用和新能源开发','',2016,'32','2016•云南模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840781,'如图是实验室用碳酸钙与稀盐酸反应制取二氧化碳并验证其性质的实验装置图，试根据题目要求回答下列问题：<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao93/1d73b28f-94d4-11e9-a7ad-b42e9921e93e_xkb21.png\" style=\"vertical-align:middle\" /><br />（1）仪器a的名称是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；装置B中发生的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．装置C中观察到的现象是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，写出装置A还可制取气体的反应方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（1）装置D中观察到下层蜡烛先熄灭，上层蜡烛后熄灭，可验证二氧化碳的性质是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．因此，二氧化碳可用于<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）实验室用装置E来收集CO<SUB>2</SUB>时，CO<SUB>2</SUB>应从<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>端（填“c”或“d”）通入．','','','','','','长颈漏斗$###$CO<SUB>2</SUB>+Ca（OH）<SUB>2</SUB>=CaCO<SUB>3</SUB>↓+H<SUB>2</SUB>O$###$石蕊试液变红色$###$CaCO<SUB>3</SUB>+2HCl═CaCl<SUB>2</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑$###$密度比空气大，不能燃烧，不支持燃烧$###$灭火$###$c','【解答】解：（1）仪器a是长颈漏斗，通过长颈漏斗可以向锥形瓶中滴加液体；装置B可以用来检验CO<SUB>2</SUB>，发生反应的化学方程式是CO<SUB>2</SUB>+Ca（OH）<SUB>2</SUB>=CaCO<SUB>3</SUB>↓+H<SUB>2</SUB>O；装置A中，碳酸钙和稀盐酸反应的化学方程式为：CaCO<SUB>3</SUB>+2HCl═CaCl<SUB>2</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑；装置C中二氧化碳进入石蕊试液中时，能和其中的水反应生成碳酸，碳酸显酸性，能使石蕊试液变红色．<br />（2）因为二氧化碳的密度比空气大，不能燃烧，不支持燃烧，所以装置D中能够观察到下层蜡烛先熄灭，上层蜡烛后熄灭；二氧化碳在生活中可用于灭火．<br />（3）因为二氧化碳能溶于水，不能用排水法收集；因为二氧化碳的密度比空气大，可以用向上排空气法收集，实验室用装置E来收集二氧化碳时，二氧化碳应从c端通入．<br />故答案为：<br />（1）长颈漏斗； CO<SUB>2</SUB>+Ca（OH）<SUB>2</SUB>=CaCO<SUB>3</SUB>↓+H<SUB>2</SUB>O．石蕊试液变红色，CaCO<SUB>3</SUB>+2HCl═CaCl<SUB>2</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑．<br />（2）密度比空气大，不能燃烧，不支持燃烧． 灭火．（3）c．','【分析】（1）要熟悉各种仪器的名称、用途及其使用方法；考虑稀盐酸和碳酸钙反应生成氯化钙、水和二氧化碳，二氧化碳能使澄清石灰水变浑浊，二氧化碳能和水反应生成碳酸，碳酸能使石蕊试液变红色．<br />（2）根据实验现象可以判断二氧化碳的性质，进一步可以判断二氧化碳有那些方面的用途．<br />（3）根据二氧化碳的密度可以判断收集二氧化碳的方法．','书写',3.00,'8716e686af19b5653566531fa3bd4856',9,400,'二氧化碳的实验室制法,二氧化碳的物理性质,二氧化碳的化学性质,书写化学方程式、文字表达式、电离方程式','宁夏',2016,'37','2016•宁夏校级一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840782,'活动探究一、实验是进行科学探究的重要方式，请根据如图甲回答问题：<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao33/1d7a6951-94d4-11e9-8c0e-b42e9921e93e_xkb63.png\" style=\"vertical-align:middle\" /><br />（1）用A图所示装置净化黄泥水时，玻璃棒的作用是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）通过B图所示酸的性质实验，可认识到：虽然酸类物质具有相似的化学性质，但由于不同的酸溶于水时电离出的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>不同，其性质也存在差异．<br />（3）进行C图所示蜡烛在氧气中燃烧实验时，可用简单方法验证蜡烛燃烧的产物．请选择一种产物简要写出其验证方法：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />（4）用D图所示装置进行蒸馏时，为使水蒸气的冷凝效果更好，可对实验装置做的一项改进是：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />活动探究二、实验是科学探究的重要方法．如图乙是测定空气中氧气含量实验的两套装置图，请结合图示回答有关问题．<br />（1）根据下表提供的实验数据，完成如表<br /><table class=\"edittable\"><TBODY><TR><td width=125>硬质玻璃管中空气的体积</TD><td width=125>反应前注射器中空气体积</TD><td width=125>反应后注射器中气体体积</TD><td width=125>实验测得空气中氧气的体积分数</TD></TR><TR><td>25mL</TD><td>15mL</TD><td>9mL</TD><td><u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD></TR></TBODY></TABLE>（2）装置一和装置二中气球的位置不同，<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“装置一”或“装置二”）更合理，理由是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />（3）若实验测得的结果偏小（氧气的体积分数小于21%），可能的原因有哪些？（列举两条）<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','引流$###$阴离子或酸根离子$###$集气瓶内壁出现水雾，证明有水生成$###$将小试管放入冷水中$###$15%$###$装置二$###$能让空气更易流通，全部通过铜粉，使氧气与铜粉充分反应$###$铜粉的量不足$###$未待装置完全冷却至室温就读数','【解答】解：活动探究一<br />（1）（1）过滤实验中玻璃棒的作用是引流．<br />故答案为：引流．<br />（2）酸类因为溶于水时电离出的阳离子相同，有形同的化学性质，又因为溶于水时电离出的阴离子或酸根离子不同，其性质也存在差异．<br />故答案为：阴离子或酸根离子．<br />（3）蜡烛燃烧的产物有水、二氧化碳等，验证有水的生成时，看到集气瓶内壁出现水雾，证明有水生成．<br />故答案为：集气瓶内壁出现水雾，证明有水生成．<br />（4）将小试管放入冷水中，会使水蒸气的冷凝效果更好．<br />故答案为：将小试管放入冷水中．<br />活动探究二<br />（1）注射器内气体减少的量就是氧气的体积，则实验测得空气中氧气的体积分数为：<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">15mL-9mL</td></tr><tr><td>25mL+15mL</td></tr></table></span>×100%=15%；<br />（2）装置二与装置一装置比较，装置二中的气球在玻璃管的后部，更易让空气流通，氧气与铜能充分反应；<br />（3）在测定空气中氧气的体积分数时，实验后发现测定氧气的体积分数低于五分之一，可能的原因有：①铜粉的量可能不足，没有将氧气消耗尽；②装置可能漏气，；③实验中可能未冷却至室温就读数等．<br />故答案为：（1）15%；<br />（2）装置二；能让空气更易流通，全部通过铜粉，使氧气与铜粉充分反应；<br />（3）铜粉的量不足；未待装置完全冷却至室温就读数（其他合理答案也可）．','【分析】活动探究一（1）过滤实验中玻璃棒的作用是引流；<br />（2）酸类因为溶于水时电离出的阳离子相同，有形同的化学性质，又因为溶于水时电离出的阴离子或酸根离子不同，其性质也存在差异．<br />（3）蜡烛燃烧的产物有水、二氧化碳等，根据特点进行验证；<br />（4）将小试管放入冷水中，会使水蒸气的冷凝效果更好．<br />活动探究二（1）注射器内气体减少的量就是氧气的量；<br />（2）从两种装置的区别来分析；<br />（3）根据测定空气中氧气的体积分数实验的原理、注意事项分析回答．','简答题',3.00,'f744db2900586b519875b922449a7ef3',9,400,'过滤的原理、方法及其应用,蒸发与蒸馏操作,蜡烛燃烧实验,空气组成的测定,酸的化学性质','新泰市',2016,'32','2016•新泰市校级模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840785,'下列观点你认为不正确的是（　　）','世界是物质的，物质是由分子构成的','宏观上看静止状态的物质从微观上看则是永恒运动的','分子虽小但也有一定的质量和体积','夏天雨后天晴，路面上的水渍很快变干，这时水就不存在了','','A|D','【解答】解：<br />A、世界是物质的，物质是由分子、原子、离子构成的，故错误；<br />B、微粒是不断运动的，观上看静止状态的物质从微观上看则是永恒运动的，故正确；<br />C、分子虽小但也有一定的质量和体积，故正确；<br />D、天雨后天晴，路面上的水渍很快变干，这时水蒸发，变为水蒸汽，水仍然存在，故错误．<br />答案：AD','【分析】A、根据物质是由分子、原子、离子构成的解答；<br />B、根据微粒是不断运动的解答；<br />C、根据分子的性质解答；<br />D、根据水的三态变化解答．','多选题',3.00,'ca43ba417f2db8195b523d631519b5bd',9,400,'物质的微粒性,分子、原子、离子、元素与物质之间的关系,分子的定义与分子的特性','',0,'37','',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840789,'小光同学在2016年合肥中考理科实验操作“氢氧化钠溶液的性质”实验中，向盛有硫酸铜溶液的试管中滴加氢氧化钠溶液，发现刚开始时无明显现象，继续滴加后才出现蓝色絮状沉淀．小光思考这是什么原因？与同学继续进行了如下实验探究：<br />[实验过程]<br /><table class=\"edittable\"><TBODY><TR><td></TD><td>实验操作</TD><td>实验现象</TD><td>实验结论</TD></TR><TR><td>甲同学</TD><td>向一洁净试管中加入2mLCuSO<SUB>4</SUB>溶液，向其中滴加<u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD><td>现象：①出现蓝色絮状沉淀<br />②反应方程式：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD><td rowSpan=2>假设成立</TD></TR><TR><td>乙同学</TD><td>向一洁净试管中加入2mLCuSO<SUB>4</SUB>溶液，<u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD><td>现象：①先无明显现象，后产生蓝色絮状沉淀；<br />②首先反应的化学方程式：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD></TR></TBODY></TABLE>[交流与反思]通守实验探究，你获得的感受是：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','氢氧化钠溶液$###$2NaOH+CuSO<SUB>4</SUB>=Cu（OH）<SUB>2</SUB>↓+Na<SUB>2</SUB>SO<SUB>4</SUB>$###$向其中滴加氢氧化钠溶液$###$2NaOH+H<SUB>2</SUB>SO<SUB>4</SUB>=Na<SUB>2</SUB>SO<SUB>4</SUB>+2H<SUB>2</SUB>O$###$把碱溶液加到盐和酸的混合溶液中，碱和酸先反应，而后再与盐反应','【解答】解：<br />甲同学：向一洁净试管中加入2mLCuSO<SUB>4</SUB>溶液，向其中滴加氢氧化钠溶液，出现蓝色絮状沉淀，反应方程式为：2NaOH+CuSO<SUB>4</SUB>=Cu（OH）<SUB>2</SUB>↓+Na<SUB>2</SUB>SO<SUB>4</SUB>．<br />乙同学：向一洁净试管中加入2mLCuSO<SUB>4</SUB>溶液，刚开始没有出现明显蓝色沉淀，是由于该硫酸铜溶液含酸性物质，酸先和氢氧化钠反应，因此刚开始并未发现预料中的蓝色沉淀现象；2NaOH+H<SUB>2</SUB>SO<SUB>4</SUB>=Na<SUB>2</SUB>SO<SUB>4</SUB>+2H<SUB>2</SUB>O．<br />把碱溶液加到盐和酸的混合溶液中，碱和酸先反应，而后再与盐反应．<br />故答案为：<br /><table class=\"edittable\"><TBODY><TR><td>验操作</TD><td>实验现象</TD><td>实验结论</TD></TR><TR><td>甲同学</TD><td>向一洁净试管中加入2mLCuSO<SUB>4</SUB>溶液，向其中滴加 氢氧化钠溶液</TD><td>现象：①出现蓝色絮状沉淀<br />②反应方程式：2NaOH+CuSO<SUB>4</SUB>=Cu（OH）<SUB>2</SUB>↓+Na<SUB>2</SUB>SO<SUB>4</SUB></TD><td rowSpan=2>假设成立</TD></TR><TR><td>乙同学</TD><td>向一洁净试管中加入2mLCuSO<SUB>4</SUB>溶液，向其中滴加氢氧化钠溶液</TD><td>现象：①先无明显现象，后产生蓝色絮状沉淀；<br />②首先反应的化学方程式：2NaOH+H<SUB>2</SUB>SO<SUB>4</SUB>=Na<SUB>2</SUB>SO<SUB>4</SUB>+2H<SUB>2</SUB>O</TD></TR></TBODY></TABLE>把碱溶液加到盐和酸的混合溶液中，碱和酸先反应，而后再与盐反应．','【分析】根据硫酸铜溶液可以和氢氧化钠溶液反应生成蓝色沉淀，可以据此写出该反应的化学方程式．','书写',3.00,'df325944430f0cd0bc643c7fec6434da',9,400,'探究酸碱的主要性质,碱的化学性质,书写化学方程式、文字表达式、电离方程式','',2016,'37','2016•合肥校级一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840792,'下列现象能说明空气中含有氧气的是（　　）','饼干放置在空气中容易受潮','澄清石灰水在空气中放置一段时间后产生浑浊现象','食物放置在空气中比放置在真空包装袋中容易腐烂变质','从冰箱内拿出的玻璃杯放置一会儿外壁出现水雾','','C','【解答】解：A、饼干放置在空气中变软，说明空气中有水；<br />B、澄清的石灰水在空气中放置一段时间后表面会出现一层浑浊是因为空气中有二氧化碳；<br />C、食物变质是因为食物中的物质和氧气发生缓慢氧化的结果，正确；<br />D、从冰箱内拿出的玻璃杯放置一会儿外壁出现水雾说明空气中有水；<br />故选C','【分析】根据空气的成分及各成分的性质与作用分析解答．空气中含有氮气、氧气、二氧化碳、稀有气体和其它气体和杂质．','选择题',3.00,'621b057bba23af14645869fdd0fca753',9,400,'空气的成分及各成分的体积分数','',2015,'32','2015•武侯区模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840795,'请从下列物质中选择填空（填字母编号）：<br />A．食盐&nbsp;&nbsp;&nbsp;&nbsp;B．明矾&nbsp;&nbsp;&nbsp;&nbsp;C．生石灰&nbsp;&nbsp;&nbsp;D．葡萄糖&nbsp;&nbsp;&nbsp;E．活性炭&nbsp;&nbsp;&nbsp;F．熟石灰<br />（1）<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>可用作冰箱的除味剂；<br />（2）能为人体提供能量的物质是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（3）常作食品干燥剂的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','E$###$D$###$C','【解答】解：<br />（1）活性炭具有吸附性，可用作冰箱的除味剂；&nbsp;&nbsp;&nbsp;<br />（2）葡萄糖能为人体提供能量；<br />（3）生石灰能与水反应生成了氢氧化钙，通常可作干燥剂．<br />故答案为：（1）E；（2）D；（3）C．','【分析】根据物质的结构决定物质的性质，物质的性质决定物质的用途，根据常见物质的性质和用途分析回答．','填空题',3.00,'9367587c91827ccde451ea7a0677895d',9,400,'生石灰的性质与用途,碳单质的物理性质及用途,生命活动与六大营养素','',2016,'32','2016•宿迁模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840799,'在生产面粉的过程中，会产生大量的面粉的极细的粉尘，当这些粉尘悬浮于空气中，并达到很高的浓度时，一旦遇到火苗、火星、电弧或适当的温度，瞬间就会燃烧起来，形成猛烈的爆炸，其威力不亚于炸弹，为防止发生爆炸，面粉厂必须张贴的标志是（　　）','<img  src=\"/tikuimages/9/2016/400/shoutiniao1/1dbbb811-94d4-11e9-a4f9-b42e9921e93e_xkb29.png\" style=\"vertical-align:middle\" />','<img  src=\"/tikuimages/9/2016/400/shoutiniao10/1dbf136e-94d4-11e9-8d00-b42e9921e93e_xkb53.png\" style=\"vertical-align:middle\" />','<img  src=\"/tikuimages/9/2016/400/shoutiniao31/1dc04bf0-94d4-11e9-bab9-b42e9921e93e_xkb18.png\" style=\"vertical-align:middle\" />','<img  src=\"/tikuimages/9/2016/400/shoutiniao46/1dc30b0f-94d4-11e9-8706-b42e9921e93e_xkb5.png\" style=\"vertical-align:middle\" />','','A','【解答】解：可燃性的气体或粉尘与氧气或空气混合后点燃易发生爆炸，面粉厂内的空气中混有可燃的面粉的粉尘，要防止引燃粉尘发生爆炸，所以面粉厂必须张贴严禁烟火标志．<br />A、图中所示标志是严禁烟火标志，是面粉厂必须张贴的安全标志，故选项正确．<br />B、图中所示标志是当心爆炸标志，面粉厂无需张贴，故选项错误．<br />C、图中所示标志是中国节能标志，面粉厂无需张贴，故选项错误．<br />D、图中所示标志是氧化剂标志，面粉厂无需张贴，故选项错误．<br />故选：A．','【分析】根据图标所表示的含义来考虑，并结合面粉厂应注意的事项进行分析判断．','选择题',3.00,'9e7a029fa815355c19a9ff04b5c6050d',9,400,'几种常见的与化学有关的图标','',2016,'32','2016•广东模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840801,'下列除杂质的方法不正确的是（括号内为杂质）（　　）','活性炭（铁粉）--用磁铁吸引','Na<SUB>2</SUB>SO<SUB>4</SUB>溶液（Na<SUB>2</SUB>CO<SUB>3</SUB>）--加适量稀硫酸','CaCl<SUB>2</SUB>溶液（HCl）--加过量碳酸钙','CaCO<SUB>3</SUB>（Na<SUB>2</SUB>CO<SUB>3</SUB>）--加适量稀盐酸','','D','【解答】解：A、铁粉能被磁铁吸引，而活性炭不能，能除去杂质且没有引入新的杂质，符合除杂原则，故选项所采取的方法正确．<br />B、Na<SUB>2</SUB>CO<SUB>3</SUB>能与适量稀硫酸反应生成硫酸钠、水和二氧化碳，能除去杂质且没有引入新的杂质，符合除杂原则，故选项所采取的方法正确．<br />C、HCl能与过量的碳酸钙反应生成氯化钙、水和二氧化碳，再过滤除去过量的碳酸钙，能除去杂质且没有引入新的杂质，符合除杂原则，故选项所采取的方法正确．<br />D、Na<SUB>2</SUB>CO<SUB>3</SUB>和CaCO<SUB>3</SUB>均能与适量稀盐酸反应，不但能把杂质除去，也会把原物质除去，不符合除杂原则，故选项所采取的方法错误．<br />故选：D．','【分析】根据原物质和杂质的性质选择适当的除杂剂和分离方法，所谓除杂（提纯），是指除去杂质，同时被提纯物质不得改变．除杂质题至少要满足两个条件：①加入的试剂只能与杂质反应，不能与原物质反应；②反应后不能引入新的杂质．','选择题',3.00,'0ff553bd87b679aeb61693226ffd5801',9,400,'物质除杂或净化的探究,常见金属的特性及其应用,酸的化学性质,盐的化学性质','',2016,'32','2016•金平区模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840806,'“水--生命之所系”．下列措施：<br />①农业上合理使用化肥和农药；<br />②生活污水集中处理后排放；<br />③工业污水处理达标后排放；<br />④应用新技术、新工艺减少生产生活中排放的污染物．<br />其中能改善水质的是（　　）','①②③④','①③④','②③④','①②④','','A','【解答】解：①农业上合理使用化肥和农药，能减少水体的污染；②生活污水集中处理后排放，能减少水体的污染；③工业污水处理达标后排放，能减少水体的污染；④应用新技术、新工艺减少生产生活中排放的污染，能减少水体的污染．<br />故选A．','【分析】根据造成水体污染的原因进行分析，工业废水、生活污水的任意排放以及农药化肥的任意施用造成了水体的污染．','选择题',3.00,'88065ef2af0a66233867178acb1a1dec',9,400,'水资源的污染与防治','',0,'37','',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840807,'配平下列化学方程式：<br />（1）<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>N<SUB>2</SUB>+<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>H<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;催化剂&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>高温高压</td></tr></table></span><u>&nbsp;&nbsp;&nbsp;&nbsp;</u>NH<SUB>3</SUB><br />（2）<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>C<SUB>2</SUB>H<SUB>6</SUB>+<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span><u>&nbsp;&nbsp;&nbsp;&nbsp;</u>CO<SUB>2</SUB><u>&nbsp;&nbsp;&nbsp;&nbsp;</u>H<SUB>2</SUB>0<br />（3）<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>Al<SUB>2</SUB>O<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;通电&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span><u>&nbsp;&nbsp;&nbsp;&nbsp;</u>Al+<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>O<SUB>2</SUB>↑<br />（4）<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>C<SUB>6</SUB>H<SUB>12</SUB>O<SUB>6</SUB>+<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td style=\"font-size: 90%\">酶</td></tr><tr><td><div hassize=\"9\"><div style=\"background: url(\'/upimages/images/part/8594L.png\') repeat-x; width:1px; height:5px;float:left;overflow:hidden\" muststretch=\"h\"></div><div style=\"background: url(\'/upimages/images/part/8594.png\') no-repeat; width:9px; height:5px;float:left;overflow:hidden\"></div></div></td></tr></table></span><u>&nbsp;&nbsp;&nbsp;&nbsp;</u>CO<SUB>2</SUB><u>&nbsp;&nbsp;&nbsp;&nbsp;</u>H<SUB>2</SUB>0．','','','','','','1$###$3$###$2$###$2$###$7$###$4$###$6$###$3$###$4$###$3$###$1$###$6$###$6$###$6','【解答】解：（1）利用最小公倍数法进行配平，以氢原子作为配平的起点，氢气、氨气前面的化学计量数分别为：3、2，最后调整氮气前面的化学计量数为1．<br />（2）本题可利用“定一法”进行配平，把C<SUB>2</SUB>H<SUB>6</SUB>的化学计量数定为1，则O<SUB>2</SUB>、CO<SUB>2</SUB>、H<SUB>2</SUB>O前面的化学计量数分别为：<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">7</td></tr><tr><td>2</td></tr></table></span>、2、3，同时扩大2倍，则C<SUB>2</SUB>H<SUB>6</SUB>、O<SUB>2</SUB>、CO<SUB>2</SUB>、H<SUB>2</SUB>O前面的化学计量数分别为2、7、4、6．<br />（3）利用最小公倍数法进行配平，以氧原子作为配平的起点，氧化铝、氧气前面的化学计量数分别为：2、3，最后调整铝前面的化学计量数为4．<br />（4）本题可利用“定一法”进行配平，把C<SUB>6</SUB>H<SUB>12</SUB>O<SUB>6</SUB>的化学计量数定为1，则O<SUB>2</SUB>、CO<SUB>2</SUB>、H<SUB>2</SUB>O前面的化学计量数分别为6、6、6．<br />故答案为：（1）1、3、2；（2）2、7、4、6；（3）2、4、3；（4）1、6、6、6．','【分析】根据质量守恒定律：反应前后各原子的数目不变，选择相应的配平方法（最小公倍数法、定一法等）进行配平即可；配平时要注意化学计量数必须加在化学式的前面，配平过程中不能改变化学式中的下标；配平后化学计量数必须为整数．','填空题',3.00,'2550564645423905ab69428306127b45',9,400,'化学方程式的配平','乳山市',2016,'35','2016春•乳山市期中',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840808,'新版《生活饮用水卫生标准》（简称新国标）中水质检测指标从原来的35项增加到106项，对供水各环节的水质提出了相应的要求．<br />（1）新国标在指标中修订了镉、铅等的限量，这里的镉、铅指的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填序号）．<br />A、原子&nbsp;&nbsp;&nbsp;B、分子&nbsp;&nbsp;&nbsp;C、元素&nbsp;&nbsp;&nbsp;D、单质<br />（2）新国标中消毒剂由1项增至4项，加入了对用臭氧、二氧化氢和氯胺消毒的规定：<br />①臭氧（O<SUB>3</SUB>）在消毒过程中转化为氧气，臭氧转化为氧气属于<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“物理”或“化学”）变化．<br />②氯胺（NH<SUB>2</SUB>Cl）由<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填数字）种元素组成，用氯胺消毒时，反应的反应方程式是NH<SUB>2</SUB>Cl+X═NH<SUB>3</SUB>+HClO，其中X的化学式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）净水器常用活性炭，主要是利用活性炭的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>性，生活中，既能降低水的硬度，又能杀菌消毒的方法是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','C$###$化学$###$三$###$H<SUB>2</SUB>O$###$吸附$###$煮沸','【解答】解：（1）物质的组成常用元素来描述，新国标在指标中修订了镉、铅等的限量，这里的镉、铅指的是元素；<br />（2）①臭氧（O<SUB>3</SUB>）在消毒过程中转化为氧气，臭氧转化为氧气，有新物质生成，属于化学变化．<br />②一氯胺是由氮、氢和氯三种元素组成的，根据反应的化学方程式：NH<SUB>2</SUB>Cl+X=NH<SUB>3</SUB>+HClO，反应物中氮、氢、氯原子个数分别为1、2、1，反应后的生成物中氮、氢、氯、氧原子个数分别为1、4、1、1，根据反应前后原子种类、数目不变，则每个X分子由2个氢原子和1个氧原子构成，则物质X的化学式为：H<SUB>2</SUB>O．<br />（3）由于活性炭具有吸附性，净水器常用活性炭，主要是利用活性炭的吸附性，生活中，既能降低水的硬度，又能杀菌消毒的方法是煮沸．<br />故答案为：（1）C（2）①化学；②三；H<SUB>2</SUB>O．（3）吸附，煮沸．','【分析】（1）根据物质的组成常用元素来描述分析；<br />（2）①根据变化的特征分析判断；<br />②由质量守恒定律：反应前后，原子种类、数目均不变，据此由反应的化学方程式推断反应物X的化学式．<br />（3）根据活性炭的吸附性和生活中降低水的硬度的方法分析回答．','书写',3.00,'a6746c12e655bfcc189b61d66d645351',9,400,'水的净化,硬水与软水,元素的概念,碳单质的物理性质及用途,化学式的书写及意义,化学变化和物理变化的判别,质量守恒定律及其应用','招远市',2016,'35','2016春•招远市期中',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840809,'几年前，一辆载满20t电石的挂车，在浙江滨江区燃起熊熊大火并伴有大量黑烟．事故路面上洒落了五六百米燃着的石头．这辆货车在雨中整整燃烧了一天，消防官兵对大火也束手无措，场面让人震撼．大家对电石的化学性质产生了浓厚的兴趣，请你一起与他们完成下列探究活动．<br />【查阅资料】电石的主要成分是碳化钙（CaC<SUB>2</SUB>），可以与水反应生成一种可燃性气体及白色固体．常见的可燃性气体中，乙炔（C<SUB>2</SUB>H<SUB>2</SUB>）燃烧时伴有大量黑烟．<br />【猜想与假设】<br />电石与水反应生成的可燃性气体是：氧气、氢气或乙炔（C<SUB>2</SUB>H<SUB>2</SUB>）；电石与水反应生成的白色固体是：氧化钙、氢氧化钙或碳酸钙．大家做出以上推断的依据是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．大家讨论后一致认为该气体不可能是氧气，原因是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，白色固体不可能是氧化钙，原因是（用化学方程式表示）<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />【实验探究】<br /><table class=\"edittable\"><TBODY><TR><td width=212>实验步骤</TD><td width=193>实验现象</TD><td width=145>实验结论</TD></TR><TR><td>（1）取适量电石加水反应，收集生成的气体，验纯后点燃．</TD><td>气体燃烧，产生黄色火焰并伴有浓烈黑烟．</TD><td>电石与水反应产生的气体是乙炔</TD></TR><TR><td>（2）取适量反应后生成的固体于试管中，滴加过量稀盐酸．</TD><td>固体溶解，<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．</TD><td>电石与水反应产生的固体不是碳酸钙</TD></TR><TR><td>（3）取适量反应后生成的固体于试管中加水，向上层清液中滴加2～3滴<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．</TD><td>溶液变红</TD><td>电石与水反应产生的固体是氢氧化钙</TD></TR></TBODY></TABLE>【反思交流】载有电石的货车可以在雨中整整燃烧一天而不熄灭，说明电石与水的反应是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“吸热”或“放热”）反应．','','','','','','质量守恒定律反应前后元素的种类不变$###$氧气不具有可燃性$###$CaO+H<SUB>2</SUB>O═Ca（OH）<SUB>2</SUB>$###$没有气泡产生$###$酚酞$###$放热','【解答】解：【猜想假设】反应物中含有Ca、C、O、H四种元素，根据质量守恒定律反应前后元素的种类不变，所以推测气体可能是氧气、氢气或乙炔（C<SUB>2</SUB>H<SUB>2</SUB>）；但由于氧气不具有可燃性，故不可能是氧气；因为氧化钙遇到水会产生氢氧化钙，所以白色固体不可能是氧化钙；<br />【实验探究】（1）资料中乙炔（C<SUB>2</SUB>H<SUB>2</SUB>）燃烧时伴有大量黑烟，而氢气燃烧不会产生黑烟，因此电石与水反应产生的气体是乙炔；<br />（2）因为结论电石与水反应产生的固体不是碳酸钙，因此加盐酸不会产生气泡；<br />（3）因为结论电石与水反应产生的固体是氢氧化钙，氢氧化钙的溶液呈现碱性，能够使酚酞试液变成红色；<br />【反思交流】根据满载电石的货车遇大雨着火的原因是电石主要成分碳化钙（CaC<SUB>2</SUB>）能与水反应生成可燃性气体，且该反应放热；<br />故答案为：【猜想假设】质量守恒定律反应前后元素的种类不变；氧气不具有可燃性；CaO+H<SUB>2</SUB>O═Ca（OH）<SUB>2</SUB>；<br />【实验探究】没有气泡产生；酚酞；<br />【反思交流】放热．','【分析】【猜想假设】根据质量守恒定律反应前后元素的种类不变及氧气不具有可燃性进行分析；<br />根据氧化钙遇到水会产生氢氧化钙分析；<br />【实验探究】根据信息结合实验的现象分析产物的成分；根据碳酸钙和氢氧化钙的性质分析实验；<br />【反思交流】根据现象分析能量的变化．','书写',3.00,'6ce7ee92bdc4f08cffd7f497fc2777b9',9,400,'实验探究物质的性质或变化规律,生石灰的性质与用途,盐的化学性质,物质发生化学变化时的能量变化,质量守恒定律及其应用,书写化学方程式、文字表达式、电离方程式','',2016,'37','2016•阳谷县二模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840810,'“归纳”“总结”是学习化学常用的方法，下列说法错误的是（　　）','通常物质燃烧的条件是：物质具有可燃性、跟氧气接触、温度达到可燃物的着火点，当这三个条件同时具备时燃烧才能发生','原子都是由质子、中子、电子三种微粒构成的','复分解反应的条件是：生成物有水、气体、沉淀三个条件至少满足一个，复分解反应才能发生','不同的物质构成它的基本微粒可能不同，分子、原子、离子都能构成物质','','B','【解答】解：A．通常物质燃烧的条件是：物质具有可燃性、跟氧气接触、温度达到可燃物的着火点，当这三个条件同时具备时燃烧才能发生，故正确；<br />B．绝大部分原子都是由质子、中子和电子构成，但有些原子没有中子，如氢原子中没有中子，故错误；<br />C．复分解反应的条件是：生成物有水、气体、沉淀三个条件至少满足一个．复分解反应才能发生，故正确；<br />D．不同的物质构成它的基本微粒可能不同，分子、原子、离子都能构成物质，故正确．<br />故答案为：B．','【分析】A．据物质燃烧的条件是：物质具有可燃性、跟氧气接触、温度达到可燃物的着火点解答；<br />B．根据原子的结构来分析；<br />C．据复分解反应的条件生成物要求必须含有沉淀，或者是气体，或者是水解答；<br />D．据构成物质的微粒可以是分子、原子、离子，但不同的物质构成它的基本微粒可能不同解答．','选择题',3.00,'ee39804f510a3366656aec52c629f50c',9,400,'复分解反应及其发生的条件,分子、原子、离子、元素与物质之间的关系,原子的定义与构成,燃烧与燃烧的条件','',2016,'37','2016•新化县二模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840817,'下列对有关事实的解释中，错误的是（　　）','生铁和钢的性能不同--含碳量不同','CO和C0<SUB>2</SUB>的化学性质不同--分子构成不同','钠原子和钠离子化学性质不同--质子数不同','金刚石和石墨的物理性质不同--原子排列方式不同','','C','【解答】解：A．生铁和钢的性能不同是因为含碳量不同，生铁的含碳量为2%～4.3%，钢中的含碳量为0.03%～2%，故A正确；<br />B．CO和CO<SUB>2</SUB>的性质不同，是由于构成它们的分子不同，说法正确，故B正确；<br />C．钠原子失去1个电子形成钠离子，钠原子和钠离子化学性质不同是由于其最外层电子数不同，故C错误；<br />D．金刚石和石墨的物理性质不同，是由于其碳原子排列组成不同，故D正确．<br />故选C．','【分析】A．根据生铁和钢都是铁的合金进行解答；<br />B．根据二氧化碳和一氧化碳的分子构成不同来分析；<br />C．根据钠原子失去1个电子形成钠离子来分析；<br />D．根据结构决定性质来分析．','选择题',3.00,'5e785bb978055d5591cf002782a6daf9',9,400,'生铁和钢,原子和离子的相互转化,分子的定义与分子的特性,碳元素组成的单质','',2016,'37','2016•长春二模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840818,'我国《水污染防治法》规定：建立饮用水水源保护区制度，防止水源枯竭和水体污染，禁止在饮用水水源保护区内设置排污口．下列说法正确的是（　　）','水资源是取之不尽、用之不竭的','保护水资源需要合理使用农药、化肥','只有工业污染和农业污染会造成水体污染','无色透明的工业废水可直接排放在饮用水水源保护区内','','B','【解答】解：A、水资源丰富，但是淡水资源十分紧缺，所以我们要节约用水，故A错误；<br />B、合理施用农药、化肥能减少水体污染，故B正确；<br />C、工业污染和农业污染会造成水体污染，生活污水也会造成水资源的污染，故C错误；<br />D、无色透明的工业废水也可能含有有害、有毒物质，所以不能直接排放在饮用水水源保护区内，故D错误．<br />故选：B．','【分析】A、根据水资源丰富，但是淡水资源十分紧缺进行解答；<br />B、根据合理施用农药、化肥能减少水体污染进行解答；<br />C、根据水污染的途径进行解答；<br />D、根据无色透明的工业废水也可能含有有害、有毒物质进行解答．','选择题',3.00,'812ab6bfbdaa4a127feb34513a010688',9,400,'水资源状况,水资源的污染与防治,保护水资源和节约用水','',2016,'32','2016•海安县模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840828,'<img  src=\"/tikuimages/9/2015/400/shoutiniao54/1e29e330-94d4-11e9-885d-b42e9921e93e_xkb52.png\" style=\"vertical-align:middle;FLOAT:right;\" />（2015秋•孝义市校级期中）结合如图所示常用的仪器，回答有关问题：<br />（1）写出标号仪器的名称：C<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（2）如图仪器中，能在酒精灯火焰上直接加热的有<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填标号）；<br />（3）选择图中的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>和<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填标号）组合起来，可以量取一定量的液体体积数；<br />（4）用于过滤的仪器有（写出所有需要的仪器名称）<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','量筒$###$A$###$C$###$F$###$铁架台、玻璃棒、漏斗、烧杯','【解答】解：图中A是试管、B是漏斗、C是量筒、D是酒精灯、E是铁架台、F是胶头滴管、G是集气瓶．<br />（1）图中C是量筒；故填：量筒．<br />（2）试管是可以直接加热的；故填：A．<br />（3）量筒可量取一定液体的体积，并用胶头滴管定容．故填：C；F．<br />（4）过滤所用仪器有：铁架台、玻璃棒、漏斗、烧杯．故填：铁架台、玻璃棒、漏斗、烧杯．','【分析】（1）熟悉常见仪器，了解名称；<br />（2）试管是可以直接加热的；<br />（3）量筒可量取一定液体的体积，并用胶头滴管定容．<br />（4）根据过滤所用仪器分析．','填空题',3.00,'71088e62c2dd50d9d0bfa9d50909d43a',9,400,'用于加热的仪器,测量容器-量筒,过滤的原理、方法及其应用,常用仪器的名称和选用','',2015,'35','2015秋•孝义市校级期中',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840829,'下列图象能正确反映对应变化关系的是（　　）','<img  src=\"/tikuimages/9/2016/400/shoutiniao62/1e2eec40-94d4-11e9-ab18-b42e9921e93e_xkb4.png\" style=\"vertical-align:middle\" />加热一定量的高锰酸钾固体','<img  src=\"/tikuimages/9/2016/400/shoutiniao56/1e30e80f-94d4-11e9-be56-b42e9921e93e_xkb11.png\" style=\"vertical-align:middle\" />两份完全相同的过氧化氢溶液分解','<img  src=\"/tikuimages/9/2016/400/shoutiniao47/1e31ab61-94d4-11e9-9bf7-b42e9921e93e_xkb57.png\" style=\"vertical-align:middle\" />向一定量铁粉中加入硫酸铜溶液','<img  src=\"/tikuimages/9/2016/400/shoutiniao26/1e33ce40-94d4-11e9-addc-b42e9921e93e_xkb68.png\" style=\"vertical-align:middle\" />加热一定量的饱和石灰水','','B','【解答】解：A、加热一定量的高锰酸钾固体，固体中氧元素的质量分数不会变成0，错误；<br />B、两份完全相同的过氧化氢溶液分解，最后产生的氧气的质量相等，有催化剂反应的快，无催化剂产生的慢，正确；<br />C、向一定量铁粉中加入硫酸铜溶液，<br />Fe+CuSO<SUB>4</SUB>=FeSO<SUB>4</SUB>+Cu<br />56&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 64<br />固体的质量增加，错误；<br />D、加热一定量的饱和石灰水，氢氧化钙的溶解度随温度的升高而减小，故溶液中的溶质析出，溶液质量减小，错误；<br />故选B．','【分析】A、根据加热高锰酸钾的反应解答；<br />B、根据过氧化氢分解生成氧气的质量关系解答；<br />C、根据铁与硫酸铜的反应解答；<br />D、根据氢氧化钙的溶解度与温度的变化关系解答．','选择题',3.00,'92db3705e4afcc8536127a46fb38d40a',9,400,'催化剂的特点与催化作用,固体溶解度的影响因素,金属的化学性质,盐的化学性质','',2016,'37','2016•丰台区一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840832,'屠呦呦是第一位获得诺贝尔科学奖项的中国本土科学家．1972年，屠哟哟和她的同事在青蒿中提取到了一种分子式为C<SUB>15</SUB>H<SUB>22</SUB>O<SUB>5</SUB>的无色结晶体，一种熔点为156℃〜157℃的活性成份，他们将这种无色的结晶体物质命名为青蒿素．青蒿素为一具有“高&nbsp;效、速效、低毒，优点的新结构类型抗疟药，对各型疟疾特别是抗恶性疟疾有特效．关&nbsp;于青蒿素下列说法正确的是（　　）','属于氧化物','由碳元素、氢元素、氧元素组成','282g青蒿素中含有86g碳元素','一个青蒿素分子中含有5个氧分子','','B','【解答】解：A．由青蒿素的化学式C<SUB>15</SUB>H<SUB>22</SUB>O<SUB>5</SUB>可知，青蒿素含三种元素，而氧化物中只能含两种元素，不属于氧化物，故错误；<br />B．由青蒿素的化学式C<SUB>15</SUB>H<SUB>22</SUB>O<SUB>5</SUB>可知，青蒿素含碳、氢、氧三种元素，故正确；<br />C.282g青蒿素中含有碳元素的质量为：282g×<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">12×15</td></tr><tr><td>12×15+1×22+16×5</td></tr></table>×100%</span>=180g，故错误；<br />D．分子是由原子构成的，一个青蒿素分子中含有5个氧原子，故错误．<br />故选B．','【分析】A．根据氧化物的概念来分析；<br />B．根据物质的组成来分析；<br />C．根据化合物中元素的质量来分析；<br />D．根据分子的结构来分析．','选择题',3.00,'a999be613c8c6e7252a8ae4779d07986',9,400,'从组成上识别氧化物,化学式的书写及意义,化合物中某元素的质量计算','',2015,'37','2015秋•滑县校级月考',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840835,'对于图示理解错误的是（　　）<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao77/1e3f3ff0-94d4-11e9-9d45-b42e9921e93e_xkb59.png\" style=\"vertical-align:middle\" />','原子可结合成分子，分子聚集成物质','可用<img  src=\"/tikuimages/9/2016/400/shoutiniao62/1e43d3cf-94d4-11e9-9296-b42e9921e93e_xkb3.png\" style=\"vertical-align:middle\" />表示氢分子的形成过程','物质都是由分子构成','原子是化学变化中的最小微粒','','C','【解答】解：A、据图可以看出，原子可结合成分子，分子聚集成物质，正确；<br />B、可用<img  src=\"/tikuimages/9/2016/400/shoutiniao82/1e47a45e-94d4-11e9-b69d-b42e9921e93e_xkb87.png\" style=\"vertical-align:middle\" />表示氢分子的形成过程，正确；<br />C、物质不一定是由分子构成的，错误；<br />D、原子是化学变化中的最小微粒，正确；<br />故选C．','【分析】根据分子和原子的区别和联系进行分析解答即可．','选择题',3.00,'a9092ddb5ef245769f98a636c930e8e3',9,400,'分子、原子、离子、元素与物质之间的关系,分子和原子的区别和联系','',2016,'37','2016•娄星区一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840844,'<img  src=\"/tikuimages/9/2016/400/shoutiniao81/1e60356e-94d4-11e9-99b6-b42e9921e93e_xkb95.png\" style=\"vertical-align:middle;FLOAT:right;\" />（2016•寿光市校级模拟）某同学发现，上个月做实验用的NaOH溶液忘记了盖瓶盖，对于该溶液是否变质，同学们开始实验探究．<br />【猜想假设】<br />猜想（1）：该溶液没有变质，为NaOH溶液<br />猜想（2）：该溶液全部变质，为NaCO<SUB>3</SUB>溶液<br />猜想（3）：该溶液部分变质，为NaOH和Na<SUB>2</SUB>CO<SUB>3</SUB>的混合溶液<br />【査阅资料】Na<SUB>2</SUB>CO<SUB>3</SUB>溶液呈碱性<br />【设计方案】请你完善下表几组同学探讨的设计方案<br /><table class=\"edittable\"><TBODY><TR><td width=142>&nbsp;</TD><td width=142>&nbsp;实验操作</TD><td width=142>&nbsp;可能出现的现象与结论</TD><td width=142>&nbsp;同学评价</TD></TR><TR><td>&nbsp;第1组</TD><td>&nbsp;<img  src=\"/tikuimages/9/2016/400/shoutiniao32/1e625851-94d4-11e9-bf48-b42e9921e93e_xkb65.png\" style=\"vertical-align:middle\" /></TD><td>&nbsp;若溶液变红，则猜想（2）不成立</TD><td>&nbsp;第2组同学认为：此方案结论不正确，理是<br /><u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD></TR><TR><td>&nbsp;第2组</TD><td>&nbsp;<img  src=\"/tikuimages/9/2016/400/shoutiniao23/1e6369c0-94d4-11e9-9444-b42e9921e93e_xkb81.png\" style=\"vertical-align:middle\" /></TD><td>&nbsp;若产生白色沉淀，反应方程式是<br /><u>&nbsp;&nbsp;&nbsp;&nbsp;</u>则猜想（1）不成立</TD><td>&nbsp;第3组同学认为：此方案还不能确定猜想（2）还是猜想（3）成立</TD></TR><TR><td>&nbsp;第3组</TD><td>&nbsp;<img  src=\"/tikuimages/9/2016/400/shoutiniao77/1e667700-94d4-11e9-b350-b42e9921e93e_xkb78.png\" style=\"vertical-align:middle\" /></TD><td>&nbsp;若滤液不变红，则猜想（2）成立，若滤液变红则猜想（3）成立</TD><td>&nbsp;第1组同学认为：不需过滤也能达到实验目的，更简单的操作方法是：<br /><u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD></TR></TBODY></TABLE>【探宄实验】综合各小组的方案后，动手实验<br />【结论与反思】NaOH浴液易与空气中CO<SUB>2</SUB>反应而变质，所以要密封保存<br />【拓展延伸】同学们设计了下列两套装置进行实验用饺头滴管吸取某种液体，锥形瓶中充入一种气体或放入一种固体物质，挤压胶头滴管一段时间后，两装置中气球明显胀大<br />（1）甲装置中可能发生反应的方程式是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />（2）若乙装置中头滴管吸取的是稀盐酸，则锥形瓶中放入的固体可能是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','碳酸钠溶液能使酚酞试液变红色$###$Na<SUB>2</SUB>CO<SUB>3</SUB>+CaCl<SUB>2</SUB>═CaCO<SUB>3</SUB>↓+2NaCl$###$第2组实验结束后，静置一会儿后，滴加酚酞试液，酚酞试液变红色，说明猜想（3）成立$###$2NaOH+CO<SUB>2</SUB>═Na<SUB>2</SUB>CO<SUB>3</SUB>+H<SUB>2</SUB>O$###$铁','【解答】解：【设计方案】实验过程如下表所示：<table class=\"edittable\"><TBODY><TR><td width=64></TD><td width=190>实验操作</TD><td width=161>可能出现的现象与结论</TD><td width=154>同学评价</TD></TR><TR><td>第1组</TD><td><img  src=\"/tikuimages/9/2016/400/shoutiniao15/1e6b800f-94d4-11e9-b364-b42e9921e93e_xkb19.png\" style=\"vertical-align:middle\" /></TD><td>若溶液变红，则猜想（2）不成立</TD><td>第2组同学认为：此方案结论不正确，理由是：碳酸钠溶液能使酚酞试液变红色</TD></TR><TR><td>第2组</TD><td><img  src=\"/tikuimages/9/2016/400/shoutiniao94/1e6da2f0-94d4-11e9-95e0-b42e9921e93e_xkb23.png\" style=\"vertical-align:middle\" /></TD><td>若产生白色沉淀，反应方程式是若产生白色沉淀，反应方程式是<br />Na<SUB>2</SUB>CO<SUB>3</SUB>+CaCl<SUB>2</SUB>═CaCO<SUB>3</SUB>↓+2NaCl <br />则猜想（1）不成立&nbsp;</TD><td>第3组同学认为：此方案还不能确定猜想（2）还是猜想（3）成立．</TD></TR><TR><td>第3组</TD><td><img  src=\"/tikuimages/9/2016/400/shoutiniao74/1e6f9ec0-94d4-11e9-ad78-b42e9921e93e_xkb22.png\" style=\"vertical-align:middle\" /></TD><td>若滤液不变红，则猜想（2）成立；若滤液变红，则猜想（3）成立．</TD><td>第1组同学认为：不需过滤也能达到实验目的，更简单的操作方法是：第2组实验结束后，静置一会儿后，滴加酚酞试液，酚酞试液变红色，说明猜想（3）成立</TD></TR></TBODY></TABLE>【拓展延伸】（1）甲装置中气球膨胀，说明锥形瓶中的气体减少，二氧化碳能和氢氧化钠溶液反应，因此液体可能是氢氧化钠溶液，气体可能是二氧化碳，发生反应的方程式是：2NaOH+CO<SUB>2</SUB>═Na<SUB>2</SUB>CO<SUB>3</SUB>+H<SUB>2</SUB>O．<br />故填：2NaOH+CO<SUB>2</SUB>═Na<SUB>2</SUB>CO<SUB>3</SUB>+H<SUB>2</SUB>O．<br />（2）若乙装置中胶头滴管吸取的是稀盐酸，则锥形瓶中放入的固体可能是铁，因为铁能和稀盐酸反应生成氢气，导致气球膨胀．<br />故填：铁．','【分析】【设计方案】氢氧化钠溶液和碳酸钠溶液都能够使酚酞试液变红色；<br />碳酸钠能和氯化钙反应生成白色沉淀碳酸钙和氯化钠；<br />【拓展延伸】氢氧化钠和二氧化碳反应生成碳酸钠和水；<br />稀盐酸和铁反应生成氯化亚铁和氢气．','书写',3.00,'dde40edcd12b8e687e5aecf3774e5433',9,400,'药品是否变质的探究,碱的化学性质,盐的化学性质,书写化学方程式、文字表达式、电离方程式','寿光市',2016,'32','2016•寿光市校级模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840845,'氮气和氧气在放电条件下可以发生反应．用“<img  src=\"/tikuimages/9/2016/400/shoutiniao45/1e734840-94d4-11e9-818e-b42e9921e93e_xkb25.png\" style=\"vertical-align:middle\" />”和“<img  src=\"/tikuimages/9/2016/400/shoutiniao66/1e7718cf-94d4-11e9-afa8-b42e9921e93e_xkb72.png\" style=\"vertical-align:middle\" />”分别表示氮原子和氧原子，如图是此反应的微观模拟图．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao48/1e79ff00-94d4-11e9-bf45-b42e9921e93e_xkb9.png\" style=\"vertical-align:middle\" /><br />（1）请将B图补充完整，使其正确．<br />（2）化学反应的实质就是分子分成原子，原子又结合成<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）氧气属于单质的微观解释是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','新分子$###$含有一种（或氧）分子，分子中含有一种（或氧）原子','【解答】解：（1）观察微观示意图可知，有一个氮分子没有参加反应，根据反应前后原子的种类和个数都不变，应该补上一个一氧化氮分子；<br />（2）化学反应的实质就是分子分成原子，原子又结合成新分子的过程；<br />（3）氧气分子是由氧原子一种原子构成的分子，因此属于单质；<br />故答案为：<br />（1）如图，补出一个一氧化氮分子．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao20/1e7d5a5e-94d4-11e9-87cb-b42e9921e93e_xkb9.png\" style=\"vertical-align:middle\" /><br />（2）新分子；<br />（3）含有一种（或氧）分子，分子中含有一种（或氧）原子．','【分析】（1）依据化学反应的实质分析解答即可；<br />（2）根据生成物分子的结构及分子的种类分析解答；<br />（3）根据单质的定义解答．','填空题',3.00,'1b9ba73d791f1ea392b91f6574336222',9,400,'单质和化合物的判别,微粒观点及模型图的应用,化学反应的实质','',2016,'37','2016•道外区二模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840847,'下列物质暴露于空气中，其质量改变和空气中的成分没有关系的是（　　）','生石灰','铁粉','浓硫酸','浓盐酸','','D','【解答】解：A、生石灰能与空气中的水蒸气反应，错误；<br />B、铁粉能与空气中氧气和水反应，错误；<br />C、浓硫酸会吸收空气中的水分，错误；<br />D、浓盐酸挥发导致质量变化，与空气中的成分没有关系，正确；<br />故选D．','【分析】根据已有的物质的性质以及质量的变化进行分析解答即可．','选择题',3.00,'9f65f525bece4cf3d2af3bba3503bcd8',9,400,'金属锈蚀的条件及其防护,生石灰的性质与用途,空气中常见酸碱盐的质量或性质变化及贮存法','',2015,'37','2015秋•忻州校级月考',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840848,'<img  src=\"/tikuimages/9/2016/400/shoutiniao30/1e841121-94d4-11e9-ab45-b42e9921e93e_xkb40.png\" style=\"vertical-align:middle;FLOAT:right\" />（2016•常州模拟）为创建环保节能城市，我市部分道路两侧使用太阳能路灯（如图所示）．<br />（1）图中③使用的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>材料；<br />（2）太阳能路灯的广泛使用能有效提高空气质量，这是因为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（3）白天阳光下太阳能电板中发生的能量直接转化形式：太阳能转化为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>能；<br />（4）铝合金露置在空气中表面会形成保护膜，其反应方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','合成$###$减少化石燃料燃烧$###$电$###$4Al+3O<SUB>2</SUB>=2Al<SUB>2</SUB>O<SUB>3</SUB>','【解答】解：（1）塑料属于合成材料；<br />（2）太阳能路灯的广泛使用能有效提高空气质量，这是因为减少化石燃料燃烧；<br />（3）白天阳光下太阳能电板中发生的能量直接转化形式：太阳能转化为电能；<br />（4）铝和氧气反应会生成致密的氧化铝薄膜，化学方程式为：4Al+3O<SUB>2</SUB>=2Al<SUB>2</SUB>O<SUB>3</SUB>．<br />故答案为：（1）合成；<br />（2）减少化石燃料燃烧；<br />（3）电；<br />（4）4Al+3O<SUB>2</SUB>=2Al<SUB>2</SUB>O<SUB>3</SUB>．','【分析】（1）根据塑料属于合成材料进行分析；<br />（2）根据太阳能路灯的广泛使用能有效提高空气质量，这是因为减少化石燃料燃烧进行分析；<br />（3）根据白天阳光下太阳能电板中发生的能量直接转化形式：太阳能转化为电能进行分析；<br />（4）根据铝和氧气反应会生成致密的氧化铝薄膜进行分析．','书写',3.00,'5de05bcbf4fdd31798cb7f8dffbcfb20',9,400,'书写化学方程式、文字表达式、电离方程式,资源综合利用和新能源开发,合成材料的使用及其对人和环境的影响','',2016,'32','2016•常州模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840852,'下列气体与空气混合后遇明火，可能发生爆炸的是（　　）','氧气','氮气','一氧化碳','二氧化碳','','C','【解答】解：根据爆炸是由于急速的燃烧在有限的空间而引起的，因此要满足燃烧的条件，气体必须是可燃性气体遇明火，<br />A、氧气具有助燃性，不具有可燃性；故A不符合题意；<br />B、氮气不具有助燃性，也不具有可燃性；故B不符合题意；<br />C、一氧化碳具有可燃性，因此氢气与空气混合后遇明火，可能发生爆炸；故C符合题意；<br />D、二氧化碳不能燃烧，也不支持燃烧；故B不符合题意．<br />故选C．','【分析】根据燃烧和发生爆炸的条件（在有限的空间内，可燃气体或粉尘与空气混合，达到爆炸极限，遇到明火；解答此题．','选择题',3.00,'d200e735784b9286734317351f2024d8',9,400,'燃烧、爆炸、缓慢氧化与自燃','',2016,'37','2016春•北京校级月考',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840854,'下列实验方案不可行的是（　　）','用灼烧的方法区别腈纶制品和羊毛制品','用酚酞试液区分氢氧化钠溶液和碳酸钠溶液','用高温灼烧氧化铁和木炭粉混合物的方法制取少量铁','用加水溶解的方法区别NaOH、Na<SUB>2</SUB>SO<SUB>4</SUB>、NH<SUB>4</SUB>NO<SUB>3</SUB>三种白色固体','','B','【解答】解：A、羊毛的主要成分是蛋白质，燃烧后有烧焦羽毛的味道，所以可用灼烧的方法区别腈纶制品和羊毛制品，故A正确；<br />B、氢氧化钠和碳酸钠都显碱性，都能使酚酞试液变成红色，无法鉴别，故B错误；<br />C、高温灼烧氧化铁和木炭粉混合物能生成铁，故C正确；<br />D、氢氧化钠固体溶于水放热，硝酸铵溶于水吸热，而硫酸钠溶于水没有明显的能量变化，所以可用加水溶解的方法区别NaOH、Na<SUB>2</SUB>SO<SUB>4</SUB>、NH<SUB>4</SUB>NO<SUB>3</SUB>三种白色固体，故D正确．<br />故选：B．','【分析】A、根据羊毛的主要成分是蛋白质，燃烧后有烧焦羽毛的味道进行解答；<br />B、根据氢氧化钠和碳酸钠都显碱性进行解答；<br />C、根据高温灼烧氧化铁和木炭粉混合物能生成铁，但是该实验方案比较复杂进行解答；<br />D、根据氢氧化钠固体溶于水放热，硝酸铵溶于水吸热，而硫酸钠溶于水没有明显的能量变化进行解答．','选择题',3.00,'77f0d6c81e9d8fd4e99511643500c65f',9,400,'化学实验方案设计与评价,酸、碱、盐的鉴别,碳的化学性质,棉纤维、羊毛纤维和合成纤维的鉴别','',2016,'32','2016•长春模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840857,'<img  src=\"/tikuimages/9/2016/400/shoutiniao52/1e96fcde-94d4-11e9-a84a-b42e9921e93e_xkb9.png\" style=\"vertical-align:middle;FLOAT:right\" />（2016•河北区一模）某固体混合物A中可能含硫酸钠、碳酸钠、硝酸钡、氯化铵等物质中的若干种．按如图所示进行实验，出现的现象如图所述（实验过程中所有发生的反应都恰好完全反应）．<br />根据实验过程和发生的现象做出判断，填写以下空白：<br />（1）现象X是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，气体E的化学式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）固体混合物A中，肯定存在的物质是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（写化学式）．<br />（3）写出步骤①中发生反应的化学方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（4）仅根据上述实验现象，还不能确定混合物A的组成．为确定混合物A的成分，可用沉淀G再进行实验．方法是取沉淀G少许，向其中加入足量稀硝酸，若沉淀全部溶解，则固体混合物A中不含<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（化学式）．','','','','','','石蕊试液变红色$###$NH<SUB>3</SUB>$###$Na<SUB>2</SUB>CO<SUB>3</SUB>、NH<SUB>4</SUB>Cl$###$Na<SUB>2</SUB>CO<SUB>3</SUB>+2HCl=2NaCl+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑$###$Ba（NO<SUB>3</SUB>）<SUB>2</SUB>、Na<SUB>2</SUB>SO<SUB>4</SUB>','【解答】解：碳酸根离子和氢离子反应会生成二氧化碳气体，氢氧根离子和铵根离子反应会生成氨气，硫酸根离子和钡离子反应会生成硫酸钡沉淀，混合物A中加入盐酸会生成气体B，所以混合物A中一定含有碳酸钠，气体B是二氧化碳，二氧化碳溶于水生成碳酸，碳酸能使紫色石蕊变红色；混合物A和氢氧化钠反应会生成气体C，所以混合物A中含有氯化铵，E是氨气，氨气溶于水形成氨水，氨水能使酚酞变红色；C、D溶液和氯化钡反应会生成白色沉淀，碳酸钠和氯化钡也会生成白色的碳酸钡沉淀，所以混合物A中可能含有硫酸钠，碳酸钠、硫酸钠和硝酸钡不能共存所以混合物A中一定不含硝酸钡，所以<br />（1）现象X是石蕊试液变红色，气体E的化学式为NH<SUB>3</SUB>；<br />（2）通过推导可知，固体混合物A中，肯定存在的物质是Na<SUB>2</SUB>CO<SUB>3</SUB>、NH<SUB>4</SUB>Cl；<br />（3）步骤①中发生的反应是碳酸钠和盐酸反应生成氯化钠、水和二氧化碳，化学方程式为：Na<SUB>2</SUB>CO<SUB>3</SUB>+2HCl=2NaCl+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑；<br />（4）仅根据上述实验现象，还不能确定混合物A的组成．为确定混合物A的成分，可用沉淀G再进行实验．方法是取沉淀G少许，向其中加入足量稀硝酸，若沉淀全部溶解，不含硫酸钡沉淀，则固体混合物A中不含Ba（NO<SUB>3</SUB>）<SUB>2</SUB>、Na<SUB>2</SUB>SO<SUB>4</SUB>．<br />故答案为：（1）石蕊试液变红色，NH<SUB>3</SUB>；<br />（2）Na<SUB>2</SUB>CO<SUB>3</SUB>、NH<SUB>4</SUB>Cl；<br />（3）Na<SUB>2</SUB>CO<SUB>3</SUB>+2HCl=2NaCl+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑；<br />（4）Ba（NO<SUB>3</SUB>）<SUB>2</SUB>、Na<SUB>2</SUB>SO<SUB>4</SUB>．','【分析】根据碳酸根离子和氢离子反应会生成二氧化碳气体，氢氧根离子和铵根离子反应会生成氨气，硫酸根离子和钡离子反应会生成硫酸钡沉淀等知识进行分析．','书写',3.00,'34a6ff8828189105401feab4150f2b49',9,400,'盐的化学性质,铵态氮肥的检验,物质的鉴别、推断,书写化学方程式、文字表达式、电离方程式','',2016,'37','2016•河北区一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840858,'水是生命之源，水与人类生活和生产密切相关．<br />（1）自来水属于<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“纯净物”或“混合物”）．<br />（2）电解水的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）爱护水资源，人人有责，下列属于“国家节水标志的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填字母）．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao16/1e99e30f-94d4-11e9-b204-b42e9921e93e_xkb22.png\" style=\"vertical-align:middle\" />','','','','','','纯净物$###$2H<SUB>2</SUB>O<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;通电&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>↑+O<SUB>2</SUB>↑$###$B','【解答】解：（1）自来水中常含有可溶性的杂质，属于混合物．<br />（2）电解水生成了氢气和氧气，反应的化学方程式为2H<SUB>2</SUB>O<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;通电&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>↑+O<SUB>2</SUB>↑．<br />（3）由图标的含义可知，A．是塑料制品循环使用标志；B．是节水标志；C．是禁止烟火标志；D．是腐蚀品标志．<br />故答为：（1）混合物；（2）2H<SUB>2</SUB>O<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;通电&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>↑+O<SUB>2</SUB>↑；（3）B．','【分析】（1）根据自来水的组成分析类别；<br />（2）根据电解水的反应写出化学方程式；<br />（3）根据常见图标的含义分析回答．','书写',3.00,'0966108165a019af53f243a64bd45757',9,400,'电解水实验,纯净物和混合物的判别,书写化学方程式、文字表达式、电离方程式,几种常见的与化学有关的图标','',2016,'37','2016•宝丰县二模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840861,'如图是初中化学部分重要实验的相关图示，试回答下列问题：<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao31/1ea9e89e-94d4-11e9-b2e9-b42e9921e93e_xkb8.png\" style=\"vertical-align:middle\" /><br />（1）图A是一组对比实验，实验目的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）图B所示蜡烛在氧气中燃烧实验中，可用简单方法验证蜡烛燃烧的产物．请选择一种产物简要写出其验证方法<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）图C是测定空气中氧气含量的实验．通过该实验可得结论是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，该装置中，气球的作用是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（4）图D是模拟工业炼铁的实验装置，玻璃管中发生反应的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>装置后面气球的作用<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','探究氯化钠在水和汽油中的溶解性$###$集气瓶内壁出现水雾说明有水生成或向反应后的集气瓶中倒入澄清石灰水，石灰水变浑浊说明生成二氧化碳$###$氧气约占空气体积的五分之一$###$通过气球的胀缩促进气体的流动，使氧气充分反应$###$3CO+Fe<SUB>2</SUB>O<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2Fe+3CO<SUB>2</SUB>$###$收集多余的一氧化碳，防止污染空气','【解答】解：<br />（1）图A是一组对比实验，实验目的是探究氯化钠在水和汽油中的溶解性；<br />（2）蜡烛燃烧的产物有水、二氧化碳等，验证有水的生成时，看到集气瓶内壁出现水雾，证明有水生成．或向反应后的集气瓶中倒入澄清石灰水，石灰水变浑浊说明生成二氧化碳．<br />（3）图C是测定空气中氧气含量的实验．通过该实验可得结论是氧气约占空气体积的五分之一；该装置中，气球的作用是过气球的胀缩促进气体的流动，使氧气充分反应；<br />（4）氧化铁和一氧化碳反应，生成铁和二氧化碳，玻璃管中发生反应的化学方程式为：3CO+Fe<SUB>2</SUB>O<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2Fe+3CO<SUB>2</SUB>；<br />答案：<br />（1）探究氯化钠在水和汽油中的溶解性；<br />（2）集气瓶内壁出现水雾说明有水生成或向反应后的集气瓶中倒入澄清石灰水，石灰水变浑浊说明生成二氧化碳；<br />（3）氧气约占空气体积的五分之一；通过气球的胀缩促进气体的流动，使氧气充分反应；<br />（4）3CO+Fe<SUB>2</SUB>O<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2Fe+3CO<SUB>2； </SUB>收集多余的一氧化碳，防止污染空气．','【分析】（1）根据图示分析解答；<br />（2）根据蜡烛燃烧的产物有水、二氧化碳等，根据特点进行验证；<br />（3）根据氧气约占空气体积的五分之一&nbsp;解答；<br />（4）根据反应原理写出反应的化学方程式解答．','书写',3.00,'7f3c1e6f796a7c5d15d13af4532e1bd2',9,400,'蜡烛燃烧实验,空气组成的测定,物质的溶解性及影响溶解性的因素,一氧化碳还原氧化铁,书写化学方程式、文字表达式、电离方程式','',2016,'32','2016•烟台校级模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840864,'目前很多地区空气污染指数超标，原因是（　　）','人的呼吸','植物的光合作用、','太阳能','工厂的废气','','D','【解答】解：A、人的呼吸不会产生空气污染物，不会造成空气污染指数超标；<br />B、植物的光合作用不会产生空气污染物，不会造成空气污染指数超标；<br />C、太阳能是清洁能源，不会产生空气污染物，不会造成空气污染指数超标；<br />D、工厂废气中含有大量的空气污染物，会造成空气污染指数超标；<br />故选D．','【分析】知道空气质量日报中污染指数数值越大，污染越严重，空气质量状况越差，所以要从空气的污染途径考虑本题．','选择题',3.00,'6d96f8d34c219dc6003620f053b3d58a',9,400,'空气的污染及其危害','',2015,'37','2015秋•保定校级月考',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840869,'某兴趣小组同学在实验室将一定量的稀硫酸加入到盛有氢氧化钠溶液的烧杯中，未看到明显的现象．部分同学产生疑问：酸和碱到底能否发生反应？<br />（1）该小组同学又做了如下实验：<br /><table class=\"edittable\"><TBODY><TR><td width=159>实验步骤</TD><td width=157>实验现象</TD><td width=184>实验结论及反应方程式</TD></TR><TR><td>向滴有酚酞溶液的稀氢氧化钠溶液的试管中，加入稀硫酸，并振荡</TD><td>观察到溶液由<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />变为无色．</TD><td>氢氧化钠和硫酸发生了反应．反应的化学方程式为<br /><u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．</TD></TR></TBODY></TABLE>（2）同学们对（1）实验后溶液中的溶质组成展开探究，做出如下猜想：<br />小明的猜想是：只有Na<SUB>2</SUB>SO<SUB>4</SUB>；<br />小亮的猜想是：有Na<SUB>2</SUB>SO<SUB>4</SUB>和NaOH；<br />小丽的猜想是：有Na<SUB>2</SUB>SO<SUB>4</SUB>和H<SUB>2</SUB>SO<SUB>4</SUB>；<br />你认为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>的猜想不合理．理由是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />若要验证其余两种猜想中的一种是正确的，请完成下面的实验方案．<br /><table class=\"edittable\"><TBODY><TR><td width=199>实验步骤</TD><td width=124>预期实验现象</TD><td width=170>实验结论</TD></TR><TR><td>取（1）实验后溶液少量于试管中，加入<u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD><td><br /><u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD><td><br /><u>&nbsp;&nbsp;&nbsp;&nbsp;</u>的猜想正确</TD></TR></TBODY></TABLE>','','','','','','红色$###$H<SUB>2</SUB>SO<SUB>4</SUB>+2NaOH═Na<SUB>2</SUB>SO<SUB>4</SUB>+2H<SUB>2</SUB>O$###$小亮$###$如果有NaOH存在，含有酚酞的溶液不能为无色$###$锌粒$###$有气泡产生<br />（或没有气泡产生）$###$小丽<br />（或小明）','【解答】解：（1）氢氧化钠溶液能够使酚酞变红，所以在加入稀硫酸之前溶液呈红色，当加入稀硫酸，由于氢氧化钠和稀硫酸反应生成了硫酸钠和水，而硫酸钠为中性的溶液，所以可以观察到溶液的红色消失，该反应的化学方程式为：H<SUB>2</SUB>SO<SUB>4</SUB>+2NaOH═Na<SUB>2</SUB>SO<SUB>4</SUB>+2H<SUB>2</SUB>O；<br />（2）溶液变为了无色，说明不含有氢氧化钠，故可以判断小亮的猜想是错误的，而要验证小明和小丽的猜想只需验证是否含有硫酸即可，根据硫酸的性质，可以选择活泼的金属来进行验证，例如锌，如果观察到有气泡冒出，则说明溶液中含有硫酸，反之则没有硫酸；<br />故答案为：（1）<br /><table class=\"edittable\"><TBODY><TR><td width=159>实验步骤</TD><td width=157>实验现象</TD><td width=184>实验结论及反应方程式</TD></TR><TR><td>向滴有酚酞溶液的稀氢氧化钠溶液的试管中，加入稀硫酸，并振荡</TD><td>观察到溶液由 红&nbsp;<br />色变为无色．</TD><td>氢氧化钠和硫酸发生了反应．反应的化学方程式为<br />H<SUB>2</SUB>SO<SUB>4</SUB>+2NaOH═Na<SUB>2</SUB>SO<SUB>4</SUB>+2H<SUB>2</SUB>O．</TD></TR></TBODY></TABLE>（2）小亮；如果有NaOH存在，含有酚酞的溶液不能为无色；<br /><table class=\"edittable\"><TBODY><TR><td width=103>实验步骤</TD><td width=138>预期实验现象</TD><td width=82>实验结论</TD></TR><TR><td>锌粒</TD><td>有气泡产生<br />（或没有气泡产生）</TD><td>小丽<br />（或小明）</TD></TR></TBODY></TABLE>（合理即可）．','【分析】（1）氢氧化钠溶液能够使酚酞变红，氢氧化钠和稀硫酸反应生成了硫酸钠和水，可以据此完成该题的解答；<br />（2）溶液变为了无色，说明不含有氢氧化钠，故可以判断小亮的猜想是错误的，而要验证小明和小丽的猜想只需验证是否含有硫酸即可，可以据此解答该题．','书写',3.00,'e3df08a0bb1d6b09472c3350de30029e',9,400,'探究酸碱的主要性质,酸的化学性质,碱的化学性质,中和反应及其应用,书写化学方程式、文字表达式、电离方程式','定州市',2016,'32','2016•定州市校级模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840877,'下列反应属于置换反应的是（　　）','一氧化碳与氧化铜在加热条件下反应','镁条投入稀硫酸','在过氧化氢溶液中加入二氧化锰','在熟石灰中加入稀盐酸','','B','【解答】解：A、一氧化碳与氧化铜在加热条件下反应生成铜和二氧化碳，该反应的反应物均是化合物，不属于置换反应，故选项错误．<br />B、镁条与稀硫酸反应生成硫酸镁和氢气，该反应是一种单质和一种化合物反应生成另一种单质和另一种化合物的反应，属于置换反应，故选项正确．<br />C、过氧化氢在二氧化锰的催化作用下生成水和氧气，该反应符合“一变多”的特征，属于分解反应，故选项错误．<br />D、熟石灰与稀盐酸反应生成氯化钙和水，该反应是两种化合物相互交换成分生成两种新的化合物的反应，属于复分解反应，故选项错误．<br />故选：B．','【分析】置换反应是一种单质和一种化合物反应生成另一种单质和另一种化合物的反应，据此进行分析判断．','选择题',3.00,'20832b3f7a75e81ee84d50e69914e9da',9,400,'置换反应及其应用','',2016,'32','2016•海珠区模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840882,'某研究性学习小组欲利用下列装置进行相关气体制取的探究，请你分析并填空．<br /><img  src=\"/tikuimages/9/2015/400/shoutiniao89/1f03ef80-94d4-11e9-9b98-b42e9921e93e_xkb1.png\" style=\"vertical-align:middle\" /><br />（1）图中标有a和b所指的仪器名称是：a<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，b<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）利用上述B、C装置的组合可以制取的一种气体是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；写出制取该气体的化学方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．为防止生成的气体从长颈漏斗逸出，添加的液体直至<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）实验室用A装置制取氧气的化学方程式：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；用<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>把药品装入试管底部，先给试管<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>再固定加热．若用C收集氧气，实验结束时要先<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','试管$###$集气瓶$###$氧气$###$2H<SUB>2</SUB>O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span>&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O+O<SUB>2</SUB>↑$###$浸没长颈漏斗末端管口$###$2KClO<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span></td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>△</td></tr></table></span>2KCl+3O<SUB>2</SUB>↑$###$药匙（或纸槽）$###$预热$###$将导管从水槽中移出','【解答】解：（1）试管是常用的反应容器，集气瓶是收集气体的仪器；<br />（2）B装置属于固液常温型，C装置是用排水法收集，如果用双氧水和二氧化锰制氧气就不需要加热，且氧气不易溶于水，所以BC组合可用分解过氧化氢的方法来制取氧气，反应的化学方程式为2H<SUB>2</SUB>O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span>&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O+O<SUB>2</SUB>↑；为防止生成的气体从长颈漏斗逸出，添加的液体应浸没长颈漏斗末端管口；<br />（3）A装置为固体加热型，试管口无棉花，说明是用加热氯酸钾和二氧化锰混合物的方法制取氧气，氯酸钾在二氧化锰的催化作用下生成氯化钾和氧气，反应的化学方程式为：2KClO<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span></td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>△</td></tr></table></span>2KCl+3O<SUB>2</SUB>↑，取用粉末状药品可用药匙或纸槽，加热时要先给试管预热，防止试管受热不均炸裂，实验结束应先移出导管，后熄灭酒精灯，防止水倒流，使试管炸裂；<br />故答案为：（1）①试管；&nbsp;&nbsp;&nbsp;②集气瓶；&nbsp;&nbsp;&nbsp;<br />（2）2H<SUB>2</SUB>O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span>&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O+O<SUB>2</SUB>↑；&nbsp;浸没长颈漏斗末端管口；<br />（3）2KClO<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span></td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>△</td></tr></table></span>2KCl+3O<SUB>2</SUB>↑；药匙（或纸槽）；&nbsp;&nbsp;&nbsp;预热；&nbsp;将导管从水槽中移出．','【分析】（1）根据仪器的图形和用途判断名称；<br />（2）根据发生装置和收集装置的特点判断可制取的气体，为防止生成的气体从长颈漏斗下端逸出，长颈漏斗下端应伸入液面以下；<br />（3）A装置为固体加热型，试管口无棉花，说明是用加热氯酸钾和二氧化锰混合物的方法制取氧气，并据制取氧气的注意事项分析解答．','书写',3.00,'ee4fc2bb850f13694ff29d5885eb0d6e',9,400,'常用气体的发生装置和收集装置与选取方法,实验室制取氧气的反应原理,制取氧气的操作步骤和注意点,二氧化碳的实验室制法,书写化学方程式、文字表达式、电离方程式','冷水江市',2015,'32','2015•冷水江市校级模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840894,'利用下面装置进行实验．实验前K<SUB>1</SUB>、K<SUB>2</SUB>、K<SUB>3</SUB>均已关闭．<br /><table class=\"edittable\"><TBODY><TR><td width=164>内容<br />装置</TD><td width=241>【实验1】制备气体</TD><td width=284>【实验2】测定气体含量</TD></TR><TR><td><img  src=\"/tikuimages/9/2015/400/shoutiniao25/1f2d2261-94d4-11e9-842d-b42e9921e93e_xkb66.png\" style=\"vertical-align:middle\" /></TD><td>Ⅰ．打开K<SUB>1</SUB>，用注射器向盛有锌粒的A中注入稀硫酸，直至液面浸没下端导管口<br />Ⅱ．在K<SUB>1</SUB>上方导管口收集气体</TD><td>Ⅰ．A（容积350mL）中为用排空气法收集的CO<SUB>2</SUB>，B中装满水．用注射器向A中注入15mLNaOH溶液（足量），充分反应<br />Ⅱ．打开K<SUB>2</SUB>和K<SUB>3</SUB></TD></TR></TBODY></TABLE>（1）检查装置气密性：保持K<SUB>1</SUB>关闭，打开K<SUB>2</SUB>、K<SUB>3</SUB>，向B中加水至液面浸没下端导管口，用手捂住A瓶外壁，说明装置在左侧气密性良好的现象是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；用同样原理可以检查装置另一侧的气密性．<br />（2）实验1中，锌与稀硫酸反应的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；气体收集完毕后，在不拆卸装置的情况下，使A中未反应的稀硫酸大部分转移到B中的操作是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）实验2中，当B中液面不再变化时，测得B中减少了160mL水，则A中CO<SUB>2</SUB>的体积分数约为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>%．','','','','','','B装置左侧导管口有气泡冒出$###$Zn+H<SUB>2</SUB>SO<SUB>4</SUB>═ZnSO<SUB>4</SUB>+H<SUB>2</SUB>↑$###$打开k<SUB>2</SUB>，k<SUB>3</SUB>，关闭k<SUB>1</SUB>$###$50','【解答】解：（1）检查装置气密性：保持K<SUB>1</SUB>关闭，打开K<SUB>2</SUB>、K<SUB>3</SUB>，向B中加水至液面浸没下端导管口，用手捂住A瓶外壁，A中气体体积膨胀，看到B装置左侧导管口有气泡冒出；用同样原理可以检查装置另一侧的气密性．故填：B装置左侧导管口有气泡冒出；<br />（2）锌与稀硫酸反应生成硫酸锌和氢气，打开k<SUB>2</SUB>，k<SUB>3</SUB>，关闭k<SUB>1</SUB>，导致装置A中气体压强增大，稀硫酸被气体压入装置B中，故填：Zn+H<SUB>2</SUB>SO<SUB>4</SUB>═ZnSO<SUB>4</SUB>+H<SUB>2</SUB>↑．打开k<SUB>2</SUB>，k<SUB>3</SUB>，关闭k<SUB>1</SUB>；<br />（3）二氧化碳被氢氧化钠溶液完全吸收，气体减少的气体为：160mL+15mL=175mL．则A中CO<SUB>2</SUB>的体积分数约为：<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">175mL</td></tr><tr><td>350mL</td></tr></table>×100%</span>=50%；故填：50．<br />故答案为：<br />（1）B装置左侧导管口有气泡冒出．（2）Zn+H<SUB>2</SUB>SO<SUB>4</SUB>═ZnSO<SUB>4</SUB>+H<SUB>2</SUB>↑； 打开k<SUB>2</SUB>，k<SUB>3</SUB>，关闭k<SUB>1</SUB>．（3）50．','【分析】（1）根据检查装置气密性的原理和方法来分析；<br />（2）锌与稀硫酸反应生成硫酸锌和氢气，通过改变装置内的压强的方法来分析；<br />（3）根据液体减少的体积即为被吸收的二氧化碳来分析．','书写',3.00,'9c7cba4620f2923647d9719a5017fb89',9,400,'检查装置的气密性,二氧化碳的化学性质,氢气的制取和检验,书写化学方程式、文字表达式、电离方程式','',2015,'37','2015秋•淮安校级月考',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840895,'下列依据实验目的所设计的实验操作中，正确的是（　　）<br /><table class=\"edittable\"><TBODY><TR><td width=55>选项</TD><td width=283>实验目的</TD><td width=224>实验操作</TD></TR><TR><td>A</TD><td>检验NaCl固体中含有Na<SUB>2</SUB>CO<SUB>3</SUB></TD><td>加水溶解</TD></TR><TR><td>B</TD><td>鉴别NaOH溶液和Na<SUB>2</SUB>CO<SUB>3</SUB></TD><td>加酚酞溶液</TD></TR><TR><td>C</TD><td>除去CaO固体中的CaCO<SUB>3</SUB></TD><td>加入稀盐酸</TD></TR><TR><td>D</TD><td>除去食盐中的泥沙</TD><td>加水溶解过滤</TD></TR></TBODY></TABLE>','A','B','C','D','','D','【解答】解：A、氯化钠、碳酸钠均易溶于水，检验NaCl中含有Na<SUB>2</SUB>CO<SUB>3</SUB>，不能用加水溶解的方法，故选项依据实验目的所设计的实验操作错误．<br />B、NaOH和Na<SUB>2</SUB>CO<SUB>3</SUB>溶液都显碱性，能使酚酞试液变红色；出现相同的现象，不可以鉴别，故选项依据实验目的所设计的实验操作错误．<br />C、CaO和CaCO<SUB>3</SUB>均能与稀盐酸反应，不但能把杂质除去，也会把原物质除去，不符合除杂原则，故选项依据实验目的所设计的实验操作错误．<br />D、泥沙难溶于水，而氯化钠溶于水，所以可用溶解、过滤、蒸发的方法除去NaCl固体中的泥沙，故选项依据实验目的所设计的实验操作正确．<br />故选：D．','【分析】A、根据氯化钠、碳酸钠均易溶于水，进行分析判断；<br />B、根据两种物质与同种试剂反应产生的不同现象来鉴别它们，若两种物质与同种物质反应的现象相同，则无法鉴别它们．<br />C、除杂质题至少要满足两个条件：①加入的试剂只能与杂质反应，不能与原物质反应；②反应后不能引入新的杂质．<br />D、根据泥沙难溶于水，而氯化钠溶于水进行解答．','选择题',3.00,'48b14f5b028ca115c59f77a13898fe58',9,400,'化学实验方案设计与评价,混合物的分离方法,盐的化学性质','',2016,'37','2016•上饶三模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840897,'下列变化过程中，对应关系正确的是（　　）<br /><table class=\"edittable\"><TBODY><TR><td width=170><br /><img  src=\"/tikuimages/9/2016/400/shoutiniao68/1f3586cf-94d4-11e9-86ac-b42e9921e93e_xkb2.png\" style=\"vertical-align:middle\" /></TD><td width=151><img  src=\"/tikuimages/9/2016/400/shoutiniao56/1f397e70-94d4-11e9-9835-b42e9921e93e_xkb38.png\" style=\"vertical-align:middle\" /></TD><td width=151><img  src=\"/tikuimages/9/2016/400/shoutiniao92/1f3b2c21-94d4-11e9-83e1-b42e9921e93e_xkb44.png\" style=\"vertical-align:middle\" /></TD><td width=161><img  src=\"/tikuimages/9/2016/400/shoutiniao59/1f3bef70-94d4-11e9-923a-b42e9921e93e_xkb23.png\" style=\"vertical-align:middle\" /></TD></TR><TR><td>A．稀释10%的NaOH和10% 的H<SUB>2</SUB>SO<SUB>4</SUB></TD><td>B．向一定量的二氧化锰中加入过氧化氢溶液</TD><td>C．在一定量AgNO<SUB>3</SUB>和Cu（NO<SUB>3</SUB>）<SUB>2</SUB>的混合溶液中加入铁粉</TD><td>D．将水通电电解一段时间</TD></TR></TBODY></TABLE>','A','B','C','D','','B|D','【解答】解：A、氢氧化钠溶液稀释，溶液的碱性减弱，pH逐渐减小，错误；<br />B、向一定量的二氧化锰中加入过氧化氢溶液，二氧化锰是该反应的催化剂，其质量不变，正确；<br />C、在一定量AgNO<SUB>3</SUB>和Cu（NO<SUB>3</SUB>）<SUB>2</SUB>的混合溶液中加入铁粉，最开始时溶液中的溶质是硝酸银和硝酸铜两种，不是三种，错误；<br />D、电解水时得到氢气和氧气的体积比是2：1，正确；<br />故选BD．','【分析】A、根据酸碱溶液的稀释pH的变化解答；<br />B、根据二氧化锰是过氧化氢分解的催化剂解答；<br />C、根据金属与盐溶液的反应解答；<br />D、根据电解水的知识解答．','多选题',3.00,'670e1b4a82b0d726e8ff866859450eca',9,400,'催化剂的特点与催化作用,电解水实验,金属的化学性质,酸碱溶液的稀释,溶液的酸碱性与pH值的关系','招远市',2016,'32','2016•招远市模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840899,'下列实验操作，正确的是（　　）','<img  src=\"/tikuimages/9/2016/400/shoutiniao23/1f4ae38f-94d4-11e9-aff3-b42e9921e93e_xkb64.png\" style=\"vertical-align:middle\" /><br />加热液体','<img  src=\"/tikuimages/9/2016/400/shoutiniao32/1f4feca1-94d4-11e9-9b63-b42e9921e93e_xkb29.png\" style=\"vertical-align:middle\" /><br />取用固体','<img  src=\"/tikuimages/9/2016/400/shoutiniao49/1f51c161-94d4-11e9-819a-b42e9921e93e_xkb99.png\" style=\"vertical-align:middle\" /><br />验满氧气','<img  src=\"/tikuimages/9/2016/400/shoutiniao16/1f536f0f-94d4-11e9-8e7f-b42e9921e93e_xkb83.png\" style=\"vertical-align:middle\" /><br />滴加液体','','B','【解答】解：A、给试管中的液体加热时，用酒精灯的外焰加热试管里的液体，且液体体积不能超过试管容积的<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">1</td></tr><tr><td>3</td></tr></table></span>，图中液体超过试管容积的<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">1</td></tr><tr><td>3</td></tr></table></span>，图中所示操作错误．<br />B、取用固体粉末状药品时，应用药匙取用，瓶塞要倒放，图中所示操作错误．<br />C、检验氧气是否收集满时，应将带火星的木条放在集气瓶口，不能伸入瓶中，图中所示操作错误．<br />D、使用胶头滴管滴加少量液体的操作，注意胶头滴管不能伸入到试管内或接触试管内壁，应垂直悬空在试管口上方滴加液体，防止污染胶头滴管，图中所示操作错误．<br />故选：B．','【分析】A、根据给试管中的液体加热的方法进行分析判断．<br />B、根据固体药品的取用方法进行分析判断．<br />C、根据氧气的验满方法进行分析判断．<br />D、根据使用胶头滴管滴加少量液体的方法进行分析判断．','选择题',3.00,'5c26548c8c87eeb82d8bae79300d7fbd',9,400,'固体药品的取用,液体药品的取用,给试管里的液体加热,氧气的检验和验满','',2016,'37','2016•长春二模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840907,'根据表中数据，回答下列问题：<br /><table class=\"edittable\"><TBODY><TR><td width=103 colSpan=2>&nbsp;温度/℃</TD><td width=51>&nbsp;0</TD><td width=51>&nbsp;10</TD><td width=51>&nbsp;20</TD><td width=51>&nbsp;30</TD><td width=51>&nbsp;40</TD><td width=51>&nbsp;50</TD><td width=51>60</TD><td width=51>70</TD><td width=51>&nbsp;80</TD></TR><TR><td rowSpan=2>&nbsp;溶解度/g</TD><td>&nbsp;KNO<SUB>3</SUB></TD><td>&nbsp;13.3</TD><td>&nbsp;20.5</TD><td>&nbsp;31.6</TD><td>45.8</TD><td>&nbsp;63.9</TD><td>&nbsp;85.5</TD><td>&nbsp;110</TD><td>&nbsp;138</TD><td>&nbsp;169</TD></TR><TR><td>&nbsp;NaCl</TD><td>&nbsp;35.7</TD><td>&nbsp;35.8</TD><td>&nbsp;36.0</TD><td>&nbsp;36.3</TD><td>&nbsp;36.6</TD><td>&nbsp;37.0</TD><td>&nbsp;37.3</TD><td>&nbsp;37.8</TD><td>38.4</TD></TR></TBODY></TABLE>（1）若要比较KNO<SUB>3</SUB>和NaCl在水中的溶解能力，需要控制水的质量和<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>均相同．<br />（2）30℃时，各取4.5%KNO<SUB>3</SUB>固体和NaCl固体，分别加入到10g水中，充分溶解后，达到饱和状态的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>溶解，该温度下，<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（选填“能”或“不能”）配制出45.8%的KNO<SUB>3</SUB>的溶液．<br />（3）根据上表提供的数据可绘制出KNO<SUB>3</SUB>和NaCl的溶解度曲线，两条曲线的交点所在的温度范围是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填序号）．<br />A．大于0℃，小于10℃B．大于20℃，小于30℃<br />C．大于40℃，小于50℃D．大于60℃，小于70℃<br />（4）将20℃时KNO<SUB>3</SUB>和NaCl的饱和溶液升温到80℃（不考虑水分的蒸发），溶质量分数<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>溶液．','','','','','','温度$###$NaCl$###$不能$###$B$###$NaCl','【解答】解：（1）影响固体溶解度的外界因素主要是温度，因此要比较KNO<SUB>3</SUB>与&nbsp;NaCl在水中的溶解能力，需要控制的变量是水的质量和温度；<br />故答案为：温度；<br />（2）30℃时NaCl、KNO<SUB>3</SUB>固体的溶解度分别为36.3g、45.8g，则该温度下，10g水中分别溶解3.63g氯化钠或4.58g硝酸钾就能达到饱和状态；30℃时，各取NaCl、KNO<SUB>3</SUB>固体4.5g分别加入10g水中，充分溶解后达到饱和状态的是氯化钠溶液，不能配制出45.8%的KNO<SUB>3</SUB>的溶液；<br />故答案为：NaCl；不能；<br />（3）由表中的硝酸钾和氯化钠不同温度下的溶解度数据可知：若两物质的溶解度相等，此时所对应的温度范围在20℃～40℃．<br />故答案为：B；<br />（4）将20℃的KNO<SUB>3</SUB>饱和溶液升温到80℃，硝酸钾的溶解度随温度的升高而增大，溶质量分数不变是NaCl，<br />故答案为：NaCl．','【分析】（1）根据固体的溶解度的因素来分析；<br />（2）根据30℃时两种物质的溶解度来分析；<br />（3）对比硝酸钾和氯化钾不同温度下的溶解度，分析溶解度相同时对应的温度范围；<br />（4）将20℃的KNO<SUB>3</SUB>饱和溶液升温到80℃，硝酸钾的溶解度随温度的升高而增大．','填空题',3.00,'297dabd64ba0c48e8e3a49246bf8b064',9,400,'饱和溶液和不饱和溶液,固体溶解度的影响因素,物质的溶解性及影响溶解性的因素,溶质的质量分数、溶解性和溶解度的关系','',2016,'37','2016•余干县三模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840909,'<img  src=\"/tikuimages/9/2016/400/shoutiniao10/1f7a30f0-94d4-11e9-b105-b42e9921e93e_xkb39.png\" style=\"vertical-align:middle;FLOAT:right\" />（2016•安徽三模）同学们通过查阅资料知道，过氧化氢分解反应的催化剂除二氧化锰外，还有氯化镁溶液、氯化铜溶液中的金属离子和土豆中含有的过氧化氢酶以及氧化铜等．<br />方案Ⅰ：分别取20mL&nbsp;5%的过氧化氢溶液于3个大小相同的锥形瓶中．向其中两个锥形瓶中分别加入含相同数目金属离子的氯化铁溶液和氯化铜溶液，分别连接传感器测体系压强（如图所示）．测定数据如下表所示：<br /><table class=\"edittable\"><TBODY><TR><td width=160>时间/s<br />试剂<br />压强/KPa</TD><td width=69>0</TD><td width=69>60</TD><td width=69>120</TD><td width=71>180</TD><td width=71>240</TD><td width=71>300</TD></TR><TR><td>20mL5%的过氧化氢溶液</TD><td>101.86</TD><td>102.99</TD><td>103.42</TD><td>103.67</TD><td>103.99</TD><td>104.00</TD></TR><TR><td>加入氯化铁溶液</TD><td>102.21</TD><td>105.35</TD><td>115.40</TD><td>129.69</TD><td>145.52</TD><td>163.99</TD></TR><TR><td>加入氯化铜溶液</TD><td>101.50</TD><td>102.73</TD><td>107.53</TD><td>114.78</TD><td>122.49</TD><td>130.39</TD></TR></TBODY></TABLE>（1）加入含相同数目金属离子的氯化铁溶液和氯化铜溶液，该步操作的实验目的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）根据上表数据得出的结论是（写一个）<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />方案Ⅱ：从同一部位取2个质量均为1.7g的正方体土豆块，将其中一块切成若干片，片与片相连不断开，同时分别放入盛30mL&nbsp;5%过氧化氢溶液的50mL量筒中．均看到有气泡生成，一段时间，只有土豆片不断上升．通过实验得出的结论是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）请写出该反应的化学方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />[实验反思]<br />（4）通过实验一和实验二说明，测量不同的物理量可判断过氧化氢的分解速率．除此之外，本实验还可以用其他方法比较过氧化氢的分解速率，下列说法中不合理的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填选项序号）．<br />A．测量完全反应后气体的总体积&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;B．测量相同时间内产生气体的体积<br />C．测量相同时间内体系压强的变化&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;D．测量相同时间内溶液质量的变化<br />（5）小明发现实验过程中锥形瓶外壁发烫，而过氧化氢的分解速率开始时会逐渐增大．查阅资料后得知：过氧化氢分解时会放出大量的热，你能不能据此解释其分解速率发生变化的原因？<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />（6）你还能想到对过氧化氢溶液的分解速率产生影响的其他因素吗？请选择一种，设计实验加以验证<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','控制金属离子的数目，对比它们对过氧化氢分解速率的影响$###$氯化铁溶液催化产生氧气的速率比氯化铜溶液快$###$反应物和催化剂的接触面接越大，反应速率越快$###$2H<SUB>2</SUB>O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;催化剂&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O+O<SUB>2</SUB>↑$###$A$###$过氧化氢分解时放出热量，温度升高，反应速率加快$###$过氧化氢溶液的浓度影响分解速率：<br />实验步骤：室温下，取相同质量的二氧化锰粉末分别加入两支规格相同的试管中，再分别同时加入相同体积、浓度不同的过氧化氢溶液，连接收集装置；<br />实验现象：浓度大的过氧化氢溶液产生气泡的速率较快，相同时间内浓度大的过氧化氢溶液收集装置中收集的气体体积大；<br />实验结论：其它条件相同时，过氧化氢溶液的浓度越大，分解速率越快','【解答】解：（1）加入含相同数目金属离子的氯化铁溶液和氯化铜溶液，该步操作的实验目的是：通过对比试验，控制金属离子的数目，对比它们对过氧化氢分解速率的影响．<br />故填：控制金属离子的数目，对比它们对过氧化氢分解速率的影响．<br />（2）根据上表数据可知，相同时间内，加入氯化铁粉末的体系压强大于加入氯化铜粉末的体系压强，即相同时间内，加入氯化铁粉末的体系中生成气体的体积大于加入氯化铜粉末的体系中生成气体的体积，因此氯化铁溶液催化产生氧气的速率比氯化铜溶液快；<br />成若干片的体积比不切的体积大，一段时间后，切成片的土豆片表面附着较多的气泡并不断上升，说明切成片的土豆表面产生的气体多，由此可知，反应物和催化剂的接触面接越大，反应速率越快．<br />故填：氯化铁溶液催化产生氧气的速率比氯化铜溶液快；反应物和催化剂的接触面接越大，反应速率越快．<br />（3）过氧化氢在催化剂的作用下分解生成水和氧气的化学方程式为：2H<SUB>2</SUB>O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;催化剂&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O+O<SUB>2</SUB>↑．<br />故填：2H<SUB>2</SUB>O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;催化剂&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O+O<SUB>2</SUB>↑．<br />（4）A．测量完全反应后产生气体的总体积，能够比较过氧化氢质量的大小，但是不能比较过氧化氢分解的速率，该选项做法不合理；<br />B．测量产生一定体积气体所需要的时间，可以判断过氧化氢分解的速率快慢，该选项做法合理；<br />C．测量相同时间内生成气体的体积，可以比较过氧化氢分解的速率大小，该选项做法合理；　　　<br />D．测量相同时间内锥形瓶内物质总质量的变化，可以判断气体减少的快慢，从而可以判断过氧化氢分解的速率快慢，该选项做法合理．<br />故填：A．<br />（5）因为过氧化氢分解时放出热量，导致溶液温度升高，从而加快了过氧化氢的分解．<br />故填：过氧化氢分解时放出热量，温度升高，反应速率加快．<br />（6）过氧化氢溶液的浓度能够影响分解速率：<br />实验步骤：室温下，取相同质量的二氧化锰粉末分别加入两支规格相同的试管中，再分别同时加入相同体积、浓度不同的过氧化氢溶液，连接收集装置；<br />实验现象：浓度大的过氧化氢溶液产生气泡的速率较快，相同时间内浓度大的过氧化氢溶液收集装置中收集的气体体积大；<br />实验结论：其它条件相同时，过氧化氢溶液的浓度越大，分解速率越快．<br />故填：过氧化氢溶液的浓度影响分解速率：<br />实验步骤：室温下，取相同质量的二氧化锰粉末分别加入两支规格相同的试管中，再分别同时加入相同体积、浓度不同的过氧化氢溶液，连接收集装置；<br />实验现象：浓度大的过氧化氢溶液产生气泡的速率较快，相同时间内浓度大的过氧化氢溶液收集装置中收集的气体体积大；<br />实验结论：其它条件相同时，过氧化氢溶液的浓度越大，分解速率越快．','【分析】（1）对比试验中，一定要控制好变量和常量；<br />（2）根据表中数据可以判断氯化铁溶液和氯化铜溶液对过氧化氢分解的催化效率；<br />（3）根据实验现象可以判断实验结论；<br />过氧化氢在催化剂的作用下分解生成水和氧气；<br />（4）比较过氧化氢分解速率的方法很多，要注意理解；<br />（5）反应物浓度越大，温度越高，反应速率越快；<br />（6）过氧化氢溶液的浓度能够影响过氧化氢分解的速率．','书写',3.00,'798fd9705d9a14e7852f658859c11b31',9,400,'影响化学反应速率的因素探究,催化剂的特点与催化作用,书写化学方程式、文字表达式、电离方程式','',2016,'37','2016•安徽三模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840911,'下列有关物质除杂、分离、区分所用的试剂或方法都正确的（　　）<br /><table class=\"edittable\"><TBODY><TR><td width=67>选项</TD><td width=309>实验目的</TD><td width=191>所用试剂或方法</TD></TR><TR><td>A</TD><td>除去NaCl溶液中少量Na<SUB>2</SUB>CO<SUB>3</SUB></TD><td>适量澄清石灰水、过滤</TD></TR><TR><td>B</TD><td>除去生石灰中含有杂质石灰石</TD><td>水或稀盐酸</TD></TR><TR><td>C</TD><td>从高锰酸钾制氧气的残余物中分离出MnO<SUB>2</SUB></TD><td>加水溶解、过滤、洗涤</TD></TR><TR><td>D</TD><td>区分失去标签的稀盐酸和稀硫酸</TD><td>紫色石蕊或铁粉</TD></TR></TBODY></TABLE>','A','B','C','D','','C','【解答】解：A、Na<SUB>2</SUB>CO<SUB>3</SUB>能够与氢氧化钙反应产生碳酸钙沉淀和氢氧化钠，会增加新的杂质，故方法错误；<br />B、不管水，还是盐酸都会与生石灰反应，在除杂的同时，所加的试剂不能与物质反应，故方法错误；<br />C、二氧化锰不溶于水，过滤可以得到二氧化锰，故方法正确；<br />D、稀盐酸和稀硫酸都能使紫色石蕊试液变红色，且都能和铁粉反应生成气体，现象相同，不能鉴别，故方法错误；<br />故选C．','【分析】A、根据Na<SUB>2</SUB>CO<SUB>3</SUB>能够与氢氧化钙反应产生碳酸钙沉淀和氢氧化钠分析；<br />B、根据盐酸和水会与生石灰反应进行分析；<br />C、根据二氧化锰不溶于水进行分析；<br />D、根据指示剂、金属与酸反应的现象进行分析．','选择题',3.00,'3d1b30aa07459985550307a4497707b2',9,400,'化学实验方案设计与评价,混合物的分离方法,盐的化学性质,酸、碱、盐的鉴别','',2016,'32','2016•广东模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840913,'（1）铁生锈的条件是：铁与<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>、<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>接触（铁锈的主要成分：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>）<br />（2）防止铁制品生锈的措施：<br />①<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>②<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />（3）保护金属资源的途径：<br />①<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>②<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>③<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>④<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','水$###$氧气$###$氧化铁$###$刷漆$###$涂油$###$防止金属锈蚀$###$废旧金属的回收利用$###$有计划、合理地开采金属$###$寻找金属的代用品','【解答】解：<br />（1）铁在与水和氧气并存时易生锈，铁锈的主要成分是氧化铁；<br />（2）防锈就是破坏铁生锈的条件，可以在铁制品表面刷漆、涂油等；<br />（3）保护金属资源的途径有：①防止金属锈蚀；②废旧金属的回收利用；③有计划、合理地开采金属；④寻找金属的代用品．<br />故答案为：<br />（1）水；氧气；氧化铁；<br />（2）刷漆，涂油；<br />（3）①防止金属锈蚀；②废旧金属的回收利用；③有计划、合理地开采金属；④寻找金属的代用品．','【分析】（1）根据已有的铁生锈的条件进行分析解答；<br />（2）铁在与水和氧气并存时易生锈，防锈就是破坏铁生锈的条件，据此解答；<br />（3）根据金属资源的保护措施分析回答．','填空题',3.00,'b4ed9f7302cc6b4ee634577e0a1be059',9,400,'金属锈蚀的条件及其防护,铁锈的主要成分,金属资源的保护','',0,'37','',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840915,'下面是几种实验室制取气体的发生装置和收集装置．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao37/1f8c595e-94d4-11e9-b442-b42e9921e93e_xkb17.png\" style=\"vertical-align:middle\" /><br />（1）实验室用高锰酸钾制取氧气时，可选用的发生装置是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填字母），反应的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；实验中为防止高锰酸钾粉末进行导管，应采取的措施是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；如果用E装置收集氧气，那么检验氧气是否收集满的方法是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）现有如下材料：<br />①实验室用FeS固体与H<SUB>2</SUB>SO<SUB>4</SUB>溶液在常温下反应制取H<SUB>2</SUB>S气体．<br />②通常状况下，H<SUB>2</SUB>S是一种密度比空气的密度大的有毒气体，能溶于水，水溶液能使紫色石蕊试液变红，能与碱溶液反应．<br />若在实验室中用上述装置制备并收集一瓶H<SUB>2</SUB>S气体，可选用的发生装置是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填字母），F装置有多种功能，若用F装置收集H<SUB>2</SUB>S气体，气体应从<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（选填“a”或“b”）端通入若为符合“绿色化学”理念，则应在F装置内盛放<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填物质名称）以吸收多余的H<SUB>2</SUB>S气体，H<SUB>2</SUB>S气体应从<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（选填“a”或“b”）端通入F装置中．','','','','','','B$###$2KMnO<SUB>4</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>K<SUB>2</SUB>MnO<SUB>4</SUB>+MnO<SUB>2</SUB>+O<SUB>2</SUB>↑$###$在试管口放一团棉花$###$将带火星的木条放于集气瓶口，若木条复燃则收集满了$###$A$###$a$###$氢氧化钠溶液$###$a','【解答】解：（1）加热高锰酸钾制取氧气属于固体加热型，故选发生装置B，加热高锰酸钾生成锰酸钾、二氧化锰和氧气，反应方程式是2KMnO<SUB>4</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>K<SUB>2</SUB>MnO<SUB>4</SUB>+MnO<SUB>2</SUB>+O<SUB>2</SUB>↑；试管口要放一团棉花，防止加热时高锰酸钾粉末进入导管；验满氧气的方法是将带火星的木条放于集气瓶口，观察木条是否复燃进行判断；<br />（2）用FeS固体与H<SUB>2</SUB>SO<SUB>4</SUB>溶液在常温下反应制取H<SUB>2</SUB>S气体属于固液常温型，故选发生装置A，H<SUB>2</SUB>S是一种密度比空气的密度大的有毒气体，所以应从长管进气，短管便于排尽空气；硫化氢气体溶于水，其水溶液能使紫色石蕊试液变红，能与碱溶液反应，所以可用氢氧化钠吸收多余的硫化氢气体，洗气装置要长进短出，让气体充分与氢氧化钠溶液接触；<br />故答案为：（1）B；2KMnO<SUB>4</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>K<SUB>2</SUB>MnO<SUB>4</SUB>+MnO<SUB>2</SUB>+O<SUB>2</SUB>↑；在试管口放一团棉花；将带火星的木条放于集气瓶口，若木条复燃则收集满了；<br />（2）A；a；氢氧化钠溶液；a．','【分析】（1）加热高锰酸钾制取氧气属于固体加热型，故选发生装置B，并据反应原理书写方程式；试管口要放一团棉花，防止加热时高锰酸钾粉末进入导管；验满氧气的方法是将带火星的木条放于集气瓶口，观察木条是否复燃进行判断；<br />（2）据反应物状态和反应条件选择发生装置，据气体密度确定进气口，硫化氢气体溶于水，其水溶液能使紫色石蕊试液变红，能与碱溶液反应，所以可用氢氧化钠吸收多余的硫化氢气体，洗气装置长进短出．','书写',3.00,'ec92fc71813bdb085bee9fad485b0bb5',9,400,'氧气的制取装置,氧气的收集方法,氧气的检验和验满,制取氧气的操作步骤和注意点,书写化学方程式、文字表达式、电离方程式','',2016,'37','2016•余干县三模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840917,'下列实验中无明显现象的是（　　）','向H<SUB>2</SUB>SO<SUB>4</SUB>溶液中滴加Ba（OH）<SUB>2</SUB>溶液','向NaOH溶液中通入CO<SUB>2</SUB>','向滴有酚酞的NaOH溶液中滴加盐酸','将（NH<SUB>4</SUB>）<SUB>2</SUB>SO<SUB>4</SUB>和熟石灰混合研磨','','B','【解答】解：A、向H<SUB>2</SUB>SO<SUB>4</SUB>溶液中滴加Ba（OH）<SUB>2</SUB>溶液会产生硫酸钡沉淀，错误；<br />B、向NaOH溶液中通入CO<SUB>2</SUB>生成的碳酸钠易溶于水，无明显现象，正确；<br />C、向滴有酚酞的NaOH溶液中滴加盐酸会观察到红色逐渐变浅直至褪去，有明显的现象，错误；<br />D、将（NH<SUB>4</SUB>）<SUB>2</SUB>SO<SUB>4</SUB>和熟石灰混合研磨有刺激性的气味产生，错误；<br />故选B．','【分析】根据物质间的反应以及反应的实验现象进行分析解答即可．','选择题',3.00,'b777ec3e35c51dc14a05acd86f44b2f3',9,400,'碱的化学性质,中和反应及其应用,铵态氮肥的检验','',2016,'37','2016•龙岗区二模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840918,'分类、类推是学习化学的重要方法．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao65/1f9a3c0f-94d4-11e9-8190-b42e9921e93e_xkb86.png\" style=\"vertical-align:middle\" /><br />（1）请根据图标的作用．自拟分类标准，将下列常见的标志平均分为两类：<br />所选取的三个标志是（填序号）<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，你的分类标准为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）小冰翻阅姐姐的高一教材，对教材中这个规律产生了兴趣：“同温同压下，粒子数相同的任何气体都有相同的体积”．也就是说，相同条件下的气体体积比等于分子个数比．联想电解水的实验，明白了为什么产生氢气体积是氧气体积的二倍．请你类推，空气的成分中，氮气和氧气的分子个数比约为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，如果完全燃烧42体积的氢气，至少需要<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>体积的空气．','','','','','','①③⑤$###$与燃烧爆炸有关的标志$###$78：21$###$100','【解答】解：（1）据图可以看出，①③⑤都是与燃烧爆炸有关的标志，②④⑥都是化学危险品标志，故填：①③⑤；与燃烧爆炸有关的标志；<br />（2）空气中氮气和氧气的体积比为78：21，故分子个数比为78：21，根据反应的化学方程式，<br />2H<SUB>2</SUB>+O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O<br />每2个氢分子消耗1个氧分子，故完全燃烧42体积的氢气，需要氧气的体积为21体积，故需要空气的体积为100；<br />故填：78：21；100．','【分析】根据化学图标的意义以及物质反应化学方程式中微粒数量关系进行分析解答即可．','填空题',3.00,'c008a39d5c31faf84b205e1c3cad6c6b',9,400,'空气的成分及各成分的体积分数,分子的定义与分子的特性,几种常见的与化学有关的图标','',2016,'37','2016•道外区二模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840921,'燃气灶能效标准将于2015年4月1日正式实施，这标志着与市民生活息息相关的燃气灶将进入节能时代，如图是一款燃气灶的示意图：<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao93/1fa93030-94d4-11e9-8476-b42e9921e93e_xkb14.png\" style=\"vertical-align:middle\" /><br />（1）所示物质中含有的金属单质是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填一种即可）．<br />（2）燃气灶所用燃气包括煤气、天然气、液化石油气等，天然气属于<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填”可再生”或”不可再生”）能源．写出天然气的主要成分燃烧的化学方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />（3）熄灭燃气灶的方法是关闭阀门，其灭火的原理是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','铜火圈$###$不可再生$###$CH<SUB>4</SUB>+2O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>CO<SUB>2</SUB>+2H<SUB>2</SUB>O$###$撤离可燃物','【解答】解：<br />（1）铜火圈属于金属材料；<br />（2）天然气的主要成分是甲烷（或CH<SUB>4</SUB>），在空气中燃烧可生成水和二氧化碳，化学反应式是：CH<SUB>4</SUB>+2O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>CO<SUB>2</SUB>+2H<SUB>2</SUB>O．<br />（3）关闭天然气阀门熄灭燃气灶，其灭火原理是撤离可燃物；<br />答案：<br />（1）铜火圈；<br />（2）不可再生；&nbsp; CH<SUB>4</SUB>+2O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>CO<SUB>2</SUB>+2H<SUB>2</SUB>O；<br />（3）撤离可燃物．','【分析】（1）根据铜火圈属于金属材料解答；<br />（2）根据天然气的主要成分是甲烷，在空气中燃烧可生成水和二氧化碳解答；<br />（3）根据燃烧的条件选择灭火的措施．','书写',3.00,'13e283278b50f4fcf415db7abd112809',9,400,'单质和化合物的判别,书写化学方程式、文字表达式、电离方程式,常用燃料的使用与其对环境的影响,灭火的原理和方法,常见能源的种类、能源的分类','',2016,'37','2016•肥西县一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840924,'<img  src=\"/tikuimages/9/2016/400/shoutiniao18/1fb53e21-94d4-11e9-84ba-b42e9921e93e_xkb78.png\" style=\"vertical-align:middle;FLOAT:right\" />（2016•武进区模拟）在金属活动性顺序中，活动性较强的金属一般可以将位于其后面的金属从它们的盐溶液中置换出来．张老师在做钠与硫酸铜溶液反应的演示实验时，观察到的现象是：钠块在液面上很快融成一个银白色小球，并四处游动，溶液中出现蓝色絮状沉淀，小球很快停在絮状沉淀集中处燃烧起来，并伴随有轻微的爆炸声，却始终没有发现在钠表面有红色固体析出．对此，同学们展开了如下探究：<br />[提出问题]为何钠不能置换出硫酸铜溶液里的铜呢？为何会出现蓝色絮状沉淀？<br />[猜想假设]查阅资料可知，钠可能先与盐溶液中的水发生反应，然后再与硫酸铜发生反应．<br />[设计实验]<br />①先切一小块钠（绿豆大），用刺了小孔的铝箔包好；<br />②用镊子夹住，放在图中的试管口下（试管已装满水）；<br />③等试管集满气体时，用大拇指堵住试管口并从溶液中取出，移近酒精灯点燃；<br />④最后向烧杯中滴加无色的酚酞试液．<br />[实验现象]反应过程中，有气泡逸出；试管中的气体点燃时能听到轻微的爆鸣声；滴加酚酞溶液后，溶液呈红色．<br />[实验结论]金属钠与水反应，生成了氢气和氢氧化钠．<br />[进一步推论]将实验与老师的演示实验相比较，可推理出钠与硫酸铜反应时确实是由钠与水先反应，然后生成的氢氧化钠与硫酸铜产生了蓝色絮状沉淀．<br />根据以上实验过程，回答下列问题：<br />（1）请根据张老师的演示实验现象，归纳金属钠的物理性质<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（至少答两点）；<br />（2）钠粒用刺了小孔的铝箔包好，并用镊子夹住的目的是：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（3）金属钠与水反应生成的气体应该是氢气，而不可能是甲烷或者-氧化碳，理由是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（4）请写出硫酸铜与氢氧化钠反应的化学方程式：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（5）根据实验结论，有同学认为实验步骤③中有不符合安全要求的操作，你能找出来吗？<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','银白色或熔点低或密度比水小$###$便于收集气体$###$根据质量守恒定律，元素的种类保持不变$###$2NaOH+CuSO<SUB>4</SUB>=Cu（OH）<SUB>2</SUB>↓+Na<SUB>2</SUB>SO<SUB>4</SUB>$###$氢氧化钠溶液具有腐蚀性，不能将手伸入溶液中','【解答】解：（1）由题意可知，“钠块在液面上很快融成一个银白色小球，并四处游动，溶液中出现蓝色絮状沉淀，小球很快停在絮状沉淀集中处燃烧起来”，可归纳金属钠的物理性质：银白色或熔点低或密度比水小；<br />（2）钠粒用刺了小孔的铝箔包好，并用镊子夹住的目的是：便于收集气体；<br />（3）金属钠与水反应生成的气体应该是氢气，而不可能是甲烷或者一氧化碳，理由是：根据质量守恒定律，元素的种类保持不变；<br />（4）硫酸铜与氢氧化钠反应符合复分解反应的特征，“内项结合，外项结合”，故可以是先此反应的化学方程式为：2NaOH+CuSO<SUB>4</SUB>=Cu（OH）<SUB>2</SUB>↓+Na<SUB>2</SUB>SO<SUB>4</SUB>；<br />（5）实验步骤③“等试管集满气体时，用大拇指堵住试管口并从溶液中取出”，此操作不符合安全要求，因为：氢氧化钠溶液具有腐蚀性，不能将手伸入溶液中．<br />故答为：（1）银白色或熔点低或密度比水小；<br />（2）便于收集气体；<br />（3）根据质量守恒定律，元素的种类保持不变；<br />（4）2NaOH+CuSO<SUB>4</SUB>=Cu（OH）<SUB>2</SUB>↓+Na<SUB>2</SUB>SO<SUB>4</SUB>；<br />（5）氢氧化钠溶液具有腐蚀性，不能将手伸入溶液中．','【分析】（1）由题意“银白色小球”，“在液面上”，在水面上燃烧，可推测其性质；<br />（2）了解钠粒用刺了小孔的铝箔包好，并用镊子夹住的目的；<br />（3）根据质量守恒定律，可知金属钠与水反应生成的产物；<br />（4）硫酸铜与氢氧化钠反应符合复分解反应的特征，根据复分解反应的特点是先化学方程式；<br />（5）了解氢氧化钠的性质，可知实验步骤③中有不符合安全要求．','书写',3.00,'c9ed0cd979348a3501ded53cc1db8bcf',9,400,'实验探究物质的性质或变化规律,碱的化学性质,化学性质与物理性质的差别及应用,质量守恒定律及其应用,书写化学方程式、文字表达式、电离方程式','',2016,'32','2016•武进区模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840925,'请按要求从下列物质中①C&nbsp;②NaHCO<SUB>3</SUB>③CH<SUB>4</SUB>④N<SUB>2</SUB>选择合适的物质，将其序号填写在下列横线上：<br />（1）空气中含量最多的物质是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（2）最简单的有机化合物<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（3）能治疗胃酸过多的物质是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（4）可用于高炉炼铁的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','④$###$③$###$②$###$①','【解答】解：（1）空气中含量最多的物质是氮气，其次是氧气；<br />（2）最简单的有机化合物是甲烷；<br />（3）能治疗胃酸过多的物质是碳酸氢钠；<br />（4）碳具有还原性，可用于高炉炼铁．<br />故填：（1）④（2）③（3）②（4）①．','【分析】物质的性质决定物质的用途，根据已有的物质的性质进行分析解答即可．','填空题',3.00,'e265d02f666e4400799ea3e403b0d61d',9,400,'空气的成分及各成分的体积分数,铁的冶炼,常用盐的用途,甲烷、乙醇等常见有机物的性质和用途','',2016,'37','2016•工业园区一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840926,'下列有关金属材料的叙述错误的是（　　）','铝块能制成铝箔是利用了铝的延展性','生铁可以完全溶解在足量的稀盐酸中','合金硬度一般比组成它的纯金属大','回收利用废金属可减少对环境的污染','','B','【解答】解：A、铝块能制成铝箔是利用了铝的延展性，故A正确；<br />B、生铁的成分中含有碳，不能与盐酸反应，故B错误；<br />C、合金硬度一般比组成它的纯金属大，故C正确；<br />D、回收利用废金属可减少对环境的污染，又可以节约资源，故D正确．<br />故选B．','【分析】A、根据铝的性质分析；<br />B、根据生铁的成分分析；<br />C、根据合金的性能分析；<br />D、根据金属资源的保护措施分析．','选择题',3.00,'f1f93fefeb05b5ecdd2b43a1d85cacb6',9,400,'金属的物理性质及用途,合金与合金的性质,金属的化学性质,金属的回收利用及其重要性','太仓市',2016,'32','2016•太仓市模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840931,'元素周期表是学习和研究化学的重要工具．请根据如表（元素周期表的部分内容）回答有关问题：<br /><table class=\"edittable\"><TBODY><TR><td width=59>族<br />周期</TD><td>I A</TD><td>II A</TD><td>IIIA</TD><td>IVA</TD><td>VA</TD><td>VIA</TD><td>VIIA</TD><td>0 </TD></TR><TR><td>2</TD><td>3&nbsp; Li<br />锂<br />6.941</TD><td>4&nbsp; Be<br />铍<br />9.012</TD><td>5&nbsp; B<br />硼<br />10.81</TD><td>6&nbsp; C<br />碳<br />12.01</TD><td>7&nbsp; N<br />氮<br />14.01</TD><td>8&nbsp; O<br />氧<br />16.00</TD><td>9&nbsp; F<br />氟<br />19.00</TD><td>10 Ne <br />氖<br />20.18</TD></TR><TR><td>3</TD><td>11 Na<br />钠<br />22.99</TD><td>12 Mg<br />镁<br />24.31</TD><td>13 Al<br />铝<br />26.98</TD><td>14&nbsp; Si<br />硅<br />28.09</TD><td>15&nbsp; P<br />磷<br />30.97</TD><td>16&nbsp; S<br />硫<br />32.06</TD><td>17 Cl<br />氯<br />35.45</TD><td>18 Ar<br />氩<br />39.95</TD></TR></TBODY></TABLE>（1）请从如表中查出关于锂元素的一条信息：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）在元素周期表中，同一族（纵行）的元素具有相似的化学性质．则下列各组元素具有相似化学性质的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填标号）．<br />a．C和Ne&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;b．Be和Mg&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;c．Al和Si&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;d．F和Cl<br />（3）第9号氟元素和氢元素形成的化合物的水溶液氢氟酸（HF），可用于玻璃雕刻，其主要原理是氢氟酸与玻璃的主要成分二氧化硅（SiO<SUB>2</SUB>）发生反应，生成四氟化硅气体（SiF<SUB>4</SUB>）和水，该反应的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','锂的相对原子质量为6.941$###$bd$###$SiO<SUB>2</SUB>+4HF=SiF<SUB>4</SUB>↑+2H<SUB>2</SUB>O','【解答】解：<br />（1）从元素周期表可以看出，锂原子中质子数是3，元素符号是Li，锂的相对原子质量为6.941，元素属于非金属元素等；<br />（2）根据题意，同一族（纵行）的元素具有相似的化学性质，Be和Mg、F和Cl属于同一族（纵行），化学性质相似．<br />（3）氢氟酸与玻璃的主要成分二氧化硅（SiO<SUB>2</SUB>）发生反应，生成四氟化硅气体（SiF<SUB>4</SUB>）和水，该反应的化学方程式为：SiO<SUB>2</SUB>+4HF=SiF<SUB>4</SUB>↑+2H<SUB>2</SUB>O．<br />答案：<br />（1）锂的相对原子质量为6.941（合理给分）；<br />（2）bd；<br />（3）SiO<SUB>2</SUB>+4HF=SiF<SUB>4</SUB>↑+2H<SUB>2</SUB>O．','【分析】（1）根据原子结构示意图提供的信息进行分析；<br />（2）根据题意，同一族（纵行）的元素具有相似的化学性质，据此进行分析解答；<br />（3）根据反应原理写出反应的化学方程式．','书写',3.00,'73b43e0d6376571df8a91a8eb2c2971b',9,400,'核外电子在化学反应中的作用,元素周期表的特点及其应用,书写化学方程式、文字表达式、电离方程式','恩施',2016,'32','2016•恩施州模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840936,'如图是氨气和氧气在点燃条件下，发生化学反应时部分微观示意图．下列说法正确的是（　　）<img  src=\"/tikuimages/9/2016/400/shoutiniao54/1fd7ba40-94d4-11e9-930a-b42e9921e93e_xkb74.png\" style=\"vertical-align:middle\" />','反应中各元素的化合价均发生变化','该反应属于置换反应','生成物中两种物质的质量比为3：1','该反应中分子、原子都发生了改变','','B','【解答】解：观察微观示意图可知反应物是氨气和氧气，生成物是水合氮气，微粒个数比为：4：3：6：2，故反应的方程式为：4NH<SUB>3</SUB>+3O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>6H<SUB>2</SUB>O+2N<SUB>2</SUB>；<br />A、由反应的化学方程式可知，反应中氢元素的化合价没有发生变化，故不正确；<br />B、该反应是一种单质和一种化合物反应生成另一种单质和另一种化合物，属于置换反应；故正确；<br />C、生成物中两种物质的质量比为：[6×（1×2+16）]：[2×14×2]=27：14；故C不正确；<br />D、化学反应的过程是分子分成原子，原子重新组合得到新的分子，观察微观示意图可知该反应的原子没有发生改变；故D不正确；<br />故选B．','【分析】A、置换反应为一种单质与一种化合物生成一种单质和一种化合物的反应；利用分子模型判断物质类别；<br />B、单质中元素化合价为0，单质中元素变成化合物时，元素的化合价一定会发生变化；<br />C、反应的微观图并未表示出恰好完全反应时的情况，因此，不能根据题图判断反应前后原子数目是否改变；<br />D、因微观图并未表示出恰好完全反应时的情况，要得到A和B两物质在完全反应时的计量数比，需要先把反应进行配平．','选择题',3.00,'95977362453a172c7370c18e34ca1a37',9,400,'微粒观点及模型图的应用,化合价规律和原则,反应类型的判定','丹阳市',2016,'32','2016•丹阳市模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840939,'法拉第曾以《蜡烛的故事》为题向青少年连续开展多次报告．下列列举了报告中涉及的问题及对问题的回答，其中“对问题回答”属于“设计实验方案”的是（　　）<br /><table class=\"edittable\"><TBODY><TR><td width=49>选项</TD><td width=195>问题</TD><td width=265>对问题的回答</TD></TR><TR><td>A</TD><td>吹灭蜡烛产生的白烟是什么？</TD><td>主要成分是烛油蒸气</TD></TR><TR><td>B</TD><td>火焰为什么向上？</TD><td>热气流上升，形成对流</TD></TR><TR><td>C</TD><td>火焰不同部位温度高低如何比较？</TD><td>用一张纸在火焰中心一掠，观察纸上留下的火痕特点</TD></TR><TR><td>D</TD><td>火焰明亮的原因是什么？</TD><td>与火焰中有碳颗粒有关，碳颗粒会影响火焰的明亮程度</TD></TR></TBODY></TABLE>','A','B','C','D','','C','【解答】解：A、主要成分是烛油蒸气，属于解释与结论，故选项错误．<br />B、热气流上升，形成对流，属于解释与结论，故选项错误．<br />C、用一张纸在火焰中心一掠，观察纸上留下的火痕特点，属于设计实验方案，故选项正确．<br />D、与火焰中有碳颗粒有关，碳颗粒会影响火焰的明亮程度，属于实验现象的分析，故选项错误．<br />故选：C．','【分析】科学探究的主要环节有提出问题→猜想与假设→制定计划（或设计方案）→进行实验→收集证据→解释与结论→反思与评价→拓展与迁移，据此结合题意进行分析判断．','选择题',3.00,'01d2c1deb2e5d327793427ceedb3392c',9,400,'科学探究的基本环节','江阴市',2016,'32','2016•江阴市模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840957,'下列图示与对应的叙述相符的是（　　）','<img  src=\"/tikuimages/9/2016/400/shoutiniao51/203da800-94d4-11e9-9c64-b42e9921e93e_xkb13.png\" style=\"vertical-align:middle\" />表示向稀盐酸中不断加水','<img  src=\"/tikuimages/9/2016/400/shoutiniao65/20404011-94d4-11e9-bc82-b42e9921e93e_xkb93.png\" style=\"vertical-align:middle\" />表示探究过氧化氢制氧气的反应中二氧化锰的作用','<img  src=\"/tikuimages/9/2016/400/shoutiniao24/20432640-94d4-11e9-b9c4-b42e9921e93e_xkb94.png\" style=\"vertical-align:middle\" />表示向盐酸和氯化镁的混合溶液中滴加氢氧化钠溶液','<img  src=\"/tikuimages/9/2016/400/shoutiniao79/204485cf-94d4-11e9-8916-b42e9921e93e_xkb17.png\" style=\"vertical-align:middle\" />表示20℃时，向一定量的接近饱和的硝酸钾溶液中加入硝酸钾固体','','B','【解答】解：A、向稀盐酸中不断加水，酸性减弱，pH增大，但是不会大于等于7，错误；<br />B、二氧化锰催过氧化氢的分解具有催化作用，能加速反应的进行，正确；<br />C、向盐酸和氯化镁的混合溶液中滴加氢氧化钠溶液，盐酸和氢氧化钠先反应，开始不会产生沉淀，错误；<br />D、20℃时，向一定量的接近饱和的硝酸钾溶液中加入硝酸钾固体，形成饱和溶液时溶质质量分数不再改变，错误；<br />故选B．','【分析】A、根据盐酸加水稀释pH的变化解答；<br />B、根据二氧化锰的催化作用解答；<br />C、根据氢氧化钠与盐酸和氯化镁的混合物的反应解答；<br />D、根据饱和溶液的形成解答．','选择题',3.00,'436a90672398284968f459086a57dacc',9,400,'催化剂的特点与催化作用,溶质的质量分数,碱的化学性质,酸碱溶液的稀释','',2016,'32','2016•相城区模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840968,'某校化学兴趣小组在老师指导下，进行了“影响过氧化氢分解速率因素”的实验探究．请你帮助回答下列问题．<br />（1）取两支大试管，分别倒入适量的5%、15%的过氧化氢溶液，再加入等量的MnO<SUB>2</SUB>，各收集一小试管气体，发现浓度大的15%的先收集满．<br />（2）用试管进行排水法集气时，主要出现了如图所示的3种操作，其中比较合理的操作是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．（（填序号）<br /><img  src=\"/tikuimages/9/2013/400/shoutiniao69/20766b40-94d4-11e9-9c1e-b42e9921e93e_xkb25.png\" style=\"vertical-align:middle\" /><br /><br />（3）兴趣小组还进行了如下探究，以下是他们探究的主要过程：<br />【假设】过氧化氢的分解速率与催化剂的种类有关吗？<br />【实验方案】取a、b两支试管加入等体积5%的过氧化氢溶液，再分别加入少许MnO<SUB>2</SUB>粉末、CuO粉末，发现a&nbsp;中比b中产生气泡多且快．随即用带火星的木条分别悬空伸入试管内，发现a中火星复燃，b中火星仅发亮但不复燃．<br />【结论】从该探究过程得出的结论是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />【反思】过氧化氢在常温下分解缓慢，加入MnO<SUB>2</SUB>粉末或CuO粉末后反应均明显加快，但若要证明MnO<SUB>2</SUB>或CuO是过氧化氢反应的催化剂，还需要增加一些实验证明它们在化学反应前后的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>是否改变．','','','','','','A$###$催化剂的种类$###$质量和化学性质','【解答】解：（2）用排水法收集氧气，伸入试管内的弯管不宜太长，否则不利于观察氧气的排出速率，且从水槽中移出试管时也不方便．故填：A；<br />（2）取a、b两支试管加入等体积6%的过氧化氢溶液，再分别加入少许MnO<SUB>2</SUB>粉末、CuO粉末，发现a&nbsp;中比b中产生气泡多且快．随即用带火星的木条分别悬空伸入试管内，发现a中火星复燃，b中火星仅发亮但不复燃；在实验条件中只有催化剂不同，对应的反应速率不同，说明影响反应速率的是催化剂种类．<br />故填：催化剂的种类；<br />对于MnO<SUB>2</SUB>或CuO是否是催化剂，在改变反应速率的情况下，就要紧扣催化剂的定义：一个改变（速率）两个不变（质量和化学性质）来分析．这需要通过进一步实验来论证．故填：质量和化学性质．','【分析】根据用排水法收集氧气的注意事项进行分析解答．<br />控制实验条件作对比实验，催化剂的种类和实验结果的对应关系．<br />根据催化剂的特征来分析．','填空题',3.00,'21f1e064cd4f7c497bbe2b4356b0b72c',9,400,'影响化学反应速率的因素探究,氧气的收集方法,催化剂的特点与催化作用','',2013,'32','2013•梅县区模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840971,'根据下列物质的性质，写出其对应的一种或几种用途：<br />（1）金刚石的硬度很大：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（2）石墨的导电性能好：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（3）活性炭的吸附性强：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（4）一氧化碳碳具有还原性：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（5）常温下，碳的化学性质不活泼：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','切割玻璃$###$干电池的电极$###$冰箱去味剂$###$冶炼金属$###$制成碳素墨水','【解答】解：（1）金刚石硬度大，用于切割玻璃，故填：切割玻璃；<br />（2）石墨导电性良好，可以用作干电池的电极，故填：干电池的电极；<br />（3）活性炭吸附性强，可以用于吸附色素和异味，故填：冰箱去味剂；<br />（4）CO具有还原性，可以用于冶炼金属，故填：冶炼金属；<br />（5）常温下碳的化学性质不活泼，受日光照射或与空气、水分接触，都不容易起变化，用墨书写或绘制字画等；故填：制成碳素墨水．','【分析】物质的性质决定物质的用途，根据已有的物质的性质进行分析解答即可．','填空题',3.00,'c29c35e04722311713645753a613ce83',9,400,'一氧化碳的化学性质,碳单质的物理性质及用途,碳的化学性质','',2016,'37','2016春•北京校级月考',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840983,'根据如图所示的实验回答问题：<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao24/20a06170-94d4-11e9-84ca-b42e9921e93e_xkb14.png\" style=\"vertical-align:middle\" /><br />（1）给氯酸钾和二氧化锰的混合物充分加热制得所需的氧气后，为了分离、回收反应后固体中的二氧化锰和氯化钾，应将反应后的混合物依次经过操作（甲图中的）：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>、<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，再将得到的黑色固体用蒸馏水冲洗、晾干可得到二氧化锰，然后将得到的液体进行操作；<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>可得到氯化钾固体．<br />图甲中某图操作有一处错误，指出该图操作的具体错误：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；图甲-C中玻璃棒的作用是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）实验室可用图乙能制取H<SUB>2</SUB>、O<SUB>2</SUB>、CO<SUB>2</SUB>中哪两种气体？<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．怎样通过实验证明此气体是这两种气体中的哪一种气体？<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．如果该气体是光合作用的产物，写出用此图发生装置制取该气体的化学方程式：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）对于图丙：罩上烧杯的作用是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；对照①和③能够得到可燃物燃烧需要的条件之一是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','A$###$C$###$B$###$没有用玻璃棒搅拌$###$引流$###$O<SUB>2</SUB>、CO<SUB>2</SUB>；$###$用带火星的木条放在集气瓶内，如果复燃，则证明是氧气$###$2H<SUB>2</SUB>O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span>&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O+O<SUB>2</SUB>↑$###$防止生成的五氧化二磷污染空气$###$氧气','【解答】解：滤渣是氯化钾、二氧化锰的混合物，氯化钾可溶于水，但二氧化锰不溶于水，所以先把混合物溶解，然后过滤，就可以得到氯化钾溶液，再用蒸发结晶的方法，可得到氯化钾晶体．图甲中没有用玻璃棒搅拌，图甲-C中玻璃棒的作用是引流；<br />（2）实验室制取H<SUB>2</SUB>、O<SUB>2</SUB>、CO<SUB>2</SUB>的反应原理都是固体和液体在常温下反应，属于固液常温型，氢气的密度比空气的小，用向下排空气法收集；氧气、二氧化碳的密度比空气的大，用向上排空气法收集；实验室可用图乙能制取O<SUB>2</SUB>、CO<SUB>2</SUB>；用带火星的木条放在集气瓶内，如果复燃，则证明是氧气，如果该气体是光合作用的产物，说明是氧气，实验室用二氧化锰和过氧化氢制取氧气，属于固液常温型，故选发生装置B，过氧化氢在二氧化锰的催化作用下生成水和氧气，反应的化学方程式为：2H<SUB>2</SUB>O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span>&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O+O<SUB>2</SUB>↑；<br />（3）磷燃烧生成五氧化二磷，易污染空气，对于图丙：罩上烧杯的作用是防止生成的五氧化二磷污染空气；对照①和③能够得到可燃物燃烧需要的条件之一是需要氧气．<br />答案：<br />（1）A；C；B；没有用玻璃棒搅拌；引流；<br />（2）O<SUB>2</SUB>、CO<SUB>2</SUB>；用带火星的木条放在集气瓶内，如果复燃，则证明是氧气；2H<SUB>2</SUB>O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span>&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O+O<SUB>2</SUB>↑；<br />（3）防止生成的五氧化二磷污染空气；氧气．','【分析】（1）实验室分离混合物的基本操作有：过滤、蒸发．做题时需要根据不同混合物的特点具体分析．<br />（2）根据实验室制取H<SUB>2</SUB>、O<SUB>2</SUB>、CO<SUB>2</SUB>的反应原理及气体的收集方法解答；<br />（3）根据燃烧的条件分析解答．','书写',3.00,'79f6436c6329ae34ca39ac4292c6d6e1',9,400,'混合物的分离方法,常用气体的发生装置和收集装置与选取方法,书写化学方程式、文字表达式、电离方程式,燃烧与燃烧的条件','定州市',2016,'32','2016•定州市校级模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840984,'下列化学反应属于置换反应的是（　　）','CH<SUB>4</SUB>+2O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>CO<SUB>2</SUB>+2H<SUB>2</SUB>O','3Fe+2O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>&nbsp;Fe<SUB>3</SUB>O<SUB>4</SUB>','8Al+3Fe<SUB>3</SUB>O<SUB>4</SUB>&nbsp;<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>&nbsp;4Al<SUB>2</SUB>O<SUB>3</SUB>+9Fe','2H<SUB>2</SUB>O<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;通电&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>↑+O<SUB>2</SUB>↑','','C','【解答】解：A、CH<SUB>4</SUB>+2O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>CO<SUB>2</SUB>+2H<SUB>2</SUB>O，该反应的生成物均是化合物，不属于置换反应，故选项错误．<br />B、3Fe+2O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span> Fe<SUB>3</SUB>O<SUB>4</SUB>，该反应符合“多变一”的特征，属于化合反应，故选项错误．<br />C、8Al+3Fe<SUB>3</SUB>O<SUB>4</SUB> <span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span> 4Al<SUB>2</SUB>O<SUB>3</SUB>+9Fe，该反应是一种单质和一种化合物反应生成另一种单质和另一种化合物的反应，属于置换反应，故选项正确．<br />D、2H<SUB>2</SUB>O<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;通电&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>↑+O<SUB>2</SUB>↑，该反应符合“一变多”的特征，属于分解反应，故选项错误．<br />故选：C．','【分析】置换反应是一种单质和一种化合物反应生成另一种单质和另一种化合物的反应，据此进行分析判断．','选择题',3.00,'36fa729f254195c97aaceda34402c503',9,400,'置换反应及其应用','',2016,'37','2016•番禺区一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840997,'下列不属于化学研究范畴的是（　　）','纳米铜的性质','C<SUB>60</SUB>的组成','单晶硅的结构','计算机软件的研发','','D','【解答】解：A、纳米铜的性质，属于研究物质的性质，属于化学研究的范畴，故选项错误．<br />B、C<SUB>60</SUB>的组成，属于研究物质的组成，属于化学研究的范畴，故选项错误．<br />C、单晶硅的结构，属于研究物质的构成，属于化学研究的范畴，故选项错误．<br />D、计算机软件的研发，属于信息技术领域研究的范畴，不不属于化学研究范畴，故选项正确．<br />故选：D．','【分析】根据化学的定义和研究内容进行分析判断，化学是一门在分子、原子的层次上研究物质的性质、组成、结构及其变化规律的科学，研究对象是物质，研究内容有组成、结构、性质、变化、用途等．','选择题',3.00,'a417eeeefe2f98fe155f3e422f6c9ab7',9,400,'化学的研究领域','',2016,'32','2016•云南模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1840999,'金属钛被称为铁和铝之后崛起的“第三金属”，钛的硬度大、熔点高、常温下耐酸碱、耐腐蚀，是航海、化工、医疗上不可缺少的材料．钛铁矿（主要成分FeTiO<SUB>3</SUB>）是工业上冶炼金属钛的主要原料，制备金属钛的一种工艺流程如图所示（部分产物略）．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao99/20dc31ee-94d4-11e9-a8e4-b42e9921e93e_xkb49.png\" style=\"vertical-align:middle\" /><br />（1）X为可燃性氧化物，步骤①反应的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）步骤②分离出TiCl<SUB>4</SUB>的方法，是利用了TiCl<SUB>4</SUB>与FeCl<SUB>3</SUB>的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>不同．<br />（3）步骤④的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，该反应说明Mg与Ti的金属活动性强弱关系为：Ti<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>&nbsp;Mg（填“＜”、“＞”或“=”）．<br />（4）用上述冶炼方法得到的金属钛中会混有少量金属镁，由金属钛的性质可知，除去这种杂质可以用<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填试剂名称）．<br />（5）上述工艺具有成本低、可以用低品位的矿物为原料等优点，但依据绿色低碳理念，你认为该流程中存在的不足之处是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','2FeTiO<SUB>3</SUB>+6C+7Cl<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>6CO+2TiCl<SUB>4</SUB>+2FeCl<SUB>3</SUB>$###$沸点$###$TlCl<SUB>4</SUB>+2Mg<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2MgCl<SUB>2</SUB>+Ti$###$＜$###$稀盐酸$###$耗能太高','【解答】解：（1）钛铁矿、碳和盐酸在高温的条件下反应生成一氧化碳、四氯化钛和氯化铁，化学方程式为：2FeTiO<SUB>3</SUB>+6C+7Cl<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>6CO+2TiCl<SUB>4</SUB>+2FeCl<SUB>3</SUB>；<br />（2）步骤②通过蒸馏分离出TiCl<SUB>4</SUB>的方法，是利用了TiCl<SUB>4</SUB>与FeCl<SUB>3</SUB>的沸点不同；<br />（3）四氯化钛和镁在高温的条件下反应生成氯化镁和钛，化学方程式为：TlCl<SUB>4</SUB>+2Mg<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2MgCl<SUB>2</SUB>+Ti，该反应说明Mg与Ti的金属活动性强弱关系为：Ti＜Mg；<br />（4）镁会与酸反应，钛不会与酸反应，所以除去钛杂质可以用稀盐酸；<br />（5）绿色化学的反应理念是原子利用率100%，没有有害物质生成，所以该流程中存在的不足之处是耗能太高．<br />故答案为：（1）2FeTiO<SUB>3</SUB>+6C+7Cl<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>6CO+2TiCl<SUB>4</SUB>+2FeCl<SUB>3</SUB>；&nbsp;<br />（2）沸点；<br />（3）TlCl<SUB>4</SUB>+2Mg<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2MgCl<SUB>2</SUB>+Ti，＜；<br />（4）稀盐酸；<br />（5）耗能太高．','【分析】（1）根据题中的反应物、生成物以及反应条件下书写化学方程式；<br />（2）根据步骤②通过蒸馏分离出TiCl<SUB>4</SUB>的方法，是利用了TiCl<SUB>4</SUB>与FeCl<SUB>3</SUB>的沸点不同进行分析；<br />（3）根据四氯化钛和镁在高温的条件下反应生成氯化镁和钛进行分析；<br />（4）根据镁会与酸反应，钛不会与酸反应进行分析；<br />（5）根据绿色化学的反应理念进行分析．','书写',3.00,'507bdddad90e2ffce3a6da00c57f5c58',9,400,'金属活动性顺序及其应用,常见金属的冶炼方法,书写化学方程式、文字表达式、电离方程式','',2016,'32','2016•广东模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841004,'下列关于实验现象的描述中正确的是（　　）','红磷在空气中燃烧，产生大量白色烟雾','淀粉溶液遇加碘食盐变蓝色','氯化铵与氢氧化钙混合研磨，产生使湿润红色石蕊试纸变蓝的气体','铁钉伸入盛有硫酸铜的溶液中，铁钉表面出现气泡，溶液由蓝色变为浅绿色','','C','【解答】解：A、红磷在空气中燃烧，产生大量的白烟，而不是白色烟雾，故选项说法错误．<br />B、淀粉遇碘变蓝色，加碘食盐中不含碘，淀粉溶液遇加碘食盐不变色，故选项说法错误．<br />C、氯化铵属于铵态氮肥，与氢氧化钙混合研磨，产生使湿润红色石蕊试纸变蓝的气体，故选项说法正确．<br />D、铁钉伸入盛有硫酸铜的溶液中，铁钉表面出现红色物质，溶液由蓝色变为浅绿色，故选项说法错误．<br />故选：C．','【分析】A、根据红磷在空气中燃烧的现象进行分析判断．<br />B、根据淀粉遇碘变蓝色的特性，进行分析判断．<br />C、根据铵态氮肥与碱性物质混合研磨后能放出有刺激性气味的气体，进行分析判断．<br />D、根据金属的化学性质，进行分析判断．','选择题',3.00,'e294b94e7ec64118c20889cc9b950886',9,400,'氧气与碳、磷、硫、铁等物质的反应现象,金属的化学性质,铵态氮肥的检验,鉴别淀粉、葡萄糖的方法与蛋白质的性质','',2016,'37','2016春•长沙校级月考',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841005,'初中化学中，学习过的物质有：①酒精②Ca（OH）<SUB>2</SUB>&nbsp;&nbsp;&nbsp;&nbsp;③NaCl&nbsp;&nbsp;&nbsp;④HCl&nbsp;&nbsp;⑤木炭&nbsp;&nbsp;⑥不锈钢&nbsp;&nbsp;选择其中物质序号填空：<br />（1）属于合金的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（2）作为新能源开发的燃油替代产品是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（3）两物质的溶液混合，可发生中和反应的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','⑥$###$①$###$②④','【解答】解：（1）属于合金的是不锈钢．故填：⑥．<br />（2）作为新能源开发的燃油替代产品是酒精．故填：①．<br />（3）两物质的溶液混合，可发生中和反应的是氢氧化钙和盐酸．故填：②④．','【分析】不锈钢属于合金；酒精是一种比较清洁的能源；酸和碱反应生成盐和水，属于中和反应．','填空题',3.00,'465ee294fd3d789bb07f2423cde6a6c5',9,400,'合金与合金的性质,中和反应及其应用,资源综合利用和新能源开发','',2016,'32','2016•海南校级模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841013,'下列说法完全正确的是（　　）','氖气的化学式--Ne<SUB>2 </SUB>硅的元素符号--Si&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 2个氮气分子--2N<SUB>2</SUB>','决定元素和原子种类--质子数&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 决定元素的化学性质--最外层电子数&nbsp;&nbsp;&nbsp; 决定元素周期表元素排列顺序--质子数','铁--由分子构成&nbsp;&nbsp;&nbsp;二氧化碳--由原子构成&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 氯化钠--由离子构成','分子--化学变化中的最小粒子&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 原子--保持物质化学性质的最小粒子&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 离子--带电的原子','','B','【解答】解：A、因稀有气体都是单原子构成的，则化学式可用元素符号来表示，氖气的化学式应为Ne，2个氮气分子可表示为2N<SUB>2</SUB>，故A错误；<br />B、因质子数不同，则元素不同，质子数决定元素的种类，根据质子数等于原子序数及原子结构可知元素的性质和在周期表的排列顺序，则说法正确，故B正确；<br />C、因铁是金属，是由铁原子直接构成的，二氧化碳是气体，是由分子直接构成的，故C错误；<br />D、根据化学变化的实质和分子的性质可知，化学变化中的最小微粒是原子，保持物质化学性质的最小微粒是分子，故D错误；<br />故选B','【分析】根据对化学用语的使用及物质的构成和微观粒子的性质来分析解答．','选择题',3.00,'7d18cdf6cc154a2823cf9ee915ab1794',9,400,'分子、原子、离子、元素与物质之间的关系,原子的定义与构成,核外电子在化学反应中的作用,分子的定义与分子的特性,元素的概念,元素周期表的特点及其应用,化学符号及其周围数字的意义','',2015,'37','2015秋•威海校级月考',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841014,'同学们利用如图甲装置进行氧气实验与二氧化碳有关性质的探究．<br />实验一：<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao33/2104535e-94d4-11e9-8148-b42e9921e93e_xkb62.png\" style=\"vertical-align:middle\" /><br />（1）写出装置A中发生反应的化学方程式：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）装置C中的实验现象是：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，说明氧气的化学性质<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）装置D中紫色石蕊试液变红色，说明溶液从中性变为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，间接证明了<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />实验二：<br />同学们利用如图乙装置验证CO<SUB>2</SUB>能与NaOH反应．（装置气密性良好）<br />（1）打开K<SUB>1</SUB>、K<SUB>3</SUB>，关闭K<SUB>2</SUB>，当D中液体变红色时，可确定C中<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）C中收集满CO<SUB>2</SUB>后，关闭K<SUB>1</SUB>，打开K<SUB>2</SUB>，将注射器中5mL浓NaOH溶液推入C中，观察到D中液体流入C中，说明CO<SUB>2</SUB>与NaOH发生了反应．<br />①同学们发现此装置有明显不足，认为应该在<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填装置字母序号）之间增加洗气瓶E，其作用是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />②继续完善实验，还应补充一个对比实验，即将C装置中的浓NaOH溶液改为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','2H<SUB>2</SUB>O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span>&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O+O<SUB>2</SUB>↑$###$木炭剧烈燃、烧发出白光$###$助燃性$###$酸性$###$二氧化碳能水化合生成了碳酸$###$收集满CO<SUB>2</SUB>$###$A、B$###$吸收HCl$###$水','【解答】解：（1）二氧化碳溶于水生成碳酸，碳酸能使紫色石蕊试液变红，所以当观察到装置D中的现象是紫色石蕊试液变红．可确定装置C收集满CO<SUB>2</SUB>；故填：紫色石蕊试液变红；<br />（2）<br />①二氧化碳中常含有氯化氢气体和水蒸气，所以应该先用饱和NaHCO<SUB>3</SUB>溶液除去氯化氢，所以应该在A、B之间加入洗气瓶E；故<br />②二氧化碳和NaOH溶液反应生成碳酸钠和水，C内压强减小，所以实验方案是：将注射器中的5mL水压入到装置C中，重复实验，观察到液体倒吸的量小于推入NaOH实验时的量，说明CO<SUB>2</SUB>与NaOH发生反应．<br />答案：<br />（1）紫色石蕊溶液变红<br />（2）<br />①A、B&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;除去HCl气体<br />②将注射器中的5mL水压入到装置C中，重复实验，观察到液体倒吸的量小于推入NaOH<br />实验时的量，说明CO<SUB>2</SUB>与NaOH发生反应．','【分析】（1）根据二氧化碳溶于水生成碳酸，碳酸能使紫色石蕊试液变红进行解答；<br />（2）①根据二氧化碳中常含有氯化氢气体和水蒸气，所以应该先用洗气瓶E除去氯化氢进行解答；<br />②根据二氧化碳和NaOH溶液反应生成碳酸钠和水，C内压强减小进行解答；','书写',3.00,'935f6e796f1f6fda85b642aa2235c2c8',9,400,'探究氧气的性质,探究二氧化碳的性质,书写化学方程式、文字表达式、电离方程式','',2016,'37','2016•杨浦区二模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841020,'实验室有一瓶无色溶掖，标签已模糊，为探究该溶液中的溶质究竟是什么物质，某兴趣小组的同学进行了一系列实验：<br />（1）小张同学先取该无色溶液少许，滴入紫色石蕊试液，石蕊试液色变红，此时可得出的结论是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）小王同学发现尽管试剂标签已模糊，但依稀可见“SO<SUB>4</SUB>”字样．于是，他设计了甲、乙两个方案，并分别取样品试验、记录相关现象．<br /><img  src=\"/tikuimages/9/0/400/shoutiniao92/21221491-94d4-11e9-967e-b42e9921e93e_xkb81.png\" style=\"vertical-align:middle\" /><br />上述方案中，能确定该无色溶液中含有硫酸根的方案是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />方案乙中，加入足量稀盐酸的作用是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','该溶液呈酸性$###$方案甲、方案乙$###$排除碳酸根离子和银离子的干扰','【解答】解：（1）酸性溶液能使紫色石蕊试液变红，所以取该无色溶液少许，滴入紫色石蕊试液，石蕊试液变红，此时可得出的结论是该溶液呈酸性；故填：该溶液呈酸性；<br />（2）硫酸根离子和硝酸钡反应生成硫酸钡沉淀，且硫酸钡白色沉淀不溶于稀硝酸，所以方案甲可行；样品中加入足量稀盐酸没有明显的现象，说明样品中不含有碳酸根离子和银离子，再加入氯化钡溶液产生白色沉淀，说明生成了硫酸钡白色沉淀，所以方案乙也可行；故填：方案甲、方案乙；排除碳酸根离子和银离子的干扰．','【分析】（1）根据酸性溶液能使紫色石蕊试液变红进行解答；<br />（2）根据硫酸根离子和硝酸钡反应生成硫酸钡沉淀，且硫酸钡白色沉淀不溶于稀硝酸，样品中加入足量稀盐酸没有明显的现象，说明样品中不含有碳酸根离子和银离子，再加入氯化钡溶液产生白色沉淀，说明生成了硫酸钡白色沉淀进行解答．','填空题',3.00,'893b6204c9c3719beaa2ce20632f64be',9,400,'缺失标签的药品成分的探究,酸碱指示剂及其性质,盐的化学性质','',0,'37','',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841022,'<img  src=\"/tikuimages/9/2016/400/shoutiniao53/2128f25e-94d4-11e9-bd6e-b42e9921e93e_xkb69.png\" style=\"vertical-align:middle;FLOAT:right\" />（2016•西城区一模）学习酸碱盐知识后，同学们知道碳酸钠溶液与氢氧化钙溶液能发生反应，可观察到溶液变浑浊．甲组同学进行如图1所示的实验，却未观察到预期现象．<br />【提出问题】未观察到浑浊的原因是什么？<br />【猜想和假设】<br />①与氢氧化钙溶液的浓度有关．若使用更大浓度的氢氧化钙溶液，会迅速产生浑浊．<br />②与碳酸钠溶液的浓度有关．若使用更大浓度的碳酸钠溶液，会迅速产生浑浊．<br />经讨论，同学们认为猜想①不合理，其理由是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />【进行实验】乙组同学针对猜想②进行实验．<br /><table class=\"edittable\"><TBODY><TR><td width=99>实验目的</TD><td width=141>实验操作</TD><td width=246 colSpan=2>实验现象</TD></TR><TR><td rowSpan=5>探究猜想②</TD><td rowSpan=5>取4支试管，向其中分别加入…</TD><td>碳酸钠溶液浓度/%</TD><td>是否浑浊</TD></TR><TR><td>10</TD><td>不浑浊</TD></TR><TR><td>5</TD><td>不浑浊</TD></TR><TR><td>1</TD><td>浑浊</TD></TR><TR><td>0.5</TD><td>浑浊</TD></TR></TBODY></TABLE>【解释与结论】<br />（1）补全上述实验操作：取4支试管，向其中分别加入<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）氢氧化钙溶液与碳酸钠溶液混合出现浑浊的化学方程式是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）乙组实验证明：猜想②不成立，碳酸钠溶液浓度在0.5～10%的范围内，能否出现浑浊与碳酸钠溶液的浓度有关，其关系是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />【实验反思】<br />（1）丙组同学对乙组所做实验中未出现浑浊的原因进行探究．设计了多组实验，其中部分同学的实验过程及结果如图2<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao24/212b8a6e-94d4-11e9-b222-b42e9921e93e_xkb30.png\" style=\"vertical-align:middle\" /><br />丙组同学的实验目的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）依据乙、丙两组的实验探究，丁组同学仍使用甲组的仪器和药品对甲组实验进行了改进，当滴入几滴某溶液后，溶液迅速变浑浊．其实验操作是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','饱和石灰水已是该温度下质量分数最大的溶液$###$2mL浓度为10%、5%、1%、0.5%的碳酸钠溶液，再分别滴加5滴饱和石灰水$###$Ca（OH）<SUB>2</SUB>+Na<SUB>2</SUB>CO<SUB>3</SUB>CaCO<SUB>3</SUB>↓+2NaOH$###$碳酸钠溶液浓度大，不易产生浑浊$###$探究碳酸钙能否溶解在碳酸钠溶液中$###$取2mL饱和石灰水于试管中，滴加碳酸钠溶液','【解答】解：<br />【猜想和假设】与氢氧化钙溶液的浓度有关．若使用更大浓度的氢氧化钙溶液，会迅速产生浑浊，猜想错误，因为饱和石灰水已是该温度下质量分数最大的溶液；<br />【解释与结论】<br />（1）根据图示和表格可知：取4支试管，向其中分别加入2&nbsp;mL浓度为10%、5%、1%、0.5%的碳酸钠溶液，再分别滴加5滴饱和石灰水；<br />（2）氢氧化钙溶液与碳酸钠溶液反应生成碳酸钙沉淀和氢氧化钠，反应的化学方程式是：Ca（OH）<SUB>2</SUB>+Na<SUB>2</SUB>CO<SUB>3</SUB>CaCO<SUB>3</SUB>↓+2NaOH；<br />（3）乙组实验证明：猜想②不成立，碳酸钠溶液浓度在0.5～10%的范围内，能否出现浑浊与碳酸钠溶液的浓度有关，其关系是碳酸钠溶液浓度大，不易产生浑浊；<br />【实验反思】<br />（1）根据图示分析丙组同学的实验目的是探究碳酸钙能否溶解在碳酸钠溶液中<br />（2）依据乙、丙两组的实验探究，丁组同学仍使用甲组的仪器和药品对甲组实验进行了改进，当滴入几滴某溶液后，溶液迅速变浑浊．其实验操作是取2&nbsp;mL饱和石灰水于试管中，滴加碳酸钠溶液．<br />答案：<br />[猜想与假设]饱和石灰水已是该温度下质量分数最大的溶液；<br />[解释与结论]（1）2&nbsp;mL浓度为10%、5%、1%、0.5%的碳酸钠溶液，再分别滴加5滴饱和石灰水；<br />（2）Ca（OH）<SUB>2</SUB>+Na<SUB>2</SUB>CO<SUB>3</SUB>CaCO<SUB>3</SUB>↓+2NaOH；<br />（3）碳酸钠溶液浓度大，不易产生浑浊；<br />[实验反思]（1）探究碳酸钙能否溶解在碳酸钠溶液中；<br />（2）取2&nbsp;mL饱和石灰水于试管中，滴加碳酸钠溶液．','【分析】【猜想和假设】根据饱和石灰水已是该温度下质量分数最大的溶液解答；<br />【解释与结论】<br />（1）根据图示分析解答；<br />（2）根据氢氧化钙溶液与碳酸钠溶液反应生成碳酸钙沉淀和氢氧化钠解答；<br />（3）根据碳酸钠溶液浓度大，不易产生浑浊解答；<br />【实验反思】<br />（1）根据图示分析丙组同学的实验目的是探究碳酸钙能否溶解在碳酸钠溶液中解答；<br />（2）根据氢氧化钙溶液与碳酸钠溶液反应解答．','书写',3.00,'7d108156cfa01c9a4d64a60d3ad946c5',9,400,'实验探究物质变化的条件和影响物质变化的因素,盐的化学性质,书写化学方程式、文字表达式、电离方程式','',2016,'37','2016•西城区一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841033,'化学概念的正确理解，是学好化学的关键，下列说法错误的是（　　）','一种元素与一种元素的根本区别是质子数不同','电解水的实验说明在化学反应中，分子改变了，但原子种类和数目没变','O<SUB>2</SUB>，CO<SUB>2</SUB>，SO<SUB>2</SUB>中都含有O元素','相对原子质量就是原子的质量','','D','【解答】解：A、元素是具有相同质子数的一类原子的总称，所以不同元素的本质区别是质子数不同，正确．<br />B、根据质量守恒定律可知：在化学反应前后，物质的总质量不变，元素的种类不变，元素的质量不变，原子的种类不变，原子的个数不变，原子的质量不变．正确．<br />C、根据化学式分析可知，O<SUB>2</SUB>，CO<SUB>2</SUB>，SO<SUB>2</SUB>中都含有O元素，故正确．<br />D、相对原子质量是原子的绝对质量对C-12的质量的1/12的相对值，故错误．<br />故选D','【分析】A、根据元素的定义进行分析．<br />B、根据化学反应的实质及原子、分子的概念进行解答．<br />C、根据物质的元素组成进行分析．<br />D、根据相对原子质量的定义进行分析．','选择题',3.00,'67bcc94d39811f7806c1fd0fd2dde18f',9,400,'电解水实验,元素的概念,物质的元素组成,相对原子质量的概念及其计算方法','',2016,'35','2016春•凉州区期中',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841039,'地壳中含量最高的元素是（　　）','铁','氧','铝','硅','','B','【解答】解：地壳中各元素及含量由多到少的顺序是氧、硅、铝、铁、钙，所以地壳中含量最高的元素是氧元素．<br />故选B．','【分析】根据地壳中各元素及含量由多到少的顺序是氧、硅、铝、铁、钙进行分析．','选择题',3.00,'48a396e1fab93c334c452b8564e115a4',9,400,'地壳中元素的分布与含量','',2016,'37','2016•太原二模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841041,'在10℃时，把36克NaNO<SUB>3</SUB>放在45克水中，完全溶解后溶液恰好饱和，则该温度下NaNO<SUB>3</SUB>的溶解度是（　　）','80克','20克','36克','72克','','A','【解答】解：在10℃时，把36克NaNO<SUB>3</SUB>放在45克水中，完全溶解后溶液恰好饱和，则该温度下100g水中最多能溶解硝酸钠的质量为36g×<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">100g</td></tr><tr><td>45g</td></tr></table></span>=80g，则该温度下NaNO<SUB>3</SUB>的溶解度是80g．<br />故选：A．','【分析】溶解度是在一定温度下，某固体溶质在100g溶剂里达到饱和状态所溶解的溶质质量；溶解度定义中的四要素：一定温度、100g溶剂、溶液达到饱和状态、溶解的质量及单位克．','选择题',3.00,'2efa03afc1037b083ddbe5753cc63256',9,400,'固体溶解度的概念','',2013,'37','2013•古浪县校级三模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841048,'下列说法正确的是（　　）','使用金属来制作炊具，是利用了其化学性质','自然界的水都或多或少含有杂质，所以最健康的饮用水就是蒸馏水','空气中含有的元素质量分数最大的是氧','地表上能够以游离态存在的金属元素只能是很不活泼的少数几种','','D','【解答】解：A、用金属制作炊具是利用了金属铝的延展性和传热性，属于物理性质的应用，故A说法错误；<br />B、蒸馏水不含人体必需的微量元素（或矿物质），故B说法错误；<br />C、空气主要是由氮气和氧气组成的，其中氮气约占空气体积的五分之四，故空气中体积分数最大的气体为氮气，故C说法错误；<br />D、地表上能够以游离态存在的金属元素只能是很不活泼的少数几种，多数金属性质活泼，容易被氧化，在自然界中以化合态存在，故D说法正确；<br />故选D．','【分析】A、根据金属的性质进行分析；<br />B、根据蒸馏水不含人体必需的微量元素（或矿物质），进行分析判断；<br />C、根据空气的组成进行分析判断；<br />D、根据金属在自然界中的存在形式进行分析．','选择题',3.00,'1e5f4ee931dd6f0af1f94378a895bdcb',9,400,'水的性质和应用,金属的物理性质及用途,金属元素的存在及常见的金属矿物,物质的元素组成','',2016,'32','2016•沈阳模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841065,'小亮在实验室发现一瓶无色溶液，其标签破损，从残余部分只能看出溶质质量分数为10%，具体是什么物质无法辨认．老师告诉他，这瓶溶液的溶质可能是氢氧化钠、氯&nbsp;化钠、氢氧化钙或碳酸钠中的一种．<br />（1）小亮用洁净干燥的玻璃棒蘸取该溶液滴到温润的pH试纸上，测得pH=10，他判断该溶液的溶质不可能是氯化钠．对他的操作方法、测量数据及结论的评价，正确的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />A．方法正确，数据正确，且结论正确&nbsp;&nbsp;&nbsp; B．方法不正确，数据偏小，但结论正确<br />C．方法不正确，数据偏大，结论不正确&nbsp;&nbsp; D．方法正确，数据正确，但结论不正确<br />（2）小亮查阅资料发现：氢氧化钙常温下的溶解度为0.18g后，他认为该溶液不可能是氢氧化钙，理由是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）为了确定该溶液的成分，小亮同学继续进行下列实验．<br />[设计实验方案]方案甲：选择氯化钙溶液来确定该溶液的成份；<br />方案乙：选择稀盐酸来确定该溶液的成份．<br />[进行实验]在两个方案中选一个方案，填写步骤、现象、结论．<br /><table class=\"edittable\"><TBODY><TR><td width=155>实验步骤</TD><td width=144>实验现象</TD><td width=249>实验结论</TD></TR><TR><td>取少量溶液于试管中，滴加<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．</TD><td><u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD><td>该溶液是碳酸钠溶液．<br />有关反应的化学方程式为：<br /><u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．</TD></TR></TBODY></TABLE>[实验反思]①有同学认为还可以用一种常见的碱溶液来确定是碳酸钠，那么它是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>溶液．有关反应的化学方程式为：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />②小亮经过反思，向同学们提出如下建议，你认为不合理的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填序号）．<br />A．倾倒液体时应注意保护标签不被腐蚀<br />B．要有严谨的科学实验的态度<br />C．无法辨认的药品，直接倒掉，无需保留．','','','','','','B$###$常温下不可能得到质量分数为10%的氢氧化钙溶液$###$氯化钙溶液（或稀盐酸）$###$产生白色沉淀（或有气泡冒出）$###$CaCl<SUB>2</SUB>+Na<SUB>2</SUB>CO<SUB>3</SUB>═CaCO<SUB>3</SUB>↓+2NaCl（或Na<SUB>2</SUB>CO<SUB>3</SUB>+2HCl═2NaCl+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑）$###$Ca（OH）<SUB>2</SUB>$###$Na<SUB>2</SUB>CO<SUB>3</SUB>+Ca（OH）<SUB>2</SUB>═CaCO<SUB>3</SUB>↓+2NaOH$###$C','【解答】解：（1）测定溶液的酸碱度时，不可把pH试纸用水湿润，这样测定的碱溶液的酸碱度会因加水稀释而偏小，但根据测得的酸碱度为10判断溶液呈碱性还是正确的；故选B；<br />（2）常温下氢氧化钙的溶解度为0.18g，其饱和溶液中溶质质量分数=<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">0.18g</td></tr><tr><td>0.18g+100g</td></tr></table></span>×100%＜0.18%，而该溶液溶质质量分数为10%，所以不可能是氢氧化钙；<br />（3）氯化钙和稀盐酸都可以用来鉴别氢氧化钠和碳酸钠，氯化钙与碳酸钠反应产生碳酸钙白色沉淀，方程式为CaCl<SUB>2</SUB>+Na<SUB>2</SUB>CO<SUB>3</SUB>═CaCO<SUB>3</SUB>↓+2NaCl，稀盐酸与碳酸钠反应产生二氧化碳气体，方程式为Na<SUB>2</SUB>CO<SUB>3</SUB>+2HCl═2NaCl+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑；<br />[实验反思]①根据碳酸钠溶液与澄清石灰水混合可使溶液因生成碳酸钙沉淀而变浑浊的现象，利用氢氧化钙溶液检验碳酸钠，化学方程式为Na<SUB>2</SUB>CO<SUB>3</SUB>+Ca（OH）<SUB>2</SUB>═CaCO<SUB>3</SUB>↓+2NaOH；<br />②对于标签被腐蚀的试剂，本着节约原则，应尽可能确定药品的组成，不可随意扔弃，既不安全又造成浪费；故C不可取；<br />故答案为：<br />（1）常温下不可能得到质量分数为10%的氢氧化钙溶液；<br />（2）B；<br />（3）[进行实验]实验步骤：氯化钙溶液（或稀盐酸）<br />实验现象：产生白色沉淀（或有气泡冒出）<br />CaCl<SUB>2</SUB>+Na<SUB>2</SUB>CO<SUB>3</SUB>═CaCO<SUB>3</SUB>↓+2NaCl（或Na<SUB>2</SUB>CO<SUB>3</SUB>+2HCl═2NaCl+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑）<br />[实验反思]①Ca（OH）<SUB>2</SUB>；Na<SUB>2</SUB>CO<SUB>3</SUB>+Ca（OH）<SUB>2</SUB>═CaCO<SUB>3</SUB>↓+2NaOH；②C．','【分析】（1）分析测定溶液酸碱度的操作，指出其中的问题并判断对结果的影响；<br />（2）根据溶解度与饱和溶液的溶质质量分数的关系，由氢氧化钙常温下的溶解度进行判断；<br />（3）[进行实验]根据碳酸钠与稀硫酸或氯化钡溶液反应所出现的现象，完成检验碳酸钠的实验方案的填写；<br />[实验反思]①根据碳酸钠的性质，判断能与碳酸钠反应且有明显现象出现的碱，写出反应的化学方程式；<br />②根据节约药品的原则，评价对于无法辨认的药品的处理方法．','书写',3.00,'257badd6c6f915bd7ff3b8794021ec1c',9,400,'缺失标签的药品成分的探究,溶液的酸碱度测定,溶质的质量分数、溶解性和溶解度的关系,溶液的酸碱性与pH值的关系,盐的化学性质,书写化学方程式、文字表达式、电离方程式','',2016,'37','2016•邻水县二模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841074,'如表为氯化铵和氯化钠在不同温度时的溶解度．下列说法正确的是（　　）<br /><table class=\"edittable\"><TBODY><TR><td width=102>温度/℃</TD><td width=52>0</TD><td width=53>10</TD><td width=51>20</TD><td width=50>30</TD><td width=52>40</TD><td width=53>50</TD></TR><TR><td>NH<SUB>4</SUB>Cl溶解度/g</TD><td>29.4</TD><td>33.3</TD><td>37.2</TD><td>41.4</TD><td>45.8</TD><td>50.4</TD></TR><TR><td>NaCl溶解度/g</TD><td>35.7</TD><td>35.8</TD><td>36.0</TD><td>36.3</TD><td>36.6</TD><td>37.0</TD></TR></TBODY></TABLE>','20℃时，氯化钠饱和溶液的溶质质量分数为36%','由表中数据可知，溶解度受温度变化影响较大的物质是氯化钠','氯化钠与氯化铵溶解度相等的温度在10℃-20℃之间','50℃时，100g水中加入50.0g氯化铵配成溶液，再冷却到20℃，能够析出13.2g固体','','C','【解答】解：A、20℃时氯化钠的溶解度是36g，其涵义是20℃时，100g水中最多溶解36g氯化钠，形成饱和溶液136g，溶液达到饱和状态，此温度下，该饱和溶液中溶质的质量分数为：<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">36g</td></tr><tr><td>136g</td></tr></table></span>×100%≈26.5%．故错误；<br />B、根据表中信息可知：溶解度随温度变化较大的物质是NH<SUB>4</SUB>Cl，故错误；<br />C、硝酸铵和氯化钠溶解度相同的温度既是判断该段中溶解度的数值是否存在相等的可能，经分析知在10～20℃时这两种物质存在溶解度相同的可能．<br />D、50℃时氯化铵的溶解度为50.4g，向烧杯中加入100g水和50.0g氯化铵配成50℃的溶液，氯化铵全部溶解，20℃时氯化铵的溶解度为37.2g，即100g水中最多溶解37.2g，故向烧杯中加入100g水和50.0g氯化铵配成50℃的溶液，再冷却到20℃，烧杯中析出固体为：50g-37.2g=12.8g．故错误．<br />故选：C．','【分析】A、根据饱和溶液溶质的质量分数公式分析；<br />B、根据表中信息分析解答．<br />C、硝酸铵和氯化钠溶解度相同的温度即是判断该段中溶解度的数值是否存在相等的可能；<br />D、根据表中氯化铵的溶解度解答．','选择题',3.00,'a569c7283bb69bbb50fae38a73453f70',9,400,'固体溶解度的影响因素,溶质的质量分数、溶解性和溶解度的关系','',2016,'37','2016•东城区一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841075,'王岚同学在做“盐酸中和氢氧化钠溶液”实验时，滴加盐酸前忘了加入酚酞，导致无法判断中和反应进行的程度，于是她利用实验台上的实验用品对溶液酸碱性进行探究：<br />实验用品：盐酸、氢氧化钠溶液、酚酞、碳酸钠溶液、锌粒、氧化铜、石蕊、试管、药匙、胶头滴管<br />探究实验一：<table class=\"edittable\"><TBODY><TR><td width=189>&nbsp;实验操作</TD><td width=189>&nbsp;实验现象</TD><td width=189>&nbsp;实验结论</TD></TR><TR><td>&nbsp;用洁净试管取该溶液1-2mL，滴入1-2滴无色酚酞试液，振荡．</TD><td>&nbsp;酚酞试液变<u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD><td>&nbsp;溶液呈碱性</TD></TR></TBODY></TABLE>若无色酚酞不变色，王岚对溶液的酸碱性继续探究：<br />探究实验二：<table class=\"edittable\"><TBODY><TR><td width=189>&nbsp;实验操作</TD><td width=189>&nbsp;实验现象</TD><td width=189>&nbsp;实验结论</TD></TR><TR><td><u>&nbsp;&nbsp;&nbsp;&nbsp;</u>&nbsp;</TD><td><u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD><td rowSpan=4>&nbsp;溶液呈酸性</TD></TR><TR><td><u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD><td><u>&nbsp;&nbsp;&nbsp;&nbsp;</u>&nbsp;</TD></TR><TR><td><u>&nbsp;&nbsp;&nbsp;&nbsp;</u>&nbsp;</TD><td><u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD></TR><TR><td><u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD><td><u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD></TR></TBODY></TABLE>','','','','','','红色$###$取样品，滴入碳酸钠溶液$###$有气泡产生$###$取样品，加入锌粒$###$有气泡产生$###$取样品，滴入紫色石蕊$###$溶液变红色$###$取样品，加入氧化铜$###$固体溶解，溶液变蓝色','【解答】解：探究实验一<br /><table class=\"edittable\"><TBODY><TR><td width=189>&nbsp;实验操作</TD><td width=189>&nbsp;实验现象</TD><td width=189>&nbsp;实验结论</TD></TR><TR><td>&nbsp;用洁净试管取该溶液1-2mL，滴入1-2滴无色酚酞试液，振荡．</TD><td>&nbsp;酚酞试液变红色</TD><td>&nbsp;溶液呈碱性</TD></TR></TBODY></TABLE>探究实验二<br /><table class=\"edittable\"><TBODY><TR><td width=189>&nbsp;实验操作</TD><td width=189>&nbsp;实验现象</TD><td width=189>&nbsp;实验结论</TD></TR><TR><td>取样品，滴入碳酸钠溶液&nbsp;</TD><td>有气泡产生</TD><td rowSpan=4>&nbsp;溶液呈酸性</TD></TR><TR><td>取样品，加入锌粒</TD><td>有气泡产生&nbsp;</TD></TR><TR><td>取样品，滴入紫色石蕊&nbsp;</TD><td>溶液变红色</TD></TR><TR><td>取样品，加入氧化铜</TD><td>固体溶解，溶液变蓝色</TD></TR></TBODY></TABLE>故答案为：<br />探究实验一<br /><table class=\"edittable\"><TBODY><TR><td width=189>&nbsp;实验操作</TD><td width=189>&nbsp;实验现象</TD><td width=189>&nbsp;实验结论</TD></TR><TR><td>&nbsp;</TD><td>&nbsp;红色</TD><td>&nbsp;</TD></TR></TBODY></TABLE>探究实验二<br /><table class=\"edittable\"><TBODY><TR><td width=189>&nbsp;实验操作</TD><td width=189>&nbsp;实验现象</TD><td width=189>&nbsp;实验结论</TD></TR><TR><td>取样品，滴入碳酸钠溶液&nbsp;</TD><td>有气泡产生</TD><td rowSpan=4></TD></TR><TR><td>取样品，加入锌粒</TD><td>有气泡产生&nbsp;</TD></TR><TR><td>取样品，滴入紫色石蕊&nbsp;</TD><td>溶液变红色</TD></TR><TR><td>取样品，加入氧化铜</TD><td>固体溶解，溶液变蓝色</TD></TR></TBODY></TABLE>','【分析】探究实验一：根据无色的酚酞溶液遇中性、酸溶液都不变色，遇碱溶液变成红色进行分析；<br />探究实验二：根据碳酸盐遇酸生成二氧化碳气体，活泼金属能与酸反应生成氢气，石蕊遇酸变红色，金属氧化物和溶于酸进行分析．','填空题',3.00,'ba746a1338d5a2daa7d8b3d717c13560',9,400,'中和反应及其应用,溶液的酸碱性测定','',2016,'37','2016•嘉祥县一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841083,'下列有关原子的说法中错误的是（　　）','原子之间也存在间隔','原子总是不断运动着的','原子可以构成分子','原子很小，不可以再分','','D','【解答】解：A、原子之间也存在间隔，正确；<br />B、原子总是不断运动着的，正确；<br />C、原子可以构成分子，正确；<br />D、原子很小，可以再分，错误；<br />故选D．','【分析】在化学变化中分子可分，原子不可分．原子是构成物质的一种微粒，原子不是保持物质性质的一种微粒．','选择题',3.00,'e8c5d19c96d2dafa6638434ed150e024',9,400,'原子的定义与构成,分子和原子的区别和联系','',2011,'37','2011秋•蓝山县校级月考',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841090,'下列写法中，既能表示一种元素，又能表示一种物质化学式的是（　　）','Na','H','Cl<SUB>2</SUB>','2N','','A','【解答】解：元素符号能表示一种元素，还能表示该元素的一个原子；化学式能表示一种物质，当元素符号又是化学式时，就同时具备了上述三层意义．<br />A、Na属于金属元素，可表示钠元素，表示一个钠原子，还能表示钠这一纯净物，故选项符合题意．<br />B、H属于气态非金属元素，可表示氢元素，表示一个氢原子，但不能表示一种物质，故选项不符合题意．<br />C、该符号是氯气的化学式，不是元素符号，故选项不符合题意．<br />D、该符号是可表示2个氮原子，不是元素符号，故选项不符合题意．<br />故选：A．','【分析】根据化学式与元素符号的含义进行分析解答，金属、大多数固体非金属等都是由原子直接构成的，故它们的元素符号，既能表示一个原子，又能表示一种元素，还能表示一种物质．','选择题',3.00,'e2eeb532f8cce3ecc289abd3f1cf8628',9,400,'元素的符号及其意义,化学式的书写及意义','',2016,'37','2016•河西区一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841091,'“节能减排，从我做起”．下列同学的实验操作不符合节约原则的是（　　）','碧碧做完实验后，将剩余的金属和溶液回收','爽爽用点滴板代替试管，进行酸与指示剂反应','敏敏用高锰酸钾制氧气时，先点燃酒精灯，再去组装仪器','灵灵将配好的氯化钠溶液装入试剂瓶盖好瓶塞，贴标签备用','','C','【解答】解：A、碧碧做完实验后，将剩余的金属和溶液回收，符合节约原则，故选项错误．<br />B、点滴板所需药品用量少，爽爽用点滴板代替试管，进行酸与指示剂反应，符合节约原则，故选项错误．<br />C、敏敏用高锰酸钾制氧气时，先点燃酒精灯，再去组装仪器，会浪费酒精，不符合节约原则，故选项正确．<br />D、在实验室练习配制一定质量分数的溶液时，所得溶液不能倒掉，应该贴上标签备用，符合节约原则，故选项错误．<br />故选：C．','【分析】A、根据回收多余的药品符合节约原则，进行分析判断．<br />B、根据点滴板所需药品用量少，进行分析判断．<br />C、根据敏敏用高锰酸钾制氧气时，先点燃酒精灯，会浪费酒精，进行分析判断．<br />D、根据实验室练习配制一定质量分数的溶液时，所得溶液不能倒掉，应该贴上标签备用，进行分析判断．','选择题',3.00,'a0c35815118f325869b8917d2eb51a81',9,400,'实验操作注意事项的探究,制取氧气的操作步骤和注意点','',2016,'37','2016•重庆校级一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841092,'<img  src=\"/tikuimages/9/2016/400/shoutiniao70/21d6224f-94d4-11e9-ad2e-b42e9921e93e_xkb67.png\" style=\"vertical-align:middle;FLOAT:right\" />（2016•海口模拟）某同学利用如图所示装置探究燃烧条件：<br />将液体a滴入瓶中并与固体b接触，能观察到白磷燃烧，则瓶内发生反应的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；该实验事实说明燃烧所需要的条件之一是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','2H<SUB>2</SUB>O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span>&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O+O<SUB>2</SUB>↑$###$与氧气接触','【解答】解：将液体a滴入瓶中并与固体b接触，能观察到白磷燃烧，说明液体a与固体b接触产生了氧气，过氧化氢在二氧化锰的催化作用下生成水和氧气，反应的化学方程式为：2H<SUB>2</SUB>O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span>&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O+O<SUB>2</SUB>↑．<br />白磷是可燃物，温度达到了白磷的着火点，煤产生气体前不能燃烧；将液体a滴入瓶中并与固体b接触，能观察到白磷燃烧，该实验事实说明燃烧所需要的条件之一是与氧气接触．<br />故答案为：2H<SUB>2</SUB>O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span>&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O+O<SUB>2</SUB>↑；与氧气接触．','【分析】根据将液体a滴入瓶中并与固体b接触，能观察到白磷燃烧，说明液体a与固体b接触产生了氧气，进行分析解答．','书写',3.00,'a134d5353f8f898d053195cd1cca6ce9',9,400,'燃烧的条件与灭火原理探究,书写化学方程式、文字表达式、电离方程式','',2016,'32','2016•海口模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841104,'水是一种最常见的液体．<br />（1）下列物质放入水中，能形成溶液的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填字母序号）．<br />A．面粉&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;B．碳酸钙&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;C．食盐&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;D．植物油<br />（2）实验中常用到10%的稀盐酸，将100g&nbsp;36%的浓盐酸配制成10%的稀盐酸，需要加水<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>&nbsp;g．','','','','','','C$###$260','【解答】解：（1）A、面粉不溶于水，与水混合形成的是悬浊液，故A错；<br />B、碳酸钙微溶于水，与水混合形成的是悬浊液，故B错；<br />C、食盐易溶于水，形成均一、稳定的混合物，属于溶液，故C正确；<br />D、植物油不溶于水，与水混合形成的是乳浊液，故D错．<br />故选C．<br />（2）加水前后溶液中溶质的质量不变，<br />设加水的质量为x<br />100g×36%=10%×（100g+x）<br />x=260g<br />故答案为：（1）C（2）260','【分析】（1）本题考查溶液的概念，在一定条件下溶质分散到溶剂中形成的是均一稳定的混合物．<br />（2）根据溶质质量分数公式进行计算，稀释前后溶液中溶质的质量不变分析．','填空题',3.00,'8a3fa3acd0ee692960619e04739a3635',9,400,'溶液的概念、组成及其特点,用水稀释改变浓度的方法','',2016,'37','2016•石景山区一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841106,'氧化汞受热分解生成汞和氧气的反应前后，发生改变的是（　　）','分子种类','原子种类','元素种类','原子数目','','A','【解答】解：氧化汞受热分解生成汞和氧气的反应前后，是因为氧化汞分子分裂成了汞原子和氧原子，然后氧原子重新组合形成氧分子，大量的氧分子聚集成氧气，汞原子直接聚集成金属汞．<br />A、氧化汞受热分解生成汞和氧气的反应前后，分子种类发生了改变，故选项正确．<br />B、氧化汞受热分解生成汞和氧气的反应前后，原子的种类不变，故选项错误．<br />C、氧化汞受热分解生成汞和氧气的反应前后，元素的种类不变，故选项错误．<br />D、氧化汞受热分解生成汞和氧气的反应前后，原子的数目不变，故选项错误．<br />故选：A．','【分析】化学变化的实质是分子分成原子，原子再重新组合成新分子，在反应前后，原子的种类和数目都不变，据此进行分析判断．','选择题',3.00,'632724cee4a021984b8ff76ca9f27921',9,400,'化学反应的实质','高安市',2016,'37','2016•高安市二模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841109,'甲、乙两个学习小组为了测定M物质的溶解度，分别设计了如下方案：<br />甲组方案：<br /><img  src=\"/tikuimages/9/0/400/shoutiniao68/22017811-94d4-11e9-a43d-b42e9921e93e_xkb56.png\" style=\"vertical-align:middle\" /><br />由甲组实验方案，同学们可知：<br />（1）烧杯②中溶液的质量为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>g．<br />（2）20℃时M物质的溶解度为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>g．<br />乙组方案：<br /><img  src=\"/tikuimages/9/0/400/shoutiniao41/22041021-94d4-11e9-926b-b42e9921e93e_xkb23.png\" style=\"vertical-align:middle\" /><br />由乙组方案，同学们可知：<br />（3）烧杯①中的溶液为M的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“饱和”或“不饱和”）溶液．<br />（4）乙组方案的三个烧杯中的溶液，其溶质的质量分数相同的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填序号）','','','','','','120$###$30$###$不饱和$###$②③','【解答】解：（1）烧杯②中溶液的质量为溶质质量+溶剂质量=20g+100g=120g；<br />（2）根据在20℃时M物质的溶解度为该温度下，100g溶剂中最多溶解溶质的质量为：40g-10g=30g，所以溶解度是30g；<br />（3）根据第二次蒸发10g水析出晶体3g，所以第一次蒸发水，才析出晶体1g，所以烧杯①中的溶液为M的不饱和溶液；<br />（4）饱和溶液溶质质量分数计算，由于②③溶液都是饱和溶液，溶质质量分数相同．<br />故答案为：（1）120g；（2）30；（3）不饱和；（4）②③．','【分析】（1）烧杯②中溶液的质量为溶质质量+溶剂质量；（2）根据在20℃时M物质的溶解度为该温度下，100g溶剂中最多溶解溶质的质量；（3）根据蒸发溶剂析出溶质质量考虑；（4）根据饱和溶液溶质质量分数计算方法考虑．','填空题',3.00,'d7ef7bb96cb4929a091e75b6d9bf8bc6',9,400,'饱和溶液和不饱和溶液,固体溶解度的概念,溶质的质量分数','',0,'37','',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841134,'下列图示的实验操作正确的是（　　）','<img  src=\"/tikuimages/9/2016/400/shoutiniao54/2256c400-94d4-11e9-a84e-b42e9921e93e_xkb64.png\" style=\"vertical-align:middle\" /><br />取固体氯化钠','<img  src=\"/tikuimages/9/2016/400/shoutiniao32/22595c0f-94d4-11e9-ad0a-b42e9921e93e_xkb1.png\" style=\"vertical-align:middle\" /><br />闻气体气味','<img  src=\"/tikuimages/9/2016/400/shoutiniao87/225b7ef0-94d4-11e9-a9a8-b42e9921e93e_xkb79.png\" style=\"vertical-align:middle\" /><br />移开蒸发皿','<img  src=\"/tikuimages/9/2016/400/shoutiniao63/225e3e11-94d4-11e9-a76a-b42e9921e93e_xkb48.png\" style=\"vertical-align:middle\" /><br />稀释浓硫酸','','B','【解答】解：A、固体药品的取用，瓶塞应倒放在桌面上故A错误；<br />B、正确闻气体气味的方法应为“用手轻轻扇动，让少量气体飘入鼻孔”，而不能把鼻孔凑到容器瓶口直接闻，这样易吸入过多气体，可能会使人中毒，故B正确；<br />C、取拿蒸发皿要用坩埚钳取拿，不能用手拿，故C错误；<br />D、稀释浓硫酸时，要把浓硫酸缓缓地沿器壁注入水中，同时用玻璃棒不断搅拌，以使热量及时地扩散；一定不能把水注入浓硫酸中，防止水浮在浓硫酸的水面上沸腾溅出，故D错误；<br />故答案选：B．','【分析】A、根据固体药品的取用，瓶塞倒放在桌面上进行分析；<br />B、根据闻气体气味的正确操作方法进行分析；<br />C、根据取拿蒸发皿要用坩埚钳取拿进行分析；<br />D、根据稀释浓硫酸时的注意事项进行分析．','选择题',3.00,'726e17b17bc3147166de243daf88af96',9,400,'固体药品的取用,浓硫酸的性质及浓硫酸的稀释,蒸发与蒸馏操作','',2016,'32','2016•安阳模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841142,'<img  src=\"/tikuimages/9/2016/400/shoutiniao92/227a2a80-94d4-11e9-838d-b42e9921e93e_xkb69.png\" style=\"vertical-align:middle;FLOAT:right;\" />（2016春•福清市校级月考）实验探究研究性学习小组选择“H<SUB>2</SUB>O<SUB>2</SUB>生成氧气的快慢与什么因素有关”的课题进行【假设】H<SUB>2</SUB>O<SUB>2</SUB>生成O<SUB>2</SUB>的快慢与催化剂的种类有关．<br />【实验方案】常温下，在两瓶相同体积的H<SUB>2</SUB>O<SUB>2</SUB>溶液中分别加入相同质量MnO<SUB>2</SUB>和红砖粉，测量各生成一瓶（相同体积）O<SUB>2</SUB>所需要的时间．<br />【进行实验】如图是他们进行实验的装置图写出反应的化学方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，实验中B处宜采用的气体收集方法是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />【实验记录】<br /><table class=\"edittable\"><TBODY><TR><td width=224>实验编号</TD><td width=224>1</TD><td width=224>2</TD></TR><TR><td>反应物</TD><td>6%　H<SUB>2</SUB>O<SUB>2</SUB></TD><td>6%　H<SUB>2</SUB>O<SUB>2</SUB></TD></TR><TR><td>催化剂</TD><td>1g红砖粉</TD><td>1g　MnO<SUB>2</SUB></TD></TR><TR><td>时间</TD><td>152秒</TD><td>35秒</TD></TR></TBODY></TABLE>【结论】该探究过程中得出的结论是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />【反思】H<SUB>2</SUB>O<SUB>2</SUB>生成O<SUB>2</SUB>　的快慢还与哪些因素有关？请你帮助他们继续探究．<br />【假设】<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />【实验方案】<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','2H<SUB>2</SUB>O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span>&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O+O<SUB>2</SUB>↑$###$排水法$###$相同条件下H<SUB>2</SUB>O<SUB>2</SUB>生成O<SUB>2</SUB>的快慢与催化剂的种类有关$###$过氧化氢生成氧气的快慢与过氧化氢溶液的浓度有关$###$同温度下取两份质量相等、浓度不同的过氧化氢溶液，分别加入质量相等的同种催化剂，测量收集一试管气体所需的时间','【解答】解：【进行实验】过氧化氢溶液和二氧化锰混合生成水和氧气，反应的化学方程式2H<SUB>2</SUB>O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span>&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O+O<SUB>2</SUB>↑；因为需要观察收集一瓶气体所用的时间，用排水法更好一些；故填：2H<SUB>2</SUB>O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span>&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O+O<SUB>2</SUB>↑；排水法；<br />【结论】<br />根据控制变量法，分析表中的有关数据，可知分别用1g红砖粉和1gMnO<SUB>2</SUB>做催化剂，来催化反应时所用的时间分别是152s和35s；由此可知，用1gMnO<SUB>2</SUB>做催化剂要比用1g红砖粉做催化剂时，化学反应速率要快．因此，由实验现象可以得出的实验结论是：H<SUB>2</SUB>O<SUB>2</SUB>生成氧气的快慢与催化剂种类有关．故填：相同条件下H<SUB>2</SUB>O<SUB>2</SUB>生成O<SUB>2</SUB>的快慢与催化剂的种类有关；<br />【假设】本题中只是设计了催化剂对反应的影响，过氧化氢溶液的浓度也可能影响反应的速率；故填：过氧化氢生成氧气的快慢与过氧化氢溶液的浓度有关；<br />【实验方案】在温度、相同质量的相同催化剂的条件下，采用等质量但不同的浓度的过氧化氢溶液来进行实验，然后测量收集相同体积的气体所需的时间，从而得出结论．故填：同温度下取两份质量相等、浓度不同的过氧化氢溶液，分别加入质量相等的同种催化剂，测量收集一试管气体所需的时间．','【分析】【进行实验】根据过氧化氢溶液和二氧化锰混合生成水和氧气以及氧气的密度和溶解性选择收集方法进行解答；<br />【结 论】根据表中提供的数据可以进行相关方面的判断；<br />【假 设】根据能够影响化学反应速率的因素来作出假设；<br />【实验方案】根据作出的假设采用控制变量法来进行设计实验进行验证．','书写',3.00,'744b9521cd697925ed3239e098da9dcf',9,400,'影响化学反应速率的因素探究,催化剂的特点与催化作用,书写化学方程式、文字表达式、电离方程式','福清市',2016,'37','2016春•福清市校级月考',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841146,'<img  src=\"/tikuimages/9/2016/400/shoutiniao66/228dd98f-94d4-11e9-90f5-b42e9921e93e_xkb25.png\" style=\"vertical-align:middle;FLOAT:right\" />（2016•太原二模）为探究化学反应后的质量关系，某实验小组用如图装置进行了实验，有关反应的化学反应方程式是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．若将该实验中的Na<SUB>2</SUB>CO<SUB>3</SUB>溶液换成其他试剂也能得出满意的结论，你认为该试剂可以是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />实验结束后，烧杯中剩余溶液的溶质是什么？同学们进行了深入探究．<br />【作出猜想】反应后溶液的溶质是Na<SUB>2</SUB>CO<SUB>3</SUB>和NaOH．<br />【设计方案】李欢认为上述猜想中的一种物质不需要验证，理由是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />张华设计的实验方案是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，即可证明上述猜想的正确．<br /><br />【进行实验】同学们按张华的方案进行了实验，证明了上述猜想的正确．<br />王乐收到启发选择了张华所选试剂不同类别的物质进行检验，也得出了正确结论．王乐所选试剂是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />【反思扩展】同学们讨论后认为，检验无色溶液中溶质的成分，实验设计思路是：向无色溶液中加入的试剂应该<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','Na<SUB>2</SUB>CO<SUB>3</SUB>+Ca（OH）<SUB>2</SUB>═CaCO<SUB>3</SUB>↓+2NaOH$###$硫酸铜$###$氢氧化钠是生成物$###$取少量反应后的溶液，向其中倾倒足量的稀盐酸，观察到有大量气泡产生$###$CaCl<SUB>2</SUB>溶液$###$能与被检验物质反应产生明显现象','【解答】解：<br />氢氧化钙能与碳酸钠反应生成碳酸钙沉淀和氢氧化钠，反应的化学方程式为：Na<SUB>2</SUB>CO<SUB>3</SUB>+Ca（OH）<SUB>2</SUB>═CaCO<SUB>3</SUB>↓+2NaOH；硫酸铜和氢氧化钙反应生成氢氧化铜沉淀和硫酸钙；<br />【设计方案】<br />氢氧化钙能与碳酸钠反应生成碳酸钙沉淀和氢氧化钠，反应后溶液的溶质一定含有氢氧化钠，理由是氢氧化钠是生成物；<br />取少量反应后的溶液，向其中倾倒足量的稀盐酸，观察到有大量气泡产生，于是得出该溶液中溶质是NaOH、Na<SUB>2</SUB>CO<SUB>3</SUB>的结论，证实了自己的猜想．<br />【进行实验】取反应后的溶液少许于试管中，向试管中加入足量的CaCl<SUB>2</SUB>（呈中性）溶液，振荡．若产生白色沉淀，则证明含有碳酸钠；<br />【反思扩展】检验无色溶液中溶质的成分，实验设计思路是：向无色溶液中加入的试剂应该能与被检验物质反应产生明显现象．<br />答案：<br />Na<SUB>2</SUB>CO<SUB>3</SUB>+Ca（OH）<SUB>2</SUB>═CaCO<SUB>3</SUB>↓+2NaOH；硫酸铜；<br />【设计方案】氢氧化钠是生成物；取少量反应后的溶液，向其中倾倒足量的稀盐酸，观察到有大量气泡产生；<br />【进行实验】CaCl<SUB>2</SUB>（呈中性）溶液；<br />【反思扩展】能与被检验物质反应产生明显现象．','【分析】根据氢氧化钙能与碳酸钠反应生成碳酸钙沉淀和氢氧化钠来分析；根据硫酸铜和氢氧化钙反应生成氢氧化铜沉淀和硫酸钙来分析；<br />【设计方案】根据氢氧化钠是生成物解答；根据碳酸钠和盐酸反应生成的二氧化碳解答；<br />【进行实验】氯化钙能与碳酸钠反应生成碳酸钙沉淀和氯化钠来分析；<br />【反思扩展】根据检验无色溶液中溶质的成分，实验设计思路是：向无色溶液中加入的试剂应该能与被检验物质反应产生明显现象解答．','书写',3.00,'66f990c18f57b8dca0e4500abf0e716a',9,400,'质量守恒定律的实验探究,书写化学方程式、文字表达式、电离方程式','',2016,'37','2016•太原二模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841151,'为探究一瓶敞放的氢氧化钠固体的变质情况，同学们提出了下列猜想：<br />A只有氢氧化钠固体<br />B只有碳酸钠固体<br />C你认为还可能是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>固体<br />①取少量该固体样品置于试管中，向其中加入一种无色溶液，发现有气泡产生，说明该样品中含有碳酸钠，由此可确定该固体已发生变质．该固体变质的化学方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；则无色溶液可能是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />②为探究该固体中是否还有未变质的氢氧化钠，同学们又进行了如下表所示的实验．已知碳酸钠的水溶液呈碱性，它的存在会对氢氧化钠的检验造成干扰．请根据下表部分物质的溶解性表（20℃）所提供的信息，将如表填写完整．<br /><table class=\"edittable\"><TBODY><TR><td width=131>阴离子<br />阳离子</TD><td width=78>OH<SUP>-</SUP></TD><td width=79>NO<SUB>3</SUB><SUP>-</SUP></TD><td width=80>Cl<SUP>-</SUP></TD><td width=78>SO<SUB>4</SUB><SUP>2-</SUP></TD><td width=77>CO<SUB>3</SUB><SUP>2-</SUP></TD></TR><TR><td>H<SUP>+</SUP></TD><td></TD><td>溶、挥</TD><td>溶、挥</TD><td>溶</TD><td>溶、挥</TD></TR><TR><td>Na<SUP>+</SUP></TD><td>溶</TD><td>溶</TD><td>溶</TD><td>溶</TD><td>溶</TD></TR><TR><td>Ba<SUP>2+</SUP></TD><td>溶</TD><td>溶</TD><td>溶</TD><td>不溶</TD><td>不溶</TD></TR></TBODY></TABLE><table class=\"edittable\"><TBODY><TR><td width=83>&nbsp;实验目的</TD><td width=157>实验操作</TD><td width=78>现象</TD><td width=241>结论或化学方程式</TD></TR><TR><td>除去碳酸钠</TD><td>取少量该固体样品溶于水配成溶液，滴加适量的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>溶液．充分反应后过滤</TD><td>有白色沉淀生成</TD><td>有关反应的化学方程式为<br /><u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD></TR><TR><td>检验是否含有氢氧化钠</TD><td>在滤液中滴加酚酞溶液</TD><td><u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD><td>该样品中含有氢氧化钠</TD></TR></TBODY></TABLE>','','','','','','NaOH和Na<SUB>2</SUB>CO<SUB>3</SUB>的混合物$###$2NaOH+CO<SUB>2</SUB>═Na<SUB>2</SUB>CO<SUB>3</SUB>+H<SUB>2</SUB>O$###$稀盐酸或稀硫酸、稀硝酸$###$BaCl<SUB>2</SUB>$###$Na<SUB>2</SUB>CO<SUB>3</SUB>+BaCl<SUB>2</SUB>═BaCO<SUB>3</SUB>↓+2NaCl$###$变红','【解答】解：氢氧化钠若是部分变质，则是氢氧化钠和碳酸钠的混合物，故填：NaOH和Na<SUB>2</SUB>CO<SUB>3</SUB>的混合物；<br />（1）氢氧化钠与二氧化碳反应生成碳酸钠而变质，碳酸钠能与酸反应产生气体，加入的无色液体就可以是稀盐酸、稀硫酸或是稀硝酸，故填：2NaOH+CO<SUB>2</SUB>═Na<SUB>2</SUB>CO<SUB>3</SUB>+H<SUB>2</SUB>O；稀盐酸或稀硫酸、稀硝酸；&nbsp;&nbsp;<br />（2）要检验氢氧化钠是否存在，可以加入氯化钡除去碳酸钠，然后加入酚酞试液，溶液变红，故填：BaCl<SUB>2</SUB>，Na<SUB>2</SUB>CO<SUB>3</SUB>+BaCl<SUB>2</SUB>═BaCO<SUB>3</SUB>↓+2NaCl；变红．','【分析】根据氢氧化钠变质生成碳酸钠进行分析解答，检验变质就是检验碳酸钠的存在，检验变质的程度就是检验是否含有氢氧化钠，据此解答．','书写',3.00,'0f71b57c77fe2f91711cda15bbea2c8d',9,400,'药品是否变质的探究,碱的化学性质,盐的化学性质,书写化学方程式、文字表达式、电离方程式','',2016,'32','2016•夹江县模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841154,'<img  src=\"/tikuimages/9/2016/400/shoutiniao73/22a4e400-94d4-11e9-9e44-b42e9921e93e_xkb94.png\" style=\"vertical-align:middle;FLOAT:right;\" />（2016•寿光市校级模拟）人类的生产生活离不开金属材料，随着科技水平的不断提高，金属及其合金在日常生活中扮演着越来越重要的角色．<br />（1）人们习惯上把金、银、铜、铁、锡五种金属统称为“五金”，在“五金”顺序中把金属<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>的位置移到最后，正好符合由弱到强的顺序．<br />（2）汽车车体多用钢材制造．表面喷漆不仅美观，而且可有效防止与<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>接触而生锈．<br />（3）工业上用CO还原赤铁矿冶炼金属铁的化学方程为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．西汉时期中国古老的冶铜方法是“湿法冶铜”主要的反应原理是铁与硫酸铜溶液反应，写出反应的化学方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（4）在实验室中探究铝、铜的金属活动性顺序，除铝、铜外，还需要用到的试剂是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（5）铁锅、铝锅是生活常用的炊具，如图是某锅的示意图．<br />①该锅含有的金属单质<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（写化学式）．<br />②炒菜时铁锅中的油着火可用锅盖盖灭，其原理是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />③铜也易生锈，铜锈的主要成分是碱式碳酸铜（Cu<SUB>2</SUB>（OH）<SUB>2</SUB>CO<SUB>3</SUB>），是铜与空气中的氧气、水和<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>共同作用的结果．','','','','','','铁$###$水和氧气$###$Fe<SUB>2</SUB>O<SUB>3</SUB>+3CO<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2Fe+3CO<SUB>2</SUB>$###$Fe+CuSO<SUB>4</SUB>=FeSO<SUB>4</SUB>+Cu$###$稀盐酸（或稀硫酸或硫酸铜溶液）$###$Fe$###$使油与氧气隔绝$###$CO<SUB>2</SUB>','【解答】解：（1）人们习惯上把金、银、铜、铁、锡五种金属统称为“五金”，由金属活动顺序表可知，在“五金”顺序中把金属铁的位置移到最后，正好符合由弱到强的顺序．<br />（2）汽车车体多用钢材制造．表面喷漆不仅美观，而且可有效防止与水和氧气接触而生锈．<br />（3）工业上用CO还原赤铁矿冶炼金属铁的化学方程为：Fe<SUB>2</SUB>O<SUB>3</SUB>+3CO<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2Fe+3CO<SUB>2</SUB>．西汉时期中国古老的冶铜方法是“湿法冶铜”主要的反应原理是铁与硫酸铜溶液反应，反应的化学方程式：Fe+CuSO<SUB>4</SUB>=FeSO<SUB>4</SUB>+Cu．<br />（4）在实验室中探究铝、铜的金属活动性顺序，除铝、铜外，还需要用到的试剂是：稀盐酸（或稀硫酸或硫酸铜溶液）．<br />（5）①该锅含有的金属单质是铁，化学式是：Fe．<br />②炒菜时铁锅中的油着火可用锅盖盖灭，其原理是使油与氧气隔绝．<br />③铜也易生锈，铜锈的主要成分是碱式碳酸铜（Cu<SUB>2</SUB>（OH）<SUB>2</SUB>CO<SUB>3</SUB>），是铜与空气中的氧气、水和CO<SUB>2</SUB> 共同作用的结果．<br />故答为：（1）铁；（2）水和氧气；（3）Fe<SUB>2</SUB>O<SUB>3</SUB>+3CO<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2Fe+3CO<SUB>2</SUB>；&nbsp;Fe+CuSO<SUB>4</SUB>=FeSO<SUB>4</SUB>+Cu；（4）稀盐酸或稀硫酸或硫酸铜溶液；（5）①Fe；&nbsp;&nbsp;②使油与氧气隔绝；&nbsp;&nbsp;③CO<SUB>2</SUB>．','【分析】（1）根据金属活动顺序表分析回答；<br />（2）根据铁生锈的条件分析；<br />（3）根据炼铁的原理和“湿法冶铜”主要的反应原理写出反应的化学方程式；<br />（4）根据金属与酸、盐的反应比较金属的活动性；<br />（5）①根据铁锅中成分主要是铁分析回答；<br />②根据灭火的原理分析回答；<br />③根据质量守恒定律和空气的成分分析回答．','书写',3.00,'b030f80d3b571d58f11d78e1d5da7583',9,400,'金属的化学性质,金属活动性顺序及其应用,铁的冶炼,金属锈蚀的条件及其防护,书写化学方程式、文字表达式、电离方程式,灭火的原理和方法','寿光市',2016,'32','2016•寿光市校级模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841155,'二氧化氮是大气的污染物之一．煤燃烧会产生一部分二氧化氮气体，它会与空气中的氧气、水蒸气发生反应形成酸雨，NO<SUB>2</SUB>+O<SUB>2</SUB>+H<SUB>2</SUB>O═HNO<SUB>3</SUB>，配平后各物质的化学计量数分别是（　　）','1，1，1，2','2，1，1，2','4，1，2，4','6，2，3，6','','C','【解答】解：本题可利用“定一法”进行配平，把HNO<SUB>3</SUB>的化学计量数定为1，则NO<SUB>2</SUB>、O<SUB>2</SUB>、H<SUB>2</SUB>O前面的化学计量数分别为：1、<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">1</td></tr><tr><td>4</td></tr></table></span>、<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">1</td></tr><tr><td>2</td></tr></table></span>，同时扩大4倍，则NO<SUB>2</SUB>、O<SUB>2</SUB>、H<SUB>2</SUB>O、HNO<SUB>3</SUB>前面的化学计量数分别为4、1、2、4．<br />故选：C．','【分析】根据质量守恒定律：反应前后各原子的数目不变，选择相应的配平方法（最小公倍数法、定一法等）进行配平即可；配平时要注意化学计量数必须加在化学式的前面，配平过程中不能改变化学式中的下标；配平后化学计量数必须为整数．','选择题',3.00,'da011a081ff32fff1f31780ba6b9131f',9,400,'化学方程式的配平','',2015,'33','2015秋•甘肃校级期末',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841161,'下列基本实验操作正确的是（　　）','<img  src=\"/tikuimages/9/2016/400/shoutiniao92/22c0d070-94d4-11e9-b6d5-b42e9921e93e_xkb76.png\" style=\"vertical-align:middle\" /><br />液体读数','<img  src=\"/tikuimages/9/2016/400/shoutiniao48/22c452e1-94d4-11e9-8d9f-b42e9921e93e_xkb86.png\" style=\"vertical-align:middle\" /><br />摆放试管','<img  src=\"/tikuimages/9/2016/400/shoutiniao34/22c675c0-94d4-11e9-b439-b42e9921e93e_xkb89.png\" style=\"vertical-align:middle\" /><br />洗涤试管','<img  src=\"/tikuimages/9/2016/400/shoutiniao46/22c76021-94d4-11e9-be0f-b42e9921e93e_xkb53.png\" style=\"vertical-align:middle\" /><br />倾倒液体','','B','【解答】解：A、用量筒量取液体读数时，视线要与凹液面的最低处保持水平，操作错误．<br />B、试管洗干净后要倒放在试管架上，操作正确；<br />C、洗涤试管的工具是试管刷，不用试管刷时用手腕振动而不上下移动，操作错误；<br />D、倾倒液体试剂瓶口紧贴试管口，标签向着手心，瓶塞要倒放，操作错误．<br />故选B．','【分析】A、根据量筒读数的正确方法判断；<br />B、根据摆放试管的方法判断；<br />C、根据洗涤试管的正确方法判断；<br />D、根据倾倒液体的基本操作判断；','选择题',3.00,'54dfd0ad02d0696089fdabf4552f5668',9,400,'测量容器-量筒,液体药品的取用,玻璃仪器的洗涤','宣威市',2016,'37','2016•宣威市校级一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841174,'KOH溶液中含有H<SUB>2</SUB>O、K<SUP>+</SUP>、OH<SUP>-</SUP>、能使无色酚酞试液变成红色．某化学兴趣小组想探究KOH溶液使无色酚酞试液变成红色的原因．<br />【提出问题】KOH溶液中的什么成分使无色酚酞试液变红？<br />【猜想】猜想1：KOH溶液中的H<SUB>2</SUB>O使无色酚酞试液变红．<br />猜想2：KOH溶液中的K<SUP>+</SUP>使无色酚酞试液变红．<br />猜想3：KOH溶液中的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填化学式）使无色酚酞试液变红．<br />【实验探究】<br /><table class=\"edittable\"><TBODY><TR><td width=350>实验操作</TD><td width=137>反应②<br />实验现象</TD><td width=165>结论</TD></TR><TR><td>①用试管取少量蒸馏水，滴入1-2滴无色酚酞试液</TD><td><br />试管内溶液不变色</TD><td><br />猜想1不成立</TD></TR><TR><td>②用试管取少量KCl溶液，滴入1-2滴无色酚酞试液</TD><td><br /><u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD><td><br />猜想2不成立</TD></TR><TR><td><br />③<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，滴入1-2滴无色酚酞试液</TD><td><br />试管内溶液变红</TD><td><br /><u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD></TR></TBODY></TABLE>【讨论反思】有同学认为猜想①不需要实验验证就可以排除，你认为该同学理由是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />【拓展】①向Ba（OH）<SUB>2</SUB>溶液中滴入几滴无色酚酞试液，观察到的现象是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />②根据所学有关碱的知识KOH固体应<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>保存．','','','','','','OH<SUP>-</SUP>$###$试管内溶液不变色$###$用试管取少量氢氧化钾溶液，$###$猜想3成立$###$酚酞试液本身含有水分子$###$溶液变为红色$###$密封','【解答】解：【猜想】结合猜想1和猜想2根据氢氧化钾溶液中含有的粒子可以作出如下猜想：是氢氧根离子使酚酞变红的，其符号为：OH<SUP>-</SUP>；<br />【实验探究】②猜想②不成立，说明溶液没有变色，故实验②的现象为溶液不变色；<br />③根据上述两个实验可以知道水分子和钾离子都不能使酚酞变红，所以实验③中可以直接加入氢氧化钾溶液，观察到溶液变为红色，从而可以判断猜想3成立；<br />【讨论反思】酚酞试液本身就是含有水分子，且氯化钾溶液中含有水，实验②就能同时验证水和钠离子对酚酞变色的影响；<br />【拓展】①氢氧化钡溶液中含有氢氧根离子，同样可以使酚酞变红；<br />②氢氧化钾为碱，能够和空气中的二氧化碳反应而发生变质，所以应该密封保存．<br />故答案为：猜想3：OH<SUP>-</SUP>；<br /><table class=\"edittable\"><TBODY><TR><td width=188>实验操作</TD><td width=188>反应②<br />实验现象</TD><td width=188>结论</TD></TR><TR><td>①用试管取少量蒸馏水，滴入1-2滴无色酚酞试液</TD><td><br />试管内溶液不变色</TD><td><br />猜想1不成立</TD></TR><TR><td>②用试管取少量KCl溶液，滴入1-2滴无色酚酞试液</TD><td><br /><br />试管内溶液不变色</TD><td><br />猜想2不成立</TD></TR><TR><td>③用试管取少量氢氧化钾溶液，滴入1-2滴无色酚酞试液</TD><td><br />试管内溶液变红</TD><td>猜想3成立</TD></TR></TBODY></TABLE>【讨论反思】酚酞试液本身含有水分子<br />【拓展】①溶液变为红色；&nbsp;<br />②密封．','【分析】【猜想】结合猜想1和猜想2根据氢氧化钾溶液中含有的粒子作出合理的猜想即可；<br />【实验探究】根据碱的性质进行分析，能使酚酞变红的粒子是OH<SUP>-</SUP>，设计实验时要排除水分子和K<SUP>+</SUP>的干扰；<br />【讨论反思】酚酞试液本身就是含有水分子，且实验②就能同时验证水和钠离子对酚酞变色的影响，可以据此解答该题；<br />【拓展】①氢氧化钡溶液中含有氢氧根离子，同样可以使酚酞变红，可以据此解答该题；<br />②氢氧化钾为碱，能够和空气中的二氧化碳反应而发生变质，所以应该密封保存，可以据此解答该题','填空题',2.00,'f9581dfacc57bb8f1dfbb0677f4a65d6',9,400,'探究酸碱的主要性质,碱的化学性质','',2016,'35','2016春•苏州校级期中',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841179,'人类文明进步与材料发展关系十分密切．<br />（1）日前新疆地质工作者发现一处特大型金矿--卡特巴阿苏金矿，储量有望超百吨，潜在经济价值近200亿元．从金属活动性顺序的角度，你认为金在自然界中是以<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“单质”“化合物”之一）形式存在的．<br />（2）众多的金属材料中，铁、铝、铜及其合金一直在人类生产、生活中占据着主导地位．<br />①单质铁的金属活动性<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>单质铝的金属活动性（填“大于”“小于”“等于”之一）<br />②将铁粉加入到过量的硫酸铜溶液中，反应结束后，溶液中含有的金属离子是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填离子符号）．<br />（3）钢筋容易锈蚀，请用化学方程式表示用盐酸除去铁锈（铁锈的主要成分是Fe<SUB>2</SUB>O<SUB>3</SUB>）的反应原理<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','单质$###$小于$###$Fe<SUP>2+</SUP>、Cu<SUP>2+</SUP>$###$Fe<SUB>2</SUB>O<SUB>3</SUB>+6HCl═2FeCl<SUB>3</SUB>+3H<SUB>2</SUB>O','【解答】解：（1）金在地壳中的含量虽然很少，但人类很早就发现了单质金，是因为金很稳定，自然界中存在金单质；<br />（2）①在金属活动性顺序中，铝的位置排在铁的前面，活动性比铁强；<br />②铁排在铜的前面，故铁可以把硫酸铜中的铜置换出来，生成亚铁盐，因为过量的硫酸铜溶液，故溶液中仍有铜离子，故反应结束后，溶液中含有的金属离子为：Fe<SUP>2+</SUP>、Cu<SUP>2+</SUP>；<br />（3）铁锈的主要成分是氧化铁，与盐酸反应生成氯化铁和水，反应的化学方程式是：Fe<SUB>2</SUB>O<SUB>3</SUB>+6HCl═2FeCl<SUB>3</SUB>+3H<SUB>2</SUB>O．<br />故答案为：（1）单质；（2）①小于；②Fe<SUP>2+</SUP>、Cu<SUP>2+</SUP>；（3）Fe<SUB>2</SUB>O<SUB>3</SUB>+6HCl═2FeCl<SUB>3</SUB>+3H<SUB>2</SUB>O．','【分析】（1）根据自然界中金的存在形式进行分析；<br />（2）①根据金属活动性顺序进行分析判断；<br />②根据金属活动性顺序进行分析判断；<br />（3）根据铁锈的主要成分是氧化铁，与盐酸反应生成氯化铁和水，写出反应的化学方程式即可．','书写',3.00,'e713131272ac1d69afec611228397a00',9,400,'金属的化学性质,金属元素的存在及常见的金属矿物,酸的化学性质,书写化学方程式、文字表达式、电离方程式','',2016,'37','2016•市中区二模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841180,'把足量锌粒加入有盐酸的试管中，发现随着反应的进行，产生气体的速度逐渐加快，一段时间后逐渐减慢，最后停止产生气体，在反应过程中，溶液的温度也先随之升高，最后下降到室温．于是，同学们设计实验来探究“一定质量的锌粒和同体积盐酸反应快慢的影响因素”．<br />测得实验数据如表：<br />&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 一定质量的锌粒与同体积盐酸溶液反应实验数据表<br /><table class=\"edittable\"><TBODY><TR><td width=87>盐酸溶液<br />浓度</TD><td width=157>反应开始时酸溶液的<br />温度</TD><td width=192>反应开始到2分钟产生的<br />气体体积</TD><td width=154>反应结束共产生<br />气体体积</TD></TR><TR><td>5%</TD><td>20℃</TD><td>10mL</TD><td>60mL</TD></TR><TR><td>10%</TD><td>20℃</TD><td>19mL</TD><td>118mL</TD></TR><TR><td>15%</TD><td>20℃</TD><td>28mL</TD><td>175mL</TD></TR><TR><td>5%</TD><td>35℃</TD><td>28mL</TD><td>60mL</TD></TR><TR><td>10%</TD><td>35℃</TD><td>72mL</TD><td>118mL</TD></TR><TR><td>15%</TD><td>35℃</TD><td>102mL</TD><td>175mL</TD></TR></TBODY></TABLE>（1）锌粒和盐酸反应的化学化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）分析实验数据得出：一定质量的锌粒和同体积盐酸反应，影响反应快慢的主要因素有<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>、<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）可以通过比较哪些数据来确定锌和酸反应的快慢<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（4）过量锌粒加入盐酸中，反应开始的一段时间，反应速率逐渐加快原因<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','Zn+2HCl═ZnCl<SUB>2</SUB>+H<SUB>2</SUB>↑$###$盐酸溶液浓度$###$反应时溶液温度$###$反应开始到2分钟产生的气体体积（或相同时间内产生的气体体积）$###$金属锌和盐酸反应放出热量，加快反应速率','【解答】解：（1）锌和盐酸反应生成氯化锌和氢气，反应的化学化学方程式为Zn+2HCl═ZnCl<SUB>2</SUB>+H<SUB>2</SUB>↑；故填：Zn+2HCl═ZnCl<SUB>2</SUB>+H<SUB>2</SUB>↑；<br />（2）可以通过比较相同温度、时间内，不同的盐酸浓度产生的气体体积来确定金属和酸反应的快慢；通过在相同的盐酸浓度、时间内，在不同的温度下比较产生的气体体积来确定金属和酸反应的快慢；故填：盐酸溶液浓度；反应时溶液温度；<br />（3）可以通过比较反应开始到2分钟产生的气体体积（或相同时间内产生的气体体积）实验数据来确定金属和酸反应的快慢；故填：反应开始到2分钟产生的气体体积（或相同时间内产生的气体体积）；<br />（4）金属锌和盐酸反应放出热量，反应开始的一段时间，盐酸浓度在降低，但反应温度在上升，所以温度对反应速率的影响大于盐酸溶液浓度对反应速率的影响．故填：金属锌和盐酸反应放出热量，加快反应速率．','【分析】根据影响化学方应的因素有温度，反应物浓度，有无催化剂，接粗面积大小，从表中可得到本实验的影响反应速率的因素有盐酸溶液浓度、反应时溶液温度进行解答．','书写',3.00,'b228511cab7ad6e192571c4cf3e843c1',9,400,'影响化学反应速率的因素探究,金属的化学性质,书写化学方程式、文字表达式、电离方程式','',2016,'37','2016•番禺区一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841181,'对化学概念的准确理解是正确运用的前提，下列有关认识不正确的是（　　）','氧化反应一定是化合反应','复分解反应中各元素的化合价一定不变','置换反应一定有单质生成','有盐和水生成的反应不一定是中和反应','','A','【解答】解：A、根据氧化反应是指物质与氧发生的反应，化合反应即“多变一”，如 蜡烛+氧气→点燃二氧化碳+水，是氧化反应，但不是化合反应，故A错误；<br />B、复分解反应是指两种化合物相互交换成分生成另外的两种化合物的反应，所以复分解反应中各元素的化合价一定不变，故B正确；<br />C、置换反应是一种单质和化合物反应生成另外的单质和化合物的反应，所以置换反应一定有单质生成，故C正确；<br />D、中和反应是指酸和碱反应生成盐和水的反应，但是有盐和水生成的反应不一定是中和反应，例如氧化铁和盐酸生成氯化铁和水，不属于中和反应，故D正确．<br />故选：A．','【分析】A、根据氧化反应是指物质与氧发生的反应，化合反应即“多变一”，进行判断；<br />B、根据复分解反应是指两种化合物相互交换成分生成另外的两种化合物的反应进行解答；<br />C、根据置换反应是一种单质和化合物反应生成另外的单质和化合物的反应进行解答；<br />D、根据中和反应是指酸和碱反应生成盐和水的反应进行解答．','选择题',3.00,'694da15b727ab33b8f90c56f032b59de',9,400,'中和反应及其应用,复分解反应及其发生的条件,化合反应及其应用,置换反应及其应用,氧化反应','',2016,'35','2016春•太康县期中',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841189,'指出下列操作导致的后果．<br />（1）把块状固体药品直接丢入试管底部，后果是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）使用胶头滴管后，未经清洗就吸取别的试剂，后果是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','试管破裂$###$污染别的试剂','【解答】解：（1）取用块状固体药品时，应先将试管横放，用镊子把固体药品放在试管口，再让试管慢慢地竖起来，如果把块状固体药品直接丢入试管底部，会砸裂试管底部．<br />（2）使用胶头滴管后，未经清洗就吸取别的试剂，后果是污染别的试剂．<br />故答案为：（1）试管破裂；（2）污染别的试剂．','【分析】（1）根据取用块状固体药品的注意事项进行分析．<br />（2）根据使用胶头滴管的注意事项进行分析．','填空题',3.00,'26f6805a881c2bcdea88159a7bce6394',9,400,'固体药品的取用,液体药品的取用','',2015,'35','2015秋•德州校级期中',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841190,'自然界中二氧化碳的循环<br />下列五幅图片是二氧化碳在大气圈和水圈循环的示意图，图中的数据表示二氧化碳的相对量，没有具体单位．请你仔细分析图示，结合已有知识回答下列问题：<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao25/23473e80-94d4-11e9-ad17-b42e9921e93e_xkb99.png\" style=\"vertical-align:middle\" /><br />（1）将五幅图中所有吸收二氧化碳的相对量总和与所有释放二氧化碳的相对量总和比较，你的发现是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）图（Ⅰ）所示土壤中释放出二氧化碳的途径中包括：土壤中的碳酸盐与酸（盐酸）作用：土壤中的碳酸盐受热分解，请你各举一例，写出化学方程式：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>、<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）图（Ⅴ）中植物吸收和释放二氧化碳相对量相差较大的主要原因是：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（4）能源是社会发展的基础．化石燃料是目前人类最重要的能源物质，请回答：<br />①下列燃料中，属于化石燃料的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填字母符号）<br />A．汽油&nbsp;&nbsp;&nbsp; B．液化石油气&nbsp;&nbsp;&nbsp; C．柴油&nbsp;&nbsp;&nbsp; D．木炭&nbsp;&nbsp;&nbsp; E．天然气&nbsp;&nbsp;&nbsp; F．焦炭&nbsp;&nbsp;&nbsp; G．柴草&nbsp;&nbsp;&nbsp; H．酒精<br />②煤燃烧的产物有二氧化碳、二氧化硫、二氧化氮和水等，由此你认为煤中一定含有<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>元素，推断的依据是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />③化石燃料加工主要有以下几种方法，其中主要发生化学变化的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填字母编号）．<br />A．石油的裂解&nbsp;&nbsp;&nbsp; B．石油的分馏&nbsp;&nbsp; C．煤的焦化&nbsp;&nbsp;&nbsp; D．煤的气化&nbsp;&nbsp;&nbsp; E．煤的液化<br />④汽油是石油分馏所得的产物之一，其成分之一为C<SUB>7</SUB>H<SUB>16</SUB>，写出该物质完全燃烧的化学方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（5）为减少空气污染和温室气体排放，经研究发现不含碳元素的NH<SUB>3</SUB>燃烧产物没有污染，且释放大量能量，请写出NH<SUB>3</SUB>燃烧的化学方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（6）燃烧是一类重要的化学反应，对于燃烧的下列认识正确的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填字母序号）<br />A．通常所说的燃烧是可燃物跟氧气发生的剧烈的发光、发热的氧化反应<br />B．通过控制温度（着火点以上或以下）可以使可燃物与氧气的燃烧反应发生或停止<br />C．在燃烧过程中，可燃物中储存的能量以光能或热能的形式释放出来<br />D．在人类社会的发展进程中燃烧利大于弊<br />E．物质发生燃烧时都会产生火焰．','','','','','','释放的二氧化碳相对量总和大于吸收的二氧化碳$###$CaCO<SUB>3</SUB>+2HCl=CaCl<SUB>2</SUB>+CO<SUB>2</SUB>↑+H<SUB>2</SUB>O$###$CaCO<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>CaO+CO<SUB>2</SUB>↑$###$被吸收的二氧化碳转化成为有机物构成了植物体$###$ABCEF$###$C、N、S、H$###$质量守恒定律$###$ACDE$###$C<SUB>7</SUB>H<SUB>16</SUB>+11O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>7CO<SUB>2</SUB>+8H<SUB>2</SUB>O$###$4NH<SUB>3</SUB>+3O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>6H<SUB>2</SUB>O+2N<SUB>2</SUB>$###$ABCD','【解答】解：（1）将五幅图中所有吸收二氧化碳的相对量总和与所有释放二氧化碳的相对量总和比较，我们会发现是释放的二氧化碳相对量总和大于吸收的二氧化碳．<br />（2）碳酸钙是土壤中常见的碳酸盐，可分别写出碳酸钙与盐酸、碳酸钙分解的化学方程式．<br />①CaCO<SUB>3</SUB>+2HCl=CaCl<SUB>2</SUB>+CO<SUB>2</SUB>↑+H<SUB>2</SUB>O；②CaCO<SUB>3</SUB>&nbsp;<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>CaO+CO<SUB>2</SUB>↑；<br />（3）图（Ⅴ）中植物吸收和释放二氧化碳相对量相差较大的主要原因是被吸收的二氧化碳转化成为有机物构成了植物体；<br />（4）①化石燃料主要包括煤、石油和天然气．煤是复杂的混合物，主要含有碳元素，此外还含有氢元素和少量氮、硫、氧等元素以及无机矿物质；石油也是混合物，主要含有碳元素和氢元素；天然气主要是由碳和氢组成的气态碳氢化合物，其中最主要的是甲烷；从石油中得到的石油产品，汽油、柴油、液化石油气也属于化石燃料；故属于化石燃料的是ABCEF；<br />②煤燃烧时生成：二氧化碳、一氧化碳、二氧化硫、二氧化氮、水．根据质量守恒定律：反应前后元素和种类不变．<br />由反应后的生成物可知：生成物中的元素为：C、N、S、H、O，则反应物中也应该由这几种元素组成；由于氧气是由氧元素组成的，则煤中一定有C、N、S、H四种元素，可能含有氧元素；<br />③有新物质生成的变化为化学变化，石油的裂解、煤的焦化、煤的气化和煤的液化过程中都有新物质生成，故都是化学变化；<br />④据质量守恒定律写该物质完全燃烧的化学方程式为 C<SUB>7</SUB>H<SUB>16</SUB>+11O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>7CO<SUB>2</SUB>+8H<SUB>2</SUB>O；<br />（5）NH<SUB>3</SUB>燃烧生成水和氮气，方程式为：4NH<SUB>3</SUB>+3O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>6H<SUB>2</SUB>O+2N<SUB>2</SUB>．<br />（6）A、通常所说的燃烧是可燃物跟氧气发生的剧烈的发光、发热的氧化反应，故正确；<br />B、通过控制温度（着火点以上或以下）可以使可燃物与氧气的燃烧反应发生或停止，故正确；<br />C、在燃烧过程中，可燃物中储存的能量以光能或热能的形式释放出来，故正确；<br />D、在人类社会的发展进程中燃烧利大于弊，故正确；<br />E、可燃物燃烧时都能发光放热，但不一定都有火焰，故错误．<br />故答案为：<br />（1）释放的二氧化碳相对量总和大于吸收的二氧化碳；<br />（2）①CaCO<SUB>3</SUB>+2HCl=CaCl<SUB>2</SUB>+CO<SUB>2</SUB>↑+H<SUB>2</SUB>O；&nbsp;②CaCO<SUB>3</SUB>&nbsp;<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>CaO+CO<SUB>2</SUB>↑；<br />（3）被吸收的二氧化碳转化成为有机物构成了植物体；<br />（4）①ABCEF；②C、N、S、H；&nbsp;&nbsp;质量守恒定律；③ACDE；④C<SUB>7</SUB>H<SUB>16</SUB>+11O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>7CO<SUB>2</SUB>+8H<SUB>2</SUB>O；<br />（5）4NH<SUB>3</SUB>+3O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>6H<SUB>2</SUB>O+2N<SUB>2</SUB>．<br />（6）ABCD','【分析】（1）看图分析找规律；<br />（2）碳酸钙是土壤中常见的碳酸盐，据碳酸盐与酸反应的规律些化学方程式；<br />（3）据植物吸收二氧化碳后转变成葡萄糖分析；<br />（4）①从化石燃料主要包括煤、石油和天然气去分析解答；<br />②从煤燃烧时生成：二氧化碳、一氧化碳、二氧化硫、二氧化氮、水．根据质量守恒定律：反应前后元素和种类不变去分析；<br />③从在石油加工过程中，通过分馏，依据各成分沸点不同将他们分离，可得到不同的产品，使石油得到充分利用．煤的综合利用是将煤隔绝空气加强热得到很多有用的物质，如：焦炭、煤焦油、煤气等，由于产生了新的物质，所以发生的化学变化去分析解答．<br />④据质量守恒定律写化学方程式；<br />（5）根据反应物、生成物和条件书写方程式．<br />（6）根据燃烧的定义、条件进行分析解答','书写',3.00,'d5b88daa1a4503175e45860c8b1a409c',9,400,'自然界中的碳循环,化学变化和物理变化的判别,质量守恒定律及其应用,书写化学方程式、文字表达式、电离方程式,燃烧与燃烧的条件,化石燃料及其综合利用','乳山市',2016,'35','2016春•乳山市期中',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841203,'<img  src=\"/tikuimages/9/2016/400/shoutiniao23/236b684f-94d4-11e9-998c-b42e9921e93e_xkb12.png\" style=\"vertical-align:middle;FLOAT:right\" />（2016•甘肃模拟）实验室有一瓶无色溶液，由于受到腐蚀，标签破损（如图所示），某兴趣小组同学对其成分进行探究．<br /><br />【提出问题】这瓶溶液是什么呢？<br />【猜想与假设】经了解得知，这瓶无色溶液可能是碳酸钠、氯化钠、硫酸钠、硝酸钠中的某一种物质的溶液．<br />【交流讨论】为了确定该溶液的成分，他们进行了以下分析：<br />（1）甲同学认为一定不可能是氯化钠溶液，其根据是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）乙同学根据标签上化学式中的数字2，认为也不可能是硝酸钠溶液．<br />【设计实验】为了确定该溶液究竟是剩余两种盐中的哪一种溶液，他们设计了如下实验：<br /><table class=\"edittable\"><TBODY><TR><td>实验操作</TD><td>实验现象</TD></TR><TR><td>①取少量该容器于试管中，逐滴滴加氯化钡溶液<br />②静置后，倒去上层清液，向沉淀中滴加<u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD><td>①<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />②沉淀完全消失，产生大量气泡</TD></TR></TBODY></TABLE>【实验结论】该瓶无色溶液是碳酸钠溶液，其生成白色沉淀的化学方程式是：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />【实验反思】<br />（1）丙同学认为②中设计的实验还可简化，同样能达到鉴别的目的，丙同学设计的实验方案是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（要求写出实验操作及实验现象）<br />（2）由该实验可知取用化学试剂时应注意<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','氯化钠中不含有氧元素$###$稀盐酸$###$产生白色沉淀$###$Na<SUB>2</SUB>CO<SUB>3</SUB>+BaCl<SUB>2</SUB>=BaCO<SUB>3</SUB>↓+2NaCl$###$取少量溶液于试管中，向溶液中加入适量的盐酸，若有气泡产生，则为碳酸钠溶液$###$标签向着手心','【解答】解：【交流讨论】氯化钠是由钠元素和氯元素组成的，不含有氧元素，故一定不会是氯化钠溶液，故填：氯化钠中不含有氧元素；<br />【设计实验】硫酸钠能与氯化钡反应生成硫酸钡沉淀，但是硫酸钡沉淀不溶于盐酸，滴加盐酸不会出现溶解的情况，故不会是硫酸钠；碳酸钠与氯化钡湖反应生成碳酸钡沉淀，碳酸钡沉淀能溶于盐酸且产生气体，故该瓶溶液是碳酸钠溶液，故填：稀盐酸，产生白色沉淀；<br />【实验结论】碳酸钠与氯化钡湖反应生成碳酸钡沉淀和氯化钠；故填：Na<SUB>2</SUB>CO<SUB>3</SUB>+BaCl<SUB>2</SUB>=BaCO<SUB>3</SUB>↓+2NaCl；<br />【实验反思】（1）碳酸钠能与盐酸反应生成二氧化碳气体，而硫酸钠不能与盐酸反应，故可以直接向溶液中加入盐酸，观察是否产生气泡，故填：取少量溶液于试管中，向溶液中加入适量的盐酸，若有气泡产生，则为碳酸钠溶液；<br />（2）由该实验可知取用化学试剂时应注意标签向着手心，否则会腐蚀标签；故填：标签向着手心．','【分析】根据缺损标签的有关元素组成以及物质的性质来推断试剂的成分，碳酸盐能与盐酸反应生成二氧化碳气体，据此解答即可．','书写',3.00,'0ab2a593ee5f4df1d6024d50799805c9',9,400,'缺失标签的药品成分的探究,盐的化学性质,书写化学方程式、文字表达式、电离方程式','',2016,'32','2016•甘肃模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841209,'在一次实验课上，各小组同学在不同地点找来的小石块与稀盐酸反应制取二氧化碳．此反应的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />小明发现，相邻小组气体产生的速率比自己小组的快．小明将这一情况报告老师，老师鼓励他们对此问题共同探究．<br />【提出问题】影响二氧化碳气体产生速率的因素是什么？<br />【作出猜想】<br />①不同地点石块中碳酸钙含量不同；<br />②所用盐酸的溶质质量分数不同；<br />③还可能是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />【设计实验】（1）验证猜想①是否成立，要设计的实验：分别取大小相同、质量相同、<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>的石块，加入质量相同、<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>的稀盐酸进行实验．<br />（2）验证猜想②是否成立，对盐酸的要求是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />【交流讨论】实验中可以通过观察气泡产生的剧烈程度，粗略地比较反应速率．若要做到精确比较，应该测量的实验数据是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','CaCO<SUB>3</SUB>+2HCl=CaCl<SUB>2</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑$###$石块大小不同（或石块质量不同、盐酸质量不同）$###$产地不同$###$溶质质量分数相同$###$溶质质量分数不同$###$相同时间内收集的二氧化碳体积（或收集相同体积二氧化碳所需的时间）','【解答】解：小石块与稀盐酸反应制取二氧化碳的化学方程式为 CaCO<SUB>3</SUB>+2HCl=CaCl<SUB>2</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑．<br />【作出猜想】③影响二氧化碳产生速率的因素还可能是石块大小不同（或石块质量不同、盐酸质量不同）．<br />【设计实验】（1）验证猜想①是否成立，要设计的实验：其他量都要相同，但要取产地不同的石块，进行实验．<br />（2）验证猜想②是否成立，对盐酸的要求是溶质质量分数不同．<br />【交流讨论】实验中可以通过观察气泡产生的剧烈程度，粗略地比较反应速率．若要做到精确比较，应该测量的实验数据是 相同时间内收集的二氧化碳体积（或收集相同体积二氧化碳所需的时间）．<br />故答案为：CaCO<SUB>3</SUB>+2HCl=CaCl<SUB>2</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑；<br />【作出猜想】石块大小不同（或石块质量不同、盐酸质量不同）；<br />【设计实验】（1）产地不同；溶质质量分数相同；（2）溶质质量分数不同；<br />【交流讨论】相同时间内收集的二氧化碳体积（或收集相同体积二氧化碳所需的时间）．','【分析】根据反应物书写化学方程式；<br />【作出猜想】③从药品的表面积和酸的质量分数考虑；<br />【设计实验】（1）控制不同大小的石块这一个变量；<br />（2）控制盐酸的溶质质量分数这个变量；<br />【交流讨论】精确实验结果就要通过准确的数据来证明，故应该准确确定相同时间内二氧化碳的量．','书写',3.00,'09ad2779799025a72ed26c49ed868447',9,400,'影响化学反应速率的因素探究,书写化学方程式、文字表达式、电离方程式','',2016,'37','2016•马鞍山二模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841212,'下列有关水的说法正确的是（　　）','只有水能作溶液中的溶剂','由氢、氧两种元素组成的物质只有水','自然界中的水经沉淀、过滤、吸附、杀菌后可得到纯净的水','水可以灭火，主要利用了降低温度到可燃物着火点以下的原理','','D','【解答】解：A、水是最常用的溶剂，汽油、酒精也可以做溶剂，故选项说法错误；<br />B、水和过氧化氢均由氢元素和氧元素组成；故选项说法错误；<br />C、自然界的水经过沉淀、过滤、吸附、杀菌等操作，只能除去一些难溶性和可溶性的杂质，不能除去可溶性钙镁化合物，不能得到纯净的水，故选项说法错误；<br />D、水可以灭火，是利用了降低温度到可燃物着火点以下的原理；故选项说法正确．<br />故选D．','【分析】A、根据常见的溶剂进行分析判断；<br />B、根据水和过氧化氢均由氢元素和氧元素组成进行分析；<br />C、根据自然界的水经过沉淀、过滤、吸附、杀菌等操作，只能除去一些难溶性和可溶性的杂质，不能除去可溶性钙镁化合物，进行分析判断；<br />D、根据灭火原理进行分析判断即可．','选择题',3.00,'820933ac923641c9c81cfd4eddf0b089',9,400,'水的净化,常见的溶剂,物质的元素组成,灭火的原理和方法','',2016,'37','2016•孝义市二模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841214,'四硫化钠（Na<SUB>2</SUB>S<SUB>4</SUB>）固体可替代红磷测定空气中氧气的体积分数，反应原理为：2Na<SUB>2</SUB>S<SUB>4</SUB>+O<SUB>2</SUB>+2H<SUB>2</SUB>O═8S↓+4NaOH．<br />【查阅资料】<br />①Na<SUB>2</SUB>S<SUB>4</SUB>受热时，会与空气中的水蒸气反应，生成少量有毒气体硫化氢（水溶液呈酸性）．<br />②硫代硫酸钠（Na<SUB>2</SUB>S<SUB>2</SUB>O<SUB>3</SUB>）可溶于水，常温下与NaOH溶液不反应．<br />【实验过程】<br />①取0.5g碳酸钠、0.2g硫粉混合后置于试管中，加入（如图1所示，夹持装置已略去），制得Na<SUB>2</SUB>S<SUB>4</SUB>，反应为：4Na<SUB>2</SUB>CO<SUB>3</SUB>+12S+X<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2Na<SUB>2</SUB>S<SUB>4</SUB>+2Na<SUB>2</SUB>S<SUB>2</SUB>O<SUB>3</SUB>+4CO<SUB>2</SUB>，X的化学式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，蘸有浓NaOH溶液的棉花的作用是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br /><img  src=\"/tikuimages/9/2014/400/shoutiniao57/2381fd8f-94d4-11e9-97ce-b42e9921e93e_xkb86.png\" style=\"vertical-align:middle\" /><br />②冷却后，取下棉花，放置一段时间，再向该试管中加入10mL水，迅速塞紧橡胶塞，充分振荡．测量液面至橡胶塞下沿的距离，记录数据h<SUB>1</SUB>（如图2所示）．<br />③将该试管插入水中（如图3所示），拔下橡胶塞，观察到<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，塞紧橡胶塞．将试管取出，倒转过来，测量液面至橡胶塞下沿的距离，记录数据h<SUB>2</SUB>．理论上h<SUB>2</SUB>：h<SUB>1</SUB>=<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />④按照①～③再重复实验2次．3次实验数据如下表所示．<br /><table class=\"edittable\"><TBODY><TR><td width=119></TD><td width=119>第1次</TD><td width=119>第2次</TD><td width=119>第3次</TD></TR><TR><td>h<SUB>1</SUB>/cm</TD><td>11.0</TD><td>11.4</TD><td>11.6</TD></TR><TR><td>h<SUB>2</SUB>/cm</TD><td>8.7</TD><td>9.1</TD><td>9.2</TD></TR></TBODY></TABLE>【解释与结论】<br />根据第1次实验数据，计算空气中氧气的体积分数为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>%（结果精确到0.1%）．<br />【反思与评价】<br />若实验过程②中，取下棉花后，未放置一段时间即进行后续操作，会影响测定结果，请说明有何影响，并阐述理由：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','O<SUB>2</SUB>$###$吸收生成的二氧化碳气体和有毒的硫化氢气体$###$水进入试管内，进入水的体积约为试管内空气体积的五分之一，有黄色固体生成$###$4：5$###$20.9$###$偏小，因为在（1）中的反应后试管中的氧气基本耗尽，二氧化碳虽然被吸收，但是取下棉花后试管中二氧化碳的量没有多少改变（原本没有，后来只有0.04%），可以不考虑啊，比起氧气直接可以忽略嘛（氧气原本没有，后来20%），若不放置一段时间则试管中氧气含量比空气中偏少，导致h2偏大，所以最终结果应是偏小的．','【解答】解：【实验过程】①根据化学反应前后原子的种类和数目不变和4Na<SUB>2</SUB>CO<SUB>3</SUB>+12S+X<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2Na<SUB>2</SUB>S<SUB>4</SUB>+2Na<SUB>2</SUB>S<SUB>2</SUB>O<SUB>3</SUB>+4CO<SUB>2</SUB>，可知X的化学式为O<SUB>2</SUB>；二氧化碳和NaOH溶液反应生成碳酸钠和水，Na<SUB>2</SUB>S<SUB>4</SUB>受热时，会与空气中的水蒸气反应，生成少量有毒气体硫化氢（水溶液呈酸性），所以蘸有浓NaOH溶液的棉花的作用是吸收生成的二氧化碳气体和有毒的硫化氢气体；故填：O<SUB>2</SUB>；吸收生成的二氧化碳气体和有毒的硫化氢气体；<br />③四硫化钠和氧气、水反应生成硫和氢氧化钠，会使得试管内压强变小，将该试管插入水中（如图3所示），拔下橡胶塞，观察到水进入试管内，进入水的体积约为试管内空气体积的五分之一，有黄色固体生成；空气中氧气约占空气体积的五分之一，将试管取出，倒转过来，测量液面至橡胶塞下沿的距离，记录数据h<SUB>2</SUB>．理论上h<SUB>2</SUB>：h<SUB>1</SUB>=4：5；故填：水进入试管内，进入水的体积约为试管内空气体积的五分之一，有黄色固体生成；4：5；<br />④【解释与结论】空气中氧气的体积分数=<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">11.0-8.7</td></tr><tr><td>11.0</td></tr></table></span>×100%=20.9%；故填：20.9；<br />【反思与评价】未放置一段时间会导致试管内的空气成分与外边的空气成分可能有差异，偏小，因为在（1）中的反应后试管中的氧气基本耗尽，二氧化碳虽然被吸收，但是取下棉花后试管中二氧化碳的量没有多少改变（原本没有，后来只有0.04%），可以不考虑啊，比起氧气直接可以忽略嘛（氧气原本没有，后来20%），若不放置一段时间则试管中氧气含量比空气中偏少，导致h2偏大，所以最终结果应是偏小的．<br />答案：<br />【实验过程】<br />①O<SUB>2</SUB>；吸收生成的二氧化碳气体和有毒的硫化氢气体；<br />③水进入试管内，进入水的体积约为试管内空气体积的五分之一，有黄色固体生成；4：5；<br />【解释与结论】20.9<br />【反思与评价】<br />偏小，因为在（1）中的反应后试管中的氧气基本耗尽，二氧化碳虽然被吸收，但是取下棉花后试管中二氧化碳的量没有多少改变（原本没有，后来只有0.04%），可以不考虑啊，比起氧气直接可以忽略嘛（氧气原本没有，后来20%），若不放置一段时间则试管中氧气含量比空气中偏少，导致h<SUB>2</SUB>偏大，所以最终结果应是偏小的．','【分析】【实验过程】①根据化学反应前后原子的种类和数目不变、二氧化碳和NaOH溶液反应以及Na<SUB>2</SUB>S<SUB>4</SUB>受热时，会与空气中的水蒸气反应，生成少量有毒气体硫化氢（水溶液呈酸性）进行解答；<br />③根据四硫化钠和氧气、水反应生成硫和氢氧化钠，会使得试管内压强变小以及空气中氧气约占空气体积的五分之一进行解答；<br />④【解释与结论】根据空气中氧气的体积分数=<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><span><span>h</span><span style=\"vertical-align:sub;font-size:90%\">1</span></span>-<span><span>h</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span></td></tr><tr><td style=\"padding-top:1px;font-size:90%\"><span><span>h</span><span style=\"vertical-align:sub;font-size:90%\">1</span></span></td></tr></table></span>×100%进行解答；<br />【反思与评价】根据未放置一段时间会导致试管内的空气成分与外边的空气成分可能有差异进行解答．','填空题',3.00,'f53b335cecd5ded5d42537dfe6be15b8',9,400,'测定空气里氧气含量的探究,质量守恒定律及其应用','山南地区',2014,'37','2014•山南地区校级三模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841223,'一个碳原子的质量是1.993×10<SUP>-23</SUP>千克，碳的相对原子质量为（　　）','12克','12千克','12','1.993×10<SUP>-23</SUP>千克','','C','【解答】解：A、相对原子质量是个比值，是一个比值，单位是“1”，常省略不写，该选项的单位是g，故选项错误．<br />B、相对原子质量是个比值，是一个比值，单位是“1”，常省略不写，该选项的单位是kg，故选项错误．<br />C、相对原子质量是个比值，是一个比值，单位是“1”，常省略不写，故选项正确．<br />D、相对原子质量是个比值，是一个比值，单位是“1”，常省略不写，该选项的单位是kg，故选项错误．<br />故选：C．','【分析】国际上是以一种碳原子的质量的<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">1</td></tr><tr><td>12</td></tr></table></span>作为标准，其他原子的质量跟它相比较所得到的比值，就是该原子的相对原子质量；相对原子质量是个比值，是一个比值，单位是“1”，常省略不写，据此进行分析判断．','选择题',3.00,'2a02c4041299deb7e512c6ac2f2966fe',9,400,'相对原子质量的概念及其计算方法','',0,'37','',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841224,'如图是某化工厂设计的基础化工生产流程，回答问题．<img  src=\"/tikuimages/9/2016/400/shoutiniao75/23b2354f-94d4-11e9-b116-b42e9921e93e_xkb44.png\" style=\"vertical-align:middle\" /><br />（1）白色固体A中，可用作某些气体干燥剂的成分是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（2）按生产流程，该化工厂最终生产出的产品有纯净碳酸钙粉末，还有<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（3）如果③反应Na<SUB>2</SUB>CO<SUB>3</SUB>溶液的用量不足，向无色溶液B中通入二氧化碳气体，会观察到的现象是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（4）如果③反应NaCO<SUB>3</SUB>溶液使用过得，对无色溶液B的有关说法中，正确的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />A．溶液中含有的阳离子全部是钠离子<br />B．向其中滴加稀盐酸，一定有气泡产生<br />C．可用来鉴别CaCl<SUB>2</SUB>溶液和Ba（OH）<SUB>2</SUB>溶液<br />D．向其中滴加石灰水到不再产生沉淀并过滤，滤液中只含有一种溶质．','','','','','','氧化钙$###$氢氧化钠和二氧化碳$###$溶液变浑浊$###$AD','【解答】解：（1）碳酸钙高温煅烧生成二氧化碳和氧化钙，氧化钙能吸水，所以可用干燥剂；故填：氧化钙；<br />（2）碳酸钙高温煅烧生成二氧化碳和氧化钙，氧化钙和水反应生成氢氧化钙，氢氧化钙和碳酸钠溶液反应生成碳酸钙沉淀和氢氧化钠，所以按生产流程，该化工厂最终生产出的产品有纯净碳酸钙粉末，还有氢氧化钠和二氧化碳；故填：氢氧化钠和二氧化碳；<br />（3）如果③反应Na<SUB>2</SUB>CO<SUB>3</SUB>溶液的用量不足，那么氢氧化钙会有剩余，所以向无色溶液B中通入二氧化碳气体，会观察到的现象是溶液变浑浊；故填：溶液变浑浊；<br />（4）如果③反应NaCO<SUB>3</SUB>溶液使用过量，那么无色溶液B中含有氢氧化钠和碳酸钠．<br />A．溶液中含有的阳离子全部是钠离子，故A正确；<br />B．向其中滴加稀盐酸，如果盐酸的量少，不一定会放出二氧化碳气体，所以不一定有气泡产生，故B错误；<br />C．碳酸钠和氯化钙、氢氧化钡都会生成白色沉淀，所以不能用来鉴别CaCl<SUB>2</SUB>溶液和Ba（OH）<SUB>2</SUB>溶液，故C错误；<br />D．氢氧化钙和碳酸钠反应生成碳酸钙沉淀和氢氧化钠，所以向其中滴加石灰水到不再产生沉淀并过滤，滤液中只含有一种溶质，就是氢氧化钠，故D正确．<br />故选：AD．','【分析】根据碳酸钙高温煅烧生成二氧化碳和氧化钙，氧化钙和水反应生成氢氧化钙，氢氧化钙和碳酸钠溶液反应生成碳酸钙沉淀和氢氧化钠进行解答．','填空题',3.00,'0351859bd15976e380b5c1ebe8f3f284',9,400,'碳酸钙、生石灰、熟石灰之间的转化,盐的化学性质,物质的相互转化和制备','',2016,'37','2016•绿园区一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841235,'区别下列物质的方法中，不能将本组待区别的物质区别开的是（　　）<br /><table class=\"edittable\"><TBODY><TR><td width=47>选项</TD><td width=182>待区别的物质</TD><td width=322>区别方法</TD></TR><TR><td>A</TD><td>活性炭和二氧化锰</TD><td>分别投入滴有一滴红墨水的水中，观察现象</TD></TR><TR><td>B</TD><td>氯酸钾和氯化钾</TD><td>看外观，颜色</TD></TR><TR><td>C</TD><td>氯化钠与氢氧化钠溶液</TD><td>分别滴加酚酞试剂，观察现象</TD></TR><TR><td>D</TD><td>人呼出气体、氧气和空气</TD><td>分别将燃着的木条伸入瓶中，观察现象</TD></TR></TBODY></TABLE>','A','B','C','D','','B','【解答】解：A、活性炭具有吸附性，二氧化锰不具有吸附性，分别投入滴有一滴红墨水的水中，观察现象，能使颜色消失的是活性炭，可以鉴别，故选项错误．<br />B、氯酸钾和氯化钾均是白色固体，看外观、颜色不能鉴别，故选项正确．<br />C、氯化钠与氢氧化钠溶液分别显中性、酸性，分别滴加酚酞试剂，变红色的是氢氧化钠溶液，不变色的是氯化钠溶液，可以鉴别，故选项错误．<br />D、氧气能支持燃烧，分别将燃着的木条伸入瓶中，能使木条燃烧更旺的是氧气，能使木条熄灭的是人呼出气体，木条正常燃烧的是空气，可以鉴别，故选项错误．<br />故选：B．','【分析】根据两种物质与同种试剂反应产生的不同现象来鉴别它们，若两种物质与同种物质反应的现象相同，则无法鉴别它们．','选择题',3.00,'08d1a03e27b0856ff080dd1e2b244fcb',9,400,'吸入空气与呼出气体的比较,酸、碱、盐的鉴别,物质的鉴别、推断','',2016,'32','2016•兰州模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841243,'海洋中蕴藏着丰富的化学资源，如NaCl、MgCl<SUB>2</SUB>、CaCl<SUB>2</SUB>等物质．从海水中提取金属镁的主要流程如土1：<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao36/23fcf9f0-94d4-11e9-867e-b42e9921e93e_xkb85.png\" style=\"vertical-align:middle\" /><br />（1）电解熔融的无水MgCl<SUB>2</SUB>可以得到金属镁．该反应属于基本反应类型中的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>反应．<br />（2）将海水中的镁提取出来，首先要让海水中的镁离子形成沉淀而富集起来．结合表中数据分析，为了使氯化镁转化为沉淀，应加入的溶液A可能是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />室温下几种物质的溶解度<br /><table class=\"edittable\"><TBODY><TR><td width=71>物质</TD><td width=65>MgCO<SUB>3</SUB></TD><td width=57>CaCO<SUB>3</SUB></TD><td width=76>Mg（OH）<SUB>2</SUB></TD><td width=67>Ca（OH）<SUB>2</SUB></TD></TR><TR><td>溶解度/g</TD><td>0.01</TD><td>0.0013</TD><td>0.0029</TD><td>0.16</TD></TR></TBODY></TABLE>（3）结合MgCl<SUB>2</SUB>的溶解度曲线如图2分析，操作I：蒸发浓缩、<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>、过滤．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao70/24002e40-94d4-11e9-959d-b42e9921e93e_xkb55.png\" style=\"vertical-align:middle\" />','','','','','','分解$###$Ca（OH）<SUB>2</SUB>溶液（或NaOH溶液）$###$降温结晶','【解答】解：（1）据反应反应的概念，电解熔融的无水MgCl<SUB>2</SUB>可以得到金属镁和氯气，反应属于一变多，故此反应属于基本反应类型中的分解反应；<br />（2）氯化镁转化为氢氧化镁需要加入碱液以提供氢氧根，此处选择过量的Ca（OH）<SUB>2</SUB>溶液（或NaOH溶液）进行反应，目的是让所取海水中的氯化镁全部转化为氢氧化镁沉淀；<br />（3）由于氯化镁的溶解度随温度的升高而增大，所以，从氯化镁溶液中取得氯化镁晶体，应选择降低温度结晶的方法．<br />故答案为：（1）分解；&nbsp;（2）Ca（OH）<SUB>2</SUB>溶液（或NaOH溶液）；（3）降温结晶．','【分析】（1）据反应类型的判定即化合反应、分解反应、置换反应和复分解反应的判定解答；<br />（2）氯化镁转化为氢氧化镁需要加入碱液，验证MgCl<SUB>2</SUB>已完全转化为Mg（OH）<SUB>2</SUB>的方法是取滤液滴入Ca（OH）<SUB>2</SUB>溶液（或NaOH溶液）是否有沉淀生成；<br />（3）根据氯化镁的溶解度随温度的变化情况分析结晶的方法．','填空题',3.00,'722dbfcaa616b1e1d4f7d32a6fc78574',9,400,'固体溶解度曲线及其作用,盐的化学性质,反应类型的判定,对海洋资源的合理开发与利用','',2016,'37','2016•西城区一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841253,'水是生命之源，下列有关水的说法正确的是（　　）','利用肥皂水可以区分硬水和软水','利用活性炭的吸附性，可以吸附水中不溶性杂质','75%的医用酒精中，酒精是溶剂','为节约用水，可用工业废水浇灌农田','','A','【解答】解：A、利用肥皂水可以区分硬水和软水，正确；<br />B、活性炭吸附的是水中的色素和异味，不能除去水中的不溶性杂质，错误；<br />C、75%的医用酒精中，酒精是溶质，错误；<br />D、工业废水中含有有害物质，不能用于浇灌农田，错误；<br />故选A．','【分析】根据硬水和软水的鉴别方法、活性炭的吸附溶液组成的判断以及工业废水中含有有害物质解答即可．','选择题',3.00,'e88fbc51e824846e9d5ec0b76d1475cb',9,400,'硬水与软水,水资源的污染与防治,溶液、溶质和溶剂的相互关系与判断,碳单质的物理性质及用途','',2016,'32','2016•抚顺模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841254,'掌握一些家庭安全知识和紧急灭火措施，就能避免火灾．下列做法不正确的是（　　）','炒菜时油锅着火，可以立即盖上锅盖','厨房内闻到较浓的煤气味，立即关闭阀门并开窗通风','家用电器着火，应立即用水扑灭','如果不小心洒在桌面的酒精着火，可以立即用湿布盖灭','','C','【解答】解：A、用锅盖盖灭，隔绝了氧气使油熄灭，故对．<br />B、煤气是可燃性气体，与空气混合达到一定程度，遇明火有爆炸的危险，所以厨房内闻到较浓的煤气味，立即关闭阀门并开窗通风，故对．<br />C、电器着火若用水扑灭极易造成电线短路和人体触电，所以电器着火应先切断电源，再灭火，故错．<br />D、如果不小心洒在桌面的酒精着火，可立即用湿布扑灭，是隔绝空气和降低温度灭火，故对．<br />故选C．','【分析】根据灭火的原理：（1）清除可燃物或使可燃物与其他物品隔离，（2）隔绝氧气或空气，（3）使温度降到可燃物的着火点以下，根据日常经验和燃烧的物质的性质作出正确的判断．','选择题',3.00,'f6d0fd2ac6de2a49496f544930adf57f',9,400,'灭火的原理和方法,防范爆炸的措施','',2016,'32','2016•营口模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841260,'除去下列物质中混有的少量杂质（括号内为杂质），拟定的实验方案可行的是（　　）','CuO（C）--在空气中灼烧','CaCO<SUB>3</SUB>固体（CaO固体）--高温煅烧','NaCl溶液（Na<SUB>2</SUB>CO<SUB>3</SUB>）--加入适量的澄清石灰水，过滤','CO<SUB>2</SUB>气体（HCl气体）--依次通过足量的NaOH溶液和浓硫酸','','A','【解答】解：A、C在空气中燃烧生成二氧化碳气体，能除去杂质且没有引入新的杂质，符合除杂原则，故选项所采取的方法正确．<br />B、CaCO<SUB>3</SUB>固体高温煅烧生成氧化钙和二氧化碳，反而会把原物质除去，不符合除杂原则，故选项所采取的方法错误．<br />C、碳酸钠能与适量的澄清石灰水反应生成碳酸钙沉淀和氢氧化钠，能除去杂质但引入了新的杂质氢氧化钠，不符合除杂原则，故选项所采取的方法错误．<br />D、CO<SUB>2</SUB>和HCl气体均能与NaOH溶液反应，不但能把杂质除去，也会把原物质除去，不符合除杂原则，故选项所采取的方法错误．<br />故选：A．','【分析】根据原物质和杂质的性质选择适当的除杂剂和分离方法，所谓除杂（提纯），是指除去杂质，同时被提纯物质不得改变．除杂质题至少要满足两个条件：①加入的试剂只能与杂质反应，不能与原物质反应；②反应后不能引入新的杂质．','选择题',3.00,'795c2b340e9bbe5891d6b11531393b68',9,400,'物质除杂或净化的探究,常见气体的检验与除杂方法,盐的化学性质,碳的化学性质','',2016,'37','2016•南京一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841263,'实验室中的试剂常因与空气中的一些成分作用而发生变化，下列以试剂在空气中发生变化的分析不正确的是（　　）','铁粉生锈与空气中的水蒸气和氧气有关','氢氧化钠潮解变质与空气中的水和二氧化碳有关','氧化钙的变质与空气中的水和二氧化碳有关','浓盐酸暴露在空气中变稀与空气中的水蒸气有关','','D','【解答】解：A、铁粉生锈的条件是与水和氧气共同接触，故A正确；<br />B、潮解的定义是物质吸收空气中水分而溶解的过程，氢氧化钠变质的过程是氢氧化钠与空气中二氧化碳反应生成碳酸钠的过程，故B正确；<br />C、氧化钙的变质是氧化钙与空气中水反应生成氢氧化钙，氢氧化钙又与空气中的二氧化碳反应生成碳酸钙，故C正确；<br />D、浓盐酸变稀是因为浓盐酸具有挥发性，挥发出氯化氢，使溶质减少而变稀，与空气中的水蒸气无关，故D错误．<br />故选D．','【分析】A、根据铁粉生锈的条件来回答；<br />B、根据潮解的定义和氢氧化钠变质的过程来考虑；<br />C、氧化钙变质的过程来考虑；<br />D、根据浓盐酸变稀的原因考虑．','选择题',3.00,'98036983f97aeb2d7d3038880e265b98',9,400,'金属锈蚀的条件及其防护,空气中常见酸碱盐的质量或性质变化及贮存法','',2016,'37','2016•黑龙江一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841267,'下列实验操作错误的是（　　）','<img  src=\"/tikuimages/9/2016/400/shoutiniao66/24382e30-94d4-11e9-9c94-b42e9921e93e_xkb4.png\" style=\"vertical-align:middle\" /><br />滴加液体','<img  src=\"/tikuimages/9/2016/400/shoutiniao82/24398dc0-94d4-11e9-9953-b42e9921e93e_xkb3.png\" style=\"vertical-align:middle\" /><br />测溶液pH','<img  src=\"/tikuimages/9/2016/400/shoutiniao69/243ac640-94d4-11e9-8a96-b42e9921e93e_xkb97.png\" style=\"vertical-align:middle\" /><br />取用粉末状固体','<img  src=\"/tikuimages/9/2016/400/shoutiniao91/243d5e4f-94d4-11e9-bcc9-b42e9921e93e_xkb53.png\" style=\"vertical-align:middle\" /><br />浓硫酸稀释','','B','【解答】解：A、使用胶头滴管滴加少量液体的操作，注意胶头滴管不能伸入到试管内或接触试管内壁，应垂直悬空在试管口上方滴加液体，防止污染胶头滴管，图中所示操作正确．<br />B、用pH试纸测定未知溶液的pH时，正确的操作方法为用玻璃棒蘸取少量待测液滴在干燥的pH试纸上，与标准比色卡对比来确定pH．不能用水湿润pH试纸，否则稀释了待测溶液，使溶液的酸碱性减弱，测定结果不准确，图中所示操作错误．<br />C、取用粉末状药品，试管横放，用药匙或纸槽把药品送到试管底部，图中所示操作正确．<br />D、稀释浓硫酸时，要把浓硫酸缓缓地沿器壁注入水中，同时用玻璃棒不断搅拌，以使热量及时地扩散；一定不能把水注入浓硫酸中；图中所示操作正确．<br />故选：B．','【分析】A、根据使用胶头滴管滴加少量液体的方法进行分析判断．<br />B、根据用pH试纸测定未知溶液的pH的方法进行分析判断．<br />C、根据向试管中装粉末状固体药品的方法进行分析判断．<br />D、根据浓硫酸的稀释方法（酸入水，沿器壁，慢慢倒，不断搅）进行分析判断．','选择题',3.00,'536bd324e2e18db6c0eb93055b4ce255',9,400,'固体药品的取用,液体药品的取用,浓硫酸的性质及浓硫酸的稀释,溶液的酸碱度测定','',2016,'32','2016•重庆校级模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841274,'通过近一年的化学学习，同学们已能用所学的知识回答一些问题：<br />（1）煤、石油、<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>常称为化石燃料；<br />（2）天然水中含有许多杂质，可利用吸附、沉淀、过滤和蒸馏等方法净化，其中净化程度最高的方法是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>、硬水给生活和生产带来很多麻烦，生活中可用<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>来区别硬水和软水；<br />（3）高炉炼铁（以赤铁矿为例）过程中还原成铁的主要原理<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','天然气$###$蒸馏$###$肥皂水$###$主要是利用CO的还原性，在高温下和氧化铁反应生成铁和二氧化碳，反应的化学方程式为3CO+Fe<SUB>2</SUB>O<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2Fe+3CO<SUB>2</SUB>','【解答】解：（1）煤、石油、天然气属于化石燃料；<br />故答案为：天然气．<br />（2）蒸馏之后的水是纯水，所以蒸馏的方法得到的水的净化程度最高；硬水和软水的区别在于所含的钙镁离子的多少，钙镁离子可以和肥皂水产生浮渣，所以可以用肥皂水来区分硬水和软水．<br />故答案为：蒸馏；肥皂水．<br />（3）高炉炼铁（以赤铁矿为例）过程中还原成铁的主要反应原理，主要是利用CO的还原性，在高温下和氧化铁反应生成铁和二氧化碳，反应的化学方程式为3CO+Fe<SUB>2</SUB>O<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2Fe+3CO<SUB>2</SUB>．故答案为：主要是利用CO的还原性，在高温下和氧化铁反应生成铁和二氧化碳，反应的化学方程式为3CO+Fe<SUB>2</SUB>O<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2Fe+3CO<SUB>2</SUB>．','【分析】（1）煤、石油、天然气属于化石燃料；<br />（2）蒸馏之后得到的水是纯水，净化程度最高；区分硬水和软水常用的方法是用肥皂水；<br />（3）高炉炼铁（以赤铁矿为例）过程中还原成铁的主要反应原理，主要是利用CO的还原性，在高温下和氧化铁反应生成铁和二氧化碳，写出反应的化学方程式即可．','书写',3.00,'c3a4f360b262a7c01d1db049f6f9fcbe',9,400,'硬水与软水,铁的冶炼,书写化学方程式、文字表达式、电离方程式,化石燃料及其综合利用','山南地区',2014,'37','2014•山南地区校级二模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841277,'金属材料不但与人们的生活息息相关，而且是重要的战略物资．金属矿物质的储量有限，而且不能再生．请写出保护金属资源有效途径<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．（一点合理即可）','','','','','','金属防锈（回收利用废旧金属、合理开采金属资源、寻找金属替代品）','【解答】解：保护金属资源可以从金属防锈、回收利用废旧金属、合理开采金属资源、寻找金属替代品，故填：金属防锈（回收利用废旧金属、合理开采金属资源、寻找金属替代品）．','【分析】根据保护金属资源的措施进行分析解答即可．','填空题',3.00,'f3a3502ab59c8680304c53ff77248991',9,400,'金属资源的保护','',2016,'37','2016•黑龙江二模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841281,'2013年5月14日11时到2013年5月15日10时台州市环保局监测：台州市空气质量指数98，空气质量级别Ⅱ级，空气质量良，空气首要污染物为可吸入颗粒物．下列有关空气各成分的说法正确的是（　　）','氧气的化学性质比较活泼，属于可燃物','洁净的空气是纯净物','氮气的化学性质不活泼，可用于食品防腐','二氧化碳在空气中含量增多会引起温室效应，属于空气污染物','','C','【解答】解：A、氧气的化学性质比较活泼，具有助燃性，不能做可燃物；故不正确；<br />B、洁净的空气中含有氮气和氧气等多种物质，是混合物，故错误；<br />C、氮气的化学性质在常温下很稳定，可用作保护气、防腐剂等；故正确；<br />D、二氧化碳无毒，但含量过多会引起温室效应，但是没有列入空气污染物监测范围．故不正确；<br />故选C．','【分析】A、氧气的化学性质比较活泼，具有助燃性；<br />B、根据纯净物的概念进行分析；<br />C、氮气的化学性质在常温下很稳定，可用作防腐剂；<br />D、二氧化碳不属于空气污染物．','选择题',3.00,'05d3014a1fe0ef80c5a1acca6a51d335',9,400,'空气的成分及各成分的体积分数,空气的污染及其危害,氧气的化学性质,纯净物和混合物的判别','',2013,'37','2013春•台州校级月考',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841282,'2015年我国科学家屠呦呦获诺贝尔医学奖，她发现并提取出青蒿素（化学式：C<SUB>15</SUB>H<SUB>22</SUB>O<SUB>5</SUB>）用于治疗疟疾，挽救了数百万患者的生命．有关青蒿素的说法正确的是（　　）','青蒿素是由三种原子构成','青蒿素中碳、氢、氧元素的质量比为15：22：5','青蒿素是有机高分子化合物','青蒿素中氢元素的质量分数最低','','D','【解答】解：A．青蒿素是由分子构成的而不是由原子构成的，故错误；<br />B．青蒿素中C、H、0三种元素的质量比为（12×15）：（1×22）：（16×5）=90：11：40，故错误；<br />C．有机高分子化合物是指相对分子质量很大的有机物，可达几万至几十万，青蒿素的相对分子质量为12×15+1×22+16×5=282，不能达到几万，故错误；<br />D．青蒿素中C、H、O三种元素的质量比为（12×15）：（1×22）：（16×5）=90：11：40，可见其中氢元素的质量分数最小，故正确．<br />故选D．','【分析】A．根据物质的结构来分析；<br />B．根据元素的质量比来分析；<br />C．根据有机高分子化合物的概念来分析；<br />D．根据化合物中元素的质量比来分析．','选择题',3.00,'400d3d2656b4de07c4d7a7fef36c61ae',9,400,'有机物的特征、分类及聚合物的特性,化学式的书写及意义,元素质量比的计算,元素的质量分数计算','',2016,'32','2016•云南模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841302,'在一次科学实践活动课上，同学们做了“会动的鸡蛋”系列趣味实验．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao6/24980170-94d4-11e9-a71e-b42e9921e93e_xkb57.png\" style=\"vertical-align:middle\" /><br />（1）向装有干冰的集气瓶中加入热水（如图A），观察到有大量白雾产生，熟鸡蛋在瓶口跳动，该实验说明干冰具有的性质<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）向集满二氧化碳气体的瓶中加入饱和的氢氧化钠溶液（如图B），观察到瓶口的熟鸡蛋被吞入瓶中，其原因是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）将生鸡蛋放入盛有水的烧杯中，鸡蛋沉入杯底（如图C），向水中不断地加入食盐，观察到鸡蛋逐渐上浮，由此说明食盐溶液的密度与其溶质质量分数的关系是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','干冰易升华，升华吸热$###$二氧化碳与氢氧化钠反应，瓶内压强减小，大气压将鸡蛋压入瓶中$###$食盐溶液的溶质质量分数越大，溶液密度越大','【解答】解：（1）向装有干冰的集气瓶中加入热水，观察到有大量白雾产生，熟鸡蛋在瓶口跳动，该实验说明干冰具有升华吸热的性质；<br />（2）向集满二氧化碳的瓶中加入饱和的氢氧化钠溶液，观察到瓶口的熟鸡蛋被吞入瓶中，是因为氢氧化钠与二氧化碳反应，导致瓶内压强变小；<br />（3）将生鸡蛋放入盛有水的烧杯中，鸡蛋沉入杯底，向水中不断地加入食盐，观察到鸡蛋逐渐上浮，说明食盐溶液的溶质质量分数越大，其密度越大，故答案：<br />（1）干冰易升华，升华吸热；<br />（2）二氧化碳与氢氧化钠反应，瓶内压强减小，大气压将鸡蛋压入瓶中；<br />（3）食盐溶液的溶质质量分数越大，溶液密度越大．','【分析】根据已有的干冰升华吸热，氢氧化钠能与二氧化碳反应以及溶液密度增大，物质所受的浮力增大的知识进行分析解答即可．','简答题',3.00,'cdd831367f8226bf50621aee1e204278',9,400,'二氧化碳的物理性质,溶质的质量分数,碱的化学性质,反应现象和本质的联系','',2016,'37','2016•大兴区一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841308,'根据所学知识并结合下列仪器，回答有关问题：<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao12/24bc042e-94d4-11e9-a565-b42e9921e93e_xkb13.png\" style=\"vertical-align:middle\" /><br />（1）写出图1中标有字母的仪器的名称：a<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）实验室用A装置制取氧气，反应的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；如要制取并收集二氧化碳，可选用的装置组合是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，反应的化学方程式是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，用D装置来检验二氧化碳，装置内发生的化学方程式是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）实验室制取氢气的化学方程式是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，如果用如图1放置的E装置收集该气体，则气体从<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>端进入（填“b”或“c”）；实验室常用装置C代替装置B制取氢气，该装置的优点是通过开关活塞来控制反应的开始和停止，图2和此装置原理相同的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao1/24c0e630-94d4-11e9-a7bb-b42e9921e93e_xkb11.png\" style=\"vertical-align:middle\" /><br />（4）在分子运动实验中，为帮助同学解决认识中的误区，爱实验的小赵改进了实验酚酞+棉花氨水+棉花浓氨水酚酞溶液AB装置，如图3所示：实验过程中能观察到A试管中的实验现象是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，约4分钟后才会观察到B中现象，为能快速观察B中的现象，同学们提出用热毛巾捂住B试管，依据是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','长颈漏斗$###$2KMnO<SUB>4</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>K<SUB>2</SUB> MnO<SUB>4</SUB>+MnO<SUB>2</SUB>+O<SUB>2</SUB>↑$###$BE或CE$###$CaCO<SUB>3</SUB>+2HCl═CaCl<SUB>2</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑$###$Ca（OH）<SUB>2</SUB>+CO<SUB>2</SUB>=CaCO<SUB>3</SUB>↓+H<SUB>2</SUB>O$###$Zn+H<SUB>2</SUB>SO<SUB>4</SUB>═ZnSO<SUB>4</SUB>+H<SUB>2</SUB>↑$###$c$###$ABD$###$棉花上的酚酞很快变红了$###$温度升高，分子的运动速率变快','【解答】解：（1）实验仪器a的名称是长颈漏斗；<br />（2）高锰酸钾在加热的条件下，生成锰酸钾和二氧化锰和氧气，或氯酸钾和二氧化锰做催化剂，在加热的条件下生成氯化钾和氧气，配平即可．故答案为：2KMnO<SUB>4</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>K<SUB>2</SUB> MnO<SUB>4</SUB>+MnO<SUB>2</SUB>+O<SUB>2</SUB>↑；实验室用大理石与稀盐酸制取二氧化碳，二氧化碳的密度大于空气的密度，二氧化碳能溶于水，所以只能用向上排空气法收集，．<br />反应原理用化学方程式表示是 CaCO<SUB>3</SUB>+2HCl═CaCl<SUB>2</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑；氢氧化钙和二氧化碳反应生成碳酸钙沉淀和水；反应的化学方程式为：Ca（OH）<SUB>2</SUB>+CO<SUB>2</SUB>=CaCO<SUB>3</SUB>↓+H<SUB>2</SUB>O；<br />（3）实验室制取氢气常采用锌和稀硫酸来制取氢气，锌和稀硫酸在常温下反应生成硫酸锌和氢气，反应的化学方程式为：Zn+H<SUB>2</SUB>SO<SUB>4</SUB>═ZnSO<SUB>4</SUB>+H<SUB>2</SUB>↑．氢气的密度比空气小，如果用如图1放置的E装置收集该气体，则气体从c端进入；实验室常用装置C代替装置B制取氢气，该装置的优点是通过开关活塞来控制反应的开始和停止，图2和此装置原理相同的是ABD；<br />（4）浓氨水具有挥发性．是由于浓氨水中的氨气分子运动的快．酚酞溶液中酚酞运动慢，A试管中的棉花先变红；为能快速观察B中的现象，同学们提出用热毛巾捂住B试管，依据是温度升高，分子的运动速率变快．<br />故答案为：（1）长颈漏斗；（2）2KMnO<SUB>4</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>K<SUB>2</SUB> MnO<SUB>4</SUB>+MnO<SUB>2</SUB>+O<SUB>2</SUB>↑、BE或CE、CaCO<SUB>3</SUB>+2HCl═CaCl<SUB>2</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑、Ca（OH）<SUB>2</SUB>+CO<SUB>2</SUB>=CaCO<SUB>3</SUB>↓+H<SUB>2</SUB>O；<br />（3）Zn+H<SUB>2</SUB>SO<SUB>4</SUB>=Zn&nbsp;SO<SUB>4</SUB>+H<SUB>2</SUB>↑，C，ABD；<br />（4）棉花上的酚酞很快变红了、温度升高，分子的运动速率变快．','【分析】（1）熟悉常见仪器，根据仪器的形状和用途判断名称；<br />（2）根据发生装置的特点可判断反应物状态和反应条件；根据收集装置可判断气体的性质，综合后可判断制取的气体；熟记实验室制取二氧化碳的化学反应式；<br />（3）实验室制取氢气常采用锌和稀硫酸来制取氢气，锌和稀硫酸在常温下反应生成硫酸锌和氢气；氢气的密度比空气小，能溶于水；分液漏斗可控制反应的发生于停止；<br />（4）浓氨水具有挥发性．浓氨水中的氨气分子运动的快．酚酞溶液中酚酞运动慢，A试管中的棉花先变红；','书写',3.00,'64344caf89c9f632ae668d2450cf742c',9,400,'常用气体的发生装置和收集装置与选取方法,二氧化碳的实验室制法,二氧化碳的检验和验满,氢气的制取和检验,书写化学方程式、文字表达式、电离方程式','江阴市',2016,'32','2016•江阴市模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841316,'下列变化中属于分解反应的是（　　）','碳酸→二氧化碳+水','锌+硫酸→硫酸锌+氢气','氧化钙+水→氢氧化钙','食盐水<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td style=\"font-size: 90%\">蒸发</td></tr><tr><td><div hassize=\"9\"><div style=\"background: url(\'/upimages/images/part/8594L.png\') repeat-x; width:1px; height:5px;float:left;overflow:hidden\" muststretch=\"h\"></div><div style=\"background: url(\'/upimages/images/part/8594.png\') no-repeat; width:9px; height:5px;float:left;overflow:hidden\"></div></div></td></tr></table></span>食盐+水','','A','【解答】解：A、碳酸→二氧化碳+水，该反应符合“一变多”的特征，属于分解反应，故选项正确．<br />B、锌+硫酸→硫酸锌+氢气，该反应的反应物是两种，不符合“一变多”的特征，不属于分解反应，故选项错误．<br />C、氧化钙+水→氢氧化钙，该反应符合“多变一”的特征，属于化合反应，故选项错误．<br />D、食盐水<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td style=\"font-size: 90%\">蒸发</td></tr><tr><td><div hassize=\"9\"><div style=\"background: url(\'/upimages/images/part/8594L.png\') repeat-x; width:1px; height:5px;float:left;overflow:hidden\" muststretch=\"h\"></div><div style=\"background: url(\'/upimages/images/part/8594.png\') no-repeat; width:9px; height:5px;float:left;overflow:hidden\"></div></div></td></tr></table></span>食盐+水，没有新物质生成，属于物理变化，故选项错误．<br />故选：A．','【分析】分解反应：一种物质反应后生成两种或两种以上的物质，其特点可总结为“一变多”；据此进行分析判断．','选择题',3.00,'0049d6bd91ea4149e90d74e49e91cdc9',9,400,'分解反应及其应用','',2015,'35','2015秋•周村区校级期中',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841319,'现有以下实验仪器，请按要求填空：<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao85/24fdef30-94d4-11e9-8afc-b42e9921e93e_xkb52.png\" style=\"vertical-align:middle\" /><br />（1）配制一定质量分数的氯化钠溶液，需用到的仪器有<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>、<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>、<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填字母），还缺少的仪器是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>、<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填名称）．<br />（2）某同学利用仪器A与E组成一个贮气装置．当装满水用排水法收集甲烷（甲烷难溶于水、比空气轻）时，气体应从仪器A的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>通入；集满甲烷后将其导出使用时，水应该从仪器A的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>导入．','','','','','','C$###$D$###$F$###$玻璃棒$###$胶头滴管$###$b$###$a','【解答】解：（1）配制一定质量分数的氯化钠溶液，需用到的仪器有C、D、F，还缺少的仪器是玻璃棒、胶头滴管；（2）某同学利用仪器A与E组成一个贮气装置．当装满水用排水法收集甲烷（甲烷难溶于水、比空气轻）时，气体应从仪器A的b通入，利用排水收集气体；集满甲烷后将其导出使用时，水应该从仪器A的a导入．<br />故答为：（1）C、D、F；&nbsp;玻璃棒、胶头滴管；<br />&nbsp;（2）b；a．','【分析】（1）配制一定质量分数的氯化钠溶液，需用到的仪器有量筒、烧杯、玻璃棒、托盘天平；<br />（2）某同学利用仪器A与E组成一个贮气装置．当装满水用排水法收集甲烷（甲烷难溶于水、比空气轻）时，根据气体的性质选择导管．该装置还有其他用途，可用于气体导出．','填空题',3.00,'20a998e81d26e497e974c3e69cae28ed',9,400,'分离物质的仪器,一定溶质质量分数的溶液的配制','',2016,'32','2016•双牌县模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841322,'在硝酸钾、硫酸铵、磷酸钙、尿素[CO（NH<SUB>2</SUB>）<SUB>2</SUB>]几种化肥中，不属于盐类的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，属丁复合肥的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，可以利用简单物理方法与其他几种化肥区分开的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','尿素$###$硝酸钾$###$磷酸钙','【解答】解：尿素不属于盐类；硝酸钾中含有钾和氮元素，属于复合肥、硫酸铵中含有氮元素，属于氮肥、磷酸钙中含有磷元素，属于磷肥、尿素[CO（NH<SUB>2</SUB>）<SUB>2</SUB>]中含有氮元素属于氮肥；磷矿粉是灰白色的，其它都是白色的晶体．<br />故答案为：尿素；硝酸钾；磷酸钙．','【分析】盐是指电离出金属离子和酸根离子的化合物；含有氮元素的肥料称为氮肥；含有磷元素的肥料称为磷肥；含有钾元素的肥料称为钾肥；同时含有氮、磷、钾三种元素中的两种或两种以上的肥料称为复合肥，据此分析；磷矿粉是灰白色的，其它都是白色的晶体．','填空题',3.00,'0183fd8d56cbb0694d854c032e07a811',9,400,'常见化肥的种类和作用,化肥的简易鉴别,常见的氧化物、酸、碱和盐的判别','',2016,'37','2016•卧龙区一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841326,'<img  src=\"/tikuimages/9/2016/400/shoutiniao33/251065c0-94d4-11e9-99bb-b42e9921e93e_xkb27.png\" style=\"vertical-align:middle;FLOAT:right\" />（2016•甘肃模拟）马蹄飞燕为东汉青铜器，1969年出土于甘肃省武威雷台的东汉墓，现藏于甘肃省博物馆，1983年10月，马蹄飞燕被国家旅游局确定为中国旅游标志，1986年被定为国宝级文物．<br />（1）马蹄飞燕全身都有绿色铜锈，铜锈的化学式为[Cu<SUB>2</SUB>（OH）<SUB>2</SUB>CO<SUB>3</SUB>]，据此推断铜生锈与空气中的水蒸气、<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>、<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>有关．<br />（2）请你写出一条保护和节约金属资源的有效方法<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）除铜锈和除铁锈的试剂相同，据此可选用<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>除去铜锈．<br />（4）青铜是一种合金，一般情况下合金与其组成金属相比具有的一个优点是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','氧气$###$二氧化碳$###$防止生锈$###$稀盐酸$###$熔点低或硬度大或抗腐蚀性能强','【解答】解：（1）由题意可知：铜在潮湿空气中会锈蚀，铜锈的化学成分是碱式碳酸铜，碱式碳酸铜是由铜、氢、氧、碳四种元素的组成的物质．由质量守恒定律及空气的成分可知，推测铜生锈与空气中的氧气、水有关，还与二氧化碳有关．故填：氧气；二氧化碳；<br />（2）防止金属锈蚀是保护和节约金属资源的有效的方法，故填：防止生锈；<br />（3）根据铜绿的组成可知，若要除去铜钱表面的铜绿，可选用的试剂为稀盐酸；故填：稀盐酸；<br />（4）一般说来合金的硬度一般比各成分金属大，多数合金的熔点低于组成它的成分金属的熔点，抗腐蚀性能增强；故填：熔点低或硬度大或抗腐蚀性能强．','【分析】（1）根据质量守恒定律及空气的成分进行分析；<br />（2）根据保护金属资源的方法进行分析解答即可；<br />（3）根据铜绿的组成及性质选用试剂除去铜绿；<br />（4）根据合金的优点进行分析、考虑，从而得出正确的结论．','填空题',3.00,'168a1d7a757f25853db93db94a2a701e',9,400,'合金与合金的性质,金属锈蚀的条件及其防护,金属资源的保护','',2016,'32','2016•甘肃模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841331,'下列有关CO<SUB>2</SUB>和CO的叙述，正确的是（　　）','CO<SUB>2</SUB>、CO都能用于灭火','二者的组成元素相同，密度也相同','CO<SUB>2</SUB>能使澄清石灰水变浑浊，而CO不能','CO、CO<SUB>2</SUB>在空气中含量高都能使人死亡，因为它们都有毒性','','C','【解答】解：A、二氧化碳能灭火，一氧化碳不能灭火，故选项错误；<br />B、二者的组成元素相同，密度不相同，故选项错误；<br />C、CO<SUB>2</SUB>能使澄清石灰水变浑浊，而CO没有此性质，故选项正确；<br />D、一氧化碳有毒，会污染空气，因此需要进行尾气处理，故选项错误．<br />故选C．','【分析】一氧化碳的化学性质是：可燃性、还原性和毒性．二氧化碳的化学性质有：既不能燃烧也不能支持燃烧，也不供给呼吸；能与水反应生成碳酸；能使澄清的石灰水变浑浊．','选择题',3.00,'328b48b7c1674dbaee3bed9a0952f42e',9,400,'二氧化碳的物理性质,二氧化碳的化学性质,一氧化碳的物理性质,一氧化碳的化学性质,一氧化碳的毒性','',2015,'37','2015秋•唐山校级月考',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841336,'空气是人类宝贵的自然资源，下列有关空气的说法正确的是（　　）','洁净的空气是纯净物','空气中的氧气可以供给呼吸、燃烧、炼钢、化工生产等','新鲜空气中不含二氧化碳','分离液态空气得到氧气和氮气的过程中发生了化学反','','B','【解答】解：A．洁净的空气是由氮气、氧气、稀有气体等多种气体组成的，属于混合物，故错误；<br />B．空气中的氧气可以供给呼吸、燃烧、炼钢、化工生产等，故正确；<br />C．新鲜的空气中含有氮气、氧气、稀有气体、二氧化碳等多种物质，故错误；<br />D．空气主要是由氮气和氧气组成的，分离液态空气法得到氮气和氧气属于物理变化，故错误．<br />故选B．','【分析】A．根据空气的组成来分析；<br />B．根据空气的用途来分析；<br />C．根据空气的组成来分析；<br />D．根据工业制氧气的原理来分析．','选择题',3.00,'d1dfda5f05a9e113749d9a6bc8a3569e',9,400,'空气的成分及各成分的体积分数,氧气的用途,常见的溶剂,纯净物和混合物的判别','',2016,'32','2016•河北模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841343,'化学课外小组的同学想要用一氧化碳气体（含有少量杂质二氧化碳和水蒸气）测定样品中氧化铁的质量分数，装置如图所示，回答下列问题：<img  src=\"/tikuimages/9/2016/400/shoutiniao64/2574a5cf-94d4-11e9-854a-b42e9921e93e_xkb57.png\" style=\"vertical-align:middle\" /><br />（1）乙装置的作用是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）丙装置中反应的化学方程式是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）丁装置中的现象是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（4）戊装置的作用是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（5）要生产1000t含杂质3%的生铁，需要含Fe<SUB>2</SUB>O<SUB>3</SUB>90%的铁矿石<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>t（保留整数）．','','','','','','吸收水蒸气$###$3CO+Fe<SUB>2</SUB>O<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2Fe+3CO<SUB>2</SUB>$###$石灰水变浑浊$###$处理尾气$###$1540','【解答】解：（1）浓硫酸能吸收气体中的水分，所以乙装置的作用是：吸收水蒸气；<br />（2）一氧化碳和氧化铁在高温的条件下反应生成铁和二氧化碳，化学方程式是：3CO+Fe<SUB>2</SUB>O<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2Fe+3CO<SUB>2</SUB>；<br />（3）生成的二氧化碳能使澄清石灰水变浑浊，所以丁装置中的现象是：石灰水变浑浊；<br />（4）一氧化碳有毒，不能直接排放到空气中，需进行尾气处理，所以戊装置的作用是处理尾气；<br />（5）需要含Fe<SUB>2</SUB>O<SUB>3</SUB>90%的铁矿石x<br />3CO+Fe<SUB>2</SUB>O<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2Fe+3CO<SUB>2</SUB>，<br />&nbsp;&nbsp;&nbsp; 160&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 112<br />&nbsp;&nbsp; 90%×x&nbsp;&nbsp;&nbsp;&nbsp; 1000t×（1-3%）<br />&nbsp;&nbsp; <span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">160</td></tr><tr><td>90%×x</td></tr></table></span>=<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">112</td></tr><tr><td style=\"padding-top:1px;font-size:90%\">1000t×(1-3%)</td></tr></table></span><br />&nbsp;&nbsp;&nbsp; x=1540t<br />故答案为：（1）吸收水蒸气；<br />（2）3CO+Fe<SUB>2</SUB>O<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2Fe+3CO<SUB>2</SUB>；<br />（3）石灰水变浑浊；<br />（4）处理尾气；<br />（5）1540t．','【分析】（1）根据浓硫酸有吸水性，能做干燥剂进行分析；<br />（2）根据一氧化碳和氧化铁在高温的条件下反应生成铁和二氧化碳进行分析；<br />（3）根据二氧化碳能使澄清石灰水变浑浊进行分析；<br />（4）根据一氧化碳有毒，不能直接排放到空气中，需进行尾气处理进行分析；<br />（5）根据一氧化碳和氧化铁在高温的条件下反应生成铁和二氧化碳，依据题中的数据进行计算．','书写',3.00,'bda5de8485bb6962a6f66a713b154ad9',9,400,'常见气体的检验与除杂方法,一氧化碳还原氧化铁,含杂质物质的化学反应的有关计算,书写化学方程式、文字表达式、电离方程式','',2016,'37','2016•河北区一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841344,'如图实验操作错误的是（　　）','<img  src=\"/tikuimages/9/2016/400/shoutiniao12/257ac04f-94d4-11e9-95e9-b42e9921e93e_xkb30.png\" style=\"vertical-align:middle\" /><br />滴加液体','<img  src=\"/tikuimages/9/2016/400/shoutiniao12/257cbc1e-94d4-11e9-8583-b42e9921e93e_xkb0.png\" style=\"vertical-align:middle\" /><br />&nbsp;&nbsp;&nbsp; 添加酒精','<img  src=\"/tikuimages/9/2016/400/shoutiniao88/257fc95e-94d4-11e9-b9e9-b42e9921e93e_xkb73.png\" style=\"vertical-align:middle\" /><br />&nbsp;&nbsp;&nbsp; 加热液体','<img  src=\"/tikuimages/9/2016/400/shoutiniao48/2580b3c0-94d4-11e9-ac9a-b42e9921e93e_xkb92.png\" style=\"vertical-align:middle\" /><br />&nbsp;&nbsp;&nbsp;&nbsp; 放置试管','','B','【解答】解：A、胶头滴管不能伸入试管内，应垂直悬空滴加，故A正确；<br />B、不能向燃着的酒精灯内添加酒精，以免引起失火，故B错误；<br />C、给试管中的液体加热时，用酒精灯的外焰加热试管里的液体，且液体体积不能超过试管容积的<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">1</td></tr><tr><td>3</td></tr></table></span>，故C正确；<br />D、洗涤完试管要倒扣在试管架上，故D正确．<br />故答案为：B．','【分析】A、根据使用胶头滴管滴加少量液体的方法进行分析判断．<br />B、根据酒精灯的使用方法进行分析判断．<br />C、根据给试管中的液体加热的方法进行分析判断．<br />D、根据洗涤完试管要倒扣在试管架上分析判断．','选择题',3.00,'73431eebd76bd7f3e58bb758162a36c5',9,400,'加热器皿-酒精灯,液体药品的取用,给试管里的液体加热,玻璃仪器的洗涤','',2016,'37','2016•扬州校级二模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841358,'吸烟有害健康，香烟燃烧产生的烟气中含有尼古丁（化学式为C<SUB>10</SUB>H<SUB>14</SUB>N<SUB>2</SUB>）、焦油和CO等物质．下列说法中正确的是（　　）','尼古丁、焦油和CO都是有机化合物','尼古丁中有10个碳原子、14个氢原子和2个氮原子','尼古丁中氢元素的质量分数为17.3%','CO与血液中的血红蛋白结合，会造成人体缺氧','','D','【解答】解：A．CO虽然含有碳元素，但是其性质与无机物类似，归为无机物，故错误；<br />B．尼古丁是由分子构成的，而不是由原子直接构成的，故错误；<br />C．尼古丁中氢元素的质量分数为<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">1×14</td></tr><tr><td>12×10+1×14+14×2</td></tr></table>×100%</span>≈8.6%，故错误；<br />D．一氧化碳极易与血液里的血红蛋白结合，使人体缺氧、窒息，甚至死亡，故正确．<br />故选D．','【分析】A．根据有机物的概念来分析；<br />B．根据尼古丁的结构来分析；<br />C．根据化合物中元素的质量分数来分析；<br />D．一氧化碳极易与血液里的血红蛋白结合，使人体缺氧．','选择题',3.00,'caadc75aa5d6d3f5ba940e1fab2c2dbd',9,400,'一氧化碳的毒性,有机物与无机物的区别,化学式的书写及意义,元素的质量分数计算','昆山市',2016,'37','2016•昆山市二模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841363,'2015年扬州市按空气质量标准评价达标天数比例为67.3%，共238天，空气质量轻度污染占25.1%，中度污染占5.6%，重度污染占2.0%．下列做法不利于改善扬州空气质量指数的是（　　）','焚烧秸秆增加肥力','城区严禁燃放烟花爆竹','大力发展电力公交','关闭燃煤小锅炉','','A','【解答】解：A、大量燃烧植物秸秆，会造成空气的污染，不利于保护环境，故符合题意；<br />B、城区严禁燃放烟花爆竹，利于改善扬州空气质量指数，故不符合题意；<br />C、大力发展电力公交，有利于保护大气环境，提高空气的质量，故不符合题意；<br />D、关闭燃煤小锅炉，有利于保护大气环境，提高空气的质量，故不符合题意；<br />故选：A．','【分析】空气污染指标是衡量一个地区空气质量的重要指标，污染指数越高空气质量越差．','选择题',3.00,'c2ba8b0faca545ac8e089553bbe8a6f8',9,400,'防治空气污染的措施','仪征市',2016,'37','2016•仪征市一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841367,'下列实验数据合理的是（　　）','用托盘天平称得某纯碱样品的质量为10.57&nbsp;g','用10&nbsp;mL量筒量取了6.53&nbsp;mL水','用20&nbsp;g氯化钠和100&nbsp;g水配制的120&nbsp;g食盐水，其溶质质量分数为20%','用pH试纸测得某地雨水的pH为5','','D','【解答】解：A、托盘天平用于粗略称量药品的质量，能准确到0.1g，不能精确到0.01g，不能用托盘天平称得某纯碱样品的质量为10.57g，故选项实验数据不合理．<br />B、选取量筒时，尽量选用能一次量取的最小规格的量筒，应使用10mL的量筒；且10mL量筒的精确度为0.1mL，不能用10mL量筒量取6.53mL的水，故选项实验数据不合理．<br />C、用20g氯化钠和100g水配制的120g食盐水，其溶质质量分数为<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">20g</td></tr><tr><td>120g</td></tr></table>×</span>100%＜20%，故选项实验数据不合理．<br />D、由于标准比色卡上的数字只有整数，可用用pH试纸测得某地雨水的pH为5，故选项实验数据合理．<br />故选：D．','【分析】A、托盘天平用于粗略称量药品的质量，能准确到0.1g，不能精确到0.01g．<br />B、根据10mL量筒的精确度为0.1mL，据此进行分析判断．<br />C、根据溶质质量分数=<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">溶质质量</td></tr><tr><td>溶液质量</td></tr></table></span>×100%，进行分析判断．<br />D、pH试纸上的标准比色卡上的数字只有整数，即使用pH试纸所测得的溶液酸碱度为整数．','选择题',3.00,'39ab6e1952ce46f37ba539cd2786fdb9',9,400,'实验数据处理或者误差分析的探究,测量容器-量筒,称量器-托盘天平,溶液的酸碱度测定,溶质的质量分数','',2016,'37','2016•江门一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841378,'<img  src=\"/tikuimages/9/2016/400/shoutiniao16/25e2f800-94d4-11e9-9f0d-b42e9921e93e_xkb30.png\" style=\"vertical-align:middle;FLOAT:right;\" />（2016•荔湾区模拟）小明同学为了研究影响铁生锈的因素，进行了如下实验：<br />在A、B、C、D四只烧瓶中分别放入干燥的细铁丝、浸过食盐水的细铁丝、浸过清水的细铁丝、食盐水及细铁丝（使细铁丝完全浸没在食盐水中）（四只烧瓶中细铁丝的质量均相同），然后装配成如图所示的四套装置，每隔0.5小时测量导管中水面上升的高度，结果如下表所示：<br /><table class=\"edittable\"><TBODY><TR><td width=59>烧瓶号</TD><td width=213><br />时间/h<br />导管中水面上升高度/cm</TD><td width=33>0</TD><td width=39>0.5</TD><td width=39>1.0</TD><td width=39>1.5</TD><td width=39>2.0</TD><td width=39>2.5</TD><td width=39>3.0</TD></TR><TR><td>A瓶</TD><td>盛干燥的细铁丝</TD><td>0</TD><td>0</TD><td>0</TD><td>0</TD><td>0</TD><td>0</TD><td>0</TD></TR><TR><td>B瓶</TD><td>盛浸过食盐水的细铁丝</TD><td>0</TD><td>0.4</TD><td>1.2</TD><td>3.4</TD><td>5.6</TD><td>7.6</TD><td>9.8</TD></TR><TR><td>C瓶</TD><td>盛浸过清水的细铁丝</TD><td>0</TD><td>0</TD><td>0</TD><td>0.3</TD><td>0.8</TD><td>2.0</TD><td>3.5</TD></TR><TR><td>D瓶</TD><td>盛完全浸没在食盐水中的细铁丝</TD><td>0</TD><td>0</TD><td>0</TD><td>0</TD><td>0</TD><td>0</TD><td>0</TD></TR></TBODY></TABLE>请回答：<br />（1）以上实验中，铁生锈的速率最大的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填烧瓶号）．<br />（2）导管中水面上升的原因是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）以上实验说明，影响铁生锈的因素是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（4）为了防止钢铁锈蚀，人们常采用在其表面涂刷矿物油或镀上其他金属等覆盖保护膜的方法，这些方法的共同原理是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','B$###$烧瓶中的氧气参加反应，气体减少，压强降低$###$水和氧气，食盐能够促进铁生锈$###$隔绝氧气和水','【解答】解：（1）由表中数据可知，B瓶导管中水面上升的高度最大，说明消耗的氧气最多，生锈的速率最大．<br />故填：B．<br />（2）铁生锈时，消耗烧瓶中的氧气，氧气减少后，烧瓶中的压强减小，在外界大气压的作用下，导管中的水上升，形成一段液柱．<br />故填：烧瓶中的氧气参加反应，气体减少，压强降低．<br />（3）A瓶中的铁只与氧气接触，不容易生锈；D瓶中的铁只与水接触，不容易生锈；C瓶中的铁与水和氧气接触，容易生锈；通过对比可知，铁与水和氧气同时接触时才容易生锈．<br />B瓶中的铁与水和氧气同时接触，同时存在食盐，生锈速率大于C瓶中的铁，说明食盐可以促进铁生锈；<br />以上实验说明，影响铁生锈的因素是水和氧气，食盐能够促进铁生锈．<br />故填：水和氧气，食盐能够促进铁生锈．<br />（4）使钢铁与水和氧气隔绝，可以防止生锈；在钢铁表面涂刷矿物油或镀上其他金属等覆盖保护膜的方法可以防止生锈，这些方法的共同原理是隔绝氧气和水．<br />故填：隔绝氧气和水．','【分析】从导管中水面上升的高度可以判断铁生锈的条件、生锈的速率，根据铁生锈的条件可以分析出防止铁生锈的方法．','填空题',3.00,'5f7e192c3f25c9e890bebe8c33b7b1c3',9,400,'探究金属锈蚀的条件','',2016,'32','2016•荔湾区模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841386,'下列说法正确的是（　　）','蜡烛刚刚熄灭时产生的白烟是二氧化碳气体','氢氧化钠固体易潮解，可做某些气体的干燥剂','二氧化碳不供给呼吸，不宜作为植物的养料','大量使用化石燃料会造成大气污染，应该停止对化石燃料的开采和利用','','B','【解答】解：A、蜡烛刚刚熄灭时产生白烟，白烟是石蜡蒸气凝成的石蜡固体，故选项说法错误．<br />B、氢氧化钠固体易潮解，具有吸水性，可做某些气体的干燥剂，故选项说法正确．<br />C、二氧化碳不供给呼吸，但能参与植物的光合作用，可作为植物的养料，故选项说法错误．<br />D、大量使用化石燃料会造成大气污染，但现阶段当今世界的主要燃料是化石燃料，停止对化石燃料的开采和利用是不现实的，故选项说法错误．<br />故选：B．','【分析】A、根据蜡烛刚刚熄灭时的现象，进行分析判断．<br />B、根据氢氧化钠固体易潮解，进行分析判断．<br />C、根据二氧化碳的用途，进行分析判断．<br />D、根据现阶段当今世界的主要燃料是化石燃料，进行分析判断．','选择题',3.00,'edd2e6c0cd87dad09952b9878fc47d3d',9,400,'蜡烛燃烧实验,二氧化碳的用途,常见碱的特性和用途,化石燃料及其综合利用','',2016,'32','2016•漳州模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841398,'随着雾霾天气的频繁出现，人们对空气质量的关注度越来越高了，各地都发布空气质量报告，以提高公民环保意识和自觉改变行为的主动性．<br />（1）下列属于空气质量报告中的主要内容是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填字母序号）；<br />①首要污染物②空气污染指数③空气质量状况④空气质量级别<br />A．①②B．②③C．①②④D．①②③④<br />（2）空气质量报告中，计入空气污染指数的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />①悬浮颗粒（粉尘）②臭氧③二氧化碳④一氧化碳<br />⑤可吸入颗粒物⑥二氧化硫⑦稀有气体⑧氮的氧化物（二氧化氮等）<br />A．①②④⑥⑦B．①②④⑤⑥⑧C．①④⑤⑥⑧D．①②③④⑥<br />（3）SO<SUB>2</SUB>是导致酸雨的有害气体之一．SO<SUB>2</SUB>中硫元素的化合价为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />A．+2B．+4C．+6D．-2．<br />（4）PM2.5是指大气中直径不超过2.5μm的可吸入颗粒物，也称为可入肺颗粒物，对人体健康和大气环境质量的影响更大．下列措施对PM2.5的治理能起到积极作用的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />①城市增加楼顶绿化&nbsp;②加强建筑工地、道路扬尘监管<br />③发展城市公共轨道交通&nbsp;④严厉查处焚烧垃圾、秸秆<br />A．①②B．①②③C．①③④D．①②③④<br />（5）经研究发现，PM2.5的增加是来自煤等化石燃料的燃烧，为减少污染，提高煤的利用率，可将煤转化为可燃性气体，此过程是碳与水蒸汽在高温的反应，其微观示意图如下：<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao97/263584d1-94d4-11e9-9472-b42e9921e93e_xkb34.png\" style=\"vertical-align:middle\" /><br />写出该反应的化学方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（6）医生建议雾霾天气最好减少外出，若要外出，必须戴好口罩．常用口罩的孔径如下：①普通16层纱布口罩的孔径在100微米左右&nbsp;②单层无纺布口罩的孔径在10微米左右<br />③N95专业口罩的孔径在0.1微米左右．使用口罩最好选择<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填序号），其中口罩的作用相当于水净化中的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>操作．','','','','','','D$###$B$###$B$###$D$###$C+H<SUB>2</SUB>O<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>CO+H<SUB>2</SUB>$###$③$###$过滤','【解答】解：（1）空气质量报告的主要内容包括：空气污染指数；首要污染物；空气质量级别；空气质量状况等．<br />故选：D；<br />（2）目前计入空气污染指数的项目暂定为：二氧化硫，一氧化碳，二氧化氮，可吸入颗粒物和臭氧等．<br />故选：B；<br />（3）根据在化合物中正负化合价代数和为零，可得SO<SUB>2</SUB>中硫元素的化合价为：x+（-2）×2=0，解答x=+4；<br />故选B．<br />（4）PM2.5是指大气中直径小于或等于2.5微米的颗粒物，也称为可入肺颗粒物可以知道对PM2.5的治理即是为了降低可吸入颗粒物的排放，①城市增加楼顶绿化&nbsp;②加强建筑工地、道路扬尘监管<br />③发展城市公共轨道交通&nbsp;④严厉查处焚烧垃圾、秸秆，以上都可降低可吸入颗粒物的排放；<br />故选：D；<br />（5）根据信息“碳与水的反应”和观察图示可知，碳与水的反应，生成一氧化碳和氢气，反应方程式是：C+H<SUB>2</SUB>O<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>CO+H<SUB>2</SUB>（反应在高温下进行，所以水为水蒸气状态，生成的氢气和一氧化碳都不能加气体符号）；<br />故答案为：C+H<SUB>2</SUB>O<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>CO+H<SUB>2</SUB>．<br />（6）本题中的三种口罩中，第三种口罩质量轻，是高科技．口罩的作用是过滤出空气中的可吸入颗粒物．<br />故答案为：③；过滤．','【分析】（1）空气质量报告的主要内容包括：空气污染指数；首要污染物；空气质量级别；空气质量状况等．<br />（2）根据计入空气污染指数的项目的种类分析，目前计入空气污染指数的项目暂定为：二氧化硫，一氧化碳，二氧化氮，可吸入颗粒物．<br />（3）根据在化合物中正负化合价代数和为零，进行解答本题．<br />（4）运用题中信息PM2.5是指大气中直径小于或等于2.5微米的颗粒物，也称为可入肺颗粒物可以知道对PM2.5的治理即是为了降低可吸入颗粒物的排放解答．<br />（5）观察图示，写出化学方程式，判断反应类型．<br />（6）从过滤的有关知识分析．','书写',3.00,'a1900b518358b632a6a64eeaf97bceea',9,400,'过滤的原理、方法及其应用,空气的污染及其危害,防治空气污染的措施,微粒观点及模型图的应用,有关元素化合价的计算,书写化学方程式、文字表达式、电离方程式','招远市',2016,'35','2016春•招远市期中',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841409,'我们的初中化学实验中有很多都与气压变化有关，请回答下列问题：<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao48/266cc170-94d4-11e9-b44a-b42e9921e93e_xkb61.png\" style=\"vertical-align:middle\" /><br />（1）用如图A装置测定空气中氧气的 体积分数，请写出化学方程式：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，充分反应后，待钟罩内液面不再变化，应先<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，再读数，否则测得结果将<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">1</td></tr><tr><td>5</td></tr></table></span>（填“大于”、“等于”或“小于”）．<br />（2）实验室可用注射器连接到装置 B 的导管处，检查装置B的气密性．实验步骤：向锥形瓶中加水至<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，向后拉动注射活塞，若观察到<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>时，则说明装置B气密性良好．<br />（3）实验室利用装置C制备H<SUB>2</SUB>，化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，相比较装置B，它的优点是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，若选用装置D、E测定生成H<SUB>2</SUB>的体积，仪器连接顺序为a→<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>→<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>→d．<br />（4）装置F和G可用于验证NH<SUB>3</SUB>极易溶于水，F装置的实验现象是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，G装置的实验现象是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />（5）在装置H中，某样品投入到水中，右侧U型管出现液面左低右高现象，则该样品不可能是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />①CaO&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;②NaOH固体&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;③浓硫酸&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;④NH<SUB>4</SUB>NO<SUB>3</SUB>固体．','','','','','','4P+5O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2P<SUB>2</SUB>O<SUB>5</SUB>$###$冷却后$###$小于$###$浸没长颈漏斗末端$###$长颈漏斗末端有气泡逸出$###$Zn+H<SUB>2</SUB>SO<SUB>4</SUB>═ZnSO<SUB>4</SUB>+H<SUB>2</SUB>↑$###$能随时控制反应进行或停止$###$c$###$b$###$形成喷泉$###$气球胀大$###$④','【解答】解：<br />（1）盛在A中的物质常用红磷；红磷燃烧生成五氧化二磷，化学方程式为：4P+5O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2P<SUB>2</SUB>O<SUB>5</SUB>．充分反应后，待钟罩内液面不再变化，应先冷却后再读数，否则测得结果将小于<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">1</td></tr><tr><td>5</td></tr></table></span>；<br />（2）用注射器C检查装置E的气密性时，向锥形瓶中加入少量水至浸没长颈漏斗末端，将注射器F连接到装置B的导管口处，缓慢拉动注射器的活塞，观察到长颈漏斗末端有气泡逸出，表示装置B的气密性良好；<br />（3）实验室用锌粒和稀硫酸反应制取氢气，反应的化学方程式为：Zn+H<SUB>2</SUB>SO<SUB>4</SUB>═ZnSO<SUB>4</SUB>+H<SUB>2</SUB>↑．相比较装置B，它的优点是能随时控制反应进行或停止；若选用装置D、E测定生成H<SUB>2</SUB>的体积，仪器连接顺序为a→c→b→d；<br />（4）氨气易溶于水，F装置内气压变小，实验现象是形成喷泉；G装置的实验现象是气球胀大；<br />（5）右边支管的红墨水液面上升，说明装置内气体压强增大，由于热胀冷缩的原理，可知物质溶于水放热，使瓶内气体膨胀，压强增大．<br />A、CaO溶于水放出大量的热，正确；<br />&nbsp;B、氢氧化钠固体溶于水放出大量的热，正确；<br />C、浓硫酸溶于水放出大量的热，正确；<br />D、硝酸铵溶于水温度降低．错误；<br />故选B．<br />答案：<br />（1）4P+5O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2P<SUB>2</SUB>O<SUB>5</SUB>；冷却后；小于；<br />（2）浸没长颈漏斗末端；长颈漏斗末端有气泡逸出；<br />（3）Zn+H<SUB>2</SUB>SO<SUB>4</SUB>═ZnSO<SUB>4</SUB>+H<SUB>2</SUB>↑；能随时控制反应进行或停止；c；b；<br />（4）形成喷泉；气球胀大；<br />（5）④．','【分析】（1）空气中氧气含量的测定实验的原理是通过可燃物燃烧消耗氧气，从而通过空气体积变化来得到氧气的含量，所以要让氧气尽可能被完全消耗是实验成败的关键，用红磷测定空气组成的实验原理、操作、现象、结论和注意事项，来正确解答本题；<br />（2）根据用注射器F检查装置E的气密性时，向锥形瓶中加入少量水至浸没长颈漏斗下端，将注射器F连接到装置B的导管口处，缓慢拉动注射器F的活塞，观察现象；<br />（3）根据不同的实验装置，功能不同；氢气的密度比水的小解答；<br />（4）根据氨气易溶于水解答；<br />（5）根据物质溶于水时的吸热和放热现象进行分析，U形管右侧的液面明显上升，说明瓶内的气体压强增大；氢氧化钠溶于水时放热．','书写',3.00,'5c0c5c3a77e376fce2cc77586d5d7e62',9,400,'量气装置,检查装置的气密性,空气组成的测定,溶解时的吸热或放热现象,氢气的制取和检验,生石灰的性质与用途,书写化学方程式、文字表达式、电离方程式','',2016,'37','2016•扬州校级二模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841415,'新发布的《环境空气质量标准》自2016年1月1日起在全国实施，新标准增加了PM2.5监测指标．下列措施不利于减少PM2.5污染的是（　　）','增加私家车用量，提高生活质量','大力发展风能，减少火力发电','农作物秸秆多翻土还田，不露天焚烧','居民多用天然气，少用煤作燃料','','A','【解答】解：A、增加私家车用量，会加重空气污染，增加PM2.5，故符合题意；<br />B、大力发展风能，减少火力发电，可减少空气污染，利于PM2.5治理，故不符合题意；<br />C、农作物秸秆多翻土还田，不露天焚烧，可减少空气污染，利于PM2.5治理，故不符合题意；<br />D、居民多用天然气，少用煤作燃料，可减少空气污染，利于PM2.5治理，故不符合题意；<br />故选A．','【分析】PM2.5是指大气中直径不超过2.5μm的颗粒物，主要来源是化石燃料的燃烧和扬尘，根据PM2.5的来源进行分析判断所采取的做法是否合理．','选择题',3.00,'9b76a8be4ca7abfa6711026e7f163759',9,400,'防治空气污染的措施','',2016,'37','2016•宛城区一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841421,'实验室常用图1装置制取气体，请你根据所学知识回答下列问题．<br />（1）仪器a的名称是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，a应在固定试管之<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（选填“前”或“后”）放置．<br />（2）实验室制取氧气时，常用A装置制取并在试管口塞一团棉花，其化学反应方程式是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；若选用B装置制取，其化学反应方程式是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．选用的收集装置是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填字母序号）若改用如图2装置收集气体，则气体应从<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>进（填“a”或“b”）<br />（3）甲同学在实验室制取二氧化碳的研究中，进行了如下实验：<br /><table class=\"edittable\"><TBODY><TR><td width=114>&nbsp;药品实验编号</TD><td width=114>&nbsp;甲</TD><td width=114>&nbsp;乙</TD><td width=114>&nbsp;丙</TD><td width=114>&nbsp;丁</TD></TR><TR><td>&nbsp;大理石</TD><td>&nbsp;mg，块状</TD><td>&nbsp;mg，块状</TD><td>&nbsp;mg，粉末状</TD><td>&nbsp;mg，粉末状</TD></TR><TR><td>&nbsp;盐酸（过量）</TD><td>&nbsp;wg，稀盐酸</TD><td>&nbsp;wg，浓盐酸</TD><td>&nbsp;wg，稀盐酸</TD><td>&nbsp;wg，浓盐酸</TD></TR></TBODY></TABLE>Ⅰ．上述实验中反映的化学方程式是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />Ⅱ．若要研究盐酸浓度大小对反应的影响，可选择实验甲与<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>对照（选填实验编号）．<br />Ⅲ．除盐酸的浓度外，上述实验研究的另一个影响反应的因素是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />Ⅳ．研究发现酸的浓度越大，产生的气体的速度越快，与甲比较，对丁分析正确的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />A．反应更为剧烈&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; B．最终剩余溶液的质量更小<br />C．产生的二氧化碳的质量更大&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; D．粉末状大理石利用率更高<br />（4）乙同学也取了一定浓度的盐酸与石灰石反应制取气体，并将生成的气体通入澄清石灰水中，未见变浑浊．为探究其原因，进行了如下过程：<br />【作出猜想】<br />A．石灰水已经完全变质；&nbsp; B．气体中有挥发出来的HCl气体．<br />【实验探究】<br />①取少量石灰水于试管中，并向试管中滴加几滴无色酚酞试液，振荡，观察到溶液变红色．于是排除了猜想A，你认为排除猜想A的理由是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />②乙同学想探究所得气体中是否有挥发出来的HCl气体，设计如下实验．请你帮他完成实验，填写以下表格：<br /><table class=\"edittable\"><TBODY><TR><td width=287>实验步骤</TD><td width=86>实验现象</TD><td width=97>实验结论</TD></TR><TR><td><u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD><td><u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD><td>气体中有HCl</TD></TR></TBODY></TABLE>【查阅资料】<br />碳酸钠溶液中通入二氧化碳发生反应：Na<SUB>2</SUB>CO<SUB>3</SUB>+CO<SUB>2</SUB>+H<SUB>2</SUB>O=2NaHCO<SUB>3</SUB><br />【拓展延伸】<br />如果用上述原料制得纯净的二氧化碳气体，所选仪器的连接顺序为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填字母）．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao32/26909d1e-94d4-11e9-a5b6-b42e9921e93e_xkb5.png\" style=\"vertical-align:middle\" />','','','','','','酒精灯$###$前$###$2KMnO<SUB>4</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>K<SUB>2</SUB>MnO<SUB>4</SUB>+MnO<SUB>2</SUB>+O<SUB>2</SUB>↑$###$2H<SUB>2</SUB>O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span>&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O+O<SUB>2</SUB>↑$###$C$###$b$###$CaCO<SUB>3</SUB>+2HCl=CaCl<SUB>2</SUB>+CO<SUB>2</SUB>↑+H<SUB>2</SUB>O$###$乙$###$固体反应物的颗粒大小（或反应物的接触面积）；$###$AB$###$酚酞试液变红，说明溶液中存在碱性物质Ca（OH）<SUB>2</SUB>$###$将生成的气体通入AgNO<SUB>3</SUB>溶液中$###$产生白色沉淀$###$BHEC','【解答】解：（1）通过分析题中所指仪器的作用可知，a是酒精灯，连接仪器时应该按照从下到上的顺序，所以酒精灯应在固定试管之前放置；<br />（2）A装置制取并在试管口塞一团棉花，说明是用加热高锰酸钾的方法制取氧气，反应的化学方程式是2KMnO<SUB>4</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>K<SUB>2</SUB>MnO<SUB>4</SUB>+MnO<SUB>2</SUB>+O<SUB>2</SUB>↑，B装置属于固液常温型，用过氧化氢制取氧气可用此发生装置，反应的化学方程式是：2H<SUB>2</SUB>O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span>&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O+O<SUB>2</SUB>↑；氧气的密度比空气大，不易溶于水，所以收集氧气可以用排水法和向上排空气法进行收集，选用的收集装置是C，氧气密度比水小，所以改用图2装置收集氧气，则气体应从b进入；<br />（3）Ⅰ、碳酸钙和盐酸反应生成氯化钙、水和二氧化碳，化学方程式为：CaCO<SUB>3</SUB>+2HCl=CaCl<SUB>2</SUB>+CO<SUB>2</SUB>↑+H<SUB>2</SUB>O；<br />Ⅱ、对比实验数据可见，甲和乙均为mg块状大理石，wg盐酸，但是盐酸的浓度不同；<br />Ⅲ、甲和丙对比，乙和丁对比可知，实验研究的另一个影响反应的因素是固体反应物的颗粒大小（或反应物的接触面积）；故填：<br />Ⅳ、甲和丁对比可知，丁中反应之间的接触面积大，盐酸浓度大，所以反应的速率更快；<br />丁为浓盐酸，具有挥发性，所以溶液的质量更小；<br />因为并不知道大理石与盐酸是否完全反应还是哪一种物质有剩余，故无法判断产生二氧化碳的质量粉末状大理石利用率．<br />故填：ab；<br />（4）【实验探究】①碱能使酚酞变红色，所以排除猜想A的理由是：酚酞试液变红，说明溶液中存在碱性物质Ca（OH）<SUB>2</SUB>；<br />②氯离子和银离子会生成氯化银沉淀，氯化氢溶于水形成盐酸，所以<br /><table class=\"edittable\"><TBODY><TR><td width=225>实验步骤</TD><td width=120>实验现象</TD><td width=99>实验结论</TD></TR><TR><td>将生成的气体通入AgNO<SUB>3</SUB>&nbsp;溶液中</TD><td>产生白色沉淀</TD><td>气体中有HCl</TD></TR></TBODY></TABLE>【拓展延伸】实验室制取二氧化碳的反应物是固体和液体，反应条件生成物，生成的二氧化碳中含有氯化氢气体、水蒸气，需要先用碳酸氢钠溶液将氯化氢除去，再用浓硫酸干燥，然后用向上排空气法收集二氧化碳，所选仪器的连接顺序为BHEC；<br />故答案为：（1）酒精灯；前；<br />（2）2KMnO<SUB>4</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>K<SUB>2</SUB>MnO<SUB>4</SUB>+MnO<SUB>2</SUB>+O<SUB>2</SUB>↑；2H<SUB>2</SUB>O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span>&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O+O<SUB>2</SUB>↑；C，b；<br />（3）Ⅰ、CaCO<SUB>3</SUB>+2HCl=CaCl<SUB>2</SUB>+CO<SUB>2</SUB>↑+H<SUB>2</SUB>O；<br />Ⅱ、乙；<br />Ⅲ、固体反应物的颗粒大小（或反应物的接触面积）；<br />Ⅳ、AB；<br />（4）【实验探究】①酚酞试液变红，说明溶液中存在碱性物质Ca（OH）<SUB>2</SUB>；<br />②<table class=\"edittable\"><TBODY><TR><td width=225>实验步骤</TD><td width=120>实验现象</TD><td width=99>实验结论</TD></TR><TR><td>将生成的气体通入AgNO<SUB>3</SUB>&nbsp;溶液中</TD><td>产生白色沉淀</TD><td>气体中有HCl</TD></TR></TBODY></TABLE>【拓展延伸】BHEC．','【分析】（1）根据实验室常用仪器的名称和连接仪器 的顺序进行分析；<br />（2）A装置制取并在试管口塞一团棉花，说明是用加热高锰酸钾的方法制取氧气，据反应原理书写方程式，B装置属于固液常温型，用过氧化氢制取氧气可用此发生装置，据反应原理书写方程式；并根据氧气的密度比空气大，不易溶于水，氧气密度比水小进行分析；<br />（3）大理石的主要成分是碳酸钙，与稀盐酸反应生成氯化钙、水和二氧化碳；根据实验数据分析比较解答；<br />（4）【实验探究】①根据碱能使酚酞变红色进行分析；<br />②根据氯离子和银离子会生成氯化银沉淀进行分析；<br />【拓展延伸】根据实验室制取二氧化碳的反应物是固体和液体，反应条件生成物，生成的二氧化碳中含有氯化氢气体、水蒸气，需要先用碳酸氢钠溶液将氯化氢除去，再用浓硫酸干燥，然后用向上排空气法收集二氧化碳进行分析．','书写',3.00,'b288615b6d1d922b1271edc85c2b8ef1',9,400,'实验探究物质变化的条件和影响物质变化的因素,常用气体的发生装置和收集装置与选取方法,常见气体的检验与除杂方法,实验室制取氧气的反应原理,二氧化碳的实验室制法,书写化学方程式、文字表达式、电离方程式','寿光市',2016,'32','2016•寿光市模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841422,'小明在实验室发现一瓶盛有无色溶液的试剂瓶，其标签破损，从残余部分只能看出溶质质量分数为10%，具体是什么物质无法辨认．老师告诉他，这瓶溶液可能是氢氧化钠、氯化钠、氢氧化钙或碳酸钠中的一种．<br />（1）小明查阅氢氧化钙常温下的溶解度为0.18g后，认为该溶液不可能是氢氧化钙，理由是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）小明取少量样品于试管中，滴加<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>试液，试液变红色，该溶液不可能是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />（填写化学式）．<br />（3）为了确定该溶液的成分，小明同学继续进行下列实验，请一起参与，并填写下列实验报告．<br /><table class=\"edittable\"><tbody><tr><td width=\"197\">实验步骤</td><td width=\"197\">实验现象</td><td width=\"197\">实验结论</td></tr><tr><td>取少量溶液于试管中，滴加<br /><u>&nbsp;&nbsp;&nbsp;&nbsp;</u></td><td><u>&nbsp;&nbsp;&nbsp;&nbsp;</u></td><td>该溶液是碳酸钠溶液．<br />有关反应的化学方程式为：<br /><u>&nbsp;&nbsp;&nbsp;&nbsp;</u></td></tr></tbody></table>','','','','','','Ca（OH）<sub>2</sub>微溶于水，其溶液质量分数不可能为10%$###$无色酚酞$###$NaCl$###$氯化钙溶液$###$产生白色沉淀$###$CaCl<sub>2</sub>+Na<sub>2</sub>CO<sub>3</sub>=CaCO<sub>3</sub>↓+2Na','【解答】解：（1）氢氧化钙的溶解度为0.18g，其饱和溶液溶质质量分数为：<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">0.18g</td></tr><tr><td>0.18g+100g</td></tr></table>×100%≈0.2%</span>，不能达到10%，故填：Ca（OH）<sub>2</sub>微溶于水，其溶液质量分数不可能为10%；<br />（2）碳酸钠和氢氧化钠的水溶液呈碱性，能使酚酞试液变红，而氯化钠的水溶液呈中性，不能使酚酞试液变红，故可以加入酚酞试液，溶液变红，说明不是氯化钠溶液，故填：无色酚酞，NaCl；<br />（3）根据结论，溶液是碳酸钠，则加入会产生白色沉淀，碳酸钠能与氯化钙反应生成碳酸钙沉淀和氯化钠，故填：氯化钙溶液，产生白色沉淀，CaCl<sub>2</sub>+Na<sub>2</sub>CO<sub>3</sub>=CaCO<sub>3</sub>↓+2NaCl．','【分析】根据已有的物质的溶解度、溶质质量分数的计算以及物质的性质进行分析解答即可．','书写',3.00,'2864975d3abca852162978021d143bc0',9,400,'缺失标签的药品成分的探究,溶质的质量分数、溶解性和溶解度的关系,碱的化学性质,盐的化学性质,书写化学方程式、文字表达式、电离方程式','',2016,'37','2016•德城区一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841425,'下面是初中化学中的三个小实验，请按要求完成下列各小题：<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao1/26a388e1-94d4-11e9-b07c-b42e9921e93e_xkb62.png\" style=\"vertical-align:middle\" /><br />（1）探究铁生锈的条件时，按所设计的方案进行实验，一周后观察到试管<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>中的铁钉有明显的锈蚀．通过实验探究，你可以得出铁生锈的条件是：铁要同时接触<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）探究质量守恒定律时，实验探究小组的同学按如图所选用的药品和称量方法进行实验，结果反应前后天平不平衡，分析原因是：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．因此必须对该实验装置进行改进，使反应在<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>的装置中实验，才能观察到反应前后质量守恒．<br />（3）探究二氧化碳的性质时，将CO<SUB>2</SUB>通入紫色的石蕊试液中，观察到的现象是紫色石蕊试液变<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，说明CO<SUB>2</SUB>的性质是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（用化学方程式表示）．','','','','','','B$###$与氧气、水同时接触（或在潮湿的空气中）$###$反应生成的CO<SUB>2</SUB>气体逸出烧杯$###$密闭容器$###$红$###$CO<SUB>2</SUB>+H<SUB>2</SUB>O═H<SUB>2</SUB>CO<SUB>3</SUB>','【解答】解：（1）A试管中没有水，管口闭塞，管内只有空气，铁钉不易生锈．B试管中铁钉浸在水里，管口未做任何处理，故铁钉同时和氧气、水接触，符合钢铁锈蚀条件，故最易生锈的是B试管；C试管中植物油不溶与水，且密度比水小，故可浮在水面，起到了隔绝空气的作用，铁钉虽浸在水里，但不同时具备氧气和水这两个条件，故铁钉也不易生锈．故填：B；与氧气、水同时接触（或在潮湿的空气中）；<br />（2）反应后天平不平衡的原因是盐酸与碳酸钠反应生成氯化钠、水、二氧化碳，而烧杯是敞开的，所以生成的二氧化碳逸散到空气中使质量减小；若反应物不变，要使天平在反应后仍然保持平衡，可在烧杯口处套一个气球，使之在密封的容器内进行实验．故填：反应生成的CO<SUB>2</SUB>气体逸出烧杯；密闭容器；<br />（3）二氧化碳能与水反应生成碳酸，呈酸性，能使紫色石蕊试液变红，反应的化学方程式为CO<SUB>2</SUB>+H<SUB>2</SUB>O═H<SUB>2</SUB>CO<SUB>3</SUB>．<br />答案：<br />（1）B；与氧气、水同时接触（或在潮湿的空气中）；<br />（2）反应生成的CO<SUB>2</SUB>气体逸出烧杯；密闭容器；<br />（3）红；CO<SUB>2</SUB>+H<SUB>2</SUB>O═H<SUB>2</SUB>CO<SUB>3</SUB>','【分析】（1）根据铁生锈的条件分析ABC中最易生锈的；植物油不溶于水可以隔绝空气；根据ABC的对比分析铁生锈的条件．<br />（2）根据盐酸与碳酸钠反应生成了二氧化碳气体，逸散到了空气中会使烧杯中物质的质量减少来分析解答．考虑密封装置．<br />（3）根据二氧化碳的性质进行分析，二氧化碳能与水反应生成碳酸，呈酸性，能使紫色石蕊试液变红．','书写',3.00,'83d2b5ccbd0f67287012bbcbe00f459e',9,400,'质量守恒定律的实验探究,探究金属锈蚀的条件,探究二氧化碳的性质,书写化学方程式、文字表达式、电离方程式','',2016,'37','2016•盘龙区一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841432,'<img  src=\"/tikuimages/9/2016/400/shoutiniao72/26bb569e-94d4-11e9-a182-b42e9921e93e_xkb16.png\" style=\"vertical-align:middle;FLOAT:right\" />（2016•洛阳二模）如图是某食盐说明书，KIO<SUB>3</SUB>中碘元素的化合价为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，微量元素碘缺乏和过量都会引起<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，每袋食盐中含碘的质量为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','+5$###$甲状腺肿大$###$10mg','【解答】解：根据在化合物中正负化合价代数和为零，钾元素的化合价为+1价，氧元素的化合价为-2价，设碘酸钾（KIO<SUB>3</SUB>）中碘元素的化合价为x，则（+1）+x+（-2）×3=0，解得x=+5；人体缺碘或摄入过量都会导致甲状腺肿大；每袋食盐中含碘的质量为<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">20mg</td></tr><tr><td>1000g</td></tr></table>×500g</span>=10mg．<br />故答案为：+5；甲状腺肿大；10mg．','【分析】根据在化合物中正负化合价代数和为零，进行解答．根据碘元素与人体健康的关系来分析；根据标签信息来分析．','填空题',3.00,'f71b5ed81cc6c979e9efc64573a1bedb',9,400,'有关元素化合价的计算,标签上标示的物质成分及其含量,人体的元素组成与元素对人体健康的重要作用','',2016,'37','2016•洛阳二模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841434,'金属在生产、生活中应用广泛．<br />（1）在汽车电路中，经常用铜作导线，这是利用了铜的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；汽车车体表面喷漆不仅美观，而且可有效防止与<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>接触而生锈．<br />（2）某兴趣小组的同学从实验室收集到含有H<SUB>2</SUB>SO<SUB>4</SUB>、FeSO<SUB>4</SUB>、CuSO<SUB>4</SUB>的废液，他们想从中回收金属铜和一些盐类物质，设计了如下实验方案（部分产物省略）．请结合实际方案回答：<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao60/26d2fd4f-94d4-11e9-99ff-b42e9921e93e_xkb15.png\" style=\"vertical-align:middle\" /><br />操作①中发生反应的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（写1个），A～F中含铁元素的有（填序号）<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）某钢铁厂每天需消耗980t含Fe<SUB>2</SUB>O<SUB>3</SUB>80%的赤铁矿，该厂理论上可日产含铁98%的生铁的质量是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>t．','','','','','','导电性$###$氧气和水蒸气$###$Fe+H<SUB>2</SUB>SO<SUB>4</SUB>=FeSO<SUB>4</SUB>+H<SUB>2</SUB>↑或Fe+CuSO<SUB>4</SUB>=Cu+FeSO<SUB>4</SUB>$###$ABCDE$###$560','【解答】解：（1）铜用作导线，是因为铜具有良好的导电性；铁在与氧气和水蒸气同时接触时会生锈，在车体表面喷漆可以防止铁与氧气、水蒸气接触而生锈；故填：导电性；氧气和水蒸气；<br />（2）由物质C为浅绿色，可知C为硫酸亚铁溶液，B为铁，为能把硫酸及硫酸铜全部反应掉，则铁应过量，反应的方程式为：Fe+H<SUB>2</SUB>SO<SUB>4</SUB>=FeSO<SUB>4</SUB>+H<SUB>2</SUB>↑，Fe+CuSO<SUB>4</SUB>=Cu+FeSO<SUB>4</SUB>；则D为铁和铜的混合物，向D中加盐酸将铁反应掉，剩余F为铜，E为硫酸亚铁溶液，故含铁元素的物质有ABCDE；<br />故答案为：Fe+H<SUB>2</SUB>SO<SUB>4</SUB>=FeSO<SUB>4</SUB>+H<SUB>2</SUB>↑或Fe+CuSO<SUB>4</SUB>=Cu+FeSO<SUB>4</SUB>；ABCDE；<br />（3）解：设理论上可日产含Fe98%的生铁的质量为x．<br />Fe<SUB>2</SUB>O<SUB>3</SUB>+3CO<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2Fe+3CO<SUB>2</SUB><br />160&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;112<br />980t×80%&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;98%x&nbsp;&nbsp;<br /><span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">160</td></tr><tr><td>112</td></tr></table>=<table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">980t×80%</td></tr><tr><td>98%x</td></tr></table></span><br />x=560t<br />故填：560．','【分析】（1）根据铜用作导线，是因为铜具有良好的导电性进行解答；根据铁在与氧气和水蒸气同时接触时会生锈，在车体表面喷漆可以防止铁与氧气、水蒸气接触而生锈进行解答；<br />（2）根据物质间的相互反应进行分析判断；<br />（3）由赤铁矿石的质量、氧化铁的质量分数、生铁中铁的质量分数，根据赤铁矿炼铁的化学方程式可以列式计算出炼出生铁的质量．','书写',3.00,'aa5934865868fb182a4b70ea8626aacb',9,400,'金属的物理性质及用途,金属的化学性质,含杂质物质的化学反应的有关计算,金属锈蚀的条件及其防护,书写化学方程式、文字表达式、电离方程式','',2016,'37','2016•河西区一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841440,'物质的性质决定其用途．下列叙述错误的是（　　）','“洗洁精”有乳化功能，可用“洗洁精”洗涤餐具上的油污','氮气化学性质不活泼，可将氮气充入食品包装袋内延长食品保质期','熟石灰能和酸发生反应，可用熟石灰改良酸性土壤','甲醛能使蛋白质失去生理活性，可用甲醛溶液浸泡水产品防腐','','D','【解答】解：A、洗洁精是洗涤剂，有乳化作用，能将大的油滴分散成细小的油滴随水冲走，可用“洗洁精”洗涤餐具上的油污，故选项说法正确．<br />B、氮气化学性质不活泼，可充入食品包装袋内延长食品保质期，故选项说法正确．<br />C、氢氧化钙具有碱性，能和酸发生反应，可用熟石灰改良酸性土壤，故选项说法正确．<br />D、甲醛有毒，能破坏人体蛋白质的结构，使蛋白质变质，该做法会危害人体健康，故选项说法错误．<br />故选D．','【分析】A、根据洗洁精具有乳化作用进行分析．<br />B、根据氮气化学性质不活泼进行分析判断．<br />C、根据氢氧化钙具有碱性，能中和酸性土壤进行分析．<br />D、根据甲醛能破坏蛋白质的结构进行分析判断．','选择题',3.00,'b201a0d6a023f6d9f7e640a2df4dc853',9,400,'常见气体的用途,乳化现象与乳化作用,中和反应及其应用,亚硝酸钠、甲醛等化学品的性质与人体健康','',2016,'37','2016•平南县二模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841455,'实验室有下列多种加热的方式．<br /><img  src=\"/tikuimages/9/0/400/shoutiniao69/271eac4f-94d4-11e9-8856-b42e9921e93e_xkb2.png\" style=\"vertical-align:middle\" /><br />（1）写出标有序号仪器的名称：①<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，②<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，③<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）甲中至少有三处错误，它们是①<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，②<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，③<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）按照乙方式操作的实验是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />A．镁带在空气中燃烧&nbsp;&nbsp;&nbsp;&nbsp; B．硫在空气中燃烧<br />C．氢气在空气中燃烧&nbsp;&nbsp;&nbsp; D．加热KMnO<SUB>4</SUB>制取氧气<br />（4）水浴加热是今后学习的一种加热方式，你认为丁与丙比较，正确的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />A．丁温度更高&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; B．丁受热更均匀&nbsp;&nbsp;&nbsp;&nbsp; C．丁更好控制温度．','','','','','','燃烧匙$###$坩埚钳$###$石棉网$###$试管内液体太多$###$手不能握住试管夹的短柄$###$应该使酒精灯的外焰进行加热$###$B$###$C','【解答】解：（1）通过分析题中所指仪器的作用可知，①是燃烧匙，②是坩埚钳，③是石棉网；<br />（2）使用酒精灯的外焰加热，试管内液体的体积最好不要超过试管体积的<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">1</td></tr><tr><td>3</td></tr></table></span>，所以甲中至少有三处错误是：试管内液体太多，手不能握住试管夹的短柄，应该使酒精灯的外焰进行加热；<br />（3）A、镁带在空气中燃烧应该使用坩埚钳，故A错误；<br />B、硫在空气中燃烧，可以使用燃烧匙，故B正确；<br />C、氢气在空气中燃烧，不能使用燃烧匙，故C错误；<br />D、加热KMnO<SUB>4</SUB>制取氧气，应该使用试管，故D错误．<br />故选：B；<br />（4）水浴加热可以更好的控制反应的温度，故选：C．<br />故答案为：（1）燃烧匙，坩埚钳，石棉网；<br />（2）试管内液体太多，手不能握住试管夹的短柄，应该使酒精灯的外焰进行加热；<br />（3）B；<br />（4）C．','【分析】（1）根据实验室常用仪器的名称和题中所指仪器的作用进行分析；<br />（2）根据酒精灯的正确使用方法以及给试管加热的正确操作进行分析；<br />（3）根据实验乙是给粉末加热进行分析；<br />（4）根据水浴加热的优点进行分析．','填空题',3.00,'77712352f4f252982b346f69d033f7c7',9,400,'用于加热的仪器,加热器皿-酒精灯,给试管里的液体加热','',0,'37','',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841456,'下列关于水的说法不正确的是（　　）','水是一种常见的溶剂','长期饮用蒸馏水不利于人体健康','水能与多种物质发生反应','过滤能除去天然水中可溶性的杂质','','D','【解答】解：A、可以作溶剂的物质很多，水是一种常见的溶剂，能溶解多种物质，故A正确；<br />B、人体所需的无机盐，矿物质主要来源于水，而蒸馏水中不含无机盐和矿物质，所以长期饮用蒸馏水不利于人体健康；故B正确；<br />C、水的性质比较活泼，能和多种物质反应，如二氧化碳，氧化钙，活泼的金属等，故C正确；<br />D、过滤是用来除去难溶性杂质的，不能除去可溶性的杂质；故D错误；<br />故选D．','【分析】A、根据水是常见的溶剂进行分析；<br />B、根据人体所需的无机盐，矿物质主要来源于水进行分析；<br />C、根据水的性质，水能和多种物质反应进行分析；<br />D、根据过滤的原理进行分析．','选择题',3.00,'efb728369be125c4ac05c3667ee2a99d',9,400,'过滤的原理、方法及其应用,水的性质和应用,常见的溶剂','湘乡市',2016,'35','2016春•湘乡市校级期中',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841461,'<img  src=\"/tikuimages/9/2016/400/shoutiniao21/273149f0-94d4-11e9-ba74-b42e9921e93e_xkb74.png\" style=\"vertical-align:middle;FLOAT:right\" />某兴趣小组设计的小实验，装置如图所示．打开止水夹，将A滴入试管①中，A与B接触后，在试管②中的导管口处观察到有气泡冒出．下列液体A和固体B的组合不可能出现上述现象的是（　　）','水和烧碱','稀硫酸和镁条','稀盐酸和小苏打','水和硝酸铵','','D','【解答】解：A、烧碱溶于水会放出大量的热，有气泡产生；<br />B、镁条和稀硫酸反应能产生气体，有气泡产生；<br />C、稀盐酸能与小苏打碳酸氢钠反应生成气体，有气泡产生；<br />D、硝酸铵溶于水温度降低，没有气泡产生；<br />故选D．','【分析】在试管②中的导管口处观察到有气泡冒出有两种可能：第一：物质溶于水放出热量，使试管内气体受热膨胀，压强增大从试管②中的导管口处冒出；第二：固体与液体反应产生气体，气体从试管②中的导管口处冒出．','选择题',3.00,'f1cca5d0a2a119d7b6e1964d28513683',9,400,'溶解时的吸热或放热现象,碱的化学性质,反应现象和本质的联系','',2016,'37','2016•海淀区一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841465,'二氧化碳能产生温室效应，还能作化工原料．<br />（1）化石燃料燃烧会产生大量的CO<SUB>2</SUB>，天然气（CH<SUB>4</SUB>）充分燃烧的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）膜分离是一种常用的分离技术．二氧化硅（SiO<SUB>2</SUB>）能用于制造CO<SUB>2</SUB>分离膜，CO<SUB>2</SUB>通过此膜后被氨水吸收（如图1所示），转化为可作氮肥的NH<SUB>4</SUB>HCO<SUB>3</SUB>．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao68/273fa1d1-94d4-11e9-95de-b42e9921e93e_xkb46.png\" style=\"vertical-align:middle\" /><br />①NH<SUB>4</SUB>HCO<SUB>3</SUB>固体中氮元素的质量分数为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>%（计算结果精确到0.1%）．<br />（3）CO<SUB>2</SUB>形成的超临界CO<SUB>2</SUB>流体可用于从香兰草豆荚粉中提取香兰素（C<SUB>8</SUB>H<SUB>8</SUB>O<SUB>3</SUB>）（如图2）．<br />①香兰素的乙醇溶液中的溶剂是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />②由香兰素的乙醇溶液得到香兰素的过程中，发生的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>变化（填物理或化学）．<br />③以上提取香兰素的过程中，可循环使用的物质有无水乙醇和<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />④完全燃烧19g香兰素所生成的一氧化碳与完全燃烧<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>g无水乙醇（俗名酒精，C<SUB>2</SUB>H<SUB>5</SUB>OH）所生成的二氧化碳的质量相等．','','','','','','CH<SUB>4</SUB>+2O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>CO<SUB>2</SUB>+2H<SUB>2</SUB>O$###$17.1$###$乙醇$###$物理$###$CO<SUB>2</SUB>$###$23','【解答】解：<br />（1）天然气的主要成分是甲烷，甲烷燃烧的化学方程式为：CH<SUB>4</SUB>+2O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>CO<SUB>2</SUB>+2H<SUB>2</SUB>O．故答案为：CH<SUB>4</SUB>+2O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>CO<SUB>2</SUB>+2H<SUB>2</SUB>O．<br />（2）①碳酸氢铵中氮元素的质量分数=<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">14</td></tr><tr><td>14+1×4+12+16×3</td></tr></table></span>×100%≈17.7%；故填：+4；17.7%；<br />（3）①香兰素的乙醇溶液中的溶剂是乙醇；故填：乙醇；<br />②由香兰素的乙醇溶液得到香兰素的过程中，发生的是物理变化；<br />③根据图示可知可循环使用的物质有CO<SUB>2</SUB>，无水乙醇；故填：CO<SUB>2</SUB>；<br />④二氧化碳的质量相等也就是所含碳元素的质量相等．根据质量守恒定律可知，也就是香兰素中碳元素的质量与乙醇中碳元素的质量相等．<br />设乙醇的质量为x，则<br />19g×<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">12×8</td></tr><tr><td>12×8+1×8+16×3</td></tr></table></span>×100%=x×<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">12×2</td></tr><tr><td>12×2+1×6+16</td></tr></table></span>×100%<br />x=23g<br />故填：23．<br />答案：<br />（1）CH<SUB>4</SUB>+2O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>CO<SUB>2</SUB>+2H<SUB>2</SUB>O；<br />（2）①17.7%<br />（3）①乙醇；②物理；③CO<SUB>2</SUB>；④23．','【分析】（1）根据反应物和生成物及其质量守恒定律可以书写化学方程式．<br />（2）①根据二氧化硅的化学式为SiO<SUB>2</SUB>，利用氧化物中氧元素的化合价及化合物中元素的正负化合价的代数和为0来计算其中硅元素的化合价；<br />②根据化合物中元素的质量分数进行分析；<br />（3）①根据溶液的组成解答；<br />②根据由香兰素的乙醇溶液得到香兰素的过程中没有新物质生成解答；<br />③根据题中信息解答；<br />④根据化学式进行计算即可．','填空题',3.00,'798be2ba180048cf3791eac4f15deda8',9,400,'溶液、溶质和溶剂的相互关系与判断,物质的相互转化和制备,元素的质量分数计算,化合物中某元素的质量计算,化学变化和物理变化的判别,常用燃料的使用与其对环境的影响','',2016,'32','2016•相城区模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841470,'某同学梳理并归纳了以下知识点：①质子数相同的粒子一定属于同种元素②加入活性炭能降低水的硬度；③可以氮气为原料制取硝酸和化肥；④最外层电子数为8的粒子一定是稀有气体的原子；其中正确的是（　　）','①④','①③','②③','只有③','','D','【解答】解：①质子数相同的粒子一定属于同种元素错误，如果是原子就正确，故①错误；<br />②降低水的硬度的方法是：在生活中是加热煮沸，在实验室中是蒸馏，加入活性炭活性炭吸附除去水中的颜色和气味，不能降低水的硬度，故②错误；<br />③可以氮气为原料制取硝酸和化肥正确，故③正确；<br />④最外层电子数为8的粒子，可能是稀有气体元素的原子也可能是阳离子或阴离子，如钠离子、氯离子等．故④错误；<br />故选：D．','【分析】①质子数相同的两种粒子不一定属于同种元素，例如水分子和氨分子，<br />②降低水的硬度的方法；<br />③氮气的用途；<br />④最外层电子数为8的粒子不一定是稀有气体的原子，也可能是离子；','选择题',3.00,'abde16428ee60f2828a348406a14c070',9,400,'常见气体的用途,硬水与软水,核外电子在化学反应中的作用,元素的概念','',2016,'37','2016春•巴彦淖尔校级月考',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841484,'在点燃条件下，甲、乙反应的微观示意图如下．下列说法错误的是（　　）<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao31/278695de-94d4-11e9-a97e-b42e9921e93e_xkb2.png\" style=\"vertical-align:middle\" />','反应中甲、乙的分子数比为1：1','一定条件下，丁可转化为乙','该反应属于置换反应','反应中只有氢的化合价不变','','A','【解答】解：由反应的结构示意图和模型表示的原子种类，可判断甲为NH<SUB>3</SUB>，乙为O<SUB>2</SUB>，丙为N<SUB>2</SUB>，丁为H<SUB>2</SUB>O，反应的化学方程式为：4NH<SUB>3</SUB>+3O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2N<SUB>2</SUB>+6H<SUB>2</SUB>O，由此可知：<br />A、由反应的方程式可知，氨气分子和氧气分子的个数比是4：3，故A说法错误；<br />B、因为乙为O<SUB>2</SUB>，丁为H<SUB>2</SUB>O，一定条件下，丁可转化为乙，故B说法正确；<br />C、由化学方程式可知，反应物是一种化合物和一种单质，生成物是新的单质和新的化合物，反应符合置换反应的特点，因此该反应为置换反应，故C说法正确；<br />D、反应过程中，氮元素的化合价由-3→0，氧元素由0→-2，氢元素的化合价都是+1价，所以反应中氮和氧的化合价发生改变，只有氢的化合价不变，故D说法正确．<br />故选：A．','【分析】根据反应的结构示意图和模型表示的原子种类，可判断甲为NH<SUB>3</SUB>，乙为O<SUB>2</SUB>，丙为N<SUB>2</SUB>，丁为H<SUB>2</SUB>O，反应的化学方程式为4NH<SUB>3</SUB>+3O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2N<SUB>2</SUB>+6H<SUB>2</SUB>O，据此分析有关的问题．','选择题',3.00,'d13443355c9806720328961874d6abf7',9,400,'微粒观点及模型图的应用,化合价规律和原则,反应类型的判定','扬中市',2016,'37','2016•扬中市一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841489,'铝的元素符号是（　　）','Cl','Cl<SUB>2</SUB>','Al','AL','','C','【解答】解：A、铝的元素符号是Al，Cl是氯元素的元素符号，故选项错误．<br />B、铝的元素符号是Al，Cl<SUB>2</SUB>是氯气的化学式，故选项错误．<br />C、铝的元素符号是Al，故选项正确．<br />D、铝的元素符号是Al，而不是AL，故选项错误．<br />故选：C．','【分析】书写元素符号时应注意：①有一个字母表示的元素符号要大写；②由两个字母表示的元素符号，第一个字母大写，第二个字母小写．','选择题',3.00,'c9abb6ddd3c6e217624fe4cf502c84ad',9,400,'元素的符号及其意义','',2016,'37','2016•桂林二模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841490,'随着车辆的增加，汽车尾气带给城市的污染越来越严重了，用燃气来代替燃油可以大大减少汽车尾气的污染．某市的加气站某日发生了爆炸，请你分析爆炸的原因．为防止此类事件的再次发生，请你提出一条合理化建议．','','','','','','','【解答】解：凡是可燃性的气体，如果混有空气，点燃时都可能发生爆炸．燃气爆炸也就是甲烷燃烧的结果．<br />其原因是甲烷在空气中的含量达到了其爆炸极限并且遇到了明火，所以防止明火．','【分析】可以根据爆炸的含义分析燃气爆炸的原因．','解答题',3.00,'7c5f334a10170f9e9a04982e46a7756d',9,400,'燃烧、爆炸、缓慢氧化与自燃','',0,'37','',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841495,'<img  src=\"/tikuimages/9/2016/400/shoutiniao77/27abaa0f-94d4-11e9-81ae-b42e9921e93e_xkb5.png\" style=\"vertical-align:middle;FLOAT:right;\" />（2016•本溪一模）某同学在实验室发现了一瓶标签残缺的无色溶液（如图甲所示），为确认其中的溶质，他设计并进行了如下探究活动，请回答下列问题．<br />【猜想与假设】溶质可能为NaCl、NaOH、Na<SUB>2</SUB>CO<SUB>3</SUB>和NaHCO<SUB>3</SUB>中的一种．<br />【资料查阅】上述四种物质的相关信息如表：<br /><table class=\"edittable\"><TBODY><TR><td width=181>物质</TD><td width=85>NaCl</TD><td width=85>NaOH</TD><td width=88>Na<SUB>2</SUB>CO<SUB>3</SUB></TD><td width=81>NaHCO<SUB>3</SUB></TD></TR><TR><td>常温下的溶解度/g</TD><td>36</TD><td>109</TD><td>21.5</TD><td>9.6</TD></TR><TR><td>常温下某稀溶液的pH</TD><td>7</TD><td>13</TD><td>11</TD><td>9</TD></TR></TBODY></TABLE>【探究过程】如图乙所示，在操作①后可确定溶质不是NaCl，则其实验现象应是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；在进行操作②时有无色无味的气体产生，由此又可排除假设物质中的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />【探究结论】你认为该溶液中的溶质可能是上述假设物质中的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，你的判断依据是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />【探究反思】若上述探究结论是正确的，操作②产生的气体应是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（写化学式），实验室检验该气体的实验操作及现象是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','石蕊试液变成蓝色$###$氢氧化钠$###$碳酸钠$###$碳酸氢钠饱和溶液的质量分数不可能达到10%，而碳酸钠溶液可以$###$CO<SUB>2</SUB>$###$将气体通入石灰水，石灰水变浑浊','【解答】解：【探究观过程】NaCl、NaOH、Na<SUB>2</SUB>CO<SUB>3</SUB>和NaHCO<SUB>3</SUB>中只有氯化钠溶液为中性，而其余的溶液都显碱性，若溶质不是NaCl，则滴加石蕊试液会变成蓝色；在进行操作②滴加稀盐酸时有无色无味的气体产生，说明不是氢氧化钠，因为氢氧化钠和盐酸反应无明显现象；<br />故答案为：石蕊试液变成蓝色；氢氧化钠；<br />【探究结论】根据计算可以知道在20℃时饱和碳酸氢钠的质量分数为：<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">9.6g</td></tr><tr><td>9.6g+100g</td></tr></table></span>×100%=8.8%，而题目中标注的溶液的质量分数为10%，所以这瓶试剂不可能的是碳酸氢钠溶液，而碳酸钠溶液的溶质质量分数可以等于10%，因此可能是碳酸钠；<br />故答案为：碳酸钠；碳酸氢钠饱和溶液的质量分数不可能达到10%，而碳酸钠溶液可以；<br />【探究反思】假设碳酸钠成立，则碳酸钠和盐酸反应会产生二氧化碳，检验二氧化碳可以将气体通入石灰水，石灰水变浑浊；<br />故答案为：CO<SUB>2</SUB>；将气体通入石灰水，石灰水变浑浊；','【分析】【探究过程】根据三种物质溶液的酸碱性及对石蕊试液的作用分析；根据氢氧化钠不能与盐酸反应产生气体解答；<br />【探究结论】利用碳酸钠、碳酸氢钠溶液中溶质溶解度与饱和溶液中溶质的质量分数的关系，判断溶液的可能性；<br />【探究反思】根据碳酸钠能够和稀盐酸反应生成，及二氧化碳能够使澄清的石灰水变浑浊进行二氧化碳的检验进行分析；','填空题',3.00,'a7fd10b08e7da2322db48fd4519cd28d',9,400,'缺失标签的药品成分的探究,碱的化学性质,盐的化学性质','',2016,'37','2016•本溪一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841497,'物质的用途与性质密切相关．下列叙述错误的是（　　）','熟石灰能和酸发生反应，所以常用熟石灰改良酸性土壤','洗洁精有乳化功能，所以常用洗洁精洗涤餐具上的油污','甲醛能与蛋白质发生反应，所以甲醛可用于水产食品的保鲜','氮气化学性质不活泼，所以在食品包装中充入氮气用于防腐','','C','【解答】解：A、熟石灰能和酸发生中和反应，所以常用熟石灰改良酸性土壤，故A叙述正确；<br />B、洗洁精有乳化功能，所以常用洗洁精洗涤餐具上的油污，故B叙述正确；<br />C、甲醛有毒，甲醛不可用于水产食品的保鲜，故C叙述错误；<br />D、氮气化学性质不活泼，所以在食品包装中充入氮气用于防腐，故D叙述正确．<br />故选C．','【分析】A、根据熟石灰的性质和用途分析；<br />B、根据洗洁精有乳化功能分析；<br />C、根据甲醛有毒分析；<br />D、根据氮气的稳定性分析．','选择题',2.00,'650077fd75081b44c9931b62d1f0a27a',9,400,'常见气体的用途,乳化现象与乳化作用,中和反应及其应用,亚硝酸钠、甲醛等化学品的性质与人体健康','',2016,'37','2016•郑州二模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841501,'通过化学启蒙学习，同学们对组成万物的基本物质有了进一步了解．如图是某陨石样品中各种元素含量的比例图<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao58/27b40e80-94d4-11e9-a6a8-b42e9921e93e_xkb60.png\" style=\"vertical-align:middle\" /><br />（1）该陨石中含<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>种非金属元素<br />（2）钠离子的结构示意图为<img  src=\"/tikuimages/9/2016/400/shoutiniao94/27b7df0f-94d4-11e9-8874-b42e9921e93e_xkb38.png\" style=\"vertical-align:middle\" />，则x=<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />（3）此陨石样品中不可能含有的化合物是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />A．CaCO<SUB>3</SUB>&nbsp;&nbsp;B．K<SUB>2</SUB>SO<SUB>4</SUB>&nbsp;&nbsp;C．Al<SUB>2</SUB>O<SUB>3</SUB>&nbsp;&nbsp;D．SiO<SUB>2</SUB>．','','','','','','3$###$11$###$A','【解答】解：（1）根据化学元素汉字名称的偏旁可辨别元素的种类，通过归纳，金属元素名称一般有“金”字旁；固态非金属元素名称有“石”字旁；气态非金属元素名称有“气”字头，可知图中表示出的非金属元素有3种；<br />（2）钠离子是钠原子失去一个电子形成的，故x=2+8+1=11；故填：11；<br />（3）根据质量守恒定律，化学反应前后元素的种类不变，该陨石样品中不含碳元素，故不可能含有的化合物是CaCO<SUB>3</SUB>；<br />故答案为：（1）3；（2）11；（3）A．','【分析】（1）根据化学元素汉字名称的偏旁可辨别元素的种类，通过归纳，金属元素名称一般有“金”字旁；固态非金属元素名称有“石”字旁；气态非金属元素名称有“气”字头，据此进行分析；<br />（2）根据离子形成的过程来分析；<br />（3）根据质量守恒定律进行分析．','填空题',3.00,'bd8c8205121253e9d83f58ede524bb91',9,400,'金属元素的存在及常见的金属矿物,原子结构示意图与离子结构示意图,元素的简单分类','',2016,'37','2016•重庆校级一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841529,'初中化学兴趣活动小组的活动丰富多彩，下面是他们以“化学与社会发展“为主理的两次活动记录：<br />活动1：在观看了“西气东输”与“石油分馏”等视频后，通过小组讨论，同学们认识到：<br />①<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>、石油、天然气是人类社会重要的自然资源，天然气的主要成分是甲烷，甲烷完全燃烧的化学方程式是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />②现在化石燃料等不可再生能源面临枯竭，化石燃料对环境的影响也不容忽视．所以开发和利用风能，<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>等新能源成为越来越迫切的要求．<br />③人们将化石燃料通过化学工艺可制造和合成价值更高的药物、化学纤维、塑料料盒、合成橡胶等，其中塑料属于<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>材料，随意丢弃会造成“白色污染”．<br />活动2：同学们展示了各自收集到的有关“化学物质与健康“的相关资料及一些常见食品，如：高钙牛奶、加碘食盐等．并针对“化学制品对人体有益有害”展开了激烈的辩论，从而认识到钙、铁锌等元素对人体健康有重要作用，如缺铁元素会引起缺铁性贫血，缺钙元素会引起<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>还认识到一氧化碳、甲醛等物质对人体健康的危害．<br />通过这样的社团活动，大家形成了“保护自然，关爱生命”的共识．','','','','','','煤$###$CH<SUB>4</SUB>+2O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>CO<SUB>2</SUB>+2H<SUB>2</SUB>O$###$太阳能$###$有机合成$###$骨质疏松','【解答】解：①煤、石油、天然气是人类社会重要的自然资源．天然气的主要成分是甲烷，甲烷和氧气点燃会生成水和二氧化碳，化学方程式为：CH<SUB>4</SUB>+2O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>CO<SUB>2</SUB>+2H<SUB>2</SUB>O；<br />（2）开发和利用风能、太阳能等新能源成为越来越迫切的要求；<br />（3）塑料属于有机合成材料；<br />通过分析钙元素在人体所起的作用，人体缺钙会引起骨质疏松．<br />故答案为：（1）煤；CH<SUB>4</SUB>+2O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>CO<SUB>2</SUB>+2H<SUB>2</SUB>O；（2）太阳能；（3）有机合成；（4）骨质疏松．','【分析】①根据常见的三种化石燃料是煤、石油、天然气进行分析；根据甲烷和氧气点燃会生成水和二氧化碳进行分析；<br />②根据太阳能、风能属于清洁能源进行分析；<br />④根据材料的分类进行分析；<br />根据钙元素在人体所起的作用进行分析．','填空题',3.00,'6454653b1f6f10579bad2a7e6b16e32c',9,400,'常用燃料的使用与其对环境的影响,资源综合利用和新能源开发,白色污染与防治,人体的元素组成与元素对人体健康的重要作用','',2016,'37','2016•昆明一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841539,'南海是中国四大海域中最大、最深、自然资源最为丰富的海域，南海不但资源丰富，还是亚太地区海运的“咽喉要道”，这里有世界上最繁忙的航线，因此引起周边国家的觊觎．<br />（1）海水属于<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“纯净物”或“混合物”）；海水晒盐属于混合物分离操作的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填序号）<br />　　　　A、蒸馏　　　 B、蒸发　　　 C、过滤　　　 D、沉淀<br />（2）南海油气资源丰富，石油加工可得到石油液化气，假设某石油液化气的主要成分为C<SUB>4</SUB>H<SUB>10</SUB>，写出其完全燃烧的化学方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（3）在“南海一号”打捞出来的文物中，“鎏金龙手镯”光彩夺目，完好无损；铜镜表面有铜锈；铁器则锈迹斑斑，残存很少，这说明金、铜、铁这三种金属的活动顺序由强到弱的顺序是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（4）炼铁的原理是利用一氧化碳与氧化铁反应，反应的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','混合物$###$B$###$2C<SUB>4</SUB>H<SUB>10</SUB>+13O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>8CO<SUB>2</SUB>+10H<SUB>2</SUB>O$###$铁、铜、金$###$3CO+Fe<SUB>2</SUB>O<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2Fe+3CO<SUB>2</SUB>','【解答】解：（1）海水中含有水、氯化钠、氯化镁等物质属于混合物；海水晒盐主要是通过阳光和风力使水分蒸发，使晶体结晶析出的过程，所以蒸发就是混合物的分离；<br />（2）某石油液化气的主要成分为C<SUB>4</SUB>H<SUB>10</SUB>，完全燃烧的燃烧生成二氧化碳和水，燃烧的化学方程式为：2C<SUB>4</SUB>H<SUB>10</SUB>+13O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>8CO<SUB>2</SUB>+10H<SUB>2</SUB>O．<br />（3）金属活动性越强，越容易生锈，由，“金龙纹手镯”光彩夺目，完好无损；铜镜表面有铜锈；铁器则锈迹斑斑，残存很少，可知铁最活泼，铜次之，金最不活泼；<br />（4）炼铁的原理中反应物是一氧化碳和氧化铁，生成物是铁和二氧化碳，反应的化学方程式为为；3CO+Fe<SUB>2</SUB>O<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2Fe+3CO<SUB>2</SUB><br />故答案为：（1）混合物；&nbsp;B；（2）2C<SUB>4</SUB>H<SUB>10</SUB>+13O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>&nbsp;8CO<SUB>2</SUB>+10H<SUB>2</SUB>O；（3）铁、铜、金；（4）3CO+Fe<SUB>2</SUB>O<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2Fe+3CO<SUB>2</SUB>．','【分析】（1）根据海水的成分结合海水晒盐过程考虑混何物分离操作；<br />（2）根据反应物、反应条件、生成物写出化学方程式；<br />（3）根据金属是否容易生锈分析三种金属的活动顺序即可；<br />（4）根据反应物是一氧化碳和氧化铁，生成物是铁和二氧化碳写出化学方程式即可．','书写',3.00,'f0126e28b2aaac0096ea3e6fbc36a2a9',9,400,'金属活动性顺序及其应用,铁的冶炼,海水晒盐的原理和过程,纯净物和混合物的判别,书写化学方程式、文字表达式、电离方程式,海洋中的资源','',2016,'32','2016•兰州模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841542,'<img  src=\"/tikuimages/9/2016/400/shoutiniao54/28158f70-94d4-11e9-b08b-b42e9921e93e_xkb81.png\" style=\"vertical-align:middle;FLOAT:right\" />（2016•黑龙江一模）A、B、C三种固体物质的溶解度曲线如图所示．<br />（1）在t<SUB>1</SUB>℃时，欲将C物质的饱和溶液变为不饱和溶液，可采用的方法是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（2）现有A、B、C三种物质的浓溶液，适用于海水晒盐原理进行结晶的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（3）若将A的饱和溶液从50℃降温到20℃时，一定不会发生改变的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />A．溶解度&nbsp; B．溶剂的质量&nbsp; C．溶液的质量&nbsp; D．溶质的质量分数．','','','','','','降温（或加溶剂）$###$B$###$B','【解答】解：由于C物质的溶解度随温度的升高而减小，故在t<SUB>1</SUB>℃时，欲将C物质的饱和溶液变为不饱和溶液，可采用的方法是加入溶剂或降低温度；<br />（2）如果溶解度受温度影响较大的物质从溶液中结晶析出的方法是冷却热饱和溶液，如果溶解度受温度影响不大的物质从溶液中结晶析出的方法是蒸发溶剂，由图示可知只有B物质的溶解度受温度影响较小，用蒸发溶剂的方法析出晶体，所以适用于海水晒盐原理进行结晶的是B物质；<br />（3）若将A的饱和溶液从t<SUB>2</SUB>℃降温到t<SUB>1</SUB>℃时，一定不会发生改变的是溶剂质量，溶解度变小；溶质质量和溶质质量分数也都变小．<br />（1）（1）降温（或加溶剂）；（2）B；（3）B．','【分析】根据固体物质的溶解度曲线可以：①查出甲乙物质在某温度下的溶解度，②比较不同的物质在t<SUB>2</SUB>℃时的溶解度大小；③可以确定甲物质的饱和溶液变为不饱和溶液可采用的方法；④根据甲的溶解度随着温度的升高而增大且受温度的影响较大，判断提纯甲可采用的方法．','填空题',3.00,'2184ad2c161209cbfc20cf75d1976dd7',9,400,'饱和溶液和不饱和溶液相互转变的方法,固体溶解度曲线及其作用,晶体和结晶的概念与现象,海水晒盐的原理和过程','',2016,'37','2016•黑龙江一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841551,'按要求填空：<br />（1）分别写出两个含有“钅”字旁、“石”字旁和“气”字头的元素名称，同时将其相应的元素符号写在（　　）内：<br />①“钅”<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>）；<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>）；<br />②“石”<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>）；<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>）；<br />③“气”<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>）；<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>）；<br />（2）写出由①②③中六种元素中的两种元素组成的纯净物的化学式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>、<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>、<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（写出三种即可）；再写出由上述六种元素中的一种元素组成的纯净物的化学式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（写出一种即可）．','','','','','','镁$###$Mg$###$铁$###$Fe$###$碳$###$C$###$硫$###$S$###$氧$###$O$###$氢$###$H$###$CO<SUB>2</SUB>$###$MgO$###$CH<SUB>4</SUB>$###$O<SUB>2</SUB>','【解答】解：（1）书写元素符号时应注意：①有一个字母表示的元素符号要大写；②由两个字母表示的元素符号，第一个字母大写，第二个字母小写．<br />①“钅”镁（Mg）； 钠（Na）；<br />②“石”碳（C）； 硫（S）；<br />③“气”氧（O）； 氢（H ）；<br />（2）二氧化碳只有一种物质组成，是纯净物；氧化镁是由一种物质组成的，属于纯净物；甲烷是由一种物质组成的，属于纯净物；氧气是由一种物质组成的，属于纯净物；<br />故答案为；（1）①镁；Mg； 钠；Na；②碳；C； 硫；S；③氧；O； 氢；H；（2）CO<SUB>2</SUB>；MgO；CH<SUB>4</SUB>；O<SUB>2</SUB>．','【分析】（1）根据元素符号的名称、书写原则进行分析解答即可；正确书写元素符号，可以概括为“一大二小”；<br />（2）根据纯净物的定义与化学式的书写方法进行分析．','书写',3.00,'b77b27384e0a28f702680a0fd5c11396',9,400,'元素的符号及其意义,化学式的书写及意义','',2015,'35','2015秋•孝义市校级期中',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841554,'按要求填空：<br />（1）使用最多的金属材料是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（2）正常雨水的pH<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>7（填“大于”、“等于”或“小于”）；<br />（3）在化石燃料中，<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>是比较清洁的燃料；<br />（4）煤气中毒是指<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填物质名称）使人中毒．','','','','','','钢铁$###$小于$###$天然气$###$一氧化碳','【解答】解：（1）钢铁是使用最多的金属材料，故填：钢铁；<br />（2）正常雨水因为二氧化碳与水反应生成碳酸而呈酸性，其pH小于7，故填：小于；<br />（3）在化石燃料中，天然气燃烧的产物对环境的污染小，故填：天然气；&nbsp;&nbsp;<br />（4）一氧化碳是煤气的主要成分，会引起煤气中毒，故填：一氧化碳．','【分析】物质的性质决定物质的用途，根据已有的物质的性质进行分析解答即可．','填空题',3.00,'b7cea7bdad002ae5337a494530efa731',9,400,'一氧化碳的毒性,金属材料及其应用,溶液的酸碱性与pH值的关系,常用燃料的使用与其对环境的影响','安陆市',2016,'32','2016•安陆市模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841556,'我国海岸线曲折．海洋资源十分丰富．<img  src=\"/tikuimages/9/2016/400/shoutiniao2/2834b030-94d4-11e9-b7af-b42e9921e93e_xkb10.png\" style=\"vertical-align:middle\" /><br />（1）如图1A所示，从海水中获得淡水的方法是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，你认为该技术的优点是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．如图1B所示，是恒温下模拟海水晒盐过程的示意图，与丙烧杯烧杯溶液的溶质质量分数一定相同的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填序号，下同）．<br />（2）从海水中提取金属镁，可按图2流程进行：<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao84/28372130-94d4-11e9-a2f1-b42e9921e93e_xkb46.png\" style=\"vertical-align:middle\" /><br />①写出步骤Ⅱ发生反应的化学方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />②根据图2流程图，下列说法正确的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />A．步骤Ⅰ通过一步反应即可实现<br />B．步骤Ⅱ、Ⅲ、Ⅳ的目的是从海水中提纯氧化镁<br />C．步骤Ⅲ发生了中和反应<br />D．步骤Ⅴ中化学能转化为电能<br />③该流程中采取的一种降低成本且减少污染的措施是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）我国著名化学家侯德榜发明的“联合制碱法”主要工艺流程示意图如图3：<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao27/283859b0-94d4-11e9-b22d-b42e9921e93e_xkb63.png\" style=\"vertical-align:middle\" /><br />①为了除去粗盐水中可溶性杂质，可加入下列物质，利用过滤等操作进行除杂，则加入下列三种物质的先后顺序为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />a．稍过量的Na<SUB>2</SUB>CO<SUB>3</SUB>溶液　 b．适量的盐酸　 c．稍过量的Ba（OH）<SUB>2</SUB>溶液<br />②流程一，氨化溶液后，溶液的pH<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>7（填“＞”、“＜”或“═”），从而有利于吸收气体A<br />③流程一反应的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','蒸馏法$###$节约能源、减少成本$###$丁$###$Ca（OH）<SUB>2</SUB>+MgCl<SUB>2</SUB>=Mg（OH）<SUB>2</SUB>↓+CaCl<SUB>2</SUB>$###$BC$###$将有毒的氯气转化为盐酸，既降低成本又减少污染$###$cab$###$＞$###$NH<SUB>3</SUB>+NaCl+H<SUB>2</SUB>O+CO<SUB>2</SUB>=NaHCO<SUB>3</SUB>+NH<SUB>4</SUB>Cl','【解答】解：（1）如图Ⅰ所示，从海水中获得淡水的方法是蒸馏法，使用该技术的优点是节约能源、减少成本．恒温下模拟海水晒盐过程的示意图，丙和丁都是饱和溶液，故与丙烧杯烧杯溶液的溶质质量分数一定相同的是丁，故填：蒸馏法；节约能源、减少成本；丁．<br />（2）①氢氧化钙能与氯化镁反应生成氢氧化镁沉淀和氯化钙，故填：Ca（OH）<SUB>2</SUB>+MgCl<SUB>2</SUB>=Mg（OH）<SUB>2</SUB>↓+CaCl<SUB>2</SUB>．<br />②A．步骤Ⅰ通过碳酸钙高温分解生成氧化钙和水、氧化钙能与水反应生成氢氧化钙，不能通过一步反应实现，错误；<br />B．步骤Ⅱ、Ⅲ、Ⅳ的目的是从海水中提纯氧化镁，正确；<br />C．步骤Ⅲ是氢氧化镁与盐酸反应，发生的是中和反应，正确；<br />D．步骤Ⅴ中是电能转化为化学能，错误；<br />故填：BC；<br />③该流程中采取的一种降低成本且减少污染的措施是将有毒的氯气转化为盐酸，既降低成本又减少污染，故填：将有毒的氯气转化为盐酸，既降低成本又减少污染；<br />（3）为了除去粗盐水中可溶性杂质，可加入稍过量的Ba（OH）<SUB>2</SUB>溶液、稍过量的Na<SUB>2</SUB>CO<SUB>3</SUB>溶液、适量的盐酸，利用过滤等操作进行除杂，故加入下列三种物质的先后顺序为 cab，故填：cab；<br />②流程一，氨化溶液后，溶液呈碱性，溶液的pH大于7，从而有利于吸收气体A，故填：＞；<br />③流程一反应的化学方程式为NH<SUB>3</SUB>+NaCl+H<SUB>2</SUB>O+CO<SUB>2</SUB>=NaHCO<SUB>3</SUB>+NH<SUB>4</SUB>Cl，故填：NH<SUB>3</SUB>+NaCl+H<SUB>2</SUB>O+CO<SUB>2</SUB>=NaHCO<SUB>3</SUB>+NH<SUB>4</SUB>Cl．','【分析】根据海水淡化的方法、海水制镁的方法和流程以及海水制碱的知识进行分析解答即可．','书写',3.00,'362d8496ecd4bca542f22d7edc13c7a2',9,400,'溶质的质量分数,溶液的酸碱性与pH值的关系,盐的化学性质,纯碱的制取,物质发生化学变化时的能量变化,书写化学方程式、文字表达式、电离方程式,对海洋资源的合理开发与利用','招远市',2016,'32','2016•招远市模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841559,'如图是几种常见操作．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao54/283facb0-94d4-11e9-8b51-b42e9921e93e_xkb64.png\" style=\"vertical-align:middle\" /><br />①仪器a的名称是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；②实验C进行的操作是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />③操作有错误的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填字母）．','','','','','','酒精灯$###$加热液体$###$BD','【解答】解：<br />①根据常见仪器可知：仪器a的名称是酒精灯；<br />②实验C进行的操作是加热液体；<br />③A、稀释浓硫酸时，将浓硫酸倒入水中，并用玻璃棒搅拌，故正确；<br />B、氧气的验满时，应将带火星的木条放在集气瓶口部，木条复燃，说明已满，图中将带火星的木条伸入了集气瓶内，故错误；<br />C、给试管内的液体加热时，液体的体积不能超过了试管体积的<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">1</td></tr><tr><td>3</td></tr></table></span>，加热时，应是使试管于桌面成45°角，这样可使药品受热面积大，受热均匀；为避免液体沸腾溅出伤人，加热时切不可让试管口对着有人的方向，图中操作正确；<br />D、二氧化碳易溶于水，不能用排水法收集，故错误．<br />答案：①酒精灯；②加热液体；③BD．','【分析】①根据常见仪器分析解答．<br />②根据实验C进行的操作是加热液体解答．<br />③A、根据稀释浓硫酸的方法解答；<br />B、根据氧气验满的方法判断；<br />C、根据液体加热的方法判断；<br />D、根据二氧化碳的性质分析解答．','填空题',3.00,'d54398ec41d6c41ce4a378e665f375fc',9,400,'给试管里的液体加热,浓硫酸的性质及浓硫酸的稀释,常用气体的收集方法,氧气的检验和验满','',2016,'37','2016•重庆校级一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841561,'下列图象分别与选项中的操作相对应，其中不合理的是（　　）','<img  src=\"/tikuimages/9/2016/400/shoutiniao97/2842e100-94d4-11e9-a8f7-b42e9921e93e_xkb24.png\" style=\"vertical-align:middle\" /><br />向相同质量的锌和铁中分别加入足量的溶质质量分数相同的稀盐酸','<img  src=\"/tikuimages/9/2016/400/shoutiniao96/2848864f-94d4-11e9-a10b-b42e9921e93e_xkb82.png\" style=\"vertical-align:middle\" /><br />向一定量二氧化锰固体中加入一定量过氧化氢溶液','<img  src=\"/tikuimages/9/2016/400/shoutiniao37/2849becf-94d4-11e9-8925-b42e9921e93e_xkb89.png\" style=\"vertical-align:middle\" /><br />向一定量碳酸钠溶液中滴入水','<img  src=\"/tikuimages/9/2016/400/shoutiniao9/284be1b0-94d4-11e9-a510-b42e9921e93e_xkb81.png\" style=\"vertical-align:middle\" /><br />向一定质量AgNO<SUB>3</SUB>和Cu（NO<SUB>3</SUB>）<SUB>2</SUB>的混合溶液中加入过量的Zn','','C','【解答】解：A、向相同质量的锌和铁中分别加入足量的溶质质量分数相同的稀盐酸，反应速率相同，但是等质量的铁与酸反应生成的氢气多，故A正确；<br />B、催化剂在化学反应前后质量和化学性质不变，所以二氧化锰的质量随着反应的进行，质量不变，故B正确<br />C、碳酸钠加水稀释的过程中，pH值逐渐减小，但是不会小于7，故C错误；<br />D、锌和硝酸银反应会生成硝酸锌和银，锌和硝酸铜反应会生成硝酸锌和铜，每65份质量的锌会置换出216份质量的银，每65份质量的锌会置换出64份质量的铜，所以溶液质量先减小，然后增大，故D正确．<br />故选：C．','【分析】A、根据向相同质量的锌和铁中分别加入足量的溶质质量分数相同的稀盐酸，反应速率相同，但是等质量的铁与酸反应生成的氢气多进行分析；<br />B、根据催化剂的定义进行分析；<br />C、根据碳酸钠加水稀释的过程中，pH值逐渐减小，但是不会小于7进行分析；<br />D、根据锌和硝酸银反应会生成硝酸锌和银，锌和硝酸铜反应会生成硝酸锌和铜进行分析．','选择题',3.00,'cc6dff212239584fc79d7343610ec4fa',9,400,'催化剂的特点与催化作用,金属的化学性质,酸碱溶液的稀释,溶液的酸碱性与pH值的关系','',2016,'37','2016•宛城区一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841566,'<img  src=\"/tikuimages/9/2012/400/shoutiniao19/28624fe1-94d4-11e9-9f8a-b42e9921e93e_xkb64.png\" style=\"vertical-align:middle;FLOAT:right;\" />（2012秋•阜南县校级期中）如图所示装置，有洗气，储气等多种用途．<br />（1）若用此装置收集密度比空气大的气体，则从<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>端进气．<br />（2）瓶内装满水，用排水法收集气体，应从<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>&nbsp;端进气．','','','','','','a$###$b','【解答】解：（1）收集的气体的密度比空气的密度大，所以若用排空气法收集气体时，应用向上排空气法，气体从装置a端通入；<br />（2）气体的密度比水的密度小，所以若用排水法收集气体时，瓶内先装满水，气体从装置b端通入．<br />故填：（1）a；（2）b；','【分析】（1）根据气体的密度比空气的密度大，应用向上排空气法进行分析；<br />（2）根据气体的密度比水的密度小进行解答．','填空题',3.00,'c2bbeaf997b99f5eb67a7712ecef8a24',9,400,'常用气体的收集方法','',2012,'35','2012秋•阜南县校级期中',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841579,'建立宏观、微观和符号之间的联系是化学学科的特点．<br />（1）图1表示水、二氧化碳和氧气的循环图．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao98/28903dae-94d4-11e9-bcd5-b42e9921e93e_xkb3.png\" style=\"vertical-align:middle\" /><br />①从宏观角度看，图1中三种物质的组成中都含有<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />从微观角度看，图1中三种物质都是由<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>构成的．<br />②试写出图1中c对应的化学方程式：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />③图2为图1中b对应一种反应的微观示意图．由此图你能获得的信息有<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（任写一点）．<br />（2）图3表示氯和溴元素（Br）的原子结构示意图，由图可知，溴在元素周期表中位于第<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>周期，氯和溴具有相似化学性质是因为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，氯气常用于自来水消毒，是因为氯气能与水反应生成盐酸和次氯酸（HClO），写出溴单质（Br<SUB>2</SUB>）与氢氧化钠溶液反应的化学方程式：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','氧元素$###$分子$###$2H<SUB>2</SUB>O<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;通电&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>↑+O<SUB>2</SUB>↑$###$化学反应前后原子的种类和数目不变$###$四$###$原子的最外层电子数相同$###$Br<SUB>2</SUB>+2NaOH═NaBr+NaBrO+H<SUB>2</SUB>O','【解答】解：（1）①从宏观角度看，图1中三种物质的组成中都含有氧元素．从微观角度看，图1中三种物质都是由分子 构成的．<br />②图1中c对应的反应可以是水通电生成了氧气和氢气，反应的化学方程式是：2H<SUB>2</SUB>O<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;通电&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>↑+O<SUB>2</SUB>↑．<br />③图2为图1中b对应一种反应的微观示意图．由微粒的变化可知，化学反应前后原子的种类和数目不变等．<br />（2）图3表示氯和溴元素（Br）的原子结构示意图，由图可知，溴原子的核外有四个电子层，该元素周期表中位于第四周期，氯和溴具有相似化学性质是因为原子的最外层电子数相同，氯气常用于自来水消毒，是因为氯气能与水反应生成盐酸和次氯酸（HClO），则溴单质（Br<SUB>2</SUB>）与氢氧化钠溶液反应的化学方程式为：Br<SUB>2</SUB>+2NaOH═NaBr+NaBrO+H<SUB>2</SUB>O．<br />故答为：（1）①氧元素，分子&nbsp;&nbsp;&nbsp;②2H<SUB>2</SUB>O<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;通电&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>↑+O<SUB>2</SUB>↑&nbsp;③化学反应前后原子的种类和数目不变；（2）原子的最外层电子数相同&nbsp;&nbsp;Br<SUB>2</SUB>+2NaOH═NaBr+NaBrO+H<SUB>2</SUB>O','【分析】（1）①根据物质的组成和构成分析回答；<br />②根据水通电生成了氧气分析回答；<br />③根据微粒的变化分析分子、原子的变化等；<br />（2）在原子中，电子层数等于周期数，最外层电子数决定了元素性质，根据氯气能与水的反应，写出溴单质（Br<SUB>2</SUB>）与氢氧化钠溶液反应的化学方程式．','书写',3.00,'271ce2e91789c38b5912b8441def1f50',9,400,'物质的相互转化和制备,分子、原子、离子、元素与物质之间的关系,微粒观点及模型图的应用,原子结构示意图与离子结构示意图,物质的元素组成,书写化学方程式、文字表达式、电离方程式','',2016,'32','2016•海安县模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841591,'下列对实验分析所得出的结论错误的是（　　）<br /><table class=\"edittable\"><TBODY><TR><td width=30></TD><td width=215>实验操作</TD><td width=172>现象</TD><td width=162>结论</TD></TR><TR><td>A</TD><td>向浸没在热水中的白磷通入氧气</TD><td>通入氧气后白磷燃烧</TD><td>氧气是燃烧需要的条件之一</TD></TR><TR><td>B</TD><td>点燃某气体，在火焰上方罩冷而干燥的烧杯</TD><td>烧杯内壁有无色液滴产生</TD><td>该气体可能是H<SUB>2</SUB>、CH<SUB>4</SUB>或CO中的一种</TD></TR><TR><td>C</TD><td>50℃时，将33gKCl固体加入100g蒸馏水中，完全溶解后缓慢降温</TD><td>20℃时继续降温有晶体析出</TD><td>20℃时，KCl的溶解度为33g</TD></TR><TR><td>D</TD><td>用两支大小相同的注射器分别吸入等体积的空气和水，堵住注射器末端小孔，慢慢推栓塞</TD><td>装有空气的注射器内物质易被压缩</TD><td>气体物质微粒间空隙较大；液体物质微粒间空隙较小</TD></TR></TBODY></TABLE>','A','B','C','D','','B','【解答】解：A、向浸没在热水中的白磷通氧气，白磷可以燃烧，说明燃烧需要有氧气，即氧气是燃烧需要的条件之一，故选项实验结论正确．<br />B、点燃某气体，在火焰上方罩冷而干燥的烧杯，烧杯内壁有无色液滴产生，氢气、甲烷燃烧均有水生成，该气体该气体可能是H<SUB>2</SUB>、CH<SUB>4</SUB>，一氧化碳燃烧生成二氧化碳，没有水生成，该气体不可能是一氧化碳，故选项实验结论错误．<br />C、50℃时，将33gKCl固体加入100g蒸馏水中，完全溶解后缓慢降温，降温至20℃时开始有晶体析出，说明20℃时，100g水中最多溶解氯化钾33g，即20℃时，KCl的溶解度为33g，故选项实验结论正确．<br />D、用两支大小相同的注射器分别吸入等体积的空气和水，堵住注射器末端小孔，慢慢推栓塞，装有空气的注射器内物质易被压缩，气体容易压缩，说明气体物质微粒间空隙较大；液体物质微粒间空隙较小，故选项实验结论正确．<br />故选：B．','【分析】A、根据燃烧需要同时满足三个条件：①可燃物、②氧气或空气、③温度要达到着火点，进行分析判断．<br />B、根据烧杯内壁有无色液滴产生，说明有水生成，进行分析判断．<br />C、根据溶解度是在一定温度下，某固体溶质在100g溶剂里达到饱和状态所溶解的溶质质量，进行分析判断．<br />D、根据分子之间存在间隔进行分析判断．','选择题',3.00,'301f8bfb80d7496632a350288b139246',9,400,'化学实验方案设计与评价,常见气体的检验与除杂方法,固体溶解度的概念,燃烧与燃烧的条件','',2016,'37','2016•苏州一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841595,'根据下图所示实验分析得出的结论不正确的是（　　）','<img  src=\"/tikuimages/9/2016/400/shoutiniao28/28c49421-94d4-11e9-83eb-b42e9921e93e_xkb55.png\" style=\"vertical-align:middle\" />&nbsp;<br />蜡烛燃烧生成二氧化碳','<img  src=\"/tikuimages/9/2016/400/shoutiniao47/28c7051e-94d4-11e9-886c-b42e9921e93e_xkb61.png\" style=\"vertical-align:middle\" /><br />铁能在氧气中燃烧','<img  src=\"/tikuimages/9/2016/400/shoutiniao9/28c8b2cf-94d4-11e9-b596-b42e9921e93e_xkb5.png\" style=\"vertical-align:middle\" /><br />白磷的着火点比红磷低','<img  src=\"/tikuimages/9/2016/400/shoutiniao91/28c94f0f-94d4-11e9-8799-b42e9921e93e_xkb95.png\" style=\"vertical-align:middle\" /><br />说明稀硫酸与氢氧化钠反应','','D','【解答】解：A、蜡烛燃烧生成二氧化碳和水蒸气，二氧化碳能使澄清石灰水变浑浊，实验过程中澄清石灰水变浑浊，说明蜡烛燃烧生成了二氧化碳，该选项说法正确；<br />B、实验过程中，铁丝剧烈燃烧，火星四射，铁在氧气中燃烧生成四氧化三铁，该选项说法正确；<br />C、实验过程中，白磷燃烧，红磷不能燃烧，说明白磷的着火点比红磷低，该选项说法正确；<br />D、虽然氢氧化钠和稀硫酸反应生成硫酸钠和水，但是实验过程中无明显现象，因此不能说明稀硫酸与氢氧化钠反应，该选项说法不正确．<br />故选：D．','【分析】蜡烛燃烧生成二氧化碳和水蒸气，二氧化碳能使澄清石灰水变浑浊；<br />铁在氧气中燃烧生成四氧化三铁；<br />根据实验现象可以判断可燃物的着火点高低；<br />氢氧化钠和稀硫酸反应生成硫酸钠和水．','选择题',3.00,'dd90f67b49a0a0dd3805bd9c2c9d7840',9,400,'化学实验方案设计与评价,蜡烛燃烧实验,氧气的化学性质,中和反应及其应用,燃烧与燃烧的条件','',2016,'37','2016•宝安区二模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841597,'久置空气中的氢氧化钠固体是否变质，进行了如下探究．<br />（1）久置空气中的氢氧化钠固体：①吸收水份而潮解；②吸收<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>而变质；<br />（2）为了验证氢氧化钠是否变质，分别选用下面三种试剂进行实验：<br />&nbsp;&nbsp;&nbsp; 一、稀盐酸；二、氢氧化钙溶液；三、氯化钙溶液<br />①其中一定要将氢氧化钠样品配成溶液再进行实验的试剂是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，经检验该样品已经变质；<br />②试剂三实验中发生反应的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />③实验后将所有物质倒入一个洁净的烧杯中，得到无色澄清溶液，此溶液中一定含有的溶质是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','CO<SUB>2</SUB>$###$二、三（或氢氧化钙溶液、氯化钙溶液）$###$Na<SUB>2</SUB>CO<SUB>3</SUB>+CaCl<SUB>2</SUB>═CaCO<SUB>3</SUB>↓+2NaCl$###$NaCl、CaCl<SUB>2</SUB>','【解答】解：<br />（1）因固体氢氧化钠易吸水而潮解，与二氧化碳反应生成碳酸钠和水，变质时氢氧化钠中混有碳酸钠；<br />（2）①检验氢氧化钠溶液中是否含有碳酸钠，可向溶液中加入适量稀盐酸或稀硫酸，有气泡冒出即可检验含有碳酸钠；可向溶液中滴加某些碱溶液如氢氧化钙或氢氧化钡溶液，有沉淀产生即可检验含有碳酸钠；可向溶液中加入某些盐溶液，如氯化钙或硝酸钙，有沉淀产生即可检验含有碳酸钠；<br />其中一定要将氢氧化钠样品配成溶液再进行实验的试剂是二、三（或氢氧化钙溶液、氯化钙溶液）；<br />②氢氧化钙溶液和和碳酸钠反应生成碳酸钙沉淀和氢氧化钠；氯化钙溶液和碳酸钠反应生成碳酸钙沉淀和氯化钠，反应的化学方程式为Na<SUB>2</SUB>CO<SUB>3</SUB>+CaCl<SUB>2</SUB>═CaCO<SUB>3</SUB>↓+2NaCl；<br />③实验后将所有物质倒入一个洁净的烧杯中，得到无色澄清溶液，此溶液中一定含有的溶质是NaCl、CaCl<SUB>2</SUB><br />故答案为：<br />（1）CO<SUB>2</SUB><br />（2）<br />①二、三（或氢氧化钙溶液、氯化钙溶液）；<br />②Na<SUB>2</SUB>CO<SUB>3</SUB>+CaCl<SUB>2</SUB>═CaCO<SUB>3</SUB>↓+2NaCl&nbsp;<br />③NaCl、CaCl<SUB>2</SUB>．','【分析】本题考查氢氧化钠的性质，固体氢氧化钠易潮解，能与空气中的二氧化碳发生化学反应，生成碳酸钠，因此检验是否变质即检验是否存在碳酸根离子；据此进行分析解答．','书写',3.00,'4f6f76c069bcb389fec44dc501f6e665',9,400,'药品是否变质的探究,碱的化学性质,盐的化学性质,书写化学方程式、文字表达式、电离方程式','',2015,'37','2015•长春二模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841599,'对比归纳是学习化学的重要方法，某同学设计了如图甲、乙所示的两组对比实验，请你回答相关问题：<img  src=\"/tikuimages/9/0/400/shoutiniao87/28cd94d1-94d4-11e9-a85a-b42e9921e93e_xkb16.png\" style=\"vertical-align:middle\" /><br />（1）如图甲，分别将硫在空气和氧气中燃烧，发现硫在氧气里燃烧比在空气里燃烧更剧烈，则说明影响硫燃烧剧烈程度的因素是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）①通过如图乙所示A试管和B试管的对比实验，发现B中固体可溶，A中几乎不溶，该对比实验说明了影响物质溶解性的因素是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />②该同学又补充了如图乙中试管C所示实验，他想探究的影响因素是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','氧气的含量$###$溶剂的性质$###$溶质的性质','【解答】解：（1）由实验的过程可知，由于氧气的含量不同，硫在氧气中燃烧比在空气中燃烧剧烈，实验的现象不同，说明了响硫燃烧剧烈程度的因素是氧气的含量．<br />（2）①通过图乙所示A试管和B试管的对比实验，发现B中固体可溶，A中几乎不溶，该对比实验说明了同种溶质在不同溶剂中的溶解性不同，说明了影响物质溶解性的因素是溶剂的性质；<br />②该同学又补充了如图乙中试管C所示实验，通过A、C对比可以看出，不同的溶质在同一种溶剂中的溶解性不同，他想探究的影响因素是溶质的性质．<br />故答为：（1）氧气的含量；（2）①溶剂的性质；②溶质的性质．','【分析】（1）根据氧气的含量、实验的现象分析回答；<br />（2）根据不同的溶质和不同的溶剂分析物质的溶解性．','填空题',3.00,'fae35ec1717b6d951f68c5eb174f2a25',9,400,'氧气的化学性质,物质的溶解性及影响溶解性的因素','',0,'37','',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841602,'元素周期表是学习和研究化学的重要工具．从微观的角度了解物质及其变化，有助于认识物质变化的本质．回答下列问题：<br />（1）与元素的化学性质的关系最为密切的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填字母）．<br />A．元素的相对原子质量&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;B．元素的核电荷数<br />C．原子的核外电子数&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;D．原子的最外层电子数<br />（2）图1是元素周期表中硒元素的部分信息，图2是硒原子的结构示意图．据图填空：<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao65/28d1da8f-94d4-11e9-8164-b42e9921e93e_xkb89.png\" style=\"vertical-align:middle\" /><br />①图2中m的值为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，n的值为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />②硒元素位于元素周期表中的第<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>周期．<br />③硒原子在化学反应中容易<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“得到”或“失去”）电子．硒化钠是由钠元素和硒元素组成的化合物，硒化钠的化学式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）如图3是两个化学反应的微观示意图．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao4/28d3130f-94d4-11e9-82dd-b42e9921e93e_xkb70.png\" style=\"vertical-align:middle\" /><br />①反应A的微观实质是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />②反应B的微观实质是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（写粒子符号）结合成水分子．','','','','','','D$###$34$###$6$###$四$###$得到$###$Na<SUB>2</SUB>Se$###$化学变化中，分子变而原子不变（分子改变，原子重新组合）$###$H<SUP>+</SUP>与OH<SUP>-</SUP>','【解答】解：<br />（1）元素性质与原子核外电子的排布，特别是最外层上的电子数目有密切关系，决定元素化学性质的是最外层电子数．<br />A、与元素的化学性质关系最为密切的是原子的最外层电子数，而不是元素的相对原子质量，故选项错误．<br />B、与元素的化学性质关系最为密切的是原子的最外层电子数，而不是元素的核电荷数，故选项错误．<br />C、与元素的化学性质关系最为密切的是原子的最外层电子数，而不是原子的核外电子数，故选项错误．<br />D、与元素的化学性质关系最为密切的是原子的最外层电子数，故选项正确．<br />（2）①根据元素周期表中的一格可知，左上角的数字为34，表示原子序数为34；根据原子序数=核电荷数=质子数，则图2中m的值为34；当质子数=核外电子数，为原子，则34=2+8+18+n，n=6．故填：34；6<br />②周期数=原子核外电子层数，硒原子的核外有4个电子层，则硒元素位于元素周期表中的第四周期．故填：四；<br />③硒原子的最外层电子数为6，在化学反应中易得到2个电子而形成带2个单位负电荷的阴离子，化合价的数值等于离子所带电荷的数值，且符号一致，则该元素的化合价为-2价；钠元素显+1价，则硒化钠的化学式为Na<SUB>2</SUB>Se．故填：Na<SUB>2</SUB>Se．<br />（3）由图1反应可知，该反应过程中氢原子和氧原子结合成水分子，氢气和氧气反应生成了水；由图2微粒的变化可知，图2反应的实质是氢离子和氢氧根离子结合成水分子．<br />答案：<br />（1）D；&nbsp;&nbsp;&nbsp;&nbsp;<br />（2）①34；&nbsp;6；&nbsp;②四&nbsp; ③得到；&nbsp;Na<SUB>2</SUB>Se；<br />（3）①化学变化中，分子变而原子不变（分子改变，原子重新组合）；②H<SUP>+</SUP>与OH<SUP>-</SUP>．','【分析】（1）根据元素的化学性质跟它的原子的最外层电子数目关系非常密切，决定元素化学性质的是最外层电子数，据此进行分析解答；<br />（2）①根据图中元素周期表可以获得的信息：左上角的数字表示原子序数；原子序数=核电荷数=质子数，结合当质子数=核外电子数，为原子，进行分析解答．<br />②根据周期数=原子核外电子层数进行分析解答；<br />③根据化合物的化学式的写法来分析．<br />（3）根据图1氢气燃烧的微观示意图，分析解答；观察微粒的变化，分析分子的形成．','填空题',3.00,'939bd8d67e7523512848b8924bace44b',9,400,'中和反应及其应用,微粒观点及模型图的应用,核外电子在化学反应中的作用,原子结构示意图与离子结构示意图,元素周期表的特点及其应用','',2016,'32','2016•如东县模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841604,'在工业上用纯碱和石灰石为原料制备烧碱，主要流程如图所示：<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao97/28d954a1-94d4-11e9-903a-b42e9921e93e_xkb23.png\" style=\"vertical-align:middle\" /><br />（1）在①-④四个过程中，发生了化合反应的为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填序号）．<br />（2）③中实验操作过程的名称为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，经搅拌后，得到石灰水<br />（3）NaCl、KNO<SUB>3</SUB>、Ca（OH）<SUB>2</SUB>三物质的溶解度曲线图分别如图一、图二所示，请回答：<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao1/28db5070-94d4-11e9-a882-b42e9921e93e_xkb6.png\" style=\"vertical-align:middle\" /><br />①由图一可知，20℃时，KNO<SUB>3</SUB>溶液的溶质质量分数NaCl溶液的溶质质量分数<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“大于”“小于”“等于”“无法确定”之一）．<br />②由图一可知，将&nbsp;20℃时&nbsp;KNO<SUB>3</SUB>饱和溶液，升温到&nbsp;50℃时，溶液的溶质质量分数<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“变大”“变小”“不变”之一）．<br />③如图三所示，20℃时，将盛有饱和石灰水的小试管放入盛水的烧杯中，向水中加入氢氧化钠固体，结合图二分析石灰水中可出现的现象<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','②$###$溶解$###$无法确定$###$不变$###$变浑浊','【解答】解：（1）在①～④四个过程中，氧化钙和水反应生成氢氧化钙，属于化合反应．<br />（2）③中实验操作过程的名称为溶解，经搅拌后，得到石灰水．<br />（3）①因为不知道硝酸钾和氯化钠的水溶液中的溶质质量和溶剂质量，因此20℃时，无法判断KNO<SUB>3</SUB>溶液的溶质质量分数和NaCl溶液的溶质质量分数的大小．<br />②由图二可知，硝酸钾的溶解度随着温度升高而增大，因此将20℃时KNO<SUB>3</SUB>饱和溶液，升温到50℃时，溶质质量和溶液质量都不变，溶液的溶质质量分数不变．<br />③氢氧化钙的溶解度随着温度升高而减小，20℃时，将盛有饱和石灰水的小试管放入盛水的烧杯中，向水中加入氢氧化钠固体，氢氧化钠溶于水放热，导致溶液温度升高，氢氧化钙的溶解度减小，溶液变浑浊．<br />答案：<br />（1）②；<br />（2）溶解；<br />（3）①无法确定；②不变；③变浑浊．','【分析】（1）根据高温条件下，碳酸钙分解生成氧化钙和二氧化碳，氧化钙和水反应生成氢氧化钙，氢氧化钙和碳酸钠反应生成碳酸钙沉淀和氢氧化钠解答；<br />（2）根据溶液中溶质质量分数=<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">溶质质量</td></tr><tr><td>溶液质量</td></tr></table></span>×100%；根据物质的溶解度曲线可以判断相关方面的问题．','填空题',3.00,'c01bfbd54c9d7b9372c37a2b12cdb65c',9,400,'溶解时的吸热或放热现象,固体溶解度曲线及其作用,溶质的质量分数,溶质的质量分数、溶解性和溶解度的关系,碳酸钙、生石灰、熟石灰之间的转化,盐的化学性质,物质的相互转化和制备','',2016,'37','2016•青岛一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841607,'<img  src=\"/tikuimages/9/2016/400/shoutiniao20/28e5146e-94d4-11e9-99de-b42e9921e93e_xkb6.png\" style=\"vertical-align:middle;FLOAT:right\" />（2016•松北区二模）某化学活动小组在一次实验中进行了如图所示实验：<br />实验后，小组同学充分振荡U形玻璃管，将U形玻璃管内的物质过滤后，向滤液中滴加无色酚酞试液，滤液不变色．同学们对滤液所含的阴离子进行了进一步的探究活动．<br />【提出问题】滤液中含有什么阴离子？<br />【猜想与假设】A同学：含有SO<SUB>4</SUB><SUP>2-</SUP><br />B同学：含有Cl<SUP>-</SUP><br />C同学：含有（1）<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填离子符号）<br />请你评价A同学的猜想：（2）<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />【实验探究】B同学为了验证自己的猜想，取滤液于试管中，向其中滴加足量的硝酸银溶液，观察到有白色沉淀产生．由此B同学得出结论：自己的猜想成立．<br />【评价与交流】请你评价B同学的结论：（3）<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．同学们经过讨论后一致认为：B同学在验证自己的猜想时，若将所加试剂调换成适量的另一种试剂（4）<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填化学式），过滤后，再向所得滤液中滴加（5）<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，既能验证自己的猜想，也能确定滤液中的离子成分．<br />同学们利用该方案进行实验，最终确定了滤渡中的离子成分．<br />【归纳与总结】通过上述实验我们能够知道：在确定该实验后所得溶液中的阴离子成分时，一般应加入（6）<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，再验证（7）<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，这样就可以确定溶液中的阴离子成分．','','','','','','SO<SUB>4</SUB><SUP>2-</SUP>、Cl<SUP>-</SUP>$###$CuCl<SUB>2</SUB>为溶于水的生成物，所以滤液中一定含有Cl<SUP>-</SUP>$###$不正确，SO<SUB>4</SUB><SUP>2-</SUP>和Ag<SUP>+</SUP>结合会形成微溶于水的Ag<SUB>2</SUB>SO<SUB>4</SUB>，当有大量Ag<SUB>2</SUB>SO<SUB>4</SUB>存在时会形成白色沉淀$###$Ba（NO<SUB>3</SUB>）<SUB>2</SUB>$###$硝酸银溶液$###$验证并除去SO<SUB>4</SUB><SUP>2-</SUP>的溶液$###$Cl<SUP>-</SUP>的存在','【解答】解：<br />实验后，小组同学充分振荡U形玻璃管，将U形玻璃管内的物质过滤后，向滤液中滴加无色酚酞试液，滤液不变色．说明滤液中无氢氧化钡溶液，同学们对滤液所含的阴离子进行了进一步的探究活动．<br />【猜想与假设】根据“A同学：含有SO<SUB>4</SUB><SUP>2-</SUP>；B同学：含有Cl<SUP>-</SUP>”，则推测C同学：含有SO<SUB>4</SUB><SUP>2-</SUP>、Cl<SUP>-</SUP>；评价A同学的猜想：CuCl<SUB>2</SUB>为溶于水的生成物，所以滤液中一定含有Cl<SUP>-</SUP>．<br />【实验探究】B同学为了验证自己的猜想，取滤液于试管中，向其中滴加足量的硝酸银溶液，观察到有白色沉淀产生．由此B同学得出结论：自己的猜想成立．B同学的结论：不正确，SO<SUB>4</SUB><SUP>2-</SUP>和Ag<SUP>+</SUP>结合会形成微溶于水的Ag<SUB>2</SUB>SO<SUB>4</SUB>，当有大量Ag<SUB>2</SUB>SO<SUB>4</SUB>存在时会形成白色沉淀．同学们经过讨论后一致认为：B同学在验证自己的猜想时，若将所加试剂调换成适量的另一种试剂Ba（NO<SUB>3</SUB>）<SUB>2</SUB>，过滤后，再向所得滤液中滴加硝酸银溶液，既能验证自己的猜想，也能确定滤液中的离子成分．<br />【归纳与总结】通过上述实验我们能够知道：在确定该实验后所得溶液中的阴离子成分时，一般应加入验证并除去SO<SUB>4</SUB><SUP>2-</SUP>的溶液，再验证Cl<SUP>-</SUP>的存在，这样就可以确定溶液中的阴离子成分．<br />故答案为：<br />（1）SO<SUB>4</SUB><SUP>2-</SUP>、Cl<SUP>-</SUP>；（2）CuCl<SUB>2</SUB>为溶于水的生成物，所以滤液中一定含有Cl<SUP>-</SUP>．<br />（3）不正确，SO<SUB>4</SUB><SUP>2-</SUP>和Ag<SUP>+</SUP>结合会形成微溶于水的Ag<SUB>2</SUB>SO<SUB>4</SUB>，当有大量Ag<SUB>2</SUB>SO<SUB>4</SUB>存在时会形成白色沉淀．<br />（4）Ba（NO<SUB>3</SUB>）<SUB>2</SUB>；（5）硝酸银溶液；（6）验证并除去SO<SUB>4</SUB><SUP>2-</SUP>的溶液；（7）Cl<SUP>-</SUP>的存在．','【分析】本题能使同学们体会到实验探究的一般过程，通过实验分析可知：如图所示的实验中，此时观察到的实验现象是：U形玻璃管左、右两端均产生白色沉淀；猜想与假设中，C同学的猜想是：含有SO<SUB>4</SUB><SUP>2-</SUP>、Cl<SUP>-</SUP>；B同学的结论不正确，因为加入硝酸银溶液，无明显现象，说明溶液中一定不含氯离子，所以B同学“自己的猜想不成立”的结论正确，但无法证明溶液中是否含有硫酸根离子；同学们经过讨论后一致认为：B同学在验证自己的猜想时，如果滴加Ba（NO<SUB>3</SUB>）<SUB>2</SUB>溶液，过滤后，再向所得滤液中滴加硝酸银溶液，既能验证自己的猜想，也能确定滤液中的离子成分；通过上述实验我们能够总结出，在确定化学反应后所得溶液中的离子成分时，一般应加入适当的、适量的试剂，验证溶液中可能存在的离子是否存在就可以确定溶液中的离子成分．','填空题',3.00,'7c65b613f9ccc3bdf0b99aa820a057b7',9,400,'实验探究物质的组成成分以及含量,常见离子的检验方法及现象','',2016,'37','2016•松北区二模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841609,'下列做法不正确的是（　　）','液化石油气泄漏时应立即关掉气瓶阀门并开窗通风','只要可燃物的温度达到了着火点，可燃物就会燃烧','燃着的酒精灯打翻烧起来时，应立即用湿抹布盖灭','应该用湿毛巾捂住口鼻并尽量贴着地面逃离火灾区','','B','【解答】解：A、液化石油气泄漏时应立即关掉气瓶阀门并开窗通风，正确；<br />B、可燃物燃烧需要温度达到可燃物的着火点并且与氧气接触，错误；<br />C、燃着的酒精灯打翻烧起来时，应立即用湿抹布盖灭，可以隔绝氧气，正确；<br />D、用湿毛巾捂住口鼻并尽量贴着地面逃离火灾区，正确；<br />故选B．','【分析】A、根据燃气泄露的处理方法解答；<br />B、根据燃烧的条件进行分析解答；<br />C、根据灭火的原理解答；<br />D、根据火灾现场的逃生方法解答．','选择题',3.00,'69f61741fcf0db71bd68c8206915d7e5',9,400,'燃烧与燃烧的条件,灭火的原理和方法,防范爆炸的措施','',2016,'37','2016•工业园区一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841612,'<img  src=\"/tikuimages/9/2016/400/shoutiniao89/28f1225e-94d4-11e9-9728-b42e9921e93e_xkb88.png\" style=\"vertical-align:middle;FLOAT:right\" />（2016•潮阳区模拟）控制变量法是学习化学常用的方法，环保是实验必须提倡的内容，请你根据如图探究燃烧条件的实验图示和资料信息回答有关问题．<br />资料信息：白磷和红磷的着火点分别是40℃、240℃．<br />（1）水中的白磷不燃烧，是因为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）罩上仪器a的作用是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）对照①和③能够获得可燃物燃烧需要的条件之一是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','没有与氧气接触$###$防止五氧化二磷逸散到空气中，污染空气$###$温度达到可燃物的着火点','【解答】解：<br />（1）水中的白磷没有与氧气接触，不会燃烧，故填：没有与氧气接触；<br />（2）罩上仪器a的作用是防止五氧化二磷逸散到空气中，污染空气；<br />（3）铜片上的红磷不燃烧是因为温度没有达到红磷的着火点，对照①和③能够获得可燃物燃烧需要的条件之一是：温度达到可燃物的着火点．<br />答案：<br />（1）没有与氧气接触；<br />（2）防止五氧化二磷逸散到空气中，污染空气；<br />（3）温度达到可燃物的着火点．','【分析】（1）根据燃烧的条件进行分析解答，燃烧需要可燃物与氧气接触且温度达到可燃物的着火点，据此解答；<br />（2）根据磷燃烧生成五氧化二磷，污染空气解答；<br />（3）根据燃烧的条件进行分析解答．','填空题',3.00,'6033c0410a146ad963b698acd5f28e18',9,400,'燃烧的条件与灭火原理探究','',2016,'32','2016•潮阳区模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841637,'下列说法中，正确的是（　　）','常温下，碱溶液的pH一定大于7，盐溶液的pH一定等于7','向某溶液中滴入BaCl<SUB>2</SUB>溶液，有白色沉淀生成，该溶液中一定有Ag<SUP>+</SUP>','区别NH<SUB>4</SUB>Cl和KCl粉末，可用熟石灰研磨并闻气味的方法','碳酸氢钠是一种盐，俗称苏打，它的水溶液呈碱性','','C','【解答】解：A、常温下，碱溶液的pH一定大于7，盐溶液的pH不一定等于7，也可能小于7或大于7，故选项说法错误．<br />B、向某溶液中滴入BaCl<SUB>2</SUB>溶液，有白色沉淀生成，该溶液中一定有Ag<SUP>+</SUP>，也可能含硫酸根离子等，故选项说法错误．<br />C、NH<SUB>4</SUB>Cl属于铵态氮肥，与熟石灰混合研磨产生有刺激性气味的气体，氯化钾不能，区别NH<SUB>4</SUB>Cl和KCl粉末，可用熟石灰研磨并闻气味的方法，故选项说法正确．<br />D、碳酸氢钠是一种盐，俗称小苏打，它的水溶液呈碱性，故选项说法错误．<br />故选：C．','【分析】A、根据常见盐的酸碱性，进行分析判断．<br />B、根据BaCl<SUB>2</SUB>溶液能与含硫酸根离子和银离子的溶液反应生成硫酸钡或氯化银沉淀等，进行分析判断．<br />C、根据铵态氮肥与碱性物质混合研磨后能放出有刺激性气味的气体，进行分析判断．<br />D、根据碳酸氢钠的俗称、酸碱性等进行分析判断．','选择题',3.00,'5a600b12c88d1819b03177314882650b',9,400,'溶液的酸碱性与pH值的关系,盐的化学性质,铵态氮肥的检验,常见离子的检验方法及现象','',2016,'35','2016春•吉安校级期中',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841640,'2015年4月17日上午，青岛一购物中心举行了一场“全民砸西瓜”活动．现场，众多青岛市民亲自销毁了“问题西瓜”事件发生后被下架的西瓜．下列有关“问题西瓜”的各项研究内容中，不属于化学研究范畴的是（　　）','问题西瓜是哪里种植发送过来的','西瓜富含哪些营养物质','残留农药的主要成分是什么','西瓜中糖类物质的组成是什么','','A','【解答】解：A、问题西瓜是哪里种植发送过来的不属于化学研究内容，正确；<br />B、西瓜富含哪些营养物质属于化学研究内容，错误；<br />C、残留农药的主要成分属于化学研究内容，错误；<br />D、西瓜中糖类物质的组成属于化学研究内容，错误；<br />故选A','【分析】根据化学的定义和研究内容进行分析判断，化学是一门研究物质的组成、结构、性质及其变化规律的科学，研究对象是物质，研究内容有组成、结构、性质、变化、用途等．','选择题',3.00,'26d5d49a4bcaee6c6a6b88a098d39c16',9,400,'化学的研究领域','',2015,'35','2015秋•开封县校级期中',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841648,'下列相关比较中，关系正确的是（　　）','金属活动性：铁＜镁','相同条件下密度：氢气＞氧气','导电能力：氢氧化钠溶液＜蔗糖溶液','着火点：纸片＞煤炭','','A','【解答】解：A、金属活动性：铁＜镁，故A关系正确；<br />B、相同条件下密度：氢气＜氧气，故B关系不正确．<br />C、导电能力：氢氧化钠溶液＞蔗糖溶液，故C关系不正确．<br />D、纸片比煤炭易燃，则说明着火点：纸片＜煤炭，故D关系不正确．<br />故选：A．','【分析】A、根据常见金属的活动进行分析判断．<br />B、根据常见气体的密度进行分析判断．<br />C、根据常见物质的导电能力分析判断．<br />D、根据常见物质的着火点进行分析判断．','选择题',3.00,'ae4682133adb3a77c4734a809853d592',9,400,'氧气的物理性质,金属活动性顺序及其应用,溶液的导电性及其原理分析,燃烧与燃烧的条件,氢气的物理性质','',2016,'32','2016•常州模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841650,'“对比实验”是化学学习中常用的方法．下列对比实验中所得结论不正确的是（　　）<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao24/2955144f-94d4-11e9-bb46-b42e9921e93e_xkb47.png\" style=\"vertical-align:middle\" />','实验A可以说明呼出气体中二氧化碳含量较高','实验B可以说明温度越高，分子运动越快','实验C可以说明燃烧的条件之一是温度要达到着火点','实验D可以说明铁生锈的条件是与氧气和水同时接触','','A','【解答】解：A、图一中将燃着的木条分别伸入盛有空气和呼出气体的集气瓶中而得出了不同的实验现象，可以看出该实验探究的是呼出气体和空气中氧气含量，而不是得出二氧化碳的含量比较，故结论错误；<br />B、图二中可以看出是在不同的温度下探究分子运动快慢的实验；故结论正确；<br />C、燃烧的条件：物质具有可燃性、与氧气接触、温度达到可燃物的着火点，通过试验可以说明燃烧的条件之一是温度达到可燃物的着火点；故结论正确；<br />D、对于实验，一段时间观察试管①中的铁钉明显锈蚀，①试管中即提供了空气，也提供了水，由此得出：铁生锈的主要条件是铁与水和空气直接接触；故结论正确；<br />故选项为：A．','【分析】A、通过比较可知图一探究的呼出气体和空气中氧气含量的实验；<br />B、通过比较探究的是温度对分子运动快慢的实验；<br />C、根据燃烧的条件进行解答；<br />D、根据铁生锈的主要条件是铁与水和空气直接接触进行解答．','选择题',3.00,'e76502ebe87fa635b79c2442644cedf4',9,400,'化学实验方案设计与评价,吸入空气与呼出气体的比较,金属锈蚀的条件及其防护,分子的定义与分子的特性,燃烧与燃烧的条件','',2016,'32','2016•海安县模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841656,'下列有关化学学科观念的说法，错误的是（　　）','结构观：构成水和过氧化氢的分子结构不同，化学性质也不同','元素观：一切物质都是由元素组成的','微粒观：氯化钠是由氯化钠分子构成的','守恒观：化学反应前后，元素的种类不变','','C','【解答】解：A、水和过氧化氢的化学性质不同，因为它们的分子构成不同，该选项说法正确；<br />B、物质是由元素组成的，属于宏观解释；故B说法正确；<br />C、氯化钠是由钠离子和氯离子构成的，故选项说法错误；<br />D、元素守恒是指在化学反应前后元素的种类与质量不变，故选项说法正确；<br />故选C','【分析】根据分子不同，化学性质不同；元素是具有相同核电荷数的一类原子的总称；氯化钠是由离子构成的以及质量守恒定律解答；','选择题',3.00,'9fbfbcbc94bfbd8ddd6c65dfd2c5efb2',9,400,'分子、原子、离子、元素与物质之间的关系,分子的定义与分子的特性,物质的元素组成,质量守恒定律及其应用','',2016,'32','2016•常州模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841660,'南昌西汉海昏侯墓历时5年的考古发掘，已出土1万余件珍贵文物，创下多个“首次”和“考古之最”．<br />（1）考古中常用碳-14测定文物的年代，已知碳-14原子中质子数为6，中子数为8，则碳-14原子的核外电子数为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）明矾常用于出土的漆木器脱水定形，明矾也常用于净水，明矾在净水中的作用是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）“低氧气调链”技术是考古中的新技术，该技术是以低氧气调技术为依托，用洁净氮气作为保护气体，并通过智能检控设备及其他环节实现对文物所处空间氧含量等指标的持续稳定凋控．氮气用作保护气的主要原因是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（4）出土的文物中有大量的青铜器，青铜器表面有一层绿色物质[铜绿，化学式为Cu<SUB>2</SUB>（0H）<SUB>2</SUB>CO<SUB>3</SUB>]．铜器表面形成铜绿，除与氧气、水有关外，还与二氧化碳有关，判断的依据是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（5）出土文物中有大量的铜钱．古代湿法冶铜的原理是“曾青（硫酸铜溶液）得铁化为铜”，该原理用化学方程式表示为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','6$###$作絮凝剂，除去水中的悬浮杂质$###$氮气的化学性质比较稳定$###$质量守恒定律$###$Fe+CuSO<SUB>4</SUB>=Cu+FeSO<SUB>4</SUB>','【解答】解：（1）因原子中，质子数等于电子数，质子数为6，则电子数为6；故填：6；<br />（2）明矾能使水中不溶性固体的小颗粒凝聚成较大的颗粒，加快不溶性固体小颗粒的沉降，除去水中悬浮物；故填：作絮凝剂，除去水中的悬浮杂质；<br />（3）氮气的化学性质比较稳定，故可用洁净氮气作为保护气体；故填：氮气的化学性质比较稳定；<br />（4）由质量守恒定律可知，反应前后元素的种类不变，所以铜器表面形成铜绿，除与氧气、水有关外，还与二氧化碳有关，故填：质量守恒定律；<br />（5）曾青得铁则化为铜是指铁与硫酸铜溶液反应生成铜和硫酸亚铁，化学方程式为Fe+CuSO<SUB>4</SUB>=Cu+FeSO<SUB>4</SUB>；故答案为：Fe+CuSO<SUB>4</SUB>=Cu+FeSO<SUB>4</SUB>．','【分析】（1）在原子中，核内质子数=核外电子数=核电荷数；<br />（2）明矾能使水中不溶性固体的小颗粒凝聚成较大的颗粒，加快不溶性固体小颗粒的沉降，故其净水的作用是除去水中悬浮物；<br />（3）根据氮气的化学性质来分析；<br />（4）根据质量守恒定律来分析；<br />（5）首先根据反应原理找出反应物、生成物、反应条件，根据化学方程式的书写方法、步骤进行书写即可．','书写',3.00,'bc1601cf2b4936a5c402bc0776ebdfe4',9,400,'常见气体的用途,水的净化,金属的化学性质,金属锈蚀的条件及其防护,原子的有关数量计算,书写化学方程式、文字表达式、电离方程式','',0,'37','',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841665,'下列关于水的认识的说法错误的是（　　）','水是一种氧化物','水是生命活动中不可缺少的物质','水由氢分子和氧原子构成','电解水时在负极得到氢气','','C','【解答】解：A、水是由氢元素和氧元素组成的化合物，属于氧化物，该选项说法正确；<br />B、水是生命活动不可缺少的物质，该选项说法正确；<br />C、水分子是由氢原子和氧原子构成的，该选项说法错误；<br />D、电解水时在负极得到氢气，正极得到氧气，该选项说法正确．<br />故选：C．','【分析】A、根据氧化物的定义进行分析；<br />B、根据水的生理功能进行分析；<br />C、根据水分子是由氢原子和氧原子构成的分析；<br />D、根据电解水的结论分析．','选择题',3.00,'c93e34ac657477be3421d1b626eb9142',9,400,'电解水实验,水的组成,水的性质和应用,从组成上识别氧化物','',2015,'37','2015秋•深圳月考',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841669,'某同学设计了以下四个实验方案，理论上正确，操作上可行，经济上合理的是（　　）','CaCO<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td style=\"font-size: 90%\">高温</td></tr><tr><td><div hassize=\"9\"><div style=\"background: url(\'/upimages/images/part/8594L.png\') repeat-x; width:1px; height:5px;float:left;overflow:hidden\" muststretch=\"h\"></div><div style=\"background: url(\'/upimages/images/part/8594.png\') no-repeat; width:9px; height:5px;float:left;overflow:hidden\"></div></div></td></tr></table></span>CaO<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td style=\"font-size: 90%\"><span><span>H</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span>O</td></tr><tr><td><div hassize=\"9\"><div style=\"background: url(\'/upimages/images/part/8594L.png\') repeat-x; width:1px; height:5px;float:left;overflow:hidden\" muststretch=\"h\"></div><div style=\"background: url(\'/upimages/images/part/8594.png\') no-repeat; width:9px; height:5px;float:left;overflow:hidden\"></div></div></td></tr></table></span>Ca（OH）<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td style=\"font-size: 90%\">N<span><span>a</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span>C<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">3</span></span>溶液</td></tr><tr><td><div hassize=\"9\"><div style=\"background: url(\'/upimages/images/part/8594L.png\') repeat-x; width:1px; height:5px;float:left;overflow:hidden\" muststretch=\"h\"></div><div style=\"background: url(\'/upimages/images/part/8594.png\') no-repeat; width:9px; height:5px;float:left;overflow:hidden\"></div></div></td></tr></table></span>NaOH溶液','H<SUB>2</SUB>O<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td style=\"font-size: 90%\">通电</td></tr><tr><td><div hassize=\"9\"><div style=\"background: url(\'/upimages/images/part/8594L.png\') repeat-x; width:1px; height:5px;float:left;overflow:hidden\" muststretch=\"h\"></div><div style=\"background: url(\'/upimages/images/part/8594.png\') no-repeat; width:9px; height:5px;float:left;overflow:hidden\"></div></div></td></tr></table></span>O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td style=\"font-size: 90%\">Mg</td></tr><tr><td><div hassize=\"9\"><div style=\"background: url(\'/upimages/images/part/8594L.png\') repeat-x; width:1px; height:5px;float:left;overflow:hidden\" muststretch=\"h\"></div><div style=\"background: url(\'/upimages/images/part/8594.png\') no-repeat; width:9px; height:5px;float:left;overflow:hidden\"></div></div></td></tr></table></span>MgO','Cu<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td style=\"font-size: 90%\">稀盐酸</td></tr><tr><td><div hassize=\"9\"><div style=\"background: url(\'/upimages/images/part/8594L.png\') repeat-x; width:1px; height:5px;float:left;overflow:hidden\" muststretch=\"h\"></div><div style=\"background: url(\'/upimages/images/part/8594.png\') no-repeat; width:9px; height:5px;float:left;overflow:hidden\"></div></div></td></tr></table></span>CuCl<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td style=\"font-size: 90%\">NaOH溶液</td></tr><tr><td><div hassize=\"9\"><div style=\"background: url(\'/upimages/images/part/8594L.png\') repeat-x; width:1px; height:5px;float:left;overflow:hidden\" muststretch=\"h\"></div><div style=\"background: url(\'/upimages/images/part/8594.png\') no-repeat; width:9px; height:5px;float:left;overflow:hidden\"></div></div></td></tr></table></span>Cu（OH）<SUB>2</SUB>','MgCl<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td style=\"font-size: 90%\">NaOH溶液</td></tr><tr><td><div hassize=\"9\"><div style=\"background: url(\'/upimages/images/part/8594L.png\') repeat-x; width:1px; height:5px;float:left;overflow:hidden\" muststretch=\"h\"></div><div style=\"background: url(\'/upimages/images/part/8594.png\') no-repeat; width:9px; height:5px;float:left;overflow:hidden\"></div></div></td></tr></table></span>Mg（OH）<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td style=\"font-size: 90%\">Na</td></tr><tr><td><div hassize=\"9\"><div style=\"background: url(\'/upimages/images/part/8594L.png\') repeat-x; width:1px; height:5px;float:left;overflow:hidden\" muststretch=\"h\"></div><div style=\"background: url(\'/upimages/images/part/8594.png\') no-repeat; width:9px; height:5px;float:left;overflow:hidden\"></div></div></td></tr></table></span>Mg','','A','【解答】解：A、碳酸钙高温下分解生成的氧化钙与水反应可生成氢氧化钙，氢氧化钙溶液与碳酸钠溶液通过复分解反应，可以得到氢氧化钠溶液，此方案理论上正确，操作也较为简单可行，又可实验纯碱制得烧碱，经济上合理；故正确；<br />B、水通电分解可生成氧气，镁与氧气反应可以生成氧化镁，因此方案在理论上正确，但通过空气获得氧气要比电解水更经济，因此该方案在经济上不够合理；故不正确；<br />C、根据金属的活动性可得知，铜不能与稀硫酸发生反应，因此该方案中铜与稀硫酸发生反应在理论上是不正确的；故不正确；<br />D、氯化镁与氢氧化钠反应生成氢氧化镁沉淀，而氢氧化镁却不能分解生成金属单质镁，此方案理论上不正确；故不正确；<br />故选A．','【分析】A、碳酸钙高温下分解生成氧化钙和二氧化碳，氧化钙与水反应可以生成氢氧化钙，氢氧化钙与纯碱反应可制得氢氧化钠，这一反应常用于工业制于氢氧化钠；<br />B、工业上通常采取液化分离空气的方法获得氧气，而电解水的方便虽然可产生氧气，但反应需要消耗大量电解，不够经济；<br />C、根据金属活动性顺序，铜位于氢之后，不能与稀硫酸发生反应；<br />D、氯化镁与氢氧化钠发生复分解反应生成氢氧化镁沉淀和氯化钠，氢氧化镁不能电解生成镁；工业上获得金属镁采取电解氯化镁的方法．','选择题',3.00,'a637a0a5ea98e8b7a0feca6f6dffa618',9,400,'化学实验方案设计与评价,金属的化学性质,碳酸钙、生石灰、熟石灰之间的转化,盐的化学性质','',2016,'37','2016•泰安一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841675,'“归纳和比较”是化学学习的主要方法之一，下列关于CO和CO<SUB>2</SUB>的比较正确的是（　　）','性质：都不能溶于水，但CO能燃烧CO<SUB>2</SUB>不能','组成：都由C和O两种元素组成，但CO比CO<SUB>2</SUB>少一个氧元素','危害：都有毒性，CO<SUB>2</SUB>还能造成温室效应','用途：CO能作燃料，也能冶炼金属；CO<SUB>2</SUB>能灭火也能用于光合作用','','D','【解答】解：A、一氧化碳难溶于水，二氧化碳能溶于水，CO能燃烧CO<SUB>2</SUB>不能，故选项说法错误．<br />B、元素只讲种类、不讲个数，一氧化碳与二氧化碳，都由C和O两种元素组成，但1个CO分子比1个CO<SUB>2</SUB>分子少一个氧原子，故选项说法错误．<br />C、一氧化碳具有毒性，二氧化碳本身没有毒性，CO<SUB>2</SUB>能造成温室效应，故选项说法错误．<br />D、一氧化碳具有可燃性、还原性，CO能作燃料，也能冶炼金属；CO<SUB>2</SUB>能灭火也能用于光合作用，故选项说法正确．<br />故选：D．','【分析】A、根据一氧化碳与二氧化碳的溶解性、化学性质，进行分析判断．<br />B、根据元素只讲种类、不讲个数，进行分析判断．<br />C、根据一氧化碳具有毒性，二氧化碳对环境的影响，进行分析判断．<br />D、根据一氧化碳具有可燃性、还原性，二氧化碳的用途，进行分析判断．','选择题',3.00,'a8537ba13cfc1897584f8c740534cb43',9,400,'二氧化碳的物理性质,二氧化碳的化学性质,二氧化碳对环境的影响,一氧化碳的物理性质,一氧化碳的化学性质,一氧化碳的毒性','',2016,'32','2016•武进区模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841684,'<img  src=\"/tikuimages/9/2016/400/shoutiniao52/29ed5cb0-94d4-11e9-9477-b42e9921e93e_xkb30.png\" style=\"vertical-align:middle;FLOAT:right\" />（2016•门头沟区一模）用如图所示装置研究可燃物燃烧的条件．<br />已知：白磷的着火点为40℃，红磷的着火点为240℃<br />（1）若在a处放置红磷，在Y型管中可以观察到的现象是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，发生反应的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）若要研究可燃物燃烧的另一个条件，应在a处加入<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','白磷燃烧，红磷不燃烧$###$4P+5O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2P<SUB>2</SUB>O<SUB>5</SUB>$###$白磷和80℃热水','【解答】解：（1）白磷的着火点为40℃，红磷的着火点为240℃，烧杯内的水是80℃热水，且都与氧气接触，所以白磷达到着火点，且与空气接触，所以能燃烧，而温度达不到红磷的着火点，故红磷不燃烧；白磷燃烧的方程式为：4P+5O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2P<SUB>2</SUB>O<SUB>5</SUB>；<br />（2）可燃物燃烧还须与空气或氧气接触，故可在a处加入被80℃热水浸没的白磷，与另一侧白磷作对比进行探究；<br />故答案为：（1）白磷燃烧，红磷不燃烧、4P+5O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2P<SUB>2</SUB>O<SUB>5</SUB>；<br />（2）白磷和80℃热水（其他合理答案可酌情给分）．','【分析】（1）据燃烧的条件分析解答：白磷的着火点为40℃，红磷的着火点为240℃，烧杯内的水是80℃热水，且都与氧气接触，所以白磷燃烧，而温度达不到红磷的着火点，故不燃烧；<br />（2）可燃物燃烧还须与空气或氧气接触，故可在a处加入白磷和80℃热水，与另一侧白磷作对比；','书写',3.00,'407a25fb87e395b9fcd7729b3cdcd15d',9,400,'燃烧的条件与灭火原理探究,书写化学方程式、文字表达式、电离方程式','',2016,'37','2016•门头沟区一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841688,'从&nbsp;C+O<sub>2</sub><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td style=\"font-size: 90%\">点燃</td></tr><tr><td><div hassize=\"9\"><div style=\"background: url(\'/upimages/images/part/8594L.png\') repeat-x; width:1px; height:5px;float:left;overflow:hidden\" muststretch=\"h\"></div><div style=\"background: url(\'/upimages/images/part/8594.png\') no-repeat; width:9px; height:5px;float:left;overflow:hidden\"></div></div></td></tr></table></span>CO<sub>2</sub>中获得以下信息：①该反应反应物是碳和氧气；②反应发生条件是点燃；③反应前后元素种类和原子个数不变；④反应前后物质种类保持不变；⑤参加反应的碳和氧气的质量比为1：1．其中正确的信息是（　　）','①②④⑤','①②③','④⑤','①②③⑤','','B','【解答】解：从C+O<sub>2</sub><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td style=\"font-size: 90%\">点燃</td></tr><tr><td><div hassize=\"9\"><div style=\"background: url(\'/upimages/images/part/8594L.png\') repeat-x; width:1px; height:5px;float:left;overflow:hidden\" muststretch=\"h\"></div><div style=\"background: url(\'/upimages/images/part/8594.png\') no-repeat; width:9px; height:5px;float:left;overflow:hidden\"></div></div></td></tr></table></span>CO<sub>2 </sub>可知该反应的反应物是碳和氧气，反应条件是点燃，反应前后元素种类和原子个数不变；反应前后物质种类保持发生了改变；参加反应的碳和氧气的质量比12：（16×2）=3：8．<br />故①②③获得信息正确．<br />故选B．','【分析】从化学方程式获得的信息主要有：反应物、生成物、反应条件、各物质间量的关系，据此结合题意进行分析判断．','选择题',3.00,'b93f37bebff5d9bdb162b3c58fb7394d',9,400,'化学方程式的概念、读法和含义','',2016,'32','2016•杨浦区模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841693,'下列图示的实验操作，正确的是（　　）','<img  src=\"/tikuimages/9/2016/400/shoutiniao59/29fdfe80-94d4-11e9-b3b7-b42e9921e93e_xkb53.png\" style=\"vertical-align:middle\" /><br />稀释浓硫酸','<img  src=\"/tikuimages/9/2016/400/shoutiniao5/29ff0ff0-94d4-11e9-9848-b42e9921e93e_xkb99.png\" style=\"vertical-align:middle\" /><br />配制氯化钠溶液','<img  src=\"/tikuimages/9/2016/400/shoutiniao40/2a00968f-94d4-11e9-ae03-b42e9921e93e_xkb69.png\" style=\"vertical-align:middle\" /><br />测定溶液pH','<img  src=\"/tikuimages/9/2016/400/shoutiniao76/2a021d30-94d4-11e9-a12f-b42e9921e93e_xkb55.png\" style=\"vertical-align:middle\" /><br />蒸发食盐水','','D','【解答】解：A、稀释浓硫酸时，要把浓硫酸缓缓地沿器壁注入水中，同时用玻璃棒不断搅拌，以使热量及时地扩散；一定不能把水注入浓硫酸中；图中所示操作错误；<br />B、配制溶液应该在烧杯中进行，量筒只能量取液体的体积，图中所示操作错误；<br />C、用pH试纸测定未知溶液的pH时，正确的操作方法为用玻璃棒蘸取少量待测液滴在干燥的pH试纸上，与标准比色卡对比来确定pH．不能将pH试纸伸入待测液中，以免污染待测液，图中所示操作错误．<br />D、蒸在蒸发皿中进行，并不断的搅拌，图中所示操作正确．<br />故选项为：D．','【分析】A、根据浓硫酸的稀释方法（酸入水，沿器壁，慢慢倒，不断搅）进行分析判断；<br />B、根据配制溶液的操作进行判断；<br />C、根据用pH试纸测定未知溶液的pH的方法进行分析判断；<br />D、根据蒸发操作进行分析判断．','选择题',3.00,'77940f1e3331c96e385af98434684617',9,400,'物质的溶解,浓硫酸的性质及浓硫酸的稀释,蒸发与蒸馏操作,溶液的酸碱度测定','靖江市',2016,'37','2016•靖江市一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841714,'某兴趣小组同学对电池中物质的成分进行探究，将电池剥开时闻到异味且发现有黑色物质．<br />【提出问题】为什么有异味？黑色物质的成分是什么？<br />【网络搜索】干电池的成分有二氧化锰、铁粉及铵盐等物质．<br />【实验探究】同学们分别设计实验进行探究．<br />实验Ⅰ．小芳取适量黑色物质与<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>混合置于研钵内研磨，嗅到氨味，再用湿润的红色石蕊试纸检测，试纸变蓝，证明黑色物质中含有铵盐．<br />实验Ⅱ．小军用一种物理方法发现黑色物质中不含铁粉，他采用的方法是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />实验Ⅲ．小明将黑色物质加入适量的水中搅拌，静置后过滤，取滤渣进行实验．<br /><table class=\"edittable\"><TBODY><TR><td width=189>实验步骤</TD><td width=189>实验现象</TD><td width=189>实验结论</TD></TR><TR><td>①取适量黑色滤渣装入试管中，再加入过氧化氢溶液</TD><td><br /><u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD><td>有气体生成</TD></TR><TR><td>②立即向上述试管中<u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD><td>木条复燃</TD><td>该气体是<br /><u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD></TR></TBODY></TABLE>上述实验中产生氧气的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．据此小明认为黑色物质中一定含有二氧化锰．<br />【交流讨论】大家对小明的结论提出了质疑．<br />（1）小兰认为若要证明黑色物质中一定含有二氧化锰，必须验证二氧化锰的质量和<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>在反应前后都没有发生变化；<br />（2）小林认为除了小兰考虑的因素外，另一个原因是氧化铜等黑色物质对过氧化氢分解也有催化作用．<br />【结论反思】<br />（1）干电池中一定含有铵盐，可能含有二氧化锰，一定不含有铁粉．<br />（2）面对网络信息，我们要去伪存真．不应盲从．','','','','','','熟石灰$###$用磁铁吸引$###$产生大量气泡$###$带火星的木条$###$氧气$###$2H<SUB>2</SUB>O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span>&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O+O<SUB>2</SUB>↑$###$化学性质','【解答】解：<br />实验Ⅰ．铵盐与碱混合会放出氨气，所以取适量黑色物质与熟石灰混合置于研钵内研磨，嗅到氨味，再用湿润的石蕊试纸检测，试纸变蓝，证明黑色物质中含有铵盐；<br />实验Ⅱ．铁能被磁铁吸引，所以用一种物理方法发现黑色物质中不含铁扮．他采用的方法是用磁铁吸引；<br />实验Ⅲ．二氧化锰和过氧化氢溶液混合会生成氧气，氧气具有助燃性，能使带火星的木条复燃，所以实验方案为：<table class=\"edittable\"><TBODY><TR><td width=232>实验步骤&nbsp;</TD><td width=232>实验现象&nbsp;</TD><td width=232>实验结论&nbsp;</TD></TR><TR><td>①取适量黑色滤渣装入试管中，再加入过氧化氢溶液&nbsp;</TD><td>产生大量气泡</TD><td>有气体生成&nbsp;</TD></TR><TR><td>②立即向试管中带火星的木条</TD><td>木条复燃&nbsp;</TD><td>该气体是氧气&nbsp;</TD></TR></TBODY></TABLE>二氧化锰和过氧化氢溶液混合会生成氧气，化学方程式为2H<SUB>2</SUB>O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span>&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O+O<SUB>2</SUB>↑；<br />【交流讨论】（1）二氧化锰能加快过氧化氢溶液分解生成氧气的速率，而本身的质量和化学性质不变，所以若要证明黑色物质中一定含有二氧化锰，必须验证二氧化锰的质量和化学性质．<br />故答案为：<br />实验Ⅰ．熟石灰；实验Ⅱ．用磁铁吸引；实验Ⅲ． <table class=\"edittable\"><TBODY><TR><td width=232>实验步骤 </TD><td width=232>实验现象 </TD><td width=232>实验结论 </TD></TR><TR><td>①</TD><td>产生大量气泡</TD><td>&nbsp;</TD></TR><TR><td>②带火星的木条</TD><td>&nbsp;</TD><td>氧气 </TD></TR></TBODY></TABLE>2H<SUB>2</SUB>O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span>&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O+O<SUB>2</SUB>↑；【交流讨论】（1）化学性质．','【分析】实验Ⅰ．根据铵盐与碱混合会放出氨气进行解答；<br />实验Ⅱ．根据铁能被磁铁吸引进行解答；<br />实验Ⅲ．根据二氧化锰和过氧化氢溶液混合会生成氧气，氧气具有助燃性，能使带火星的木条复燃进行解答；<br />【交流讨论】（1）根据二氧化锰能加快过氧化氢溶液分解生成氧气的速率，而本身的质量和化学性质不变进行解答．','书写',3.00,'37cf841902d10d3a615b91db0bcd32cf',9,400,'实验探究物质的组成成分以及含量,证明铵盐,催化剂的特点与催化作用,金属的化学性质,书写化学方程式、文字表达式、电离方程式','',2016,'32','2016•抚州校级模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841716,'<img  src=\"/tikuimages/9/2016/400/shoutiniao45/2a53bfa1-94d4-11e9-b8d7-b42e9921e93e_xkb16.png\" style=\"vertical-align:middle;FLOAT:right\" />含硫煤燃烧会产生大气污染，为防治该污染，某工厂设计了新的治污方法，同时可得到化工产品，该工艺流程如图所示，下列叙述不正确的是（　　）','该过程中可得到化工产品H<SUB>2</SUB>SO<SUB>4</SUB>','该工艺流程是除去煤燃烧时产生的SO<SUB>2</SUB>','该过程中化合价发生改变的元素为Fe和S','图中涉及到的反应之一为Fe<SUB>2</SUB>（SO<SUB>4</SUB>）<SUB>3</SUB>+SO<SUB>2</SUB>+2H<SUB>2</SUB>O═2FeSO<SUB>4</SUB>+2H<SUB>2</SUB>SO<SUB>4</SUB>','','C','【解答】解：根据工艺流程所示可知，该工艺中Fe<SUB>2</SUB>（SO<SUB>4</SUB>）<SUB>3</SUB>和H<SUB>2</SUB>O吸收SO<SUB>2</SUB>生成FeSO<SUB>4</SUB>和H<SUB>2</SUB>SO<SUB>4</SUB>，该反应的化学方程式为Fe<SUB>2</SUB>（SO<SUB>4</SUB>）<SUB>3</SUB>+2H<SUB>2</SUB>O+SO<SUB>2</SUB>=2FeSO<SUB>4</SUB>+2H<SUB>2</SUB>SO<SUB>4</SUB>，该工艺不仅吸收了二氧化硫，还得到了化工产品硫酸亚铁和硫酸，过程中化合价发生变化的元素有氧、铁、硫；<br />故选：C．','【分析】根据该工艺流程所示进行分析解答即可．','选择题',3.00,'d33220fd685cf60aca60be15dfd0d7a7',9,400,'化合价规律和原则,化石燃料及其综合利用','',2016,'37','2016•石景山区一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841717,'归纳与总结是学好化学的重要方法．下列认识中，完全正确的是（　　）','物质名称与俗称的对应关系<br />①汞--水银<br />②二氧化碳气体--干冰<br />③氢氧化钠--火碱、烧碱、苛性钠','化学上有许多的等量关系<br />①ag镁和bg氧气完全反应，一定生成（a+b）g氧化铜<br />②100mL酒精与100mL水混合，体积等于200mL<br />③参加中和反应的酸和碱的质量一定相等','有关实验现象的记录<br />①硫酸铜溶液中加入氢氧化钠溶液，产生蓝色沉淀<br />②淀粉遇到碘水变蓝色<br />③红热木炭放入盛有氧气的集气瓶中，剧烈燃烧，产生白光','为更好的利用物质，可对物质进行分离或混合<br />①将纯金属制成具有特殊性能的合金<br />②利用蒸馏等方法从海水中提取淡水<br />③向煤炭中加入石灰石制成型煤，减少空气污染','','D','【解答】解：A、二氧化碳气体不能叫干冰，而是固态二氧化碳叫干冰，故A错误；<br />B、ag镁和bg氧气完全反应，不一定生成（a+b）g氧化镁，如果恰好完全反应生成（a+b）g氧化镁，如果有一种物质过量，就不相等了；由于分子之间有间隔，100mL酒精与100mL水混合，体积小于200mL，参加中和反应的酸和碱的质量不一定相等，故B错误；<br />C、应该是碘遇到淀粉变蓝色，故C错误．<br />D、合金就是指金属与其它金属或非金属组成的具有特殊性能物质，蒸馏方法是从海水中提取淡水的主要方法，向煤炭中加入石灰石制成型煤，减少二氧化硫对空气污染．<br />故选D．','【分析】A、根据物质的俗称考虑本题；<br />B、根据质量守恒定律和分子间隔来考虑本题；<br />C、根据物质的性质和反应时的生成物来看实验现象；<br />D、根据合金性质和分离方法来考虑．','选择题',3.00,'99ab13b2985165a57478b15d0c34199d',9,400,'蒸发与蒸馏操作,防治空气污染的措施,氧气的化学性质,合金与合金的性质,碱的化学性质,分子的定义与分子的特性,物质的元素组成,质量守恒定律及其应用,鉴别淀粉、葡萄糖的方法与蛋白质的性质','',0,'37','',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841728,'现有5种物质：①浓盐酸&nbsp;②烧碱&nbsp;③食盐&nbsp;⑤浓硫酸&nbsp;⑥铁．把它们长期露置在空气中，其中质量减少的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填序号，下同），发生化学变化且质量增加的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','①$###$②⑥','【解答】解：①浓盐酸 具有挥发性，敞口放置在空气中一段时间，会使其质量减少；<br />②烧碱敞口瓶久置于空气中，易吸收空气中的水分而潮解，还能与空气中的二氧化碳发生反应生成碳酸钠而变质；<br />③食盐不具有挥发性，不与空气中的成分反应，长期露置于空气中质量不变；<br />⑤浓硫酸具有吸水性，敞口放置在空气中一段时间，会吸收空气中的水分，使其质量增加； <br />⑥铁长期露置在空气中，能与空气中的氧气、水分反应生成铁锈，质量会增加．<br />故答案为：①；②⑥．','【分析】长期露置于空气中，具有挥发性的，没有发生化学变化质量减少；发生变质即发生了化学变化，也就是与空气中的成分发生了化学反应，生成了新的物质，根据常见物质的性质进行分析解答．','填空题',3.00,'784495ce255d1db5c1586e51124e318a',9,400,'金属锈蚀的条件及其防护,空气中常见酸碱盐的质量或性质变化及贮存法','',0,'37','',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841737,'不用排水法收集二氧化碳，而用向上排空气法收集二氧化碳，原因是（　　）','二氧化碳不易溶于水，密度比空气大','二氧化碳能溶于水，密度比空气大','二氧化碳溶于水变成干冰','二氧化碳能灭火','','B','【解答】解：收集气体要根据气体的密度和溶水性，二氧化碳能溶于水，不可用排水法收集；二氧化碳密度大于空气的密度，可以用向上排空气法收集．<br />故选B．','【分析】二氧化碳能溶于水，密度大于空气的密度，所以收集二氧化碳只能用向上排空气法．','选择题',2.00,'5c895099502991bcc19a887ceb002cd6',9,400,'常用气体的收集方法','',0,'37','',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841738,'<img  src=\"/tikuimages/9/2015/400/shoutiniao55/2a9a6591-94d4-11e9-9e47-b42e9921e93e_xkb10.png\" style=\"vertical-align:middle;FLOAT:right\" />（2015秋•滨州期末）某初级中学学生发现实验室里有一瓶标签破损的无色溶液（如图所示），于是在老师的指导下进行如下探究．<br />（1）甲、乙、丙三位同学对该溶液作出如下猜想：<br />甲同学猜想：该溶液是碳酸；<br />乙同学猜想：该溶液是过氧化氢溶液；<br />丙同学猜想：该溶液是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）甲同学刚提出自己的猜想，就遭到乙、丙同学的反对，乙、丙同学反对的理由是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）他们设计如下实验方案来验证乙、丙同学的猜想．<br />乙同学取少量MnO<SUB>2</SUB>粉末于试管中，加入适量该溶液，并将带火星的木条插入试管中．如果乙同学猜想正确，可观察到的现象是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；请写出上述实验中过氧化氢分解的化学方程式：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />丙同学取少量氯化钡溶液于试管中，加入适量该溶液，如果丙同学猜想正确，可观察到的现象是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','稀硫酸$###$碳酸不稳定，容易分解$###$产生气泡，带火星的木条复燃$###$2H<SUB>2</SUB>O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span>&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>0+O<SUB>2</SUB>↑$###$有白色沉淀产生','【解答】解：（1）根据残留的标签可以判断该物质中含有两个氢原子，所以可以得出丙的猜想可能是硫酸等物质；<br />（2）碳酸不稳定，容易分解，不能够形成稳定的溶液，所以甲同学的猜想不是很合适；<br />（3）根据题意可以知道氧化铜可以作为过氧化氢的催化剂，所以如果乙同学的猜想是正确的，则会产生氧气，鉴别氧气通常用带火星的木条，所以如果带火星的木条复燃，说明该溶液是过氧化氢溶液；氯化钡可以和硫酸反应生成硫酸钡沉淀，所以如果丙同学的猜想正确，则会产生白色沉淀；<br />（4）如果乙同学的猜想是正确的，那么过氧化氢会分解而产生氧气和水．<br />故答案为：（1）稀硫酸；&nbsp;&nbsp;&nbsp;&nbsp;<br />（2）碳酸不稳定，容易分解；<br />（3）产生气泡，带火星的木条复燃；2H<SUB>2</SUB>O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span>&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>0+O<SUB>2</SUB>↑；有白色沉淀产生．','【分析】（1）根据物质的化学式来进行判断；<br />（2）根据碳酸的性质可以知道，碳酸不稳定，容易分解不能稳定的存在；<br />（3）过氧化氢可以分解产生氧气，氧气可以使带火星的木条复燃，而氯化钡可以和硫酸反应生成硫酸钡沉淀．','书写',3.00,'9b9fbb5ddaaa2371f26ba6036c91c5a0',9,400,'缺失标签的药品成分的探究,氧气的检验和验满,书写化学方程式、文字表达式、电离方程式','',2015,'33','2015秋•滨州期末',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841755,'工业上用纯碱和石灰石为原料制备烧碱的简要工艺流程如下图所示，请回答．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao36/2ae2921e-94d4-11e9-b4f2-b42e9921e93e_xkb41.png\" style=\"vertical-align:middle\" /><br />（1）某同学用湿润的PH试纸测定纯碱溶液酸碱度，其测定结果偏<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“高”或“低”）．&nbsp;&nbsp;<br />（2）煅烧石灰石向空气中排放大量的二氧化碳气体，造成的环境问题是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）步骤④分离除去碳酸钙沉淀的操作是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，实验室进行此操作所需玻璃仪器有烧杯、玻璃棒、<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（4）步骤④发生反应的化学方程式：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．该工艺流程图中没有涉及到的基本反应类型是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（5）某研究性学习小组的同学想检验烧碱溶液里是否含有纯碱．甲同学选择稀盐酸，乙同学选择一种碱溶液，丙同学也选择了一种溶液，该溶液中溶质与甲、乙同学所用溶液中溶质的类别不同，他们都能达到检验目的．丙同学选用的试剂可能是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填化学式）．','','','','','','低$###$温室效应$###$过滤$###$漏斗$###$Na<SUB>2</SUB>CO<SUB>3</SUB>+Ca（OH）<SUB>2</SUB>=2NaOH+CaCO<SUB>3</SUB>↓$###$置换反应$###$CaCl<SUB>2</SUB>','【解答】解：（1）将PH试纸湿润，实际上是将纯碱溶液稀释，碱性减弱，pH减小；<br />（2）向空气中排放大量的二氧化碳气体，造成的环境问题是温室效应；<br />（3）除去不溶性杂质的操作为过滤，验室进行此操作所需玻璃仪器有烧杯、玻璃棒、漏斗；<br />（4）由题干可知，步骤④发生的反应是氢氧化钙与碳酸钠反应生成了碳酸钙和氢氧化钠，化学方程式是：Ca（OH）<SUB>2</SUB>+Na<SUB>2</SUB>CO<SUB>3</SUB>=CaCO<SUB>3</SUB>↓+2NaOH；碳酸钙在高温下生成氧化钙和二氧化碳，属于分解反应；氧化钙和水反应生成氢氧化钙，属于化合反应；氢氧化钙和碳酸钠反应生成碳酸钙和氢氧化钠，属于复分解反应，所以改工艺流程中不涉及的基本反应类型是置换反应；<br />（5）由于碳酸钠能与可溶性的钙盐、钡盐反应，丙同学所选溶液中的溶质可能是：CaCl<SUB>2</SUB>或Ca（NO<SUB>3</SUB>）<SUB>2</SUB>或BaCl<SUB>2</SUB>或Ba（NO<SUB>3</SUB>）<SUB>2</SUB>等．<br />故答案为：（1）低；<br />（2）温室效应；<br />（3）过滤；漏斗；<br />（4）Na<SUB>2</SUB>CO<SUB>3</SUB>+Ca（OH）<SUB>2</SUB>=2NaOH+CaCO<SUB>3</SUB>↓；置换反应；<br />（5）CaCl<SUB>2</SUB>（或BaCl<SUB>2</SUB>）','【分析】（1）将PH试纸湿润，实际上是将纯碱溶液稀释，碱性减弱；<br />（2）向空气中排放大量的二氧化碳气体，造成的环境问题是温室效应；<br />（3）除去不溶性杂质的操作为过滤，验室进行此操作所需玻璃仪器有烧杯、玻璃棒、漏斗；<br />（4）根据步骤④发生的反应碳酸钠和氢氧化钙反应生成碳酸钙沉淀和氢氧化钠进行分析；根据流程图碳酸钙在高温下生成氧化钙和二氧化碳，氧化钙和水反应生成氢氧化钙，氢氧化钙和碳酸钠反应生成碳酸钙和氢氧化钠分析反应的类型；<br />（5）根据碳酸钠能与盐反应的性质进行分析．','书写',3.00,'9d57e6adca958b7fea5a32706d43fe6c',9,400,'过滤的原理、方法及其应用,溶液的酸碱度测定,酸雨的产生、危害及防治,盐的化学性质,物质的相互转化和制备,反应类型的判定,书写化学方程式、文字表达式、电离方程式','',2016,'32','2016•如东县模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841765,'回答实验室用高锰酸钾制取氧气的相关问题．<br />（1）该反应的化学方程式是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）如图所示，发生装置应选用<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填序号，下同），以下气体收集装置中不能选用的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao7/2af9c3a1-94d4-11e9-a0b5-b42e9921e93e_xkb69.png\" style=\"vertical-align:middle\" /><br />（3）为测定高锰酸钾分解生成氧气的体积，应选用如图装置<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao94/2afca9cf-94d4-11e9-9534-b42e9921e93e_xkb55.png\" style=\"vertical-align:middle\" />','','','','','','2KMnO<SUB>4</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>K<SUB>2</SUB>MnO<SUB>4</SUB>+MnO<SUB>2</SUB>+O<SUB>2</SUB>↑$###$A$###$E$###$丙','【解答】解：（1）高锰酸钾受热时能够分解生成锰酸钾、二氧化锰和氧气，化学方程式为：2KMnO<SUB>4</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>K<SUB>2</SUB>MnO<SUB>4</SUB>+MnO<SUB>2</SUB>+O<SUB>2</SUB>↑；<br />（2）高锰酸钾受热时能够分解生成锰酸钾、二氧化锰和氧气，属于固固加热型；因为氧气的密度比空气大，可以用向上排空气法收集；氧气不易溶于水，可以用排水法收集；不能用向下排空气法收集；<br />（3）据氧气的密度比水的小，应从短管进，水从长管出，故选丙；<br />故答案为：<br />（1）2KMnO<SUB>4</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>K<SUB>2</SUB>MnO<SUB>4</SUB>+MnO<SUB>2</SUB>+O<SUB>2</SUB>↑；（2）A、E；（3）丙．','【分析】（1）根据高锰酸钾受热时能够分解生成锰酸钾、二氧化锰和氧气解答；&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <br />（2）根据氧气的密度比空气的密度大，不易溶于水解答；<br />（3）根据氧气的密度比水的小解答．','书写',3.00,'12085687532183ed00a5df9335a5bac5',9,400,'量气装置,氧气的制取装置,氧气的收集方法,书写化学方程式、文字表达式、电离方程式','',2016,'37','2016•怀柔区一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841775,'氧气是人类维持生命不可缺少的物质，下列有关氧气的叙述中错误的是（　　）','氧气是一种可燃性气体','氧气约占空气体积的<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">1</td></tr><tr><td>5</td></tr></table></span>','氧气的密度比空气大','氧气是一种化学性质比较活泼的非金属单质','','A','【解答】解：A、氧气不具有可燃性，具有助燃性，故A错误；<br />B、氧气约占空气体积的<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">1</td></tr><tr><td>5</td></tr></table></span>，故B正确；<br />C、氧气的密度比空气略大，故C正确；<br />D、氧气是一种化学性质比较活泼的非金属单质，故D正确．<br />故选A．','【分析】根据空气中氧气的含量、氧化的性质和用途等分析判断．','选择题',2.00,'926a6876d0f1b345f0c2280610eb86bc',9,400,'氧气的物理性质,氧气的化学性质','',2016,'35','2016春•清河县校级期中',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841804,'生活中的下列做法错误的是（　　）','每到冬季时节，哈市的人们经常用铜质的炭火锅涮肉','用炉具清洁剂除去纯棉内衣上的油污','运动出汗过多时，常引用一些食盐水来补充由于出汗而流失的钠元素','燃气灶火焰呈现黄色时，此时应调节灶具的进风口，即加大空气的进入量','','B','【解答】解：A、铜具有良好的导热性，每到冬季时节，哈市的人们经常用铜质的炭火锅涮肉，故选项说法正确．<br />B、炉具清洁剂中含有氢氧化钠，具有强烈的腐蚀性，不能用炉具清洁剂除去纯棉内衣上的油污，故选项说法错误．<br />C、人们每天都需要摄入一些食盐来补充由于出汗而排出的氯化钠，以满足人体的正常需要，故选项说法正确．<br />D、燃气灶火焰呈现黄色时，说明燃烧不充分，此时应调节灶具的进风口，即加大空气的进入量，故选项说法正确．<br />故选：B．','【分析】A、根据铜具有良好的导热性，进行分析判断．<br />B、根据炉具清洁剂中含有氢氧化钠，具有强烈的腐蚀性，进行分析判断．<br />C、根据运动出汗过多时会生成氯化钠的流失，进行分析判断．<br />D、根据燃气灶火焰呈现黄色，说明燃烧不充分，进行分析判断．','选择题',3.00,'549f0a4d9786dda8897ddbdcfb45b023',9,400,'金属的物理性质及用途,氯化钠与粗盐提纯,盐的化学性质,完全燃烧与不完全燃烧','',2016,'37','2016•南岗区三模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841810,'置于空气中，下列物质总质量不会增加的是（　　）','生石灰','食盐','石灰水','浓硫酸','','B','【解答】解：A、生石灰易吸收空气中的水分质量增加，故错；<br />B、食盐在空气中质量几乎不会变化，故对；<br />C、石灰水在空气中会与二氧化碳反应是物质的质量增加，故错；<br />D、浓硫酸具有吸水性，能吸收空气中水蒸气，溶液质量增加，故错．<br />故选：B．','【分析】根据常见物质的物理和化学性质依次分析即可，生石灰易吸收空气中的水分质量增加，食盐在空气中质量几乎不会变化，石灰水在空气中会与二氧化碳反应，浓硫酸具有吸水性．','选择题',3.00,'b1b3319eef3d76552e97f4d979c7f86b',9,400,'生石灰的性质与用途,空气中常见酸碱盐的质量或性质变化及贮存法','',2016,'37','2016春•姜堰区校级月考',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841811,'<img  src=\"/tikuimages/9/2016/400/shoutiniao76/2b85afee-94d4-11e9-93b3-b42e9921e93e_xkb86.png\" style=\"vertical-align:middle;FLOAT:right\" />（2016•黄冈模拟）构建知识网络是化学学习中的重要方法．如图是李红同学构建碳及含碳物质间转化的知识网络，请根据图回答下列问题：<br />（1）含碳元素的单质有金刚石、石墨和足球烯（足球烯的化学式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>），它们的物理性质差异很大，原因是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）写出实现图示中③这一转化的两种途径（要求另外一种反应物类别不同）：①<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；②<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（3）二氧化碳使石蕊试液变红涉及的反应是如图示转化中<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填序号）．','','','','','','C<SUB>60</SUB>$###$碳原子的排列方式不同$###$2CO+O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2CO<SUB>2</SUB>$###$CO+CuO<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>Cu+CO<SUB>2</SUB>$###$⑦','【解答】解：<br />（1）金刚石、石墨和和足球烯（足球烯的化学式为C<SUB>60</SUB>）都是由碳元素组成的，碳原子排列方式不同，决定了物理性质不同．<br />（2）根据图式可知：实现图示中③这一转化，可以是CO在氧气中燃烧生成二氧化碳，反应的化学方程式为：2CO+O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2CO<SUB>2</SUB>；<br />CO和氧化铜反应生成铜和二氧化碳，反应的化学方程式为：CO+CuO<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>Cu+CO<SUB>2</SUB>；<br />（3）二氧化碳使石蕊试液变红，是因为二氧化碳和水反应生成碳酸，二氧化碳使石蕊试液变红涉及的反应是如图示转化中⑦．<br />答案：<br />（1）C<SUB>60</SUB>；碳原子的排列方式不同；<br />（2）①2CO+O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2CO<SUB>2</SUB>；②CO+CuO<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>Cu+CO<SUB>2</SUB>；<br />（3）⑦．','【分析】（1）根据物质的组成及构成解答．物质是由元素组成的，结构决定性质解答；<br />（2）根据图式可知：CO在氧气中燃烧生成二氧化碳；<br />（3）根据二氧化碳使石蕊试液变红，是因为二氧化碳和水反应生成碳酸解答．','填空题',3.00,'d87a91edc8655669322d77ce33a9d619',9,400,'二氧化碳的化学性质,一氧化碳的化学性质,物质的相互转化和制备,碳元素组成的单质','',2016,'32','2016•黄冈模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841827,'“低碳生活”是指返璞归真地去进行人与自然的活动，要求减少人们生活中作息时间所消耗&nbsp;的能量，从而减低碳的排放，下列活动不符合“低碳生活”的是（　　）','为了防止疾病传染，提必人们生活质量，尽量使用一次性餐具','不能长时间把“QQ”挂在线上，即用即上线','用传统发条式闹钟代替电子闹钟','减少使用私家车次数，多乘公交或骑自行车','','A','【解答】解：<br />A、一次性餐具制作过程消耗能源和资源，多作用一次性餐具，能增大消耗的能量、二氧化碳的排放；故选项不符合“低碳生活”理念．<br />B、不能长时间把“QQ”挂在线上，即用即上线，能减少电能的消耗，故选项符合“低碳生活”理念．<br />C、用传统发条式闹钟代替电子钟，能减少电能的消耗，故选项符合“低碳生活”理念．<br />D、减少使用私家车次数，多乘公交或骑自行车，能减少二氧化碳的排放；故选项符合“低碳生活”理念．<br />故选A．','【分析】根据题意：“低碳生活”是指返璞归真地去进行人与自然的活动，要求减少生活作息时间所消耗的能量，从而减低碳的排放；可以从节电、节能和回收等环节来改变生活细节，据此进行分析解答即可．','选择题',3.00,'7a09af357aa30a653ba52b15a84bd360',9,400,'自然界中的碳循环','普宁市',2015,'33','2015秋•普宁市校级期末',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841838,'配平下列化学方程式：<br />（1）<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>H<SUB>2</SUB>+<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>O<SUB>2</SUB>--<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>H<SUB>2</SUB>O<br />（2）<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>C<SUB>2</SUB>H<SUB>4</SUB>+<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>O<SUB>2</SUB>--<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>CO<SUB>2</SUB>+<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>H<SUB>2</SUB>O<br />（3）<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>CuO+<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>C--<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>Cu+<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>CO<SUB>2</SUB><br />（4）<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>Fe<SUB>2</SUB>O<SUB>3</SUB>+<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>CO--<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>Fe+<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>CO<SUB>2</SUB><br />（5）<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>Cu+<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>HNO<SUB>3</SUB>--<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>Cu（NO<SUB>3</SUB>）<SUB>2</SUB>+<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>H<SUB>2</SUB>O+<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>NO．','','','','','','2$###$1$###$2$###$1$###$3$###$2$###$2$###$2$###$1$###$2$###$1$###$1$###$3$###$2$###$3$###$1$###$2$###$1$###$1$###$2','【解答】解：（1）利用最小公倍数法进行配平，以氧原子作为配平的起点，氧气、水前面的化学计量数分别为：1、2，最后调整氢气前面的化学计量数为2．<br />（2）本题可利用“定一法”进行配平，把C<SUB>2</SUB>H<SUB>4</SUB>的化学计量数定为1，则氧气、二氧化碳、水前面的化学计量数分别为：3、2、2．<br />（3）利用最小公倍数法进行配平，以氧原子作为配平的起点，氧化铜、二氧化碳前面的化学计量数分别为：2、1，最后调整碳、铜前面的化学计量数为1、2．<br />（4）本题可从得失氧的角度配平，一个CO分子反应中获得一个氧原子变为二氧化碳分子，由于一个Fe<SUB>2</SUB>O<SUB>3</SUB>分子中氧原子的数目为3个，所以CO与CO<SUB>2</SUB>前面的化学计量数都为3，最后配平其它元素的原子，Fe<SUB>2</SUB>O<SUB>3</SUB>、Fe前面的化学计量数分别为1、2．<br />（5）本题可利用“定一法”进行配平，把Cu（NO<SUB>3</SUB>）<SUB>2</SUB>+的化学计量数定为1，则铜、硝酸、水、一氧化氮前面的化学计量数分别为：1、2、1、2．<br />故答案为：（1）2、1、2；（2）1、3、2、2；（3）2、1、2、1；（4）1、3、2、3；（5）1、2、1、1、2．','【分析】根据质量守恒定律：反应前后各原子的数目不变，选择相应的配平方法（最小公倍数法、定一法等）进行配平即可；配平时要注意化学计量数必须加在化学式的前面，配平过程中不能改变化学式中的下标；配平后化学计量数必须为整数．','填空题',3.00,'2cdcff55b445b96f2ce253acd81c738f',9,400,'化学方程式的配平','北安市',2016,'35','2016春•北安市校级期中',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841839,'举例是掌握化学规律的重要方法，下列规律所举实例错误的是（　　）<br /><table class=\"edittable\"><TBODY><TR><td>选项</TD><td>化学规律</TD><td>示例</TD></TR><TR><td>A</TD><td>物质结构决定性质</TD><td>由于碳原子排列方式不同，金刚石与石墨物理性质存在很大差异</TD></TR><TR><td>B</TD><td>现象说明微观本质</TD><td>物体热胀冷缩现象，说明构成物质的微粒间距离发生了变化</TD></TR><TR><td>C</TD><td>性质决定物质用途</TD><td>甲醛能使蛋白质变性，可用甲醛溶液浸泡海鲜食品来保鲜</TD></TR><TR><td>D</TD><td>性质决定鉴别方法</TD><td>有些物质燃烧气味不同，可用灼烧闻气味区分羊毛绒和棉线</TD></TR></TBODY></TABLE>','A','B','C','D','','C','【解答】解：A、金刚石与石墨物理性质存在很大差异，由于碳原子排列方式不同，故A正确；<br />B、物体热胀冷缩现象，说明构成物质的微粒间距离发生了变化，故B正确；<br />C、甲醛有毒，所以不能甲醛溶液浸泡海鲜食品来保鲜，故C错误；<br />D、羊毛绒和棉线燃烧后的气味不同，可以鉴别，故D正确．<br />故选：C．','【分析】A、根据金刚石与石墨物理性质存在很大差异，由于碳原子排列方式不同进行分析；<br />B、根据物体热胀冷缩现象，说明构成物质的微粒间距离发生了变化进行分析；<br />C、根据甲醛有毒进性分析；<br />D、根据羊毛绒和棉线燃烧后的气味不同进行分析．','选择题',3.00,'4aa2ea8961e71cc11bb6d9cff56a992f',9,400,'分子的定义与分子的特性,碳元素组成的单质,棉纤维、羊毛纤维和合成纤维的鉴别,亚硝酸钠、甲醛等化学品的性质与人体健康','',2016,'37','2016•安庆一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841840,'通过如表实验操作和现象能得出相应结论的是（　　）<br /><table class=\"edittable\"><TBODY><TR><td width=53></TD><td width=270>实验操作</TD><td width=177>现象</TD><td width=146>结论</TD></TR><TR><td>A</TD><td>向收集满CO<SUB>2</SUB>的软塑料瓶中加入约<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">1</td></tr><tr><td>3</td></tr></table></span>体积滴有石蕊试液的水，旋紧瓶盖，振荡</TD><td>塑料瓶变瘪，溶液变红</TD><td>CO<SUB>2</SUB>能与石蕊试液变红</TD></TR><TR><td>B</TD><td>向浸没在热水中的白磷通氧气</TD><td>白磷燃烧</TD><td>燃烧需要氧参与</TD></TR><TR><td>C</TD><td>在无色溶液中滴加氯化钡溶液</TD><td>有白色沉淀生成</TD><td>溶液中一定含有SO<SUB>4</SUB><SUP>2-</SUP></TD></TR><TR><td>D</TD><td>向盛有氢氧化钡溶液的烧杯中滴加稀盐酸</TD><td>无明显现象</TD><td>两者不发生反应</TD></TR></TBODY></TABLE>','A、','B、','C、','D、','','B','【解答】解：A、向收集满CO<SUB>2</SUB>的软塑料瓶中加入约<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">1</td></tr><tr><td>3</td></tr></table></span>体积滴有石蕊试液的水，二氧化碳能溶于水，使塑料瓶内压强减小，会观察到塑料瓶变瘪；二氧化碳与水反应生成碳酸，碳酸能使石蕊溶液变红色，不能得出CO<SUB>2</SUB>能与石蕊试液变红，故选项说法错误．<br />B、将白磷浸没在热水中，白磷不燃烧；通氧气后白磷燃烧，说明燃烧需要氧参与，故选项说法正确．<br />C、在无色溶液中滴加氯化钡溶液，有白色沉淀生成，溶液中不一定含有SO<SUB>4</SUB><SUP>2-</SUP>，也可能含有硝酸银溶液，故选项说法错误．<br />D、向盛有氢氧化钡溶液的烧杯中滴加稀盐酸，氢氧化钡溶液与稀盐酸反应生成氯化钡和水，但无明显现象，但两者已发生了化学反应，故选项说法错误．<br />故选：B．','【分析】A、根据二氧化碳溶于水，使塑料瓶内压强减小，进行分析判断．<br />B、将白磷浸没在热水中，白磷不燃烧，通氧气后白磷燃烧，进行分析判断．<br />C、根据氯化钡能与硝酸银或硫酸盐溶液等反应产生白色沉淀，进行分析判断．<br />D、根据氢氧化钡溶液与稀盐酸反应生成氯化钡和水，进行分析判断．','选择题',3.00,'2176c801d5c4a8d95f6c30f409d50c33',9,400,'化学实验方案设计与评价,证明硫酸和可溶性硫酸盐,二氧化碳的化学性质,中和反应及其应用,燃烧与燃烧的条件','',2016,'37','2016•滨海县二模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841858,'下列“课外实验”对应的结论错误的是（　　）','<img  src=\"/tikuimages/9/2016/400/shoutiniao54/2c2af0a1-94d4-11e9-9169-b42e9921e93e_xkb28.png\" style=\"vertical-align:middle\" /><br />说明蜡烛中含有炭黑','<img  src=\"/tikuimages/9/2016/400/shoutiniao26/2c2fd29e-94d4-11e9-963d-b42e9921e93e_xkb20.png\" style=\"vertical-align:middle\" /><br />说明小苏打和柠檬酸反应能产生气体','<img  src=\"/tikuimages/9/2016/400/shoutiniao49/2c3095f0-94d4-11e9-8ec7-b42e9921e93e_xkb60.png\" style=\"vertical-align:middle\" /><br />说明有些花的色素可做酸碱指示剂','<img  src=\"/tikuimages/9/2016/400/shoutiniao39/2c318051-94d4-11e9-8a94-b42e9921e93e_xkb74.png\" style=\"vertical-align:middle\" /><br />说明同种溶质在不同的溶剂里溶解性不同','','A','【解答】解：A、蜡烛燃烧过程中能够产生炭黑，说明蜡烛中含有碳元素，不能说明含有炭黑，该选项说法不正确；<br />B、小苏打是碳酸氢钠的俗称，能和柠檬酸反应生成二氧化碳气体，该选项说法正确；<br />C、酸碱性溶液能够改变花的颜色，说明有些花的色素可做酸碱指示剂，该选项说法正确；<br />D、碘在水和酒精中的溶解度不同，说明同种溶质在不同的溶剂里溶解性不同，该选项说法正确．<br />故选：A．','【分析】蜡烛燃烧过程中能够产生炭黑；<br />小苏打是碳酸氢钠的俗称，能和柠檬酸反应生成二氧化碳气体；<br />酸碱指示剂能和酸碱性溶液相互作用而变色；<br />同种溶质在不同的溶剂里溶解性不同．','选择题',3.00,'af2845dc0cbead889e4f6a10fc370fab',9,400,'化学实验方案设计与评价,蜡烛燃烧实验,物质的溶解性及影响溶解性的因素,酸碱指示剂及其性质,酸的化学性质','江阴市',2016,'35','2016春•江阴市校级期中',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841864,'据有关资料介绍，儿童缺钙会得佝偻病，成年人缺钙会得软骨病，血液中也有少量钙，它对皮肤伤口血液的凝固起着重要作用．<br /><img  src=\"/tikuimages/9/2015/400/shoutiniao62/2c4d6cc0-94d4-11e9-8bf4-b42e9921e93e_xkb61.png\" style=\"vertical-align:middle\" /><br />（1）钙原子结构示意图如图1所示，该原子的核电荷数为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，它在元素周期表中位于第<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>周期；<br />钙离子符号为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）下列元素与钙元素的化学性质相似的是如图2<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填编号）．','','','','','','20$###$4$###$Ca<SUP>2+</SUP>$###$B','【解答】解：（1）依据原子中质子数等于核电荷数等于核外电子数，所以可知该元素的核电荷数为2+8+8+2=20；由于电子层数与周期数相同，所以该元素是第四周期元素，由于该原子的最外层电子数是2，化学反应中易失掉最外层的两个电子，所以钙离子的符号为Ca<SUP>2+</SUP>；故填：20；4；Ca<SUP>2+</SUP>；<br />（2）A化学性质稳定，C易得电子，D性质稳定；由于最外层电子数相同的元素化学性质相似，观察图示可知B应该与钙元素化学性质相似，故填：B．','【分析】（1）依据原子中质子数等于核外电子数，电子层数与周期数相同以及钙离子符号写法分析解答；<br />（2）依据最外层电子数相同的元素化学性质相似分析解答．','填空题',3.00,'2123607bb4d5d3cb959544720bbbdb9a',9,400,'核外电子在化学反应中的作用,原子结构示意图与离子结构示意图','',2015,'35','2015秋•蓝山县期中',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841868,'某校化学兴趣小组的同学根据实验室提供的仪器和药品，在老师的指导下从下图中选择装置进行了氧气的制备实验．<br /><img  src=\"/tikuimages/9/0/400/shoutiniao38/2c5b7680-94d4-11e9-98ed-b42e9921e93e_xkb11.png\" style=\"vertical-align:middle\" /><br />（1）写出仪器a的名称：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）甲同学从如图中选择B、E装置的组合制取氧气，反应的化学方程式为：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．在用胶皮管连接装置B和E中的玻璃导管时，应先把玻璃管口<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，然后稍稍用力把玻璃管插入胶皮管．<br />（3）过氧化氢溶液通过分液漏斗加入的目的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，如果锥形瓶中的反应过于剧烈，应该采取的措施是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（4）图E所示实验操作中，当集气瓶内即将充满氧气时，要使集气瓶内充有尽可能多的氧气，应该进行的操作是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（5）某同学选择体积为330mL的集气瓶用排水法收集氧气，若要收集到体积分数为80%的氧气，则集气瓶预先装水的体积是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>毫升．（假设空气中氧气的体积分数为20%）<br />（6）制取氧气的实验完毕后，为使催化剂再利用，小兵同学要回收混合液中的二氧化锰，一般采用的实验操作是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','锥形瓶$###$2H<SUB>2</SUB>O<SUB>2</SUB> <span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span>&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O+O<SUB>2</SUB>↑$###$用水润湿$###$可以控制液体的滴加速率$###$减缓液体的滴加速率$###$将集气瓶竖直$###$247.5$###$过滤','【解答】解：<br />（1）仪器a是作为反应容器的锥形瓶；<br />（2）B装置属于固液常温型，过氧化氢制取氧气可用此装置，反应方程式是：2H<SUB>2</SUB>O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span>&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O+O<SUB>2</SUB>↑，在用胶皮管连接装置B和E中的玻璃导管时，应先把玻璃管口用水润湿，使之润滑，然后稍稍用力把玻璃管插入胶皮管；<br />（3）分液漏斗可逐滴滴加液体，控制液体的滴加速率；反应过程中减小浓度、换较大的锥形瓶不太现实，可以用减缓滴加速度的方法；<br />（4）要使瓶内充有尽可能多的氧气，应将集气瓶竖直；<br />（5）要考虑空气中也有氧气，氧气约占空气总体积的五分之一；<br />设通入氧气后在集气瓶中预先放水的体积为x，<br /><span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\">x+(330mL-x)×<table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">1</td></tr><tr><td>5</td></tr></table></td></tr><tr><td>330mL</td></tr></table></span>×100%=80%<br />解得x═247.5mL．<br />故在容积为330mL的集气瓶中收集一瓶氧气的体积分数为80%的气体，需要在集气瓶中预先放247.5毫升水；<br />（6）二氧化锰不溶于水，因此可用过滤的方法分离；<br />故答案为：<br />（1）锥形瓶；<br />（2）2H<SUB>2</SUB>O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span>&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O+O<SUB>2</SUB>↑；用水润湿；<br />（3）可以控制液体的滴加速率；减缓液体的滴加速率；<br />（4）将集气瓶竖直；<br />（5）247.5；<br />（6）过滤．','【分析】（1）依据常用仪器回答；<br />（2）B装置属于固液常温型，过氧化氢制取氧气可用此装置，据反应原理、装置连接分析解答；<br />（3）分液漏斗可逐滴滴加液体，控制液体的滴加速率；过氧化氢溶液浓度过大等因素会造成反应剧烈，如果反应剧烈，应采取控制滴加速度、减小浓度、换较大锥形瓶等措施；<br />（4）要使瓶内充有尽可能多的氧气，应尽可能地将瓶内的水排净，将集气瓶竖直；<br />（5）需要考虑到空气中也含有氧气，氧气约占空气总体积的五分之一，氧气的总体积应是通入的氧气的体积加上通入的空气中含有的氧气的体积；然后根据氧气的体积分数约为60%，列等式进行计算；<br />（6）二氧化锰不溶于水，因此可用过滤的方法分离．','书写',3.00,'69bafecdc16714e11c785c198c66ccf6',9,400,'混合物的分离方法,仪器的装配或连接,氧气的制取装置,氧气的收集方法,书写化学方程式、文字表达式、电离方程式','',0,'37','',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841873,'某化学兴趣小组对金属的性质做了如下探究，请你填写空格：<br />（1）用如图所示实验探究铁生锈的条件（每支试管中均放有完全相同的洁净铁片）：<br /><img  src=\"/tikuimages/9/0/400/shoutiniao80/2c670f40-94d4-11e9-8350-b42e9921e93e_xkb45.png\" style=\"vertical-align:middle\" /><br />①甲同学认为，试管A发生的现象就能够说明铁的锈蚀是铁与空气中的氧气、水蒸气共同作用的结果．乙同学不同意他的观点，认为必须全面观察试管<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（选填试管编号）发生的现象，并经过科学严谨的推理，才能得出上述结论．<br />②试管D和E实验的目的是进一步探究铁在<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>的环境中是否更容易锈蚀．<br />③为防止金属锈蚀，除了采用覆盖保护膜等措施以外，还可以制成合金．这是因为合金与纯金属相比，其组成和内部组织结构发生变化，从而引起<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>的变化．<br />（2）写出锌与稀硫酸反应的化学方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）将一定质量的铁粉放入硝酸铜、硝酸锌的混合溶液中，充分反应后过滤，所得固体中一定含有的物质是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，且反应后溶液质量比反应前<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（选填“增大”、“减小”或“不变”）．','','','','','','ABC$###$盐、酸存在$###$性质$###$Zn+H<SUB>2</SUB>SO<SUB>4</SUB>═ZnSO<SUB>4</SUB>+H<SUB>2</SUB>↑$###$Cu$###$减小','【解答】解：（1））①要证明铁的锈蚀是铁与空气中的氧气、水蒸气共同作用的结果，就要看铁分别在隔绝空气、隔绝水的条件下，和暴露在有水和空气的条件下的不同，得出结论．所以必须全面观察试管ABC发生的现象，并经过科学严谨的推理，才能得出上述结论．<br />故答案为：ABC．<br />②试管D和E实验的目的是进一步探究铁在盐、酸存在的环境中铁的锈蚀情况．<br />故答案为：盐、酸存在．<br />③金属的组成和内部组织结构发生变化，其性质等就会发生变化．<br />故答案为：性质．<br />（2）锌和稀硫酸反应方程式为：Zn+H<SUB>2</SUB>SO<SUB>4</SUB>═ZnSO<SUB>4</SUB>+H<SUB>2</SUB>↑．<br />故答案为：Zn+H<SUB>2</SUB>SO<SUB>4</SUB>═ZnSO<SUB>4</SUB>+H<SUB>2</SUB>↑．<br />（3）三种金属活动性由强到弱的顺序：锌＞铁＞铜，可知铁只能置换出铜，根据铁与硝酸铜反应的方程式：Fe+Cu（NO<SUB>3</SUB>）<SUB>2</SUB>═Cu+Fe（NO<SUB>3</SUB>）<SUB>2</SUB>．可知56份质量的铁可以置换出64份质量的铜，所以反应后溶液质量比反应前减小．<br />故答案为：Cu；减小．<br />答案：<br />（1）①ABC；②盐、酸存在；③性质；<br />（2）Zn+H<SUB>2</SUB>SO<SUB>4</SUB>═ZnSO<SUB>4</SUB>+H<SUB>2</SUB>↑；<br />（3）Cu；减小．','【分析】（1）①要证明铁的锈蚀是铁与空气中的氧气、水蒸气共同作用的结果，就要看铁分别在隔绝空气、隔绝水的条件下，和暴露在有水和空气的条件下的不同，得出结论．<br />②试管D和E实验的目的是进一步探究铁在盐、酸存在的环境中铁的锈蚀情况；<br />③金属的组成和内部组织结构发生变化，其性质等就会发生变化；<br />（2）锌和稀硫酸反应生成硫酸锌和氢气；<br />（3）根据三种金属活动性由强到弱的顺序：锌铁铜，进行分析．','书写',3.00,'c3b963b764f1ebb5373717938c459dc9',9,400,'探究金属锈蚀的条件,金属的化学性质,书写化学方程式、文字表达式、电离方程式','',0,'37','',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841874,'下列家庭小实验不能成功的是（　　）','用冷碟子放在蜡烛火焰上方制取炭黑','将橄榄油加入水中配制成溶液','用铅笔芯代替石墨试验导电性','用加热的方法给聚乙烯塑料袋封口','','B','【解答】解：A、蜡烛不充分燃烧会产生炭黑，用冷碟子放在蜡烛火焰上方制取炭黑，故选项家庭小实验能成功．<br />B、橄榄油难溶于水，将橄榄油加入水中不能配制成溶液，故选项家庭小实验不能成功．<br />C、铅笔芯中含有石墨，用铅笔芯代替石墨试验导电性，故选项家庭小实验能成功．<br />D、聚乙烯塑料具有热塑性，用加热的方法给聚乙烯塑料袋封口，故选项家庭小实验能成功．<br />故选：B．','【分析】A、根据蜡烛不充分燃烧会产生炭黑，进行分析判断．<br />B、根据橄榄油难溶于水，进行分析判断．<br />C、根据铅笔芯中含有石墨，进行分析判断．<br />D、根据聚乙烯塑料具有热塑性，进行分析判断．','选择题',3.00,'4ba57e433a677cf76c72859a644bc837',9,400,'化学实验方案设计与评价,蜡烛燃烧实验,溶液的概念、组成及其特点,碳单质的物理性质及用途,塑料及其应用','',2016,'37','2016•潮州校级一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841876,'<img  src=\"/tikuimages/9/2015/400/shoutiniao83/2c7455b0-94d4-11e9-8188-b42e9921e93e_xkb32.png\" style=\"vertical-align:middle;FLOAT:right\" />（2015•新化县校级模拟）水和溶液在生命活动和生活中起着十分重要的作用．<br />（1）海水中含有大量的氯化钠，可以通过<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>原理（填“蒸发结晶”或“降温结晶”）获得粗盐，粗盐提纯得到精盐．<br />（2）熟石灰在部分温度下的溶解度如下表所示．<br /><table class=\"edittable\"><TBODY><TR><td width=69>温度/℃</TD><td width=52>0</TD><td width=52>10</TD><td width=52>20</TD><td width=52>30</TD><td width=52>40</TD><td width=52>50</TD><td width=52>60</TD></TR><TR><td>溶解度/g</TD><td>0.18</TD><td>0.17</TD><td>0.16</TD><td>0.15</TD><td>0.14</TD><td>0.13</TD><td>0.12</TD></TR></TBODY></TABLE>①20℃时，0.74g熟石灰放入到盛有100g水的烧杯中，搅拌，所得溶液中氢氧化钙的质量与0.74g相比<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“增大”、“减小”或“不变”），溶液中的溶质可以用生石灰为原料来制取，其化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />②采用一种操作方法，将上述烧杯中剩余固体全部溶解，变为不饱和溶液．下列说法正确的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />A．溶液的质量可能不变&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;B．溶液中溶质的质量分数一定减小<br />C．溶液中溶质的质量一定增大&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;D．可以升温使之变成不饱和溶液．','','','','','','蒸发结晶$###$减小$###$CaO+H<SUB>2</SUB>O═Ca（OH）<SUB>2</SUB>$###$C','【解答】解：（1）氯化钠的溶解度受温度的影响不大，因此一般采用蒸发溶剂的方法得到氯化钠晶体；<br />（2）①由题意可知，在20℃时，氢氧化钙的溶解度是0.16g，将0.74g熟石灰放入到盛有100g水的烧杯中只能溶解0.16g，所以，所得溶液中氢氧化钙的质量与0.74g相比减小，溶液中的溶质可以用生石灰为原料来制取，其化学方程式为：CaO+H<SUB>2</SUB>O═Ca（OH）<SUB>2</SUB>．<br />②若采用一种操作方法，将上述烧杯中剩余固体全部溶解，变为不饱和溶液，由于氢氧化钙的溶解度很小，受温度的影响也很小，不能采取降温的方法，所以：<br />A、将上述烧杯中剩余固体全部溶解，溶液的质量一定增加，故A错误；<br />B、由于同温下同种溶质的饱和溶液比不饱和溶液浓，所以，变为不饱和溶液溶液中溶质的质量分数一定减小．故B正确；<br />C、由于剩余固体全部溶解，溶液中溶质的质量一定增大．故C正确；<br />D、由于氢氧化钙的溶解度随温度的升高而减小，升温不可能使之变成不饱和溶液．故D错误．<br />故答为：（1）蒸发结晶；（2）①减小，CaO+H<SUB>2</SUB>O═Ca（OH）<SUB>2</SUB>；②C．','【分析】（1）氯化钠的溶解度受温度的影响不大，因此一般采用蒸发溶剂的方法得到氯化钠晶体；<br />（2）①根据在20℃时熟石灰的溶解度分析判断；<br />②根据饱和溶液和不饱和溶液的转化关系和氢氧化钙的溶解度受温度的影响分析有关的说法．','填空题',3.00,'ad33e8d56e48b7b6d136e60b5999968f',9,400,'饱和溶液和不饱和溶液相互转变的方法,固体溶解度曲线及其作用,溶质的质量分数,生石灰的性质与用途,海水晒盐的原理和过程','',2015,'32','2015•新化县校级模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841882,'进行下列对比实验，不需要控制变量的是（　　）','用酚酞溶液鉴别稀硫酸和氢氧化钠溶液','用红磷和白磷探究可燃物燃烧的条件','用二氧化锰、氧化铜和过氧化氢溶液比较二氧化锰、氧化铜的催化效果','用镁、锌和稀盐酸比较镁、锌的金属活动性强弱','','A','【解答】解：A．酸不能使酚酞试液变色，碱能使酚酞试液变红色，无需控制变量；<br />B．探究可燃物燃烧的条件时，均与氧气接触，只是可燃物的着火点不同，属于控制变量法；<br />C．此时实验中，过氧化氢溶液的浓度、质量均相同，只是加入的催化剂不同，根据产生气泡的速率来判断，属于控制变量法；<br />D．此实验中，金属片的长度、厚薄、稀盐酸的浓度均相同，只是金属的种类不同，根据反应生成气泡的速率来判断金属的活动性，属于控制变量法．<br />故选A．','【分析】根据控制变量法进行对比要有一个变量，其它条件必须相同进行分析．','选择题',3.00,'90576f1f9cd124ad7280391077676eeb',9,400,'科学探究的基本方法,催化剂的特点与催化作用,金属活动性顺序及其应用,酸、碱、盐的鉴别,燃烧与燃烧的条件','',2016,'32','2016•乐山模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841887,'如图中与对应的实验像符合的是（　　）','<img  src=\"/tikuimages/9/2016/400/shoutiniao78/2c88ef21-94d4-11e9-8414-b42e9921e93e_xkb28.png\" style=\"vertical-align:middle\" /><br />加水稀释氢氧化钠','<img  src=\"/tikuimages/9/2016/400/shoutiniao54/2c8d34de-94d4-11e9-9630-b42e9921e93e_xkb94.png\" style=\"vertical-align:middle\" /><br />等质量等浓度的过氧化氢溶液分别制取氧气','<img  src=\"/tikuimages/9/2016/400/shoutiniao42/2c909040-94d4-11e9-9888-b42e9921e93e_xkb67.png\" style=\"vertical-align:middle\" /><br />等质量的铁粉、镁粉、锌粉分别跟同浓度的盐酸反应','<img  src=\"/tikuimages/9/2016/400/shoutiniao56/2c930140-94d4-11e9-9c48-b42e9921e93e_xkb40.png\" style=\"vertical-align:middle\" /><br />某温度下，向一定量的不饱和硝酸钾溶液中加入硝酸钾固体','','D','【解答】解：A、加水稀释氢氧化钠，氢氧化钠溶液的pH不会等于7、更不会小于7，故A错误；<br />B、催化剂不能改变生成物的量，故B错误；<br />C、等质量的铁粉、镁粉、锌粉分别跟同浓度的盐酸反应，生成氢气的质量是镁的最多，锌的最少，故C错误；<br />D、某温度下，不饱和硝酸钾溶液中加入硝酸钾固体，当达到饱和时不能再溶解，溶液的质量达到最大，故D正确．<br />故选D．','【分析】A、根据加水稀释氢氧化钠时溶液的pH的变化分析；<br />B、根据催化剂的特点分析；<br />C、根据金属与酸的反应分析；<br />D、根据某温度下，不饱和硝酸钾溶液中加入硝酸钾固体，当达到饱和时不能再溶解分析．','选择题',3.00,'1e8bd825f5beedd6c107c7008c8cbcec',9,400,'催化剂的特点与催化作用,金属的化学性质,酸碱溶液的稀释,溶液的酸碱性与pH值的关系','',2016,'37','2016•宝安区二模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841899,'分析推理是化学学习中常用的思维方法，下列说法正确的是（　　）','元素的化学性质是由原子的组外层电子数决定的，那么最外层电子数相同的粒子元素的化学性质一定相同','原子对外不显电性，所以原子中不存在带电的粒子','金属与酸发生置换反应，溶液的质量都会增加','复分解反应中有沉淀、气体或生成水，所以有上述物质生成的反应就是复分解反应','','C','【解答】解：A、元素的化学性质是由原子的组外层电子数决定的，但是最外层电子数相同的粒子元素的化学性质不一定相同，比如镁元素和氦元素最外层电子都是2，但是化学性质不同，错误；<br />B、原子对外不显电性，是因为原子中的带正电的原子核和核外带负电的电子所带的电荷数相等的缘故，错误；<br />C、金属与酸发生置换反应，溶液的质量都会增加，正确；<br />D、复分解反应中有沉淀、气体或生成水，但是有上述物质生成的反应不一定就是复分解反应，比如过氧化氢分解生成水的反应，错误；<br />故选C．','【分析】A、根据化学性质的决定因素解答；<br />B、根据原子的构成微粒解答；<br />C、根据金属与酸的反应解答；<br />D、根据复分解反应的条件解答．','选择题',3.00,'e9128c83726e87c12ee596d20356daff',9,400,'金属的化学性质,复分解反应及其发生的条件,原子的定义与构成,核外电子在化学反应中的作用','',0,'37','',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841905,'尿素【CO（NH<SUB>2</SUB>）<SUB>2</SUB>】是一种常用氮肥．下列关于尿素的说法正确的是（　　）','CO（NH<SUB>2</SUB>）<SUB>2</SUB>中氧、氢两种元素的质量比为8：1','1个分子由1个碳原子、1个氧原子、1个氮原子和2个氢原子构成','30g尿素中氮元素的质量是14g','所含氮元素的质量分数为<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">14</td></tr><tr><td style=\"padding-top:1px;font-size:90%\">12+16+(14+1×2)×2</td></tr></table></span>×100%','','C','【解答】解：A、尿素中氧、氢两种元素的质量比为16：（1×2×2）=4：1，故选项说法错误<br />B、分子是由原子构成的，一个尿素分子中含有1个碳原子、1个氧原子、2个氮原子和4个氢原子构成，故选项说法错误．<br />C、30g尿素中氮元素的质量是：30g×<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">14×2</td></tr><tr><td style=\"padding-top:1px;font-size:90%\">12+16+(14+1×2)×2</td></tr></table></span>×100%=14g，故选项正确．<br />D、尿素中氮元素的质量分数为<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">14×2</td></tr><tr><td style=\"padding-top:1px;font-size:90%\">12+16+(14+1×2)×2</td></tr></table></span>×100%，故选项说法错误．<br />故选C','【分析】A、根据化合物中各元素质量比=各原子的相对原子质量×原子个数之比，进行分析判断．<br />B、根据分子是由原子构成的进行分析判断．<br />C、根据化合物中元素的质量=化合物的质量×元素的质量分数．<br />D、根据元素的质量分数=<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">元素的相对原子质量×原子个数</td></tr><tr><td>化合物的相对分子质量</td></tr></table></span>进行解答．','选择题',3.00,'ab050b597a0a63f2013386d2457ea477',9,400,'化学式的书写及意义,元素质量比的计算,元素的质量分数计算,化合物中某元素的质量计算','',2016,'32','2016•海珠区模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841920,'下列关于资源能源叙述错误的是（　　）','煤、石油和天然气是化石燃料且不可再生，我们要节约能源','空气是一种宝贵的资源，它主要是由氧气和氮气组成的','氢能 源没有污染、燃烧放热多，来源广并且成本低，现已普及使用','使用乙醇汽油既节约石油资源、又能减少污染空气还能促进农业生产','','C','【解答】解：A、煤、石油和天然气都是不可再生的化石燃料，我们要节约，叙述正确；<br />B、空气是一种重要的自然资源，它主要是由氧气和氮气组成的，叙述正确；<br />C、氢能源生产成本高，没有普遍应用，叙述错误；<br />D、乙醇可以由粮食生产，乙醇汽油既节约石油资源，又能促进农业生产，叙述正确；<br />故选C．','【分析】根据已有的知识进行分析，化石能源是不可再生的，我们要节约；空气的主要成分是氮气和氧气；氢能源生产成本高，没有普遍应用；乙醇汽油的使用能节约能源．','选择题',3.00,'531460c759244ada1b33d9a7768c767e',9,400,'空气的成分及各成分的体积分数,常用燃料的使用与其对环境的影响,化石燃料及其综合利用,氢气的用途和氢能的优缺点','',2015,'33','2015秋•龙沙区期末',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841927,'下列有关水的叙述正确的是（　　）','所有溶液中的溶剂均为水','受热时水分子的体积逐渐变大','衣服晒干过程中水分子分解为气体','与肥皂水混合能产生大量泡沫的是软水','','D','【解答】解：A、溶剂是能溶解其他物质的物质，有溶液中的溶剂不是水，例如，碘酒中溶剂就是酒精，故A错误；<br />B、水在受热的情况下水分子获得能量，水分子运动加快，水分子之间的间隔增大，导致水的体积增大，在这个过程中水分子本身的体积大小是不变的，故B错误；<br />C、分子是在不断运动的，湿衣服晒干是因为湿衣服上的水分子运动到空气中去了，故C错误；<br />D、区分硬水和软水的方法是：用肥皂水，加入肥皂水，泡沫多的是软水，泡沫少的是硬水，故D正确．<br />故选D．','【分析】A、根据溶剂的定义进行分析；<br />B、根据分子的特征进行分析；<br />C、根据分子的基本特征进行分析；<br />D、根据区分硬水和软水的方法进行分析．','选择题',3.00,'3ef59de3259d82bbfc8302c61cba96e4',9,400,'硬水与软水,常见的溶剂,分子的定义与分子的特性','',2016,'37','2016•郑州二模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841945,'某学校化学小组的同学，取刚降地面的雨水样，用pH计（测pH的仪器）每隔几分钟测一次pH，其数据如表所示：<table class=\"edittable\"><TBODY><TR><td width=78>定时间</TD><td width=65>5：05</TD><td width=65>5：10</TD><td width=65>5：15</TD><td width=65>5：20</TD><td width=65>5：25</TD><td width=65>5：30</TD><td width=65>5：35</TD></TR><TR><td>pH</TD><td>4.95</TD><td>4.94</TD><td>4.94</TD><td>4.88</TD><td>4.86</TD><td>4.85</TD><td>4.85</TD></TR></TBODY></TABLE>（1）绘制时间-pH关系图<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao38/2d3534b0-94d4-11e9-ae21-b42e9921e93e_xkb28.png\" style=\"vertical-align:middle\" /><br />（2）在测定的时间内，酸雨的酸性是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“增强”或“减弱”）<br />（3）煤中含有的硫在燃烧时会以二氧化硫的形式排放，污染大气．<br />①硫在空气中燃烧的化学方程式是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />②回收二氧化硫可以生成硫酸，化学方程式为2SO<SUB>2</SUB>+2H<SUB>2</SUB>O+O<SUB>2</SUB>═2H<SUB>2</SUB>SO<SUB>4</SUB>，参加反应的SO<SUB>2</SUB>与生成的H<SUB>2</SUB>SO<SUB>4</SUB>的质量比是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，将6.4t的SO<SUB>2</SUB>全部转化，可得到H<SUB>2</SUB>SO<SUB>4</SUB>的质量为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>t．','','','','','','增强$###$S+O<SUB>2 </SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>SO<SUB>2</SUB>$###$32：49$###$9.8','【解答】解：<br />（1）根据表中信息，绘制时间-pH关系图如下：<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao50/2d37f3d1-94d4-11e9-98fd-b42e9921e93e_xkb47.png\" style=\"vertical-align:middle\" /><br />（2）在测定的时间内，雨水的PH逐渐减小，酸性逐渐增强．故填：增强．<br />（3）①硫在空气中燃烧的化学方程式为：S+O<SUB>2 </SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>SO<SUB>2 </SUB>．<br />②由2SO<SUB>2</SUB>+O<SUB>2</SUB>+2H<SUB>2</SUB>O=2H<SUB>2</SUB>SO<SUB>4</SUB>可知，参加反应的SO<SUB>2</SUB>与生成的H<SUB>2</SUB>SO<SUB>4</SUB>的质量比是：128：196=32：49．故填：32：49．<br />设将6.4t的SO<SUB>2</SUB>全部转化，可得到H<SUB>2</SUB>SO<SUB>4</SUB>的质量为x，<br />2SO<SUB>2</SUB>+O<SUB>2</SUB>+2H<SUB>2</SUB>O=2H<SUB>2</SUB>SO<SUB>4</SUB><br />128&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 196<br />6.4t&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; x<br /><span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">128</td></tr><tr><td>6.4t</td></tr></table>=<table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">196</td></tr><tr><td>x</td></tr></table></span><br />x=9.8t．<br />答案：<br />（1）<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao21/2d39c88f-94d4-11e9-9874-b42e9921e93e_xkb42.png\" style=\"vertical-align:middle\" /><br />（2）增强；<br />（3）①S+O<SUB>2 </SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>SO<SUB>2 </SUB>．②32：49；9.8．','【分析】（1）根据表中信息分析解答；<br />（2）PH＜5.6的雨水称为酸雨，当溶液的PH小于7时，随着PH的减小酸性增强；<br />（3）硫在空气中燃烧能生成二氧化硫，根据化学方程式可以进行相关方面的计算．','书写',3.00,'5c27f0ea7cb3f5f8511030f28daab9a5',9,400,'溶液的酸碱性与pH值的关系,酸雨的产生、危害及防治,书写化学方程式、文字表达式、电离方程式,根据化学反应方程式的计算','',2016,'37','2016•番禺区一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841963,'下列叙述正确的是（　　）','石油分馏可得到汽油、煤油等多种产品，属于化学变化','只有产生燃烧的化学反应才会放热','氧化反应都会放出热量','能量都是由化学反应产生的','','C','【解答】解：A、石油分馏可得到汽油、煤油等多种产品，这一过程中没有新物质生成，属于物理变化，故A错误．<br />B、燃烧能放出一定的热量，但在不燃烧时有可能也会放出热量，如：生石灰与水反应的过程中也会放出大量的热，故B错误．<br />C、氧化反应都会放出热量，故C正确．<br />D、能量并不都是由化学反应产生的，如风能、潮汐能等能量不是由化学反应产生的，故D错误．<br />故选C．','【分析】A、根据石油分馏属于物理变化进行分析；<br />B、根据放热反应进行分析；<br />C、根据氧化反应的特点进行分析；<br />D、高级能量并不都是由化学反应产生的进行分析．','选择题',3.00,'17ba8529fc01c2c9911ee00d53c3bac2',9,400,'物质发生化学变化时的能量变化,氧化反应,石油加工的产物','',2016,'37','2016•黑龙江一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841965,'利用海水提取粗盐的过程如下图所示，回答有关问题．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao90/2d9ef300-94d4-11e9-b9cd-b42e9921e93e_xkb91.png\" style=\"vertical-align:middle\" /><br />（1）一定质量的海水，通过贮水池引入到蒸发池，在没有引入结晶池之前的蒸发过程中，蒸发池中氯化钠的质量会<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“增大”、“不变”或“减小”）．<br />（2）粗盐中含有的难溶性杂质，在实验室里可以通过溶解、过滤、蒸发等操作将其去除，这些操作中都会用到玻璃棒，其中在过滤操作中玻璃棒的作用是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>&nbsp;（填“引流”或“加快溶解”）．<br />（3）用氯化钠固体配制100g质量分数为6%的氯化钠溶液．<br />①配制时，涉及以下实验步骤：A．称量及量取B．计算C．溶解D．装瓶贴标签．其正确的实验步骤顺序是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填字母）．<br />②将已配好的100g质量分数为6%的氯化钠溶液变成质量分数为3%的氯化钠溶液，需要再加水的体积是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>ml．','','','','','','不变$###$引流$###$BACD$###$100','【解答】解：（1）在蒸发池中，水蒸发了，溶剂的质量减少，溶质的质量不变；<br />（2）在过滤操作中玻璃棒的作用是引流；<br />（3）①配制一定质量分数溶液的步骤是：计算、称量、溶解、装瓶，所以正确的实验步骤顺序是BACD；<br />②100g质量分数的6%的氯化钠溶液中溶质质量为：100g×6%=6g，根据稀释前后溶质质量不变，设稀释后溶液的质量为x，则3%x=6g，x=200g，所以需要加水200g-100g=100g，故加水100ml；<br />故答案为：（1）不变&nbsp;&nbsp;&nbsp;&nbsp;（2）引流&nbsp;&nbsp;&nbsp;&nbsp;（3）①BACD&nbsp;&nbsp;&nbsp;&nbsp;②100．','【分析】（1）根据在蒸发池中，水蒸发了，溶剂的质量减少，溶质的质量不变进行分析；<br />（2）根据在过滤操作中玻璃棒的作用是引流进行分析；<br />（3）①根据配制一定质量分数溶液的步骤是：计算、称量、溶解、装瓶进行分析；<br />②根据溶质质量=溶液质量×溶质质量分数进行分析．','填空题',3.00,'40c61e2aa10f9c08d2267128f6f3a5f5',9,400,'一定溶质质量分数的溶液的配制,过滤的原理、方法及其应用,溶质的质量分数,氯化钠与粗盐提纯,海水晒盐的原理和过程','',2016,'35','2016春•津南区校级期中',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841972,'下列有关二氧化碳的说法错误的是（　　）','二氧化碳可用于制碳酸类饮料','吸过多的二氧化碳排放会导致温室效应','干冰可用于人工降雨','二氧化碳可用块状石灰石和稀硫酸大量制取','','D','【解答】解：A、二氧化碳可用于制碳酸类饮料，故A正确；<br />B、吸过多的二氧化碳排放会导致温室效应，故B正确；<br />C、二氧化碳的固体干冰，可以进行人工降雨，也可以做制冷剂，故C正确；<br />D、二氧化碳的制取方法，实验室中一般用石灰石或大理石与稀盐酸反应，一般不用稀硫酸，因为稀硫酸与碳酸钙反应生成微溶的硫酸钙，覆盖在大理石表面阻止反应的进行，故D错误．<br />故选D．','【分析】二氧化碳的用途有：①二氧化碳既不能燃烧也不能支持燃烧，因此二氧化碳能够灭火；②二氧化碳可以做化工原料，如制汽水等；③二氧化碳的固体干冰，可以进行人工降雨，也可以做制冷剂；④二氧化碳促进绿色植物的光合作用，因此可以做气体肥料；二氧化碳的制取方法，实验室中一般用石灰石或大理石与稀盐酸反应．','选择题',3.00,'dcce4053b1369d42fbf9bf2f191bda6e',9,400,'二氧化碳的实验室制法,二氧化碳的用途,二氧化碳对环境的影响','肥城市',2016,'37','2016•肥城市二模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841976,'以下说法错误的是（　　）','工业中利用活性炭脱色以制白糖','室内放一盆水不能防止CO中毒','金刚石和石墨的碳原子排列方式相同','CO<SUB>2</SUB>可用于灭火，CO可用作燃料','','C','【解答】解：<br />A、由于活性炭具有较强的吸附性，所以工业中常利用活性炭脱色以制白糖；正确；<br />B、由于一氧化碳不能与水反应也不会溶于水，所以室内放一盆水不能防止一氧化碳中毒；正确；<br />C、金刚石中的碳原子是正四面体结构，硬度很大；石墨中的碳原子是正六边形结构，并且形成层状，硬度较小；正是因为金刚石和石墨里碳原子的排列不同，才导致了金刚石和石墨的物理性质有很大差异；错误；<br />D、二氧化碳不燃烧也不支持燃烧，所以CO<SUB>2</SUB>可用于灭火，一氧化碳具有可燃性，所以CO可用作燃料；正确．<br />故选C．','【分析】A、依据活性炭的吸附性解决问题即可；<br />B、根据两者的碳原子排列情况分析解答；<br />C、根据一氧化碳的溶解性及性质分析解答；<br />D、依据两者的性质分析判断；','选择题',3.00,'f9e1bfced6d121da49eb8025d5fe4ced',9,400,'二氧化碳的用途,一氧化碳的毒性,碳单质的物理性质及用途,碳元素组成的单质','',2016,'37','2016•西区一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841987,'从下列物质中选择适当的物质替代括号内的物质，使之能具有类似的作用（用A、B、C、D、E填空）<br />A．稀硫酸&nbsp;&nbsp;&nbsp;&nbsp;B．小苏打&nbsp;&nbsp;&nbsp;&nbsp;C．pH试纸&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;D．硝酸铵&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;E．铁粉&nbsp;&nbsp;F&nbsp;氮气<br />（1）用（稀盐酸）除铁锈<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；&nbsp;&nbsp;&nbsp;&nbsp;<br />（2）用（尿素）给庄稼施肥<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>：<br />（3）用（生石灰）做食品干燥剂<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，<br />（4）用含（氢氧化铝）的药物治疗胃酸过多<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />（5）用（石蕊试液）检测溶液的酸碱性<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（6）用CO<SUB>2</SUB>灭火<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','A$###$D$###$E$###$B$###$C$###$F','【解答】解：（1）硫酸与盐酸都属于酸，具有相似的性质，都能与某些金属氧化物起反应．稀盐酸能除铁锈，稀硫酸也能除铁锈；<br />（2）尿素和硝酸铵都含有氮元素，都可以用于庄稼施氮肥；<br />（3）生石灰能与水反应，可用作食品干燥剂；铁粉生锈时能消耗水分，也可用作做食品干燥剂；<br />（4）氢氧化铝能与胃液中的酸反应，小苏打也能与胃液中的酸反应，可用作治疗胃酸过多；<br />（5）石蕊试液能检测溶液的酸碱性，通过pH试纸测定溶液的酸碱度，也可知道溶液的酸碱性；<br />（6）二氧化碳不支持燃烧能用于灭火，氮气也不支持燃烧，也能用于灭火；<br />故答为：A，D，E，B，C，F．','【分析】根据物质的用途，分析物质的性质，找出具体的性质相似的物质即可．','填空题',3.00,'238339803f220b5261d1bfa74d5c2c13',9,400,'常见气体的用途,溶液的酸碱性测定,常用盐的用途','',0,'37','',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841988,'洗过的玻璃仪器内壁<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>时，表示仪器已洗干净．','','','','','','附着的水既不聚成水滴$###$也不成股流下','【解答】解：洗过的玻璃仪器内壁附着的水既不聚成水滴也不成股流下时，表示仪器已洗干净．<br />故答案为：附着的水既不聚成水滴；也不成股流下．','【分析】根据实验室玻璃仪器洗净的标准，进行分析解答即可．','填空题',3.00,'2ebefad580322adf08f6b860148f1efb',9,400,'玻璃仪器的洗涤','',2011,'37','2011秋•蓝山县校级月考',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841989,'下列说法错误的是（　　）','用氢氧化钠固体干燥氧气','蒸馏是净化程度最高的净水方法','稀释浓硫酸时应把浓硫酸倒入水中','用润湿的pH试纸测定溶液的酸碱度','','D','【解答】解：A、氢氧化钠具有吸水性，又不和氧气反应，因此可用氢氧化钠固体干燥氧气，故选项说法正确．<br />B、在净化水的方法中，净化水程度最高的方法是蒸馏，故选项说法正确．<br />C、稀释浓硫酸时，要把浓硫酸缓缓地沿器壁注入水中，同时用玻璃棒不断搅拌，以使热量及时地扩散；一定不能把水注入浓硫酸中；故选项说法正确．<br />D、用pH试纸测定未知溶液的pH时，正确的操作方法为用玻璃棒蘸取少量待测液滴在干燥的pH试纸上，与标准比色卡对比来确定pH．不能用水湿润pH试纸，否则稀释了待测溶液，使溶液的酸碱性减弱，测定结果不准确，故选项说法错误．<br />故选：D．','【分析】A、根据氢氧化钠的性质进行分析判断．<br />B、根据蒸馏可以得到纯净水进行分析判断．<br />C、根据浓硫酸的稀释方法（酸入水，沿器壁，慢慢倒，不断搅）进行分析判断．<br />D、根据用pH试纸测定未知溶液的pH的方法进行分析判断．','选择题',3.00,'cbe9544357a158581b6dd066481d79de',9,400,'浓硫酸的性质及浓硫酸的稀释,溶液的酸碱度测定,水的净化,根据浓硫酸或烧碱的性质确定所能干燥的气体','',2016,'37','2016•郑州二模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1841994,'<img  src=\"/tikuimages/9/2016/400/shoutiniao88/2df74c30-94d4-11e9-a221-b42e9921e93e_xkb54.png\" style=\"vertical-align:middle;FLOAT:right\" />（2016•余干县三模）如图是某化学反应的微观示意图，从图中获得的有关信息正确的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />A．参加反应的<img  src=\"/tikuimages/9/2016/400/shoutiniao78/2dfb43d1-94d4-11e9-89a4-b42e9921e93e_xkb82.png\" style=\"vertical-align:middle\" />和<img  src=\"/tikuimages/9/2016/400/shoutiniao55/2dfdb4d1-94d4-11e9-9c55-b42e9921e93e_xkb70.png\" style=\"vertical-align:middle\" />的分子个数比为2：1<br />B．该化学反应前后分子的个数不变<br />C．该化学反应中共有三种单质<br />D．该反应的本质是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','A$###$分子发生再分，原子重新组合','【解答】解：A、由物质的微观构成可知，参加反应的<img  src=\"/tikuimages/9/2016/400/shoutiniao24/2e0073f0-94d4-11e9-ac7e-b42e9921e93e_xkb59.png\" style=\"vertical-align:middle\" />和<img  src=\"/tikuimages/9/2016/400/shoutiniao41/2e00e921-94d4-11e9-9c97-b42e9921e93e_xkb85.png\" style=\"vertical-align:middle\" />的分子个数比为2：1．故A正确；<br />B、由化学反应的微观模拟示意图可知，该化学反应前后分子的个数改变．故B错误；<br />C、由反应前后微粒的变化可知，该化学反应中共有两种单质．故C错误．<br />D、由化学反应的微观模拟示意图可知，该反应的本质是分子发生再分，原子重新组合．<br />故选A，D补充：该反应是分子发生再分，原子重新组合．','【分析】观察分析化学反应的微观模拟示意图．分析物质的微观构成，判断物质的类别，根据反应的特点分析反应的类型等．','填空题',3.00,'7a3c737a7498f9bca377c1aae02d1ae1',9,400,'单质和化合物的判别,微粒观点及模型图的应用,化学反应的实质','',2016,'37','2016•余干县三模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842001,'安全用火是每个人都应该了解的生活常识，下列做法正确的是（　　）','燃气灶火焰呈黄色时，调小进风口','炒菜时油锅着火用锅盖盖灭','液化气泄漏，立即打开排风扇','存放图书的地方起火，用泡沫灭火器扑灭','','B','【解答】解：A、燃气灶火焰呈黄色时，说明气体燃烧不充分，应调大进风口，故选项说法错误．<br />B、炒菜时油锅着火，用锅盖盖灭，利用的是隔绝氧气的灭火原理，故选项说法正确．<br />C、液化气泄漏，不能立即打开排风扇，因为打开抽油烟机会产生电火花，可能会发生爆炸，故选项说法错误．<br />D、存放图书的地方起火，不能用泡沫灭火器扑灭，因为喷出的溶液会损坏图书，故选项说法错误．<br />故选：B．','【分析】A、燃气灶火焰呈黄色时，说明气体燃烧不充分，据此进行分析判断．<br />B、根据灭火原理：①清除或隔离可燃物，②隔绝氧气或空气，③使温度降到可燃物的着火点以下，进行分析判断．<br />C、根据可燃性气体与空气混合后点燃可能发生爆炸进行分析判断．<br />D、用来扑灭图书档案、精密仪器等处的火灾不能有水，否则容易受损．','选择题',3.00,'97c44dbebb7f02b62a16ea9cc03f5d4c',9,400,'完全燃烧与不完全燃烧,灭火的原理和方法,防范爆炸的措施,几种常用的灭火器','',0,'37','',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842011,'下列相关事实用微观粒子的知识解释不正确的是（　　） <table class=\"edittable\"><TBODY><TR><td width=33>选项</TD><td width=248>事&nbsp;&nbsp; 实</TD><td width=292>解&nbsp;&nbsp; 释</TD></TR><TR><td>A</TD><td>氯化钠溶液可以导电</TD><td>氯化钠溶液中存在自由移动的离子</TD></TR><TR><td>B</TD><td>水烧开后水壶容易被顶起</TD><td>温度升高，分子间隔变大</TD></TR><TR><td>C</TD><td>金刚石比石墨硬度大</TD><td>碳原子的结构不同</TD></TR><TR><td>D</TD><td>在阳光下，湿衣服容易晾干</TD><td>分子的运动速率随温度升高而加快</TD></TR></TBODY></TABLE>','A','B','C','D','','C','【解答】解：A、氯化钠溶液可以导电，是因为溶液中存在自由移动的钠离子和氯离子，故选项解释正确．<br />B、水烧开后水壶容易被顶起，是因为温度升高，分子间的间隔变大，故选项解释正确．<br />C、金刚石比石墨硬度大，是因为碳原子的排列方式不同，故选项解释错误．<br />D、在阳光下，湿衣服容易晾干，是因为阳光下温度高，分子运动的速率加快，故选项解释正确．<br />故选：C．','【分析】根据分子的基本特征：分子质量和体积都很小；分子之间有间隔；分子是在不断运动的；同种的分子性质相同，不同种的分子性质不同，可以简记为：“两小运间，同同不不”，结合质的水溶液能够导电，应能产生自由移动离子，进行分析判断即可．','选择题',3.00,'743b78a932c1e89c7fe34c3080a55f66',9,400,'溶液的导电性及其原理分析,利用分子与原子的性质分析和解决问题','',0,'37','',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842027,'如图是我们做过的四个实验，请你回答下列问题：<br />（1）A实验瓶中放少量水的作用是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，写出该反应的化学方程式：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）B实验在烧杯中滴加酚酞溶液的目的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，可观察到的现象是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）C实验探究的问题是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（4）D实验测定结果偏小的原因可能是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，该实验可得出的结论是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao33/2e574680-94d4-11e9-b2d6-b42e9921e93e_xkb79.png\" style=\"vertical-align:middle\" />','','','','','','防止溅落的熔化物炸裂瓶底$###$3Fe+2O<SUB>2</SUB> <span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>Fe<SUB>3</SUB>O<SUB>4</SUB>$###$指示酸碱发生了中和反应$###$红色褪去$###$呼出的气体与空气中二氧化碳含量的高低$###$装置漏气、红磷不足等$###$空气的氧气约占空气体积的<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">1</td></tr><tr><td>5</td></tr></table></span>','【解答】解：（1）A实验是铁丝在氧气中燃烧生成了四氧化三铁，放出了大量的热，瓶中放少量水的作用是：防止溅落的熔化物炸裂瓶底，该反应的化学方程式：3Fe+2O<SUB>2</SUB> <span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>Fe<SUB>3</SUB>O<SUB>4</SUB>；<br />（2）B实验盐酸与氢氧化钠发生了中和反应，在烧杯中滴加酚酞溶液的目的是：指示酸碱发生了中和反应，可观察到的现象是：红色褪去，<br />（3）由图示可知，C实验探究的问题是：呼出的气体与空气中二氧化碳含量的高低；<br />（4）D实验测定空气中氧气的含量，结果偏小的原因可能是：装置漏气、红磷不足等，该实验可得出的结论是：空气的氧气约占空气体积的<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">1</td></tr><tr><td>5</td></tr></table></span>．<br />故答为：（1）防止溅落的熔化物炸裂瓶底，3Fe+2O<SUB>2</SUB> <span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>Fe<SUB>3</SUB>O<SUB>4</SUB>；<br />（2）指示酸碱发生了中和反应，红色褪去，<br />（3）呼出的气体与空气中二氧化碳含量的高低；<br />（4）是装置漏气、红磷不足等，空气的氧气约占空气体积的<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">1</td></tr><tr><td>5</td></tr></table></span>．','【分析】观察图示中实验，分析实验的目的、发生的反应、注意事项、实验的目的等，据此分析回答有关的问题．','书写',3.00,'96992d44f18287f4c0308bf541d1748b',9,400,'吸入空气与呼出气体的比较,空气组成的测定,氧气的化学性质,中和反应及其应用,书写化学方程式、文字表达式、电离方程式','宁国市',2016,'37','2016•宁国市一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842040,'如表为某校化学实验小组测定不同体积H<SUB>2</SUB>和空气混合气体点燃时实验现象：根据如表数据和现象，判断下列说法中正确的是（　　）<br /><table class=\"edittable\"><TBODY><TR><td width=119>H<SUB>2</SUB>体积分数%</TD><td width=59>90</TD><td width=47>80</TD><td width=48>70</TD><td width=48>60</TD><td width=48>50</TD><td width=58>40</TD><td width=48>30</TD><td width=48>20</TD><td width=48>10</TD><td width=65>5</TD></TR><TR><td>空气体积分数%</TD><td>10</TD><td>20</TD><td>30</TD><td>40</TD><td>50</TD><td>60</TD><td>70</TD><td>80</TD><td>90</TD><td>95</TD></TR><TR><td>点燃现象</TD><td>安静<br />燃烧</TD><td>安静<br />燃烧</TD><td>弱爆炸</TD><td>强爆炸</TD><td>强爆炸</TD><td>强爆炸</TD><td>强爆炸</TD><td>强爆炸</TD><td>弱爆炸</TD><td>不燃烧<br />不爆炸</TD></TR></TBODY></TABLE>','H<SUB>2</SUB>体积分数为10%-70%的H<SUB>2</SUB>和空气混合气体，点燃时会发生爆炸','收集的H<SUB>2</SUB>如果能安静燃烧，说明H<SUB>2</SUB>的纯度为100%','用向下排空气法收集H<SUB>2</SUB>，保持试管倒置移近火焰，如没有听到任何声音，表示收集的H<SUB>2</SUB>已纯净','氢气和空气的混合气体点燃时一定发生爆炸','','A','【解答】解：A、从题目中表格知，H<SUB>2</SUB>体积分数为10%--70%的H<SUB>2</SUB>和空气混合气体，点燃时会发生爆炸，故A正确；<br />B、收集的H<SUB>2</SUB>能安静燃烧，说明H<SUB>2</SUB>的纯度大于等于80%，故B项错误；<br />C、用向下排空气法收集H<SUB>2</SUB>，保持试管倒置移近火焰，如果没有听到任何声音，表示收集的H<SUB>2</SUB>纯度大于等于80%，故C项错误；<br />D、氢气和空气的混合气体点燃不一定发生爆炸，只有在爆炸极限范围内才会发生爆炸，故D项错误．<br />故选A．','【分析】可燃物质（可燃气体、蒸气和粉尘）与空气（或氧气）在一定的浓度范围内均匀混合，遇着火源可能会发生爆炸，这个浓度范围称为爆炸极限．可燃性混合物能够发生爆炸的最低浓度和最高浓度，分别称为爆炸下限和爆炸上限，在低于爆炸下限时不爆炸也不着火，在高于爆炸上限同样不燃不爆．因此可燃性气体在点燃前需要先检验气体的纯度，以防发生爆炸．','选择题',3.00,'c83580daf001ecf74e185249d00d902b',9,400,'燃烧和爆炸实验','',2016,'32','2016•内江模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842053,'某化学兴趣小组的同学们学习了纯碱的知识后，对著名的侯氏制碱法产生了浓厚的兴趣，决定对侯氏制碱法进行深入的探究，请你一起参与他们的探究．<br />【查阅资料】<br />Ⅰ．侯氏制碱法的原理：<br />第一步：向饱和食盐水中先通入过量氨气（溶液显碱性），再通入过量的二氧化碳，即可析出NaHCO<SUB>3</SUB>固体，反应方程式是：NaCl+CO<SUB>2</SUB>+NH<SUB>3</SUB>+H<SUB>2</SUB>O═NaHCO<SUB>3</SUB>↓+NH<SUB>4</SUB>Cl．<br />第二步：将第一步反应后所得的混合物进行过滤后，将滤渣加热，可生成纯碱、一种能使澄清石灰水变浑浊的气体和一种常见的氧化物．<br />Ⅱ．已知20℃时有关物质的溶解度如下（气体指1体积水中能溶解得气体体积）<br /><table class=\"edittable\"><TBODY><TR><td width=83>物质</TD><td width=83>NaCl</TD><td width=83>NaHCO<SUB>3</SUB></TD><td width=83>NH<SUB>4</SUB>Cl</TD><td width=83>NH<SUB>3</SUB></TD><td width=83>CO<SUB>2</SUB></TD></TR><TR><td>溶解度</TD><td>36.0g</TD><td>9.6g</TD><td>37.2g</TD><td>710</TD><td>0.9</TD></TR></TBODY></TABLE>Ⅲ．已知NH<SUB>4</SUB>Cl加热条件下易分解，反应的化学方程式是NH<SUB>4</SUB>Cl<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>NH<SUB>3</SUB>↑+HCl↑<br />【资料回顾】<br />（1）第一步中析出NaHCO<SUB>3</SUB>固体的原因是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，实验中先向饱和食盐水中通入过量氨气，再通入过量的二氧化碳，其原因是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填序号）．<br />①使CO<SUB>2</SUB>更易被吸收&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; ②NH<SUB>3</SUB>比CO<SUB>2</SUB>更易制取&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; ③CO<SUB>2</SUB>的密度比NH<SUB>3</SUB>大<br />（2）写出第二步中将滤渣加热发生反应的化学方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）该工业同时可得副产物NH<SUB>4</SUB>Cl，它在农业上可作<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>肥，写出NH<SUB>4</SUB>Cl与熟石灰固体加热的化学方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />【定性探究】<br />第一步反应所得混合物在第二步过滤后所得滤液中溶质的成分是什么？<br />【作出猜想】<br />猜想Ⅰ：NH<SUB>4</SUB>Cl<br />猜想Ⅱ：NH<SUB>4</SUB>Cl和NaHCO<SUB>3</SUB><br />猜想Ⅲ：NH<SUB>4</SUB>Cl和NaCl<br />猜想Ⅳ：NH<SUB>4</SUB>Cl、NaCl和<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />【设计实验】<br /><table class=\"edittable\"><TBODY><TR><td width=187>&nbsp;实验步骤</TD><td width=113>实验现象&nbsp;</TD><td width=202>实验结论&nbsp;</TD></TR><TR><td>&nbsp;①取少量滤液于试管中，滴加<u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD><td>产生无色无味气体&nbsp;</TD><td>证明滤液中含有的物质是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>&nbsp;</TD></TR><TR><td>&nbsp;②另取少量滤液于试管中，滴加足量稀HNO<SUB>3</SUB>后，再滴加AgNO<SUB>3</SUB>溶液</TD><td>&nbsp;产生白色沉淀</TD><td>证明滤液中含有NaCl&nbsp;</TD></TR></TBODY></TABLE>【交流讨论】小明同学认为步骤②结论不正确，他的理由是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />小张为证明滤液中是否含NaCl，设计了以下方案：另取少量滤液蒸干后充分灼烧，取灼烧后的残余固体溶于水，往其中滴加足量稀HNO<SUB>3</SUB>后，再滴加AgNO<SUB>3</SUB>溶液，产生白色沉淀．<br />【得出结论】猜想<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>成立．<br />【定量探究】<br />实际生产中得到的纯碱样品中含有少量NaCl，国际上规定纯碱的纯度（即纯碱的质量分数）≥98.0%为合格品，同学们想判断某纯碱样品是否合格，分成四个小组，各取25克该样品全部溶解于水中，向其中加入相同溶质质量分数的CaCl<SUB>2</SUB>溶液．四组加入的CaCl<SUB>2</SUB>溶液的质量与产生的沉淀的质量关系如表．<br /><table class=\"edittable\"><TBODY><TR><td width=141></TD><td width=57>第一组</TD><td width=57>第二组</TD><td width=57>第三组</TD><td width=57>第四组</TD></TR><TR><td>CaCl<SUB>2</SUB>溶液的质量（克）</TD><td>100</TD><td>200</TD><td>300</TD><td>400</TD></TR><TR><td>产生沉淀的质量（克）</TD><td>8</TD><td>16</TD><td>20</TD><td>20</TD></TR></TBODY></TABLE>（1）通过计算判断该纯碱样品是否是合格品？<br />（2）若要计算CaCl<SUB>2</SUB>溶液的溶质质量分数，可以选择第<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>组数据计算．','','','','','','相同条件下，碳酸氢钠的溶解度比氯化钠和碳酸钠小$###$①$###$2NaHCO<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>Na<SUB>2</SUB>CO<SUB>3</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑$###$氮$###$2NH<SUB>4</SUB>Cl+Ca（OH）<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>CaCl<SUB>2</SUB>+2H<SUB>2</SUB>O+2NH<SUB>3</SUB>↑$###$NaHCO<SUB>3</SUB>$###$稀盐酸$###$NaHCO<SUB>3</SUB>$###$氯化铵能和硝酸银反应生成白色沉淀氯化银$###$Ⅳ$###$一、二','【解答】解：【资料回顾】<br />（1）第一步中析出NaHCO<SUB>3</SUB>固体的原因是相同条件下，碳酸氢钠的溶解度比氯化钠和碳酸钠小；<br />实验中先向饱和食盐水中通入过量氨气，再通入过量的二氧化碳，其原因是使CO<SUB>2</SUB>更易被吸收．<br />故填：相同条件下，碳酸氢钠的溶解度比氯化钠和碳酸钠小；①．<br />（2）第二步中将滤渣加热时，碳酸氢钠分解生成碳酸钠、水和二氧化碳，发生反应的化学方程式为：2NaHCO<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>Na<SUB>2</SUB>CO<SUB>3</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑．<br />故填：2NaHCO<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>Na<SUB>2</SUB>CO<SUB>3</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑．<br />（3）该工业同时可得副产物NH<SUB>4</SUB>Cl中含有氮元素，它在农业上可作氮肥；<br />NH<SUB>4</SUB>Cl与熟石灰固体加热的化学方程式为：2NH<SUB>4</SUB>Cl+Ca（OH）<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>CaCl<SUB>2</SUB>+2H<SUB>2</SUB>O+2NH<SUB>3</SUB>↑．<br />故填：氮；2NH<SUB>4</SUB>Cl+Ca（OH）<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>CaCl<SUB>2</SUB>+2H<SUB>2</SUB>O+2NH<SUB>3</SUB>↑．<br />【作出猜想】<br />猜想Ⅰ：NH<SUB>4</SUB>Cl<br />猜想Ⅱ：NH<SUB>4</SUB>Cl和NaHCO<SUB>3</SUB><br />猜想Ⅲ：NH<SUB>4</SUB>Cl和NaCl<br />猜想Ⅳ：NH<SUB>4</SUB>Cl、NaCl和NaHCO<SUB>3</SUB>．<br />故填：NaHCO<SUB>3</SUB>．<br />【设计实验】实验过程如下表所示：<br /><table class=\"edittable\"><TBODY><TR><td width=187>实验步骤</TD><td width=113>实验现象 </TD><td width=202>实验结论 </TD></TR><TR><td>①取少量滤液于试管中，滴加稀盐酸</TD><td>产生无色无味气体 </TD><td>证明滤液中含有的物质是NaHCO<SUB>3</SUB></TD></TR><TR><td>②另取少量滤液于试管中，滴加足量稀HNO<SUB>3</SUB>后，再滴加AgNO<SUB>3</SUB>溶液</TD><td>产生白色沉淀</TD><td>证明滤液中含有NaCl </TD></TR></TBODY></TABLE>【交流讨论】<br />小明同学认为步骤②结论不正确，他的理由是氯化铵能和硝酸银反应生成白色沉淀氯化银．<br />故填：氯化铵能和硝酸银反应生成白色沉淀氯化银．<br />【得出结论】<br />另取少量滤液蒸干后充分灼烧，取灼烧后的残余固体溶于水，往其中滴加足量稀HNO<SUB>3</SUB>后，再滴加AgNO<SUB>3</SUB>溶液，产生白色沉淀，说明滤液中溶质的成分是NH<SUB>4</SUB>Cl、NaCl和NaHCO<SUB>3</SUB>，因此猜想Ⅳ成立．<br />故填：Ⅳ．<br />【定量探究】<br />（1）设碳酸钠质量为x，<br />Na<SUB>2</SUB>CO<SUB>3</SUB>+CaCl<SUB>2</SUB>═CaCO<SUB>3</SUB>↓+2NaCl，<br />&nbsp; 106&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;100<br />&nbsp;&nbsp; x&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;20g<br /><span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">106</td></tr><tr><td>x</td></tr></table></span>=<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">100</td></tr><tr><td>20g</td></tr></table></span>，<br />x=21.2g，<br />该纯碱样品中碳酸钠质量分数为：<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">21.2g</td></tr><tr><td>25g</td></tr></table></span>×100%=84.8%，<br />因此该纯碱样品不合格．<br />（2）若要计算CaCl<SUB>2</SUB>溶液的溶质质量分数，可以选择第一、二组数据计算，这是因为一、二组中的氯化钙完全反应．<br />故填：一、二．','【分析】【资料回顾】<br />相同条件下，溶解度越小，越容易从溶液中析出；<br />显碱性的溶液更容易吸收二氧化碳；<br />碳酸氢钠受热分解生成碳酸钠、水和二氧化碳；<br />氯化铵和氢氧化钙在加热条件下反应生成氯化钙、水和氨气；<br />【定性探究】<br />析出碳酸氢钠固体后的溶液仍然是碳酸氢钠的饱和溶液；<br />【设计实验】<br />碳酸氢钠和稀盐酸反应生成氯化钠、水和二氧化碳；<br />【交流讨论】<br />银离子能和氯离子结合成白色沉淀氯化银；<br />【定量探究】<br />根据提供的数据和反应的化学方程式可以进行相关方面的计算和判断．','书写',3.00,'aaacecf5574316bb5d417dcc72fb5d39',9,400,'实验探究物质的组成成分以及含量,有关溶质质量分数的简单计算,盐的化学性质,常见化肥的种类和作用,纯碱的制取,书写化学方程式、文字表达式、电离方程式,根据化学反应方程式的计算','靖江市',2016,'37','2016•靖江市二模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842075,'化学概念在逻辑上存在如图所示关系，下列概念间的关系说法正确是（　　）<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao18/2eeb7030-94d4-11e9-9e32-b42e9921e93e_xkb3.png\" style=\"vertical-align:middle\" />','纯净物与混合物属于包含关系','化合物与氧化物属于包含关系','单质与化合物属于交叉关系','金属元素和非金属元素属于并列关系','','B|D','【解答】解：A、纯净物与混合物属于并列关系，不是包含关系，故选项错误；<br />B、氧化物是指由两种元素组成的化合物中，其中一种元素是氧元素，化合物与氧化物属于包含关系，故选项正确；<br />C、单质与化合物属于并列关系，不属于交叉关系，故选项错误；<br />D、由同种元素组成的纯净物叫单质，单质又分为金属单质和非金属单质，金属元素和非金属元素属于并列关系，故选项正确；<br />故选B、D','【分析】物质分为混合物和纯净物，混合物是由两种或两种以上的物质组成；纯净物是由一种物质组成．纯净物又分为单质和化合物．由同种元素组成的纯净物叫单质，单质又分为金属单质和非金属单质；由两种或两种以上的元素组成的纯净物叫化合物．氧化物是指由两种元素组成的化合物中，其中一种元素是氧元素．','多选题',2.00,'ebbe66a3388c54f5f3e1c93ab6cb0093',9,400,'从组成上识别氧化物,纯净物和混合物的概念,单质和化合物的概念,元素的简单分类','龙口市',2016,'35','2016春•龙口市校级期中',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842099,'<img  src=\"/tikuimages/9/2015/400/shoutiniao73/2f16ed00-94d4-11e9-aa7a-b42e9921e93e_xkb9.png\" style=\"vertical-align:middle;FLOAT:right;\" />（2015•冷水江市校级一模）2015年3月22日是第二十三个“世界水日”，它的主题是“水合作”．关于水的知识有下列话题，请按要求填空：<br />（1）人类生活和工农业生产都离不开水，下图是自来水厂净化水的过程示意图：<br />（1）在这个过程中，未涉及的净水方法是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填序号）A．吸附&nbsp;&nbsp;&nbsp;B．沉淀&nbsp;&nbsp;&nbsp;C．过滤&nbsp;&nbsp;&nbsp;D．煮沸<br />（2）有关水的组成和结构的叙述中，正确的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />A．水中氢、氧元素的质量比为2：1<br />B．水是由水分子构成的<br />C．水分子是由氢分子和氧原子构成的<br />（3）注意饮水安全，保证人体健康．在天然水净化过程中，人们常用活性炭去除异味和色素，这是利用活性炭的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>性．此外还需加入二氧化氯（C1O<SUB>2</SUB>）进行杀菌和消毒，在二氧化氯（C1O<SUB>2</SUB>）中氯元素的化合价为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（4）若要测定水的酸碱度，最适宜的试剂或用品是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填序号）．<br />A．紫色石蕊溶液&nbsp;&nbsp;&nbsp;&nbsp;B．pH试纸&nbsp;&nbsp;&nbsp;&nbsp;C．石灰石&nbsp;&nbsp;&nbsp;&nbsp;D．无色酚酞溶液．','','','','','','D$###$B$###$吸附$###$+4$###$B','【解答】解：<br />（1）观察图示可知：自来水厂生产自来水时，使用的净水方法有沉淀、过滤、吸附等，而没有煮沸，故选：D；<br />（2）A、水中氢、氧元素的质量比为1×2：16=1：8，故错误；<br />B、水由水分子构成，正确；<br />C、水分子是由氢原子和氧原子构成的，故错误；<br />（3）在天然水净化过程中，人们常用活性炭去除异味和色素，这事利用活性炭的吸附性；在二氧化氯（ClO<SUB>2</SUB>）中氧元素的化合价是-2价，故氯元素的化合价为+4价；<br />（4）测定溶液的酸碱度，用pH试纸；<br />答案：<br />（1）D；&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<br />（2）B；<br />（3）吸附；+4；&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<br />&nbsp;（4）B','【分析】（1）根据自来水厂生产自来水时，使用的净水方法有沉淀、过滤、吸附等，而没有煮沸进行分析；<br />（2）依据水由氢、氧元素组成及元素质量比的算法分析，并据水由水分子构成，分子由原子构成分析解答；<br />（3）活性炭具有吸附性；化合物中正负元素化合价的总数和为0；<br />（4）测定溶液的酸碱度，用pH试纸；','填空题',3.00,'c0e65a36e6ac79168d13e9897f2d00c0',9,400,'溶液的酸碱度测定,水的组成,自来水的生产过程与净化方法,碳单质的物理性质及用途,有关元素化合价的计算','冷水江市',2015,'37','2015•冷水江市校级一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842105,'<img  src=\"/tikuimages/9/2016/400/shoutiniao20/2f239730-94d4-11e9-a679-b42e9921e93e_xkb59.png\" style=\"vertical-align:middle;FLOAT:right\" />（2016•青岛二模）某化学兴趣小组对金属的性质做了如下探究，请你参与：<br />某化学兴趣小组为探究铁、铜、锌、银的金属活动性顺序，设计了如图所示的实验（其中金属均已打磨，且其形状、大小相同；所用盐酸的溶质质量分数、用量也相同）．<br />（1）试管②中化学方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，试管②得出结论是：金属活动性<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）比较①③两只试管现象中<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>的不同，可判断出铁的活动性弱于锌．<br />（3）通过上述实验，仍不能判断出四种金属的活动性顺序，若增加一个如图试管④所示的实验可判断出四种金属的活动性顺序，其中金属A是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>、溶液B是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（4）运用“控制变量法”设计对比方案是某化学研究的重要方法．上述实验对比观察①③两只试管判断出两种金属的活动性强弱，控制的实验条件除了金属的体积、溶液的体积、溶质的质量分数以外，还控制了<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>相同．<br />（5）某同学看到试管②中溶液变成蓝色后，为节约金属将铜片取出，此时溶液中可能含有的溶质是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，经过思考他设计如下实验进行验证：<table class=\"edittable\"><TBODY><TR><td width=189>&nbsp;实验操作</TD><td width=189>&nbsp;实验现象</TD><td width=189>&nbsp;结论</TD></TR><TR><td>&nbsp;取2ml溶液于试管中，加入<u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD><td>&nbsp;<br /><u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD><td>&nbsp;溶液中含有这种溶质</TD></TR></TBODY></TABLE>该同学认为试管②中废液直接排放会造成水污染，他向试管中加入<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>进行处理，不仅消除了<br />污染还回收了金属．','','','','','','Cu+2AgNO<SUB>3</SUB>=Cu（NO<SUB>3</SUB>）<SUB>2</SUB>+2Ag$###$Cu＞Ag$###$产生气泡的快慢$###$Fe$###$硫酸铜溶液$###$溶液的种类$###$AgNO<SUB>3</SUB>$###$HCl$###$生成白色沉淀$###$铁粉','【解答】解：（1）把铜放到硝酸银溶液中，在金属活动顺序表中，铜在银的前面，能置换出硝酸银中的银，其反应的化学方程式为：Cu+2AgNO<SUB>3</SUB>=Cu（NO<SUB>3</SUB>）<SUB>2</SUB>+2Ag；由B中实验可以得出：Cu＞Ag；故答案为：Cu+2AgNO<SUB>3</SUB>=Cu（NO<SUB>3</SUB>）<SUB>2</SUB>+2Ag；Cu＞Ag；<br />（2）在金属活动顺序表中，镁在铁的前面，镁的活动性比铁强，所以反应速度镁比铁快，因此产生气泡的速度镁比铁快；故答案为：产生气泡的快慢；<br />（3）从（1）（2）的结论可得到Mg＞Fe，Cu＞Ag；只要再证明Fe＞Cu，就能证明这四种金属的活动顺序性，在硫酸铜溶液中加入铁，就能验证Fe＞Cu；故答案为：Fe；硫酸铜溶液；<br />（4）设计对比方案要控制变量，述实验对比观察①③两只试管判断出两种金属的活动性强弱，控制的实验条件除了金属的体积、溶液的体积、溶质的质量分数以外，还需要控制溶液的种类；故答案为：溶液的种类；<br />（5）把铜放到硝酸银溶液中，在金属活动顺序表中，铜在银的前面，能置换出硝酸银中的银，其反应的化学方程式为：Cu+2AgNO<SUB>3</SUB>=Cu（NO<SUB>3</SUB>）<SUB>2</SUB>+2Ag，因为Cu（NO<SUB>3</SUB>）<SUB>2</SUB>溶液是蓝色的，所以溶液变蓝色，这时可能没有完全反应，故溶液中可能仍有AgNO<SUB>3</SUB>检验银离子用氯离子，生成氯化银白色沉淀；要回收铜和银加入铁；故答案为：AgNO<SUB>3</SUB>；HCl；生成白色沉淀；铁粉．','【分析】（1）从在金属活动顺序表中，铜在银的前面，能置换出硝酸银中的银去分析；<br />（2）从在金属活动顺序表中，镁在铁的前面，镁的活动性比铁强去分析；<br />（3）从在金属活动顺序表中，铁在铜的前面，能置换出硫酸铜中的铜去分析；<br />（4）根据设计对比方案要控制变量进行分析；<br />（5）根据铜在银的前面，能置换出硝酸银中的银，银离子的检验去分析；','书写',3.00,'641adfb9618fe0be7e67efddcbc008f9',9,400,'金属活动性的探究,书写化学方程式、文字表达式、电离方程式','',2016,'37','2016•青岛二模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842112,'在4P+5O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2P<SUB>2</SUB>O<SUB>5</SUB>的反应中，<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>份质量的磷跟<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>份质量的氧气反应，生成<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>份质量的五氧化二磷．','','','','','','124$###$160$###$284','【解答】解：在4P+5O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2P<SUB>2</SUB>O<SUB>5</SUB>的反应中，每124份质量的磷与160份质量的氧气完全反应，生成284份质量的五氧化二磷．<br />故答案为：124；160；284．','【分析】化学方程式可表示反应物和生成物的质量比，利用各物质之间的质量比等于相对分子质量和的比，进行分析解答即可．','填空题',3.00,'5df32ee2b8c23741c7b77a3f2c23356e',9,400,'化学方程式的概念、读法和含义','新疆',2015,'33','2015春•新疆校级期末',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842115,'下列关于空气的说法中，错误的是（　　）','按质量计算，空气中约含氮气78%，氧气21%，其他气体和杂质约占1%','少量有害气体进入空气中，依靠大自然的自净能力，空气仍能保持洁净','空气是一种十分重要的天然资源','空气中的稀有气体一般不跟其他物质反应，曾被称为“惰性气体”','','A','【解答】解：A．空气中各成分的含量多少是按体积计算的，不是按质量计算的，故错误；<br />B．空气有自净能力，能净化空气中的少量有害气体，故正确；<br />C．空气是人类及动植物赖以生存的重要天然资源，故正确；<br />D．空气中的稀有气体化学性质稳定，一般不跟其他物质反应，曾被称“惰性气体”，故正确；<br />故选A．','【分析】空气是人类及动植物赖以生存的重要天然资源；空气有自净能力，但排放到空气中有害气体过多会严重污染空气；空气中各成分的含量是体积分数．空气中的稀有气体化学性质稳定，曾被称“惰性气体”．','选择题',3.00,'bfb2a340ac3d811b782f956438cca574',9,400,'空气的成分及各成分的体积分数,空气对人类生活的重要作用,空气的污染及其危害','大石桥市',2014,'35','2014秋•大石桥市校级期中',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842116,'现有银、铜、铁三种金属，某研究小组的同学为探究银、铜、铁的金属活动性顺序，设计了三个实验：Ⅰ．将铁片浸入稀硫酸中；Ⅱ．将银片浸入稀硫酸中；Ⅲ．将铁片浸入硫酸铜溶液中．<br />（1）实验Ⅰ的现象是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）上述三个实验还不能完全证明三种金属的活动性顺序，请你补充一个实验来达到实验目的：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（写出实验操作和现象）．根据以上探究，三种金属的活动性由强到弱的顺序是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）将银、铜、铁中的两种金属分别放入<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>溶液中，即可验证这三种金属的活动性强弱，反应的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（4）将铜片加热，铜片表面变黑．同学们设计如下实验证明此黑色物质是氧化铜．<br /><table class=\"edittable\"><TBODY><TR><td width=276>实验操作</TD><td width=276>实验现象</TD></TR><TR><td>剪下一片变黑的铜片，放入试管中，加入足量的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD><td>铜片表面的黑色物质全部消失，露出红色的铜，溶液变为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>色．</TD></TR></TBODY></TABLE>','','','','','','有气泡产生，溶液由无色逐渐变为浅绿色$###$将铜丝浸入硝酸银的溶液中，铜的表面有银白色固体析出，溶液由无色变为蓝色$###$铁＞铜＞银$###$硫酸铜$###$Fe+CuSO<SUB>4</SUB>═FeSO<SUB>4</SUB>+Cu$###$稀硫酸$###$蓝','【解答】解：（1）反应物是铁和硫酸，生成物是硫酸亚铁和氢气，实验Ⅰ的现象是有气泡产生，溶液由无色逐渐变为浅绿色；<br />，氢气后面标上上升符号；氢气能够燃烧并发出蓝色火焰，所以用燃着的木条进行检验即可；<br />（2）将铁片浸入稀硫酸中能产生氢气，说明铁排在氢的前面，将银片浸入稀硫酸中，不反应说明银排在氢的后面；将铁片浸入硫酸铜溶液中铁的表面有红色固体析出，说明铁比铜活泼，但铜和银不能比较，所以将铜丝浸入硝酸银的溶液中，铜的表面有银白色固体析出，溶液由无色变为蓝色，说明铜比银活泼；（3）金属与盐反应说明金属排在盐中金属的前面，金属与盐不反应，说明金属排在了盐中金属的后面，所以加入的盐溶液应该位于两种金属之间；由于铜在中间，所以两种金属放到硫酸铜溶液中即可；铁片和蓝色的硫酸铜反应生成浅绿色的硫酸亚铁和红色的铜，反应的化学方程式为：Fe+CuSO<SUB>4</SUB>═FeSO<SUB>4</SUB>+Cu．<br />（4）氧化铜能与硫酸反应生成硫酸铜和水，硫酸铜溶液是蓝色的，所以加入稀硫酸，如果铜片表面的黑色物质全部消失，露出红色的铜，溶液变为蓝色，说明属于氧化铜．<br />故答案为：<br />（1）有气泡产生，溶液由无色逐渐变为浅绿色；<br />（2）将铜丝浸入硝酸银的溶液中，铜的表面有银白色固体析出，溶液由无色变为蓝色；铁＞铜＞银；<br />（3）硫酸铜；Fe+CuSO<SUB>4</SUB>═FeSO<SUB>4</SUB>+Cu．<br />（4）稀硫酸；蓝．','【分析】（1）根据方程式的写法考虑，根据氢气的验证方法来回答；<br />（2）通过三个实验可知铜和银还不能比较活泼性，所以只要设计的实验能比较出二者的活泼性就行；<br />（3）根据金属与盐反应的条件考虑；<br />（4）根据氧化铜能与酸反应生成盐和水考虑．','书写',3.00,'70813cfbcaeb0b8f692eddf2f22d2e6b',9,400,'金属活动性的探究,书写化学方程式、文字表达式、电离方程式','',2016,'37','2016春•赣州校级月考',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842121,'下列物质不是由分子构成的是（　　）','水','氯化钠','二氧化碳','氧气','','B','【解答】解：A、水是由水分子构成的，故选项错误．<br />B、氯化钠是由钠离子和氯离子构成的，故选项正确．<br />C、二氧化碳是由二氧化碳分子构成的，故选项错误．<br />D、氧气属于气态非金属单质，是由氧分子构成的，故选项错误．<br />故选：B．','【分析】根据金属、大多数固态非金属单质、稀有气体等由原子构成；有些物质是由分子构成的，气态的非金属单质和由非金属元素组成的化合物，如氢气、水等；有些物质是由离子构成的，一般是含有金属元素和非金属元素的化合物，如氯化钠，进行分析判断即可．','选择题',3.00,'489f7bea3110e6f1f66e0c27f5d8c544',9,400,'物质的构成和含量分析','',2016,'37','2016•东城区一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842144,'通常情况下二氧化碳不支持燃烧，但镁带却能在二氧化碳中燃烧发出耀眼的光，2Mg+CO<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td style=\"font-size: 90%\">点燃</td></tr><tr><td><div hassize=\"9\"><div style=\"background: url(\'/upimages/images/part/8594L.png\') repeat-x; width:1px; height:5px;float:left;overflow:hidden\" muststretch=\"h\"></div><div style=\"background: url(\'/upimages/images/part/8594.png\') no-repeat; width:9px; height:5px;float:left;overflow:hidden\"></div></div></td></tr></table></span>2MgO+C，关于该反应的下列说法正确的是（　　）','二氧化碳发生氧化反应','二氧化碳是氧化剂','Mg被还原','Mg发生还原反应','','B','【解答】解：A、在2Mg+CO<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td style=\"font-size: 90%\">点燃</td></tr><tr><td><div hassize=\"9\"><div style=\"background: url(\'/upimages/images/part/8594L.png\') repeat-x; width:1px; height:5px;float:left;overflow:hidden\" muststretch=\"h\"></div><div style=\"background: url(\'/upimages/images/part/8594.png\') no-repeat; width:9px; height:5px;float:left;overflow:hidden\"></div></div></td></tr></table></span>2MgO+C中，二氧化碳失去了氧，发生了还原反应，故选项说法错误．<br />B、在2Mg+CO<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td style=\"font-size: 90%\">点燃</td></tr><tr><td><div hassize=\"9\"><div style=\"background: url(\'/upimages/images/part/8594L.png\') repeat-x; width:1px; height:5px;float:left;overflow:hidden\" muststretch=\"h\"></div><div style=\"background: url(\'/upimages/images/part/8594.png\') no-repeat; width:9px; height:5px;float:left;overflow:hidden\"></div></div></td></tr></table></span>2MgO+C中，二氧化碳失去了氧，发生了还原反应，是氧化剂，故选项说法正确．<br />C、在2Mg+CO<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td style=\"font-size: 90%\">点燃</td></tr><tr><td><div hassize=\"9\"><div style=\"background: url(\'/upimages/images/part/8594L.png\') repeat-x; width:1px; height:5px;float:left;overflow:hidden\" muststretch=\"h\"></div><div style=\"background: url(\'/upimages/images/part/8594.png\') no-repeat; width:9px; height:5px;float:left;overflow:hidden\"></div></div></td></tr></table></span>2MgO+C中，镁夺取了二氧化碳中的氧，发生了氧化反应，被氧化，故选项说法错误．<br />D、在2Mg+CO<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td style=\"font-size: 90%\">点燃</td></tr><tr><td><div hassize=\"9\"><div style=\"background: url(\'/upimages/images/part/8594L.png\') repeat-x; width:1px; height:5px;float:left;overflow:hidden\" muststretch=\"h\"></div><div style=\"background: url(\'/upimages/images/part/8594.png\') no-repeat; width:9px; height:5px;float:left;overflow:hidden\"></div></div></td></tr></table></span>2MgO+C中，镁夺取了二氧化碳中的氧，发生了氧化反应，故选项说法错误．<br />故选：B．','【分析】根据在2Mg+CO<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td style=\"font-size: 90%\">点燃</td></tr><tr><td><div hassize=\"9\"><div style=\"background: url(\'/upimages/images/part/8594L.png\') repeat-x; width:1px; height:5px;float:left;overflow:hidden\" muststretch=\"h\"></div><div style=\"background: url(\'/upimages/images/part/8594.png\') no-repeat; width:9px; height:5px;float:left;overflow:hidden\"></div></div></td></tr></table></span>2MgO+C中，镁夺取了二氧化碳中的氧，二氧化碳失去了氧，进行分析判断．','选择题',3.00,'a27c1e65049f2dbff05743eaa822c91b',9,400,'氧化反应,还原反应','',0,'37','',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842158,'化学就在我们身边，请从下列物质中选择适当的字母序号填空：A盐酸　B纯碱　C食盐　D氢氧化铜　E聚氯乙烯塑料等<br />（1）人体胃液中含有的酸是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />（2）候氏制碱法中的碱是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>　　<br />（3）“白色污染”物是指<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>　　<br />（4）生活中常用作调料的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','A$###$B$###$E$###$C','【解答】解：（1）人体胃液中含有的酸是盐酸，故填：A；&nbsp;&nbsp;<br />（2）候氏制碱法中的碱是纯碱碳酸钠，故填：B；&nbsp;&nbsp;<br />（3）白色污染是指废弃塑料造成的污染，故填：E；&nbsp;&nbsp;<br />（4）食盐在生活中常用于调味品，故填：C．','【分析】物质的性质决定物质的用途，根据已有的物质的性质进行分析解答即可．','填空题',3.00,'181553950a03622e1301232d27274f3b',9,400,'酸碱盐的应用,白色污染与防治','',2016,'32','2016•夹江县模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842159,'<img  src=\"/tikuimages/9/0/400/shoutiniao48/2fe1de1e-94d4-11e9-aa2c-b42e9921e93e_xkb30.png\" style=\"vertical-align:middle;FLOAT:right;\" />通过如图所示的反应，回答下列问题<br />①丙物质属于<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填物质分类）<br />②该反应的基本反应类型为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>反应<br />③用分子和原子的观点来解释化学反应的实质：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','单质$###$分解$###$分子的破裂，原子的重新组合','【解答】解：①由物质的微观构成可知，丙物质的分子是由同种的原子构成的，属于单质；<br />②该反应由一种物质生成了两种物质，基本反应类型为分解反应；<br />③由微粒的变化可知，化学反应的实质是：分子的破裂，原子的重新组合．<br />故答为：①单质；②分解；③分子的破裂，原子的重新组合．','【分析】观察反应的微观模型图，根据微粒的构成分析物质的类别，根据反应的特点判断反应的类型，根据微粒的变化分析化学反应的实质．','填空题',3.00,'ff3df9dbdd5cf0f6ac581588ac9e4721',9,400,'单质和化合物的判别,微粒观点及模型图的应用,化学反应的实质,反应类型的判定','',0,'37','',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842172,'以下测量数据合理的是（　　）','用托盘天平称取80kg的食盐','用pH试纸测得某溶液的pH为7.82','用普通温度计测得室温为26.68℃','用10mL量筒取7.5mL水','','D','【解答】解：A、托盘天平用于粗略称量少量药品的质量，称取80kg的食盐不能使用托盘天平，故选项实验数据不合理．<br />B、由于标准比色卡上的数字只有整数，不能用广泛pH试纸测得某溶液的PH为7.82，故选项测量数据不合理．<br />C、普通温度计能准确到0.1℃，用普通温度计无法显示室温为26.68℃，故选项实验数据不合理．<br />D、量筒量程选择的依据有两点：一是保证测量一次，二是量程要与液体的取用量最接近，应用10ml量筒量取7.5ml水，故选项实验数据合理．<br />故选：D．','【分析】A、托盘天平用于粗略称量少量药品的质量，进行分析判断．<br />B、pH试纸上的标准比色卡上的数字只有整数，即使用pH试纸所测得的溶液酸碱度为整数．<br />C、普通温度计能准确到0.1℃．<br />D、量筒量程选择的依据有两点：一是保证测量一次，二是量程要与液体的取用量最接近．','选择题',3.00,'0c349fe3642937998e47a5377f888501',9,400,'实验数据处理或者误差分析的探究,测量容器-量筒,称量器-托盘天平,溶液的酸碱度测定','',2016,'35','2016春•定陶县期中',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842180,'下列物质的洗涤原理相同的是（　　）<br />①用酒精洗涤衣服上的碘②用洗发液洗头发<br />③用稀盐酸洗附着在试剂瓶上的碳酸钙④生活中常用汽油洗油漆．','①②③④','①④','①②④','②③','','B','【解答】解：①用酒精洗去试管中的碘是利用碘能溶解于酒精的性质；②用洗发液洗头发是利用洗发液的乳化作用；③稀盐酸洗附着在试剂瓶上的碳酸钙是利用盐酸与碳酸钙的反应；④汽油洗油漆是利用汽油能溶解油污的性质；<br />故选B','【分析】根据已有的知识进行分析，酒精清洗碘是利用碘溶于酒精的性质；洗发液洗头发是利用洗发液的乳化作用；稀盐酸洗附着在试剂瓶上的碳酸钙是利用盐酸与碳酸钙的反应；汽油能溶解油污．','选择题',3.00,'92119ef7c670a599ef75a85e93cd8c2c',9,400,'溶解现象与溶解原理,乳化现象与乳化作用,酸的化学性质','',2015,'33','2015春•渠县校级期末',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842182,'地壳里含量最多的元素是（　　）','硅','铝','氧','铁','','C','【解答】解：地壳中含量最多的元素为氧，其它元素含量由高到低的顺序依次为硅、铝、铁等；<br />故选C．','【分析】根据地壳中元素分布图，地壳中元素含量由高到低居前四位的元素依次是氧、硅、铝、铁．','选择题',2.00,'6ecfcbf7baa4751e0980ac1c543202c2',9,400,'地壳中元素的分布与含量','',2016,'37','2016•东城区一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842188,'“家”是一个化学小世界，蕴含丰富的化学知识．<br />①以下配方可自制水培植物营养液：Ca（NO<SUB>3</SUB>）<SUB>2</SUB>&nbsp;125克、FeSO<SUB>4</SUB>&nbsp;12克、水1公斤，Ca（NO<SUB>3</SUB>）<SUB>2</SUB>和FeSO<SUB>4</SUB>均属于<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“酸”、“碱”、“盐”），Ca（NO<SUB>3</SUB>）<SUB>2</SUB>属于<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>肥（填“氮”、“磷”、“钾”）．<br />②活性炭可作家用冰箱除味剂，是因为它具有<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>性．<br />③生石灰是常用的食品干燥剂，请用化学方程式表示其干燥原理<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />④碳酸氢钠（NaHCO<SUB>3</SUB>）可用于焙制糕点的膨松剂．碳酸氢钠由<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>种元素组成，1mol碳酸氢钠约含有<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>个氧原子（用科学计数法表示）．<br />⑤今年我们遭遇了特大寒潮袭击，造成很多家庭水管爆裂，这是由于水具有<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>的性质．','','','','','','盐$###$氮$###$吸附$###$Ca0+H<SUB>2</SUB>O═Ca（0H）<SUB>2</SUB>$###$4$###$1.806×10<SUP>24</SUP>$###$变成固态时体积膨胀','【解答】解：①Ca（NO<SUB>3</SUB>）<SUB>2</SUB>和FeSO<SUB>4</SUB>均由金属离子和酸根离子构成的化合物，属于盐；Ca（NO<SUB>3</SUB>）<SUB>2</SUB>中含有农作物需求量较大的氮元素，属于氮肥，故填：盐；氮；<br />②活性炭能除异味是因为它具有吸附性，能吸附异味和色素．故填：吸附；<br />③氧化钙和水反应生成氢氧化钙，化学方程式为Ca0+H<SUB>2</SUB>O═Ca（0H）<SUB>2</SUB>；故填：Ca0+H<SUB>2</SUB>O═Ca（0H）<SUB>2</SUB>；<br />④碳酸氢钠是由钠元素、氢元素、碳元素和氧元素四种元素组成的，因1个NaHCO<SUB>3</SUB>分子中含有3个O原子，则1molNaHCO<SUB>3</SUB>中含3molO原子，O原子的个数为3×N<SUB>A</SUB>=3×6.02×10<SUP>23</SUP>=1.806×10<SUP>24</SUP>．故填：4；1.806×10<SUP>24</SUP>；<br />⑤水结冰时，质量不变，密度变小，所以体积变大，由于体积膨胀，所以水管爆裂；故填：变成固态时体积膨胀．','【分析】①根据酸碱盐的概念以及植物生长所需营养素来分析；<br />②根据活性炭的吸附性来分析；<br />③根据氧化钙和水反应生成氢氧化钙进行解答；<br />④根据物质的组成、构成来分析分子中的原子的物质的量，再由N=n×N<SUB>A</SUB>来计算微粒的个数；<br />⑤根据水的性质来分析．','书写',3.00,'f68c52808d38bdaad9910506cd1a6313',9,400,'水的性质和应用,生石灰的性质与用途,常见化肥的种类和作用,常见的氧化物、酸、碱和盐的判别,化学式的书写及意义','',2016,'37','2016•奉贤区二模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842199,'日常生活中，区别下列各组物质所用的方法，错误的是（　　）','氮气和氧气--将燃着的木条分别伸入集气瓶内，观察火焰是否熄灭','苏打和小苏打--分别与足量的白醋反应，观察是否有气泡产生','氮肥和磷肥--观察颜色','热塑性塑料和热固性塑料--加热，观察外形变化','','B','【解答】解：A、取样品，将燃着的木条分别伸入集气瓶内，木条熄灭的是氮气，木条燃烧更旺的是氧气，现象不同，可以鉴别，故A正确；<br />B、苏打和小苏打分别与足量的白醋反应，都会生成二氧化碳气体，现象相同，不能鉴别，故B错误；<br />C、取样品，观察颜色，白色晶体的是氮肥，灰白色粉末的是磷肥，现象不同，可以鉴别，故C正确；<br />D、通过加热观察外形来区别热塑性塑料和热固性塑料：由于热塑性塑料在受热时外形明显变化，而热固性塑料无明显变化，现象不同，可以鉴别，故D正确．<br />故选：B．','【分析】A、根据氮气和氧气的化学性质进行分析；<br />B、根据碳酸盐和酸反应会生成二氧化碳气体进行分析；<br />C、根据氮肥是白色晶体，磷肥是灰白色粉末进行分析；<br />D、根据热塑性塑料和热固性塑料加热后的不同现象进行分析．','选择题',3.00,'764ed5bc77c01d832979fbb784e07c17',9,400,'常见气体的检验与除杂方法,化肥的简易鉴别,酸、碱、盐的鉴别,物质的鉴别、推断,塑料及其应用','',2016,'37','2016•新化县二模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842201,'<img  src=\"/tikuimages/9/2016/400/shoutiniao87/307396cf-94d4-11e9-808c-b42e9921e93e_xkb54.png\" style=\"vertical-align:middle;FLOAT:right\" />（2016•青岛模拟）硫酸是常见的酸，也是常用的化工原料．由于浓度不同，浓硫酸与稀硫酸在性质上存在较大差异，如浓硫酸具有脱水性、吸水性和强腐蚀性等．某课外兴趣小组为了探究浓硫酸的某些特性做了如下一些实验．请结合有关实验，按要求回答下列问题：<br />（1）在洗气瓶中盛放浓硫酸，除去氢气或氧气中的水分，是利用其<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>性．<br />（2）探究浓硫酸的吸水性．兴趣小组把98%的硫酸10mL和63.3%的硫酸（用10mL98%的硫酸与10mL水配成）约20mL分别放入两个相同的大表面皿中，称量、观察、记录、分析．根据室温环境下实验的数据绘成的曲线如图：<br />①稀释浓硫酸时，应将<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>缓缓地注入盛有<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>的烧杯中并不断搅拌；<br />②由如图曲线你能得到的实验结论有<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填序号）．<br />A．只有浓硫酸具有吸水性<br />B．浓硫酸吸水能力比稀硫酸强（吸水速度快、吸水量大）<br />C．开始两者吸水速度差异比较大，随着时间的推移吸水速度越来越接近，吸水能力越来越强<br />D．硫酸浓度越稀，吸水性越弱，到一定浓度后就失去吸水性．','','','','','','吸水$###$浓硫酸$###$水$###$BD','【解答】解：（1）浓硫酸具有的脱水性与吸水性，除去氢气或氧气中的水分，是利用其吸水性，起到干燥剂的作用；<br />（2）①浓硫酸的稀释时应酸入水并不断搅拌使溶解时产生的热量尽快散失；<br />②从曲线观察两种不同的酸的吸水情况的变化可知：浓硫酸和一定浓度以上的稀硫酸都具有吸水性，故A错误；<br />浓硫酸吸水能力比稀硫酸强（吸水速度快、吸水量大），故B正确；<br />开始两者吸水速度差异比较大，随着时间的推移吸水速度越来越接近，吸水能力越来越弱，故C错误；<br />硫酸浓度越稀吸水性越弱，到一定浓度后就失去吸水性，故D正确．<br />故答案为：（1）吸水；<br />（2）①浓硫酸，水；&nbsp;②B、D．','【分析】（1）依据浓硫酸具有的脱水性与吸水性进行分析判断；<br />（2）①根据浓硫酸的稀释过程的操作要领分析解答；<br />②从曲线观察两种不同的酸的吸水情况的变化趋势分析解答','填空题',3.00,'11bfe2dbd64da9de1bcdca21e910f23f',9,400,'探究酸碱的主要性质,浓硫酸的性质及浓硫酸的稀释','',2016,'32','2016•青岛模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842203,'多角度认识溶液，有助于我们更好地了解其在生产生活中的重要作用．<br />（1）溶液的形成<br />①将适量调味品加入水中，充分搅拌后不能形成溶液的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填标号）．<br />A．食盐&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; B．蔗糖&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; C．花生油<br />②将生石灰和足量的水充分搅拌后静置，取上层溶液即为澄清石灰水．写出生石灰与水反应的化学方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）溶液的性质<br />①关于溶液的说法正确的有<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填标号）．<br />A．盐酸是纯净物&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; B．溶液是均一、稳定的<br />C．溶液的酸碱度可用石蕊溶液测定&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; D．饱和溶液不一定是浓溶液<br />②常温下，将某固态物质A溶于水，形成饱和溶液（如图1），进行以下实验．回答下列问题：<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao38/307b5f00-94d4-11e9-90f3-b42e9921e93e_xkb7.png\" style=\"vertical-align:middle\" /><br />a．该物质的溶解度随温度升高而<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“增大”或“减小”）<br />b．往溶液Ⅱ中加入少量高锰酸钾完全溶解，溶液呈紫红色．据此现象，写出对“饱和溶液”概念的理解<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />③常温下，该物质在不同质量的水中溶解达到饱和状态，溶质的质量与水的质量关系如图2所示．<br />a．常温时，该物质的溶解度是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>g．<br />b．D点时，该溶液的溶质质量分数为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（结果精确到0.1%）<br />c．将A点状态的溶液变化到C点状态的操作步骤为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />d．图中A、B、C三个点表示溶液中溶质质量分数从大到小的关系是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）溶液的用途<br />生活中溶液由许多用途，例如：碘酒能用于杀菌消毒；请再举一个生活中溶液应用的实例：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','C$###$CaO+H<SUB>2</SUB>O═Ca（OH）<SUB>2</SUB>$###$BD$###$减小$###$一定温度下，某物质的饱和溶液对于其它物质不一定是饱和的$###$40$###$28.6%$###$常温下，往A状态的溶液中加入20g溶质和20g水，使之充分溶解$###$C、A、B$###$生理盐水给病人补充水分','【解答】解：（1）①食盐和蔗糖都能溶于水，形成溶液，而花生油不溶于水，不能形成溶液，故填：C；<br />②生石灰是氧化钙的俗称，与水反应生成氢氧化钙，故填：CaO+H<SUB>2</SUB>O═Ca（OH）<SUB>2</SUB><br />（2）①A．盐酸是氯化氢气体和水的混合物，故错误；<br />B．溶液具有均一性和稳定性，故正确；<br />C．石蕊试液只能测定出溶液的酸碱性，无法测定酸碱度，测定溶液的酸碱度可用pH试纸，故错误；<br />D．饱和溶液不一定是浓溶液，如饱和的氢氧化钙溶液就是稀溶液，故正确．<br />故填：BD；<br />②a．由图象可知，温度降低烧杯中未溶解的固体溶质减少了，说明该物质的溶解度随着温度的升高而减小；故填：减小；<br />b．该物质的饱和溶液中，不能继续溶解该物质，但是可以继续溶解其他物质，故填：一定温度下，某物质的饱和溶液对于其它物质不一定是饱和的；<br />③a．由溶解度的概念可知，在常温下该物质的溶解度为40g，故填：40；<br />b．D点表示的溶液为该温度的下的饱和溶液，溶质的质量分数为：<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">40g</td></tr><tr><td>100g+40g</td></tr></table></span>×100%≈28.6%；故填：28.6%；<br />c．A到B是向溶液中加入了20g的水，由B到C是向溶液中加入了20g溶质并使之完全溶解，故填：常温下，往A状态的溶液中加入20g溶质和20g水，使之充分溶解．<br />d．A和B中，溶质的质量相等，B中的溶剂多，所以浓度B＜A；A是20g溶质溶解在80g水中，所得溶液液的溶质质量分数为：<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">20g</td></tr><tr><td>20g+80g</td></tr></table></span>×100%=20%；C点表示的溶液为该温度的下的饱和溶液，溶质的质量分数为：<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">40g</td></tr><tr><td>40g+100g</td></tr></table></span>×100%≈28.6%；故填：C、A、B；<br />（3）溶液在生产和生活中的应用非常广泛，如生理盐水给病人补充水分．故填：生理盐水给病人补充水分．<br />答案：<br />（1）①C；②CaO+H<SUB>2</SUB>O═Ca（OH）<SUB>2</SUB><br />（2）①BD；②a．减小；b．一定温度下，某物质的饱和溶液对于其它物质不一定是饱和的；<br />③a.40；b.28.6%；c．常温下，往A状态的溶液中加入20g溶质和20g水，使之充分溶解；d．C、A、B；<br />（3）生理盐水给病人补充水分','【分析】（1）根据溶液的形成以及化学方程式的书写方法来分析；<br />（2）根据溶液的组成、特征、酸碱度的测定方法来分析；<br />根据溶解度曲线以及图象信息来分析；<br />（3）根据溶液的用途来分析．','书写',3.00,'6543c1e2b67b97c256c13efb321227fc',9,400,'溶液的酸碱度测定,溶液的概念、组成及其特点,饱和溶液和不饱和溶液,浓溶液、稀溶液跟饱和溶液、不饱和溶液的关系,固体溶解度的影响因素,生石灰的性质与用途,书写化学方程式、文字表达式、电离方程式','长乐市',2016,'35','2016春•长乐市期中',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842204,'下列有关CO和CO<SUB>2</SUB>的比较中，正确的是（　　）','分子构成相同','组成元素种类相同','物理性质相同','化学性质相同','','B','【解答】解：A、CO和CO<SUB>2</SUB>的分子构成不相同，故A说法不正确；<br />B、CO和CO<SUB>2</SUB>的组成元素种类相同，故B说法正确；<br />C、CO和CO<SUB>2</SUB>的物理性质不相同，故C说法不正确；<br />D、CO和CO<SUB>2</SUB>的化学性质不相同，故D说法正确．<br />故选B．','【分析】根据CO和CO<SUB>2</SUB>的构成和性质分析判断有关的说法．','选择题',3.00,'4f3d2610e60f0cbef188559396293e8d',9,400,'二氧化碳的物理性质,二氧化碳的化学性质,一氧化碳的物理性质,一氧化碳的化学性质,分子、原子、离子、元素与物质之间的关系','',2016,'32','2016•长春模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842236,'说明硫在氧气中燃烧是化学变化的依据是（　　）','发出蓝紫色火焰','放出大量的热','硫粉减少','产生一种有刺激性气味的气体','','D','【解答】解：根据生成其他物质的变化叫化学变化，它的本质特征是有其他（新的）物质生成，硫在氧气中燃烧是否发生了化学变化的根本依据是产生一种有刺激性气味的气体；其余均是伴随的现象，辅助判断，但不是根本依据．<br />故选：D．','【分析】根据生成其他物质的变化叫化学变化，又叫化学反应，它的本质特征是有其他（新的）物质生成，进行分析解答本题．','选择题',2.00,'6d59bed5faaa338e7c973e09cbebc90c',9,400,'化学变化的基本特征','',2013,'37','2013•泾川县校级二模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842239,'有化合价升降的反应是氧化还原反应，下列属于氧化还原反应的是（　　）','H<SUB>2</SUB>SO<SUB>4</SUB>+2NaOH═Na<SUB>2</SUB>SO<SUB>4</SUB>+2H<SUB>2</SUB>O','Na<SUB>2</SUB>CO<SUB>3</SUB>+Ca（OH）<SUB>2</SUB>═CaCO<SUB>3</SUB>+2NaOH','2HCl+CaCO<SUB>3</SUB>═CaCl<SUB>2</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>','Fe+2HCl═FeCl<SUB>2</SUB>+H<SUB>2</SUB>↑','','D','【解答】解：A、H<SUB>2</SUB>SO<SUB>4</SUB>+2NaOH═Na<SUB>2</SUB>SO<SUB>4</SUB>+2H<SUB>2</SUB>O中反应物和生成物中都没有元素化合价发生变化，不属于氧化还原反应，故A错误；<br />B、NaCO<SUB>3</SUB>+Ca（OH）<SUB>2</SUB>═CaCO<SUB>3</SUB>+2NaOH中反应物和生成物中都没有元素化合价发生变化，不属于氧化还原反应，故B错误；<br />C、2HCl+CaCO<SUB>3</SUB>═CaCl<SUB>2</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑中反应物和生成物中都没有元素化合价发生变化，不属于氧化还原反应，故C错误；<br />D、Fe+2HCl═FeCl<SUB>2</SUB>+H<SUB>2</SUB>↑中铁元素的化合价由0价变为+2价，氢元素的化合价由+1价变为0价，属于氧化还原反应，故D正确．<br />故选D．','【分析】在单质中，元素化合价是0；在化合物中，元素化合价的代数和是0，可以根据这一原则求出化合物中各种元素的化合价．然后根据元素的化合价是否有升降进行判断是否是氧化还原反应．','选择题',3.00,'8f17672d2473ca1512fb5dadda99ff1b',9,400,'氧化反应,还原反应','',2016,'37','2016春•阜南县校级月考',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842243,'下列叙述错误的是（　　）','用活性炭可将硬水软化','通过相互刻画的方法，发现黄铜片的硬度比铜片大','用水基型灭火器扑灭棉布引起的失火','将尿素和熟石灰一起研磨，没有刺激性气味的气体产生','','A','【解答】解：<br />A、活性炭吸附作用，除去有臭味的物质和一些可溶性的杂质，不能软化，故A错误；<br />B、合金与纯金属比较硬度更大，故B正确；<br />C、水可以用于棉布的灭火，不能用于油电的灭火，故C正确；<br />D、尿素中不为含铵根离子，不能用碱检验，故D正确．<br />故答案为：A','【分析】A、活性炭吸附作用，除去有臭味的物质和一些可溶性的杂质，不能软化；<br />B、合金与纯金属比较硬度更大，进行解答；<br />C、水可以用于棉布的灭火；<br />D、尿素中不含铵根离子，不能用碱检验．','选择题',3.00,'3d65418a935e80c702f01e9d5c84172a',9,400,'硬水与软水,合金与合金的性质,铵态氮肥的检验,几种常用的灭火器','',2016,'37','2016•南京一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842246,'工业炼锰的反应为：8Al+3Mn<SUB>3</SUB>O<SUB>4</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>9Mn+4Al<SUB>2</SUB>O<SUB>3</SUB>．对该反应说法错误的是（　　）','属于置换反应','铝发生了还原反应','可用于工业生产，说明锰的市场价值高于铝','反应前后，锰元素的化合价降低','','B','【解答】解：A、该反应是一种单质和一种化合物反应生成另一种单质和另一种化合物的反应，属于置换反应，故选项说法正确．<br />B、铝夺取了四氧化三锰中的氧，发生了氧化反应，故选项说法错误．<br />C、可用于工业生产，说明锰的市场价值高于铝，故选项说法正确．<br />D、四氧化三锰中锰元素的化合价不为0，锰属于单质，锰元素的化合价为0，反应前后，锰元素的化合价降低，故选项说法正确．<br />故选：B．','【分析】A、置换反应是一种单质和一种化合物反应生成另一种单质和另一种化合物的反应．<br />B、根据铝夺取了四氧化三锰中的氧，进行分析判断．<br />C、根据该反应可用于工业生产，进行分析判断．<br />D、根据单质中元素的化合价为0，进行分析判断．','选择题',3.00,'ff6ad7d41a2003aaa222b41e839113b8',9,400,'常见金属的冶炼方法,化合价规律和原则,反应类型的判定','',2016,'37','2016•重庆校级一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842272,'<img  src=\"/tikuimages/9/2016/400/shoutiniao15/315e1de1-94d4-11e9-899b-b42e9921e93e_xkb46.png\" style=\"vertical-align:middle;FLOAT:right\" />（2016•寿光市模拟）1926年，我国著名化学家侯德榜先生创立了侯氏制碱法，促进了我国民族工业发展和世界制碱技术的进步，其生产过程由下列反应：<br />①NaCl+NH<SUB>3</SUB>+CO<SUB>2</SUB>+H<SUB>2</SUB>O═NaHCO<SUB>3</SUB>+NH<SUB>4</SUB>Cl<br />②2NaHCO<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;加热&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>Na<SUB>2</SUB>CO<SUB>3</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑<br />（1）侯氏制碱法所制的“碱”是指<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）工业生产过程中，铵盐水吸收二氧化碳后生成碳酸氢钠和氯化铵，在常温下，两者首先从溶液中结晶析出的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>晶体（填物质名称）．<br />（3）现有Na<SUB>2</SUB>CO<SUB>3</SUB>和NaCl的混合样品22.3g，将其放入干净的烧杯中，加一定质量的水使其完全溶解，向所得溶液中逐渐加入溶质质量分数为7.3%的稀盐酸烧杯中溶液的质量与滴入稀盐酸的质量关系曲线如图所示，试回答下列问题：<br />①当滴入上述稀盐酸至图中B点时，烧杯中溶液里的溶质为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（写化学式）．<br />②在Na<SUB>2</SUB>CO<SUB>3</SUB>和NaCl的混合物样品中，含Na<SUB>2</SUB>CO<SUB>3</SUB>的质量为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>g．<br />③当滴入上述稀盐酸至图中A点时，试通过计算，求此温度时所得不饱和溶液中溶质的质量（计算结果精确至0.1g）','','','','','','碳酸钠$###$碳酸氢钠$###$NaCl、HCl$###$10.6','【解答】解：（1）侯氏制碱法所制的“碱”是指碳酸钠；<br />（2）工业生产过程中，铵盐水吸收二氧化碳后生成碳酸氢钠和氯化铵．在常温下，由于碳酸氢钠的溶解度较小，两者首先从溶液中结晶析出的是碳酸氢钠晶体；<br />（3）①由图象可知，当滴入上述稀盐酸至图中B点时，加入的稀盐酸的量是过量的，烧杯中溶液里的溶质为NaCl、HCl．<br />②由图象可知，当滴入上述稀盐酸至图中A点时，加入的稀盐酸签好与碳酸钠反应．<br />设样品中，含Na<SUB>2</SUB>CO<SUB>3</SUB>的质量为x，生成的氯化钠的质量为y．<br />Na<SUB>2</SUB>CO<SUB>3</SUB>+2HCl═2NaCl+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑<br />106&nbsp;&nbsp;&nbsp;&nbsp; 73&nbsp;&nbsp;&nbsp; 117<br />x&nbsp;&nbsp; 100g×7.3% y<br /><span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">106</td></tr><tr><td>x</td></tr></table>=<table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">73</td></tr><tr><td>100g×7.3%</td></tr></table>=<table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">117</td></tr><tr><td>y</td></tr></table></span><br />解得：x=10.6g&nbsp;&nbsp; y=11.7g<br />③当滴入上述稀盐酸至图中A点时，所得不饱和溶液中溶质的质量为：11.7g+（22.3g-10.6g）=23.4g．<br />故答为：（1）碳酸钠；（2）碳酸氢钠；（3）①NaCl、HCl；②10.6；③所得不饱和溶液中溶质的质量为23.4g．','【分析】（1）侯氏制碱法所制的“碱”是碳酸钠；<br />（2）根据碳酸氢钠的溶解度较小分析回答；<br />（3）根据图象溶液的总质量与加入的稀盐酸的关系分析碳酸钠与稀盐酸的反应，当滴加稀盐酸溶液至图中A点时氯化钠与稀盐酸恰好反应，可以计算溶质的质量；当在B点时，稀盐酸是过量的，分析溶液中的溶质．','填空题',3.00,'f72f713c9474e34fea90fbd9a8b45b4a',9,400,'纯碱的制取,根据化学反应方程式的计算','寿光市',2016,'32','2016•寿光市模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842307,'汽车尾气中主要含有NO、CO等气体，为减轻污染给车安装“三效催化净化器”，反应为：2CO+2NO<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td style=\"font-size: 90%\">催化剂</td></tr><tr><td><div hassize=\"9\"><div style=\"background: url(\'/upimages/images/part/8594L.png\') repeat-x; width:1px; height:5px;float:left;overflow:hidden\" muststretch=\"h\"></div><div style=\"background: url(\'/upimages/images/part/8594.png\') no-repeat; width:9px; height:5px;float:left;overflow:hidden\"></div></div></td></tr></table></span>&nbsp;N<SUB>2</SUB>+2CO<SUB>2</SUB>，其中发生还原反应的是（　　）','NO','CO','N<SUB>2</SUB>','CO<SUB>2</SUB>','','A','【解答】解：在氧化还原反应中，失去氧的物质是氧化剂，得到氧的物质是还原剂，在反应2CO+2NO<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td style=\"font-size: 90%\">催化剂</td></tr><tr><td><div hassize=\"9\"><div style=\"background: url(\'/upimages/images/part/8594L.png\') repeat-x; width:1px; height:5px;float:left;overflow:hidden\" muststretch=\"h\"></div><div style=\"background: url(\'/upimages/images/part/8594.png\') no-repeat; width:9px; height:5px;float:left;overflow:hidden\"></div></div></td></tr></table></span>&nbsp;N<SUB>2</SUB>+2CO<SUB>2</SUB>中，NO失去氧变成N<SUB>2</SUB>是氧化剂，发生还原反应；而CO得到氧变成CO<SUB>2</SUB>是还原剂，发生氧化反应．<br />故选：A．','【分析】根据在反应中，一氧化碳与氧发生了氧化反应，作还原剂；一氧化氮失去了氧，发生了还原反应，作氧化剂进行解答．','选择题',3.00,'f9b4c26f23a83937227057f784d857fc',9,400,'还原反应','',2016,'37','2016•黄浦区三模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842311,'下列物质的用途和性质对应错误的是（　　）','浓硫酸做干燥剂：浓硫酸具有吸水性','氧气可用于炼钢：氧气具有可燃性','二氧化碳可用于灭火：二氧化碳不支持燃烧','氢氧化铝可用于治疗胃酸过多症：氢氧化铝具有碱性可与胃酸发生中和反应','','B','【解答】解：<br />A、浓硫酸具有吸水性，可以做干燥剂，故正确；<br />B、氧气能够支持燃烧，可用于炼钢，氧气是助燃剂，没有可燃性，故错误；<br />C、二氧化碳不支持燃烧、不燃烧，密度比空气的大，可用于灭火，故正确；<br />D、氢氧化铝具有碱性可与胃酸发生中和反应，可用于治疗胃酸过多，故正确．<br />答案：B','【分析】A、根据浓硫酸具有吸水性解答；<br />B、根据氧气能够支持燃烧解答；<br />C、根据二氧化碳不支持燃烧、不燃烧，密度比空气的大解答；<br />D、根据氢氧化铝与胃液中的盐酸生成氯化铝和水解答．','选择题',3.00,'ba52039d3aeab3e5c42482814b4d9bfc',9,400,'氧气的用途,二氧化碳的用途,酸碱盐的应用','',2016,'35','2016春•临沂期中',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842366,'利用能源、节约能源、保护环境是我们大家共同关注的问题．<img  src=\"/tikuimages/9/2016/400/shoutiniao68/32ac48c0-94d4-11e9-8652-b42e9921e93e_xkb47.png\" style=\"vertical-align:middle\" /><br />（1）充分燃烧1000g天然气和煤所产生的CO<SUB>2</SUB>和SO<SUB>2</SUB>气体的质量如上图所示，则<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>&nbsp;燃烧对环境影响较小；煤中硫元素的质量分数为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>&nbsp;&nbsp;（假设硫元素完全转化为SO<SUB>2</SUB>）．<br />（2）SO<SUB>2</SUB>的大量排放造成的环境问题有<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（写出一例）；在煤燃烧后将烟气通入吸收塔并“喷淋”石灰水进行“脱硫”可减少SO<SUB>2</SUB>的排放，石灰水需要“喷淋”的目的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，石灰水与SO<SUB>2</SUB>反应的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）目前，在汽车尾气系统中安装催化转化器可将污染物CO和NO转化，生成两种属于空气成份的气体．请写出该反应的化学方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（4）科学家采取“组合转化”技术，可将CO<SUB>2</SUB>和H<SUB>2</SUB>在一定条件下转化为乙烯（C<SUB>2</SUB>H<SUB>4</SUB>），乙烯可进一步合成为聚乙烯，聚乙烯属于<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填序号）<br />A．金属材料&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;B．无机非金属材料&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;C．合成材料&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;D．复合材料．','','','','','','天然气$###$5g$###$酸雨$###$二氧化硫被充分吸收$###$SO<SUB>2 </SUB>+Ca（OH）<SUB>2 </SUB>═CaSO<SUB>3</SUB>↓+H<SUB>2</SUB>O$###$2CO+2NO<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;催化剂&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2CO<SUB>2</SUB>+N<SUB>2</SUB>$###$C','【解答】解：<br />（1）两种燃料燃烧对环境影响较小的是天然气；由图可知1000g煤产生10g二氧化硫，其中硫元素的质量为10g×<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">32</td></tr><tr><td>64</td></tr></table></span>×100%=5g；<br />（2）SO<SUB>2</SUB>的大量排放能形成酸雨；在煤燃烧后将烟气通入吸收塔并“喷淋”石灰水进行“脱硫”可减少SO<SUB>2</SUB>的排放，石灰水需要“喷淋”的目的是二氧化硫被充分吸收；根据题意，SO<SUB>2</SUB>、石灰水、氧气反应生成硫酸钙和水，反应的化学方程式为SO<SUB>2 </SUB>+Ca（OH）<SUB>2 </SUB>═CaSO<SUB>3</SUB>↓+H<SUB>2</SUB>O；<br />（3）在汽车尾气系统中安装催化转化器可将污染物CO和NO转化，生成两种属于空气成份的气体．请写出该反应的化学方程式为：2CO+2NO<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;催化剂&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2CO<SUB>2</SUB>+N<SUB>2</SUB><br />（4）聚乙烯是塑料的主要成分，属于有机合成材料．<br />故答案为：<br />（1）天然气；5g；<br />（2）酸雨；二氧化硫被充分吸收；SO<SUB>2 </SUB>+Ca（OH）<SUB>2 </SUB>═CaSO<SUB>3</SUB>↓+H<SUB>2</SUB>O；<br />（3）2CO+2NO<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;催化剂&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2CO<SUB>2</SUB>+N<SUB>2</SUB>．<br />（4）C','【分析】（1）根据天然气和煤燃烧后的产物来分析；<br />（2）根据题意，产生的SO<SUB>2</SUB>用石灰水淋洗，外加氧气的作用，使之充分反应生成硫酸钙和水，写出反应的化学方程式即可；<br />（3）根据反应原理写出反应的化学方程式解答；<br />（4）根据聚乙烯是塑料的主要成分，属于有机合成材料解答．','书写',3.00,'52762d05ebc2f2117fb37b5ad537eda7',9,400,'酸雨的产生、危害及防治,书写化学方程式、文字表达式、电离方程式,常用燃料的使用与其对环境的影响','张家港市',2016,'32','2016•张家港市模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842375,'某化学兴趣小组的学生，发现金属R不在初中学到的金属活动性顺序表中，该小组为了了解R与常见金属铝、铜<br />的金属活动性顺序，进行如下探究活动；<br />【做出猜想】他们考虑到铝的活动性比铜强，对三种金属的活动性顺序作出如下猜想<br />猜想一：Al＞Cu＞R　　　　　　　猜想二：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>　　　　猜想三：R＞Al＞Cu<br />【查阅资料】R是一种银白色的金属，在R的盐溶液中只有+2价，R的盐溶液呈蓝色；常温下铝的表面会形成一层致密的氧化膜，硫酸铝、氯化铝、硝酸铝的溶液均为无色．<br />【实验探究】为了探究哪一种猜想成立，甲、乙、丙三位同学分别针对猜想一、猜想二、猜想三设计实验方案并展开实验探究．<br /><table class=\"edittable\"><TBODY><TR><td width=54>猜想</TD><td width=167>主要操作</TD><td width=198>主要现象</TD><td width=179>实验结论</TD></TR><TR><td>猜想一</TD><td>打磨R丝，并将其插入到硫酸铜溶液中</TD><td>R丝表面覆盖了一层红色的物质</TD><td>猜想一<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“成立”或“不成立”）</TD></TR><TR><td>猜想二</TD><td>打磨粗细相同的R丝、铝丝、铜丝，分别将它们插入到体积相同、溶质质量分数也相同的稀硫酸中</TD><td>①R丝表面产生气泡缓慢，溶液由无色逐渐变成蓝色<br />②铝丝表面产生气泡较快<br />③铜丝表面没有气泡产生</TD><td>猜想二成立．<br />R与稀硫酸反应的化学方程式为<br /><u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD></TR><TR><td>猜想三</TD><td><u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD><td>无明显现象</TD><td>猜想三成立</TD></TR></TBODY></TABLE>【交流反思】<br />究竟哪种猜想成立？三位同学经过讨论交流，发现丙同学在实验前没有打磨铝丝，这可能会导致实验结论错误．于是他们先打磨铝丝，再将其插入到丙同学做实验用过的硫酸R　溶液中，一段时间后，溶液的颜色<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，进而确认猜想三不成立，猜想二成立．<br />【归纳总结】实验探究后，他们总结得出：通过金属与<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>反应或金属与<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>可以比较金属活动性强弱．','','','','','','Al＞R＞Cu$###$不成立$###$R+H<SUB>2</SUB>SO<SUB>4</SUB>═RSO<SUB>4</SUB>+H<SUB>2</SUB>↑$###$把没有打磨的铝丝伸入RSO<SUB>4</SUB>溶液中$###$变浅，或褪去$###$酸$###$盐溶液反应','【解答】解：【做出猜想】<br />铝比铜活泼，铜可能比R活泼，铝可能比R活泼，R可能比铜活泼，R可能比铝活泼，因此可作以下猜想：<br />猜想一：Al＞Cu＞R　　　　　　　猜想二：Al＞R＞Cu　　　　猜想三：R＞Al＞Cu<br />故填：Al＞R＞Cu．<br />【实验探究】<br />实验探究如下表所示：<br /><table class=\"edittable\"><TBODY><TR><td width=54>猜想</TD><td width=167>主要操作</TD><td width=198>主要现象</TD><td width=179>实验结论</TD></TR><TR><td>猜想一</TD><td>打磨R丝，并将其插入到硫酸铜溶液中</TD><td>R丝表面覆盖了一层红色的物质</TD><td>猜想一 不成立</TD></TR><TR><td>猜想二</TD><td>打磨粗细相同的R丝、铝丝、铜丝，分别将它们插入到体积相同、溶质质量分数也相同的稀硫酸中</TD><td>①R丝表面产生气泡缓慢，溶液由无色逐渐变成蓝色<br />②铝丝表面产生气泡较快<br />③铜丝表面没有气泡产生</TD><td>猜想二成立．<br />R与稀硫酸反应的化学方程式为<br />R+H<SUB>2</SUB>SO<SUB>4</SUB>═RSO<SUB>4</SUB>+H<SUB>2</SUB>↑</TD></TR><TR><td>猜想三</TD><td>把没有打磨的铝丝伸入RSO<SUB>4</SUB>溶液中</TD><td>无明显现象</TD><td>猜想三成立</TD></TR></TBODY></TABLE>【交流反思】<br />他们先打磨铝丝，再将其插入到丙同学做实验用过的硫酸R　溶液中，一段时间后，溶液的颜色变浅，或褪去，进而确认猜想三不成立，猜想二成立．<br />故填：变浅，或褪去．<br />【归纳总结】实验探究后，他们总结得出：通过金属与酸反应或金属与盐溶液反应可以比较金属活动性强弱．<br />故填：酸；盐溶液反应．','【分析】【做出猜想】<br />铝比铜活泼，铜可能比R活泼，铝可能比R活泼，R可能比铜活泼，R可能比铝活泼；<br />【实验探究】<br />根据实验现象可以判断实验结论，根据实验结论可以判断实验现象；<br />【交流反思】<br />铝在通常情况下能够能被空气中的氧气氧化；<br />【归纳总结】<br />通过金属与酸反应或金属与盐溶液反应可以比较金属活动性强弱．','书写',3.00,'b8a83f20a0dbb0b9b107fed2d64f6666',9,400,'金属活动性的探究,书写化学方程式、文字表达式、电离方程式','',2016,'35','2016春•薛城区期中',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842378,'水和溶液在生命活动和生活中起着十分重要的作用．<br />（1）自然界的水常常遭到人为污染，污染物中氮和磷含量过高会造成水中藻类过度繁殖，使水质恶化，这里的“氮和磷”是指<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填字母）．<br />A．分子&nbsp; B．原子&nbsp; C．元素&nbsp; D．单质<br />（2）水在化学实验中的作用不可忽视．如图1中五个实验分别用到水．下列说法中正确的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填序号）．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao70/32d50670-94d4-11e9-b2c3-b42e9921e93e_xkb40.png\" style=\"vertical-align:middle\" /><br />①实验A中水做溶剂<br />②实验B中水的作用是吸收生成物，防止生成物扩散到空气中污染大气<br />③实验C中通过烧杯中水进入集气瓶中的体积可得出氧气约占空气总体积<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">1</td></tr><tr><td>5</td></tr></table></span>的结论<br />④实验D中水的作用只是提供热量<br />⑤实验E中的水仅作为反应物．<br />（3）高铁酸钾（K<SUB>2</SUB>FeO<SUB>4</SUB>）是一种新型、高效的多功能水处理剂，受热时发生如下反应：4K<SUB>2</SUB>FeO<SUB>4</SUB> <span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>aR+4K<SUB>2</SUB>O+3O<SUB>2</SUB>↑．则a、R分别是（<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>）<br />A.4&nbsp;FeO&nbsp;&nbsp;&nbsp; B.2&nbsp; Fe<SUB>2</SUB>O<SUB>3</SUB>&nbsp;&nbsp;&nbsp;&nbsp; C.3&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Fe<SUB>2</SUB>O<SUB>3</SUB>&nbsp;&nbsp;&nbsp;&nbsp; D.1&nbsp;&nbsp; Fe<SUB>3</SUB>O<SUB>4</SUB>．<br />（4）甲、乙、丙三种固体物质的溶解度曲线如图2所示，回答下列问题：<br />①t<SUB>2</SUB>℃时，甲、乙、丙三种物质的溶解度由大到小的顺序是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />②t<SUB>1</SUB>℃时，甲和乙的溶解度<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“相等”或“不相等”．）<br />③t<SUB>2</SUB>℃时，甲物质的饱和溶液中溶质与溶剂的质量比为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（写最简比）．<br />④t<SUB>3</SUB>℃时，将等质量的甲、乙、丙三种物质的饱和溶液分别同时降温至t<SUB>2</SUB>℃，所得溶液中溶质质量分数由大到小的顺序是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','C$###$①③⑤$###$B$###$丙＞甲＞乙$###$相等$###$3：10$###$丙＞甲＞乙','【解答】解：<br />（1）受了污染的水能够使藻类物质过渡繁殖是因为水中溶有一些含氮、磷元素的物质，故答案为：C；<br />（2）<br />①物质溶于水时，水做溶剂，正确；<br />②铁丝在氧气中燃烧会产生大量的热使铁丝熔融溅落，瓶底的水能防止溅落的熔融的铁丝炸裂集气瓶，错误．<br />③空气中的氧气约占空气体积的五分之一，实验C中通过烧杯中水进入集气瓶中的体积可得出氧气约占空气总体积<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">1</td></tr><tr><td>5</td></tr></table></span>的结论，正确；<br />④实验D中水的作用是提供热量和隔离空气，错误；<br />⑤二氧化碳与水反应生成了碳酸，实验E中的水仅作为反应物，正确．<br />（3）根据质量守恒定律，化学反应前后元素的种类不变，从化学方程式看，反应前有8个钾原子、4个铁原子和16个氧原子，反应后除aR外，有8个钾原子、10个氧原子，因此aR中有4个铁原子和6个氧原子，观察选项可知B正确；<br />（4）①t<SUB>1</SUB>℃时甲、乙、丙三种物质的溶解度曲线，丙的曲线在最上方，此时物质丙的溶解度最大；甲的曲线在最下方，此时物质甲的溶解度最小；因此，三种物质的溶解度由大到小的顺序为丙＞甲＞乙；<br />②在t<SUB>2</SUB>℃时，甲、乙两物质的溶解度曲线交于一点，说明在t<SUB>2</SUB>℃时，两物质的溶解度相等；<br />③据溶解度曲线知，t<SUB>2</SUB>℃时甲物质的溶解度为30g，即100g溶剂，溶解30g甲，故t<SUB>2</SUB>℃时，甲物质的饱和溶液中溶质与溶剂的质量比为30：100=3：10；<br />④别将t<SUB>3</SUB>℃时，甲、乙、丙的饱和溶液降温至t<SUB>2</SUB>℃时，甲和乙仍然是饱和溶液，由于甲的溶解度在t<SUB>2</SUB>℃时大于乙的溶解度，因此甲的溶质质量分数大于乙；丙无晶体析出，但在t<SUB>2</SUB>℃时丙中溶质最多，故所得的三种溶液中溶质的质量分数由大到小的顺序是丙＞甲＞乙．<br />答案：<br />（1）C；<br />（2）①③⑤；<br />（3）B；<br />（4）①丙＞甲＞乙；③相等；③3：10；④丙＞甲＞乙．','【分析】（1）水污染引起的藻类物质过渡的繁殖主要是因为水中溶有一些含有氮、磷、钾元素的物质；<br />（2）①根据物质溶于水时，水做溶剂分析．<br />②根据铁丝在氧气中燃烧会产生大量的热使铁丝熔融溅落，瓶底的水能防止溅落的熔融的铁丝炸裂集气瓶分析．<br />③根据空气中的氧气约占空气体积的五分之一分析．<br />④根据白磷放在水中就和空气隔离分析．<br />⑤根据二氧化碳与水反应生成碳酸分析．<br />（3）利用质量守恒定律中元素的种类没有改变，原子的个数也没有增减的知识解决；<br />（4）①利用在t<SUB>1</SUB>℃时甲、乙、丙三种物质的溶解度曲线位置的上下关系，根据曲线在上方溶解度大，判断三种物质的溶解度大小；<br />②在t<SUB>2</SUB>℃时，甲、乙两物质的溶解度曲线交于一点，说明在t<SUB>2</SUB>℃时，两物质的溶解度相等；<br />③据t<SUB>2</SUB>℃时甲物质的溶解度为30g解答此题；<br />④根据溶解度的变化规律解答．','填空题',3.00,'585ee967dc6a07a454816d2ee30dd586',9,400,'水的性质和应用,固体溶解度曲线及其作用,溶质的质量分数、溶解性和溶解度的关系,元素的概念,质量守恒定律及其应用','',2016,'37','2016•天津一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842383,'根据如图所示的实验A、B、C，回答下题<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao20/32ea1511-94d4-11e9-b410-b42e9921e93e_xkb79.png\" style=\"vertical-align:middle\" /><br />（1）实验A中，铜片上的红磷没有燃烧，说明燃烧需要<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />（2）实验B中，能充分证明质量守恒定律的现象是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />（3）实验C中的一处明显错误是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>如果量取水时，仰视读数，配得的氢氧化钠溶液中溶质的质量分数将<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“偏大”、“偏小”或“不变”），配置好的氢氧化钠溶液必须密封保存，原因是氢氧化钠会和空气中的CO<SUB>2</SUB>反应而变质．写出反应的化学方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />（4）某兴趣小组同学为了验证实验室中久置的氢氧化钠是否变质，进行了如下表的实验．请你与他们一起完成以下验证实验：<br /><table class=\"edittable\"><TBODY><TR><td width=229>设计实验</TD><td width=181>实验现象</TD><td width=128>结论</TD></TR><TR><td>为检验是否变质，取少量固体加水溶解后，滴入足量稀盐酸；</TD><td><u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD><td>已变质</TD></TR><TR><td>为进一步探究是否全部变质，另取少量固体加水溶解后，加入过量<br />的 CaCl<SUB>2</SUB>溶液，完全反应后过滤，往滤液中滴加酚酞．</TD><td><br /><u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD><td>部分变质</TD></TR></TBODY></TABLE>','','','','','','达到着火点$###$天平仍保持平衡$###$氢氧化钠放在纸片上称量$###$偏小$###$2NaOH+CO<SUB>2</SUB>═Na<SUB>2</SUB>CO<SUB>3</SUB>+H<SUB>2</SUB>O$###$产生大量的气泡$###$酚酞溶液变红色','【解答】解：（1）根据燃烧的条件可知，铜片上的红磷与空气接触，是可燃物，但不能燃烧，说明是没有达到着火点，说明燃烧需要达到着火点．<br />故答案为：达到着火点．<br />（2）硫酸铜溶液能与铁钉反应，但天平仍然能够保持平行，可以证明质量守恒定律现象；从微观角度分析质量守恒的原因是：反应前后的原子个数，质量分数，种类不变．<br />故答案为：天平仍保持平衡．<br />（3）氢氧化钠具有腐蚀性，氢氧化钠放在纸片上称量<br />若其它操作都正确，而只有在量取水时仰视读数，仰视时所读取的数据小于所量水的实际体积，即量取水的体积大于实际所需水的量．由于溶剂水的质量偏大，可判断如此操作所配溶液溶质的质量分数比实际的偏小；氢氧化钠溶液需要密封保存是因为它接触空气中的二氧化碳而变质，氢氧化钠变质是因为能与二氧化碳反应生成碳酸钠和水，故填：2NaOH+CO<SUB>2</SUB>═Na<SUB>2</SUB>CO<SUB>3</SUB>+H<SUB>2</SUB>O．<br />（4）氢氧化钠敞口放置会和空气中的CO<SUB>2</SUB>反应生成Na<SUB>2</SUB>CO<SUB>3</SUB>而变质，为检验是否变质，取少量固体加水溶解后，滴入足量稀盐酸，碳酸钠能与稀盐酸反应生成二氧化碳气体，若产生大量的气泡，说明已变质．<br />为进一步探究是否全部变质，另取少量固体加水溶解后，加入过量的氯化钙溶液，碳酸钠能够和显中性的氯化钙溶液反应生成碳酸钙沉淀和氯化钠，完全反应后过滤，往滤液中滴加酚酞溶液，若酚酞溶液变红色，说明药品存在碳酸钠和氢氧化钠，即氢氧化钠部分变质．<br />答案：<br />（1）达到着火点；<br />（2）天平仍保持平衡；<br />（3）氢氧化钠放在纸片上称量；偏小；<br />（4）<table class=\"edittable\"><TBODY><TR><td width=229>设计实验</TD><td width=181>实验现象</TD><td width=128>结论</TD></TR><TR><td>为检验是否变质，取少量固体加水溶解后，滴入足量稀盐酸；</TD><td>产生大量的气泡</TD><td>已变质</TD></TR><TR><td>为进一步探究是否全部变质，另取少量固体加水溶解后，加入过量<br />的 CaCl<SUB>2</SUB>溶液，完全反应后过滤，往滤液中滴加酚酞．</TD><td><br />酚酞溶液变红色</TD><td>部分变质</TD></TR></TBODY></TABLE>','【分析】（1）根据燃烧的条件分析；<br />（2）如果反应前后天平依然平衡则能够证明质量守恒定律；反应前后的原子个数，质量分数，种类不变．<br />（3）根据氢氧化钠具有腐蚀性分析；根据溶液配制过程中，在用量筒取水时仰视读数，所读取数据小于所量取水的实际体积，分析由此所导致所配溶液溶质的质量分数与所需要配制溶液的溶质质量分数的大小关系；<br />（4）根据氢氧化钠敞口放置会和空气中的CO<SUB>2</SUB>反应生成Na<SUB>2</SUB>CO<SUB>3</SUB>而变质，根据生成的杂质碳酸钠能与稀盐酸反应生成二氧化碳气体，根据碳酸钠能够和显中性的氯化钙溶液反应生成碳酸钙沉淀和氯化钠，可加入CaCl<SUB>2</SUB>溶液除去碳酸根离子，利用酚酞试液遇碱溶液变红色判断是否完全变质．','书写',3.00,'93823799cc88f6063ece3481523c92a8',9,400,'燃烧的条件与灭火原理探究,一定溶质质量分数的溶液的配制,碱的化学性质,盐的化学性质,质量守恒定律及其应用,书写化学方程式、文字表达式、电离方程式','',2016,'32','2016•云南模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842392,'下列说法正确的是（　　）<br />①向某无色溶液中滴加紫色石蕊试液变蓝色，说明该溶液是碱性溶液<br />②铜丝浸入硝酸汞溶液中，表面覆盖一层银白色的物质，说明铜比汞活泼<br />③在某固体中滴加稀盐酸，有气泡产生，说明该固体是碳酸盐<br />④向某无色溶液中滴加氯化钡溶液，产生白色沉淀，说明该溶液是硫酸溶液．','①②','②③','③④','①④','','A','【解答】解：①向某无色溶液中滴加紫色石蕊试液变蓝色，说明该溶液呈碱性．滴加紫色石蕊试液，石蕊变蓝色，说明该溶液显碱性，但是显碱性未必就是碱；<br />②铜丝浸入硝酸汞溶液中，表面覆盖一层银白色的物质，说明铜比汞活泼．铜丝与硝酸汞反应，证明铜比汞活泼；<br />③在某固体中滴加稀盐酸，有气泡产生，生成气体无法确定，可以是二氧化碳，也可能是二氧化硫，甚至可以是氢气，所以对固体无法下结论就一定是碳酸盐；<br />④向某无色溶液中滴加氯化钡溶液，产生白色沉淀，由于加入氯化钡，产生沉淀可以是与氯离子有关，即沉淀为氯化银，也可能是与钡离子有关，即沉淀为碳酸钡、硫酸钡或者是亚硫酸钡，故无法下结论．<br />故选A．','【分析】题目给出了四个实验的设计，在解答时可以找明显正确或者明显错误的，之后用排除法来解题，没有必要逐一验证，如④中加入氯化钡溶液，产生白色沉淀，有可能沉淀是氯化银、碳酸钡、硫酸钡，甚至还可以是亚硫酸钡，所以对该溶液无法下结论，所以④错误，答案中含有④的都排除；就是选项A①②，B②③，AB都含有②，所以②一定正确，只需要看①和③到底哪一个正确即可．','选择题',3.00,'881bb3394bb8fde36466483633d2c67d',9,400,'证明硫酸和可溶性硫酸盐,金属的化学性质,酸碱指示剂及其性质,酸的化学性质','',2016,'37','2016春•镇江校级月考',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842425,'实验室常用下列装置制取气体，根据所学知识回答下列问题．<br /><img  src=\"/tikuimages/9/2012/400/shoutiniao97/33778800-94d4-11e9-898c-b42e9921e93e_xkb93.png\" style=\"vertical-align:middle\" /><br />（1）指出A装置中的任意一种仪器的名称<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />（2）用A装置制取氧气，在加入氧气前应先检查装置的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，写出用A装置制取氧气的一个文字表达式 <br /><u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）实验室用过氧化氢溶液和二氧化锰粉末混合制取氧气，不选B选C做气体发生装置，其优点是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，用E装置收集氧气的依据是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，检验氧气的方法是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（4）汽车尾气中的一氧化氮是一种大气污染物．它是一种无色气体，难溶于水，密度比空气略大，在空气中能与氧气迅速反应生成红棕色的二氧化氮气体．实验室中制取一氧化氮采用的收集装置是（填序号）<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．选择气体收集方法，从气体性质的角度需要考虑的是：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>和<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（5）甲烷在实验室里可用排水法或向下排空气法收集，请根据甲烷的收集方法，推测该气体具有的性质<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','酒精灯$###$气密性$###$氯酸钾<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"font-size:90%;\">二氧化锰</td></tr><tr><td><div hassize=\"9\"><div style=\"background: url(\'/upimages/images/part/8594L.png\') repeat-x; width:1px; height:5px;float:left;overflow:hidden\" muststretch=\"h\"></div><div style=\"background: url(\'/upimages/images/part/8594.png\') no-repeat; width:9px; height:5px;float:left;overflow:hidden\"></div></div></td></tr><tr><td style=\"font-size:90%\">加热</td></tr></table></span>氯化钾+氧气$###$随时控制反应的发生和停止$###$氧气的密度比空气大$###$将带火星的木条升入集气瓶中，若木条复燃，则为氧气$###$D$###$密度$###$水溶性$###$难溶于水、密度比空气小','【解答】解：（1）A中有酒精灯、试管、铁架台；<br />（2）用A装置制取氧气，在加入氧气前应先检查装置的气密性；装置A是加热固体制取气体，由于试管口没有棉花团，则是加热氯酸钾和二氧化锰的混合物制取氧气，生成氧气的同时还有氯化钾生成，故其文字表达式为：氯酸钾<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"font-size:90%;\">二氧化锰</td></tr><tr><td><div hassize=\"9\"><div style=\"background: url(\'/upimages/images/part/8594L.png\') repeat-x; width:1px; height:5px;float:left;overflow:hidden\" muststretch=\"h\"></div><div style=\"background: url(\'/upimages/images/part/8594.png\') no-repeat; width:9px; height:5px;float:left;overflow:hidden\"></div></div></td></tr><tr><td style=\"font-size:90%\">加热</td></tr></table></span>氯化钾+氧气；<br />故答案为：气密性；氯酸钾<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"font-size:90%;\">二氧化锰</td></tr><tr><td><div hassize=\"9\"><div style=\"background: url(\'/upimages/images/part/8594L.png\') repeat-x; width:1px; height:5px;float:left;overflow:hidden\" muststretch=\"h\"></div><div style=\"background: url(\'/upimages/images/part/8594.png\') no-repeat; width:9px; height:5px;float:left;overflow:hidden\"></div></div></td></tr><tr><td style=\"font-size:90%\">加热</td></tr></table></span>氯化钾+氧气；<br />（3）注射器和长颈漏斗都可以通过控制加入液体的量来控制反应的发生和停止；但前者在制取氧气时，由于长颈漏斗下端要浸没在液面以下，加入液体的量较多，这时无法控制反应的停止，当第一次加入的过氧化氢溶液反应完之后，可通过加入过氧化氢溶液的量来控制反应的发生与停止；而后者由于注射器是密封的，可通过推动活塞随进控制加入过氧化氢溶液的量来达到控制反应的发生与停止的目的；<br />氧气的密度比空气大，用E装置收集氧气；<br />检验氧气的方法是：将带火星的木条升入集气瓶中，若木条复燃，则为氧气；<br />故答案为：随时控制反应的发生和停止；氧气的密度比空气大；将带火星的木条升入集气瓶中，若木条复燃，则为氧气；<br />（4）一氧化氮难溶于水，可用排水法收集；密度比空气略大，在空气中能与氧气迅速反应生成红棕色的二氧化氮气体，所以不能用排空气法收集；故实验室中制取一氧化氮采用的收集装置是D；由此可知我们在选择收集气体的方法是通常需要考虑的因素有气体的密度和溶解性，此外还要考虑气体能否与空气中的某些气体反应、能否与水反应；<br />故答案为：D；密度和溶解性；<br />（5）气体难溶于水可以用排水法收集，气体的密度决定了气体的收集方法，所以甲烷气体在实验室里可用排水法或向下排空气法收集，甲烷具有的性质是难溶于水、密度比空气小；<br />故答案为：难溶于水、密度比空气小．','【分析】（1）从仪器的图形和用途去分析解答；<br />（2）从装置A是加热固体制取气体，由于试管口没有棉花团，则是加热氯酸钾和二氧化锰的混合物制取氧气，生成氧气的同时还有氯化钾生成去分析解答；<br />（3）从注射器和长颈漏斗都可以通过控制加入液体的量来控制反应的发生和停止；但前者在制取氧气时，由于长颈漏斗下端要浸没在液面以下，加入液体的量较多，这时无法控制反应的停止，当第一次加入的过氧化氢溶液反应完之后，可通过加入过氧化氢溶液的量来控制反应的发生与停止；而后者由于注射器是密封的，可通过推动活塞随进控制加入过氧化氢溶液的量来达到控制反应的发生与停止的目的去分析解答；<br />（4）依据一氧化氮的性质分析所用收集方法；并由此分析确定收集方法所需要考虑的因素；<br />（5）根据气体难溶于水可以用排水法收集，气体的密度决定了气体的收集方法进行分析．','书写',3.00,'ab785a487ddc720e4f9f75d69e096043',9,400,'实验室制取气体的思路,氧气的制取装置,氧气的收集方法,氧气的检验和验满,书写化学方程式、文字表达式、电离方程式','',2012,'35','2012秋•阜南县校级期中',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842432,'<img  src=\"/tikuimages/9/2016/400/shoutiniao50/33934d61-94d4-11e9-9c03-b42e9921e93e_xkb61.png\" style=\"vertical-align:middle;FLOAT:right\" />（2016•合肥校级二模）据报道：5月9日上午10时许，位于铁西新区宁官附近的某淀粉厂发生粉尘爆炸．淀粉加工厂的车间里若悬浮着较多的淀粉，遇明火就有发生爆炸的危险．下面是模拟粉末爆炸实验：<br />如图所示，挤压洗耳球鼓入空气，不久会听到“砰”的一声．爆炸的气浪将纸罐盖掀起．请问：<br />（1）在上述实验中，鼓入空气的作用是：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（2）下列情况时，不会发生爆炸的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填序号）；<br />①纸罐不加盖时②酒精棉团没有点燃时③若淀粉换成干燥的煤粉<br />（3）淀粉厂为了避免爆炸事故发生，应采取的安全措施有<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（写出一种即可）．','','','','','','使淀粉充满纸罐，与氧气充分接触$###$①②$###$严禁烟火（或保持通风）','【解答】解：（1）根据可燃性气体或易燃物的粉尘在空气中的含量达到爆炸极限时，遇到明火就会剧烈燃烧，导致有限空间里气体的体积急剧膨胀，从而引起爆炸．鼓入空气的作用是：易燃物的粉尘在空气中的含量达到爆炸极限即 使面粉充满金属筒，与空气混合均匀．<br />（2）根据燃烧和发生爆炸的条件：在有限的空间内，可燃气体或粉尘与空气混合达到爆炸极限，遇到明火，可知①纸罐不加盖时，不是有限空间，不会发生爆炸；&nbsp; ②酒精棉球没有点燃时，没有遇明火不会爆炸；<br />（3）根据燃烧和发生爆炸的条件，面粉厂为了避免爆炸事故发生，应采取的安全措施有：严禁烟火（或保持通风）．<br />故答案为：（1）使淀粉充满纸罐，与氧气充分接触；<br />（2）①②；<br />（3）严禁烟火（或保持通风）．','【分析】据燃烧和发生爆炸的条件（在有限的空间内，可燃气体或粉尘与空气混合，达到爆炸极限，遇到明火剧烈燃烧）解答此题．','简答题',3.00,'200d0e85d0903c7b0641368702c977b8',9,400,'燃烧和爆炸实验','',2016,'37','2016•合肥校级二模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842437,'<img  src=\"/tikuimages/9/2016/400/shoutiniao20/33a4d991-94d4-11e9-8f35-b42e9921e93e_xkb60.png\" style=\"vertical-align:middle;FLOAT:right\" />（2016•东城区一模）2015年1月起，国内重型柴油机普推国*MERGEFORMAT IV排放标准．减排的原理为：向反应罐内喷入车用尿素[CO（NH<SUB>2</SUB>）<SUB>2</SUB>]溶液．在一定条件下，尿素先转化为NH<SUB>3</SUB>，NH<SUB>3</SUB>再与发动机排出的NO等反应生成氮气和水．<br />（1）尿素属于<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“有机物”或“无机物”）．<br />（2）车用尿素溶液还可用作氮肥．根据右图标签，计算该桶溶液能供给作物约<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>kg氮元素（结果精确到0.1kg）．<br />（3）喷入反应罐内的尿素溶液不能过多或过少．如果喷入过多，会因产生大量有刺激性气味的氨气造成<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','有机物$###$3.0$###$空气污染','【解答】解：（1）由尿素的化学式可知，尿素是一种含有碳元素的化合物，属于有机物；<br />（2）该桶溶液能供给作物单元的质量为：20kg×32.5%×<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">14×2</td></tr><tr><td>60</td></tr></table>×100%</span>≈3.0kg；<br />（3）喷入的尿素溶液过多会生成大量的氨气，排放到空气中会造成空气的污染．<br />故答为：（1）有机物；（2）3.0；（3）空气污染．','【分析】（1）含有碳元素的化合物属于有机化合物，简称有机物；不含有碳元素的化合物，属于无机化合物，简称无机物．<br />（2）根据标签中的数据和化学式的意义分析计算．<br />（3）根据氨气能污染空气分析．','填空题',3.00,'29dab2c9d99bf2576936b20507d95b69',9,400,'有机物与无机物的区别,混合物中某元素的质量计算,标签上标示的物质成分及其含量','',2016,'37','2016•东城区一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842455,'如表是某同学用多种方法鉴别物质的情况，其中两种方法一定都正确的是（　　） <table class=\"edittable\"><TBODY><TR><td width=30 rowSpan=2></TD><td width=168 rowSpan=2>需要鉴别的物质</TD><td width=244 colSpan=3>方法和所加试剂</TD></TR><TR><td colSpan=2>方法一</TD><td>方法二</TD></TR><TR><td>A</TD><td>软水和硬水</TD><td>加适量肥皂水搅拌</TD><td colSpan=2>观察颜色</TD></TR><TR><td>B</TD><td>人体吸入的空气和呼出的空气</TD><td>带火星的木条</TD><td colSpan=2>适量澄清石灰水</TD></TR><TR><td>C</TD><td>氯化钠溶液和硝酸铵</TD><td>加适量水溶解后测温度</TD><td colSpan=2>加熟石灰研磨，闻气味</TD></TR><TR><td>D</TD><td>澄清石灰水和氢氧化钠溶液</TD><td>滴加碳酸钠溶液</TD><td colSpan=2>滴加紫色石蕊试剂</TD></TR></TBODY></TABLE>','A、','B、','C、','D、','','C','【解答】解：A、软水和硬水，都为无色，无法鉴别，故A错误；<br />B、用带火星的木条，无法鉴别人体吸入的空气和呼出的空气，故错误；<br />C、氯化钠溶液溶于水，温度不变，硝酸铵溶于水吸热，使用加水溶解的方法可以鉴别；硝酸铵与氢氧化钙混合研磨会产生有刺激性气味的氨气，使用氢氧化钙可以鉴别，故C正确；<br />D、澄清石灰水和氢氧化钠溶液都为碱性，滴加紫色石蕊试剂，现象相同，无法鉴别，故D错误；<br />故选C．','【分析】根据物质的性质差异进行分析，考虑软水和硬水的鉴别、物理性质；氯化钠溶液溶于水，温度不变，硝酸铵溶于水吸热，铵态氮肥能与碱性物质混合产生有刺激性气味的氨气；考虑澄清石灰水和氢氧化钠溶液都为碱性．据此解答．','选择题',3.00,'c5fc2cc485b4340688e2516db5696a14',9,400,'吸入空气与呼出气体的比较,硬水与软水,铵态氮肥的检验,酸、碱、盐的鉴别,物质的鉴别、推断','',2016,'37','2016•天津一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842484,'下列实验方案设计可行的是（　　）','用溶解、过滤的方法分离氯化钠和硝酸钾的固体混合物','用酚酞鉴别氢氧化钠溶液和碳酸钠溶液','向氧化铜中加入适量稀硫酸恰好反应后，再加入适量氢氧化钡溶液，过滤即可制得氢氧化铜','向某无色溶液中加入足量稀盐酸无现象，再加入氯化钡溶液，若产生白色沉淀，证明原溶液中一定含有SO<SUB>4</SUB><SUP>2-</SUP>','','D','【解答】解：A、氯化钠和硝酸钾都溶于水，所以可用溶解、结晶、过滤的方法分离氯化钠和硝酸钾的固体混合物，故A错误；<br />B、氢氧化钠溶液和碳酸钠溶液都显碱性，都能使酚酞试液变红，所以不能用酚酞鉴别氢氧化钠溶液和碳酸钠溶液，故B错误；<br />C、氧化铜和硫酸反应生成硫酸铜和水，而硫酸铜和氢氧化钡溶液反应会生成氢氧化铜沉淀和硫酸钡沉淀，所以不只是能到氢氧化铜的沉淀，还有硫酸钡沉淀，故C错误；<br />D、向某无色溶液中加入足量稀盐酸无现象，说明溶液不含有碳酸根离子和银离子，再加入氯化钡溶液，若产生白色沉淀，说明生成的白色沉淀为硫酸钡沉淀，所以证明原溶液中一定含有SO<SUB>4</SUB><SUP>2-</SUP>，故D正确．<br />故选：D．','【分析】A、根据氯化钠和硝酸钾都溶于水进行解答；<br />B、根据氢氧化钠溶液和碳酸钠溶液都显碱性，都能使酚酞试液变红进行解答；<br />C、根据氧化铜和硫酸反应生成硫酸铜和水，而硫酸铜和氢氧化钡溶液反应会生成氢氧化铜沉淀和硫酸钡沉淀进行解答；<br />D、根据向某无色溶液中加入足量稀盐酸无现象，说明溶液不含有碳酸根离子和银离子，再加入氯化钡溶液，若产生白色沉淀，说明生成的白色沉淀为硫酸钡沉淀进行解答．','选择题',3.00,'c4b302d2c10a769adb11011685485788',9,400,'化学实验方案设计与评价,混合物的分离方法,证明硫酸和可溶性硫酸盐,酸的化学性质,碱的化学性质,酸、碱、盐的鉴别','',2016,'37','2016•市中区二模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842486,'化学和生产、生活紧密相关．请你用所学知识回答下列问题：<br />（1）空气由多种气体组成，它属于<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（选填“混合物”<br />或“纯净物”）．植物的光合作用是空气中<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>的主要来源．<br />（2）合金具有良好的物理、化学性能，下列物质不属于合金的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（选填序号）．<br />a．黄铜&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;b不锈钢&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;c．铁矿石<br />（3）如果锅里的油着火，可用锅盖盖灭，其原理是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（4）稀土元素铕是激光及原子能应用的重要材料．已知三氯化铕的化学式为EuCl<SUB>3</SUB>，则氧化铕的化学式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','混合物$###$氧气$###$C$###$隔绝氧气$###$Eu<SUB>2</SUB>O<SUB>3</SUB>','【解答】（1）纯净物和混合物的判别是根据纯净物和混合物的概念或特点，来对某物质进行辨析、推断，看它到底是纯净物还是混合物，从宏观角度来说，主要看它含有几种物质，含有一种的是纯净物；含有多种的是混合物，空气中含有多种气体，所以可以判断空气为混合物；而植物的光合作用能够提供氧气；故填：混合物；氧气；<br />（2）黄铜和不锈钢是合金，铁矿石的主要成分铁的化合物，不属于合金；故填：C；<br />（3）锅里的油一旦着火，可用锅盖盖灭，其原理是隔绝空气；故填：隔绝空气；<br />（4）在化合物中正负化合价的代数和为零，而氯化物中氯元素的化合价为-1，<br />设铕元素的化合价为x，则<br />x+（-1）×3=0，<br />解得：x=+3；<br />即铕元素的化合价为+3，氧元素的化合价为-2，根据化合价原则可以写出氧化铕的化学式为：Eu<SUB>2</SUB>O<SUB>3</SUB>．<br />故填：Eu<SUB>2</SUB>O<SUB>3</SUB>','【分析】（1）根据混合物的概念以及光合作用的原理来分析；<br />（2）由金属和金属或非金属熔合而成的具有金属特性的物质属于合金；<br />（3）灭火的方法有：移走可燃物，隔绝氧气，降低温度到可燃物的着火点以下；<br />（4）根据化合物中元素的化合价原则结合三氯化铕的化学式判断出铕元素的化合价，并写出氧化铕的化学式即可．','书写',3.00,'548a70fbc457f4be067d8be71a1b345d',9,400,'空气的成分及各成分的体积分数,自然界中的氧循环,合金与合金的性质,纯净物和混合物的判别,化学式的书写及意义,灭火的原理和方法','',2016,'37','2016•湘潭二模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842488,'自然界中存在氧循环和碳循环，其中能将二氧化碳转化为氧气的是（　　）','光合作用','燃料的燃烧','水的吸收','动植物的呼吸','','A','【解答】解：<br />A、植物的光合作用是吸收植物吸收二氧化碳释放氧气的过程，能将二氧化碳转化为氧气，故选项正确；<br />B、燃料的燃烧，是燃料与氧气反应生成二氧化碳，是将氧气转化为二氧化碳，故选项错误；<br />C、水能吸收部分二氧化碳，但不能放出氧气，故选项错误；<br />D、动植物的呼吸，是将氧气在体内发生缓慢氧化释放出二氧化碳，是将氧气转化为二氧化碳，故选项错误．<br />故选：A．','【分析】结合光合作用、呼吸作用、燃烧的常识等进行分析解答，自然界中由二氧化碳转化为氧气唯一的方法是光合作用，据此进行分析判断．','选择题',3.00,'2e38dd6a2e5d31a94b3a97478605b9ae',9,400,'自然界中的氧循环','',0,'37','',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842506,'我区某校进行化学实验操作考核，下列四位同学的操作中正确的是（　　）','<img  src=\"/tikuimages/9/2016/400/shoutiniao99/3471ed8f-94d4-11e9-a473-b42e9921e93e_xkb67.png\" style=\"vertical-align:middle\" /><br />检验溶液酸碱性','<img  src=\"/tikuimages/9/2016/400/shoutiniao96/3472d7f0-94d4-11e9-943d-b42e9921e93e_xkb67.png\" style=\"vertical-align:middle\" /><br />溶解固体','<img  src=\"/tikuimages/9/2016/400/shoutiniao7/347548f0-94d4-11e9-ac2e-b42e9921e93e_xkb62.png\" style=\"vertical-align:middle\" /><br />&nbsp; 倾倒液体','<img  src=\"/tikuimages/9/2016/400/shoutiniao20/3478cb61-94d4-11e9-998b-b42e9921e93e_xkb77.png\" style=\"vertical-align:middle\" /><br />&nbsp;熄灭酒精灯','','D','【解答】解：A、检验溶液酸碱性的方法，指示剂不能直接滴入试剂瓶，会污染试剂，图中所示操作错误；<br />B、溶解固体药品通常在烧杯中进行，量筒可用来量取一定量液体体积，图中所示操作错误；<br />C、向试管中倾倒液体药品时，瓶塞要倒放，标签要对准手心，瓶口紧挨，图中所示操作错误；<br />D、熄灭酒精灯要用灯帽盖灭；图中所示操作正确．<br />故选：D．','【分析】A、根据检验溶液酸碱性的方法进行分析判断；<br />B、根据溶解固体的方法进行分析判断；<br />C、根据倾倒液体的方法进行分析判断；<br />D、根据酒精灯的使用方法进行分析判断．','选择题',3.00,'6b39a51140457ec5ee72cf0932cdf6e5',9,400,'加热器皿-酒精灯,液体药品的取用,物质的溶解','',2016,'37','2016•南开区一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842521,'张红同学的新家用水了铬钢水龙头，好奇的张红按下列步骤比较金属铬（Cr）与金属铁、铜的金属活动性强弱．<br />①联系金属活动顺序张红提出了三种合理的猜想，请完成猜想3的内容：<br />猜想1：Cr＞Fe＞Cu<br />猜想2：Fe＞Cr＞Cu<br />猜想3：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />②张红将金属铁和铜分别放入硫酸亚铬（CrSO<SUB>4</SUB>）溶液中，无明显现象发生，张红得出的结论是猜想<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>成立；③如果将金属铁和铬分别放入下列<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填序号）中，也能得到②中的结论．<br />A．CuCl<SUB>2</SUB>溶液&nbsp;&nbsp; B．AgNO<SUB>3</SUB>溶液&nbsp;&nbsp; C．FeSO<SUB>4</SUB>溶液&nbsp; D．Na<SUB>2</SUB>SO<SUB>4</SUB>溶液．','','','','','','Fe＞Cu＞Cr$###$1$###$A','【解答】解：<br />①三种金属的活动性顺序可能存在三种情况：猜想1为Cr＞Fe＞Cu；&nbsp;&nbsp;&nbsp;&nbsp;猜想2为Fe＞Cr＞Cu，故猜想3答案：Fe＞Cu＞Cr；<br />②张红将金属铁和铜分别放入硫酸亚铬（CrSO<SUB>4</SUB>）溶液中，无明显现象发生，说明铁和铜不如铬活泼，张红得出的结论是猜1成立；<br />③如果将金属铁和铬分别放入下列CuCl<SUB>2</SUB>溶液中，也能得到②中的结论．<br />答案：<br />①Fe＞Cu＞Cr；<br />②1；<br />③A','【分析】①根据三种金属的活动性顺序进行猜想；<br />②根据在金属活动性顺序中，位于前面的金属能把位于后面的金属从它们化合物的溶液里置换出来解答．<br />③根据探究金属活动性的方法解答．','填空题',3.00,'25e236045c490225e6405c08a52adb4d',9,400,'金属活动性的探究','',0,'37','',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842548,'下列实验操作正确的是（　　）','<img  src=\"/tikuimages/9/2016/400/shoutiniao38/351188f0-94d4-11e9-ba69-b42e9921e93e_xkb29.png\" style=\"vertical-align:middle\" /><br />塞紧橡皮塞','<img  src=\"/tikuimages/9/2016/400/shoutiniao52/3515808f-94d4-11e9-90c7-b42e9921e93e_xkb52.png\" style=\"vertical-align:middle\" /><br />测定溶液的pH','<img  src=\"/tikuimages/9/2016/400/shoutiniao84/35170730-94d4-11e9-a2f5-b42e9921e93e_xkb26.png\" style=\"vertical-align:middle\" /><br />稀释浓硫酸','<img  src=\"/tikuimages/9/2016/400/shoutiniao53/3518dbf0-94d4-11e9-a293-b42e9921e93e_xkb61.png\" style=\"vertical-align:middle\" /><br />称取一定质量的氢氧化钠固体','','C','【解答】解：A、把橡皮塞慢慢转动着塞进试管口，切不可把试管放在桌上在使劲塞进塞子，以免压破试管，图中所示操作错误．<br />B、用pH试纸测定未知溶液的pH时，正确的操作方法为用玻璃棒蘸取少量待测液滴在干燥的pH试纸上，与标准比色卡对比来确定pH．不能将pH试纸伸入待测液中，以免污染待测液，图中所示操作错误．<br />C、稀释浓硫酸时，要把浓硫酸缓缓地沿器壁注入水中，同时用玻璃棒不断搅拌，以使热量及时地扩散；一定不能把水注入浓硫酸中；图中所示操作正确．<br />D、托盘天平的使用要遵循“左物右码”的原则，且氢氧化钠具有腐蚀性，应放在玻璃器皿中称量，图中所示操作错误．<br />故选：C．','【分析】A、根据把橡皮塞塞进试管口的方法进行分析判断．<br />B、根据用pH试纸测定未知溶液的pH的方法进行分析判断．<br />C、根据浓硫酸的稀释方法（酸入水，沿器壁，慢慢倒，不断搅）进行分析判断．<br />D、根据托盘天平的使用要遵循“左物右码”的原则、氢氧化钠具有腐蚀性，进行分析判断．','选择题',3.00,'73b57b43873676cabb85dd547576971c',9,400,'称量器-托盘天平,浓硫酸的性质及浓硫酸的稀释,仪器的装配或连接,溶液的酸碱度测定','',2016,'37','2016春•福州校级月考',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842567,'结构决定性质，性质决定用途．下列物质的性质和用途对应有误的是（　　）','稀盐酸--能与金属氧化物反应--除铁锈','苛性钠--容易吸收水分--作某些气体干燥剂','氧气--可燃性--气焊','碳酸氢钠--能与胃酸反应--治疗胃酸过多症','','C','【解答】解：<br />A、铁锈的主要成分是氧化铁，能与盐酸反应，盐酸可用于除铁锈，利用了盐酸能与碱性氧化物反应的性质，故正确；<br />B、苛性钠具有吸水性，作某些气体干燥剂，故正确；<br />C、氧气用于炼铁是利用了氧气能支持燃烧的性质．氧气没有可燃性，故错误；<br />D、碳酸氢钠能与胃酸反应，治疗胃酸过多症，故正确．<br />故选C','【分析】A、根据铁锈的主要成分是氧化铁，能与盐酸反应分析；<br />B、根据苛性钠具有吸水性解答；<br />C、根据氧气能支持燃烧的性质解答；<br />D、根据碳酸氢钠能与胃酸反应解答．','选择题',3.00,'e31fa0d59cdd9b8c0509174828712dcb',9,400,'氧气的用途,酸碱盐的应用','',2016,'32','2016•重庆校级模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842570,'下列认识或说法中正确的是（　　）','硫在氧气中燃烧发出淡蓝色火焰','燃烧一定需要氧气参加','洒出的酒精在桌面燃烧，用湿布盖灭','用汽油去除油污属于乳化现象','','C','【解答】解：A、硫粉在氧气中燃烧，发出明亮的蓝紫色火焰，故选项错误．<br />B、镁和二氧化碳在点燃的条件下生成氧化镁和碳，故不正确；<br />C、洒出酒精着火了可以用湿抹布盖灭，以隔绝氧气和降低温度，故正确．<br />D、汽油去除衣服上的油污是溶解现象，错误；<br />答案：C','【分析】A、根据硫在氧气中燃烧的现象进行分析判断．<br />B、根据一些活泼金属在二氧化碳中也可以燃烧进行分析；<br />C、根据灭火的原理进行分析来回答本题．<br />D、根据乳化现象分析；','选择题',3.00,'e494220a3a75ad3e8e20ccefd4d84eb6',9,400,'氧气的化学性质,溶解现象与溶解原理,燃烧与燃烧的条件,灭火的原理和方法','',2016,'37','2016•徐闻县一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842575,'【实验一】结合下列实验装置图回答问题<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao63/3585f59e-94d4-11e9-9229-b42e9921e93e_xkb48.png\" style=\"vertical-align:middle\" /><br />（1）用A图制取二氧化碳气体，反应的化学方程式为：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，若锥形瓶内的反应进行较长时间后，用燃着的火柴放在集气瓶口，火焰仍不熄灭，则是由于A装置存在缺陷，其缺陷为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（2）图B、C、D是探究二氧化碳化学性质的实验，C中发生反应的化学方程式是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，证明二氧化碳与氢氧化钠溶液发生反应的现象是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（3）图E和F是探究哪种离子能促进H<SUB>2</SUB>O<SUB>2</SUB>分解的实验，现象是：E中产生大量气泡，F中没有明显现象，则促进H<SUB>2</SUB>O<SUB>2</SUB>分解的离子符号是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />【实验二】探究燃烧的条件<br />实验：下图的装置可用于探究燃烧的条件<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao85/3588dbcf-94d4-11e9-aa68-b42e9921e93e_xkb88.png\" style=\"vertical-align:middle\" /><br />现象：①不通氧气时，A中白磷不燃烧<br />②不通氧气时，B中白磷不燃烧<br />③通氧气时，A中白磷不燃烧<br />④通氧气时，B中白磷燃烧<br />（1）证明可燃物燃烧，温度必须达到着火点的现象是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填序号）；<br />（2）A、B组合与C装置相比的优点是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />【实验三】老师在课堂上说：“氢氧化钡与氢氧化钙化学性质相似”．化学兴趣小组的同学决定用氢氧化钡溶液来进行验证，他们做了以下实验．<br /><table class=\"edittable\"><TBODY><TR><td width=57>序&nbsp;&nbsp;号</TD><td width=183>操&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;作</TD><td width=105>现&nbsp;&nbsp;象</TD><td width=143>结&nbsp;&nbsp;&nbsp;&nbsp;论</TD></TR><TR><td>实验一</TD><td>向氢氧化钡溶液中滴入酚酞</TD><td>溶液变红</TD><td>氢氧化钡溶液呈碱性</TD></TR><TR><td>实验二</TD><td>向实验一的溶液中通入<br />适量的二氧化碳</TD><td>产生白色沉淀，红色消失</TD><td>氢氧化钡溶液能与<br />二氧化碳反应</TD></TR><TR><td>实验三</TD><td>另取氢氧化钡溶液，滴入<br />一定量的碳酸钠溶液</TD><td>产生白色沉淀</TD><td>氢氧化钡溶液能与<br />碳酸钠溶液反应</TD></TR></TBODY></TABLE>结论：氢氧化钡的化学性质与氢氧化钙相似．<br />（1）写出实验三中发生反应的化学方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />同学们将上述反应后的废液倒入烧杯中，过滤后，得到白色沉淀和红色滤液，他们决定对沉淀和滤液进行探究．<br />探究一：<br />【提出问题】沉淀能否与酸发生反应？<br />【实验探究】向沉淀中加入足量的稀硫酸，有气泡产生，白色固体不消失．<br />【实验结论】沉淀与酸不反应．<br />（2）有同学提出，根据“白色固体不消失”这一现象所得出的结论不严谨，理由是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />探究二：<br />【提出问题】滤液中除酚酞外，还有哪些溶质？<br />【作出猜想】Ⅰ．只有氢氧化钠Ⅱ．有氢氧化钠和碳酸钠Ⅲ．有氢氧化钠和氢氧化钡<br />【实验探究】<br /><table class=\"edittable\"><TBODY><TR><td width=104>序号</TD><td width=199>操作</TD><td width=133>现象</TD><td width=113>结论</TD></TR><TR><td>实验四</TD><td>取滤液样品于试管中，<br />加入几滴稀盐酸</TD><td>没有明显现象</TD><td>猜想Ⅱ不成立</TD></TR><TR><td>实验五</TD><td>取滤液样品于试管中，<br />加入几滴碳酸钠溶液</TD><td>有白色沉淀产生，红色不消失</TD><td>猜想Ⅲ成立</TD></TR></TBODY></TABLE>（3）有同学提出：实验四由于实验操作存在问题，导致结论不准确，该实验&nbsp;&nbsp;操作上存在的问题是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />【反思交流】<br />（4）小组同学经分析、讨论后发现，只用一种溶液做一次实验，就可以验证猜想，该溶液是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（5）在废液中滴加稀盐酸至<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填写实验现象）时，表示废液已呈中性．','','','','','','CaCO<SUB>3</SUB>+2HCl=CaCl<SUB>2</SUB>+CO<SUB>2</SUB>↑+H<SUB>2</SUB>O$###$长颈漏斗未伸入液面，造成气体逸出$###$CO<SUB>2</SUB>+Ca（OH）<SUB>2</SUB>=CaCO<SUB>3</SUB>↓+H<SUB>2</SUB>O，$###$C中有白色沉淀生成$###$Fe<SUP>3+</SUP>$###$③④$###$C装置在密闭环境中进行，避免了P<SUB>2</SUB>O<SUB>5</SUB>对空气的污染$###$Na<SUB>2</SUB>CO<SUB>2</SUB>+Ba（OH）<SUB>2</SUB>═BaCO<SUB>3</SUB>↓+2NaOH$###$硫酸钡也是白色固体$###$加入的稀盐酸的量太少，现象不明显$###$稀硫酸$###$红色消失','【解答】解：实验一（1）根据反应物是石灰石和盐酸，生成物是氯化钙、水和二氧化碳，反应方程式是：CaCO<SUB>3</SUB>+2HCl═CaCl<SUB>2</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑；装置的气密性都是经过检验的，观察装置看到长颈漏斗下端的管口与大气是相通的，生成的二氧化碳会从管口逸到空气中装置的气密性都是经过检验的，观察装置看到长颈漏斗下端的管口与大气是相通的，生成的二氧化碳会从管口逸到空气中<br />（2）B中发生反应的化学方程式是：CO<SUB>2</SUB>+Ca（OH）<SUB>2</SUB>=CaCO<SUB>3</SUB>↓+H<SUB>2</SUB>O；<br />（3）根据向下：E中产生大量气泡，F中没有明显现象分析，氯离子、钠离子不能使过氧化氢分解速度加快，所以是铁离子使双氧水分解的速率加快；<br />实验二（1）能证明可燃物必须达到一定温度（着火点）的变量是：温度，有冷水，有热水，有热水的必须能够燃烧，其它变量均相同，所以③④符合这一说法<br />（2）C装置在密闭环境中进行，避免了P<SUB>2</SUB>O<SUB>5</SUB>对空气的污染；<br />实验三（1）氢氧化钡与碳酸钠反应生成氢氧化钠和碳酸钡沉淀，反应的方程式为：Na<SUB>2</SUB>CO<SUB>2</SUB>+Ba（OH）<SUB>2</SUB>═BaCO<SUB>3</SUB>↓+2NaOH；<br />（2）碳酸钡与稀硫酸反应生成硫酸钡、水、二氧化碳，硫酸钡也是白色固体；<br />（3）实验四取滤液样品于试管中，加入几滴稀盐酸，如果稀盐酸的量过少，则现象不明显；<br />（4）加入稀硫酸，如果只是红色消失，则证明有氢氧化钠，如红色消失，且有沉淀生成，则一定有氢氧化钡，如红色消失，且有气泡生成，则一定有碳酸钠．<br />（5）稀盐酸与碱完全反应溶液呈中性，酚酞试液显示无色．<br />故答案为：实验一（1）CaCO<SUB>3</SUB>+2HCl═CaCl<SUB>2</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑、长颈漏斗未伸入液面，造成气体逸出；<br />（2）CO<SUB>2</SUB>+Ca（OH）<SUB>2</SUB>=CaCO<SUB>3</SUB>↓+H<SUB>2</SUB>O；<br />（3）Fe<SUP>3+</SUP>；<br />实验二（1）③④；（2）C装置在密闭环境中进行，避免了P<SUB>2</SUB>O<SUB>5</SUB>对空气的污染；<br />实验三（1）Na<SUB>2</SUB>CO<SUB>2</SUB>+Ba（OH）<SUB>2</SUB>═BaCO<SUB>3</SUB>↓+2NaOH；（2）硫酸钡也是白色固体；（3）稀硫酸；（5）红色消失．','【分析】实验一（1）据二氧化碳的实验室制取原理及可以从装置的漏气和气密性等方面分析原因解答；<br />（2）据二氧化碳的检验方法解答；<br />（3）据已知条件分析解答；<br />实验二（1）能证明可燃物必须达到一定温度（着火点）的变量是：温度，有冷水，有热水，有热水的必须能够燃烧，其它变量均相同，所以③④符合这一说法<br />（2）C装置在密闭环境中进行，避免了P<SUB>2</SUB>O<SUB>5</SUB>对空气的污染；<br />实验三（1）氢氧化钡与碳酸钠反应生成氢氧化钠和碳酸钡沉淀；<br />（2）碳酸钡与稀硫酸反应生成硫酸钡、水、二氧化碳；<br />（3）实验四取滤液样品于试管中，加入几滴稀盐酸，如果稀盐酸的量过少，则现象不明显；<br />（4）加入稀硫酸，如果只是红色消失，则证明有氢氧化钠，如红色消失，且有沉淀生成，则一定有氢氧化钡，如红色消失，且有气泡生成，则一定有碳酸钠．<br />（5）稀盐酸与碱完全反应溶液呈中性，酚酞试液显示无色．','书写',3.00,'7e39f81811f7875e07a70f450c29a9ec',9,400,'实验探究物质的性质或变化规律,燃烧的条件与灭火原理探究,实验探究物质的组成成分以及含量,二氧化碳的实验室制法,探究二氧化碳的性质,碱的化学性质,盐的化学性质,书写化学方程式、文字表达式、电离方程式','如皋市',2016,'32','2016•如皋市模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842588,'下列实验操作正确的是（　　）','<img  src=\"/tikuimages/9/2016/400/shoutiniao40/35bdce80-94d4-11e9-bf75-b42e9921e93e_xkb23.png\" style=\"vertical-align:middle\" />','<img  src=\"/tikuimages/9/2016/400/shoutiniao74/35bfca4f-94d4-11e9-8b2b-b42e9921e93e_xkb34.png\" style=\"vertical-align:middle\" />','<img  src=\"/tikuimages/9/2016/400/shoutiniao29/35c0dbc0-94d4-11e9-81b4-b42e9921e93e_xkb27.png\" style=\"vertical-align:middle\" />','<img  src=\"/tikuimages/9/2016/400/shoutiniao95/35c1ed30-94d4-11e9-a3ef-b42e9921e93e_xkb19.png\" style=\"vertical-align:middle\" />','','A','【解答】解：A、药品的称量方法：左物右码，具有腐蚀性药品放在玻璃器皿中称量，故A正确；<br />B、pH试纸的测定方法：不能直接浸入溶液中，要用玻璃棒蘸取待测液，滴到pH试纸上，故B错；<br />C、浓硫酸的稀释方法：将浓硫酸沿着容器壁慢慢倒入水中，并用玻璃棒不断搅拌，故C错；<br />D、收集气体的注意事项：用排空气法收集气体时，导管要伸到接近集气瓶的底部，不能直接靠在集气瓶底部，故D错．<br />故选A．','【分析】A、根据药品的称量方法考虑；B、根据pH试纸的测定方法考虑；C、根据浓硫酸的稀释方法考虑；D、根据收集气体的注意事项考虑．','选择题',3.00,'49adad78dcc1fbfe5cfb65970e0eb3ac',9,400,'称量器-托盘天平,浓硫酸的性质及浓硫酸的稀释,常用气体的收集方法,溶液的酸碱度测定','',2016,'32','2016•岱岳区模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842594,'<img  src=\"/tikuimages/9/0/400/shoutiniao84/35cc4d6e-94d4-11e9-b9c2-b42e9921e93e_xkb65.png\" style=\"vertical-align:middle;FLOAT:right;\" />某化学兴趣小组的同学对空气中氧气含量的测定实验进行探究．<br />（1）如图1所示装置和药品进行实验，实验中可观察到的现象是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，待装置冷却到室温后，打开止水夹看到<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，该反应的化学式表达式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）甲同学认为可用木炭代替红磷<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>测定空气中氧气的含量，乙同学&nbsp;认为不可以，其原因是：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．乙同学用图1实验装置测定的结果是：空气中氧气含量与正常值有较明显的偏差，其原因可能是（答出一种即可）：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）丙同学用镁条代替红磷来测定空气中氧气的含量．结果却发现倒吸的水量远远超过集气瓶 的<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">1</td></tr><tr><td>5</td></tr></table></span>．小组内同学帮她仔细检查，发现装置的气密性 及操作步骤均无问题．你认为造成此现象的原因可能是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（4）小组内同学反思上述实验的探究过程后认为：用燃烧法测定空气中氧气含量的实验时，在药品的<U>选择</U>或生成物的要求上应考虑的是：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（5）小组内同学还对该实验进行了如图2改进．（已知试管容积为45mL）实验的主要操作步骤如下：<br />①点燃酒精灯．<br />②撤去酒精灯，待试管冷却后松开弹簧夹．<br />③将少量红磷平装入试管中，将20mL的注射器 活塞置于10mL刻度处，并按图2中所示的连接方式固定好，再将弹簧夹紧橡皮管．<br />④读取注射器活塞的数据．<br />你认为正确的实验操作顺序是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填序号）．注射器活塞将从10mL刻度处慢慢前移到约为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>&nbsp;mL刻度处才停止．<br />（6）由图1实验还可推知反应后剩余气体的两点性质是：①物理性质<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>&nbsp;②化学性质<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','红磷燃烧放出大量白烟$###$烧杯中的水倒流进集气瓶，且上升到刻度“1”处$###$4P+5O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2P<SUB>2</SUB>O<SUB>5</SUB>$###$红磷$###$木炭和氧气反应生成二氧化碳气体，装置内的压强基本不变$###$装置漏气$###$镁条和空气中其他成分也发生了反应$###$药品本身是固体，能且只能和空气中氧气反应，且生成物不是气体$###$③①②④$###$1$###$难溶于水$###$不燃烧、不支持燃烧','【解答】解：（1）红磷燃烧时生成大量白烟，装置冷却到室温后，打开止水夹看到烧杯中的水倒流进集气瓶，且上升到刻度“1”处，反应的化学方程式为4P+5O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2P<SUB>2</SUB>O<SUB>5</SUB><br />（2）用木炭代替红磷时，木炭和氧气反应生成二氧化碳气体，装置内的压强基本不变；空气中氧气含量与正常值有较明显的偏差，其原因可能是装置漏气、红磷不足、装置没冷却就打开止水夹等．<br />（3）镁条不但可以和氧气反应，也可以和氮气反应，所以消耗的气体的体积要大于五分之一，进入广口瓶中水的体积也会超过五分之一；<br />（4）根据药品与氧气反应的产物是否能够产生压强差可以知道在选择药品时要保证药品本身是固体，能且只能和空气中氧气反应，且生成物不是气体；<br />（5）正确的实验操作顺序是：先将少量红磷平装入试管中，将20mL的注射器活塞置于10mL刻度处，并按上图2中所示的连接方式固定好，再将弹簧夹紧橡皮管，再点燃酒精灯，再撤去酒精灯，待试管冷却后松开弹簧夹，最后读取注射器活塞的数据，即顺序是③①②④；<br />因为试管容积为45mL，所以试管中氧气的体积为45mL×<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">1</td></tr><tr><td>5</td></tr></table></span>=9mL，待试管冷却后松开弹簧夹，注射器活塞将从10mL刻度处慢慢前移到约为1mL刻度处才停止．<br />（6）由于烧杯中的水倒流进集气瓶，且上升到刻度“1”处，而不是全部，说明剩余的气体具有难溶于水的物理性质，不燃烧、不支持燃烧的化学性质．<br />故答案为：<br />（1）红磷燃烧放出大量白烟；烧杯中的水倒流进集气瓶，且上升到刻度“1”处；4P+5O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2P<SUB>2</SUB>O<SUB>5</SUB>；<br />（2）木炭和氧气反应生成二氧化碳气体，装置内的压强基本不变；装置漏气；<br />（3）镁条和空气中其他成分也发生了反应；<br />（4）药品本身是固体，能且只能和空气中氧气反应，且生成物不是气体；<br />（5）③①②④；1；<br />（6）难溶于水；不燃烧、不支持燃烧．','【分析】本题考查的内容有常规性的实验现象的描述和化学式表达式的书写，实验误差分析，在此基础上加大了深度，分析木炭能否代替红磷以及镁带代替红磷燃烧产生的截然相反地误差的原因，这要求学生要从瓶内压强的变化去分析．对于用注射器改进后的仪器其刻度更容易读取，计算部分要注意被反应消耗掉的是试管内的氧气，不包含注射器内的部分．','书写',3.00,'046a0dbb78d62209f038f4a7b6403bfa',9,400,'测定空气里氧气含量的探究,书写化学方程式、文字表达式、电离方程式','',0,'37','',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842600,'下列说法正确的是（　　）','空气是由空气分子构成的','空气中的氧气只有点燃时才能发生化学反应','用过氧化氢制取氧气的反应属于分解反应','凡是氧化反应都是化合反应','','C','【解答】解：A、空气是由氮气分子、氧气分子等不同种分子构成的，没有空气分子，故A错误；<br />B、空气中的氧气性质比较活泼，在一定条件下能与很多种物质发生反应，条件不一定是点燃，故B错误；<br />C、用过氧化氢制取氧气的反应属于分解反应，故C正确；<br />D、氧化反应不一定都是化合反应，例如甲烷燃烧生成了二氧化碳和水，不属于化合反应，故D错误．<br />故选C．','【分析】A、根据空气的构成分析；<br />B、根据氧气的性质分析；<br />C、根据用过氧化氢制取氧气反应的特点分析；<br />D、根据氧化反应和化合反应的特点分析．','选择题',3.00,'faa0b9d6b0b5237cc955f14f7a36ff60',9,400,'氧气的化学性质,分子的定义与分子的特性,化合反应及其应用,氧化反应,反应类型的判定','招远市',2016,'35','2016春•招远市期中',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842603,'下列与生产生活相关的化学知识，正确的是（　　）','硬水不利于生活生产，软化硬水是物理变化','pH小于5.6的雨水为酸雨，能使无色酚酞溶液变红','厨房里管道煤气可能有泄露，用打火机检查漏气的位置','在汽油中加入适量乙醇作为汽车燃料，可以减少尾气污染','','D','【解答】解：<br />A、硬水不利于生活生产，软化硬水有新物质生成，是化学变化，故错误；<br />B、pH小于5.6的雨水为酸雨，不能使无色酚酞溶液变红，故错误；<br />C、管道煤气具有可燃性，泄露遇明火可能发生爆炸，不能用打火机检查漏气的位置，故错误；<br />D、乙醇具有可燃性，是一种较清洁的能源，汽油中加入适量乙醇作为汽车燃料，可节省石油资源、减少污染．故正确．<br />答案：D','【分析】A、根据软化硬水的原理解答；<br />B、根据无色酚酞遇酸性溶液不变色解答；<br />C、根据可燃性气体与空气混合后点燃可能发生爆炸进行分析判断．<br />D、根据乙醇具有可燃性，是一种较清洁的能源分析．','选择题',3.00,'bf3868dc1c304d4d22562642640bf8f2',9,400,'硬水与软水,酸雨的产生、危害及防治,化学变化和物理变化的判别,常用燃料的使用与其对环境的影响,防范爆炸的措施','',2016,'32','2016•福田区模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842607,'下列有关玻璃导气管的使用图示正确的是（省略夹持和加热装置）（　　）','<img  src=\"/tikuimages/9/2016/400/shoutiniao78/35ec0a70-94d4-11e9-a3d4-b42e9921e93e_xkb30.png\" style=\"vertical-align:middle\" /><br />&nbsp;润湿后旋转套入橡胶管','<img  src=\"/tikuimages/9/2016/400/shoutiniao68/35eddf30-94d4-11e9-aa11-b42e9921e93e_xkb23.png\" style=\"vertical-align:middle\" /><br />向下排空气法收集氢气','<img  src=\"/tikuimages/9/2016/400/shoutiniao42/35f0020f-94d4-11e9-98e0-b42e9921e93e_xkb85.png\" style=\"vertical-align:middle\" /><br />&nbsp;加热高锰酸钾制氧气','<img  src=\"/tikuimages/9/2016/400/shoutiniao21/35f1afc0-94d4-11e9-9525-b42e9921e93e_xkb39.png\" style=\"vertical-align:middle\" /><br />将氨气通入酚酞试液','','A|D','【解答】解：A、导管连接胶皮管时，先把导管一端湿润，然后稍用力转动使之插入胶皮管内，图中所示装置正确．<br />B、向下排空气法收集氢气时，为利于空气的排出，导管应伸入集气瓶的底部，图中所示装置错误．<br />C、加热高锰酸钾制氧气时，导管应刚刚露出塞子即可，图中所示装置错误．<br />D、将氨气通入酚酞试液时，应将导管伸入溶液中，图中所示装置正确．<br />故选：AD．','【分析】A、根据导管连接胶皮管的方法进行分析判断．<br />B、根据向下排空气法收集气体的注意事项，进行分析判断．<br />C、根据加热高锰酸钾制氧气的注意事项，进行分析判断．<br />D、根据用溶液中通入气体的方法，进行分析判断．','多选题',3.00,'ee098e6e4650494caf14f9679ad9ad42',9,400,'仪器的装配或连接,常用气体的收集方法,制取氧气的操作步骤和注意点','仪征市',2016,'37','2016•仪征市一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842609,'新装修房室内存在甲醛（CH<SUB>2</SUB>O）、苯（C<SUB>6</SUB>H<SUB>6</SUB>）等有害物质．下列叙述正确的是（　　）','甲醛分子中含有水分子','甲醛相对分子质量为30&nbsp;g','苯中碳元素与氢元素的质量比为1：1','90&nbsp;g甲醛与39&nbsp;g苯中的碳元素质量相当','','D','【解答】解：A．甲醛分子是由碳原子和氢原子、氧原子构成的，其中不含水分子，故错误；<br />B．相对分子质量的单位不是“g”而是“1”，通常省略不写，故错误；<br />C．苯（C<SUB>6</SUB>H<SUB>6</SUB>）中，碳元素与氢元素的质量比为：（12×6）：（1×6）=12：1，故错误；<br />D.90g甲醛中含碳元素的质量为：90g×<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">12</td></tr><tr><td>12+1×2+16</td></tr></table>×100%</span>=36g；39g苯中的碳元素质量为：39g×<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">12×6</td></tr><tr><td>12×6+1×6</td></tr></table>×100%</span>=36g，故正确．<br />故选D．','【分析】A．根据甲醛的分子结构来分析；<br />B．根据相对分子质量的单位来分析；<br />C．根据化合物中元素之相比的计算方法来分析；<br />D．根据化合物中某元素的质量=该化合物的质量×该元素的质量分数，进行分析解答．','选择题',3.00,'1592863b17640144e66a56bbb72d7adf',9,400,'化学式的书写及意义,相对分子质量的概念及其计算,元素质量比的计算,化合物中某元素的质量计算','',2016,'37','2016•南通一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842621,'<img  src=\"/tikuimages/9/2016/400/shoutiniao99/363c2640-94d4-11e9-8bce-b42e9921e93e_xkb77.png\" style=\"vertical-align:middle;FLOAT:right;\" />（2016•合肥校级模拟）我校化学兴趣小组测定空气中氧气的含量，并改进教材方案．<br />【实验方案】选用实验容积为40mL的试管作反应容器，通过导管与容积为60mL且润滑性很好的针筒注射器组成如图的实验装置．（白磷所占体积与导管内的气体体积忽略不计）<br />【实验过程】（1）实验前，打开弹簧夹，将注射器的活塞前湍从20mL刻度处推至15mL刻度处，然后松手，若活塞仍能返回至20mL刻度处，则说明<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）若先夹紧弹簧夹，用酒精灯加热白磷，可观察现象<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．写出化学反应方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，燃烧结束，等到试管冷却后再松形式弹簧夹，预测活塞移动到<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>mL处．<br />（3）若不使用弹簧夹，用酒精灯加热白磷，充分反应直至燃烧结束，试管冷却．可观察到活塞的移动现象为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />【交流与反思】实际实验中，活塞停止的刻度始终大于理论刻度值．在老师的指导下，同学们利用铁锈原理再次测量空气中氧气的体积分数．<br />【查阅资料】①脱氧剂的主要成分为铁粉、少许活性炭、少量氯化钠和水．<br />②铁锈的主要成分化学式为：Fe<SUB>2</SUB>O<SUB>3</SUB>．<br />【实验原理】铁生锈消耗<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>和水，生成固体铁锈；氧化钠可以加快铁锈的速率．<br />【实验过程】（1）取一个集气瓶，其实际容积是150mL；取一个量筒，内盛有100mL的水．<br />（2）室温时，先夹紧弹簧夹，将一定量的脱氧剂悬吊在橡皮塞反面，再将橡皮塞旋紧到集气瓶口，将导气管插入量筒水中．2个小时后打开弹簧夹，观察到量筒内的现象<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）当量筒内水面高度不再变化，且温度恢复至室温时，读出水面在71mL刻度处．<br />【数据分析】由上述实验数据可以算出，空气中氧气的体积分数是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>%．理论上空气中氧气体积分数为21%，请你分析产生这一差异的可能原因是：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','装置的气密性好$###$白磷燃烧，产生大量白烟，放热$###$4P+5O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2P<SUB>2</SUB>O<SUB>5</SUB>$###$12$###$先向右移动，再向左移动$###$氧气$###$液面下降水进入到集气瓶内$###$19.3$###$集气瓶内还有氧气','【解答】解：<br />（1）打开弹簧夹，将注射器活塞前沿从20mL刻度处推至15mL刻度处，然后松开手，若活塞仍能返回至20mL刻度处，则说明整个装置都不漏气；<br />（2）若先夹紧弹簧夹，用酒精灯加热白磷，可观察到：白磷燃烧，产生大量白烟，放热；白磷燃烧的化学方程式为：4P+5O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2P<SUB>2</SUB>O<SUB>5</SUB>．这时白磷燃烧消耗掉的只是试管内的氧气为40mL×<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">1</td></tr><tr><td>5</td></tr></table></span>=8mL，所以活塞应停在20mL-8mL=12mL的刻度处；<br />（3）若不使用弹簧夹，用酒精灯加热白磷，充分反应直至燃烧结束，试管冷却．可观察到活塞的移动现象为先向右移动，再向左移动；<br />【查阅资料】<br />铁生锈消耗氧气和水，生成固体铁锈；<br />（2）如图所示的装置内一会儿观察到量筒内的现象液面下降，水进入到集气瓶内，这是因为集气瓶内压强小于外界大气压；<br />（3）当量筒内水面高度不再变化，且温度恢复至室温时，读出水面在71mL刻度处．由上述实验数据可以算出，空气中氧气的体积分数是<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">100mL-71mL</td></tr><tr><td>150mL</td></tr></table></span>×100%=19.3%；本实验的测量结果与理论值不一致，原因可能是脱氧剂中有部分铁粉在实验前已被氧化；脱氧剂中的内层的铁粉没有与氧气接触．集气瓶内还有氧气．<br />故答案为：<br />（1）装置的气密性好；<br />（2）白磷燃烧，产生大量白烟，放热；4P+5O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2P<SUB>2</SUB>O<SUB>5</SUB>；12；<br />（3）先向右移动，再向左移动；<br />【实验原理】氧气；<br />（2）液面下降水进入到集气瓶内；<br />（3）19.3%；集气瓶内还有氧气．','【分析】本题是测定氧气在空气中的体积分数的实验探究，要把空气中的氧气消耗掉，在本实验中用的是过量的白磷，燃烧后的生成物是固体，不需要占很大的空间．本实验中用的是40mL的试管与实际容积为60mL且润滑性很好的针筒注射器，并且活塞在20mL处，有助于反应结束后计算体积．要使本实验成功，必须注意以下几点：①装置的气密性好；②白磷足量；③必须冷却到室温再读数．还要注意防止试管受热不均或气体体积受热膨胀过大，从而造成试管炸裂．','书写',3.00,'fae93b81f7c369ac674bd89714f39c35',9,400,'测定空气里氧气含量的探究,书写化学方程式、文字表达式、电离方程式','',2016,'32','2016•合肥校级模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842623,'一个碳---12的原子的实际质量为m&nbsp;kg，计算实际质量为2mkg的另一种元素的原子相对原子质量<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','24','【解答】解：碳-12原子的实际质量为mkg，则实际质量为2mkg的另一种元素的原子的相对原子质量为<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">2mkg</td></tr><tr><td style=\"padding-top:1px;font-size:90%\">mkg×<table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">1</td></tr><tr><td>12</td></tr></table></td></tr></table></span>=24．<br />答：实际质量为2mkg的另一种元素的原子相对原子质量是24．','【分析】相对原子质量是以一种碳原子质量的<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">1</td></tr><tr><td>12</td></tr></table></span>标准，其他原子的质量跟它相比较所得到的比，据此进行分析解答．','填空题',3.00,'22fe50585abb26df519f3ff5ae5a14ab',9,400,'相对原子质量的概念及其计算方法','',2015,'37','2015秋•烟台校级月考',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842625,'<img  src=\"/tikuimages/9/2016/400/shoutiniao31/36443c8f-94d4-11e9-b83c-b42e9921e93e_xkb74.png\" style=\"vertical-align:middle;FLOAT:right\" />（2016•丹阳市模拟）社会发展需要充足的能源，化石燃料等不可再生能源将日趋枯竭．世界各国人民的节能意识不断增强（如图为中国节能标志），科学家也在研究开发新能源．请回答：<br />（1）有待继续开发、利用的新能源有<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（2）氢气具有热值高、原料不受限制且<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>的特点，因此被认为是最清洁的能源．','','','','','','太阳能等$###$燃烧产物是水，无污染','【解答】解：（1）有待继续开发、利用的新能源包括太阳能、风能、核能、地热能、潮汐能等．<br />（2）工业上制取氢气是通过电解水得到的，而地球上水资源丰富，可以从水中提取氢气，资源广泛；氢气的燃烧值高；氢气燃烧产物是水，不污染环境，因此被认为是最清洁的能源．<br />故答案为：（1）太阳能等；（2）燃烧产物是水，无污染．','【分析】（1）新能源是指无污染、可以持续利用的能源，包括太阳能、风能、核能、地热能、潮汐能等，进行分析解答即可．<br />（2）根据氢能源的优点，进行分析解答即可．','填空题',3.00,'15630caa47b7ddb5ea2952860f8c7287',9,400,'资源综合利用和新能源开发,氢气的用途和氢能的优缺点','丹阳市',2016,'32','2016•丹阳市模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842627,'2015年“世界水日”“中国水周”宣传主题是“节约水源、保障水安全”．下列做法符合这一主题的是（　　）','水在自然界不断循环，取之不尽，用之不竭','工厂排放的废水直接通入地下，避免污染地表水','公园浇花采用喷灌和滴灌','为了节约用水，将工厂排放的废水直接浇灌庄稼','','C','【解答】解：A、水在自然界不断循环，但分布不均匀，淡水资源是有限的，并不是取之不尽，用之不竭的，故A错误；<br />B、工厂排放的废水直接通入地下，会造成地下水的污染，故B错误；<br />C、公园浇花采用喷灌和滴灌，能节约用水，故C正确；<br />D、工厂排放的废水，含有有毒的物质，会造成农业上减产或绝产、水体的污染等，故D错误．<br />故选C．','【分析】根据水资源的分布、爱护资源的措施分析判断．','选择题',3.00,'7ed9ae6a4b19a082ed5bee5b328e2bfe',9,400,'水资源状况,水资源的污染与防治','',2015,'32','2015•当涂县模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842640,'属于浊液的是（　　）','糖水','盐水','油水','汽水','','C','【解答】解：A、糖水是均一、稳定的混合物，属于溶液，故选项错误．<br />B、盐水是均一、稳定的混合物，属于溶液，故选项错误．<br />C、油难溶于水，油放到水里是小液滴悬浮于液体中形成的混合物，油水属于乳浊液；故选项正确．<br />D、汽水是均一、稳定的混合物，属于溶液，故选项错误．<br />故选：C．','【分析】不溶性的固体小颗粒悬浮于液体中形成的混合物是悬浊液；小液滴分散到液体中形成的混合物是乳浊液；一种或几种物质分散到另一种物质里，形成的均一稳定的混合物是溶液．','选择题',3.00,'449be062859a8a0f0440f84614b52468',9,400,'悬浊液、乳浊液的概念及其与溶液的区别','',2016,'37','2016•浦东新区二模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842686,'科学研究证明，物质是原子、分子或离子等基本粒子构成的，请填空：<br />（1）应用最广泛的铁是由<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>构成的．<br />（2）空气中含量最多的氮气是由<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>构成的．<br />（3）调味用的氯化钠是由<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>构成的．','','','','','','铁原子$###$氮分子$###$氯离子和钠离子','【解答】解：（1）铁属于金属单质，是由铁原子直接构成的．<br />（2）氮气属于气态非金属单质，是由氮分子构成的．<br />（3）氯化钠是由钠离子和氯离子构成的．<br />故答案为：（1）铁原子；（2）氮分子；（3）氯离子和钠离子．','【分析】根据金属、大多数固态非金属单质、稀有气体等由原子构成；有些物质是由分子构成的，气态的非金属单质和由非金属元素组成的化合物，如氢气、水等；有些物质是由离子构成的，一般是含有金属元素和非金属元素的化合物，如氯化钠，进行分析解答即可．','填空题',3.00,'7c1640c28c01651acc2261d564318df6',9,400,'物质的构成和含量分析','',2011,'33','2011秋•湖南校级期末',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842687,'海洋资源的开发对人类社会进步起着重要的作用，请你回答问题：<br />（1）要检验海水中含量最多的离子步骤、现象和结论是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）人们在海底发现了一种新型矿产资源--天然气水合物，也叫<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，它被认为是一种清洁能源，原因是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）某厂采用下列技术进行海水淡化，过程如下：<br />步骤Ⅰ：利用风力发电机作为动力，将海水置入海水灌中，海水经过多介质过滤器、活性吸附塔等仪器．<br />步骤Ⅱ：将上步所得海水注入反渗透海水淡化机中，利用反渗透膜将海水淡化，同时得到的高盐度的浓缩海水，将高盐度的海水进一步晒制生产出海盐．<br />请你回答下列相关问题：<br />①步骤Ⅰ活性炭的主要作用是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />②步骤Ⅱ中得到淡水的原理是利用海水中各成份的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>不同分离出淡水．<br />③从能源和资源的角度来看，你认为该技术的优点分别体现在<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（答两点）．<br />④淡化海水还可采用蒸馏法，它是利用海水中各成份的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>不同分离出淡水．','','','','','','取少量海水于试管中，加入少量硝酸银溶液，若产生白色沉淀，则证明含有Cl<SUP>-</SUP>$###$可燃冰$###$燃烧后几乎不产生任何残渣和废气$###$吸附色素和异味$###$颗粒大小$###$风力发电，节约能源$###$沸点','【解答】解：（1）海水中含有的最多的离子是氯离子，要检验海水中氯离子的步骤、现象和结论是：取少量海水于试管中，加入少量硝酸银溶液，若产生白色沉淀，则证明含有Cl<SUP>-</SUP>．<br />（2）人们在海底发现了一种新型矿产资源--天然气水合物，也叫可燃冰，它被认为是一种清洁能源，原因是燃烧后几乎不产生任何残渣和废气．<br />（3）①由于活性炭具有吸附性，步骤Ⅰ活性炭的主要作用是吸附色素和异味．<br />②步骤Ⅱ中得到淡水的原理是利用海水中各成份的颗粒大小不同分离出淡水．<br />③从能源和资源的角度来看，你认为该技术的优点分别体现在风力发电，节约能源等．<br />④淡化海水还可采用蒸馏法，它是利用海水中各成份的沸点不同分离出淡水．<br />故答为：（1）取少量海水于试管中，加入少量硝酸银溶液，若产生白色沉淀，则证明含有Cl<SUP>-</SUP>；（2）可燃冰，燃烧后几乎不产生任何残渣和废气；（3）①吸附色素和异味，②颗粒大小，③风力发电，节约能源，④沸点．','【分析】（1）根据氯离子的检验方法分析回答；<br />（2）根据天然气水合物的俗称和燃烧的产物分析回答；<br />（3）①根据活性炭具有吸附性分析．<br />②根据过滤的原理分析回答．<br />③根据风力发电的特点分析回答；<br />④根据蒸馏法分离混合物的原理分析回答．','填空题',3.00,'c9cfac0e5a1e3356e3d4847f4ffc87a6',9,400,'证明盐酸和可溶性盐酸盐,对海洋资源的合理开发与利用','',2016,'32','2016•烟台校级模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842688,'物质发生化学变化的前后，总质量是否发生改变？是增加、减少还是不变？<br />小安、小静按下面的步骤进行探究：<br />[提出假设]物质发生化学变化的前后总质量不变．<br />[制定计划]小安设计的实验装置如图A所示；小莉设计的实验装置如图B所示．他们在反应前后都进行了规范的操作、正确的称量和细致的观察．<br /><img  src=\"/tikuimages/9/0/400/shoutiniao92/3709885e-94d4-11e9-96ae-b42e9921e93e_xkb32.png\" style=\"vertical-align:middle\" /><br />[收集证据]小安在整个过程中观察到的现象是（1）<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />B中物质发生反应的化学方程式是（2）<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />[交流与评价]<br />小安认为：在化学反应前后，生成物的总质量与反应物的总质量是相等的；<br />小静认为：在化学反应前后，生成物的总质量与反应物的总质量是不相等的．<br />你认为（3）的结论正确，导致另一个实验结论错误的原因是：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />请从原子的角度分析你认为正确结论的原因是：（4）<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />[拓展与延伸]使用上述实验B装置，请你选择另外两种药品进行实验达到实验目的，这两种药品可以是（5）<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>和<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．（填化学式）','','','','','','红磷燃烧，产生大量白烟，反应后，天平仍然平衡$###$Na<SUB>2</SUB>CO<SUB>3</SUB>+2HCl=2NaCl+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑$###$有气体生成，逸散到空气中$###$化学反应前后原子的种类、数目、质量都不变$###$NaCl$###$AgNO<SUB>3</SUB>','【解答】解：<br />[收集证据]<br />（1）由图示可知发生的反应是红磷燃烧，产生大量白烟，反应后，天平仍然平衡；<br />（2）根据反应物是碳酸钠和盐酸及复分解反应的原理可知，生成物是氯化钠、水和二氧化碳，故答案为：Na<SUB>2</SUB>CO<SUB>3</SUB>+2HCl=2NaCl+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑<br />（3）小安的结论是正确的，因为在质量守恒定律中，反应物的总质量等于生成物的总质量，小静的实验有气体生成，逸散到空气中，因而漏掉生成物的气体的质量，可将反应物放在密闭容器中进行实验；<br />（4）从原子的角度解释质量守恒定律，是因为反应前后原子的种类，数目，质量都不改变；故答案为：化学反应前后原子的种类、数目、质量都不变；<br />（5）择药品在敞口容器中验证质量守恒定律时，要求两种物质要能反应，注意不生成气体物质或气体参加反应，故可以选NaCl、AgNO<SUB>3</SUB>或H<SUB>2</SUB>SO<SUB>4</SUB>、Ba（OH）<SUB>2</SUB>（答案合理即可）．<br />故答案为：<br />（1）红磷燃烧，产生大量白烟，反应后，天平仍然平衡；<br />（2）Na<SUB>2</SUB>CO<SUB>3</SUB>+2HCl=2NaCl+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑；<br />（3）有气体生成，逸散到空气中；<br />（4）化学反应前后原子的种类、数目、质量都不变；<br />（5）NaCl、AgNO<SUB>3</SUB>','【分析】（1）根据红磷燃烧的现象分析解答；<br />（2）根据碳酸钠和盐酸反应生成氯化钠、水和二氧化碳解答；<br />（3）质量守恒定律适用所有的化学变化，因此在表达质量守恒定律时，一定不要漏掉反应物或生成物的质量，特别是气体物质，据此分析；<br />（4）从原子的角度解释质量守恒定律，主要是原子的种类，数目，质量都不改变角度分析；<br />（5）选择另外两种药品验证质量守恒定律时，除了两种物质要反应外，注意不生成气体物质或气体参加反应．','书写',3.00,'23917cef2499708566237bf973a4ff03',9,400,'质量守恒定律的实验探究,书写化学方程式、文字表达式、电离方程式','',0,'37','',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842695,'物质发生化学变化的本质特征是（　　）','有发光发热的现象','有气体放出','有新物质产生','有颜色变化','','C','【解答】解：化学变化的基本特征是有新物质生成，观察选项，故选C．','【分析】根据已有的知识进行分析解答，化学变化的基本特征是有新物质生成，据此解答．','选择题',2.00,'44df010e8577d265ba5bf194c85e2236',9,400,'化学变化的基本特征','',2014,'37','2014秋•宁县月考',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842706,'下列气体与空气混合后遇明火，可能发生爆炸的是<br />A．氧气B．氮气C．甲烷&nbsp;&nbsp;&nbsp;&nbsp;D．<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','一氧化碳','【解答】解：<br />氧气具有助燃性，不具有可燃性、氮气不具有可燃性，遇明火不可能发生爆炸；<br />甲烷属于可燃物，与空气混合后遇明火可能发生爆炸．<br />一氧化碳具有可燃性，与空气混合后遇明火可能发生爆炸．<br />故答案为：C；一氧化碳．','【分析】爆炸是指可燃物在有限的空间里急剧燃烧，放出大量的热，生成的气体急剧膨胀，发生爆炸','填空题',3.00,'a171f7a791d6ea0776be8a881105925a',9,400,'燃烧、爆炸、缓慢氧化与自燃','',2016,'37','2016•上饶二模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842713,'阅读材料，回答问题<br />材料1：臭氧（化学式为O<SUB>3</SUB>）是淡蓝色气体，大气中的臭氧层能有效阻挡紫外线，保护地球的生存环境，但目前南极出现臭氧层空洞，并有继续扩大的趋势．<br />材料2：复印机在工作时，会在高压放电产生一定浓度的臭氧．长期吸入大量臭氧会引起口干舌燥，咳嗽等不适症状，还可能诱发中毒性肺气肿．<br />材料3：臭氧发生器是在高压电极的作用下将空气的氧气转化为臭氧的装置．利用臭氧的强氧化性，可将其应用于游泳池、生活用水、污水的杀菌和消毒．<br />（1）请总结臭氧的有关知识<br />①物理性质：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；②化学性质：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />③用途：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（2）请从分子构成的角度，指出氧气和臭氧的不同点：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（3）写出材料3中氧气转化为臭氧的化学方程式：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','淡蓝色气体$###$强氧化性$###$游泳池、生活用水、污水的杀菌和消毒$###$构成它们各自分子的氧原子的个数不同$###$3O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;放电&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2O<SUB>3</SUB>','【解答】解：<br />（1）①物质的颜色、状态属于物质的物理性质．故填：淡蓝色气体．<br />②氧化性属于物质的化学性质．故填：具有强氧化性．<br />③用途：用于游泳池、生活用水、污水的杀菌和消毒．<br />（2）氧气和臭氧的分子构成不同，故填：构成它们各自分子的氧原子的个数不同．<br />（3）氧气转化为臭氧的化学方程式为：3O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;放电&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2O<SUB>3</SUB><br />答案：<br />（1）①淡蓝色气体；②强氧化性；③游泳池、生活用水、污水的杀菌和消毒；<br />（2）构成它们各自分子的氧原子的个数不同<br />（3）3O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;放电&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2O<SUB>3</SUB>','【分析】（1）物理性质主要包括：色态味、密度、溶解性等等；常见的化学性质有：氧化性、还原性、毒性、酸碱性、稳定性等等．<br />（2）分子由原子构成，包括原子的种类、原子的个数．<br />（3）化学方程书写基础是确定反应物和生成物，然后用化学式表示出来．','书写',3.00,'7d22b9a6369884c421f45e8419f4cf87',9,400,'分子的定义与分子的特性,氧元素组成的单质,化学性质与物理性质的差别及应用,书写化学方程式、文字表达式、电离方程式','山南地区',2014,'37','2014•山南地区校级二模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842715,'下图是某课外活动小组的同学设计的实验室制取二氧化碳并检验其性质的装置示意图，请回答下列问题：<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao26/375cff91-94d4-11e9-904c-b42e9921e93e_xkb80.png\" style=\"vertical-align:middle\" /><br />（1）代号a所指仪器的名称是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />（2）将A中产生的气体通入到B中一会儿，未见其中变浑浊．李祥同学认为这是浓盐酸挥发出的HCl气体干扰所致．他用化学方程式表示为：Ca（OH）<SUB>2</SUB>+CO<SUB>2</SUB>═CaCO<SUB>3</SUB>↓+H<SUB>2</SUB>O；CaCO<SUB>3</SUB>+2HCl═CaCl<SUB>2</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑&nbsp;你的解释还可更简捷，请用一个化学方程式表示：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（3）张华同学用烧瓶收集A中产生的气体后，向其中加入过量无色透明的M溶液组成C装置，振荡，发现烧瓶中的气球逐渐变大，且溶液始终无色透明，则M可能是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>&nbsp;（填写一种物质的化学式）；<br />（4）请设计一个方案，验证C装置中反应的生成物中含有CO<SUB>3</SUB><SUP>2-</SUP>．简述实验过程：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','试管$###$2HCl+Ca（OH）<SUB>2</SUB>=CaCl<SUB>2</SUB>+2H<SUB>2</SUB>O；$###$KOH（或NaOH）$###$向其中加入BaCl<SUB>2</SUB>溶液，若生成白色沉淀，即可证明其中含CO<SUB>3</SUB><SUP>2-</SUP>（或：向其中加入过量（或足量）的稀盐酸，若有气泡冒出，则可证明其中含CO<SUB>3</SUB><SUP>2-</SUP>','【解答】解：（1）试管是常用的反应容器；<br />（2）将A中产生的气体通入到B中一会儿，未见其中变浑浊，更简捷的解释是：盐酸和氢氧化钙溶液反应生成氯化钙和水，配平即可；反应的方程式为：2HCl+Ca（OH）<SUB>2</SUB>=CaCl<SUB>2</SUB>+2H<SUB>2</SUB>O&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<br />（3）振荡，发现烧瓶中的气球逐渐变大，且溶液始终无色透明，说明二氧化碳与M溶液反应无变化，M溶液是NaOH（或KOH）溶液；<br />（4）要证明生成物中含有CO<SUB>3</SUB><SUP>2-</SUP>，向其中加入BaCl<SUB>2</SUB>溶液或足量的稀盐酸都可，即向其中加入BaCl<SUB>2</SUB>溶液，若生成白色沉淀，即可证明其中含CO<SUB>3</SUB><SUP>2-</SUP>；故答案为：向其中加入BaCl<SUB>2</SUB>溶液，若生成白色沉淀，即可证明其中含CO<SUB>3</SUB><SUP>2-</SUP>（或：向其中加入过量（或足量）的稀盐酸，若有气泡冒出，则可证明其中含CO<SUB>3</SUB><SUP>2-</SUP>（或其他合理表述）<br />故答案为：（1）试管；<br />（2）2HCl+Ca（OH）<SUB>2</SUB>=CaCl<SUB>2</SUB>+2H<SUB>2</SUB>O；<br />（3）KOH（或NaOH）；<br />（4）向其中加入BaCl<SUB>2</SUB>溶液，若生成白色沉淀，即可证明其中含CO<SUB>3</SUB><SUP>2-</SUP>&nbsp;（或：向其中加入过量（或足量）的稀盐酸，若有气泡冒出，则可证明其中含CO<SUB>3</SUB><SUP>2-</SUP>&nbsp;（或其他合理表述）．','【分析】实验室制取CO<SUB>2</SUB>，是在常温下，用碳酸钙和盐酸互相交换成分生成氯化钙和水和二氧化碳，因此不需要加热．二氧化碳能溶于水，密度比空气的密度大，因此只能用向上排空气法收集．将A中产生的气体通入到B中一会儿，未见其中变浑浊，更简捷的解释是：盐酸和氢氧化钙溶液反应生成氯化钙和水，配平即可；振荡，发现烧瓶中的气球逐渐变大，且溶液始终无色透明，说明二氧化碳与M溶液反应无变化，M溶液是NaOH（或KOH）溶液；要证明生成物中含有CO<SUB>3</SUB><SUP>2-</SUP>，向其中加入BaCl<SUB>2</SUB>溶液或足量的稀盐酸都可．','书写',3.00,'5c0d954396da19b5be41f067216d63de',9,400,'证明碳酸盐,二氧化碳的实验室制法,二氧化碳的化学性质,书写化学方程式、文字表达式、电离方程式','',2016,'32','2016•保定模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842720,'我国科学家屠呦呦发现了青蒿素（C<SUB>15</SUB>H<SUB>22</SUB>O<SUB>5</SUB>）药物，从而拯救了百万疟疾患者的生命，因此获得2015年诺贝尔医学奖．下列关于青蒿素的说法正确的是（　　）','青蒿素是有机高分子化合物','青蒿素的相对分子质量为282','青蒿素中含量最多元素是氢元素','青蒿素在高温下性质稳定','','B','【解答】解：A．C<SUB>15</SUB>H<SUB>22</SUB>O<SUB>5</SUB>的相对分子质量为12×15+1×22+16×5=282．而有机高分子化合物的相对分子质量大到几万甚至几十万，故错误；<br />B．C<SUB>15</SUB>H<SUB>22</SUB>O<SUB>5</SUB>的相对分子质量为12×15+1×22+16×5=282．故正确；<br />C．青蒿素中碳、氢、氧三种元素的质量比为（12×15）：（1×22）：（16×5）=90：11：40，可见其中碳元素的质量分数最大，故错误；<br />D．青蒿素在该温度下易分解，化学性质不稳定，故错误．<br />故选B．','【分析】A．根据有机高分子化合物的概念来分析；<br />B．根据相对分子质量的计算方法来分析；<br />C．根据化合物中元素的质量比来分析；<br />D．根据青蒿素的性质来分析．','选择题',3.00,'0bdb667e60158bc33456c1cd9da06b1f',9,400,'有机物的特征、分类及聚合物的特性,化学式的书写及意义,相对分子质量的概念及其计算','',2016,'35','2016春•睢宁县期中',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842734,'下列说法中，不正确的是（　　）','在其它条件不变时，CO<SUB>2</SUB>的溶解度随压强的升高而增大','碳具有可燃性，在常温下其化学性质比较活泼','一氧化碳和木炭都具有还原性','CO<SUB>2</SUB>能溶于水，CO难溶于水','','B','【解答】解：A、在其它条件不变时，CO<SUB>2</SUB>的溶解度随压强的升高而增大，正确；<br />B、碳具有可燃性，碳原子最外层电子数是4个，所以在常温下其化学性质比较稳定，故错误；<br />C、一氧化碳和木炭均具有还原性，可用来冶炼金属，正确；<br />D、1体积的水大约能溶解1体积的二氧化碳，能溶于水，而一氧化碳难溶于水，正确；<br />故选：B．','【分析】A、气体的溶解度随压强的增大而增大；<br />B、据碳的化学性质分析解答；<br />C、据一氧化碳和木炭均具有还原性分析解答；<br />D、据一氧化碳和二氧化碳的溶解性分析解答．','选择题',3.00,'1d729974b7ec9f184911df25db9cb444',9,400,'二氧化碳的物理性质,一氧化碳的物理性质,一氧化碳的化学性质,气体溶解度的影响因素,碳的化学性质','',2016,'32','2016•海珠区模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842755,'下列常见仪器的使用正确的是（　　）','试管可以直接加热','量筒可用作反应的容器','温度计可用于搅拌溶液','用燃着的酒精灯引燃另一只酒精灯','','A','【解答】解：A、试管可直接加热，故正确．<br />B、量筒只能作为量液体体积的仪器，不能最为反应容器或配制的容器，故错误．<br />C、温度计只能用于测量温度，如果用于搅拌可能会打破温度计，故错误．<br />D、点燃酒精灯要用火柴点燃，禁止用酒精灯引燃另一只酒精灯，以防引起火灾，故错误；<br />故选A．','【分析】A、试管可直接加热；<br />B、量筒不可用作反应容器；<br />C、温度计只能用于测量温度；<br />D、根据酒精灯的使用方法分析．','选择题',3.00,'41062a183703ec822fb7adefcb2f1a07',9,400,'用于加热的仪器,测量容器-量筒,加热器皿-酒精灯','',2016,'37','2016•番禺区一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842768,'金属在生活中应用广泛．<br />（1）最初输水管用铸铁，目前多用塑料管和塑料金属复合管，输送热水一般用铜管．下列管材中，属于有机合成材料的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填字母序号，下同）．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao85/37f76acf-94d4-11e9-aa3f-b42e9921e93e_xkb79.png\" style=\"vertical-align:middle\" /><br />（2）铁生锈的条件是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．将生锈的铁片放在硫酸和硫酸铜的混合溶液中，可能发生的置换反应的化学方程式是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）金属的回收和利用是保护金属资源的一种有效途径．现将一定量铁粉和铜粉的混合物放入硝酸银溶液中，反应结束后，对剩余固体成分的判断正确的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />A．肯定有银　　　　　B．肯定有铜　　　　　　　　C．可能有铁　　　　　　　D．可能是银和铁两种．','','','','','','C$###$与水和氧气接触$###$Fe+H<SUB>2</SUB>SO<SUB>4</SUB>=FeSO<SUB>4</SUB>+H<SUB>2</SUB>↑$###$AC','【解答】解：（1）图示的三种管材中铸铁管、铜管都属于金属材料，而塑料管属于有机合成材料；<br />（2）铁生锈需要使铁在潮湿的空气中同时与氧气和水接触；将生锈的铁片放在硫酸和硫酸铜的混合溶液中，当铁锈反应完后，铁能与硫酸、硫酸铜发生置换反应，例如铁与硫酸发生的置换反应的方程式是：Fe+H<SUB>2</SUB>SO<SUB>4</SUB>=FeSO<SUB>4</SUB>+H<SUB>2</SUB>↑．<br />（3）根据三种金属活动性强弱关系铁＞铜＞银，将铁粉和铜粉的混合物放入硝酸银溶液中，铁粉、铜粉都能与硝酸银发生反应得到银，铁粉先反应，因此剩余固体中一定含有银；因未说明混合物的用量，因此对于固体混合物是否出现剩余则无法准确判断，只能得到剩余固体中可能含有铁、铜的结论，但根据铁和铜的活动性关系，若剩余固体中有铁，则也一定含有铜；因此肯定有银、可能有铁的判断是正确的，而肯定是铜或可能是银和铁的说法不正确；<br />故答为：（1）C；（2）与水和氧气接触，Fe+H<SUB>2</SUB>SO<SUB>4</SUB>=FeSO<SUB>4</SUB>+H<SUB>2</SUB>↑；（3）AC．','【分析】（1）合成材料包括塑料、合成纤维、合成橡胶三大类；<br />（2）根据对铁生锈的认识，说明铁生锈所需满足的条件；<br />（3）根据三种金属活动性强弱关系铁＞铜＞银，对将铁粉和铜粉的混合物放入硝酸银溶液中可能出现的变化进行分析，判断反应结束后剩余固体成分．','书写',3.00,'68bef790295e5ce3e6d4eefda795e985',9,400,'金属的化学性质,金属锈蚀的条件及其防护,书写化学方程式、文字表达式、电离方程式,复合材料、纳米材料','',2016,'35','2016春•江津区校级期中',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842801,'实验设计是化学实验的重要环节，用对比实验方法探究二氧化碳的性质，请根据下列实验要求回答相关问题，<br /><table class=\"edittable\"><TBODY><TR><td width=258>实验一</TD><td width=262>实验二</TD></TR><TR><td><img  src=\"/tikuimages/9/2016/400/shoutiniao90/389f1c80-94d4-11e9-86cb-b42e9921e93e_xkb84.png\" style=\"vertical-align:middle\" /></TD><td><img  src=\"/tikuimages/9/2016/400/shoutiniao66/38a3b05e-94d4-11e9-9ab0-b42e9921e93e_xkb17.png\" style=\"vertical-align:middle\" /></TD></TR></TBODY></TABLE>【实验一】振荡3个矿泉水塑料瓶，观察到塑料瓶变瘪的程度为A＞B＞C，其中变浑浊的瓶内发生反应的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；对比A瓶与<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“B”或“C”）瓶的实验现象，可证明CO<SUB>2</SUB>能与NaOH发生反应．<br />【实验二】观察到C装置中发生的现象是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，结论是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（用化学方程式表示）．','','','','','','Ca（OH）<SUB>2</SUB>+CO<SUB>2</SUB>=CaCO<SUB>3</SUB>↓+H<SUB>2</SUB>O$###$C$###$干燥的石蕊试纸无变化，湿润的蓝色石蕊试纸变红色$###$CO<SUB>2</SUB>+H<SUB>2</SUB>O=H<SUB>2</SUB>CO<SUB>3</SUB>','【解答】解：（1）用对比实验方法探究二氧化碳的性质：实验一中振荡3个矿泉水塑料瓶，观察到塑料瓶变瘪的程度为A＞B＞C，其中变浑浊的瓶内发生的反应是：二氧化碳与氢氧化钙反应生成碳酸钙白色沉淀和水，对比A瓶与C瓶的实验现象，可证明CO<SUB>2</SUB>能与NaOH发生反应；故答案为：Ca（OH）<SUB>2</SUB>+CO<SUB>2</SUB>=CaCO<SUB>3</SUB>↓+H<SUB>2</SUB>O；C；<br />（2）实验二观察到C装置中发生的现象是：干燥的石蕊试纸无变化，湿润的蓝色石蕊试纸变红色；因为二氧化碳与水反应生成了碳酸；故答案为：干燥的石蕊试纸无变化，湿润的蓝色石蕊试纸变红色；CO<SUB>2</SUB>+H<SUB>2</SUB>O=H<SUB>2</SUB>CO<SUB>3</SUB>；','【分析】二氧化碳的化学性质有：既不能燃烧也不能支持燃烧，也不供给呼吸；能与水反应生成碳酸；能使澄清的石灰水变浑浊．用对比实验方法探究二氧化碳的性质：<br />实验一中振荡3个矿泉水塑料瓶，观察到塑料瓶变瘪的程度为A＞B＞C，其中变浑浊的瓶内发生的反应是：二氧化碳与氢氧化钙反应生成碳酸钙白色沉淀和水，对比A瓶与C瓶的实验现象，可证明CO<SUB>2</SUB>能与NaOH发生反应．','书写',3.00,'555769d56aa96633467efa1105b04198',9,400,'探究二氧化碳的性质,书写化学方程式、文字表达式、电离方程式','',2016,'32','2016•抚州校级模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842828,'下列在化学反应只属于氧化反应但不属化合反应的是（　　）','水<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td style=\"font-size: 90%\">通电</td></tr><tr><td><div hassize=\"9\"><div style=\"background: url(\'/upimages/images/part/8594L.png\') repeat-x; width:1px; height:5px;float:left;overflow:hidden\" muststretch=\"h\"></div><div style=\"background: url(\'/upimages/images/part/8594.png\') no-repeat; width:9px; height:5px;float:left;overflow:hidden\"></div></div></td></tr></table></span>氧气+氢气','木炭+氧气<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td style=\"font-size: 90%\">点燃</td></tr><tr><td><div hassize=\"9\"><div style=\"background: url(\'/upimages/images/part/8594L.png\') repeat-x; width:1px; height:5px;float:left;overflow:hidden\" muststretch=\"h\"></div><div style=\"background: url(\'/upimages/images/part/8594.png\') no-repeat; width:9px; height:5px;float:left;overflow:hidden\"></div></div></td></tr></table></span>二氧化碳','过氧化氢<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td style=\"font-size: 90%\">催化剂</td></tr><tr><td><div hassize=\"9\"><div style=\"background: url(\'/upimages/images/part/8594L.png\') repeat-x; width:1px; height:5px;float:left;overflow:hidden\" muststretch=\"h\"></div><div style=\"background: url(\'/upimages/images/part/8594.png\') no-repeat; width:9px; height:5px;float:left;overflow:hidden\"></div></div></td></tr></table></span>水+氧气','酒精+氧气→二氧化碳+水','','D','【解答】解：A、水<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td style=\"font-size: 90%\">通电</td></tr><tr><td><div hassize=\"9\"><div style=\"background: url(\'/upimages/images/part/8594L.png\') repeat-x; width:1px; height:5px;float:left;overflow:hidden\" muststretch=\"h\"></div><div style=\"background: url(\'/upimages/images/part/8594.png\') no-repeat; width:9px; height:5px;float:left;overflow:hidden\"></div></div></td></tr></table></span>氧气+氢气，该反应符合“一变多”的特征，属于分解反应，故选项错误；<br />B、木炭+氧气<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td style=\"font-size: 90%\">点燃</td></tr><tr><td><div hassize=\"9\"><div style=\"background: url(\'/upimages/images/part/8594L.png\') repeat-x; width:1px; height:5px;float:left;overflow:hidden\" muststretch=\"h\"></div><div style=\"background: url(\'/upimages/images/part/8594.png\') no-repeat; width:9px; height:5px;float:left;overflow:hidden\"></div></div></td></tr></table></span>二氧化碳，该反应是物质和氧气发生的化学反应，属于氧化反应；但该反应符合“多变一”的特征，属于化合反应；故选项错误；<br />C、过氧化氢<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td style=\"font-size: 90%\">催化剂</td></tr><tr><td><div hassize=\"9\"><div style=\"background: url(\'/upimages/images/part/8594L.png\') repeat-x; width:1px; height:5px;float:left;overflow:hidden\" muststretch=\"h\"></div><div style=\"background: url(\'/upimages/images/part/8594.png\') no-repeat; width:9px; height:5px;float:left;overflow:hidden\"></div></div></td></tr></table></span>水+氧气，该反应符合“一变多”的特征，属于分解反应，故选项错误；<br />D、酒精+氧气<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td style=\"font-size: 90%\">点燃</td></tr><tr><td><div hassize=\"9\"><div style=\"background: url(\'/upimages/images/part/8594L.png\') repeat-x; width:1px; height:5px;float:left;overflow:hidden\" muststretch=\"h\"></div><div style=\"background: url(\'/upimages/images/part/8594.png\') no-repeat; width:9px; height:5px;float:left;overflow:hidden\"></div></div></td></tr></table></span>二氧化碳+水，该反应是物质和氧气发生的化学反应，属于氧化反应；且该反应的生成物是两种，不符合“多变一”的特征，不属于化合反应；故选项正确．<br />故选：D．','【分析】化合反应：两种或两种以上物质反应后生成一种物质的反应，其特点可总结为“多变一”；物质与氧发生的化学反应是氧化反应；据此进行分析判断．','选择题',3.00,'b0bca691a5ef81718e6c6bfc32a3af11',9,400,'化合反应及其应用,氧化反应','',0,'37','',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842831,'2010年诺贝尔化学奖授予美日三位科学家，表彰他们为研发有机合成中钯催化的交叉偶联反应而作出的杰出贡献．则下列说法中不正确的是（　　）','有机物中一定含有碳元素','含碳元素的化合物不都属于有机化合物','有机物数目庞大是因为组成元素种类多','食品包装的聚乙烯塑料袋可用加热的方法封口','','C','【解答】解：A、有机物中一定含有碳元素，正确；<br />B、含碳元素的化合物不都属于有机化合物，比如一氧化碳或是二氧化碳不是有机物，正确；<br />C、机物数目庞大不是因为组成元素种类多，而是其组成元素的原子结合的方式不同，错误；<br />D、食品包装的聚乙烯塑料具有热塑性，故聚乙烯塑料袋可用加热的方法封口，正确；<br />故选C．','【分析】根据有机物的组成、特征以及聚乙烯塑料的性质进行分析解答即可．','选择题',3.00,'1f911e2e2898c98673ea5e506e59622b',9,400,'有机物的特征、分类及聚合物的特性,塑料及其应用','',2016,'32','2016•秦皇岛模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842859,'天然气水合物（化学式可表示为：CH<SUB>4</SUB>•nH<SUB>2</SUB>O）是由天然气与水在高压低温条件下形成&nbsp;的类冰状的结晶物质．因其外观像冰一样而且遇火即可燃烧，所以又被称作“可燃冰”．天然气水合物燃烧污染比煤、石油、天然气都小得多，而且储量丰富，全球储量足够人类&nbsp;使用1000年，因而被各国视为未来石油天然气的替代能源．目前，30多个国家和地区已经进行“可燃冰”的研究与调查勘探，最近两年开采试验取得较大进展•我国计划于2015年&nbsp;在中国海域实施天然气水合物的钻探工程，将有力推动中国“可燃冰”勘探与开发的进程．根据以上内容回答5-7题（　　）','','','','','','','','','选择题',1.00,'a18b99e0746b12d7989fb82fc2ae6c2a',9,400,'燃烧、爆炸、缓慢氧化与自燃','',0,'37','',1,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842880,'下列有关燃料和燃烧的说法正确的是（　　）','用水灭火，目的是降低可燃物的着火点','在化石燃料中，天然气是比较清洁的燃料','有发光、放热现象的一定是燃烧','石油、乙醇均是可再生资源','','B','【解答】解：A、用水能够灭火，是因为水蒸发时吸收热量，能使温度降低到可燃物的着火点以下，而不是降低可燃物的着火点，故错误；<br />B、在化石燃料中，天然气燃烧产物主要是二氧化碳和水，是比较清洁的燃料，故正确；<br />C、有发光、放热现象的不一定是燃烧，例如灯泡发光放热不属于燃烧，故错误；<br />D、石油属于不可再生能源，乙醇属于可再生能源，故错误．<br />故选：B．','【分析】A、可燃物的着火点一般情况下不能改变；<br />B、根据天然气的主要成分是甲烷进行分析；<br />C、根据燃烧的概念进行分析；<br />D、石油属于不可再生能源．','选择题',3.00,'0e236a4a1189384c9429e9ab967a4d69',9,400,'常用燃料的使用与其对环境的影响,灭火的原理和方法,燃烧、爆炸、缓慢氧化与自燃,常见能源的种类、能源的分类','普宁市',2015,'37','2015•普宁市一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842882,'奠定近代化学基础的是（　　）','打造石器','发现和利用了火','元素周期律的发现','原子论和分子学说的创立','','D','【解答】解：A、打造石器，改善了人类的生活条件，不是奠定近代化学基础的理论，故选项错误．<br />B、发现和利用了火，改善了人类的生存条件，不是奠定近代化学基础的理论，故选项错误．<br />C、元素周期律的发现，使化学的学习和研究变得有规律可循，不是奠定近代化学基础的理论，故选项错误．<br />D、原子论和分子学说的创立，是认识和分析化学现象及本质的基础，是奠定了近代化学基础的理论，故选项正确．<br />故选：D．','【分析】化学的发展史可分为三个阶段：古代化学、近代化学、现代化学，在古代对化学的认识只停留在表象阶段，到了近代，道尔顿和阿伏加德罗创立了原子论-分子学说，奠定了近代化学的基础．','选择题',3.00,'acb2eb5e3e22ca26a2533db2e0ad0587',9,400,'化学的历史发展过程','',2012,'35','2012秋•阜南县校级期中',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842884,'我校兴趣小组利用以下实验装置制到并收集气体．<br />（1）写出图中标号的仪器名称：①<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；②<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）图A是制备CO<SUB>2</SUB>的装置图．实验室制取二氧化碳的原理是（用化学方程式表示）：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．该装置的特点是：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．上面B、C、D、E四套装置中，能产生与该装置相同的效果有<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填写编号）．<br />（3）通常，实验室收集二氧化碳的方法是：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，为什么：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（4）在实验室中，同学们利用E图装置制取并收集氧气，关闭K<SUB>1</SUB>打开K<SUB>2</SUB>，当量筒中水的体积为20mL时，立即关闭K<SUB>2</SUB>打开K<SUB>1</SUB>，发现量筒液面略有下降，请你分析可能的原因是：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（5）在利用图A装置制取二氧化碳时，发现不同小组产生气泡的速率不同，请你猜想可能的原因是：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，请写出验证你的猜想的实验方案：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<img  src=\"/tikuimages/9/2016/400/shoutiniao22/39cfad40-94d4-11e9-9281-b42e9921e93e_xkb25.png\" style=\"vertical-align:middle\" />','','','','','','长颈漏斗$###$锥形瓶$###$CaCO<SUB>3</SUB>+2HCl=CaCl<SUB>2</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑$###$便于控制反应的发生和停止$###$BD$###$向上排空气法$###$二氧化碳密度比空气大，二氧化碳能溶于水，能与水反应$###$氧气不易溶于水，有一部分氧气溶于水$###$大理石颗粒大小不同$###$在室温下，分别取等质量不同颗粒大小的大理石放入锥形瓶中，再分别加入相同体积、相同浓度的稀盐酸，比较收集相同体积气体所需的时间．','【解答】解：（1）通过分析题中所指仪器的作用可知，①是长颈漏斗，②是锥形瓶；<br />（2）碳酸钙和盐酸反应生成氯化钙、水和二氧化碳，化学方程式为：CaCO<SUB>3</SUB>+2HCl=CaCl<SUB>2</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑，通过装置内外压强差，可以实现固液分离，该装置的特点是：便于控制反应的发生和停止，B、C、D、E四套装置中，能产生与该装置相同的效果有BD；<br />（3）二氧化碳密度比空气大，易溶于水，易溶于水反应，所以实验室收集二氧化碳的方法是：向上排空气法；<br />（4）氧气在水中有一定的溶解能力，所以当量筒中水的体积为20mL时，立即关闭K<SUB>2</SUB>打开K<SUB>1</SUB>，发现量筒液面略有下降，可能的原因是：氧气不易溶于水，有一部分氧气溶于水；<br />（5）在利用图A装置制取二氧化碳时，发现不同小组产生气泡的速率不同，猜想可能的原因是：大理石颗粒大小不同，验证猜想的实验方案是：在室温下，分别取等质量不同颗粒大小的大理石放入锥形瓶中，再分别加入相同体积、相同浓度的稀盐酸，比较收集相同体积气体所需的时间．<br />故答案为：（1）长颈漏斗，锥形瓶；<br />（2）CaCO<SUB>3</SUB>+2HCl=CaCl<SUB>2</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑，便于控制反应的发生和停止，BD；<br />（3）向上排空气法，二氧化碳密度比空气大，易溶于水，易溶于水反应；<br />（4）氧气不易溶于水，有一部分氧气溶于水；<br />（5）大理石颗粒大小不同，在室温下，分别取等质量不同颗粒大小的大理石放入锥形瓶中，再分别加入相同体积、相同浓度的稀盐酸，比较收集相同体积气体所需的时间．','【分析】（1）根据实验室常用仪器的名称和题中所指仪器的作用进行分析；<br />（2）根据碳酸钙和盐酸反应生成氯化钙、水和二氧化碳，通过装置内外压强差，可以实现固液分离进行分析；<br />（3）根据二氧化碳密度比空气大，易溶于水，易溶于水反应进行分析；<br />（4）根据氧气在水中有一定的溶解能力进行分析；<br />（5）根据控制变量法的具体操作进行分析．','书写',3.00,'0d06f5f5abb3116da72328a31aaac42b',9,400,'量气装置,二氧化碳的实验室制法,书写化学方程式、文字表达式、电离方程式','',2016,'32','2016•合肥校级模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842905,'构成物质的微粒有多种，下列物质由离子构成的是（　　）','水银','金刚石','氯化钠','水','','C','【解答】解：A、水银是金属汞的俗称，是由汞原子直接构成的，故选项错误．<br />B、金刚石属于固态非金属单质，是由碳原子直接构成的，故选项错误．<br />C、氯化钠是由钠离子和氯离子构成的，故选项正确．<br />D、水是由水分子构成的，故选项错误．<br />故选：C．','【分析】根据金属、大多数固态非金属单质、稀有气体等由原子构成；有些物质是由分子构成的，气态的非金属单质和由非金属元素组成的化合物，如氢气、水等；有些物质是由离子构成的，一般是含有金属元素和非金属元素的化合物，如氯化钠，进行分析判断即可．','选择题',3.00,'baaf7e4bb77813e90f9710b6ecda020c',9,400,'物质的构成和含量分析','',2016,'35','2016春•平和县期中',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842907,'2015年“世界环保日”中国主题为“低碳减排，绿色生活”，这一主题旨在倡导低能耗，低排放的生活方式，下列做法不符合这一主题的是（　　）','随手关灯，使用高效节能灯泡','自备购物袋或重复使用塑料购物袋','推广利用太阳能、风能，缓解温室效应','多使用一次性筷子和一次性快餐盒','','D','【解答】解：<br />A、出门随手关灯节约用电，符合“绿色、节能、低碳”的理念．故符合题意；<br />B、常自备购物袋或重复使用塑料购物袋，可以防止白色污染，符合“绿色、节能、低碳”的理念．故符合题意；<br />C、推广利用太阳能、风能，可以节约能源，缓解温室效应，故符合题意；<br />D、经常使用一次性筷子和饭盒，浪费了资源，不符合低碳生活的理念；<br />答案：D','【分析】低碳生活”代表着更健康、更自然、更安全的生活，同时也是一种低成本、低代价的生活方式．指的是生活作息时所耗用的能量要尽量减少，特别是减少二氧化碳的排放量，减缓生态恶化；可以从节电、节能和回收等环节来改变生活细节，据此进行分析解答即可．','选择题',3.00,'58416ce6509e94fc1fdc024555878f47',9,400,'自然界中的碳循环','',2016,'37','2016•江东区一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842909,'2015年8月12日，天津港发生重大爆炸事故，报道指出其涉及过氧化物等特殊化学货物，这种化学品燃烧产生的火焰不能用水直接扑灭．<br />（1）已知过氧化钠（Na<SUB>2</SUB>O<SUB>2</SUB>）能与水和二氧化碳反应，均产生氧气．过氧化钠与水反应产生苛性钠和氧气，写出反应的方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）过氧化钠着火可用下列哪种物质灭火<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />A．干粉灭火器&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;B．水基型灭火器&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;C．沙土．','','','','','','2Na<SUB>2</SUB>O<SUB>2</SUB>+2CO<SUB>2</SUB>═2Na<SUB>2</SUB>CO<SUB>3</SUB>+O<SUB>2</SUB>↑$###$C','【解答】解：（1）过氧化钠（Na<SUB>2</SUB>O<SUB>2</SUB>）常温下可与二氧化碳反应生成碳酸钠和氧气，反应的化学方程式为：2Na<SUB>2</SUB>O<SUB>2</SUB>+2CO<SUB>2</SUB>═2Na<SUB>2</SUB>CO<SUB>3</SUB>+O<SUB>2</SUB>．<br />（2）过氧化钠，与二氧化碳、水都反应，则不能用泡沫灭火器、干粉灭火器、水灭火，加入沙土可掩盖，隔绝空气，可起到灭火的作用；<br />故答案为：（1）2Na<SUB>2</SUB>O<SUB>2</SUB>+2CO<SUB>2</SUB>═2Na<SUB>2</SUB>CO<SUB>3</SUB>+O<SUB>2</SUB>↑．（2）C','【分析】（1）过氧化钠（Na<SUB>2</SUB>O<SUB>2</SUB>）常温下可与二氧化碳反应生成碳酸钠和氧气，写出反应的化学方程式即可；<br />（2）过氧化钠，可与二氧化碳或水反应生成氧气分析解答．','书写',3.00,'8b2a5b8e12356ac0ed6178bbbf287467',9,400,'书写化学方程式、文字表达式、电离方程式,几种常用的灭火器','',2016,'37','2016春•长沙校级月考',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842911,'下列关于金属材料的说法正确的是（　　）','生铁、硬铝、焊锡都是合金','铁、铜、铝都是银白色金属','锈蚀后的金属材料不能回收','铁、铜、锌都能与稀硫酸反应','','A','【解答】解：A、合金是金属与金属或金属与非金属融合在一起形成的具有金属特性的混合物，生铁、硬铝、焊锡都属于合金，故A正确；<br />B、铜是紫红色的固体，故B错误；<br />C、金属制品腐蚀后仍然能进行回收利用，这样可大大节约金属资源，故C错误；<br />D、铜在金属活动性顺序中排在氢的后面，不能与与稀硫酸反应，故D错误；<br />故选：A．','【分析】A、根据合金的定义分析；<br />B、根据铜是紫红色进行分析；<br />C、根据锈蚀后的金属材料的回收价值进行分析；<br />D、根据金属活动性顺序进行分析．','选择题',3.00,'92de8e55cfc7a39f8dea25e1b3052883',9,400,'金属的物理性质及用途,合金与合金的性质,金属的化学性质,金属的回收利用及其重要性','',2016,'37','2016•延平区一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842920,'位于我市西南部的平阴县是全国著名的“玫瑰之乡”．玫瑰花，蔷薇科，落叶灌木．夏季开花，“香腻馥郁，愈干愈烈”．玫瑰花中含有多种挥发性物质，主要成分为芳樟醇，化学式为&nbsp;C<SUB>10</SUB>H<SUB>18</SUB>O，无色液体，既有紫丁香，铃兰与玫瑰的花香，又有木香、果香气息．在全世界每年生产的最常用和用量最大的香料中，芳樟醇几乎年年排在首位．<br />（1）用微粒的观点解释，人们闻到花香气味的主要原因：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）在芳樟醇中，碳元素与氢元素的原子个数比为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填最简整数比）<br />（3）玫瑰花的病害主要有，白粉病，锈病，黑斑病，枯枝病，花叶病等．可以在花后每隔半月喷一次一定浓度的波尔多液，连喷4次．有经验的种植者知道配好的波尔多液盛放在铁桶内会降低药效，请用化学方程式表示出其中的原因：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（4）科学的肥水管理能增强玫瑰的长势，磷酸二氢铵、尿素[CO（NH<SUB>2</SUB>）<SUB>2</SUB>]、KNO<SUB>3</SUB>等化肥的使用，提高了鲜花的产量和质量．下列有关说法中正确的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填选项序号）<br />①尿素中的氢元素的质量分数最大②尿素是一种复合肥<br />③化肥使用越多越好④铵态氮肥不能和碱性物质混合使用．','','','','','','分子是不断运动的$###$5：9$###$Fe+CuSO<SUB>4</SUB>═Cu+FeSO<SUB>4</SUB>$###$④','【解答】解：（1）春暖花开的季节里，人们在公园常能闻到怡人的花香气味，是因为花香中含有的分子是在不断运动的，向四周扩散，使人们闻到花香．故填：分子在不断的运动；<br />（2）芳樟醇的化学式为C<SUB>10</SUB>H<SUB>18</SUB>O，可见一个芳樟醇分子中含有10个碳原子、18个氢原子和1个氧原子，碳元素与氢元素的原子个数比为10：18=5：9；故填：5：9；<br />（3）铁的活动性比铜强，所以可以和硫酸铜溶液发生置换反应，该反应生成了铜和硫酸亚铁．反应的方程式为：Fe+CuSO<SUB>4</SUB>═Cu+FeSO<SUB>4</SUB>；故答案为：Fe+CuSO<SUB>4</SUB>═Cu+FeSO<SUB>4</SUB>．<br />（4）①尿素中碳、氧、氮、氢元素的质量比为12：16：（14×2）：（1×2×2）=3：4：7：1．可见其中氮元素的质量分数最大，故错误；<br />②尿素中只含营养元素中的氮元素，属于氮肥，故错误；<br />③化肥要根据实际适量、合理使用，过度使用会造成污染，故错误；<br />④铵态氮肥可与碱性物质反应生成氨气，使氮肥丧失肥效，所硝酸铵、硫酸铵等铵态氮肥不能与碱性物质混合使用，故正确；<br />故选④．','【分析】（1）根据分子是在不断的运动的特征分析回答．<br />（2）根据化学式的意义分析计算；<br />（3）根据铁能够和硫酸铜溶液反应和化学方程式的书写方法来完成该题的解答；<br />（4）根据元素的质量比、复合肥的概念、化肥的使用以及铵态氮肥的性质来分析．','书写',3.00,'5a370345b2a95af1c74177c10d3f6e06',9,400,'金属的化学性质,常见化肥的种类和作用,利用分子与原子的性质分析和解决问题,化学式的书写及意义,元素的质量分数计算,合理使用化肥、农药对保护环境的重要意义','',2016,'37','2016•青岛一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842921,'下列物质的检验或鉴别，所用试剂错误的是（　　）','检验碳酸盐--稀盐酸和Ca（OH）<SUB>2</SUB>溶液','检验淀粉--加碘食盐','鉴别NaCl和NaOH溶液--紫色石蕊试液','鉴别软水和硬水--肥皂水','','B','【解答】解：A、检验碳酸盐能与酸反应生成气体，与氢氧化钙反应生成碳酸钙白色沉淀，检验碳酸盐可用稀盐酸和Ca（OH）<SUB>2</SUB>溶液，故选项所用试剂正确．<br />B、淀粉遇碘变蓝色，加碘食盐中不含碘，不能用于检验淀粉，故选项所用试剂错误．<br />C、紫色石蕊溶液遇酸性溶液变红，遇碱性溶液变蓝，NaCl和NaOH溶液分别显中性、碱性，使石蕊溶液分别显示紫色、蓝色，能出现两种明显不同的现象，可以鉴别，故选项所用试剂正确．<br />D、硬水和软水的区别在于所含的钙镁离子的多少，可用等量的肥皂水进行鉴别，产生泡沫较多的是软水，较少的是硬水，故选项所用试剂正确．<br />故选：B．','【分析】A、根据碳酸盐能与酸反应生成气体，与氢氧化钙反应生成碳酸钙白色沉淀，进行分析判断．<br />B、根据淀粉遇碘变蓝色，进行分析判断．<br />C、根据两种物质与同种试剂反应产生的不同现象来鉴别它们，若两种物质与同种物质反应的现象相同，则无法鉴别它们．<br />D、硬水和软水的区别在于所含的钙镁离子的多少，结合其检验方法进行分析判断．','选择题',3.00,'93b431f4018c4610d8737cc65b0526ef',9,400,'证明碳酸盐,硬水与软水,酸、碱、盐的鉴别,鉴别淀粉、葡萄糖的方法与蛋白质的性质','',2016,'32','2016•泗阳县模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1842923,'<img  src=\"/tikuimages/9/2016/400/shoutiniao18/3a7b7d9e-94d4-11e9-a3ec-b42e9921e93e_xkb47.png\" style=\"vertical-align:middle;FLOAT:right\" />（2016•抚顺模拟）请回答下列有关金属的问题．<br />（1）铝是活泼金属，为什么通常铝锅却很耐腐蚀？<br />（2）如图是某探究实验装置图．一段时间后，能观察到什么现象是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．（装置气密性良好，开始时U型管两端的红墨水液面相平）<br />（3）X、Y、Z是三种金属固体，将X和Y浸入稀硫酸中，Y溶解并产生氢气，X无变化；将X和Z浸入硝酸银溶液中，X表面有银析出，而Z无变化．<br />①判断X、Y、Z和银四种金属的活动性由强带弱的顺序是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />②具体确定一种X后，写出X与硝酸银溶液反应的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（4）金属资源不可再生．请提出一种保护金属资源的方法<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','右侧液面下降，左侧液面上升$###$Y＞X＞Ag＞Z$###$Cu+2AgNO<SUB>3</SUB>═Cu（NO<SUB>3</SUB>）<SUB>2</SUB>+2Ag$###$防止金属的腐蚀等','【解答】解：（1）铝是活泼金属，通常铝锅却很耐腐蚀的原因是：铝与氧气反应，其表面生成的致密的氧化铝薄膜起保护作用．<br />（2）铁钉处在潮湿的空气中易生锈，消耗了密封试管内的氧气，使其压强减小，外界大气压不变，故U型管右侧液面下降，左侧液面上升．<br />（3）①由题意可知，Y能与稀硫酸反应而X不能，则可得Y的活动性大于氢，X的活动性小于氢，即：Y＞H＞X；X和Z浸入硝酸银溶液中，X表面有Ag析出而Z无变化，X的活动性大于Ag，而Z的活动性小于Ag，即X＞Ag＞Z；综上所述则有：Y＞X＞Ag＞Z．<br />②选用的X其活动性要排在H与Ag之间即可，如Cu、Hg等，反应的方程式是：Cu+2AgNO<SUB>3</SUB>═Cu（NO<SUB>3</SUB>）<SUB>2</SUB>+2Ag．<br />（4）金属资源不可再生．保护金属资源的方法很多，例如：防止金属的腐蚀、提倡金属的回利用等．<br />故答为：（1）铝与氧气反应，其表面生成的致密的氧化铝薄膜起保护作用；（2）右侧液面下降，左侧液面上升．（3）①Y＞X＞Ag＞Z．②Cu+2AgNO<SUB>3</SUB>═Cu（NO<SUB>3</SUB>）<SUB>2</SUB>+2&nbsp;Ag．（4）防止金属的腐蚀等．','【分析】（1）根据铝的特性分析回答；<br />（2）铁钉在潮湿的环境中易于氧气发生反应，从而使密封试管内的氧气密度与压强均发生了改变．<br />（3）根据金属活动性顺序应用，“反应则活泼，不反应则不活泼”，进行分析解答．<br />（4）根据保护金属资源的措施分析回答．','书写',3.00,'0283fd50c92d96fa9698131585a86da7',9,400,'金属的化学性质,金属活动性顺序及其应用,金属锈蚀的条件及其防护,金属资源的保护,书写化学方程式、文字表达式、电离方程式','',2016,'32','2016•抚顺模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1843031,'在牙膏中常用轻质碳酸钙粉末作摩擦剂．如图是用石灰石（杂质均不参加反应，且不溶于水）制取高纯度轻质碳酸钙的主要工艺流程，请你回答下列问题．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao6/3bb50f0f-94d4-11e9-b1c3-b42e9921e93e_xkb34.png\" style=\"vertical-align:middle\" /><br />（1）实验操作①的名称是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，进行操作①用到的玻璃仪器除玻璃棒、烧杯外，还需要<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）X中主要物质的化学式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，x与试剂②反应的方程式为：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）工业生产中，为了节约成本，会把有用的产物重新循环利用，该工艺中可循环使用的物质是<br /><u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填化学式）<br />（4）你认为用作牙膏摩擦剂牙膏摩擦剂的物质应具备什么特点呢？（回答一条即可）<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','过滤$###$漏斗$###$Ca（OH）<SUB>2</SUB>$###$Ca（OH）<SUB>2</SUB>+CO<SUB>2</SUB>=CaCO<SUB>3</SUB>↓+H<SUB>2</SUB>O$###$H<SUB>2</SUB>O$###$难溶于水等','【解答】解：（1）由工艺流程图可知，实验操作①能将固液分开，所以，实验操作①的名称是过滤，进行过滤操作用到的玻璃仪器除玻璃棒、烧杯外，还需要漏斗．<br />（2）由于碳酸钙高温分解生成了氧化钙，氧化钙能与水化合生成了氢氧化钙，氢氧化钙能与二氧化碳反应，所以X中主要物质的化学式为Ca（OH）<SUB>2</SUB>，氢氧化钙能与试剂②二氧化碳反应生成了碳酸钙和水，反应的方程式为：Ca（OH）<SUB>2</SUB>+CO<SUB>2</SUB>=CaCO<SUB>3</SUB>↓+H<SUB>2</SUB>O．<br />（3）工业生产中，为了节约成本，会把有用的产物重新循环利用，该工艺中可循环使用的物质是水，化学式是：H<SUB>2</SUB>O．<br />（4）牙膏摩擦剂是进入人口腔的物质，应该具备无毒的性质，能起到摩擦的作用，则该物质应该具备难溶于水的性质．<br />故答为：（1）过滤；漏斗；&nbsp;（2）Ca（OH）<SUB>2</SUB>；&nbsp;Ca（OH）<SUB>2</SUB>+CO<SUB>2</SUB>=CaCO<SUB>3</SUB>↓+H<SUB>2</SUB>O；（3）H<SUB>2</SUB>O；&nbsp;（4）难溶于水等．','【分析】（1）根据过滤的原理和操作分析回答；<br />（2）碳酸钙高温分解生成了氧化钙，氧化钙能与水化合生成了氢氧化钙，氢氧化钙能与二氧化碳反应，据此分析回答有关的问题；<br />（3）根据工艺流程图分析可以循环利用的物质；<br />（4）根据牙膏摩擦剂的应用分析应具备的特点','书写',3.00,'96e03cc3f36669380b67918706987c0f',9,400,'过滤的原理、方法及其应用,碳酸钙、生石灰、熟石灰之间的转化,物质的相互转化和制备,书写化学方程式、文字表达式、电离方程式','',2016,'37','2016春•蒙城县校级月考',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1843041,'某同学对一袋化肥的成分进行了探究，请你参与探究并填空：<br />提出问题：该化肥中含有什么物质？<br />收集信息：经询问得知，该化肥可能为铵态氮肥，且可能混入了其他化肥，所谓铵态氮肥指的是含有<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填离子符号）的盐．<br />提出猜想：该化肥所含的阴离子可能是Cl<SUP>-</SUP>、CO<SUB>3</SUB><SUP>2-</SUP>、SO<SUB>4</SUB><SUP>2-</SUP>中的一种或几种．<br />实验、记录与分析：<br /><table class=\"edittable\"><TBODY><TR><td width=272>实验操作步骤</TD><td width=181>实验现象</TD><td width=116>实验分析</TD></TR><TR><td>（1）取少量该化肥样品和少量熟石灰放在<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填仪器名称）中混合研磨</TD><td><u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD><td>含有铵根离子</TD></TR><TR><td>（2）另取少量该化肥样品于试管中，加入适量的水完全溶解，滴加足量的硝酸钡溶液，再滴加稀硝酸，过滤．</TD><td><u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD><td>没有存在CO<SUB>3</SUB><SUP>2-</SUP>&nbsp;存在，有SO<SUB>4</SUB><SUP>2-</SUP>存在</TD></TR><TR><td>（3）取（2）所得滤液少量于试管中，滴加硝酸银溶液和稀硝酸．</TD><td>产生白色沉淀</TD><td>有Cl<SUP>-</SUP></TD></TR></TBODY></TABLE>实验结论：若该化肥中只含有一种阳离子，则其中一定含有的物质是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（写化学式）．<br />实验反思：小红提出在步骤（2）中用氯化钡溶液代替硝酸钡溶液，小明立即否定了这种做法，理由是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','NH<SUB>4</SUB><SUP>+</SUP>$###$研钵$###$有氨味或有刺激性气味$###$有白色沉淀产生，无气泡，沉淀不溶解$###$NH<SUB>4</SUB>Cl和（NH<SUB>4</SUB>）<SUB>2</SUB>SO<SUB>4</SUB>$###$对氯离子的鉴别产生干扰','【解答】解：所谓铵态氮肥指的是含有NH<SUB>4</SUB><SUP>+</SUP>的盐．<br />【实验、记录与分析】<br />（1）取少量该化肥样品和少量熟石灰放在研钵中混合研磨，因为含有铵根离子，所以铵根离子和氢氧化钙反应产生刺激性气味的氨气，故实验现象是：有氨味或有刺激性气味；<br />（2）因为向溶液中加硝酸钡溶液，再滴加少量稀硝酸，白色沉淀不溶解，说明不是碳酸根离子，是硫酸根离子和钡离子产生的硫酸钡沉淀，因而也可证明含有硫酸根离子；<br />（3）检验氯离子可以向（2）所得滤液加入硝酸银溶液产生白色沉淀证明；<br />【实验结论】因为该化肥中只含有一种阳离子，即铵根离子，阴离子中含有硫酸根离子和氯离子，所以其中一定含有的物质是：硫酸铵和氯化铵，故答案为：NH<SUB>4</SUB>Cl和（NH<SUB>4</SUB>）<SUB>2</SUB>SO<SUB>4</SUB>；步骤（2）中的反应方程式：（NH<SUB>4</SUB>）<SUB>2</SUB>SO<SUB>4</SUB>+Ba（NO<SUB>3</SUB>）<SUB>2</SUB>=BaSO<SUB>4</SUB>↓+2NH<SUB>4</SUB>NO<SUB>3</SUB>；<br />步骤（3）中的反应方程式：NH<SUB>4</SUB>Cl+AgNO<SUB>3</SUB>═NH<SUB>4</SUB>NO<SUB>3</SUB>+AgCl↓；<br />【实验反思】因为后面还要检验氯离子的存在，所以加入的试剂不能引入氯离子，因此不能用氯化钡溶液代替硝酸钡溶液；故答案为：否；氯化钡会引入新的氯离子，对原有氯离子的检验产生干扰．<br />故答案为：<br />NH<SUB>4</SUB><SUP>+</SUP><br /><br /><table class=\"edittable\"><TBODY><TR><td width=272>实验操作步骤</TD><td width=181>实验现象</TD><td width=116>实验分析</TD></TR><TR><td>（1）取少量该化肥样品和少量熟石灰放在 研钵（填仪器名称）中混合研磨</TD><td>有氨味或有刺激性气味</TD><td>含有铵根离子</TD></TR><TR><td>（2）另取少量该化肥样品于试管中，加入适量的水完全溶解，滴加足量的硝酸钡溶液，再滴加稀硝酸，过滤．</TD><td>有白色沉淀产生，无气泡，沉淀不溶解</TD><td>没有存在CO<SUB>3</SUB><SUP>2-</SUP>&nbsp;存在，有SO<SUB>4</SUB><SUP>2-</SUP>存在</TD></TR><TR><td>（3）取（2）所得滤液少量于试管中，滴加硝酸银溶液和稀硝酸．</TD><td>产生白色沉淀</TD><td>有Cl<SUP>-</SUP></TD></TR></TBODY></TABLE>【实验结论】NH<SUB>4</SUB>Cl和（NH<SUB>4</SUB>）<SUB>2</SUB>SO<SUB>4</SUB>；<br />【实验反思】对氯离子的鉴别产生干扰','【分析】【实验、记录与分析】<br />（1）根据铵根离子能够和熟石灰中的氢氧根离子反应产生刺激性气味的氨气分析；<br />（2）根据硫酸根离子的鉴别方法和碳酸根离子的特点分析：加硝酸钡溶液，再滴加少量稀硝酸，沉淀不溶解；<br />（3）根据氯离子的鉴别方法分析；<br />【实验结论】根据实验得出的阳离子和阴离子的结论写出反应的方程式；<br />【实验反思】根据检验物质时引入的离子不能对后面的离子检验产生干扰分析．','填空题',3.00,'08a7edf359024b7e340a5299f2a2b1a8',9,400,'实验探究物质的组成成分以及含量,铵态氮肥的检验,常见离子的检验方法及现象','广水市',2016,'32','2016•广水市模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1843052,'被称为“绿色氧化剂”的过氧化氢（H<SUB>2</SUB>0<SUB>2</SUB>），其水溶液称双氧水，是一种无色液体，常用作氧化剂、消毒杀菌剂和漂白剂等，在较低温度下和少量催化剂（如MnO<SUB>2</SUB>）条件下，它能迅速分解，生成氧气和水，请回答下列问题：<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao23/3bf01c40-94d4-11e9-8f19-b42e9921e93e_xkb58.png\" style=\"vertical-align:middle\" /><br />（1）在实验室中制取气体的一般思路时，首先要选择<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，控制<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，然后根据<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>和<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，确定制取气体的发生装置，根据制取气体的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>和<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，确定制取气体的收集方法，最后要审查整个反应体系是否符合<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>、<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>的要求．<br />（2）实验室制取氧气时，从组成上看，你认为下列哪些物质能够作为实验室制取氧气的反应物？<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，理由<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />A．H<SUB>2</SUB>O<SUB>2</SUB>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; B．KClO<SUB>3</SUB>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; C．KCl&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; D．KMnO<SUB>4</SUB><br />（3）实验室中用双氧水和二氧化锰制取一瓶纯净的氧气时，应选用的发生装置是（填序号，下同）<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，收集装置是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，二氧化锰在该反应中的作用是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，反应方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，属于<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>反应．<br />（4）实验室中如果用E装置收集氧气时，<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>开始收集氧气，实验室中如果用C装置收集氧气，检验集气瓶中是否收集满氧气的方法是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（5）乙炔俗称电石气，是一种无色、无味、密度比空气小，不溶于水的气体，在氧气中燃烧能产生高温，工业上常用它来切割和焊接金属，实验室里常用电石（固体）与水反应制取乙炔，你认为制取乙炔应选择的发生装置是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，收集装置是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，请写出乙炔燃烧的方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','反应物$###$反应的条件$###$反应物的状态$###$反应条件$###$密度$###$溶解度$###$环保$###$安全$###$ABD$###$反应物中都含有氧元素$###$B$###$E$###$催化作用，加快化学反应速率$###$2H<SUB>2</SUB>O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span>&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O+O<SUB>2</SUB>↑$###$分解$###$导管口有气泡连续均匀冒出时$###$将带火星的木条平放在集气瓶口，木条复燃，证明氧气满了$###$B$###$D或E$###$2C<SUB>2</SUB>H<SUB>2</SUB>+5O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>4CO<SUB>2</SUB>+2H<SUB>2</SUB>O','【解答】解：（1）在实验室中制取气体的一般思路时，首先要选择反应物，控制反应的条件，然后根据反应物的状态和反应条件确定制取气体的发生装置，根据制取气体的密度和溶解度确定制取气体的收集方法，最后要审查整个反应体系是否符合环保、安全的要求；故答案为：反应物；反应的条件；反应物的状态；反应条件；密度；溶解度；环保；安全；<br />（2）实验室制取氧气时，从组成上看，H<SUB>2</SUB>O<SUB>2</SUB>、KClO<SUB>3</SUB>、KMnO<SUB>4</SUB>都能制氧气，因为反应物中都含有氧元素，故答案为：ABD；反应物中都含有氧元素；<br />（3）如果用双氧水和二氧化锰制氧气就不需要加热，其中二氧化锰起催化作用，加快化学反应速率；氧气的密度比空气的密度大，不易溶于水，因此能用向上排空气法和排水法收集，排水法收集的氧气比较纯净；过氧化氢在二氧化锰做催化剂的条件下生成水和氧气，要注意配平，属于分解反应；故答案为：B；E；催化作用，加快化学反应速率；2H<SUB>2</SUB>O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span>&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O+O<SUB>2</SUB>↑；分解反应；<br />（4）实验室中如果用E装置收集氧气时，导管口有气泡连续均匀冒出时开始收集氧气；氧气的验满方法是：将带火星的木条平放在集气瓶口，木条复燃，证明氧气满了；故答案为：导管口有气泡连续均匀冒出时；将带火星的木条平放在集气瓶口，木条复燃，证明氧气满了；<br />（5）实验室里常用电石（固体）与水反应制取乙炔，因此不需要加热；乙炔俗称电石气，是一种无色、无味、密度比空气小，不溶于水的气体，因此能用向下排空气法和排水法收集；乙炔和氧气在点燃的条件下生成二氧化碳和水，配平即可；故答案为：B；D或E；2C<SUB>2</SUB>H<SUB>2</SUB>+5O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>4CO<SUB>2</SUB>+2H<SUB>2</SUB>O；','【分析】在实验室中制取气体的一般思路时，首先要选择反应物，控制反应的条件，然后根据反应物的状态和反应条件确定制取气体的发生装置，根据制取气体的密度和溶解度确定制取气体的收集方法，最后要审查整个反应体系是否符合环保、安全的要求；制取装置包括加热和不需加热两种，如果用双氧水和二氧化锰制氧气就不需要加热，如果用高锰酸钾或氯酸钾制氧气就需要加热．氧气的密度比空气的密度大，不易溶于水，因此能用向上排空气法和排水法收集，排水法收集的氧气比较纯净．氧气的验满方法是：将带火星的木条平放在集气瓶口，木条复燃，证明氧气满了．实验室里常用电石（固体）与水反应制取乙炔，因此不需要加热；乙炔俗称电石气，是一种无色、无味、密度比空气小，不溶于水的气体，因此能用向下排空气法和排水法收集；乙炔和氧气在点燃的条件下生成二氧化碳和水，配平即可．','书写',3.00,'23f9c99aecdd44f200488ddd57e124b6',9,400,'实验室制取气体的思路,氧气的制取装置,氧气的收集方法,催化剂的特点与催化作用,反应类型的判定,书写化学方程式、文字表达式、电离方程式','乳山市',2016,'35','2016春•乳山市期中',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1843066,'用下列装置进行实验，不能达到实验目的是（　　）','<img  src=\"/tikuimages/9/2016/400/shoutiniao70/3c177a5e-94d4-11e9-b2ff-b42e9921e93e_xkb60.png\" style=\"vertical-align:middle\" /><br />&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 收集CO<SUB>2</SUB>','<img  src=\"/tikuimages/9/2016/400/shoutiniao18/3c1a608f-94d4-11e9-ad10-b42e9921e93e_xkb92.png\" style=\"vertical-align:middle\" /><br />&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 收集O<SUB>2</SUB>','<img  src=\"/tikuimages/9/2016/400/shoutiniao72/3c1c0e40-94d4-11e9-a6c0-b42e9921e93e_xkb3.png\" style=\"vertical-align:middle\" /><br />&nbsp;&nbsp;&nbsp; 验证CO<SUB>2</SUB>性质','<img  src=\"/tikuimages/9/2016/400/shoutiniao53/3c1caa80-94d4-11e9-b339-b42e9921e93e_xkb95.png\" style=\"vertical-align:middle\" /><br />&nbsp;&nbsp;&nbsp; 监控气体流','','B','【解答】解：A、二氧化碳密度大于空气，能达到实验目的；<br />B、氧气的密度比水的密度小，会上升，氧气从长管进入，会从短管迅速跑出，不会将水排出，收集不到氧气，故B不能达到目的；<br />C、验证二氧化碳的酸性气体可以使石蕊试液变色，故能达到实验目的；<br />D、氧气不易溶于水，进入水中会迅速溢出，所以通过气泡冒出的速度不同，来监控气体流速，故D能达到实验目的；<br />故选B','【分析】A、根据二氧化碳的密度比空气大分析；B、根据氧气的密度比水的密度小，会上升考虑；C、根据二氧化碳的性质分析；D、根据氧气不易溶于水考虑．','选择题',3.00,'aae02d95442ea90f780ae15862b737a8',9,400,'分离物质的仪器,常用气体的收集方法,二氧化碳的化学性质','临清市',2016,'32','2016•临清市模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1843137,'2015年11月2日，C919大型客机首架机正式下线．制造大飞机使用了大量的铝、铁、铜．小组同学对三种金属进行了如下探究：<br />【实验】<br /><table class=\"edittable\"><TBODY><TR><td width=95>&nbsp;实验</TD><td width=95><img  src=\"/tikuimages/9/0/400/shoutiniao39/3d1c6740-94d4-11e9-870a-b42e9921e93e_xkb4.png\" style=\"vertical-align:middle\" />&nbsp;</TD><td width=95>&nbsp;<img  src=\"/tikuimages/9/0/400/shoutiniao6/3d1eff4f-94d4-11e9-bced-b42e9921e93e_xkb2.png\" style=\"vertical-align:middle\" /></TD><td width=95><img  src=\"/tikuimages/9/0/400/shoutiniao61/3d21be70-94d4-11e9-b308-b42e9921e93e_xkb94.png\" style=\"vertical-align:middle\" />&nbsp;</TD><td width=95><img  src=\"/tikuimages/9/0/400/shoutiniao15/3d22cfe1-94d4-11e9-a8ce-b42e9921e93e_xkb4.png\" style=\"vertical-align:middle\" />&nbsp;</TD><td width=95>&nbsp;<img  src=\"/tikuimages/9/0/400/shoutiniao26/3d23ba40-94d4-11e9-b25d-b42e9921e93e_xkb82.png\" style=\"vertical-align:middle\" /></TD></TR><TR><td>&nbsp;现象</TD><td>有气泡产生，产生气泡较慢</TD><td>铁片上有红色物质附着，溶液变成浅绿色&nbsp;</TD><td>有气泡产生，产生气泡较快&nbsp;</TD><td><br /><u>&nbsp;&nbsp;&nbsp;&nbsp;</u>&nbsp;</TD><td>&nbsp;无气泡产生无明显现象</TD></TR></TBODY></TABLE>【分析】（1）实验时，需先用砂纸打磨金属片，其目的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）如果要验证铝、铁、铜三种金属活动性强弱，从优化实验的角度考虑，只需利用上述实验中的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（用实验编号填空）两组实验就能达到目的．<br />（3）将打磨后的铝片放入稀硫酸中，开始反应很慢，然后试管逐渐变热，反应逐渐加快．一段时间后，反应速率变慢，最后停止产生气泡．请结合所学知识，解释产生该现象的原因<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />【拓展】将一定质量的锌粉加入含硝酸铝、硝酸铜和硝酸银的溶液中，充分反应后过滤，然后分别向滤液、滤渣中加入稀盐酸，均无明显现象，则滤液中存在的阳离子可能是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（用离子符号表示所有可能存在的情况）．','','','','','','无现象$###$除去金属表面的氧化膜，利于反应$###$BE$###$铝易被空气中的氧气氧化，在铝条表面生成一层致密的氧化物保护膜铝易被空气中的氧气氧化，在铝条表面生成一层致密的氧化物保护膜．将铝条放入稀盐酸中，稀盐酸与铝条表面的氧化铝反应，所以气泡较少．随着氧化铝的减少，铝与盐酸接触面积增大，气泡逐渐增加并达到反应最快．后来盐酸的质量分数越来越小，反应速率减慢．直至停止$###$Al<SUP>3+</SUP>、Zn<SUP>2+</SUP>或Al<SUP>3+</SUP>、Zn<SUP>2+</SUP>、Cu<SUP>2+</SUP>或Al<SUP>3+</SUP>、Zn<SUP>2+</SUP>、Cu、Ag<SUP>+</SUP>','【解答】解：【实验】因为铜的活动性顺序位于氢的后面，因此不能和酸反应，表现为无现象；<br />【分析】（1）因为金属的表面常有一层金属氧化物会影响金属和溶液的反应，因此实验前用砂纸打磨金属片的目的是除去金属表面的氧化膜，利于反应；<br />（2）试管B中铁片上有红色物质附着，溶液变成浅绿色，说明铁比铜活泼；试管E中无气泡产生无明显现象，说明铝比铁活泼；如果要验证铝、铁、铜三种金属活动性强弱，从优化实验的角度考虑，只需利用上述实验中的BE两组实验就能达到目的．<br />（3）铝易被空气中的氧气氧化，在铝条表面生成一层致密的氧化物保护膜铝易被空气中的氧气氧化，在铝条表面生成一层致密的氧化物保护膜．将铝条放入稀盐酸中，稀盐酸与铝条表面的氧化铝反应，所以气泡较少．随着氧化铝的减少，铝与盐酸接触面积增大，气泡逐渐增加并达到反应最快．后来盐酸的质量分数越来越小，反应速率减慢．直至停止．<br />（4）锌能与硝酸银、硝酸铜依次发生反应而不能与硝酸铝反应，向滤出的固体中加入稀硫酸，没有气泡冒出，说明锌粉无剩余，可判断所加入的锌粉一定与硝酸银反应，是否与硝酸铜发生反应因不能确定锌粉的量而无法判断，因此滤液中一定含有没反应的硝酸铝，生成的硝酸锌，可能含有硝酸铜和硝酸银；<br />答案：<br />【实验】无现象；<br />【分析】（1）除去金属表面的氧化膜，利于反应；<br />（2）BE；<br />（3）铝易被空气中的氧气氧化，在铝条表面生成一层致密的氧化物保护膜铝易被空气中的氧气氧化，在铝条表面生成一层致密的氧化物保护膜．将铝条放入稀盐酸中，稀盐酸与铝条表面的氧化铝反应，所以气泡较少．随着氧化铝的减少，铝与盐酸接触面积增大，气泡逐渐增加并达到反应最快．后来盐酸的质量分数越来越小，反应速率减慢．直至停止；<br />（4）Al<SUP>3+</SUP>、Zn<SUP>2+</SUP>或Al<SUP>3+</SUP>、Zn<SUP>2+</SUP>、Cu<SUP>2+</SUP>或Al<SUP>3+</SUP>、Zn<SUP>2+</SUP>、Cu、Ag<SUP>+</SUP>','【分析】【实验】根据铜的活动性顺序位于氢的后面分析现象；<br />【分析】<br />（1）根据金属的表面常有一层金属氧化物会影响金属和溶液反应分析； <br />（2）根据验证金属活动性强弱的方法解答；<br />（3）根据实验的现象得出结论；由图中信息知开始反应速度慢，然后逐渐加快，随反应进行酸逐渐减少，速度变慢．再结合铝的化学性质解决开始快后来慢的问题；<br />（4）根据四种金属活动性强弱关系铝＞锌＞氢＞铜＞银，锌能与硝酸银、硝酸铜依次发生反应而不能与硝酸铝反应，向滤出的固体中加入稀硫酸，没有气泡冒出，说明锌粉无剩余，据此分析．','填空题',3.00,'5ac4d80fbaa346b7ae6d40de56cefbed',9,400,'金属活动性的探究','',0,'37','',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1843139,'水是生命之源，下列关于水说法均正确的一组是（　　）<br />①用肥皂水鉴别硬水和软水；<br />②冰由氧元素和氢元素组成；<br />③用滤纸过滤可以除去水中的所有杂质；<br />④电解水时生成的氢气和氧气的质量比为2：1；<br />⑤保持水化学性质的最小微粒是氢分子；<br />⑥用活性炭可以将硬水软化；<br />⑦净化水的方法中，蒸馏是单一净化程度最高的；<br />⑧水是由氢分子和氧原子组成的；<br />⑨活性炭可除去水中部分不溶性杂质和异味、色素等．','①②③⑧','⑤②①④','⑨①②⑦','⑥④⑤③','','C','【解答】解：①用肥皂水鉴别硬水和软水，遇肥皂水产生的泡沫少的是硬水，遇肥皂水产生的泡沫多的是软水，说法正确；<br />②冰由氧元素和氢元素组成，说法正确；<br />③用滤纸过滤可以除去水中的不可溶性的杂质，不能除去所有杂质，说法不正确；<br />④电解水时生成的氢气和氧气的质量比为1：8，说法不正确；<br />⑤保持水化学性质的最小微粒是水分子，说法不正确；<br />⑥活性炭不能减少水中钙、镁离子的化合物的含量，不可以将硬水软化，说法不正确；<br />⑦净化水的方法中，蒸馏是单一净化程度最高的，说法正确；<br />⑧水是由氢元素和氧元素组成的，说法不正确；<br />⑨活性炭可除去水中部分不溶性杂质和异味、色素等，说法正确．<br />由以上分析可知，C正确．<br />故选C．','【分析】根据水的组成、净化水的方法、电解水水的实验以及硬水和软水的鉴别方法等知识分析判断有关的说法．','选择题',2.00,'dc37bd75e3ad6557fc321164b472b26e',9,400,'过滤的原理、方法及其应用,电解水实验,水的组成,水的净化,硬水与软水,分子的定义与分子的特性','',2016,'37','2016春•潜江校级月考',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1843161,'下列化学反应，属于复分解反应的是（　　）','C+2CuO<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2Cu+CO<sub>2</sub>↑','2H<sub>2</sub>O<sub>2</sub><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;催化剂&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<sub>2</sub>O+O<sub>2</sub>↑','C<sub>2</sub>H<sub>5</sub>OH+3O<sub>2</sub><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2CO<sub>2</sub>+3H<sub>2</sub>O','Na<sub>2</sub>CO<sub>3</sub>+Ca（OH）<sub>2</sub>═CaCO<sub>3</sub>↓+2NaOH','','D','【解答】解：A、碳和氧化铜的反应属于置换反应，故A错误；<br />B、过氧化氢分解生成水和氧气，属于分解反应，故B错误；<br />C、酒精和氧气点燃生成二氧化碳和水，是氧化反应，故C错误；<br />D、碳酸钠和氢氧化钙反应生成碳酸钙沉淀和氢氧化钠，属于复分解反应，故D正确．<br />故选D．','【分析】复分解反应是指两种化合物相互交换成分生成两种新的化合物的反应，反应特点是：反应物和生成物都是化合物，且两种反应物相互交换成分．','选择题',3.00,'50ecc4236ffafba1daf46b6b56a2012c',9,400,'复分解反应及其应用','',2016,'37','2016•顺义区一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1843169,'下列说法不正确的是（　　）','石油是工业的血液，应保护好南海石油资源','稀土是不可再生的重要战略资源，应合理利用和保护','合金的熔点一般比纯金属高','地球上水储量是丰富的，但可利用的淡水资源是有限的','','C','【解答】解：A．石油是一种重要的资源，应保护石油资源，故正确；<br />B．稀土是不可再生的重要战略资源，应合理利用和保护，故正确；<br />C．合金的熔点比它的各成分金属的熔点低，故错误；<br />D．地球上的水储量是丰富的，但人类能直接利用的淡水资源仅占全球水量的0.3%，是有限的，故正确．<br />故选C．','【分析】A．根据资源保护来分析；<br />B．根据资源保护来分析；<br />C．合金的特点是：硬度大，熔点低，耐腐蚀；<br />D．根据人类能直接利用的淡水资源仅占全球水量的0.3%进行解答．','选择题',3.00,'3046c1ddc3dbfb9eef13487d83435a92',9,400,'水资源状况,合金与合金的性质,金属资源的保护,化石燃料及其综合利用','',2016,'32','2016•营口模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1843187,'金属材料应用广泛．<br />（1）地壳中含量最高的金属元素是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；使用最多的金属材料是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）菜刀用铁制而不用铝制是因为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；铝锅空烧容易穿孔的原因是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）铁制品锈蚀的过程，实际上是铁与空气中的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>发生化学反应，防止铁钉生锈的措施之一是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','铝$###$钢铁$###$铁的硬度比铝的大$###$铝的熔点较低$###$氧气、水蒸气$###$涂油等','【解答】解：（1）地壳中含量最高的金属元素是铝；使用最多的金属材料是钢铁．<br />（2）菜刀用铁制而不用铝制是因为铁的硬度比铝的大；铝锅空烧容易穿孔的原因是铝的熔点较低，易熔化．<br />（3）铁制品锈蚀的过程，实际上是铁与空气中的氧气、水蒸气发生化学反应，防止铁钉生锈的措施是涂油、刷漆等．<br />故答为：（1）①铝，②钢铁；（2）①铁的硬度比铝的大，②铝的熔点较低；（3）①氧气、水蒸气，②涂油等．','【分析】（1）根据地壳中元素的含量和使用最多的金属材料分析回答．<br />（2）根据铁和铝的性质和应用分析回答．<br />（3）根据铁生锈的条件和防止生锈的措施分析回答．','填空题',3.00,'e93e4db50928c9b1392ad2d518660901',9,400,'金属的物理性质及用途,金属材料及其应用,金属锈蚀的条件及其防护,地壳中元素的分布与含量','',2016,'37','2016•大连一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1843191,'配平下列化学方程式，并将等号补齐<br />（1）<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>KClO<SUB>3</SUB>&nbsp;<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span>&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>△</td></tr></table></span><u>&nbsp;&nbsp;&nbsp;&nbsp;</u>KCl+<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>O<SUB>2</SUB>↑<br />（2）<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>KMnO<SUB>4</SUB>&nbsp;<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span><u>&nbsp;&nbsp;&nbsp;&nbsp;</u>K<SUB>2</SUB>MnO<SUB>4</SUB>+<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>MnO<SUB>2</SUB>+<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>O<SUB>2</SUB>↑<br />（3）<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>H<SUB>2</SUB>O<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;通电&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span><u>&nbsp;&nbsp;&nbsp;&nbsp;</u>H<SUB>2</SUB>↑+<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>O<SUB>2</SUB>↑<br />（4）<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>H<SUB>2</SUB>O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span>&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span><u>&nbsp;&nbsp;&nbsp;&nbsp;</u>H<SUB>2</SUB>O+<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>O<SUB>2</SUB>↑<br />（5）<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>HgO<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;△&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span><u>&nbsp;&nbsp;&nbsp;&nbsp;</u>Hg+<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>O<SUB>2</SUB>↑','','','','','','2$###$2$###$3$###$2$###$1$###$1$###$1$###$2$###$2$###$1$###$2$###$2$###$1$###$2$###$2$###$1','【解答】解：（1）利用最小公倍数法进行配平，以氧原子作为配平的起点，氯酸钾、氧气前面的化学计量数分别为：2、3，最后调整氯化钾前面的化学计量数为2．<br />（2）本题可利用“定一法”进行配平，把K<SUB>2</SUB>MnO<SUB>4</SUB>的化学计量数定为1，则KMnO<SUB>4</SUB>、MnO<SUB>2</SUB>、O<SUB>2</SUB>前面的化学计量数分别为：2、1、1．<br />（3）利用最小公倍数法进行配平，以氧原子作为配平的起点，水、氧气前面的化学计量数分别为：2、1，最后调整氢气前面的化学计量数为2．<br />（4）本题可利用“定一法”进行配平，把过氧化氢的化学计量数定为1，则水、O<SUB>2</SUB>前面的化学计量数分别为：1、<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">1</td></tr><tr><td>2</td></tr></table></span>，同时扩大2倍，则过氧化氢、水、氧气前面的化学计量数分别是2、2、1．<br />（5）利用最小公倍数法进行配平，以氧原子作为配平的起点，氧化汞、氧气前面的化学计量数分别为：2、1，最后调整汞前面的化学计量数为2．<br /><br />故答案为：（1）2、2、3；（2）2、1、1、1；（3）2、2、1；（4）2、2、1；（5）2、2、1．','【分析】根据质量守恒定律：反应前后各原子的数目不变，选择相应的配平方法（最小公倍数法、定一法等）进行配平即可；配平时要注意化学计量数必须加在化学式的前面，配平过程中不能改变化学式中的下标；配平后化学计量数必须为整数．','填空题',3.00,'dc3328de073b7f274de706b774311065',9,400,'化学方程式的配平','',2016,'35','2016春•清河县校级期中',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1843197,'阅读材料，回答问题：<br />材料1．臭氧（化学式为O<SUB>3</SUB>）是淡蓝色气体，大气中的臭氧层能有效阻挡紫外线，保护地球的生存环境，但目前南极出现了臭氧层空洞，并有继续扩大的趋势．<br />材料2．复印机在工作时，会因高压放电产生一定浓度的臭氧，长期吸入大量臭氧会引起口干舌燥，咳嗽等不适症状，还可能诱发中毒性肺气肿．<br />材料3．臭氧发生器是在高压电极的作用下，将空气中的氧气转化为臭氧的装置．利用臭氧的强氧化性，可将其应用于游泳池、生活用水、污水的杀菌和消毒．<br />（1）请总结臭氧的有关知识：<br />①物理性质：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />②化学性质：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />③用途：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）请从分子构成的角度，指出氧气和臭氧的不同点：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）写出材料3中氧气转化为臭氧的化学方程式：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（4）已知磷和氧气在点燃的条件下可反应生成五氧化二磷，磷在臭氧中燃烧与在氧气中燃烧相似，生成物也相同．请写出磷在臭氧中燃烧的化学方程式：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（5）从上述材料中可见臭氧对人类有利有弊．请再举出一种物质，并说出其利弊：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','淡蓝色气体$###$具有强氧化性$###$用于游泳池、生活用水、污水的杀菌和消毒$###$构成它们各自分子的氧原子的个数不同$###$3O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;放电&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2O<SUB>3</SUB>$###$6P+5O<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>3P<SUB>2</SUB>O<SUB>5</SUB>$###$CO<SUB>2</SUB>可用于灭火但过多排放会造成温室效应','【解答】解：（1）①物质的颜色、状态属于物质的物理性质．故填：淡蓝色气体．<br />②氧化性属于物质的化学性质．故填：具有强氧化性．<br />③用途：用于游泳池、生活用水、污水的杀菌和消毒．<br />（2）氧气和臭氧的分子构成不同，故填：构成它们各自分子的氧原子的个数不同．<br />（3）氧气转化为臭氧的化学方程式为：3O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;放电&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2O<SUB>3</SUB><br />（4）磷在臭氧中燃烧生成五氧化二磷，故填：6P+5O<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>3P<SUB>2</SUB>O<SUB>5</SUB><br />（5）二氧化碳对人类既有有利的一面，也有不利的一面．例如，CO<SUB>2</SUB>可用于灭火但过多排放会造成温室效应，故填：CO<SUB>2</SUB>可用于灭火但过多排放会造成温室效应；','【分析】可以根据物质的性质和用途及其质量守恒定律等方面进行分析、判断，从而得出正确的结论．','书写',3.00,'22850f861f01a52544c3630557bed444',9,400,'氧元素组成的单质,化学性质与物理性质的差别及应用,书写化学方程式、文字表达式、电离方程式','荣成市',2016,'35','2016春•荣成市期中',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1843199,'如图是实验室常用的仪器．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao91/3de5f8d1-94d4-11e9-a030-b42e9921e93e_xkb54.png\" style=\"vertical-align:middle\" /><br />（1）图中仪器D的名称为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；（1分，其余每空2分）<br />（2）写出实验室制取二氧化碳的化学方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；若要制取并收集较多的二氧化碳，应选择的仪器有F、<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填序号）．<br />（3）利用（2）中组装的发生装置制取氧气的化学方程式是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；若要测量产生的氧气的体积，应选择下列装置中的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填序号）．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao95/3de70a40-94d4-11e9-addd-b42e9921e93e_xkb21.png\" style=\"vertical-align:middle\" />','','','','','','水槽$###$CaCO<SUB>3</SUB>+2HCl═CaCl<SUB>2</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑$###$ACEHJ$###$2H<SUB>2</SUB>O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span>&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O+O<SUB>2</SUB>↑$###$③','【解答】解：（1）图中仪器D是水槽； <br />（2）实验室制取二氧化碳的反应物是碳酸钙和盐酸，生成物是氯化钙、水、二氧化碳，方程式是：CaCO<SUB>3</SUB>+2HCl═CaCl<SUB>2</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑；该反应不需加热，要制取较多的气体，则发生装置可选较大的发生容器锥形瓶，因为题目中给出了长颈漏斗，说明用双孔塞，收集气体需要集气瓶，由于二氧化碳能溶于水不能用排水法收集，密度比空气大，可以用向上排空气法收集，需要用J直导管；<br />（3）由于该装置不需要加热，所以制取氧气的反应物应该是固液型，所以是过氧化氢溶液和二氧化锰．要测量产生O<SUB>2</SUB>体积，要从短管进气，长管出水，更利于水的排出，从而测量O<SUB>2</SUB>体积；<br />故答案为：<br />（1）水槽；（2）CaCO<SUB>3</SUB>+2HCl═CaCl<SUB>2</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑；ACEHJ；（3）2H<SUB>2</SUB>O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span>&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O+O<SUB>2</SUB>↑；③．','【分析】（1）熟记仪器名称；<br />（2）根据实验室制取二氧化碳的反应原理书写方程式，并据制取并收集二氧化碳的装置图回答选择的仪器；<br />（3）（2）中组装的发生装置适用于固体和液体不需加热制取气体，实验室用双氧水制取氧气不需加热，要测量产生O<SUB>2</SUB>体积，可利用排水法，排出水的体积即是产生O<SUB>2</SUB>体积，要从短管进气，长管出水．','书写',3.00,'ce55129bef7a6b4380f9dc379d851c8e',9,400,'量气装置,实验室制取氧气的反应原理,二氧化碳的实验室制法,书写化学方程式、文字表达式、电离方程式','',2016,'37','2016•上饶三模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1843231,'为了达到收旧利废的目的，欲从含有金属镁、铁、铜的粉末中，分离和提取出重要化工原料MgSO<SUB>4</SUB>和有关金属，其主要实验过程如下：<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao18/3e6e62ae-94d4-11e9-a9f7-b42e9921e93e_xkb94.png\" style=\"vertical-align:middle\" /><br />请回答：<br />（1）步骤①中被磁铁吸引的金属A是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）步骤③的操作是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）步骤②所涉及的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','铁$###$蒸发$###$Mg+H<SUB>2</SUB>SO<SUB>4</SUB>=MgSO<SUB>4</SUB>+H<SUB>2</SUB>↑','【解答】解：（1）根据题意：铁能够被磁铁吸引，因此A是铁；<br />（2）步骤③的操作是蒸发．<br />（3）镁和硫酸反应生成硫酸镁和氢气，其化学方程式为：Mg+H<SUB>2</SUB>SO<SUB>4</SUB>=MgSO<SUB>4</SUB>+H<SUB>2</SUB>↑．<br />故答案为：<br />（1）铁；（2）蒸发；（3）Mg+H<SUB>2</SUB>SO<SUB>4</SUB>=MgSO<SUB>4</SUB>+H<SUB>2</SUB>↑．','【分析】根据题目给出的信息：铁能够被磁铁吸引，因此A是铁；镁和铜的混合物中，加入适量的稀硫酸，镁和硫酸反应生成硫酸镁和氢气，铜不反应，因此过滤可以得到B，即金属铜；在蒸发过程中需要使用玻璃棒，其作用是搅拌作用，是为了防止局部温度过高造成液滴飞溅．','书写',3.00,'a812b84ad24817526c435c812989e3da',9,400,'过滤的原理、方法及其应用,常见金属的特性及其应用,金属的化学性质,书写化学方程式、文字表达式、电离方程式','',2016,'37','2016•通州区一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1843233,'<img  src=\"/tikuimages/9/2016/400/shoutiniao27/3e731da1-94d4-11e9-8e17-b42e9921e93e_xkb31.png\" style=\"vertical-align:middle;FLOAT:right\" />（2016•肥西县一模）牙膏是和牙刷一起用于清洁牙齿，保护口腔卫生，是对人体安全的一种日用必需品．某化学兴趣小组对某种牙膏的成分进行了如下探究：<br />探究一：牙膏的酸碱性<br />实验1：取少量牙膏放入烧杯中，加入足量的水溶解，静置，取上层清液，测定溶液的&nbsp;pH，pH<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>7（填”＞”或”＜”），牙膏呈碱性．<br />探究二：牙膏摩擦剂的成分<br />小明同学发现实验1烧杯底部有沉淀，便想探究其成分．<br />【查阅资料】牙膏成分中的固体原料为磨擦剂，它是擦去牙齿表面牙垢，减轻牙渍的成分．常见的摩擦剂有碳酸钙、二氧化硅（SiO<SUB>2</SUB>）等．<br />【猜想与假设】该牙膏摩擦剂成分可能是：①为碳酸钙；②二氧化硅；③你的猜想是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />实验2：【设计方案和实验】<br /><table class=\"edittable\"><TBODY><TR><td width=228>步骤</TD><td width=190>现象</TD><td width=163>结论</TD></TR><TR><td>①取少量沉淀，加入适量稀盐酸</TD><td>沉淀完全溶解，有气泡产生</TD><td rowSpan=2>假设<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>正确（填序号）</TD></TR><TR><td>②将产生的气体通入澄清的石灰水</TD><td><u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD></TR></TBODY></TABLE>请你写出步骤①的化学方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','＞$###$碳酸钙和二氧化硅混合物$###$①$###$澄清石灰水变浑浊$###$CaCO<SUB>3</SUB>+2HCl=CaCl<SUB>2</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑','【解答】解：<br />取少量牙膏放入烧杯中，加入足量的水溶解，静置，取上层清液，测定溶液的pH，pH＞7，牙膏呈碱性；<br />【猜想与假设】根据牙膏的组成成分可以猜想其成分可能是碳酸钙和二氧化硅混合物；<br />【设计方案和实验】<br />①取少量沉淀，A加入适量稀盐酸，沉淀完全溶解，有气泡产生，产生的气体通入澄清的石灰水，澄清石灰水变浑浊，说明没有二氧化硅，只有碳酸钙，故假设①正确；<br />碳酸钙与盐酸反应生成氯化钙、水和二氧化碳，化学方程式为：CaCO<SUB>3</SUB>+2HCl=CaCl<SUB>2</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑；<br />故答案为：<br />实验1：＞．<br />【猜想与假设】碳酸钙和二氧化硅混合物．<br /><table class=\"edittable\"><TBODY><TR><td width=228>步骤</TD><td width=190>现象</TD><td width=163>结论</TD></TR><TR><td></TD><td></TD><td rowSpan=2>&nbsp; ①</TD></TR><TR><td></TD><td>澄清石灰水变浑浊</TD></TR></TBODY></TABLE>CaCO<SUB>3</SUB>+2HCl=CaCl<SUB>2</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑．','【分析】实验1：根据牙膏呈碱性，pH＞7解答；<br />【猜想与假设】根据牙膏的成分进行猜想；<br />【设计方案和实验】根据碳酸钙与盐酸反应生成二氧化碳，二氧化碳与氢氧化钙反应生成碳酸钙沉淀解答；','书写',3.00,'d4040bc44da08138eb9a96fbe8b14d1e',9,400,'实验探究物质的组成成分以及含量,溶液的酸碱性测定,盐的化学性质,书写化学方程式、文字表达式、电离方程式','',2016,'37','2016•肥西县一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1843258,'丰富多彩的生活展示着化学与生活的完美结合．回答下列问题：<br />（1）用电力驱动代替燃油的电动汽车被称为“未来汽车”．<br />①图1中标示物<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填序号）是复合材料．<br />②生产电动汽车需要消耗大量的钢铁，写出工业上以磁铁矿（主要成分为四氧化三铁）为原料炼铁的化学反<br />应方程式：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）《生活饮用水卫生标准》中规定水的总硬度（以碳酸钙计）不超过450mg/L，测得某水样的硬度为600mg/L，则1吨该水样中含钙的质量约为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>克，生活中常用肥皂水来区分硬水和软水，若为软水，鉴别时加入肥皂水观察到的现象为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．K<SUB>2</SUB>FeO<SUB>4</SUB>是一种常用的净水剂，其中铁的化合价为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>价．<br />（3）氯化钙在工业上可用于制造防冻液等．某厂以石灰石为原料生产CaCl<SUB>2</SUB>•2H<SUB>2</SUB>O的流程如图2．原料石灰石中含有的杂质主要是Fe<SUB>2</SUB>O<SUB>3</SUB>、MgSO<SUB>4</SUB>和MgCO<SUB>3</SUB>．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao33/3ed1914f-94d4-11e9-89ff-b42e9921e93e_xkb78.png\" style=\"vertical-align:middle\" /><br />①试剂1的名称为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />②加入Ca（OH）<SUB>2</SUB>除去的杂质离子是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />③实验室进行过滤操作所用的玻璃仪器有烧杯、玻璃棒和<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />④若过滤后的滤液呈碱性，则加入适量盐酸的目的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','④$###$4CO+Fe<SUB>3</SUB>O<SUB>4</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>3Fe+4CO<SUB>2</SUB>$###$240$###$浮渣少，泡沫多$###$+6$###$氯化钡溶液$###$Fe<SUP>3+</SUP>和Mg<SUP>2+</SUP>$###$漏斗$###$除去过量的Ca（OH）<SUB>2</SUB>，提高产品纯度','【解答】解：（1）橡胶轮胎是由合成橡胶、金属材料复合而成的复合材料；在高温的条件下，一氧化碳与四氧化三铁反应生成铁和二氧化碳；故填：④；②4CO+Fe<SUB>3</SUB>O<SUB>4</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>3Fe+4CO<SUB>2</SUB>；<br />（2）1吨该水样中含钙的质量约为600mg×<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">40</td></tr><tr><td>40+12+16×3</td></tr></table>×100%</span>=240mg；不超过450mg/L，所以属于软水，加入肥皂水后浮渣少、泡沫多；钾元素显+1价，氧元素显-2价，设铁元素的化合价是x，根据在化合物中正负化合价代数和为零，可得：（+1）×2+x+（-2）×4=0，则x=+6价．<br />故填：240；浮渣少，泡沫多；+6；<br />（3）①加入试剂1是为了除去SO<SUB>4</SUB><SUP>2-</SUP>，由于溶液中含有氯离子，应选择适合合题意的试剂1是氯化钡溶液；故填：氯化钡溶液；<br />②氢氧化钙能与溶液中氯化铁、氯化镁反应生成氢氧化铁沉淀、氢氧化镁沉淀和氯化钙，因此，除去的杂质离子为镁离子、铁离子；故填：Fe<SUP>3+</SUP>和Mg<SUP>2+</SUP><br />③实验室进行过滤操作所用的玻璃仪器有烧杯、玻璃棒和漏斗，故填：漏斗；<br />④如果滤液显碱性，则前后操作中加入的氢氧化钙过量，可加入稀盐酸与氢氧化钙反应生成氯化钙和水，除去氢氧化钙而得到较为纯净的氯化钙；<br />故填：除去过量的Ca（OH）<SUB>2</SUB>，提高产品纯度．','【分析】（1）根据材料的分类以及化学方程式的写法来分析；<br />（2）根据化合物中元素的含量以及化合价的计算方法来分析；<br />（3）①根据除去SO<SUB>4</SUB><SUP>2-</SUP>常用的试剂分析回答；<br />②氢氧化钙能与溶液中氯化镁、氯化铁生成氢氧化镁、氢氧化铁的沉淀，因此，加入的氢氧化钙可除去溶液中镁离子、铁离子；<br />③根据过滤操作的实验来分析；<br />④若滤液呈碱性，可说明前面操作所加氢氧化钙过量，为得到较纯净氯化钙，应加入稀盐酸进行中和．','书写',3.00,'28beff6953b42ce46546069171bf235c',9,400,'过滤的原理、方法及其应用,硬水与软水,铁的冶炼,盐的化学性质,物质的相互转化和制备,有关元素化合价的计算,书写化学方程式、文字表达式、电离方程式,资源综合利用和新能源开发,复合材料、纳米材料','',2016,'32','2016•海安县模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1843278,'配平下列化学方程式，在横线上填上适当的化学计量数．<br />（1）<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>C<SUB>3</SUB>H<SUB>6</SUB>+<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>O<SUB>2</SUB> <span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span><u>&nbsp;&nbsp;&nbsp;&nbsp;</u>CO<SUB>2</SUB>+<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>H<SUB>2</SUB>O．<br />&nbsp;（2）<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>Fe<SUB>2</SUB>O<SUB>3</SUB>+<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>HCl═<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>FeCl<SUB>3</SUB>+<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>H<SUB>2</SUB>O．','','','','','','2$###$9$###$6$###$6$###$1$###$6$###$2$###$3','【解答】解：（1）本题可利用“定一法”进行配平，把C<SUB>3</SUB>H<SUB>6</SUB>的化学计量数定为1，则氧气、二氧化碳、水前面的化学计量数分别为：<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">9</td></tr><tr><td>2</td></tr></table></span>、3、3，同时扩大2倍，则C<SUB>3</SUB>H<SUB>6</SUB>、氧气、二氧化碳、水前面的化学计量数分别为：2、9、6、6．<br />（2）本题可利用“定一法”进行配平，把Fe<SUB>2</SUB>O<SUB>3</SUB>的化学计量数定为1，则HCl、FeCl<SUB>3</SUB>、H<SUB>2</SUB>O前面的化学计量数分别为：6、2、3．<br />故答案为：（1）2、9、6、6；（2）1、6、2、3．','【分析】根据质量守恒定律：反应前后各原子的数目不变，选择相应的配平方法（最小公倍数法、定一法等）进行配平即可；配平时要注意化学计量数必须加在化学式的前面，配平过程中不能改变化学式中的下标；配平后化学计量数必须为整数．','填空题',3.00,'661af7492068ecbc46c29dbbf916cfa1',9,400,'化学方程式的配平','',0,'37','',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1843331,'判断蜡烛燃烧是化学变化的依据是（　　）','发出黄白色火焰','固态石蜡变成了液态','生成了二氧化碳和水','放出了大量的热','','C','【解答】解：判断蜡烛燃烧燃烧是否发生了化学变化，关键是看是否有新物质的生成，选项A、B、D均是描述了蜡烛燃烧的现象，不能作为判断化学变化的依据；而选项C是说明蜡烛燃烧有新物质生成，是判断蜡烛燃烧发生化学变化的依据．<br />故选C．','【分析】化学变化是指有新物质生成的变化，化学变化常常伴随着发光、放热、颜色变化、产生气体、生成沉淀等现象，但有这些现象不一定是化学变化，化学变化的本质特征是有新物质生成，据此逐项分析判断．','选择题',3.00,'ba95fd6743e6f1c0752e9f692a5be30c',9,400,'化学变化的基本特征','',2015,'37','2015秋•广元校级月考',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1843341,'厨房中存在许多物理和化学知识，以下说法中错误的是（　　）','向切好的紫色卷心菜中倒入食醋，颜色发生变化-植物汁液与食醋发生了反应','打开香油瓶盖，香气扑鼻-分子在不断运动','菜刀钝了要磨一磨-为了减小压强','揭开正在烧水的锅盖，看到“白汽”-水蒸气液化','','C','【解答】解：A、向切好的紫色卷心菜中倒入食醋，颜色发生变化-植物汁液与食醋发生了反应，正确；<br />B、打开香油瓶盖，香气扑鼻，是因为分子在不断运动，正确；<br />C、菜刀钝了要磨一磨，是在压力一定的情况下，减小受力面积，增大压强，错误；<br />D、揭开正在烧水的锅盖，看到“白汽”，是因为水蒸气液化的缘故，正确；<br />故选C．','【分析】A、根据指示剂变色的知识解答；<br />B、根据分子的基本性质解答；<br />C、根据压强的知识解答；<br />D、根据液化的知识解答．','选择题',3.00,'cbfb24ba4cfd3d5e3bec6be2b0b95902',9,400,'酸的化学性质,物质的三态及其转化,分子的定义与分子的特性','',2016,'37','2016•石家庄一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1843387,'某物质的溶液中加入氯化钡溶液，产生了不溶于稀硝酸的白色沉淀，下列说法中正确的是（　　）','原溶液中一定含有Ag<SUP>+</SUP>','原溶液中一定含有SO<SUB>4</SUB><SUP>2-</SUP>','原溶液中一定不含CO<SUB>3</SUB><SUP>2-</SUP>','原溶液中可能含有Ag<SUP>+</SUP>或SO<SUB>4</SUB><SUP>2-</SUP>','','D','【解答】解：一种溶液中加入氯化钡溶液，产生了不溶于稀硝酸的白色沉淀，则生成的沉淀可能是硫酸钡，也可能是氯化银，则这种溶液中一定含有SO<SUB>4</SUB><SUP>2-</SUP>或Ag<SUP>+</SUP>．<br />A、这种溶液中不一定含有SO<SUB>4</SUB><SUP>2-</SUP>，也可能含有Ag<SUP>+</SUP>，故选项说法错误．<br />B、这种溶液中不一定含有SO<SUB>4</SUB><SUP>2-</SUP>，也可能含有Ag<SUP>+</SUP>，故选项说法错误．<br />C、这种溶液中可能含有SO<SUB>4</SUB><SUP>2-</SUP>，不可能含有碳酸根离子，因为碳酸根离子与氯化钡反应生成的碳酸钡沉淀，能溶于稀硝酸，故选项说法错误．<br />D、这种溶液中一定含有SO<SUB>4</SUB><SUP>2-</SUP>或Ag<SUP>+</SUP>中的一种，故选项说法正确．<br />故选：D．','【分析】根据题意，一种溶液中加入氯化钡溶液，产生了不溶于稀硝酸的白色沉淀，结合不溶于酸的沉淀有硫酸钡、氯化银，进行分析判断．','选择题',3.00,'9c6cf27c81b530d79980c7604042389f',9,400,'常见离子的检验方法及现象','',2015,'37','2015秋•滨州校级月考',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1843389,'下列关于“一定”和“不一定”叙述错误的是（　　）','在变化中有颜色改变的不一定是化学变化','在变化中有状态变化的一定是物理变化','在变化中有发光、放热现象的一定是化学变化','在化学变化中不一定发生物理变化','','A','【解答】解：A、在变化中有颜色改变的不一定是化学变化，例如氧气降温下变为淡蓝色的液氧，故说法正确；<br />B、在变化中有状态变化的不一定是物理变化，例如水通电分解为氢气和氧气，故说法错误；<br />C、在变化中有发光放热现象的不一定是化学变化，例如灯泡发光发热就属于物理变化，故说法错误；<br />D、在物理变化中不一定发生化学变化，化学变化中一定包含化学变化，故说法错误．<br />故选：A．','【分析】根据化学变化和物理变化的概念以及化学变化的特征和常常伴随的现象进行分析．','选择题',3.00,'36e2725203701bc24eb0a6ef132b50c6',9,400,'化学变化的基本特征,物理变化的特点','',0,'37','',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1843405,'下列化学反应属于化合反应的是（　　）','镁+氧气&nbsp;<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td style=\"font-size: 90%\">点燃</td></tr><tr><td><div hassize=\"9\"><div style=\"background: url(\'/upimages/images/part/8594L.png\') repeat-x; width:1px; height:5px;float:left;overflow:hidden\" muststretch=\"h\"></div><div style=\"background: url(\'/upimages/images/part/8594.png\') no-repeat; width:9px; height:5px;float:left;overflow:hidden\"></div></div></td></tr></table></span>&nbsp;&nbsp;氧化镁','高锰酸钾&nbsp;<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td style=\"font-size: 90%\">△</td></tr><tr><td><div hassize=\"9\"><div style=\"background: url(\'/upimages/images/part/8594L.png\') repeat-x; width:1px; height:5px;float:left;overflow:hidden\" muststretch=\"h\"></div><div style=\"background: url(\'/upimages/images/part/8594.png\') no-repeat; width:9px; height:5px;float:left;overflow:hidden\"></div></div></td></tr></table></span>锰酸钾+二氧化锰+氧气','盐酸+锌→氯化锌+氢气','氧化铁+一氧化碳<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td style=\"font-size: 90%\">高温</td></tr><tr><td><div hassize=\"9\"><div style=\"background: url(\'/upimages/images/part/8594L.png\') repeat-x; width:1px; height:5px;float:left;overflow:hidden\" muststretch=\"h\"></div><div style=\"background: url(\'/upimages/images/part/8594.png\') no-repeat; width:9px; height:5px;float:left;overflow:hidden\"></div></div></td></tr></table></span>&nbsp;&nbsp;铁+二氧化碳','','A','【解答】解：A、镁+氧气 <span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td style=\"font-size: 90%\">点燃</td></tr><tr><td><div hassize=\"9\"><div style=\"background: url(\'/upimages/images/part/8594L.png\') repeat-x; width:1px; height:5px;float:left;overflow:hidden\" muststretch=\"h\"></div><div style=\"background: url(\'/upimages/images/part/8594.png\') no-repeat; width:9px; height:5px;float:left;overflow:hidden\"></div></div></td></tr></table></span> 氧化镁，该反应符合“多变一”的特征，属于化合反应，故选项正确．<br />B、高锰酸钾 <span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td style=\"font-size: 90%\">△</td></tr><tr><td><div hassize=\"9\"><div style=\"background: url(\'/upimages/images/part/8594L.png\') repeat-x; width:1px; height:5px;float:left;overflow:hidden\" muststretch=\"h\"></div><div style=\"background: url(\'/upimages/images/part/8594.png\') no-repeat; width:9px; height:5px;float:left;overflow:hidden\"></div></div></td></tr></table></span>锰酸钾+二氧化锰+氧气，该反应符合“一变多”的特征，属于分解反应，故选项错误．<br />C、盐酸+锌→氯化锌+氢气，该反应的生成物是两种，不符合“多变一”的特征，不属于化合反应，故选项错误．<br />D、氧化铁+一氧化碳<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td style=\"font-size: 90%\">高温</td></tr><tr><td><div hassize=\"9\"><div style=\"background: url(\'/upimages/images/part/8594L.png\') repeat-x; width:1px; height:5px;float:left;overflow:hidden\" muststretch=\"h\"></div><div style=\"background: url(\'/upimages/images/part/8594.png\') no-repeat; width:9px; height:5px;float:left;overflow:hidden\"></div></div></td></tr></table></span> 铁+二氧化碳，该反应的生成物是两种，不符合“多变一”的特征，不属于化合反应，故选项错误．<br />故选：A．','【分析】化合反应：两种或两种以上物质反应后生成一种物质的反应，其特点可总结为“多变一”，据此进行分析判断．','选择题',3.00,'ea0fb19c9b8cd92013d968e48b5353f3',9,400,'化合反应及其应用','',2015,'35','2015秋•聊城期中',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1843416,'下列关于溶液的说法，正确的是（　　）','溶液都是无色透明的','用汽油洗去衣服上的油污是乳化现象','饱和溶液不能再溶解任何物质','硝酸铵溶解在水中，溶液温度会降低','','D','【解答】解：A、溶液可以有颜色，例如硫酸铜溶液是蓝色，故A错；<br />B、汽油洗去衣服上的油污属于溶解现象，故B错；<br />C、饱和溶液只是在该温度下，不能再溶解某一溶质的溶液，可以溶解其它物质，故C错；<br />D、硝酸铵溶解在水中，溶液温度会降低，故D正确．<br />故选D．','【分析】A、溶液可以有颜色；B、汽油洗去衣服上的油污属于溶解现象；C、饱和溶液只是在该温度下，不能再溶解某一溶质的溶液；D、硝酸铵溶解在水中，溶液温度会降低．','选择题',3.00,'e6b21e083488b00d8663080869f5df80',9,400,'溶解现象与溶解原理,溶液的概念、组成及其特点,溶解时的吸热或放热现象,饱和溶液和不饱和溶液','',2016,'37','2016•潍坊一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1843422,'通过海水晾晒可得粗盐，粗盐除NaCl外，还含有MgCl<SUB>2</SUB>、CaCl<SUB>2</SUB>、Na<SUB>2</SUB>SO<SUB>4</SUB>以及泥沙等杂质．以下是制备精盐的实验方案，各步操作流程如图：<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao92/40c3eb70-94d4-11e9-b84d-b42e9921e93e_xkb1.png\" style=\"vertical-align:middle\" /><br />（1）在第①步粗盐“溶解”操作要用玻璃棒搅拌，作用是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）第②步操作的目的是除去粗盐中的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填化学式，下同），第⑥步操作的目的是除去滤液中过量的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','搅拌，加速溶解$###$Na<SUB>2</SUB>SO<SUB>4</SUB>$###$NaOH和Na<SUB>2</SUB>CO<SUB>3</SUB>','【解答】解：（1）用玻璃棒搅拌，加速了液体的流动，使固体很快溶解；<br />（2）加入过量的氯化钡，氯离子对氯化钠来说不是杂质，钡离子可以将硫酸根离子转化为沉淀，第②步操作加入过量的BaCl<SUB>2</SUB>目的是除去粗盐中的硫酸钠，反应的化学方程式Na<SUB>2</SUB>SO<SUB>4</SUB>+BaCl<SUB>2</SUB>=BaSO<SUB>4</SUB>↓+2NaCl；最后得到的滤液中溶质除含有氯化钠外还含有碳酸钠和氢氧化钠，利用粗盐制备精盐过程的第⑥步操作中，加入适量盐酸的目的是除去多余的氢氧化钠和碳酸钠；<br />故答案为：<br />（1）搅拌，加速溶解；　（2）Na<SUB>2</SUB>SO<SUB>4</SUB>；NaOH和Na<SUB>2</SUB>CO<SUB>3</SUB>　&nbsp;（必须填写化学式）','【分析】（1）可以从粗盐溶解过程中的玻璃棒的作用分析；<br />（2）从加入的物质中的离子作用角度分析；根据滤液中溶质含有碳酸钠和氢氧化钠进行解答；','书写',3.00,'2cdf799fe6127d837b44775c1cb383dd',9,400,'物质的溶解,氯化钠与粗盐提纯,盐的化学性质,书写化学方程式、文字表达式、电离方程式','',2016,'37','2016•祁阳县二模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1843430,'<img  src=\"/tikuimages/9/2013/400/shoutiniao55/40e3a870-94d4-11e9-843b-b42e9921e93e_xkb31.png\" style=\"vertical-align:middle;FLOAT:right;\" />（2013•古浪县校级三模）如图所示，A、B为两个电极，放在盛有稀硫酸的玻璃容器内．闭合开关后，A电极上出现的气泡为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，B电极上出现的气泡为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，在这一过程中，电流表示数逐渐变<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；灯泡亮度逐渐变<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','氧气$###$氢气$###$变大$###$亮','【解答】解：A管与电源的正极相连其中的气体是氧气，能使带火星的木条复燃，B管与负极相连其中的气体是氢气，能够燃烧，氧气和氢气的体积比是1：2．电解水的过程中，随着水不断被电解，玻璃容器内的稀硫酸溶液的浓度增大，导电能力增强，电流变大，小灯泡亮度增加．<br />故答案为：氧气；氢气；变大；亮．','【分析】电解水时，与正极相连的试管产生的是氧气，与负极相连的试管产生的是氢气，氧气和氢气的体积比约为1：2；<br />纯净水几乎不导电，加入稀硫酸溶液后，能够电离出大量的氢离子、硫酸根离子，能够增强水的导电性；','填空题',3.00,'caee9e82a9facbfd3b3d2fc80ede01ac',9,400,'电解水实验,溶液的导电性及其原理分析','',2013,'37','2013•古浪县校级三模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1843440,'<img  src=\"/tikuimages/9/2016/400/shoutiniao57/410190b0-94d4-11e9-8345-b42e9921e93e_xkb90.png\" style=\"vertical-align:middle;FLOAT:right\" />（2016•平谷区一模）同学们发现将NaOH溶液和稀盐酸混合后没有明显现象，为了证明NaOH能与HCl发生中和反应，设计了如下实验：<br /><table class=\"edittable\"><TBODY><TR><td width=190>实验步骤</TD><td width=190>现象</TD><td width=190>结论</TD></TR><TR><td>实验1．取少量NaOH溶液于试管中，向其中滴加几滴无色酚酞试液，振荡．</TD><td>溶液呈<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>色</TD><td></TD></TR><TR><td>实验2．向实验1的试管中加入适量的稀盐酸，振荡．</TD><td><u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．</TD><td>NaOH与HCl能反应．</TD></TR></TBODY></TABLE>【解释与结论】写出NaOH与HCl反应的化学方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />【反思与评价】同学们对实验2反应后所得溶液的酸碱性进行讨论，小玲同学认为呈碱性，小生同学认为呈中性，小柏同学认为也可能呈酸性，最后同学们否定了小玲同学的说法，其理由是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；欲进一步确定实验2反应后所得溶液的酸碱性，其操作是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','红$###$红色逐渐消失$###$NaOH+HCl═NaCl+H<SUB>2</SUB>O$###$红色逐渐消失$###$用试管取少量所得溶液，滴入1-2滴紫色石蕊试液，振荡','【解答】解：向氢氧化钠溶液中滴入几滴酚酞试液，再逐滴加入稀盐酸，溶液由红色变为无色，则可证明二者发生了化学反应；NaOH与HCl反应的化学方程式为NaOH+HCl═NaCl+H<SUB>2</SUB>O．<br />【反思与评价】同学们对实验2反应后所得溶液的酸碱性进行讨论，小玲同学认为呈碱性，小生同学认为呈中性，小柏同学认为也可能呈酸性，最后同学们否定了小玲同学的说法，其理由是红色逐渐消失，故不可能为碱性；欲进一步确定实验2反应后所得溶液的酸碱性，其操作是用试管取少量所得溶液，滴入1-2滴紫色石蕊试液，振荡；观察实验现象及结论．<br />故答案为：<br /><table class=\"edittable\"><TBODY><TR><td width=190>实验步骤</TD><td width=190>现象</TD><td width=190>结论</TD></TR><TR><td></TD><td>溶液呈 红色</TD><td></TD></TR><TR><td></TD><td>红色逐渐消失．</TD><td>NaOH与HCl能反应．</TD></TR></TBODY></TABLE>【解释与结论】NaOH+HCl═NaCl+H<SUB>2</SUB>O．<br />【反思与评价】红色逐渐消失；用试管取少量所得溶液，滴入1-2滴紫色石蕊试液，振荡．','【分析】根据氢氧化钠溶液显碱性，可使酚酞变红，而酸性、中性溶液不能使酚酞变色设计实验；盐酸和氢氧化钠溶液反应生成了氯化钠和水，其中盐酸和氯化钠都不能使酚酞变色；可以据此解答该题；<br />测定溶液的酸碱性，可以考虑紫色石蕊试液或pH试纸，进行分析．','书写',3.00,'84369a9580c49e62206d3f9d3dcad472',9,400,'中和反应及其应用,溶液的酸碱性测定,书写化学方程式、文字表达式、电离方程式','',2016,'37','2016•平谷区一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1843515,'甲、乙两位同学分别设计下述两个实验方案，并都认为如果观察到的现象和自己设计的方案一致，即可证明溶液中含有SO<SUB>4</SUB><SUP>2-</SUP><br />甲同学的方案：溶液&nbsp;&nbsp;<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td style=\"font-size: 90%\">BaC<span><span>l</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span>溶液</td></tr><tr><td><div hassize=\"9\"><div style=\"background: url(\'/upimages/images/part/8594L.png\') repeat-x; width:1px; height:5px;float:left;overflow:hidden\" muststretch=\"h\"></div><div style=\"background: url(\'/upimages/images/part/8594.png\') no-repeat; width:9px; height:5px;float:left;overflow:hidden\"></div></div></td></tr></table></span>&nbsp;&nbsp;白色沉淀&nbsp;&nbsp;<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td style=\"font-size: 90%\">加足足量稀盐酸</td></tr><tr><td><div hassize=\"9\"><div style=\"background: url(\'/upimages/images/part/8594L.png\') repeat-x; width:1px; height:5px;float:left;overflow:hidden\" muststretch=\"h\"></div><div style=\"background: url(\'/upimages/images/part/8594.png\') no-repeat; width:9px; height:5px;float:left;overflow:hidden\"></div></div></td></tr></table></span>&nbsp;沉淀不溶解<br />乙同学的方案：溶液&nbsp;&nbsp;<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td style=\"font-size: 90%\">加足足量稀盐酸</td></tr><tr><td><div hassize=\"9\"><div style=\"background: url(\'/upimages/images/part/8594L.png\') repeat-x; width:1px; height:5px;float:left;overflow:hidden\" muststretch=\"h\"></div><div style=\"background: url(\'/upimages/images/part/8594.png\') no-repeat; width:9px; height:5px;float:left;overflow:hidden\"></div></div></td></tr></table></span>&nbsp;&nbsp;无沉淀&nbsp;&nbsp;<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td style=\"font-size: 90%\">BaC<span><span>l</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span>溶液</td></tr><tr><td><div hassize=\"9\"><div style=\"background: url(\'/upimages/images/part/8594L.png\') repeat-x; width:1px; height:5px;float:left;overflow:hidden\" muststretch=\"h\"></div><div style=\"background: url(\'/upimages/images/part/8594.png\') no-repeat; width:9px; height:5px;float:left;overflow:hidden\"></div></div></td></tr></table></span>&nbsp;&nbsp;白色沉淀<br />你认为哪种方案合理，为什么？','','','','','','','【解答】解：<br />对于甲同学，先加氯化钡会生成白色沉淀，硫酸根离子会与钡离子生成硫酸钡沉淀，银离子也会与氯离子生成沉淀，加盐酸沉淀都不溶解；<br />对于乙同学，加盐酸没有沉淀，说明了不存在银离子，再加氯化钡会生成沉淀，就只能是硫酸钡沉淀．<br />故答案为：乙；甲不能有效地排除银离子的干扰，而乙可以排除银离子和碳酸根离子的干扰．','【分析】根据氯离子会与银离子生成氯化银沉淀，先加盐酸会排除银离子的干扰进行分析．','解答题',3.00,'70d3d888ff2857a695a1fd1cabb6f375',9,400,'证明硫酸和可溶性硫酸盐','',2016,'37','2016春•常州校级月考',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1843517,'复分解反应发生的条件是有<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>、<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>或水生成．其实质是溶液中自由移动的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>数目减少．如图所示，碳酸钠和盐酸反应的类型为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>反应．图中所示反应的微观实质是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>离子（填写离子符号，下同）和<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>&nbsp;离子结合生成了化学性质不稳定的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，分解生成了<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，从而导致反应的发生．<br /><br /><img  src=\"/tikuimages/9/2015/400/shoutiniao52/4208c780-94d4-11e9-9b1a-b42e9921e93e_xkb79.png\" style=\"vertical-align:middle\" />','','','','','','气体$###$沉淀$###$离子$###$复分解$###$H<SUP>+</SUP>$###$CO<SUB>3</SUB><SUP>2-</SUP>$###$H<SUB>2</SUB>CO<SUB>3</SUB>$###$H<SUB>2</SUB>O和CO<SUB>2</SUB>','【解答】解：复分解反应发生的条件是有气体、沉淀或水生成．其实质是溶液中自由移动的离子数目减少．如图所示，碳酸钠和盐酸反应的类型为复分解反应．图中所示反应的微观实质是H<SUP>+</SUP>和CO<SUB>3</SUB><SUP>2-</SUP>&nbsp;结合生成了化学性质不稳定的H<SUB>2</SUB>CO<SUB>3</SUB>，H<SUB>2</SUB>CO<SUB>3</SUB>分解生成了H<SUB>2</SUB>O和CO<SUB>2</SUB>，从而导致反应的发生．<br />故答案为：气体，沉淀，复分解反应，H<SUP>+</SUP>，CO<SUB>3</SUB><SUP>2-</SUP>，H<SUB>2</SUB>CO<SUB>3</SUB>，H<SUB>2</SUB>O，CO<SUB>2</SUB>．','【分析】根据复分解反应的条件、盐酸与碳酸钠反应会生成氯化钠、水与二氧化碳结合图示中离子的情况完成有关的问题．','填空题',3.00,'e074b23eb131fec16ddc838b50908ecc',9,400,'复分解反应的条件与实质,反应类型的判定','',2015,'37','2015秋•滨州校级月考',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1843547,'如图是初中化学中常用的实验装置，请回答下列问题．<img  src=\"/tikuimages/9/2016/400/shoutiniao82/4275ba1e-94d4-11e9-8736-b42e9921e93e_xkb22.png\" style=\"vertical-align:middle\" /><br />（1）仪器x的名称是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（2）实验过程中经常要连接仪器．将玻璃导管插入带孔橡皮塞时，应先将要插入塞子的玻璃导管的一端<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，然后稍稍用力转动，使它插入带孔橡皮塞；<br />（3）实验室制取O<SUB>2</SUB>、CO<SUB>2</SUB>、H<SUB>2</SUB>均可选择<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填装置代号）作为气体的发生装置；<br />（4）实验室不使用催化剂就能制取氧气的反应化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；若要制取纯度较高的氧气，应采用的收集方法是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（5）请从如图1装置选择并连接成一套组合装置，用于实验室制取干燥的二氧化碳气体．<br />按照气体从左至右的流向写出所选装置接口的连接顺序：<br /><u>&nbsp;&nbsp;&nbsp;&nbsp;</u>→<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>→<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>→<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>→<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>→<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>→<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>…（可不填满，也可补充）<br />（6）氨气是一种无色、有强烈刺激性气味的气体，极易溶于水，水溶液显碱性．实验室常加热氯化铵和熟石灰两种固体混合物来制取氨气．某兴趣小组的同学设计如图2G～J所示装置对氨气的制取、性质进行探究．<br />①装置I中的现象是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />②装置H的作用是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />③根据氨气的性质，J处应选择的最佳收集装置为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；（填“K”、“L”或“M”）<br />（7）将CO<SUB>2</SUB>和O<SUB>2</SUB>通入图3进行分离，最后收集到的气体是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />A．CO<SUB>2</SUB>、O<SUB>2</SUB>&nbsp;&nbsp;&nbsp;&nbsp;B．CO<SUB>2</SUB>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;C．O<SUB>2</SUB>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;D．O<SUB>2</SUB>、H<SUB>2</SUB>O．','','','','','','集气瓶$###$用水润湿$###$B$###$2KMnO<SUB>4</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>K<SUB>2</SUB>MnO<SUB>4</SUB>+MnO<SUB>2</SUB>+O<SUB>2</SUB>↑$###$排水集气法$###$b$###$h$###$i$###$d$###$c$###$-$###$-$###$导管口有气泡冒出，酚酞溶液变红$###$防止倒吸$###$M$###$B','【解答】解：（1）仪器x是集气瓶；故填：集气瓶；&nbsp;&nbsp;<br />（2）将玻璃导管插入带孔橡皮塞时，应先将要插入塞子的玻璃导管的一端用水润湿（水起润滑作用），然后稍用力转动使之插入橡皮塞内．故答案为：用水润湿；<br />（3）实验室用过氧化氢制取氧气、大理石和稀盐酸反应制取二氧化碳、锌粒和硫酸反应制取氢气均不需加热，属于固液常温型，可用B作发生装置；故填：B；<br />（4）加热高锰酸钾制取氧气不需要催化剂，加热生成锰酸钾、二氧化锰和氧气，反应的方程式是：2KMnO<SUB>4</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>K<SUB>2</SUB>MnO<SUB>4</SUB>+MnO<SUB>2</SUB>+O<SUB>2</SUB>↑，氧气不易溶于水，所以<br />收集纯度较高的气体氧气用排水法；故填：2KMnO<SUB>4</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>K<SUB>2</SUB>MnO<SUB>4</SUB>+MnO<SUB>2</SUB>+O<SUB>2</SUB>↑；排水集气法；&nbsp;&nbsp;<br />（5）在实验室中制取二氧化碳常用大理石或石灰石和稀盐酸反应来制取，属于固液常温型，所以应该选择B装置来制取二氧化碳气体；利用浓硫酸可以干燥，使用浓硫酸进行干燥，长管进短管出；二氧化碳密度比空气的大，用C收集时，应从长管通入二氧化碳，故填：b；h；i；d；c；<br />bhidc；<br />（6）①氨气是一种无色、有强烈刺激性臭味的气体，极易溶于水．水溶液呈碱性．能使无色酚酞溶液变红；<br />②由于氨气易溶于水，装置H的作用是防止液体倒流入装置G中，使试管炸裂；<br />③氨气是一种无色、有强烈刺激性臭味的气体，为防止氨气跑到空气中去，J处应选择的最佳收集装置为M．<br />故答案为：①导管口有气泡冒出，酚酞溶液变红；&nbsp;&nbsp;<br />②防止倒吸；&nbsp;&nbsp;<br />③M；<br />（7）混合气体通入后，氢氧化钠会与二氧化碳反应生成碳酸钠，氧气不反应，稀硫酸先不与碳酸钠接触反应，所以先分离出来的只有氧气，浓硫酸有吸水性，后将稀硫酸加入到碳酸钠溶液中，稀硫酸与碳酸钠反应生成二氧化碳，被浓硫酸干燥后得到较为纯净的二氧化碳．所以将CO<SUB>2</SUB>和O<SUB>2</SUB>通入图L进行分离，最终得到的气体是二氧化碳，故选：B．','【分析】（1）熟记仪器的名称；<br />（2）根据仪器连接的方法来分析；<br />（3）据实验室制取氧气、二氧化碳、氢气的方法分析解答；<br />（4）实验室制取氧气可用加热高锰酸钾，氯酸钾和二氧化锰混合物的方法及二氧化锰催化过氧化氢的方法，只有高锰酸钾制取氧气不需要催化剂，据反应原理书写方程式，收集纯度较高的气体氧气用排水法．<br />（5）根据反应物的状态和反应条件选择发生装置，利用浓硫酸可以干燥，根据二氧化碳的密度选择收集方法，然后进行顺序连接；<br />（6）根据实验室用加热氯化铵和熟石灰两种固体的混合物来制取氨气；制取气体主要看反应物的状态和反应条件；收集气体看气体的密度与水溶性；浓硫酸具有吸水性，可以做干燥剂；<br />（7）根据氢氧化钠会与二氧化碳反应，浓硫酸有吸水性进行分析．','书写',3.00,'30e88ff1d4a48955a27bd97d72d08885',9,400,'仪器的装配或连接,常用气体的发生装置和收集装置与选取方法,常见气体的检验与除杂方法,书写化学方程式、文字表达式、电离方程式','',2016,'37','2016•潮州校级一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1843554,'水是宝贵的自然资源，以下关于水的叙述中正确的是（　　）','淡水资源取之不尽，用之不竭','水由氢气和氧气组成','用肥皂水区别不开硬水和软水','改大水漫灌为喷灌、滴灌的方法','','D','【解答】解：<br />A、地球上的总水量很大，但淡水很少、分布不均，故错误；<br />B、水由氢元素和氧元素组成的，故错误；<br />C、用肥皂水可以区别硬水和软水，泡沫多的是软水，泡沫少的是硬水，故错误；<br />D、改水漫灌为喷灌、滴灌的方法可以节约用水，故正确．<br />答案：D．','【分析】A、根据淡水资源的状况分析判断；<br />B、根据水由氢元素和氧元素组成的解答；<br />C、根据区分硬水和软水的方法解答；<br />D、根据改水漫灌为喷灌、滴灌的方法可以节约用水解答．','选择题',3.00,'d8cfcb74ab23eb7a4b2b022c5da513e5',9,400,'水的组成,硬水与软水,水资源状况,保护水资源和节约用水','',2015,'37','2015秋•承德校级月考',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1843563,'下列有关能源资源叙述中正确的是（　　）','新能源的开发和利用很重要，所以我们应该大力开发水能、电能、核能等新能源','我国的金属矿藏比较丰富，其中金、银、铜等金属都以单质的形式存在','作为高能燃料，液氢已应用于航天等领域','我们周围的空气可以用于化肥，炼钢，石油加工等，但不能用于发电','','C','【解答】解：A．水能和核能属于新能源，而电能不属于新能源，错误；<br />B．只有少量不活泼的金属（金、银、铂）以单质的形式存在，而铜还是以化合物的形式存在为主，错误；<br />C．氢能是一种高能燃料，已经应用于航天、航空，正确；<br />D．空气是一种重要的自然资源，广泛的应用于生产化肥、炼钢、石油加工、发电等，错误．<br />故选C．','【分析】A．根据新能源的分类来分析；<br />B．根据金属的储量和金属的存在形式来分析；<br />C．根据氢能的应用来分析；<br />D．根据空气的重要用途来分析．','选择题',3.00,'69866d18668d1ca36e4e80d235cfe585',9,400,'空气对人类生活的重要作用,金属元素的存在及常见的金属矿物,资源综合利用和新能源开发','',2016,'37','2016•道里区二模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1843567,'下列实验操作、现象和结论对应正确的是（　　）<br /><table class=\"edittable\"><TBODY><TR><td width=17></TD><td width=270>实验操作</TD><td width=100>现象</TD><td width=165>结论</TD></TR><TR><td>A</TD><td>向久置于空气中的NaOH溶液滴加稀H<SUB>2</SUB>SO<SUB>4</SUB></TD><td>　有气泡产生</TD><td>NaOH和H<SUB>2</SUB>SO<SUB>4</SUB>发生了化学反应</TD></TR><TR><td>B</TD><td>向某溶液中滴加BaCl<SUB>2</SUB>溶液</TD><td>　有白色沉淀</TD><td>原溶液含有SO<SUB>4</SUB><SUP>2-</SUP></TD></TR><TR><td>C</TD><td>向集满CO<SUB>2</SUB>的软塑料瓶中加入约<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">1</td></tr><tr><td>3</td></tr></table></span>体积滴有石蕊试液的水，旋紧瓶盖，振荡</TD><td>　塑料瓶变瘪，<br />溶液变红</TD><td>　CO<SUB>2</SUB>能与石蕊反应</TD></TR><TR><td>D</TD><td>将质量相同的Mg、Zn分别投入到足量的稀HCl中</TD><td>镁条表面产生气体更多、更快</TD><td>金属活动性：Mg＞Zn</TD></TR></TBODY></TABLE>','A','B','C','D','','D','【解答】解：A、向久置于空气中的NaOH溶液滴加稀H<SUB>2</SUB>SO<SUB>4</SUB>，有气泡产生，是因为碳酸钠与稀硫酸反应生成了二氧化碳，不能说明NaOH和H<SUB>2</SUB>SO<SUB>4</SUB>发生了化学反应，故选项实验操作、现象和结论对应错误．<br />B、向某溶液中滴加BaCl<SUB>2</SUB>溶液，有白色沉淀，原溶液不一定含有SO<SUB>4</SUB><SUP>2-</SUP>，也可能是硝酸银溶液等，故选项实验操作、现象和结论对应错误．<br />C、向集满CO<SUB>2</SUB>的软塑料瓶中加入约<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">1</td></tr><tr><td>3</td></tr></table></span>体积滴有石蕊试液的水，旋紧瓶盖，振荡，塑料瓶变瘪，溶液变红，是因为二氧化碳与水反应生成碳酸，碳酸能使石蕊溶液变红色，故选项实验操作、现象和结论对应错误．<br />D、将质量相同的Mg、Zn分别投入到足量的稀HCl中，镁条表面产生气体更多、更快，说明金属活动性：Mg＞Zn，故选项实验操作、现象和结论对应正确．<br />故选：D．','【分析】A、向久置于空气中的NaOH溶液滴加稀H<SUB>2</SUB>SO<SUB>4</SUB>，有气泡产生，是因为碳酸钠与稀硫酸反应生成了二氧化碳，进行分析判断．<br />B、根据氯化钡溶液能与硫酸盐、硝酸银溶液等反应生成白色沉淀，进行分析判断．<br />C、根据二氧化碳的化学性质，进行分析判断．<br />D、根据金属的活动性越强，与酸反应越剧烈，进行分析判断．','选择题',3.00,'9d8150616aa82422a08719ffcae15a65',9,400,'化学实验方案设计与评价,证明硫酸和可溶性硫酸盐,二氧化碳的化学性质,金属活动性顺序及其应用,碱的化学性质','',2016,'32','2016•深圳校级模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1843576,'<img  src=\"/tikuimages/9/2016/400/shoutiniao81/42deb51e-94d4-11e9-bcb8-b42e9921e93e_xkb8.png\" style=\"vertical-align:middle;FLOAT:right\" />如图，往容器内吹入一定量的空气，发生燃爆现象：硬纸板随热气流冲高，蜡烛熄灭．下列说法不正确的是（　　）','发生燃爆后体系的温度升高，氧气耗尽，蜡烛熄灭','蜡烛熄灭是因为蜡烛的着火点降低了','在有限的作业空间，要谨防可燃性粉尘发生燃爆事故','实验用的容器可用废旧矿泉水瓶，但不能用玻璃瓶','','B','【解答】解：A、发生爆炸时有大量的热量聚集，故温度升高，由于消耗大量的氧气，故蜡烛会熄灭，正确；<br />B、着火点是物质固有的属性，不会降低，错误；<br />C、在有限的作业空间，要谨防可燃性粉尘发生燃爆事故，正确；<br />D、实验要发生爆炸，故不能使用玻璃瓶，正确；<br />故选B．','【分析】根据已有的爆炸的条件进行分析解答，可燃物在有限的空间内急剧燃烧可能会引起爆炸，据此解答．','选择题',3.00,'78ee2c242607652029ee653d946a0945',9,400,'燃烧和爆炸实验','',2016,'32','2016•泰州模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1843577,'某化学兴趣小组对金属的性质做了如下探究，请你填写空格：<br />（1）用如图所示实验探究铁生锈的条件（每支试管中均放有完全相同的洁净铁片）：<br /><img  src=\"/tikuimages/9/2015/400/shoutiniao20/42e2378f-94d4-11e9-bc54-b42e9921e93e_xkb13.png\" style=\"vertical-align:middle\" /><br />①甲同学认为，试管A发生的现象就能够说明铁的锈蚀是铁与空气中的氧气、水蒸气共同作用的结果．乙同学不同意他的观点，认为必须全面观察试管<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（选填试管编号）发生的现象，并经过科学严谨的推理，才能得出上述结论．<br />②试管D和E实验的目的是进一步探究铁在<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>的环境中是否更容易锈蚀．<br />③为防止金属锈蚀，除了采用覆盖保护膜等措施以外，还可以制成合金．这是因为合金与纯金属相比，其组成和内部组织结构发生变化，从而引起<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>的变化．<br />（2）某同学向分别盛有等质量的铁粉和锌粒的试管中，倒入等质量、等溶质质量分数的稀硫酸，以“金属表面产生气泡的快慢”为标准来判断两种金属的活动性强弱．有的同学认为这种方案不够合理，理由是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）将一定质量的铁Z粉放入硝酸铜、硝酸锌的混合溶液中，充分反应后过滤，所得固体中一定含有的物质是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，且反应后溶液质量比反应前<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（选填“增大”、“减小”或“不变”）．','','','','','','ABC$###$盐、酸$###$化学性质$###$没控制相同的条件，颗粒大小不同$###$Cu$###$减小','【解答】解：（1）①要证明铁生锈是与水和氧气同时存在的原因，必须设计使铁只与水接触、只与氧气接触而不生锈的实验，故ABC三支试管都需要观察，故填：ABC；<br />②试管D在含有盐，试管E中含有酸，目的是探究铁在盐、酸存在的环境中是否更易生锈，故填：盐、酸；<br />③金属的组成和内部组织结构发生变化，其化学性质等就会发生变化．故答案为：化学性质．&nbsp;<br />（2）金属的颗粒大小不同，与酸的反应速度也不同，所以没控制相同的条件，颗粒大小不同，金属表面产生气泡的快慢，也会受到影响．故答案为：没控制相同的条件，颗粒大小不同．<br />（3）三种金属活动性由强到弱的顺序：锌＞铁＞铜，可知铁只能置换出铜，根据铁与硝酸铜反应的方程式：Fe+Cu（NO<SUB>3</SUB>）<SUB>2</SUB>═Cu+Fe（NO<SUB>3</SUB>）<SUB>2</SUB>．可知56份质量的铁可以置换出64份质量的铜，所以反应后溶液质量比反应前减小．故答案为：Cu；减小．','【分析】（1）根据已有的铁生锈的条件进行分析解答，铁在与水和氧气并存时易生锈，防锈就是破坏铁生锈的条件，据此解答．金属的组成和内部组织结构发生变化，其性质等就会发生变化；<br />（2）金属的颗粒大小不同，与酸的反应速度也不同；<br />（3）根据三种金属活动性由强到弱的顺序：锌铁铜，进行分析．','填空题',3.00,'34b8746e23449632e614aa11f35b7ae1',9,400,'探究金属锈蚀的条件,金属的化学性质','冷水江市',2015,'32','2015•冷水江市校级模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1843589,'海水中含有MgCl<SUB>2</SUB>和NaCl等多种物质，利用海水提取金属镁的工业流程如图所示：<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao39/42f94200-94d4-11e9-b1e6-b42e9921e93e_xkb69.png\" style=\"vertical-align:middle\" /><br /><br />（1）在①～⑤的每个反应装置内都发生了化学反应，发生分解反应的装置有<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填序号）．<br />（2）写出装置④内发生反应的化学方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）MgCl<SUB>2</SUB>溶液在进入结晶池之前需先加热浓缩，加热过程中，MgCl<SUB>2</SUB>溶液中溶质质量分数会逐渐<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（4）利用海水除了可以提取镁，还可以有很多开发利用的价值，请你说出其中一点<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','①⑤$###$2HCl+Mg（OH）<SUB>2</SUB>=MgCl<SUB>2</SUB>+H<SUB>2</SUB>O$###$增大$###$海水晒盐','【解答】解：（1）据图可以看出，①是碳酸钙高温分解生成氧化钙和二氧化碳，②是氧化钙和水发生化合反应，③是氢氧化钙和氯化镁发生复分解反应，④是氢氧化镁和盐酸发生复分解反应，⑤是氯化镁分解生成镁和氯气，故填：①⑤；<br />（2）氢氧化镁和盐酸反应生成氯化镁和水，故填：2HCl+Mg（OH）<SUB>2</SUB>=MgCl<SUB>2</SUB>+H<SUB>2</SUB>O；<br />（3）MgCl<SUB>2</SUB>溶液在进入结晶池之前需先加热浓缩，加热过程中，MgCl<SUB>2</SUB>溶液中溶质质量分数会由于水分的减少而增大，故填：增大；&nbsp;&nbsp;&nbsp;<br />（4）海水中的资源很多，可以利用海水晒盐、淡化海水，故填：海水晒盐．','【分析】根据已有的反应的类化学方程式的书写、加热蒸发溶质质量分数的变化以及海水资源的利用进行分析解答即可．','书写',3.00,'d868ae12c7cfff1df56cc19957c56972',9,400,'盐的化学性质,分解反应及其应用,书写化学方程式、文字表达式、电离方程式,对海洋资源的合理开发与利用','',2016,'37','2016•东城区一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1843602,'下列实验操作中正确的是（　　）','给试管内的液体加热时，试管口对着人','把烧杯放在铁圈上直接加热','用天平称量时，把称量物放在左盘，砝码放右盘','用漏斗过滤时液面高于滤纸的边缘','','C','【解答】解：A、给试管内液体加热的注意事项是：试管内液体量不能超过试管容积的三分之一，试管口不能朝着有人的地方，夹试管时从底部往上套，套到距试管口三分之一处．试管与桌面呈45度角．开始加热时要预热，再在不停移动试管使液体中下部均匀受热．故A不正确；<br />B、加热烧杯时，要垫石棉网，不可直接加热，故B错；<br />C、人们在操作时，一般右手比较灵活，所以左物右码方便人们用右手添加砝码，故C正确；<br />D、过滤实验中液面不能高于滤纸边缘，否则过滤的液体将浑浊，故D错误．<br />故选C．','【分析】A、根据给试管内的液体加热的注意事项考虑；<br />B、根据加热烧杯时，不可直接加热考虑；<br />C、根据天平的使用方法考虑；<br />D、根据过滤实验中液面不能高于滤纸边缘考虑','选择题',3.00,'34d9c9b24f74bb0e39bab065e89cec8c',9,400,'用于加热的仪器,称量器-托盘天平,给试管里的液体加热,过滤的原理、方法及其应用','',2011,'37','2011秋•蓝山县校级月考',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1843606,'某学生做完实验后，采用以下方式清洗试管，不能达到目的是（　　）','用稀硫酸清洗用一氧化碳还原氧化铁后留下的固体物质','内壁不易洗掉的物质用试管刷猛力刷洗','用稀盐酸清洗久盛石灰水留下的白膜','用稀硫酸清洗用氢气还原氧化铜后留下的固体物质','','B|D','【解答】解：A、一氧化碳还原氧化铁后留下的固体物质是铁，能与稀硫酸反应生成硫酸亚铁和氢气，可用稀硫酸清洗，能能达到目的，故选项错误．<br />B、内壁不易洗掉的物质用试管刷猛力刷洗，容易损坏试管，不能达到目的，故选项正确．<br />C、久盛石灰水留下的白膜是氢氧化钙与空气中的二氧化碳反应生成的碳酸钙沉淀，碳酸钙不溶于水，但能与酸反应，可用稀盐酸清洗，能能达到目的，故选项错误．<br />D、氢气还原氧化铜后留下的固体物质是铜，铜不与稀硫酸反应，不能用稀硫酸清洗，不能达到目的，故选项正确．<br />故选：BD．','【分析】A、一氧化碳还原氧化铁后留下的固体物质是铁，据此进行分析判断．<br />B、猛力刷洗试管，容易损坏试管．<br />C、久盛石灰水留下的白膜是氢氧化钙与空气中的二氧化碳反应生成的碳酸钙沉淀，据此进行分析判断．<br />D、氢气还原氧化铜后留下的固体物质是铜，据此进行分析判断．','多选题',3.00,'6a1cc8ba2cb16e79409c7a45a478fac8',9,400,'玻璃仪器的洗涤,酸的化学性质','',0,'37','',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1843612,'<img  src=\"/tikuimages/9/0/400/shoutiniao77/432c38e1-94d4-11e9-bd1d-b42e9921e93e_xkb4.png\" style=\"vertical-align:middle;FLOAT:right;\" />许多科学概念在逻辑上存在着包含、并列或交叉的关系，如图所示，下列概念间关系说法正确的是（　　）','功与功率属于包含关系','汽化与蒸发属于并列关系','纯净物和混合物属于交叉关系','化合物和氧化物属于包含关系','','D','【解答】解：A．功是力和作用力方向位移的乘积，功率是单位时间内做的功，功和功率应该是并列关系．故错误；<br />B．汽化包括蒸发和沸腾，所以汽化和蒸发属于包含关系，故错误；<br />C．物质按含有物质种类的多少可分为纯净物与混合物，是并列关系，故错误；<br />D．化合物有多种元素组成，其中氧化物是含有氧元素和另外一种元素的化合物，是包含关系，故正确．<br />故选D．','【分析】应用各知识点的概念，理解概念间相关的关系，结合图示所提供的关系意义，分析相关的选项从而判断正确与否．','选择题',3.00,'84a91dd972d5de613de6c995e54f367a',9,400,'从组成上识别氧化物,纯净物和混合物的概念,单质和化合物的概念','',0,'37','',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1843627,'<img  src=\"/tikuimages/9/2016/400/shoutiniao70/435e1e4f-94d4-11e9-aa24-b42e9921e93e_xkb66.png\" style=\"vertical-align:middle;FLOAT:right;\" />（2016•建邺区一模）2015年11月首架C919大飞机下线，标志着我国跻身飞机研制先进国家行列．<br />（1）机身蒙皮使用的是铝锂合金材料．下列属于铝锂合金性质的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填字母）．<br />A．密度大&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;B．硬度大&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;C．抗腐蚀<br />（2）①锂元素的某种粒子结构示意图为<img  src=\"/tikuimages/9/2016/400/shoutiniao6/43606840-94d4-11e9-800e-b42e9921e93e_xkb37.png\" style=\"vertical-align:middle\" />，该粒子是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />A．分子&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;B．原子&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;C．阳离子&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;D．阴离子<br />②锂原子在化学反应中容易<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“得到”或“失去”）电子，锂是一种活泼金属．<br />（3）小明利用硝酸银溶液、硫酸铝溶液和光亮的铜丝，设计了如下实验比较铝、银、铜的金属活动性强弱，请你和他一起完成：<br /><table class=\"edittable\"><tbody><tr><td width=\"152\">实验步骤</td><td width=\"114\">现象</td><td width=\"133\">结论</td><td width=\"114\">发生反应的化学方程式</td></tr><tr><td>取一根光亮的铜丝插入到硫酸铝溶液中</td><td>铜丝表面：<br /><u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；</td><td rowspan=\"2\">三种金属的金属活动性由强到弱的<br />顺序为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u></td><td rowspan=\"2\"><u>&nbsp;&nbsp;&nbsp;&nbsp;</u></td></tr><tr><td>另取一根光亮的铜丝插入到硝酸银溶液中</td><td>铜丝表面：<br /><u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．</td></tr></tbody></table>','','','','','','BC$###$C$###$失去$###$无现象$###$Al、Cu、Ag$###$Cu+2AgNO<sub>3</sub>=Cu（NO<sub>3</sub>）<sub>2</sub>+2Ag$###$有银白色物质析出','【解答】解：<br />（1）机身蒙皮使用的是铝锂合金材料．铝锂合金性质的是硬度大、抗腐蚀；<br />（2）该粒子的质子数＞核外电子数，为阳离子，还原成原子之后的最外层电子数为3，所以属于金属元素，该元素的原子在化学反应中易失去电子．<br />（3）取一根光亮的铜丝插入到硫酸铝溶液中，铜位于铝的后面，不能置换铝，故无现象；另取一根光亮的铜丝插入到硝酸银溶液中，观察铜片表面有银白色固体析出，溶液由无色变为蓝色，说明铜的活泼性大于银，即银、铝、铜三种金属的活动性强弱依次为：Al＞Cu＞Ag．&nbsp;发生反应的化学方程式为：<br />Cu+2AgNO<sub>3</sub>=Cu（NO<sub>3</sub>）<sub>2</sub>+2Ag．<br />答案：<br />（1）BC；&nbsp;&nbsp;<br />（2）①C；&nbsp;②失去；<br />&nbsp;（3）<br /><table class=\"edittable\"><tbody><tr><td width=\"83\">实验步骤</td><td width=\"133\">现象</td><td width=\"95\">结论</td><td width=\"208\">发生反应的化学方程式</td></tr><tr><td></td><td>无现象</td><td rowspan=\"2\">Al、Cu、Ag</td><td rowspan=\"2\">Cu+2AgNO<sub>3</sub>=Cu（NO<sub>3</sub>）<sub>2</sub>+2Ag</td></tr><tr><td></td><td>有银白色物质析出</td></tr></tbody></table>','【分析】（1）根据合金的特性解答；<br />（2）根据原子中核内质子数=核外电子数，阳离子中质子数＞核外电子数，阴离子中质子数＜核外电子数；一般来说金属元素的原子最外层电子数少于4个，在化学变化中易失去电子，非金属元素的原子最外层电子数等于或多于4个，在化学反应中容易得到电子，可以据此来解答该题；<br />（3）验证三种金属活动性强弱关系时，通常采取“三取中”的方法：取中间金属的单质或两端金属的盐溶液反应或取中间金属的盐溶液与两端金属的单质反应．','书写',3.00,'97f37b392619f1018ddd9be8856b60b2',9,400,'金属活动性的探究,合金与合金的性质,原子结构示意图与离子结构示意图,书写化学方程式、文字表达式、电离方程式','',2016,'37','2016•建邺区一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1843638,'<img  src=\"/tikuimages/9/2016/400/shoutiniao17/437f3ade-94d4-11e9-8a0e-b42e9921e93e_xkb78.png\" style=\"vertical-align:middle;FLOAT:right\" />（2016•马鞍山二模）资料显示：碳还原四氧化三铁制得铁的温度在700℃～800℃．某校化学兴趣小组在探究用碳冶炼铁的原理的实验时，取一定质量的碳粉和四氧化三铁粉的混合物，在700℃--800℃的高温条件下使其充分<br />反应，设计的实验装置如图：<br />（1）写出碳还原四氧化三铁的化学方程式：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，该实验中观察到的实验现象是：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>对反应后试管中剩余的固体是什么，兴趣小组的同学做出了如下猜想：甲同学认为是铁粉，乙同学认为是碳粉和铁粉的混合物，丙同学的猜想是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）乙同学设计一个用化学方法证明自己的猜想是否正确，他的方法是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，<br />理由是：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（3）丙同学认为要验证自己的猜想是否正确可用磁铁去吸引，你认为丙同学的实验方案是否可行：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，理由是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />请你设计一个实验证明丙同学的猜想：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','2C+Fe<SUB>3</SUB>O<SUB>4</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>3Fe+2CO<SUB>2</SUB>↑$###$澄清的石灰水变浑浊$###$四氧化三铁粉和铁粉$###$取试管里的固体剩余物加入足量的稀盐酸充分反应，看是否有固体剩余$###$铁跟稀盐酸反应，碳跟稀盐酸不反应$###$不可行$###$四氧化三铁也能被磁铁吸引$###$取试管里剩余固体少许，加入足量碳粉，放入如图所示装置加热至700℃～800℃，若澄清石灰水变浑浊，证明丙同学的猜想正确．','【解答】解：（1）在高温条件下，碳还原四氧化三铁生成了铁和二氧化碳，反应的化学方程式：2C+Fe<SUB>3</SUB>O<SUB>4</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>3Fe+2CO<SUB>2</SUB>↑，该实验中观察到的实验现象是：澄清的石灰水变浑浊，对反应后试管中剩余的固体是什么，兴趣小组的同学做出了如下猜想：甲同学认为是铁粉，乙同学认为是碳粉和铁粉的混合物，丙同学的猜想是四氧化三铁粉和铁粉．<br />（2）乙同学设计一个用化学方法证明自己的猜想是否正确，他的方法是取试管里的固体剩余物加入足量的稀盐酸充分反应，看是否有固体剩余，理由是：铁跟稀盐酸反应，碳跟稀盐酸不反应；<br />（3）丙同学认为要验证自己的猜想是否正确可用磁铁去吸引，丙同学的实验方案不可行，理由是：四氧化三铁也能被磁铁吸引．由于在高温条件下，碳还原四氧化三铁生成了铁和二氧化碳，所以丙同学的猜想验证的方法是：取试管里剩余固体少许，加入足量碳粉，放入如图所示装置加热至700℃～800℃，若澄清石灰水变浑浊，证明丙同学的猜想正确．<br />故答为：（1）2C+Fe<SUB>3</SUB>O<SUB>4</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>3Fe+2CO<SUB>2</SUB>↑；&nbsp;澄清的石灰水变浑浊；&nbsp;四氧化三铁粉和铁粉．（2）取试管里的固体剩余物加入足量的稀盐酸充分反应，看是否有固体剩余；铁跟稀盐酸反应，碳跟稀盐酸不反应．<br />（3）不可行，四氧化三铁也能被磁铁吸引；取试管里剩余固体少许，加入足量碳粉，放入如图所示装置加热至700℃～800℃，若澄清石灰水变浑浊，证明丙同学的猜想正确．','【分析】（1）根据碳与四氧化三铁的反应分析回答有关问题；<br />（2）根据铁能与盐酸反应，碳不能与盐酸反应分析回答；<br />（3）根据铁、四氧化三铁都能被磁铁吸引，四氧化三铁能与碳反应分析回答．','书写',3.00,'5754d1aa2a8c977ecd2ce28d4a733e25',9,400,'常见金属的特性及其应用,碳的化学性质,书写化学方程式、文字表达式、电离方程式','',2016,'37','2016•马鞍山二模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1843642,'实验室里制取并收集二氧化碳气体时一般有以下操作步骤：①检查装置的气密性&nbsp;&nbsp;②连接装置&nbsp;&nbsp;③加入大理石&nbsp;&nbsp;&nbsp;④加入稀盐酸&nbsp;&nbsp;⑤收集&nbsp;&nbsp;⑥验满．其中排列顺序正确的是（　　）','①②③④⑤⑥','②①④③⑤⑥','②①③④⑤⑥','②③④①⑤⑥','','C','【解答】解：制取二氧化碳的操作步骤是：连（连接仪器、组装实验装置）→查（检查装置的气密性）→加（向广口瓶中装入大理石）→倒（向漏斗中注入稀盐酸）→收集（用向上排空气法收集）→验满（将燃着的木条放在集气瓶口）．则正确的操作顺序为：②①③④⑤⑥．<br />故选：C．','【分析】根据实验室制取二氧化碳的步骤（连→查→加→倒→收→验）进行分析解答即可．','选择题',3.00,'287c229014f9c60636a4f5bc610d2c27',9,400,'制取二氧化碳的操作步骤和注意点','',2014,'33','2014秋•宁波期末',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1843645,'下面是某小组同学探究实验室制取二氧化碳的反应原理的过程．<br />【查阅资料】a．碳酸盐与酸在常温下反应可得到二氧化碳；<br />b．碳酸氢铵受热分解NH<SUB>4</SUB>HCO<SUB>3</SUB>（固）<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>NH<SUB>3</SUB>+CO<SUB>2</SUB>↑+H<SUB>2</SUB>O<br />【讨论】（1）小静认为用碳酸氢铵受热分解制取二氧化碳不合适，理由是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />（2）小欣认为碳酸盐与酸反应，所用的酸不宜选浓盐酸的原因是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />【提出问题】选择哪种碳酸盐与酸的反应制二氧化碳合适？<br />小组的同学用碳酸钠粉末、块状大理石、稀盐酸、稀硫酸这些药品，探究适合实验室制二氧化碳的药品．<br />【实验与结论】<br /><table class=\"edittable\"><TBODY><TR><td width=121>实验内容</TD><td width=142>实验现象</TD><td width=199>结论</TD></TR><TR><td><img  src=\"/tikuimages/9/2016/400/shoutiniao46/438ecb40-94d4-11e9-8677-b42e9921e93e_xkb56.png\" style=\"vertical-align:middle\" /></TD><td>剧烈反应，迅速放出大量气泡</TD><td>反应速率过快，气体不便收集，不宜实验室制取二氧化碳</TD></TR><TR><td><img  src=\"/tikuimages/9/2016/400/shoutiniao3/4391ff91-94d4-11e9-b1af-b42e9921e93e_xkb77.png\" style=\"vertical-align:middle\" /></TD><td>产生气泡，速率迅速减慢，反应几乎停止</TD><td>无法持续产生二氧化碳，不能用于实验室制取二氧化碳</TD></TR><TR><td><img  src=\"/tikuimages/9/2016/400/shoutiniao83/4393ad40-94d4-11e9-a283-b42e9921e93e_xkb90.png\" style=\"vertical-align:middle\" /></TD><td><u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD><td><u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD></TR></TBODY></TABLE>（1）根据探究结论，实验室制取二氧化碳的反应原理为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />（用化学方程式表示）<br />（2）实验室制二氧化碳的发生装置还可用于制取氧气，制取氧气的药品是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填药品的化学式）','','','','','','碳酸氢铵受热分解产生产生多种气体，使制取的二氧化碳不纯$###$因浓盐酸挥具有挥发性，使收集到的二氧化碳不纯$###$产生大量气泡$###$反应速率适中，气体便于收集，适宜实验室制取二氧化碳$###$CaCO<SUB>3</SUB>+2HCl═CaCl<SUB>2</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑$###$H<SUB>2</SUB>O<SUB>2</SUB>、MnO<SUB>2</SUB>','【解答】解：（1）碳酸氢铵受热分解产生氨气、水和二氧化碳，使制取的二氧化碳不纯；<br />（2）因浓盐酸挥具有挥发性，使收集到的二氧化碳不纯，所以浓盐酸不适合用于实验室制取CO<SUB>2</SUB>；<br />【实验与结论】B中反应产生硫酸钙，覆盖在碳酸钙表面，阻止反应；C中稀盐酸与大理石反应产生大量气泡；反应速率适中，气体便于收集，适宜实验室制取二氧化碳；<br />（1）制取二氧化碳的反应原理为：CaCO<SUB>3</SUB>+2HCl═CaCl<SUB>2</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑<br />（2）实验室制二氧化碳的发生装置还可用于制取氧气，制取氧气的药品是过氧化氢溶液和二氧化锰，化学式分别是：H<SUB>2</SUB>O<SUB>2</SUB>、MnO<SUB>2</SUB>；<br />答案：<br />（1）碳酸氢铵受热分解产生产生多种气体，使制取的二氧化碳不纯；<br />（2）因浓盐酸挥具有挥发性，使收集到的二氧化碳不纯；<br />【实验与结论】<br />产生大量气泡；反应速率适中，气体便于收集，适宜实验室制取二氧化碳；<br />（1）CaCO<SUB>3</SUB>+2HCl═CaCl<SUB>2</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑；<br />（2）H<SUB>2</SUB>O<SUB>2</SUB>、MnO<SUB>2</SUB>','【分析】（1）根据碳酸氢铵受热分解产生氨气、水和二氧化碳解答；<br />（2）根据浓盐酸挥具有挥发性；<br />【实验与结论】根据实验室制取气体的反应原理进行分析解答本题．<br />根据实验室用过氧化氢溶液和二氧化锰反应子氧气属于固液常温型解答．','书写',3.00,'e5380040f233d5705b58cb86c70b219c',9,400,'制取气体的反应原理的探究,二氧化碳的实验室制法,书写化学方程式、文字表达式、电离方程式','宁国市',2016,'37','2016•宁国市一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1843646,'用下列物质的字母序号填空：<br />A、二氧化硫&nbsp;B、盐酸&nbsp;C、氧氢化钙D、硝酸钾E、淀粉F、金刚石<br />（1）属于复合肥料的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（2）用于刻划玻璃的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（3）属于天然有机高分子的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（4）可作改良酸性土壤的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','D$###$F$###$E$###$C','【解答】解：（1）属于复合肥料的是硝酸钾；<br />（2）用于刻划玻璃的是金刚石；<br />（3）属于天然有机高分子的是淀粉；<br />（4）可作改良酸性土壤的是氢氧化钙，<br />故答案为：D；F；E；C．','【分析】根据物质的性质进行分析解答即可．','填空题',3.00,'768c96e3bfe4e97cf4504207d55235d2',9,400,'常见碱的特性和用途,常见化肥的种类和作用,有机物的特征、分类及聚合物的特性,碳单质的物理性质及用途','',2016,'37','2016•泗县二模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1843651,'<img  src=\"/tikuimages/9/2016/400/shoutiniao8/43a0cc9e-94d4-11e9-ae7a-b42e9921e93e_xkb67.png\" style=\"vertical-align:middle;FLOAT:right\" />（2016春•阜南县校级月考）如图所示，打开分液漏斗，使其中的无色液体与烧瓶中的固体接触反应，可观察到尖嘴导管口处有水喷出，请分别写出一个符合图中现象和下列要求的化学方程式：<br />（1）分解反应<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />（2）置换反应<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />（3）化合反应<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />（4）复分解反应<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','2H<SUB>2</SUB>O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span>&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O+O<SUB>2</SUB>↑$###$Zn+H<SUB>2</SUB>SO<SUB>4</SUB>═ZnSO<SUB>4</SUB>+H<SUB>2</SUB>↑$###$CaO+H<SUB>2</SUB>O═Ca（OH）<SUB>2</SUB>$###$CaCO<SUB>3</SUB>+2HCl═CaCl<SUB>2</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑','【解答】解：（1）过氧化氢溶液和二氧化锰固体混合能生成氧气，使瓶内的压强变大，该反应属于分解反应，反应的化学方程式为：2H<SUB>2</SUB>O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span>&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O+O<SUB>2</SUB>↑；<br />（2）置换反应是一种单质和一种化合物生成另一种单质和另一种化合物的反应，锌是固体单质，与稀硫酸反应生成氢气，使瓶内的压强变大，该反应属于置换反应，反应的化学方程式为：Zn+H<SUB>2</SUB>SO<SUB>4</SUB>═ZnSO<SUB>4</SUB>+H<SUB>2</SUB>↑；<br />（3）氧化钙与水反应放出大量的热，瓶中的空气受热膨胀使气球胀大，该反应属于化合反应，反应的化学方程式为：CaO+H<SUB>2</SUB>O═Ca（OH）<SUB>2</SUB>；<br />（4）复分解反应是两种化合物相互交换成分生成另外两种化合物的反应，碳酸钙和盐酸反应生成氯化钙、水和二氧化碳气体，使瓶内的压强变大，该反应属于复分解反应，反应的化学方程式为：CaCO<SUB>3</SUB>+2HCl═CaCl<SUB>2</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑；<br />故答案为：（1）2H<SUB>2</SUB>O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span>&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O+O<SUB>2</SUB>↑；<br />（2）Zn+H<SUB>2</SUB>SO<SUB>4</SUB>═ZnSO<SUB>4</SUB>+H<SUB>2</SUB>↑；<br />（3）CaO+H<SUB>2</SUB>O═Ca（OH）<SUB>2</SUB>；<br />（4）CaCO<SUB>3</SUB>+2HCl═CaCl<SUB>2</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑．','【分析】无色液体与烧瓶中的固体接触反应，可观察到尖嘴导管口处有水喷出，说明瓶内的压强变大；据此结合四种基本的反应类型进行分析解答．','书写',3.00,'3e7d2844bd9ee47c664856d6faa1c5fb',9,400,'反应现象和本质的联系,书写化学方程式、文字表达式、电离方程式','',2016,'37','2016春•阜南县校级月考',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1843672,'如图所示实验操作中正确的是（　　）','<img  src=\"/tikuimages/9/2016/400/shoutiniao50/43fa5e4f-94d4-11e9-a990-b42e9921e93e_xkb46.png\" style=\"vertical-align:middle\" /><br />&nbsp;&nbsp; 测溶液的pH','<img  src=\"/tikuimages/9/2016/400/shoutiniao15/43fef230-94d4-11e9-be3d-b42e9921e93e_xkb63.png\" style=\"vertical-align:middle\" /><br />&nbsp;&nbsp; 浓硫酸稀释','<img  src=\"/tikuimages/9/2016/400/shoutiniao62/43ffb580-94d4-11e9-ad80-b42e9921e93e_xkb37.png\" style=\"vertical-align:middle\" /><br />&nbsp;&nbsp; CO<SUB>2</SUB>的验满','<img  src=\"/tikuimages/9/2016/400/shoutiniao91/440078d1-94d4-11e9-89d4-b42e9921e93e_xkb29.png\" style=\"vertical-align:middle\" /><br />&nbsp; 配制NaCl溶液','','B','【解答】解：A、用pH试纸测定未知溶液的pH时，正确的操作方法为用玻璃棒蘸取少量待测液滴在干燥的pH试纸上，与标准比色卡对比来确定pH．不能将pH试纸伸入待测液中，以免污染待测液，图中所示操作错误．<br />B、稀释浓硫酸时，要把浓硫酸缓缓地沿器壁注入水中，同时用玻璃棒不断搅拌，以使热量及时地扩散；一定不能把水注入浓硫酸中；图中所示操作正确．<br />C、检验二氧化碳是否收集满时，应将燃着的木条放在集气瓶口，不能伸入瓶中，图中所示操作错误．<br />D、配制NaCl溶液应在烧杯中进行，并能在量筒内配制溶液，图中所示操作错误．<br />故选：B．','【分析】A、根据用pH试纸测定未知溶液的pH的方法进行分析判断．<br />B、根据浓硫酸的稀释方法（酸入水，沿器壁，慢慢倒，不断搅）进行分析判断．<br />C、根据二氧化碳气体的验满方法进行分析判断．<br />D、根据溶解操作的方法进行分析判断．','选择题',3.00,'d9fd2fabde497c394f83e34e045f751d',9,400,'物质的溶解,浓硫酸的性质及浓硫酸的稀释,溶液的酸碱度测定,二氧化碳的检验和验满','',2016,'32','2016•宜昌模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1843754,'A～G七种物质在一定条件下能发生如图所示的化学反应，其中F是自然界中最常见的液体，D在反应③中可以重复使用，C、E都是气体单质．<br /><img  src=\"/tikuimages/9/2015/400/shoutiniao3/44eea95e-94d4-11e9-8eef-b42e9921e93e_xkb37.png\" style=\"vertical-align:middle\" /><br />（1）试推断有关物质的名称：B<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>、E<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>、G<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）上述关系中属于分解反应的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填化学反应的编号）','','','','','','锰酸钾$###$氢气$###$过氧化氢$###$①③④','【解答】解：根据F是自然界中最常见的液体，能够通电分解产生C和E，C、E都是气体，D在反应③中作催化剂使用，可以产生C，因此F是水，C是氧气，E是氢气；A分解产生B、氧气和D，B可溶于水，D不溶于水，因此A是高锰酸钾，B是锰酸钾，D是二氧化锰，则G是过氧化氢，带入验证符合转化关系，因此：<br />（1）根据分析，B是锰酸钾，E是氢气，G是过氧化氢；<br />（2）反应①是高锰酸钾分解产生锰酸钾、二氧化锰和氧气，属于分解反应；<br />反应②是氢气和氧气点燃产生水，属于化合反应；<br />反应③是过氧化氢分解产生水和氧气，属于分解反应；<br />反应④是水通电产生氢气和氧气，属于分解反应；<br />故上述关系中属于分解反应的是①③④．<br />故答案为：<br />（1）锰酸钾、氢气、过氧化氢．（2）①③④．','【分析】根据F是自然界中最常见的液体，能够通电分解产生C和E，C、E都是气体，D在反应③中作催化剂使用，可以产生C，因此F是水，C是氧气，E是氢气；A分解产生B、氧气和D，B可溶于水，D不溶于水，因此A是高锰酸钾，B是锰酸钾，D是二氧化锰，则G是过氧化氢，带入验证解答该题．','推断题',3.00,'2709d5b0c2940cef6c76f97e41517575',9,400,'物质的鉴别、推断,分解反应及其应用','',2015,'35','2015秋•合肥校级期中',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1843776,'<img  src=\"/tikuimages/9/2016/400/shoutiniao86/451d3370-94d4-11e9-843c-b42e9921e93e_xkb34.png\" style=\"vertical-align:middle;FLOAT:right\" />（2016•淮安模拟）小明对铁的锈蚀进行如下探究，室温时，将包有样品的滤纸包用大头针固定在橡胶塞上，迅速塞紧，装置如图，观察到量筒内水沿导管慢慢进入广口瓶（净容积为146mL）．当温度恢复至室温，且量筒内水面高度不变时读数（此时瓶内氧气含量近似为零）．记录起始和最终量筒的读数以及所需时间如表．<br /><table class=\"edittable\"><TBODY><TR><td width=54>序号</TD><td width=180>样品</TD><td width=111>量筒起始<br />读数/mL</TD><td width=112>量筒最终<br />读数/mL</TD><td width=105>所需时间<br />/min</TD></TR><TR><td>1</TD><td>1g铁粉、0.2g碳和10滴水</TD><td>100</TD><td>70</TD><td>约120</TD></TR><TR><td>2</TD><td>lg铁粉、0.2g碳、10滴水和少量NaCl</TD><td>100</TD><td>70</TD><td>约70</TD></TR><TR><td>3</TD><td><u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD><td>100</TD><td>70</TD><td>约480</TD></TR></TBODY></TABLE>（1）实验①和②说明NaCl可以<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>铁锈蚀的速率．<br />（2）实验开始后，广口瓶内温度有所上升，说明铁的锈蚀过程是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>　（填“放热”或“吸热”）过程．<br />（3）实验结束后取出滤纸包，观察到有红棕色物质生成，该物质的化学式是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（4）已知含碳能够加快铁的生锈速度，小明想通过实验①和③是探究碳对铁锈蚀速率的影响，请在表格空白处填写实验③的样品组成．<br />（5）该装置还可用于测量空气中氧气的含量，根据上述数据计算氧气的体积含量是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（小数点后保留1位）．','','','','','','1g铁粉、10滴水$###$加快$###$放热$###$Fe<SUB>2</SUB>O<SUB>3</SUB>•xH<SUB>2</SUB>O$###$20.5%','【解答】解：（1）实验①和②之间唯一的变量为是否有NaCl，根据含有NaCl的②组实验，反应所需的时间较短，得到NaCl能加快铁锈蚀的速率；<br />（2）通过温度升高，可以直接得出铁锈蚀的过程是放热的过程；<br />（3）根据题干“小明对铁的锈蚀进行如下探究”及“红棕色物质生成”，我们可以判定此物质为铁锈，主要成分为Fe<SUB>2</SUB>O<SUB>3</SUB>•xH<SUB>2</SUB>O；<br />（4）“实验①和③是探究碳对铁锈蚀速率的影响”，所以该对照组间唯一的变量应为是否含有碳，因为实验①还有碳，则实验③不含碳，而其它的因素应完全相同，所以实验③的样品组成为1g铁粉、10滴水；<br />（5）因为量筒内液体减少的体积即为广口瓶内所含有的全部氧气的体积，所以氧气的体积含量：30ml/146ml×100%=20.5%．<br />该题答案为：（1）加快 （2）放热 （3）Fe<SUB>2</SUB>O<SUB>3</SUB>•xH<SUB>2</SUB>O （4）1g铁粉、10滴水&nbsp; （5）20.5%','【分析】通过铁生锈的探究实验，分析缓慢氧化过程中的能量变化及铁锈的主要成分．从探究实验中，对照试验组之间只能有一个变量，从而根据唯一变量所出现的不同结果，得出科学的结论．分析表中数据，根据唯一变量是碳，得出碳对铁生锈的影响．通过空气体积的变化得出，空气中氧气的体积分数．','填空题',3.00,'0dadefd56b62c2bdfb09aa856d193471',9,400,'探究金属锈蚀的条件,物质发生化学变化时的能量变化','',2016,'32','2016•淮安模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1843780,'下列有关能源和资源的说法错误的是（　　）','在汽油中加入适量的乙醇可以一定程度减少汽车尾气的污染','因制取成本高和贮存困难，氢气作为燃料和化学电源暂时还未能应用','金属的回收利用，既可以节约金属资源又可以节约能源，还可以减少对环境的污染','空气的成分按体积计大约是：氧气21%、氮气78%、稀有气体等其他成分1%','','B','【解答】解：<br />A、使用车用乙醇汽油，在大大减少汽车尾气中一氧化碳、碳氢化合物排放同时，还使尾气中氮氧化物、酮类等污染物浓度明显降低，故正确；<br />B、制取成本高和贮存困难，作为燃料氢气还不能广泛应用，不是还未能应用，故错误；<br />C、金属的回收利用，既可以节约金属资源又可以节约能源，还可以减少对环境的污染，故正确；<br />D、空气的成分按体积计大约是：氧气21%、氮气78%、稀有气体等其他成分1%，故正确．<br />答案：B','【分析】A、根据乙醇汽油，大大减少汽车尾气中一氧化碳、碳氢化合物排放同时，还使尾气中氮氧化物、酮类等污染物浓度明显降低解答；<br />B、根据制取成本高和贮存困难，作为燃料氢气还不能广泛应用解答；<br />C、根据金属回收利用的意义解答；<br />D、根据空气的成分解答．','选择题',3.00,'b8530ebed495f89ad1b5391b186c43dc',9,400,'空气的成分及各成分的体积分数,金属的回收利用及其重要性,常用燃料的使用与其对环境的影响,氢气的用途和氢能的优缺点','',2016,'37','2016•南岗区二模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1843786,'我国古代掌握了多种金属冶炼技术．<br />（1）春秋时期，发明了冶铁技术，其方法是将赤铁矿（主要成分为Fe<SUB>2</SUB>O<SUB>3</SUB>）与木炭交错堆积，利用自然风力进行燃烧，木炭不完全燃烧产生的CO将矿石中的Fe<SUB>2</SUB>O<SUB>3</SUB>还原成铁，该反应的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）西汉时期，刘安《淮南万毕术》记载有“曾青得铁则化为铜”，其含义是把铁片放入硫酸铜溶液中置换出铜，该反应的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）2015年，海昏侯墓出土了大量的青铜器，其表面有绿色的铜锈，其主要成分为碱式碳酸铜[Cu<SUB>2</SUB>（OH）<SUB>2</SUB>CO<SUB>3</SUB>]．试推测铜生锈是铜与氧气、水、<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>共同作用的结果．','','','','','','3CO+Fe<SUB>2</SUB>O<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2Fe+3CO<SUB>2</SUB>$###$CuSO<SUB>4</SUB>+Fe=FeSO<SUB>4</SUB>+Cu$###$二氧化碳','【解答】解：（1）一氧化碳能与氧化铁反应生成铁和二氧化碳，故填：3CO+Fe<SUB>2</SUB>O<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2Fe+3CO<SUB>2</SUB>；<br />（2）铁能与硫酸铜反应生成硫酸亚铁和铜，故填：CuSO<SUB>4</SUB>+Fe=FeSO<SUB>4</SUB>+Cu；&nbsp;&nbsp;&nbsp;&nbsp;<br />（3）碱式碳酸铜[Cu<SUB>2</SUB>（OH）<SUB>2</SUB>CO<SUB>3</SUB>]的成分中含有碳元素，空气中含有碳元素的物质是二氧化碳，故铜生锈与氧气、水和二氧化碳有关，故填：二氧化碳．','【分析】根据一氧化碳还原氧化铁的化学方程式的书写、铁与硫酸铜反应化学方程式的书写以及铜生锈的因素进行分析解答即可．','书写',3.00,'6253fc0f1f113b9cd603c1cc10fe14c0',9,400,'金属的化学性质,常见金属的冶炼方法,金属锈蚀的条件及其防护,书写化学方程式、文字表达式、电离方程式','',2016,'37','2016•东城区一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1843793,'水是人们生活、生产各方面都离不开的重要物质．下列有关水的说法错误的是…（　　）','水是一种最常用的溶剂','用肥皂水鉴别硬水和软水','自然界的水都含有杂质','自来水生产使用活性炭主要作用是除去水中的不溶性杂质','','D','【解答】解：A、溶剂是一种可以溶化固体，液体或气体溶质的液体，继而成为溶液．在日常生活中水是一种最常见的溶剂．故A正确；<br />B、向等量的两种水样中分别加入等量的肥皂水，振荡，软水中泡沫多，硬水中泡沫少，且操作方法简单易行，故B正确；<br />C、自然界中的水通常含有一些可溶性和不溶性杂质．故C正确；<br />D、自来水生产过程中，可用过滤的方法除去水中的不溶性杂质，活性炭具有吸附性，主要作用是除去水中的异味和色素．故D错误；<br />故选D．','【分析】A、根据在日常生活中水是一种最常见的溶剂进行分析；<br />B、根据鉴别硬水和软水的方法进行分析；<br />C、根据自然界中的水都含有杂质进行分析；<br />D、根据活性炭的吸附作用进行分析．','选择题',3.00,'e1dce61c60155895917e27d187df5dfd',9,400,'自来水的生产过程与净化方法,硬水与软水,常见的溶剂','',2016,'35','2016春•嘉峪关校级期中',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1843796,'近几年和国某些城市酸雨污染较为严重，主要是因为大量燃烧含硫量高的煤而形成的，此外，种机动车排放的尾气也是形成酸雨的重要原因．<br />【实验目的】通过实验证明煤中含有碳元素和硫元素．<br />【查阅资料】（1）二氧化硫能使高锰酸钾溶液褪色（由紫红色变为无色）．<br />&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;（2）二氧化硫和二氧化碳一样，也能使澄清石灰水变浑浊．<br />【实验探究】点燃一小块煤收集气体并将生成的气体分别通入下列溶液中：<br /><table class=\"edittable\"><TBODY><TR><td width=190>实验</TD><td width=190>现象</TD><td width=190>结论</TD></TR><TR><td>酸性高锰酸钾溶液</TD><td>紫红色变为无色</TD><td rowSpan=2>证明煤中含有碳元素和硫元素</TD></TR><TR><td>澄清石灰水</TD><td>澄清变浑浊</TD></TR></TBODY></TABLE>请写出硫燃烧的化学方程式：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />【反思评价】实验方案不合理，其理由是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />【改进实验】如图所示实验（部分装置在图中略去），仪器a的名称：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao37/453d1780-94d4-11e9-a10a-b42e9921e93e_xkb5.png\" style=\"vertical-align:middle\" /><br />【现象与结论】B、C、D中的现象分别是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，证明煤中含有碳元素和硫元素．<br />【反思交流】如图C装置的作用是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','S+O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>SO<SUB>2</SUB>$###$二氧化硫也能使澄清石灰水变浑浊$###$酒精灯$###$B中高锰酸钾溶液褪色，C中高锰酸钾溶液不变色或颜色变浅，说明二氧化硫被完全吸收，D中澄清石灰水变浑浊$###$除去过多的二氧化硫或排除二氧化硫的干扰','【解答】解：硫燃烧生成二氧化硫，其化学方程式为：S+O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>SO<SUB>2</SUB>，故填：S+O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>SO<SUB>2</SUB>．<br />【反思评价】实验方案不合理，其理由是二氧化硫也能使澄清石灰水变浑浊．<br />故填：二氧化硫也能使澄清石灰水变浑浊．<br />【改进实验】如图所示实验中仪器a是酒精灯，故填：酒精灯．<br />【现象与结论】证明煤中含有碳元素和硫元素，则硫元素转化生成二氧化硫，碳元素转化生成二氧化碳，故B中高锰酸钾溶液褪色，C中高锰酸钾溶液不变色或颜色变浅，说明二氧化硫被完全吸收，D中澄清石灰水变浑浊．故填：B中高锰酸钾溶液褪色，C中高锰酸钾溶液不变色或颜色变浅，说明二氧化硫被完全吸收，D中澄清石灰水变浑浊．<br />【反思交流】二氧化硫能使高锰酸钾溶液褪色（由紫红色变为无色），故C装置的作用是：除去过多的二氧化硫或排除二氧化硫的干扰，故填：除去过多的二氧化硫或排除二氧化硫的干扰．','【分析】根据硫燃烧能生成二氧化硫、二氧化硫的性质以及物质间反应的实验现象进行分析解答即可．','填空题',3.00,'2f5cba50f68513ec83e6a0ee85ac0484',9,400,'质量守恒定律的实验探究,常见气体的检验与除杂方法,质量守恒定律及其应用','',2016,'32','2016•南陵县模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1843816,'下列仪器可以用酒精灯直接加热的是（　　）','烧杯','量筒','燃烧匙','蒸发皿','','C|D','【解答】解：A、烧杯加热需要垫石棉网，不能直接加热，故A错误；<br />B、量筒不能加热，故B错误；<br />C、燃烧匙可以在酒精灯上直接加热，故C正确；<br />D、蒸发皿可以在酒精灯上直接加热，故D正确．<br />故选CD．','【分析】熟悉常用仪器的用途及注意事项才能很好的回答本题．','多选题',3.00,'258c3424f1234c9fce8d8653bc4d0690',9,400,'用于加热的仪器','',2015,'35','2015秋•周村区校级期中',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1843862,'<img  src=\"/tikuimages/9/2016/400/shoutiniao14/4636448f-94d4-11e9-b15d-b42e9921e93e_xkb52.png\" style=\"vertical-align:middle;FLOAT:right\" />（2016•河西区一模）含氟物质的合成，对推动社会进步发挥着巨大作用．<br />（1）如图是氟元素在元素周期表中的相关信息，画出氟元素的原子结构示意图<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，由氟原子的结构示意图知，氟原子在化学反应中易<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（选填“失”或“得”）电子．<br />（2）氟的化学性质与下列<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填序号）元素的化学性质相似．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao49/463a8a4f-94d4-11e9-aa88-b42e9921e93e_xkb45.png\" style=\"vertical-align:middle\" /><br />（3）氟是人体必须的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（选填“常量”或“微量”）元素．氟化钠常用作牙膏添加剂，能有效预防龋齿．氟化钠和氯化钠都是由<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（选填“分子”、“原子”或“离子”）构成．<br />（4）HF的水溶液叫氢氟酸，写出氢氟酸与SiO<SUB>2</SUB>反应生成SiF<SUB>4</SUB>气体和水的化学方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','<img  src=\"/tikuimages/9/2016/400/shoutiniao64/463d978f-94d4-11e9-92ba-b42e9921e93e_xkb20.png\" style=\"vertical-align:middle\" />$###$得$###$E$###$微量$###$离子$###$4HF+SiO<SUB>2</SUB>=SiF<SUB>4</SUB>↑+2H<SUB>2</SUB>O','【解答】解：（1）氟是9号元素，其原子结构示意图为<img  src=\"/tikuimages/9/2016/400/shoutiniao58/4641410f-94d4-11e9-98ef-b42e9921e93e_xkb89.png\" style=\"vertical-align:middle\" />；其最外层有7个电子，在化学变化中易得到电子；故填：<img  src=\"/tikuimages/9/2016/400/shoutiniao65/46433ce1-94d4-11e9-9920-b42e9921e93e_xkb56.png\" style=\"vertical-align:middle\" />；得；<br />（2）由氟原子结构可知，其最外层有7个电子，易得一个电子，与E化学性质相似；故填：E；<br />（3）氟是人体必须的微量元素．氟化钠常用作牙膏添加剂，能有效预防龋齿．氟化钠和氯化钠都是由离子构成．故填：微量；离子；<br />（4）氢氟酸与玻璃的主要成分二氧化硅发生反应，生成四氟化硅气体和水，反应的化学方程式为：4HF+SiO<SUB>2</SUB>=SiF<SUB>4</SUB>↑+2H<SUB>2</SUB>O．<br />故填：4HF+SiO<SUB>2</SUB>=SiF<SUB>4</SUB>↑+2H<SUB>2</SUB>O．','【分析】（1）根据原子结构示意图的画法以及最外层电子数来分析；<br />（2）根据最外层电子数决定元素的化学性质来分析；<br />（3）根据构成物质的粒子来分析．<br />（4）首先根据反应原理找出反应物、生成物、反应条件，根据化学方程式的书写方法、步骤进行书写即可．','书写',3.00,'3c1ca3da2e4d163618b0a0d3888d52ee',9,400,'分子、原子、离子、元素与物质之间的关系,核外电子在化学反应中的作用,原子结构示意图与离子结构示意图,元素周期表的特点及其应用,书写化学方程式、文字表达式、电离方程式,矿物质与微量元素','',2016,'37','2016•河西区一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1843866,'下列物质由分子构成的是（　　）','氯化钠','铜','晶体硅','冰','','D','【解答】解：A、氯化钠是由钠离子和氯离子构成的，故选项错误．<br />B、铜属于金属单质，是由铜原子直接构成的，故选项错误．<br />C、晶体硅属于固态非金属单质，是由硅原子构成的，故选项错误．<br />D、冰是固态的水，是由水分子构成的，故选项正确．<br />故选：D．','【分析】根据金属、大多数固态非金属单质、稀有气体等由原子构成；有些物质是由分子构成的，气态的非金属单质和由非金属元素组成的化合物，如氢气、水等；有些物质是由离子构成的，一般是含有金属元素和非金属元素的化合物，如氯化钠，进行分析判断即可．','选择题',3.00,'4eecf6257feba21e3b1da1298bd5a393',9,400,'物质的构成和含量分析','',2016,'37','2016•南京一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1843870,'金属盒金属制品与我们的生活息息相关，铁、铜、银及其合金使用非常广泛．<br />（1）上述三种元素在地壳中含量最多的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）要验证上述三种元素的活动性顺序，用一种金属盒两种常见的溶液就可完成，这两种溶液的溶质是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，写出对应的化学反应方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）金属和金属制品除了正常消耗外，还有很多因腐蚀而消耗，防止钢铁生锈的措施有（填两种）<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','铁$###$硝酸亚铁和硝酸银$###$Cu+2AgNO<SUB>3</SUB>=Cu（NO<SUB>3</SUB>）<SUB>2</SUB>+2Ag$###$涂油、刷漆','【解答】解：（1）在铁、铜、银上述三种元素中，在地壳中含量最多的是铁；<br />（2）要验证上述三种元素的活动性顺序，用一种金属和两种常见的溶液就可完成，这两种溶液的溶质是氯化亚铁、硝酸银，铜能与硝酸银反应不能与氯化亚铁反应，可比较三种金属的活动性．反应的方程式是：Cu+2AgNO<SUB>3</SUB>=Cu（NO<SUB>3</SUB>）<SUB>2</SUB>+2Ag；<br />（3）防止钢铁生锈的措施有很多，例如：涂油、刷漆、保持铁制品表面干燥、洁净等．<br />故答为：（1）铁；（2）氯化亚铁、硝酸银，Cu+2AgNO<SUB>3</SUB>=Cu（NO<SUB>3</SUB>）<SUB>2</SUB>+2Ag；（3）涂油、刷漆等．','【分析】（1）根据在地壳中元素的含量分析判断；<br />（2）验证三种金属活动性强弱时，通常采取“三取中”的方法，即取中间金属单质与两端的金属的盐溶液反应或取中间金属的盐溶液与两端金属的单质反应；<br />（3）根据防止铁生锈的措施分析回答．','填空题',3.00,'ae151d50b5f08fb111df09b1c3aa103a',9,400,'金属活动性顺序及其应用,金属锈蚀的条件及其防护,地壳中元素的分布与含量','',2016,'37','2016•邻水县三模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1843872,'下列说法不正确的是（　　）','过氧化氢与水的化学性质不同是因为组成它们的分子结构不同','盐酸与硫酸化学性质相似是由于溶液中所含的阳离子全部是H<SUP>+</SUP>','NaCl水溶液能够导电是由于水溶液中含有大量自由移动的离子','金刚石与石墨物理性质差异较大是由于所含碳原子的结构不同','','D','【解答】解：A、过氧化氢与水化学性质不同，是因为它们分子的构成不同，不同种的分子性质不同，故选项解释正确．<br />B、盐酸与硫酸化学性质相似的原因是在水溶液中都含有H<SUP>+</SUP>，故选项解释正确．<br />C、NaCl水溶液能够导电是由于水溶液中含有大量自由移动的离子，故选项解释正确．<br />D、金刚石与石墨物理性质差异较大的原因是由于碳原子的排列方式不同，故选项解释错误．<br />故选：D','【分析】A、根据分子的基本特征：同种物质的分子性质相同，不同物质的分子性质不同，结合事实进行分析判断即可．<br />B、根据酸是电离时产生的阳离子全部是氢离子的化合物，据此进行分析判断．<br />C、根据NaCl水溶液能够导电的原因进行分析判断．<br />D、根据金刚石与石墨中碳原子的排列方式不同，进行分析判断．','选择题',3.00,'b0d89f22e519d4ec7e2bfc2a14106a2b',9,400,'酸的化学性质,溶液的导电性及其原理分析,分子的定义与分子的特性,碳元素组成的单质','',2016,'37','2016•工业园区一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1843895,'<img  src=\"/tikuimages/9/2016/400/shoutiniao75/469c0b40-94d4-11e9-a015-b42e9921e93e_xkb85.png\" style=\"vertical-align:middle;FLOAT:right\" />（2016•沈河区一模）牙膏是人们日常生活中的必需品，在牙膏的生产中需要添加摩擦剂，常见牙膏中的摩擦剂一般有CaCO<SUB>3</SUB>、Al（OH）<SUB>3</SUB>、SiO<SUB>2</SUB>等．某化学兴趣小组的同学针对牙膏中的摩擦剂进行了一系列的探究活动，请你参与并回答相关问题：<br />探究一、如何验证牙膏中含有碳酸盐，请简述实验操作，现象及结论<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />探究二、查阅资料得知：牙膏中的摩擦剂之一碳酸钙可以用石灰石来制备，制备流程如图所示：<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao2/469e2e21-94d4-11e9-92eb-b42e9921e93e_xkb24.png\" style=\"vertical-align:middle\" /><br />（1）在“煅烧炉”中发生反应的化学方程式是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）在“沉淀池”中生石灰与水反应属于<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（“放热”或“吸热”）反应．<br />（3）投入到“反应池”的石灰乳是不均一、不稳定的混合物，属于<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（选填“溶液”、“悬浊液”或“乳浊液”）．<br />（4）反应池中发生的反应是Ca（OH）<SUB>2</SUB>+Na<SUB>2</SUB>CO<SUB>3</SUB>═CaCO<SUB>3</SUB>↓+2NaOH，若在实验室里分离出“反应池”的沉淀物，该操作名称是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（5）为了达到节能减排、降低成本的目的，该流程向反应池所加的“碳酸钠溶液”可以用图中的某种物质来代替，该物质是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填物质名称）．<br />探究三：小组同学经过实验检测出某品牌牙膏中摩擦剂成分是碳酸钙后，又设计实验对其含量进行以下探究：<br />[查得资料]<br />①碱石灰是氢氧化钠和氧化钙的固体混合物；<br />②该品牌牙膏中其它成分遇到稀硫酸时无气体产生；<br />③氢氧化钡与氢氧化钙具有相似的化学性质，都能与二氧化碳发生化学反应．<br />[设计实验]利用如图所示装置（图中夹持仪器略去）进行实验，充分反应后测定C中生成BaCO<SUB>3</SUB>沉淀的质量，以确定碳酸钙的质量分数．（已知：充分搅拌后的牙膏样品中的碳酸钙粉末能与足量的稀硫酸完全反应）<br />依据图示回答下列问题：<br />（1）该装置中氢氧化钠溶液的作用是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（2）实验过程中需持续缓缓通入空气，其作用是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />[解释与结论]实验测得如表数据：<br /><table class=\"edittable\"><TBODY><TR><td>实验次数</TD><td>样品质量（g）</TD><td>所用稀硫酸质量（g）</TD><td>所得沉淀质量（g）</TD></TR><TR><td>第一次</TD><td>8.00</TD><td>50</TD><td>1.95</TD></TR><TR><td>第二次</TD><td>8.00</TD><td>50</TD><td>1.98</TD></TR><TR><td>第三次</TD><td>8.00</TD><td>50</TD><td>1.9</TD></TR></TBODY></TABLE>欲计算样品中碳酸钙的含量，应如何选择数据<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（选项“第一次”、“第二次”、“第三次”或者“三次数据的平均值”）．<br />[反思与评价]<br />&nbsp; 反思实验过程，小刚同学提出应采取必要措施提高测定准确度，他提出的下列各项措施中，不能提高测定准确度的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填字线序号）．<br />a．连接A、B、C、D装置后，再通入空气<br />b．在加入稀硫酸前，排净装置内的CO<SUB>2</SUB>气体<br />c．放慢滴加稀硫酸的速度<br />d．在A～B之间增添盛有浓硫酸的洗气装置．','','','','','','取少量牙膏于试管中，加入足量稀盐酸，若有气泡产生，再将气体通入澄清石灰水中，若石灰水变浑浊，证明牙膏中含有碳酸盐$###$CaCO<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>CaO+CO<SUB>2</SUB>↑$###$放热$###$悬浊液$###$过滤$###$二氧化碳$###$吸收空气中的二氧化碳$###$使反应产生的二氧化碳全部被氢氧化钡溶液吸收$###$三次数据的平均值$###$ad','【解答】解：探究一：碳酸盐会与酸反应生成二氧化碳，二氧化碳能使澄清石灰水变浑浊，所以验证牙膏中含有碳酸盐的实验操作过程是：取少量牙膏于试管中，加入足量稀盐酸，若有气泡产生，再将气体通入澄清石灰水中，若石灰水变浑浊，证明牙膏中含有碳酸盐；<br />探究二：（1）碳酸钙在高温的条件下生成氧化钙和二氧化碳，化学方程式为：CaCO<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>CaO+CO<SUB>2</SUB>↑；<br />（2）氧化钙和水反应生成氢氧化钙，放出大量的热，所以该反应属于放热反应；<br />（3）石灰乳是不均一、不稳定的混合物，溶质是固体氢氧化钙，属于悬浊液；<br />（4）过滤可以将不溶性的固体从溶液中分离出来，所以若在实验室里分离出“反应池”的沉淀物的操作名称是过滤；<br />（5）氢氧化钙会与碳酸盐、二氧化碳反应生成碳酸钙沉淀，所以该流程向反应池所加的“碳酸钠溶液”可以用图中的二氧化碳来代替；<br />探究三【设计实验】（1）氢氧化钠会与二氧化碳反应生成碳酸钠和水，所以装置中氢氧化钠溶液的作用是吸收空气中的二氧化碳；<br />（2）实验过程中需持续缓缓通入空气，其作用是使反应产生的二氧化碳全部被氢氧化钡溶液吸收；<br />【解释与结论】实验过程中为了减小误差，常采用多次测量取平均值的方法进行计算；<br />【反思与评价】该实验的反应原理是：充分反应后测定C中生成BaCO<SUB>3</SUB>沉淀的质量，以确定碳酸钙的质量分数，所以<br />a．连接A、B、C、D装置后，再通入空气，不能提高测定准确度，故a正确；<br />b．在加入稀硫酸前，排净装置内的CO<SUB>2</SUB>气体，能提高测定准确度，故b错误；<br />c．放慢滴加稀硫酸的速度，能提高测定准确度，故c错误；<br />d．在A～B之间增添盛有浓硫酸的洗气装置，不能提高测定准确度，故d正确．<br />故选：ad．<br />故答案为：探究一：取少量牙膏于试管中，加入足量稀盐酸，若有气泡产生，再将气体通入澄清石灰水中，若石灰水变浑浊，证明牙膏中含有碳酸盐；<br />探究二：（1）CaCO<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>CaO+CO<SUB>2</SUB>↑；<br />（2）放热；<br />（3）悬浊液；<br />（4）过滤；<br />（5）二氧化碳；<br />探究三【设计实验】（1）吸收空气中的二氧化碳；<br />（2）使反应产生的二氧化碳全部被氢氧化钡溶液吸收；<br />【解释与结论】三次数据的平均值；<br />【反思与评价】ad．','【分析】探究一：根据碳酸盐会与酸反应生成二氧化碳，二氧化碳能使澄清石灰水变浑浊进行分析；<br />探究二：（1）根据碳酸钙在高温的条件下生成氧化钙和二氧化碳进行分析；<br />（2）根据氧化钙和水反应生成氢氧化钙，放出大量的热进行分析；<br />（3）根据石灰乳是不均一、不稳定的混合物，溶质是固体氢氧化钙，属于悬浊液进行分析；<br />（4）根据过滤可以将不溶性的固体从溶液中分离出来进行分析；<br />（5）根据氢氧化钙会与碳酸盐、二氧化碳反应生成碳酸钙沉淀进行分析；<br />探究三【设计实验】（1）根据氢氧化钠会与二氧化碳反应生成碳酸钠和水进行分析；<br />（2）根据实验过程中需持续缓缓通入空气，其作用是使反应产生的二氧化碳全部被氢氧化钡溶液吸收进行分析；<br />【解释与结论】根据实验过程中为了减小误差，常采用多次测量取平均值的方法进行计算；<br />【反思与评价】根据该实验的反应原理是：充分反应后测定C中生成BaCO<SUB>3</SUB>沉淀的质量，以确定碳酸钙的质量分数进行分析．','书写',3.00,'7c2e7962015264d7ea8f66c5850d1e59',9,400,'实验探究物质的组成成分以及含量,证明碳酸盐,悬浊液、乳浊液的概念及其与溶液的区别,物质的相互转化和制备,物质发生化学变化时的能量变化,书写化学方程式、文字表达式、电离方程式','',2016,'37','2016•沈河区一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1843910,'请从①熟石灰，②食盐，③硝酸钾，④氮气，⑤聚乙烯塑料袋，⑥水果．中选择适当的物质填空（填序号）：<br />（1）生活中常用做调味品的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（2）随便废弃后会带来“白色污染”的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（3）用于改良酸性土壤的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（4）膨化食品常包装，充入的气体可能是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（5）农业上用作复合肥的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（6）日常食物中，能提供大量维生素的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','②$###$⑤$###$①$###$④$###$③$###$⑥','【解答】解：（1）生活中常用做调味品的是食盐，故填：②；<br />（2）随便废弃后会带来“白色污染”的是废弃塑料，故填：⑤；<br />（3）熟石灰氢氧化钙具有碱性，用于改良酸性土壤，故填：①；<br />（4）氮气的化学性子稳定，常用于膨化食品常包装充入的气体，故填：④；<br />（5）硝酸钾含有氮元素和钾元素，农业上用作复合肥，故填：③；<br />（6）日常食物中，能提供大量维生素的是水果，故填：⑥．','【分析】物质的性质决定物质的用途，根据已有的物质的性质进行分析解答即可．','填空题',3.00,'63ff8d9d3c27a69c224616fe3a1ed30f',9,400,'酸碱盐的应用,常见化肥的种类和作用,塑料制品使用的安全,食品、药品与健康食品中的有机营养素','',2016,'37','2016•青岛二模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1843995,'<img  src=\"/tikuimages/9/2016/400/shoutiniao74/47f2c19e-94d4-11e9-a04c-b42e9921e93e_xkb82.png\" style=\"vertical-align:middle;FLOAT:right;\" />（2016•蜀山区二模）饮用可乐时，小华对“二氧化碳的溶解性”产生了兴趣，并进行了探究．<br />【查阅资料】二氧化碳能溶于水．在通常情况下，1体积的水约能溶解1体积的二氧化碳．二氧化碳溶于水，既有物理变化，又有化学变化，发生化学变化的化学方程式：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />【提出问题】在实验室条件下，1体积的水究竟能溶解多少体积的二氧化碳呢？<br />【进行实验】<br />（1）将蒸馏水煮沸后，置于细口瓶中，盖上瓶塞，冷却至室温，备用．将蒸馏水煮沸的目的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．盖上瓶塞后再冷却的原因是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）制取二氧化碳，用向上排空法收集，验满后盖上玻璃片，备用．检验二氧化碳是否集满的操作方法是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />&nbsp;（3）实验①：取2支医用注射器分别抽取10mL空气和10mL备用的蒸馏水，用胶管连接（如图所示），交替推动2支注射器的活塞，反复多次．<br />实验②：取2支医用注射器分别抽取10mL现制的二氧化碳气体和10mL备用的蒸馏水，用胶管连接（如图所示），交替推动2支注射器的活塞，反复多次至体积不再减少为止．反复交替推动注射器活塞的作用是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（4）记录数据<br /><table class=\"edittable\"><TBODY><TR><td width=71>实验编号</TD><td width=153>抽取气体的体积</TD><td width=127>抽取蒸馏水的体积</TD><td width=183>充分混合后，气、液总体积</TD></TR><TR><td>①</TD><td>10mL空气</TD><td>10mL</TD><td>20mL</TD></TR><TR><td>②</TD><td>10mL现制的二氧化碳</TD><td>10mL</TD><td>12mL</TD></TR></TBODY></TABLE>【得出结论】由上表所列数据可知，在实验室条件下，空气难溶于水，二氧化碳的溶解性为1体积水最多能溶解<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>体积的二氧化碳．<br />【注意：若答对以下小题奖励4分，化学试卷总分不超过60分】<br />【反思质疑】小芳对实验结论产生了疑问：实验②中不能再溶解的气体中是否还含有二氧化碳气体？如收集二氧化碳时空气未完全排尽，也可能导致实验②的残留气体是空气，就不能再溶，则实验结论不可信．请你设计一个方案探究以上所得实验结论是否可信．写清操作步骤、现象及结论．<br /><table class=\"edittable\"><TBODY><TR><td width=152>&nbsp;操作步骤</TD><td width=152>实验现象&nbsp;</TD><td width=152>实验结论&nbsp;</TD></TR><TR><td><u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD><td><u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD><td><u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD></TR></TBODY></TABLE>','','','','','','CO<SUB>2</SUB>+H<SUB>2</SUB>O=H<SUB>2</SUB>CO<SUB>3</SUB>$###$气体的溶解度随温度的升高而减小，赶走蒸馏水中溶解的气体，减小对实验数据的影响$###$防止气体进入，影响实验$###$将燃着的木条放在集气瓶口，如果熄灭，则收集满$###$让气体和水充分溶解$###$0.8$###$将气体推至一注射器内，气液分离，取下盛有液体的注射器并挤出液体，重新抽取一定体积的蒸馏水，再将两注射器连接，反复交替推动活塞．$###$①气体体积减少（总体积减少），②若气体体积不变（总体积不变）．$###$①实验结论可信②实验结论不可信．','【解答】解：【查阅资料】水和二氧化碳反应生成碳酸，故答案：CO<SUB>2</SUB>+H<SUB>2</SUB>O=H<SUB>2</SUB>CO<SUB>3</SUB>；<br />【进行实验】<br />（1）气体的溶解度随温度的升高而减小，赶走蒸馏水中溶解的气体，减小对实验数据的影响；防止气体进入，影响实验；<br />（2）二氧化碳不燃烧也不支持燃烧，验满将燃着的木条放在集气瓶口，故答案：将燃着的木条放在集气瓶口，如果熄灭，则收集满；<br />（3）反复交替推动注射器活塞的作用是让气体和水充分溶解，故答案：让气体和水充分溶解；<br />（4）表中数据可知二氧化碳的溶解性为1体积水最多能溶解0.8；<br />【反思质疑】将气体推至一注射器内，气液分离，取下盛有液体的注射器并挤出液体，重新抽取一定体积的蒸馏水，再将两注射器连接，反复交替推动活塞．若气体体积减少（总体积减少），说明残留的气体中有二氧化碳，实验结论可信．（若气体体积不变（总体积不变），说明残留的气体中没有二氧化碳，实验结论不可信．）将气体推至一注射器内，气液分离，取下盛有气体的注射器，将气体通入澄清的石灰水中．若澄清的石灰水变浑浊，说明残留的气体中有二氧化碳，实验结论可信．<br />故答案：将气体推至一注射器内，气液分离，取下盛有液体的注射器并挤出液体，重新抽取一定体积的蒸馏水，再将两注射器连接，反复交替推动活塞．若气体体积减少（总体积减少），说明残留的气体中有二氧化碳，实验结论可信．（若气体体积不变（总体积不变），说明残留的气体中没有二氧化碳，实验结论不可信．','【分析】【查阅资料】根据水和二氧化碳反应生成碳酸解答；<br />【进行实验】<br />（1）根据气体的溶解度随温度的升高而减小分析；<br />（2）根据二氧化碳的性质确定方法；<br />（3）根据反复交替推动注射器活塞的作用是让气体和水充分溶解分析；<br />（4）根据表中数据可知二氧化碳的溶解性为1体积水最多能溶解0.8分析；<br />【反思质疑】将气体推至一注射器内，气液分离，取下盛有液体的注射器并挤出液体，重新抽取一定体积的蒸馏水，再将两注射器连接，反复交替推动活塞．若气体体积减少（总体积减少），说明残留的气体中有二氧化碳，实验结论可信．（若气体体积不变（总体积不变），说明残留的气体中没有二氧化碳，实验结论不可信．）将气体推至一注射器内，气液分离，取下盛有气体的注射器，将气体通入澄清的石灰水中．若澄清的石灰水变浑浊，说明残留的气体中有二氧化碳，实验结论可信','填空题',3.00,'108deec92ae409b84e1755d37b03e214',9,400,'二氧化碳的检验和验满,探究二氧化碳的性质','',2016,'37','2016•蜀山区二模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1843996,'纳米材料具有特殊的性质和功能，纳米二氧化钛（TiO<SUB>2</SUB>）参与的光催化反应可使吸附在其表面的甲醛等物质被氧化，降低空气中有害物质的浓度．正钛酸（H<SUB>4</SUB>TiO<SUB>4</SUB>）在一定条件下分解失水可制得纳米TiO<SUB>2</SUB>．下列说法不正确的是（　　）','纳米TiO<SUB>2</SUB>属于氧化物','纳米TiO<SUB>2</SUB>与TiO<SUB>2</SUB>的性质、功能完全相同','纳米TiO<SUB>2</SUB>添加到墙面涂料中，可减少污染','制备纳米TiO<SUB>2</SUB>的反应：H<SUB>4</SUB>TiO<SUB>4</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;一定条件&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>TiO<SUB>2</SUB>+2H<SUB>2</SUB>O','','B','【解答】解：A．纳米TiO<SUB>2</SUB>是由两种元素组成的且有一种是氧元素的化合物，属于氧化物，故选项说法正确．<br />B．纳米材料具有特殊的性质和功能，纳米TiO<SUB>2</SUB>与TiO<SUB>2</SUB>的性质、功能不完全相同，故选项说法错误．<br />C．纳米二氧化钛（TiO<SUB>2</SUB>）参与的光催化反应可使吸附在其表面的甲醛等物质被氧化，降低空气中有害物质的浓度，纳米TiO<SUB>2</SUB>添加到墙面涂料中，可减少污染，故选项说法正确．<br />D．正钛酸（H<SUB>4</SUB>TiO<SUB>4</SUB>）在一定条件下分解失水可制得纳米TiO<SUB>2</SUB>，反应的化学方程式为H<SUB>4</SUB>TiO<SUB>4</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;一定条件&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>TiO<SUB>2</SUB>+2H<SUB>2</SUB>O，故选项说法正确．<br />故选：B．','【分析】A．根据氧化物是只含有两种元素且其中一种元素是氧元素的化合物，进行分析判断．<br />B．根据纳米材料具有特殊的性质和功能，进行分析判断．<br />C．根据题意，纳米二氧化钛（TiO<SUB>2</SUB>）参与的光催化反应可使吸附在其表面的甲醛等物质被氧化，降低空气中有害物质的浓度，进行分析判断．<br />D．根据题意，正钛酸（H<SUB>4</SUB>TiO<SUB>4</SUB>）在一定条件下分解失水可制得纳米TiO<SUB>2</SUB>，进行分析判断．','选择题',3.00,'1e499063774a1c10af6525392344a717',9,400,'从组成上识别氧化物,复合材料、纳米材料','',2016,'37','2016•黄冈二模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1844020,'下列叙述错误的是（　　）','日常生活中可用肥皂水鉴别硬水和软水','向酸性土壤里撒熟石灰，可调节土壤的pH','铝比铁具有更好的抗腐蚀性能，是因为铁比铝更活泼','用汽油、加了洗涤剂的水分别除去衣服上的油污，两者去污原理不同','','C','【解答】解：A、肥皂水可以鉴别硬水和软水，泡沫多的是软水，泡沫少的是硬水，故正确；<br />B、熟石灰是碱性物质，可改良酸性土壤，改变土壤的pH，故正确；<br />C、铝在空气中易被氧气氧化，表面形成致密的氧化铝保护膜，故耐腐蚀，铝的金属活动性比铁活泼，故不正确；<br />D、汽油除去衣服上的油污是溶解，洗涤剂除去衣服上的油污是溶解乳化，二者原理不同，故正确．<br />故选：C．','【分析】A、根据用肥皂水可以鉴别硬水和软水解答；<br />B、根据熟石灰是碱性物质，可改良酸性土壤分析；<br />C、根据铝、铁与氧气反应生成的氧化物的特点分析判断；<br />D、根据汽油除去衣服上的油污是溶解，洗涤剂除去衣服上的油污是溶解乳化分析．','选择题',3.00,'a02cfbeebce79f20d69e36c9545576e4',9,400,'硬水与软水,溶解现象与溶解原理,乳化现象与乳化作用,金属的化学性质,中和反应及其应用','',2016,'32','2016•濮阳校级模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1844036,'下面的化学实验操作都正确的一组是（　　）<br />①给试管里的液体加热时，试管应与桌面垂直．<br />②给试管中的固体加热时试管口应略向下倾斜．<br />③用酒精灯给物质加热时，应该用外焰．<br />④玻璃仪器洗过后，如果内壁的水既不成水滴也不成股流下，才算洗干净了．','④①②','②①③','③①④','②③④','','D','【解答】解：①给试管里的液体加热时，试管应与桌面呈45度角，此时与火焰接触面积最大，故①错；<br />②给试管中的固体加热时试管口应略向下倾斜，故②正确；<br />③酒精灯的火焰分为：外焰、内焰、焰心，其中外焰温度最高，所以一般用外焰加热，故③正确；<br />④玻璃仪器洗净的标志是：玻璃仪器内壁附着的水既不聚集成水滴也不成股流下，或均匀覆盖着一层水膜，故④正确．<br />故选D．','【分析】①根据给试管里的液体加热需要考虑试管与酒精灯火焰的接触面积要大；②根据给试管中的固体加热时试管口应略向下倾斜考虑；③根据酒精灯火焰的温度考虑；④根据玻璃仪器洗净的标志考虑．','选择题',3.00,'81c18da22ace51873294a7b37f4b1ee9',9,400,'加热器皿-酒精灯,给试管里的固体加热,给试管里的液体加热,玻璃仪器的洗涤','',2014,'37','2014秋•宁县月考',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1844063,'一种铁原子的原子核内有26个质子和30个中子，该原子的核外电子数为（　　）','56','30','26','4','','C','【解答】解：在原子中，原子序数=质子数=核电荷数=核外电子数，已知一种铁原子，原子核内有26个质子，因此该铁原子的核外电子数为26．<br />故选：C．','【分析】根据在原子中，原子序数=质子数=核电荷数=核外电子数，进行解答．','选择题',3.00,'f1b941f3cab1f1f802a2ddd6c6f17106',9,400,'原子的有关数量计算','',2016,'37','2016•顺义区一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1844068,'<img  src=\"/tikuimages/9/2016/400/shoutiniao3/48dd6fc0-94d4-11e9-b011-b42e9921e93e_xkb24.png\" style=\"vertical-align:middle;FLOAT:right\" />（2016•吉林一模）图1是氢弹的结构图，如表是氢弹相关原子的结构．<br />（1）钋元素是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>元素（填“金属”或“非金属”）；<br />（2）氢原子和氚原子属于同种元素的原子依据是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>相同；<br />（3）图2中表示氘原子的结构示意图的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br /><table class=\"edittable\"><TBODY><TR><td>原子</TD><td>质子数</TD><td>中子数</TD><td>电子数</TD></TR><TR><td>氢</TD><td>1</TD><td>0</TD><td>1</TD></TR><TR><td>氘</TD><td>1</TD><td>1</TD><td>1</TD></TR><TR><td>氚</TD><td>1</TD><td>2</TD><td>1</TD></TR></TBODY></TABLE>','','','','','','金属$###$质子数$###$A','【解答】解：（1）钋元素属于金属元素，故填：金属；<br />（2）氢原子和氚原子属于同种元素的原子是因为二者的质子数相同，故填：质子数；<br />（3）图2中表示氘原子的结构示意图的是A，故填：A．','【分析】根据已有的元素的类别、元素种类的决定因素以及原子结构示意图的知识进行分析解答即可．','填空题',3.00,'65fdcd18c0849991c86a212a373f8f87',9,400,'原子的定义与构成,原子结构示意图与离子结构示意图,元素的简单分类','',2016,'37','2016•吉林一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1844071,'下列反应既是复分解反应，又是中和反应的是（　　）','BaCl<SUB>2</SUB>+H<SUB>2</SUB>SO<SUB>4</SUB>═BaSO<SUB>4</SUB>↓+2HCl','Mg（OH）<SUB>2</SUB>+2HCl=MgCl<SUB>2</SUB>+2H<SUB>2</SUB>O','CuO+H<SUB>2</SUB>SO<SUB>4</SUB>=CuSO<SUB>4</SUB>+H<SUB>2</SUB>O','SO<SUB>2</SUB>+2NaOH=Na<SUB>2</SUB>SO<SUB>3</SUB>+H<SUB>2</SUB>O','','B','【解答】解：A、BaCl<SUB>2</SUB>+H<SUB>2</SUB>SO<SUB>4</SUB>═BaSO<SUB>4</SUB>↓+2HCl，该反应是两种化合物相互交换成分生成两种新的化合物的反应，属于复分解反应；但不是酸与碱的反应，不属于中和反应，故选项错误．<br />B、Mg（OH）<SUB>2</SUB>+2HCl=MgCl<SUB>2</SUB>+2H<SUB>2</SUB>O，该反应是两种化合物相互交换成分生成两种新的化合物的反应，属于复分解反应；且反应物是酸和碱，生成物是盐和水，属于中和反应；故选项正确．<br />C、CuO+H<SUB>2</SUB>SO<SUB>4</SUB>=CuSO<SUB>4</SUB>+H<SUB>2</SUB>O，该反应是两种化合物相互交换成分生成两种新的化合物的反应，属于复分解反应；但不是酸与碱的反应，不属于中和反应，故选项错误．<br />D、SO<SUB>2</SUB>+2NaOH=Na<SUB>2</SUB>SO<SUB>3</SUB>+H<SUB>2</SUB>O，该反应不是两种化合物相互交换成分生成两种新的化合物的反应，不属于复分解反应，故选项错误．<br />故选：B．','【分析】复分解反应是两种化合物相互交换成分生成两种新的化合物的反应；中和反应是酸与碱作用生成盐和水的反应，反应物是酸和碱，生成物是盐和水，据此进行分析判断．','选择题',3.00,'61bd4055518eba71528dd017cc89316a',9,400,'中和反应及其应用,复分解反应及其应用','',0,'37','',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1844086,'小明发现家中铜制水阀门上有绿色的铜锈，想对铜生锈的原因进行探究．<br />【查阅资料】铜锈为绿色，其主要成分是Cu<SUB>2</SUB>（OH）<SUB>2</SUB>CO<SUB>3</SUB>．<br />【提出猜想】小明根据铜绿的组成元素推测，铜生锈可能是铜与氧气以及<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>、<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>共同作用的结果．<br />【实验设计】小明与同学一起设计如下图所示4个实验．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao78/490bd2c0-94d4-11e9-91fc-b42e9921e93e_xkb91.png\" style=\"vertical-align:middle\" /><br />【实验记录】若干天后，小明发现只有“实验四”出现绿色铜锈，其它实验中均无铜锈出现．<br />【分析与评价】<br />（1）<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>与<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填实验序号）对比，可以说明“与氧气接触”是铜生锈的必要条件．<br />（2）“实验二”与“实验四”对比可以得出的结论是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）实验二、实验三中所使用的蒸馏水是经煮沸迅速冷却的蒸馏水，其目的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />【实验结论】小明的猜想是正确的．','','','','','','二氧化碳$###$水$###$实验三$###$实验四$###$与二氧化碳接触是铜生锈的必要条件$###$除去氧气和二氧化碳','【解答】解：[猜想]根据铜绿的主要成分是Cu<SUB>2</SUB>（OH）<SUB>2</SUB>CO<SUB>3</SUB>，可知铜生锈可能是铜、水、二氧化碳、水蒸气共同作用的结果．<br />【分析与评价】<br />实验一中有氧气和二氧化碳；实验二中有水、氧气；实验三中有氧化碳和水；实验四中有氧气、水和二氧化碳；<br />（1）实验三&nbsp;和实验四；对比，可以说明“与氧气接触”是铜生锈的必要条件．<br />（2）“实验二”与“实验四”对比可以得出的结论是与二氧化碳接触是铜生锈的必要条件；<br />（3）实验二、实验三中所使用的蒸馏水是经煮沸迅速冷却的蒸馏水，其目的是除去除去氧气和二氧化碳．<br />答案：<br />【提出猜想】二氧化碳；&nbsp;&nbsp;&nbsp;水；<br />【分析与评价】<br />（1）实验三；&nbsp;&nbsp;&nbsp;实验四；<br />（2）与二氧化碳接触是铜生锈的必要条件．<br />（3）除去氧气和二氧化碳','【分析】[猜想]根据铜锈的化学式判断；<br />【分析与评价】根据各试管内的条件判断，实验一中有氧气和二氧化碳；实验二中有水、氧气；实验三中有氧化碳和水；实验四中有氧气、水和二氧化碳．','填空题',3.00,'3a2457b3940a628c9ae40dc2eb15c7a8',9,400,'探究金属锈蚀的条件','',2016,'37','2016•延平区一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1844126,'学以致用．亚文同学在下列化学知识的应用中，你认为不正确的是（　　）','浓硫酸沾到皮肤上，立即用大量水冲洗','酒精在桌面燃起来用嘴吹灭','用肥皂水区分蒸馏水和硝酸镁溶液','进入久未开肩的菜窖前，应先做灯火实验','','B','【解答】解：A、浓硫酸具有较强的腐蚀性，不慎沾到皮肤上，应立即用大量清水冲洗，后涂上3%～5%的碳酸氢钠溶液，故选项说法正确．<br />B、酒精在桌面燃起来，应用湿抹布铺盖，不能用嘴吹灭，故选项说法错误．<br />C、硝酸镁溶液中含有可溶性镁离子，可用肥皂水区分蒸馏水和硝酸镁溶液，产生泡沫较多的是蒸馏水，较少的是硝酸镁溶液，故选项说法正确．<br />D、久未开启的菜窖中容易积聚大量的二氧化碳，二氧化碳不支持燃烧，进入久未开启的菜窖前可以先做灯火实验以检验二氧化碳含量是否过高，故选项说法正确．<br />故选：B．','【分析】A、根据浓硫酸不慎沾到皮肤上的处理方法，进行分析判断．<br />B、根据酒精在桌面燃起来的处理方法，进行分析判断．<br />C、根据硝酸镁溶液中含有可溶性镁离子，进行分析判断．<br />D、久未开启的菜窑中可能含有大量的不能供给呼吸的二氧化碳．','选择题',3.00,'2425b1c312aaff624caed4bb14593678',9,400,'常见的意外事故的处理方法,二氧化碳的化学性质,硬水与软水,灭火的原理和方法','',2016,'35','2016春•重庆校级期中',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1844213,'某校化学研究性学习小组的同学在老师指导下开展了一系列探究物质组成的实验活动，请按要求回答问题：<br />探究活动一：探究空气中氧气的体积分数<br />（1）选用汞、磷、铜等物质均可以完成该实验，选择反应物的主要依据是（填序号）<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />A．反应物是固体&nbsp;&nbsp;&nbsp;&nbsp; B．在空气中反应且只消耗氧气<br />C．燃烧有明显现象&nbsp;&nbsp;&nbsp; D．生成物不是气体<br />（2）用甲、乙两套装置进行实验（红磷充足、装置不漏气）．<br /><img  src=\"/tikuimages/9/2015/400/shoutiniao68/4a87010f-94d4-11e9-8989-b42e9921e93e_xkb58.png\" style=\"vertical-align:middle\" /><br />①红磷燃烧的化学方程式是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，下列有关分析合理的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />A．都只能在实验结束后，推测出氮气是无色气体<br />B．甲中燃烧匙伸入集气瓶太慢，测得空气中氧气的体积分数将偏大<br />C．乙中瘪气球可以防止燃烧放热使橡胶塞弹出<br />D．乙比甲更节约能源<br />②利用乙装置进行实验时，先关闭弹簧夹，加热使合力反应，待装置冷却至室温后打开弹簧夹，反应前后各数据及结论如下，请分析后将玻璃管中原有的空气体积填入空格内：<br /><table class=\"edittable\"><TBODY><TR><td width=116 rowSpan=2>玻璃管中原有<br />空气的体积</TD><td width=229 colSpan=2>注射器中气体的体积</TD><td width=198 rowSpan=2>结论</TD></TR><TR><td>反应前</TD><td>反应后</TD></TR><TR><td><br /><u>&nbsp;&nbsp;&nbsp;&nbsp;</u>mL</TD><td>40 mL</TD><td>30 mL</TD><td>氧气约占空气体积的五分之</TD></TR></TBODY></TABLE>探究活动二、探究水的组成<br />（1）如图1，是水通电分解的示意图，此实验可以探究水的组成．在实验过程中，观察到ab两玻璃管中产生气体的体积比为1：2，则a玻璃管连接的是电源的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>极．<br />（2）还能说明组成水的元素种类的实验是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />探究活动三、探究不同物质的组成元素<br />为探究不同物质中可能含有相同元素，设计了如下两个实验：<br />（1）如图2，取少量白砂糖、面粉，分别放在燃烧匙中，在酒精灯上加热，直至完全烧焦，燃烧匙中所剩物质均呈黑色，由此可知白砂糖、面粉中都有<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>元素；<br />（2）如图3，加热碱式碳酸铜的实验过程中，观察到什么现象时也可说明碱式碳酸铜中含有与白砂糖、面粉相同的某种元素？<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />以上实验可以用来推断物质的组成元素，试用微粒的观点分析设计这些实验的依据是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br /><img  src=\"/tikuimages/9/2015/400/shoutiniao58/4a8ad19e-94d4-11e9-9c1e-b42e9921e93e_xkb38.png\" style=\"vertical-align:middle\" />','','','','','','ABD$###$4P+5O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2P<SUB>2</SUB>O<SUB>5</SUB>$###$BC$###$50$###$正$###$氢气燃烧生成水$###$碳$###$澄清石灰水变浑浊$###$反应前后原子种类不变','【解答】解：探究活动一<br />（1）选用汞、磷、铜等物质均可以完成该实验，选择反应物的主要依据是反应物是固体，在空气中反应且只消耗氧气，生成物不是气体．<br />故填：ABD．<br />（2）①红磷燃烧的化学方程式是：4P+5O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2P<SUB>2</SUB>O<SUB>5</SUB>；<br />A．在实验结前就能够推测出氮气是无色气体，该选项说法不正确；<br />B．甲中燃烧匙伸入集气瓶太慢时，会导致部分空气受热膨胀逸出，从而导致测得空气中氧气的体积分数偏大，该选项说法正确；<br />C．乙中瘪气球可以防止燃烧放热使橡胶塞弹出，该选项说法正确；<br />D．乙中红磷燃烧后可以熄灭酒精灯，该选项说法不正确．<br />故填：4P+5O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2P<SUB>2</SUB>O<SUB>5</SUB>；BC．<br />②因为氧气约占空气总体积的五分之一，注射器中的气体减少10mL，说明玻璃管中的氧气是10mL，因此玻璃管中原有空气的体积为：<br />10mL÷<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">1</td></tr><tr><td>5</td></tr></table></span>=50mL，<br />故填：50．<br />探究活动二<br />（1）在实验过程中，观察到ab两玻璃管中产生气体的体积比为1：2，则a玻璃管连接的是电源的正极．<br />故填：正．<br />（2）还能说明组成水的元素种类的实验是氢气燃烧生成水，氢气和氧气反应生成水，说明水中含有氢元素和氧元素．<br />故填：氢气燃烧生成水．<br />探究活动三<br />（1）取少量白砂糖、面粉，分别放在燃烧匙中，在酒精灯上加热，直至完全烧焦，燃烧匙中所剩物质均呈黑色，由此可知白砂糖、面粉中都有碳元素．<br />故填：碳．<br />（2）加热碱式碳酸铜的实验过程中，观察到澄清石灰水变浑浊，说明反应生成了二氧化碳，进一步说明碱式碳酸铜中含有碳元素；<br />以上实验可以用来推断物质的组成元素，设计这些实验的依据是反应前后原子种类不变．<br />故填：澄清石灰水变浑浊；反应前后原子种类不变．','【分析】探究活动一<br />选择反应物的主要依据是反应物是固体，在空气中反应且只消耗氧气，生成物不是气体；<br />红磷在空气中燃烧生成五氧化二磷；<br />氧气约占空气总体积的五分之一；<br />探究活动二<br />电解水时，正极产生的是氧气，负极产生的是氢气，氧气和氢气的体积比约为1：2；&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <br />氢气燃烧生成水；<br />探究活动三<br />二氧化碳能使澄清石灰水变浑浊；<br />化学反应前后元素种类不变，原子种类和总个数不变．','填空题',3.00,'b8b5d371a2a0926b31673bd3a9ff9d4c',9,400,'测定空气里氧气含量的探究,质量守恒定律的实验探究,电解水实验','',2015,'33','2015秋•沈河区期末',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1844224,'<img  src=\"/tikuimages/9/2016/400/shoutiniao16/4aab03d1-94d4-11e9-bda5-b42e9921e93e_xkb97.png\" style=\"vertical-align:middle;FLOAT:right;\" />如图所示的装置中，胶头滴管中吸入某种液体，平底烧瓶中充入（或放入）另一种物质，挤压胶头滴管加入液体，一段时间后装置中的气球都明显胀大（忽略液体体积对气球体积的影响）．则滴管和烧瓶中所用试剂可能是（　　）','NaOH溶液和CO<SUB>2</SUB>','双氧水和MnO<SUB>2</SUB>','稀硫酸和铜片','稀盐酸和石灰石','','A','【解答】解：<br />A、瓶中加入烧碱和二氧化碳，二氧化碳与烧碱反应生成碳酸钠，气体减少，烧瓶内压强小于外界压强，气球胀大，故A正确；<br />B、瓶中加入双氧水和二氧化锰，反应生成氧气，使烧瓶内的压强增大，烧瓶内压强大于外界，气球变小，故B错误；<br />C、瓶中加入铜和稀硫酸，铜在金属活动性表中排在氢后面，不能与稀硫酸反应，压强不变，故C错误；<br />D、瓶加入稀盐酸和石灰石，反应生成二氧化碳，使烧瓶内的压强增大，烧瓶内压强大于外界，气球变小，故D错误．<br />故选A．','【分析】装置气球膨大原因是烧瓶内的气体反应或者温度降低，使烧瓶内的压强小于外界压强．','选择题',3.00,'fe27a8618c20968c2bcc6740c2fc6207',9,400,'催化剂的特点与催化作用,酸的化学性质,碱的化学性质,反应现象和本质的联系','',2016,'37','2016•淮安校级一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1844233,'铁锈的主要成分是&nbsp;（写化学式）．<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>将一根生锈的铁钉&nbsp;插入足量的稀硫酸中，首先看到<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>、<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，反应方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，一段时间后，又看到<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，反应的方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','Fe<SUB>2</SUB>O<SUB>3</SUB>$###$锈迹溶解$###$溶液呈黄色$###$Fe<SUB>2</SUB>O<SUB>3</SUB>+6HCl═2FeCl<SUB>3</SUB>+3H<SUB>2</SUB>O$###$有气泡产生，溶液呈浅绿色$###$Fe+2HCl═FeCl<SUB>2</SUB>+H<SUB>2</SUB>↑','【解答】解：铁锈的主要成分是氧化铁．当将带锈迹的铁钉放入过量的盐酸中，首先看到锈迹溶解，溶液呈黄色；<br />这是因为氧化铁和稀盐酸反应生成氯化铁和水，反应的化学方程式为：Fe<SUB>2</SUB>O<SUB>3</SUB>+6HCl═2FeCl<SUB>3</SUB>+3H<SUB>2</SUB>O．<br />当氧化铁完全反应后，铁又和稀盐酸反应生成氯化亚铁和氢气．反应的化学方程式为：Fe+2HCl═FeCl<SUB>2</SUB>+H<SUB>2</SUB>↑．<br />故答案为：Fe<SUB>2</SUB>O<SUB>3</SUB>；锈迹溶解，溶液呈黄色；Fe2O3+6HCl═2FeCl3+3H2O；有气泡产生，溶液呈浅绿色；Fe+2HCl═FeCl<SUB>2</SUB>+H<SUB>2</SUB>↑．','【分析】铁锈的主要成分是氧化铁；氯化铁溶液的颜色是黄色的，氯化亚铁溶液的颜色是浅绿色的；根据反应物和生成物及其质量守恒定律可以书写化学方程式','书写',3.00,'df7c13b45c4f926a79eb9b12317c62d8',9,400,'铁锈的主要成分,酸的化学性质,书写化学方程式、文字表达式、电离方程式','',2016,'37','2016春•沈阳月考',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1844240,'下列关于水的组成的说法中，不正确的是（　　）','水由水分子构成','水由氢原子和氧原子构成','水由氢、氧两种元素组成','一个水分子由两个氢原子和一个氧原子构成','','B','【解答】解：A、水是由水分子构成的，故A正确；<br />B、水分子是由氢原子和氧原子构成的，水是由水分子构成的，故B说法不正确；<br />C、水是由氢元素和氧元素组成的，故C说法正确；<br />D、一个水分子是由两个氢原子和一个氧原子构成，故D说法正确；<br />故选B','【分析】A、根据水的构成分析；<br />B、根据水的构成和水分子的构成分析判断；<br />C、根据水是由氢元素和氧元素组成的分析判断；<br />D、根据分子是由原子构成的分析判断．','选择题',3.00,'7cf9abd90753074fd277f3d3558f3209',9,400,'水的组成','',2016,'37','2016•顺义区一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1844254,'<img  src=\"/tikuimages/9/2016/400/shoutiniao9/4aed6400-94d4-11e9-8531-b42e9921e93e_xkb3.png\" style=\"vertical-align:middle;FLOAT:right\" />（2016•临清市二模）某化学兴趣小组设计了一组“吹气球”实验，三套装置如图：<br />（1）甲装置：向a中加水会看到气球胀大，一段时间内气球大小没有变化，说明该装置气密性<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）乙装置：<br />①若生成H<SUB>2</SUB>使气球胀大，则分液漏斗中所装的液体物质是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（写出一种具体物质的化学式）．<br />②若锥形瓶中装有NaOH固体，分液漏斗中加入少量水，水与NaOH固体接触后，则气球胀大的主要原因<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）丙装置：若锥形瓶中盛满CO<SUB>2</SUB>，欲使气球胀大，则分液漏斗中的液体可以是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>或<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','好$###$H<SUB>2</SUB>SO<SUB>4</SUB>$###$氢氧化钠溶于水放出大量的热$###$氢氧化钠$###$氢氧化钡','【解答】解：（1）向a中加水会看到气球胀大，一段时间内气球大小没有变化，说明该装置气密性良好．<br />（2）①该装置是固液在常温下制取气体的装置，若生成H<SUB>2</SUB>使气球胀大，则分液漏斗中所装的液体物质是稀硫酸，化学式是：H<SUB>2</SUB>SO<SUB>4</SUB>；<br />②若锥形瓶中装有NaOH固体，分液漏斗中加入少量水，水与NaOH固体接触后，则气球胀大的主要原因是：氢氧化钠溶于水放出大量的热．<br />（3）丙中CO<SUB>2</SUB>气体被吸收时，装置内压强减小，内部的气球就会胀大，氢氧化钠、氢氧化钡等溶液可吸收二氧化碳气体．<br />故答案为：（1）好；（2）①H<SUB>2</SUB>SO<SUB>4</SUB>，②氢氧化钠溶于水放出大量的热；（3）氢氧化钠，氢氧化钡．','【分析】（1）根据仪器的名称和检查气密性的方法进行分析；<br />（2）根据氢气的实验室制法和氢氧化钠溶于水放出大量的热进行分析；<br />（3）丙中CO<SUB>2</SUB>气体被吸收时，装置内压强减小，内部的气球就会胀大，据此分析．','简答题',3.00,'7635281ee28806178137081a6d504226',9,400,'检查装置的气密性,二氧化碳的化学性质,溶解时的吸热或放热现象,氢气的制取和检验,反应现象和本质的联系','临清市',2016,'37','2016•临清市二模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1844256,'目前，教科书中测定空气里氧气含量的方法，无法满足严谨的科学要求，利用NO和氢氧化钠溶液可以更准确地测定空气中O<SUB>2</SUB>的含量．<br />【查阅资料】①NO是一种无色无味，难溶于水的有毒气体．相同条件下，2体积的NO与1体积的O<SUB>2</SUB>完全反应，生成2体积的NO<SUB>2</SUB>（化学方程式为：2NO+O<SUB>2</SUB>═2NO<SUB>2</SUB>）<br />②氢氧化钠溶液吸收NO<SUB>2</SUB>生成NaNO<SUB>2</SUB>、NaNO<SUB>3</SUB>和H<SUB>2</SUB>O<br />③2FeSO<SUB>4</SUB>+3H<SUB>2</SUB>SO<SUB>4</SUB>+2NaNO<SUB>2</SUB>═Fe<SUB>2</SUB>（SO<SUB>4</SUB>）<SUB>3</SUB>+2NaHSO<SUB>4</SUB>+2H<SUB>2</SUB>O+2NO↑<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao28/4af3f3b0-94d4-11e9-a15d-b42e9921e93e_xkb50.png\" style=\"vertical-align:middle\" /><br />（1）写出仪器d的名称：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（2）若要调节A中试管与酒精灯相对高度，应调节螺旋<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“a”、“b”或“c”）．<br />（3）写出氢氧化钠溶液吸收NO<SUB>2</SUB>反应的化学方程式：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（4）实验室用稀硫酸、NaNO<SUB>2</SUB>饱和溶液和FeSO<SUB>4</SUB>•7H<SUB>2</SUB>O固体反应制备NO，应选用的发生装置是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填序号）．用E装置收集纯净的NO气体，而不用C或D装置的理由是：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（5）如图F所示，向收集有60mLNO气体的量筒中，缓慢推入20mL的空气，用橡胶塞塞住量筒口，小心混合，使气体充分反应．1～2min后，在液面下移开橡胶塞．2min后，观察和记录最终量筒中气体的体积为67.4mL．则20mL空气中含氧气<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>mL，操作时存在的主要安全隐患<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（6）收集60mLNO气体（密度约为1.27g/L），至少需要固体NaN0<SUB>2</SUB>的质量是多少？（请写出计算过程，计算结果精确到0.01g）','','','','','','水槽$###$b$###$2NaOH+2NO<SUB>2</SUB>=NaNO<SUB>2</SUB>+NaNO<SUB>3</SUB>+H<SUB>2</SUB>O$###$B$###$一氧化氮能与氧气反应且密度与空气接近$###$4.2$###$氢氧化钠溶液对皮肤具有腐蚀性','【解答】解：（1据图可以看出，仪器d是水槽，故填：水槽；<br />（2）若要调节A中试管与酒精灯相对高度，应调节螺旋b，故填：b；<br />（3）氢氧化钠溶液吸收NO<SUB>2</SUB>反应生成亚硝酸钠、硝酸钠和水，故填：2NaOH+2NO<SUB>2</SUB>=NaNO<SUB>2</SUB>+NaNO<SUB>3</SUB>+H<SUB>2</SUB>O．<br />（4）实验室用稀硫酸、NaNO<SUB>2</SUB>饱和溶液和FeSO<SUB>4</SUB>•7H<SUB>2</SUB>O固体反应制备NO，是固液常温型反应，应选用的发生装置是B；用E装置收集纯净的NO气体，而不用C或D装置是因为一氧化氮能与氧气反应且密度与空气接近．故填：B；一氧化氮能与氧气反应且密度与空气接近；<br />（5）2NO+O<SUB>2</SUB>═2NO<SUB>2 </SUB><br />&nbsp;&nbsp;&nbsp;&nbsp; 2&nbsp;&nbsp;&nbsp; 1&nbsp;&nbsp; 2&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<br />2NaOH+2NO<SUB>2</SUB>=NaNO<SUB>2</SUB>+NaNO<SUB>3</SUB>+H<SUB>2</SUB>O<br />减少的气体的体积为：60mL+20mL-67.4mL=12.6mL<br />氧气的体积占减少的气体体积的三分之一，故氧气的体积为：12.6mL×<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">1</td></tr><tr><td>3</td></tr></table></span>=4.2mL；<br />由于氢氧化钠具有极强的腐蚀性，因此在实验中要注意氢氧化钠对皮肤的腐蚀性；<br />故填：4.2；氢氧化钠溶液对皮肤具有腐蚀性；<br />（6）60mL一氧化氮的质量为：60×10<SUP>-3</SUP>L×1.27g/L=0.08g<br />设至少需要亚硝酸钠的质量为x<br />2FeSO<SUB>4</SUB>+3H<SUB>2</SUB>SO<SUB>4</SUB>+2NaNO<SUB>2</SUB>═Fe<SUB>2</SUB>（SO<SUB>4</SUB>）<SUB>3</SUB>+2NaHSO<SUB>4</SUB>+2H<SUB>2</SUB>O+2NO↑<br />&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 138&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 60<br />&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; x&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 0.08g<br /><span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">138</td></tr><tr><td>60</td></tr></table>=<table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">x</td></tr><tr><td>0.08g</td></tr></table></span>&nbsp;&nbsp; x=0.18g<br />答：至少需要固体亚硝酸钠的质量为0.18g．','【分析】据图即可知道有关仪器的名称；根据反应的化学方程式的书写方法书写反应的化学方程式；根据气体的性质确定收集气体的方法；根据测定氧气含量的方法以及氢氧化钠的腐蚀性解答，根据反应的化学方程式结合数据计算即可．','书写',3.00,'081d80daff9f86aac35dd5f5da5528e1',9,400,'测定空气里氧气含量的探究,书写化学方程式、文字表达式、电离方程式,根据化学反应方程式的计算','',2016,'32','2016•莆田模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1844263,'下列属于分解反应的是（　　）','C+2CuO<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>&nbsp;2Cu+CO<SUB>2</SUB>↑','2Mg+O<SUB>2</SUB>&nbsp;<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>&nbsp;2MgO','Ba（OH）<SUB>2</SUB>+2HCl═BaCl<SUB>2</SUB>+2H<SUB>2</SUB>O','NH<SUB>4</SUB>Cl&nbsp;<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;加热&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>&nbsp;NH<SUB>3</SUB>↑+HCl↑','','D','【解答】解：A、C+2CuO<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span> 2Cu+CO<SUB>2</SUB>↑，该反应是一种单质和一种化合物反应生成另一种单质和另一种化合物的反应，属于置换反应，故选项错误．<br />B、2Mg+O<SUB>2</SUB> <span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span> 2MgO，该反应符合“多变一”的特征，属于化合反应，故选项错误．<br />C、Ba（OH）<SUB>2</SUB>+2HCl═BaCl<SUB>2</SUB>+2H<SUB>2</SUB>O，该反应是两种化合物相互交换成分生成两种新的化合物的反应，属于复分解反应，故选项错误．<br />D、NH<SUB>4</SUB>Cl <span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;加热&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span> NH<SUB>3</SUB>↑+HCl↑，该反应符合“一变多”的特征，属于分解反应，故选项正确．<br />故选：D．','【分析】分解反应：一种物质反应后生成两种或两种以上的物质，其特点可总结为“一变多”；据此进行分析判断．','选择题',3.00,'35b5c7cb7db1185ff2ad534846435f0c',9,400,'分解反应及其应用','冷水江市',2015,'32','2015•冷水江市校级模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1844271,'如图所示装置有多种用途，下列说法错误的是（　　）<br /><img  src=\"/tikuimages/9/0/400/shoutiniao53/4b3f066e-94d4-11e9-b7ef-b42e9921e93e_xkb64.png\" style=\"vertical-align:middle\" />','洗气：除去CO中的HCl，装置内应盛H<SUB>2</SUB>O','检验：证明CO中混有CO<SUB>2</SUB>，装置内应盛澄清石灰水','集气：若用排空气法收集比空气轻的气体时，气体应从装置的A端通入','若要测定实验室制得的少量0<SUB>2</SUB>的体积，则应在瓶内先装满水，再将O<SUB>2</SUB>从B端通入','','C','【解答】解：（1）HCl易溶于H<SUB>2</SUB>O，故除去CO中的HCl，装置内应盛H<SUB>2</SUB>O；正确；<br />（2）二氧化碳能使澄清石灰水变浑浊，故证明CO中混有CO<SUB>2</SUB>，装置内应盛澄清石灰水，正确；<br />（3）该装置收集密度比空气密度小的气体时应短进长出，故错误；<br />（4）用该装置测量气体体积就是将装置中装满的水全部排出该装置需要短进长出；<br />故答案：C','【分析】（1）根据HCl易溶于H<SUB>2</SUB>O分析解答；<br />（2）根据二氧化碳与澄清石灰水的反应分析；<br />（3）根据利用该装置收集密度比空气密度小的气体时要短进长出分析；<br />（4）根据用该装置测量气体体积就是将装置中装满的水全部排出该装置需要短进长出解答；','选择题',3.00,'dbf100ed7e9b8e5f0ffb95f274e3f97d',9,400,'分离物质的仪器,常见气体的检验与除杂方法,二氧化碳的检验和验满','',0,'37','',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1844311,'<img  src=\"/tikuimages/9/2016/400/shoutiniao91/4bbf32f0-94d4-11e9-9fb1-b42e9921e93e_xkb35.png\" style=\"vertical-align:middle;FLOAT:right\" />（2016•马鞍山二模）某化学兴趣小组的同学设计了如图所示实验装置，进行气体的制取与收集实验．<br />（1）图中①的仪器名称是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）装置中有一处设计明显错误，请改正：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）用改正后的装置进行如下实验：李军同学想利用该装置制取和收集干燥纯净的氧气，写出甲中发生反应的化学方程式：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，该反应的基本反应类型是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，乙中应盛放的试剂是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．收集气体所采用的方法是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（4）工业上制氧气是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“物理”或“化学”）变化．','','','','','','集气瓶$###$乙中应将进气管插入溶液中，瓶内出气导管刚好露出橡皮塞$###$2H<SUB>2</SUB>O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span>&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O+O<SUB>2</SUB>↑$###$分解反应$###$浓硫酸$###$向上排空气法$###$物理','【解答】解：（1）装置有一处设计明显错误，改正为：乙装置中应该进气管长，出气管短或左长右短；故答案为：乙装置中应该进气管长，出气管短（或左长右短）；<br />（2）如果用双氧水和二氧化锰制氧气就不需要加热，过氧化氢在二氧化锰做催化剂的条件下生成水和氧气，要注意配平；反应的方程式为：2H<SUB>2</SUB>O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span>&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O+O<SUB>2</SUB>↑，此反应为分解反应；氧气可以用浓硫酸干燥；丙所用的是向上排空气法；<br />（3）工业上制氧气的方法：在低温条件下加压，使空气转变为液态空气，然后蒸发．由于液态氮的沸点比液态氧的沸点低，因此氮气首先从液态空气中蒸发出来，剩下的主要是液态氧，在整个过程中没有生成其它物质，只是状态的变化，属于物理变化．<br />故答案为：①浓硫酸；将带火星的木条平放在集气瓶口，木条复燃，证明氧气满了<br />故答案为：（1）集气瓶；（2）乙中应将进气管插入溶液中，瓶内出气导管刚好露出橡皮塞；<br />（3）2H<SUB>2</SUB>O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span>&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O+O<SUB>2</SUB>↑、分解反应、浓硫酸、向上排气法；<br />（4）物理','【分析】装置有一处设计明显错误，改正为：乙装置中应该进气管长，出气管短或左长右短；制取装置包括加热和不需加热两种，如果用双氧水和二氧化锰制氧气就不需要加热，氧气的密度比空气的密度大，不易溶于水，因此能用向上排空气法和排水法收集；氧气可以用浓硫酸干燥．根据物理变化和化学变化的本质区别进行分析．','书写',3.00,'c885bfd5691e07c19d2c89441fb8fefa',9,400,'氧气的工业制法,氧气的制取装置,氧气的收集方法,反应类型的判定,书写化学方程式、文字表达式、电离方程式','',2016,'37','2016•马鞍山二模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1844313,'生活中的食醋（含醋酸）除作调味剂外，还有很多用途．下列不属于食醋用途的是（　　）','鉴别黄金饰品中是否含有铜','除去菜刀上的铁锈','清洗溅到皮肤上的碱液','除去水壶壁上的水垢','','A','【解答】解：A．铜的金属活动性比氢弱，不能与醋酸反应，故选项实验不能达到目的．<br />B．铁锈的主要成分是氧化铁，能与酸反应，使用食醋可除去菜刀上的铁锈，故选项实验能达到目的．<br />C．酸与碱能发生中和反应，使用食醋清洗洒落在皮肤上的碱液，故选项实验能达到目的．<br />D．水垢（主要含CaCO<SUB>3</SUB>）能与醋酸反应生成易溶于水的醋酸钙，可除去热水瓶中的水垢，故选项实验能达到目的．<br />故选A．','【分析】食醋中含有醋酸，醋酸属于酸，酸能与活泼金属、金属氧化物、碱、碳酸盐等反应，据此进行分析判断．','选择题',3.00,'c891d7600fdfd80acb423281c6994d7c',9,400,'醋酸的性质及醋酸的含量测定','',2016,'37','2016•苏州一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1844317,'以下说法错误的是（　　）','空气中氧气约占<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">1</td></tr><tr><td>5</td></tr></table></span>体积','氧气能使带火星的木条复燃','水中的生物能依靠微溶于水中的氧气而生存','氧气能支持燃烧，可作燃料','','D','【解答】解：A、空气中氧气约占<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">1</td></tr><tr><td>5</td></tr></table></span>体积，故A说法正确；<br />B、氧气有助燃性，能使带火星的木条复燃，故B说法正确；<br />C、氧气能供给呼吸，水中的生物能依靠微溶于水中的氧气而生存，故C说法正确；<br />D、氧气能支持燃烧，可作助燃剂，不具有可燃性，不可助燃料，故D说法错误．<br />故选D．','【分析】根据空气的成分及各成分的性质及其应用分析判断有关的说法．','选择题',2.00,'1837dcd1ed191e51d386fca3172db5ff',9,400,'空气的成分及各成分的体积分数,氧气的物理性质,氧气的化学性质','',2016,'37','2016•西区一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1844380,'构建安全体系，建设小康社会，关系国计民生．<br />（1）据报道，按我国《食品添加剂使用卫生标准》在豆芽上喷洒适量“速长王”即&nbsp;4-氯苯氧乙酸钠（化学式：C<SUB>8</SUB>H<SUB>6</SUB>O<SUB>3</SUB>ClNa）并不会对人体造成任何危害．在4-氯苯氧乙酸钠中，所含非金属元素共有<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>种．<br />（2）在日常生活中，隔夜的剩菜中亚硝酸钠（NaNO<SUB>2</SUB>）的含量会增加，地沟油、霉变花生中都含有黄曲霉素（C<SUB>17</SUB>H<SUB>12</SUB>O<SUB>6</SUB>），而烧烤中苯并芘（C<SUB>20</SUB>H<SUB>12</SUB>）的含量也很高，常食用上述物质会危害身体健康．在黄曲霉素中，氢元素与碳元素的原子个数之比为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）汽车尾气、工业尾气中的二氧化氮是一种气体污染物，在一定条件下氨气和二氧化氮反应能转化为无害的N<SUB>2</SUB>和H<SUB>2</SUB>O，试写出上述反应的化学方程式：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（4）公共场合禁止吸烟已经成为共识．烟雾中含有一氧化碳、尼古丁（C<SUB>10</SUB>H<SUB>14</SUB>N<SUB>2</SUB>）、焦油、汞、铅等，都会影响吸烟者的健康，也给被动吸烟的人带来危害．下列有关说法中，正确的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填写序号A、B、C、D之一）<br />①吸烟有害身体健康&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;②焦油能诱发细胞病变&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;③在尼古丁中氮元素的含量最高<br />④煤气中毒是由一氧化碳引起的&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;⑤汞和铅属于人体中的常量元素<br />A．①④⑤B．①②③C．①②④D．②③⑤','','','','','','4$###$12：17$###$6NO<SUB>2</SUB>+8NH<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;一定条件&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>7N<SUB>2</SUB>+12H<SUB>2</SUB>O$###$C','【解答】解：（1）在4-氯苯氧乙酸钠（C<SUB>8</SUB>H<SUB>6</SUB>O<SUB>3</SUB>ClNa）中，所含非金属元素有碳、氢、氧、氮共有4种；<br />（2）原子个数比等于化学式的角标之比，所以在黄曲霉素中，氢元素与碳元素的原子个数之比为12：17；<br />（3）二氧化氮和氨气在一定条件下反应生成水和氮气，化学方程式为：6NO<SUB>2</SUB>+8NH<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;一定条件&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>7N<SUB>2</SUB>+12H<SUB>2</SUB>O；<br />（4）①吸烟有害身体健康，故正确；<br />②焦油能诱发细胞病变，故正确；<br />③在尼古丁中碳元素元素的含量最高，故C错误；<br />④煤气中毒是由一氧化碳引起的，故正确；<br />⑤汞和铅属于人重金属元素，故错误．<br />故选：C．<br />故答案为：（1）4；<br />（2）12：17；<br />（3）6NO<SUB>2</SUB>+8NH<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;一定条件&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>7N<SUB>2</SUB>+12H<SUB>2</SUB>O；<br />（4）C．','【分析】（1）根据物质的化学式分析物质的元素组成；<br />（2）根据原子个数比等于化学式的角标之比进行分析；<br />（3）根据二氧化氮和氨气在一定条件下反应生成水和氮气进行分析；<br />（4）①根据吸烟有害身体健康进行分析；<br />②根据焦油能诱发细胞病变进行分析；<br />③根据在尼古丁中各元素的含量进行分析；<br />④根据煤气中毒是由一氧化碳引起的进行分析；<br />⑤根据汞和铅重金属元素进行分析．','书写',3.00,'4c7807272cd3e931039041a5e9d57511',9,400,'元素的简单分类,化学式的书写及意义,元素的质量分数计算,书写化学方程式、文字表达式、电离方程式,烟的危害性及防治','',2016,'37','2016•高新区一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1844402,'<img  src=\"/tikuimages/9/2015/400/shoutiniao50/4cc53140-94d4-11e9-a27e-b42e9921e93e_xkb23.png\" style=\"vertical-align:middle;FLOAT:right;\" />（2015秋•临汾校级月考）A、B、C、D中均含有一种相同的元素，常温下，A为无色液体，生活中最常见的金属与A、B等物质作用，发生一系列复杂的化学反应，生成一种红棕色的固体物质C，C-D用于冶金工业，它们之间有如图所示的<br />（1）写出物质B的一种用途<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）物质C与D反应的化学方程式是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）写出图中的一个化合反应<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（4）写出（3）中反应的微观实质<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','支持燃烧$###$3CO+Fe<SUB>2</SUB>O<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2Fe+3CO<SUB>2</SUB>$###$2H<SUB>2</SUB>+O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O$###$分子的破裂，原子的重新组合','【解答】解：根据“A、B、C、D中均含有一种相同的元素，常温下，A为无色液体”，根据“生活中最常见的金属与A、B等物质作用，发生一系列复杂的化学反应，生成一种红棕色的固体物质C”，A和B会相互转化，B会生成C，所以A是水，B是氧气，C是氧化铁，D是一氧化碳，代入检验，符合题意．<br />，C与D发生的反应是一氧化碳和氧化铜反应生成铜和二氧化碳，该反应反应不是置换反应；<br />（1）根据分析，物质B为氧气，支持燃烧或供给呼吸；<br />（2）C-D是氧化铁和一氧化碳在高温的条件下生成铁和二氧化碳，反应的化学方程式为：3CO+Fe<SUB>2</SUB>O<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2Fe+3CO<SUB>2</SUB>．<br />（3）图中的一个化合反应为2H<SUB>2</SUB>+O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O或2CO+O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2CO<SUB>2</SUB>．<br />（4）（3）中反应的微观实质：分子的破裂，原子的重新组合．<br />故答案为：<br />（1）支持燃烧．（2）3CO+Fe<SUB>2</SUB>O<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2Fe+3CO<SUB>2</SUB>．<br />（3）2H<SUB>2</SUB>+O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O．（4）分子的破裂，原子的重新组合．','【分析】根据“A、B、C、D中均含有一种相同的元素，常温下，A为无色液体”，根据“生活中最常见的金属与A、B等物质作用，发生一系列复杂的化学反应，生成一种红棕色的固体物质C”，A和B会相互转化，B会生成C，所以A是水，B是氧气，C是氧化铁，D是一氧化碳，然后将推出的物质进行验证即可．','书写',3.00,'b4d6f1f87538623f02ef696977c3987b',9,400,'物质的鉴别、推断,化合反应及其应用,书写化学方程式、文字表达式、电离方程式','',2015,'37','2015秋•临汾校级月考',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1844427,'下列关于资源和能源的叙述错误的是（　　）','煤、石油、沼气、风能等都是可再生能源','氢气燃烧后不污染空气，是理想的清洁、高能燃料','在金属资源中，铝元素是地壳中含量最多的金属元素','维护我国海洋权益，也是保护化石燃料和金属资源的表现','','A','【解答】解：A．煤、石油属于不可再生能源，故错误；<br />B．氢气燃烧后不污染空气，是理想的清洁、高能燃料，故正确；<br />C．在金属资源中，铝元素是地壳中含量最多的金属元素，故正确；<br />D．海洋资源中也包括丰富的矿物资源，有常规的化石燃料、新型的矿产资源等，因此维护我国海洋权益，也是保护化石燃料和金属资源的表现，故正确．<br />故选A．','【分析】A．根据能源的分类进行分析；<br />B．根据氢气的特点进行分析；<br />C．根据地壳中金属资源的含量进行分析；<br />D．根据海洋资源中也包括丰富的矿物资源分析．','选择题',3.00,'bfa23c650d5bffb662e9792b46aef6d8',9,400,'地壳中元素的分布与含量,海洋中的资源,氢气的用途和氢能的优缺点,常见能源的种类、能源的分类','',2016,'32','2016•沙湾区模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1844463,'<img  src=\"/tikuimages/9/2015/400/shoutiniao70/4d75204f-94d4-11e9-9010-b42e9921e93e_xkb48.png\" style=\"vertical-align:middle;FLOAT:right\" />在集气瓶中有气体，胶头滴管中有液体，如果把液体滴入瓶中，气球会逐渐胀大，则气体与液体分别可能是（　　）','O<SUB>2</SUB>、稀盐酸','二氧化碳气体、澄清石灰水','CO、NaOH溶液','氢气、水','','B','【解答】解：A、氧气与稀盐酸不反应，压强不变，则气球不变，故A错误；<br />B、二氧化碳能与澄清石灰水反应而使瓶内额压强变小，气球变大，故B正确；<br />C、一氧化碳与氢氧化钠溶液不反应，压强不变，则气球不变，故C错误；<br />D、氢气与水不反应，也难溶于水，压强不变，则气球不变，故D错误；<br />故选：B．','【分析】根据已有的知识进行分析，气球胀大则是因为瓶内的压强减小，可能是物质接触消耗瓶内的气体，也可能是温度降低导致气体遇冷收缩．','选择题',3.00,'3b678c5932b4b20ec733263655637eb8',9,400,'二氧化碳的化学性质,酸的化学性质,反应现象和本质的联系,氢气的物理性质','',2015,'37','2015秋•永年县月考',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1844470,'学习和研究化学，有许多重要方法．下列方法中所举示例错误的是（　　）<br /><table class=\"edittable\"><TBODY><TR><td width=51>选项</TD><td width=80>方法</TD><td width=439>示例</TD></TR><TR><td>A</TD><td>实验法</TD><td>用新配制的氢氧化铜悬浊液检验葡萄糖</TD></TR><TR><td>B</TD><td>归纳法</TD><td>水、二氧化碳是分子构成的，归纳出所有物质都是由分子构成的</TD></TR><TR><td>C</TD><td>分类法</TD><td>根据组成物质的元素种类，将纯净物分为单质和化合物</TD></TR><TR><td>D</TD><td>对比法</TD><td>用等体积的水和空气进行压缩，得出气体与液体微粒间的空隙不同</TD></TR></TBODY></TABLE>','A','B','C','D','','B','【解答】解：A、含有醛基的物质与新制的氢氧化铜悬浊液共热时，能产生红色沉淀；故A正确；<br />B、铁是由铁原子构成的，氯化钠是由氯离子和钠离子构成的，故B错误；<br />C、根据组成纯净物的元素的种类，有同一种元素组成的纯净物，属于单质，由两种或两种以上的元素组成的纯净物，属于化合物，故C正确；<br />D、用等体积的水和空气进行压缩，空气比水容易被压缩，得出气体的间隔大，液体的间隔小，故D正确．<br />故选：B．','【分析】A、根据含有醛基的物质与新制的氢氧化铜悬浊液共热时产生红色沉淀分析；<br />B、根据构成物质的微粒进行分析；<br />C、根据纯净物的分类分析；<br />D、根据分子的性质进行分析．','选择题',3.00,'9599d15a12828a9fa5092ae6a2e5ebaa',9,400,'科学探究的基本方法','',2016,'32','2016•泰州模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1844514,'阅读下面科普短文．<br />19世纪初，铝是比黄金还贵的金属，主要是由于铝的化学性质很活泼，在自然界以化合物形式存在．当时采用钠和氯化铝在一定条件下发生置换反应得到铝，因生产成本高，所以铝十分珍贵．<br />19世纪末，霍尔应用电解熔融金属化合物的方法制备金属．他在用氧化铝制备金属铝的过程中，发现氧化铝的熔点很高（2050℃），很难达到熔融状态，必须物色一种能够溶解氧化铝而又能降低其熔点的材料，实验过程中发现冰晶石（Na<SUB>3</SUB>AlF<SUB>6</SUB>）能起到这种作用．<br />霍尔在坩埚中，把氧化铝溶解在10%～15%的熔融的冰晶石里进行电解，发现有小球状银白色的铝生成，冰晶石在电解过程中不被分解，并有足够的流动性，有利于电解的进行．这种廉价炼铝方法的发现，使铝成为广泛应用的金属材料，至今仍在使用．<br />根据文章内容，回答下列问题：<br />（1）金属铝的物理性质<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）冰晶石中氟元素的化合价为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）用金属钠制取铝的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（4）霍尔用电解法制取铝的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />（5）电解法制铝的过程中，冰晶石的作用是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','熔点高$###$-1$###$3Na+AlCl<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>3NaCl+Al$###$2Al<SUB>2</SUB>O<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;通电&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>4Al+3O<SUB>2</SUB>↑$###$催化','【解答】【解答】解：（1）由题意可知，铝的熔点高，是不需要通过化学变化表现出来的性质，属于物理性质；<br />（2）在Na<SUB>3</SUB>AlF<SUB>6</SUB>中，钠、铝元素的化合价分别是+1价和+3价，根据在化合物中正负化合价的代数和为零的原则，可设氟元素的化合价为x，则+1×3+（+3）+x×6=0，则氟元素的化合价为-1价；<br />（3）钠与氯化铝在一定条件下发生置换反应，反应的方程式是：3Na+AlCl<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>3NaCl+Al；<br />（4）电解氧化铝的方法制取单质铝的化学方程式为：2Al<SUB>2</SUB>O<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;通电&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>4Al+3O<SUB>2</SUB>↑；<br />（5）根据信息：冰晶石在电解过程中不被分解，并有足够的流动性，有利于电解的进行．可知冰晶在电解法制铝的过程中起到催化作用．<br />故答案为：（1）熔点高；（2）-1；（3）3Na+AlCl<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>3NaCl+Al；（4）2Al<SUB>2</SUB>O<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;通电&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>4Al+3O<SUB>2</SUB>↑；<br />（5）催化．','【分析】（1）物质的物理性质是指不需要通过化学变化表现出来的性质，物质的物理性质有：颜色、状态、气味、密度、熔点、沸点、溶解性等方面．<br />（2）根据化合价的规律，计算氟元素的化合价．<br />（3）根据钠与氯化铝在一定条件下发生置换反应得到铝写出化学方程式；<br />（4）根据信息：把氧化铝溶解在10%～15%的熔融的冰晶石里进行电解，发现有小球状银白色的铝生成，写出化学方程式；<br />（5）根据信息：冰晶石在电解过程中不被分解，并有足够的流动性，有利于电解的进行．分析解答．','书写',3.00,'f5132f1e5053c0a34421dbfaadadff0f',9,400,'金属的化学性质,常见金属的冶炼方法,有关元素化合价的计算,书写化学方程式、文字表达式、电离方程式','',2015,'35','2015秋•朝阳区校级期中',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1844548,'<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellspacing=\"0\" cellpadding=\"0\"><tr><td>&nbsp;</td><td style=\"line-height: normal; padding-left: 1px; font-size: 90%\"><div mathtag=\"msubsup_sup\" hassize=\"-1\">235</div><div mathtag=\"msubsup_sub\">92</div></td></tr></table></span>U是重要的核工业原料，已知<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellspacing=\"0\" cellpadding=\"0\"><tr><td>&nbsp;</td><td style=\"line-height: normal; padding-left: 1px; font-size: 90%\"><div mathtag=\"msubsup_sup\" hassize=\"-1\">235</div><div mathtag=\"msubsup_sub\">92</div></td></tr></table></span>U中质子数为92，中子数为143，则<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellspacing=\"0\" cellpadding=\"0\"><tr><td>&nbsp;</td><td style=\"line-height: normal; padding-left: 1px; font-size: 90%\"><div mathtag=\"msubsup_sup\" hassize=\"-1\">235</div><div mathtag=\"msubsup_sub\">92</div></td></tr></table></span>U的核外电子数是（　　）','143','92','235','51','','B','【解答】解：因为原子中：核电荷数=核内质子数=核外电子数，已知<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellspacing=\"0\" cellpadding=\"0\"><tr><td>&nbsp;</td><td style=\"line-height: normal; padding-left: 1px; font-size: 90%\"><div mathtag=\"msubsup_sup\" hassize=\"-1\">235</div><div mathtag=\"msubsup_sub\">92</div></td></tr></table></span>U原子核内有92个质子，根据原子中，质子数=核外电子数可知，<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellspacing=\"0\" cellpadding=\"0\"><tr><td>&nbsp;</td><td style=\"line-height: normal; padding-left: 1px; font-size: 90%\"><div mathtag=\"msubsup_sup\" hassize=\"-1\">235</div><div mathtag=\"msubsup_sub\">92</div></td></tr></table></span>U的核外电子数为92．<br />故选：B．','【分析】根据原子中：核电荷数=核内质子数=核外电子数，结合题意进行分析解答．','选择题',3.00,'11fa6ff5593894be74088444e4902ce6',9,400,'原子的有关数量计算','',2016,'37','2016•扬州校级一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1844575,'钙是人体必需的常量元素，每日要摄取足量的钙，才能避免佝倭病或骨质疏松，因此有时需要服用补钙扪满足人体需求．如表是一种补钙剂说明书的一部分．<br /><table class=\"edittable\"><TBODY><TR><td width=569>金钙尔奇<br />主要原料：碳酸钙（C&nbsp;a&nbsp;C0<SUB>3</SUB>）、氧化镁、硫酸锌、硫酸铜、维生素等<br />含量：每片含：钙2&nbsp;8&nbsp;0m&nbsp;g&nbsp;镁99.7mg锌3.08mg&nbsp;铜0.51mg等<br />用量：每日2次，一次1片</TD></TR></TBODY></TABLE>根据说明书中的信息回答下列问题．<br />（1）说明书中的钙、镁、锌＞铜指的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />A、分子B、原子C、元素D、单质<br />（2）碳酸钙中钙元素的质量分数为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）每片金钙尔奇中含碳酸钙的质量为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>mg．<br />（4）市场上另一种补钙产品葡萄糖酸钙含片，每片含葡萄糖酸钙（C<SUB>12</SUB>H<SUB>24</SUB>O<SUB>15</SUB>Ca）0.5g．若按照金钙尔奇说明书每日的补钙量，服用葡萄糖酸钙含片，则应一日3次，每次应服用<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>片．','','','','','','C$###$40%$###$700$###$4','【解答】解：（1）物质是由元素组成的，这里的钙、镁、锌、铜指的是元素，观察选项，故选C；<br />（2）碳酸钙中钙元素的质量分数为：<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">40</td></tr><tr><td>40+12+16×3</td></tr></table>×100%=40%</span>；故填：40%；<br />（3）每片金钙尔奇中含碳酸钙的质量为：280mg÷40%=700mg；故填：700；<br />（4）每天服用的金钙尔奇中含钙元素的质量为280mg×2=560mg；每片葡萄糖酸钙片中，钙元素的质量为：0.5g×<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">40</td></tr><tr><td style=\"padding-top:1px;font-size:90%\">40+(12×6+1×11+16×7)×2</td></tr></table>×100%</span>≈47mg．<br />则服用葡萄糖酸钙片，一日3次，每次：560mg÷47mg÷3≈4．故填：4．','【分析】（1）物质是由元素组成，据此解答；<br />（2）根据化合物中元素的质量分数进行分析解答；<br />（3）根据一定质量的化合物中某元素的质量计算方法来分析；<br />（4）根据钙元素的质量相等来分析．','填空题',3.00,'b9c218af69afca4224423d3b59fe77a1',9,400,'元素的概念,元素的质量分数计算,化合物中某元素的质量计算,标签上标示的物质成分及其含量','',2015,'33','2015春•渠县校级期末',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1844599,'<img  src=\"/tikuimages/9/2016/400/shoutiniao53/4f08918f-94d4-11e9-822d-b42e9921e93e_xkb37.png\" style=\"vertical-align:middle;FLOAT:right\" />如图是地壳中元素含量分布示意图，能表示铝元素含量的是（　　）','1','2','3','4','','C','【解答】解：地壳中各元素的含量由多到少的顺序（前四位）排列依次是氧、硅、铝、铁，从图中看，3区域代表的元素含量在地壳中占第三位，故图中3区域代表的元素是铝．<br />故选C．','【分析】根据地壳里所含各种元素的含量（质量分数），结合图表进行分析解答．','选择题',3.00,'a3ac7aaec457192fb04840cecb012696',9,400,'地壳中元素的分布与含量','',2016,'37','2016•朝阳区一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1844634,'碳化硅陶瓷基复合材料是一种新型热结构材料．它比铝还轻，比钢还强，比碳化硅陶瓷更耐高温、更抗氧化烧蚀，而且克服了陶瓷的脆性．<br />（1）在这种新型材料的上述性能中，涉及的物理性质有<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；涉及的化学性质有<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．（各填一个）<br />（2）在空气中，碳化硅能与熔融的氢氧化钠反应：SiC+2NaOH+2O<SUB>2</SUB>═Na<SUB>2</SUB>SiO<SUB>3</SUB>+X+H<SUB>2</SUB>O，其中X的化学式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，Na<SUB>2</SUB>SiO<SUB>3</SUB>中硅元素的化合价为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','密度小（或硬度大、熔点高）$###$抗氧化性（或热稳定性）$###$CO<SUB>2</SUB>$###$+4','【解答】解：（1）由题意“它比铝还轻，比钢还强，比碳化硅陶瓷更耐高温、更抗氧化烧蚀”，可知这种新型材料的上述性能中，涉及的物理性质有密度小（或硬度大、熔点高）；涉及的化学性质有具有抗氧化性（或具有热稳定性）；<br />（2）在空气中，碳化硅能与熔融的氢氧化钠反应：SiC+2NaOH+2O<SUB>2</SUB>═Na<SUB>2</SUB>SiO<SUB>3</SUB>+X+H<SUB>2</SUB>O，根据质量守恒定律，反应前后元素种类不变，原子个数相同，则可计算其中X的化学式为CO<SUB>2</SUB>；已知钠元素的化合价为+1价，氧元素的化合价为-2价，根据化合价的原则，则可求Na<SUB>2</SUB>SiO<SUB>3</SUB>中硅元素的化合价为+4．<br />故答案为：（1）密度小（或硬度大、熔点高），抗氧化性（或热稳定性）；<br />（2）CO<SUB>2</SUB>，+4．','【分析】（1）根据题意“它比铝还轻，比钢还强，比碳化硅陶瓷更耐高温”，可知这种新型材料涉及的物理性质；由“比碳化硅陶瓷更耐高温、更抗氧化烧蚀”，可知涉及的化学性质；<br />（2）在空气中，碳化硅能与熔融的氢氧化钠反应：SiC+2NaOH+2O<SUB>2</SUB>═Na<SUB>2</SUB>SiO<SUB>3</SUB>+X+H<SUB>2</SUB>O，根据质量守恒定律，可求其中X的化学式；已知钠元素的化合价为+1价，氧元素的化合价为-2价，根据化合价的原则，则可求Na<SUB>2</SUB>SiO<SUB>3</SUB>中硅元素的化合价．','填空题',3.00,'966c3e75b2fe456d2dcf93df8c4c444f',9,400,'有关元素化合价的计算,化学性质与物理性质的差别及应用,质量守恒定律及其应用,新材料的开发与社会发展的密切关系','',0,'37','',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1844694,'2015年8月12日晚11时许，天津滨海新区第五大街与跃进路交叉口的一处集装箱码头发生爆炸．在仓库内的危险品主要是三大类，一是具氧化性物质，是硝酸铵、硝酸钾；二是易燃物体，主要是金属钠和金属镁；三是剧毒物质，以氰化钠为主．<br />（1）题中涉及到的物质中可做复合肥的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填化学式）．<br />（2）金属钠燃烧不能用水灭火，因为金属钠会跟水反应生成氢氧化钠和氢气，请写出该反应的化学方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（3）爆炸是可燃物在有限的空间燃烧造成的，关于燃烧的有关说法错误的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填序号）．<br />①可燃物温度达到着火点，就会发生燃烧&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;②燃烧一定是化学变化<br />③沙土扑灭金属燃烧的原理是使可燃物隔绝空气&nbsp;&nbsp;&nbsp;④燃烧一定会引起爆炸．','','','','','','KNO<SUB>3</SUB>$###$2Na+2H<SUB>2</SUB>O=2NaOH+H<SUB>2</SUB>↑$###$①④','【解答】解：（1）硝酸钾中含有农作物需求量较大的氮元素与钾元素，属于复合肥，故填：KNO<SUB>3</SUB>；<br />（2）金属钠与水反应生成氢氧化钠和氢气，化学方程式为2Na+2H<SUB>2</SUB>O=2NaOH+H<SUB>2</SUB>↑；故填：2Na+2H<SUB>2</SUB>O=2NaOH+H<SUB>2</SUB>↑；<br />（3）可燃物的温度达到着火点，但是若不与氧气接触就不会发生燃烧；燃烧一定属于化学变化；沙土覆盖在可燃物的表面隔绝了氧气，达到了灭火的目的；只有急速的燃烧发生在有限的空间内，热量来不及散失，才可能会引发爆炸．<br />故填：①④．','【分析】（1）根据复合肥的概念来分析；<br />（2）根据反应物、生成物和反应条件及质量守恒定律的原则，写出正确的化学方程式；<br />（3）燃烧通常是指可燃物与氧气发生的发光、放热的剧烈的氧化反应；<br />发生燃烧必须具备三个基本条件：要有可燃物；（2）要有氧气；（3）可燃物的温度必需达到其自身的着火点．','书写',3.00,'82fd91a75d0dfb067ba6ef7025e49425',9,400,'常见化肥的种类和作用,书写化学方程式、文字表达式、电离方程式,灭火的原理和方法,燃烧、爆炸、缓慢氧化与自燃,易燃物和易爆物安全知识','',2016,'32','2016•新昌县模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1844759,'CO<SUB>2</SUB>和O<SUB>2</SUB>是自然界中不可缺少的两种气体，它们的相同点是（　　）','都是无色无味','都能供给呼吸','都能灭火','都能支持燃烧','','A','【解答】解：A、CO<SUB>2</SUB>和O<SUB>2</SUB>都是无色无味气体，故A正确；<br />B、O<SUB>2</SUB>能供给呼吸，CO<SUB>2</SUB>不能供给呼吸，故B错误；<br />C、CO<SUB>2</SUB>不燃烧不支持燃烧，密度比空气大，能用于灭火，O<SUB>2</SUB>有助燃性，不能用于灭火，故C错误；<br />D、O<SUB>2</SUB>有助燃性，能支持燃烧，CO<SUB>2</SUB>不燃烧不支持燃烧，密度比空气大，可用于灭火，故D错误．<br />故选A．','【分析】根据氧气的二氧化碳的性质和用途分析判断有关的说法．','选择题',3.00,'1c699db17ca92160f98f12b40aa21f8d',9,400,'氧气的物理性质,氧气的化学性质,氧气的用途,二氧化碳的化学性质,一氧化碳的物理性质','',2016,'32','2016•双牌县模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1844779,'铝在氧气中燃烧生成三氧化二铝，则参加反应的铝、氧气，生成的三氧化二铝的质量比是（　　）','27：32：102','54：32：102','54：24：102','108：96：204','','D','【解答】解：铝和氧气燃烧的化学方程式为：<br />4Al+3O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2Al<SUB>2</SUB>O<SUB>3</SUB><br />108 &nbsp;96&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 204<br />在此反应中铝、氧气和氧化铝的质量比为108：96：204=9：8：17．<br />故选：D．','【分析】首先正确写出铝和氧气反应的化学方程式，利用各物质之间的质量比等于相对分子质量和的比，进行分析解答即可．','选择题',3.00,'3e1eb76eccde1652c7521a9cd53c72eb',9,400,'常见化学反应中的质量关系','',2016,'35','2016春•淄博期中',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1844790,'下列实验的错误操作与“可能产生的后果”不一致的是（　　）','<img  src=\"/tikuimages/9/2016/400/shoutiniao87/50ee8f9e-94d4-11e9-94a5-b42e9921e93e_xkb69.png\" style=\"vertical-align:middle\" /><br />　　　 标签受损','<img  src=\"/tikuimages/9/2016/400/shoutiniao40/50f3e6cf-94d4-11e9-9772-b42e9921e93e_xkb13.png\" style=\"vertical-align:middle\" /><br />&nbsp;&nbsp;&nbsp;&nbsp;受热仪器破裂','<img  src=\"/tikuimages/9/2016/400/shoutiniao8/50fa014f-94d4-11e9-aaad-b42e9921e93e_xkb67.png\" style=\"vertical-align:middle\" /><br />　　溶液蒸不干','<img  src=\"/tikuimages/9/2016/400/shoutiniao61/50fd35a1-94d4-11e9-be21-b42e9921e93e_xkb91.png\" style=\"vertical-align:middle\" /><br />　　 读数不准确','','C','【解答】解：A、图中操作标签没有向着手心，液体会流出腐蚀标签，图示的“错误操作”与相对应选项的“可能产生后果”一致，故选项不符合题意．<br />B、试管口没有略向下倾斜，冷凝水倒流到热的试管底部使试管炸裂，图示的“错误操作”与相对应选项的“可能产生后果”一致，故选项不符合题意．<br />C、蒸发溶液时，应用玻璃棒进行搅拌，图中无玻璃棒，则液体沸腾，液体溅出，图示“错误操作”与相对应选项“可能产生后果”不一致，故选项符合题意．<br />D、用量筒量取液体时，若俯视读数，则量取的液体的体积偏小，图示“错误操作”与相对应选项“可能产生后果”一致，故选项不符合题意．<br />故选C．','【分析】A、根据液体药品的取用方法进行分析判断．<br />B、根据给试管中的固体加热的注意事项进行分析判断．<br />C、根据蒸发溶液时用玻璃棒进行搅拌进行分析判断．<br />D、根据量筒读数时视线要与凹液面的最低处保持水平进行分析判断．','选择题',3.00,'de6899eef29fc07adf87288a9ca92192',9,400,'测量容器-量筒,液体药品的取用,给试管里的固体加热,蒸发与蒸馏操作','',2016,'32','2016•泗阳县模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1844800,'近日，中外科学家开发出一种新型光催化剂-碳纳米点一氮化碳纳米复合物，它可以利用太阳能高效并完全地分解水，整个过程分为两个阶段：第一步，氮化碳分解水生成过氧化氢和氢气；第二步，碳纳米点将过氧化氢分解成水和氧气．下列有关该过程的说法正确的是（　　）','碳纳米点一氮化碳纳米复合物是一种新型化合物','氮化碳（C<SUB>3</SUB>N<SUB>4</SUB>）中碳、氧质量比为3：4','该新型光催化剂的质量在反应前后发生了变化','以上两个反应前后原子总数都不变','','D','【解答】解：A、碳纳米管是由碳元素组成的碳的单质．不是化合物，故A说法错误；<br />B、氮化碳（C<SUB>3</SUB>N<SUB>4</SUB>）中碳、氧质量比为：12×3：14×4=9：14；故B说法不正确；<br />C、催化剂的质量在反应前后不变化，故C说法不正确；<br />D 化学反应前后原子总数都不变，说法正确；<br />故选D．','【分析】要解本题需掌握：纳米材料的应用，以及碳纳米管的特点．纳米碳管只有碳原子构成，是单质．','选择题',3.00,'936fe95642d5d9ada1ab1b5b4d580b82',9,400,'催化剂的特点与催化作用,单质和化合物的判别,元素质量比的计算,质量守恒定律及其应用,新材料的开发与社会发展的密切关系','',0,'37','',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1844809,'一盆嫩绿的绿色植物密封在塑料袋内，在暗处放置24小时后，将袋中的气体通入澄清的石灰水中，发现<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>现象，说明植物在暗处也进行呼吸作用，属于<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>变化．（填物理或化学）','','','','','','石灰水变浑浊$###$化学','【解答】解：植物在光照下会进行光合作用吸收二氧化碳，而光合作用的进行需要光；一盆嫩绿的绿色植物密封在塑料袋内，在暗处放置24小时后，植物不能进行光合作用，还能进行呼吸作用产生二氧化碳，二氧化碳会使澄清的石灰水变浑浊．因此将袋中的气体通入澄清的石灰水中，石灰水变浑浊，这个现象说明了植物在暗处也进行呼吸作用释放二氧化碳．该反应是物质与氧气发生的反应，属于化学反应中的氧化反应．<br />故答案为：石灰水变浑浊；化学．','【分析】根据植物呼吸作用的原理：植物细胞利用氧，将有机物分解成二氧化碳和水，并且将储存在有机物中的能量释放出来，供给生命活动的需要的过程；进行分析解答．','填空题',3.00,'560db35118f9593f293ea5b23fc99ee8',9,400,'化学变化和物理变化的判别,光合作用与呼吸作用','',2015,'37','2015秋•保定校级月考',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1844834,'科学严谨的态度、安全规范的操作是成功进行化学实验的保障．<br />（1）化学实验安全第一，下列实验要特别注意安全的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填序号）．<br />A．取用大理石&nbsp;&nbsp; B．称量粗盐&nbsp;&nbsp; C．稀释浓硫酸&nbsp;&nbsp; D．研碎胆矾<br />（2）如图是初中化学中的几种常见操作．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao94/51b89661-94d4-11e9-a4ce-b42e9921e93e_xkb64.png\" style=\"vertical-align:middle\" /><br />写出图中标号①的仪器名称：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；上述操作中错误的一项是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；请写出D图中制取氧气的化学方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）根据图1读出液体的体积为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>mL；图2读出精盐质量是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>g．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao92/51bba3a1-94d4-11e9-b69c-b42e9921e93e_xkb79.png\" style=\"vertical-align:middle\" /><br />（4）实验室用图3所示的装置制取和收集某种气体，由装置可推知该气体的物理性质是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，制取气体的反应一定是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填序号）．<br />A．不需要加热&nbsp;&nbsp;&nbsp;&nbsp; B．需要催化剂&nbsp;&nbsp;&nbsp;&nbsp; C．制取氧气的反应&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; D．反应物都是固体．','','','','','','C$###$酒精灯$###$C$###$2H<SUB>2</SUB>O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span>&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O+O<SUB>2</SUB>↑$###$4.7$###$13.2$###$密度比空气小$###$D','【解答】解：（1）稀释浓硫酸时，要把浓硫酸缓缓地沿器壁注入水中，同时用玻璃棒不断搅拌，以使热量及时地扩散；一定不能把水注入浓硫酸中；如果水加入浓硫酸时，水的密度小于浓硫酸的密度，在浓硫酸的上面，浓硫酸溶解放出大热量，使水沸腾，液体就会喷出伤人，因此，稀释浓硫酸时要特别注意安全；<br />故填：C；<br />（2）图中的实验仪器是酒精灯；<br />A、给试管内的液体加热时，液体的体积不能超过了试管体积的1/3，加热时，应是使试管于桌面成45°角，这样可使药品受热面积大，受热均匀；为避免液体沸腾溅出伤人，加热时切不可让试管口对着有人的方向，图中操作正确；<br />B、用细口瓶向试管中倾倒液体时，应先拿下瓶塞，倒放在桌上，然后拿起瓶子，标签要对着手心，瓶口要紧挨着试管口，试管稍倾斜，使液体缓慢地倒入试管，图中操作正确；<br />C、二氧化碳的验满方法是把燃烧的木条放在集气瓶瓶口，若木条立即熄灭，说明装满，图中操作错误．<br />D、双氧水在二氧化锰的催化作用下生成水和氧气，氧气的密度大于空气，所以又可选向上排空气法收集，图中操作正确；<br />过氧化氢在二氧化锰的催化作用下生成水和氧气，反应的化学方程式为：2H<SUB>2</SUB>O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span>&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O+O<SUB>2</SUB>↑；<br />故填：酒精灯；C；2H<SUB>2</SUB>O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span>&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O+O<SUB>2</SUB>↑；<br />（3）量筒度数时，视线要与量筒内凹液面的最低处保持水平；药品的质量为托盘上砝码的质量与横梁上游码质量之和；<br />故填：4.7；13.2；<br />（4）用该装置来制取气体，说明反应物是固体加热制取气体，但不能说明一定使用了催化剂；用向下排空气法来收集，说明该气体的密度比空气小，不会是氧气．<br />故填：密度比空气小；D．','【分析】（1）根据实验存在的危险性来分析；<br />（2）根据常见仪器进行分析；<br />A、根据液体加热的方法判断；<br />B、根据液体药品的取用方法判断；<br />C、根据二氧化碳验满的方法判断；<br />D、根据氧气的制取方法判断；<br />根据过氧化氢在二氧化锰的催化作用下生成水和氧气进行分析；<br />（3）根据量筒和托盘天平的度数方法来分析；<br />（4）根据制取气体的装置与收集气体的方法来分析．','填空题',3.00,'f5cde13581c0b92f9a92bc57bd9f1a5a',9,400,'测量容器-量筒,称量器-托盘天平,固体药品的取用,液体药品的取用,给试管里的液体加热,浓硫酸的性质及浓硫酸的稀释,实验室制取气体的思路,氧气的制取装置,二氧化碳的检验和验满','',2016,'37','2016•临朐县二模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1844867,'熟练掌握化学实验基本技能是顺利完成化学实验的保障，请根据我们己经学过的“药品的取用、物质的加热、物质的称量和仪器的连接与洗涤”等相关知识，完成以下各题．<br />首先，每一项准确、规范的操作背后，都有其原因和道理，想学好化学得注重手脑结合，例如：<br />（1）给试管里的液体，进行较短时间的加热，可以选择酒精灯和试管夹．加热前试管夹要从试管<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（选填“口部”或“底部”）套入，这样操作的目的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；试管夹要夹在试管的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，加热时，手持长柄，不能按着短柄，以防<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）给试管里的固体，进行较长时间的加热，可以选择酒精灯和<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，加热前先将试管<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，然后再固定加热，这样操作的目的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；加热时，试管口要<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，原因是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；实验结束时，要等试管<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，才能用冷水冲洗，试管洗净的标准是，水在试管内壁既不<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，也不成股流下．<br />其次，在刚接触基本操作训练时难免会犯错误，我们要养成分析与推理的习惯，例如：<br />（3）要较准确的称量一定量的固体药品，用到的主要仪器为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，药品和砝码的位置要求是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，要是位置颠倒会出现的影响是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（多选少选均不得分）<br />A．没有任何影响，上述要求只是一种习惯；<br />B．在不拨动游码的时候，位罝颠倒不影响称量结果<SUB>：</SUB><br />C．位置颠倒后，实际所称物质的质量为砝码质量加上游码所处刻度值；<br />D．位置颠倒后，实际所称物质的质量为砝码质量减去游码所处刻度值；<br />（4）要较准确的量取一定量的液体试剂，经常用到量筒和<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，操作时<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>要与液体的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>保持水平，仰视或俯视操作的影响是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />A．测量某液体的体积时，仰视读数会偏大：<br />B．测量某液体的体积时，俯视读数会偏大；<br />C．量取一定量的液体试剂时，仰视操作，所取液体会偏多；<br />D．量取一定量的液体试剂时，俯视操作，所取液体会偏多．<br />再者，交流与合作是学习化学必不可少的环节，帮同伴指出其操作中的错误，并助其改正，也需耍规范的语言．例如：<img  src=\"/tikuimages/9/2015/400/shoutiniao27/526fff80-94d4-11e9-b113-b42e9921e93e_xkb48.png\" style=\"vertical-align:middle;FLOAT:right;\" /><br />指出错误的语言一般为：没有…，或者不应…<br />如图1，错误一：手心没有对着标签；<br />错误二：试管不应直立，等等．<br />改正错误的语言一般为：应该…，要…<br />接图1，改正一：手心应该对着标签；<br />改正二：试管要稍微倾斜．<br />（5）请仿照以上语言格式，指出图1中的另两处错误，并改正．<br />错误三：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；错误四：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />改正三：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；改正四：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（6）过滤是分离混合物常用的方法之一，请仔细观察图2，指出最明显的两处错误：<br />错误一：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />错误一：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />经过初步改正之后，如果利用该装置进行过滤，得到的滤液仍显浑浊，请你推测其原因<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（写出一条即可）．','','','','','','底部$###$防止污物掉入试管$###$中上部$###$试管滑落$###$铁架台$###$均匀受热$###$防止试管炸裂$###$略向下倾斜$###$防止冷凝水倒流炸裂试管$###$冷却$###$聚成水滴$###$托盘天平$###$左物右码$###$BD$###$胶头滴管$###$视线$###$凹液面最低处$###$BC$###$瓶口没有紧挨试管口$###$瓶塞没有倒置桌面$###$瓶口要紧挨试管口$###$瓶塞应该倒置桌面$###$没有用玻璃棒引流$###$漏斗下端管口没有紧靠烧杯内壁$###$滤纸破损','【解答】解：（1）给试管里的液体，进行较短时间的加热，可以选择酒精灯和试管夹．加热前试管夹要从试管底部套入，这样操作的目的是防止污物掉入试管；试管夹要夹在试管的中上部，加热时，手持长柄，不能按着短柄，以防试管滑落；<br />（2）给试管里的固体，进行较长时间的加热，可以选择酒精灯和铁架台，加热前先将试管均匀受热，然后再固定加热，这样操作的目的是防止试管炸裂；加热时，试管口要略向下倾斜，原因是防止冷凝水倒流，炸裂试管；实验结束时，要等试管冷却，才能用冷水冲洗，试管洗净的标准是，水在试管内壁既不，也不成股流下聚成水滴．<br />（3）要较准确的称量一定量的固体药品，用到的主要仪器为托盘天平，药品和砝码的位置要求是左物右码，要是位置颠倒会出现的影响是：在不拨动游码的时候，位罝颠倒不影响称量结果，位置颠倒后，实际所称物质的质量为砝码质量减去游码所处刻度值；<br />（4）要较准确的量取一定量的液体试剂，经常用到量筒和胶头滴管，操作时视线要与液体的凹液面最低处保持水平，仰视或俯视操作的影响是BC；<br />（5）错误三：瓶口没有紧挨试管口；错误四：瓶塞没有倒置桌面；<br />改正三：瓶口要紧挨试管口；改正四：；错误三：瓶塞应该倒置桌面；<br />（6）错误一：没有用玻璃棒引流；<br />错误一：漏斗下端管口没有紧靠烧杯内壁；<br />经过初步改正之后，如果利用该装置进行过滤，得到的滤液仍显浑浊，其原因可能是滤纸破损或漏斗内液面高于滤纸边缘．<br />故答案为：（1）底部；防止污物掉入试管；中上部；试管滑落；&nbsp;（2）铁架台；均匀受热；&nbsp;防止试管炸裂；略向下倾斜；防止冷凝水倒流，炸裂试管；&nbsp;冷却；&nbsp;聚成水滴；&nbsp;（3）托盘天平；左物右码；BD；（4）胶头滴管；视线；凹液面最低处；BC；（5）瓶口没有紧挨试管口；瓶塞没有倒置桌面；瓶口要紧挨试管口；瓶塞应该倒置桌面；（6）没有用玻璃棒引流；漏斗下端管口没有紧靠烧杯内壁；滤纸破损．','【分析】（1）根据给试管里的液体加热时的注意事项去分析解答；<br />（2）根据给试管里的液体加热时的注意事项去分析解答；<br />（3）根据托盘天平的使用方法和注意事项进行分析；<br />（4）根据量液时，量筒必须放平，视线要与量筒内液体的凹液面最低处保持水平去分析解答；<br />（5）根据液体药品的取用方法进行分析；<br />（6）根据过滤的注意事项进行分析．','填空题',3.00,'b74b02fae839bb14d2aad513107bf4a0',9,400,'测量容器-量筒,称量器-托盘天平,液体药品的取用,给试管里的固体加热,给试管里的液体加热,过滤的原理、方法及其应用,玻璃仪器的洗涤','',2015,'35','2015秋•周村区校级期中',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1844869,'联合国为纪念化学所取得的成就以及对人类文明的贡献，将2011年定为“国际化学年”，以提高民众对化学知识的重视．下列有关认识，科学的是（　　）','氧气具有氧化性','“绿色食品”就是绿颜色的食品','稀有气体不能与任何物质反应','太阳能是一种无污染的燃料','','A','【解答】解：<br />A、氧气化学性质活泼，具有氧化性，故正确；<br />B、“绿色食品”不是指绿颜色的食品，是指无污染的食品．故错误；<br />C、稀有气体化学性质稳定，一般不能与物质反应，不是任何物质，故错误；<br />D、太阳是新能源，不是燃料，故错误．<br />答案：A','【分析】A、根据氧气的性质解答；<br />B、根据“绿色食品”是指无污染的食品解答；<br />C、根据稀有气体化学性质稳定解答；<br />D、根据太阳是新能源解答．','选择题',3.00,'d916f4aecb4b6f5c8c145e6056a3aef4',9,400,'绿色化学,氧气的化学性质,资源综合利用和新能源开发','',2011,'37','2011秋•马鞍山月考',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1844870,'能源利用和环境保护是人类共同关注的问题．<br />（1）家用燃料的发展历程如图所示，括号内的物质是该燃料的主要成分．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao71/527c829e-94d4-11e9-a899-b42e9921e93e_xkb57.png\" style=\"vertical-align:middle\" /><br />下列说法不正确的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填字母）．<br />A．煤、石油、天然气都是化石燃料<br />B．液化石油气是石油化工的一种产品<br />C．可燃冰将成为未来新能源，其中主要含有丙烷水合物<br />D．等质量的甲烷和丙烷完全燃烧，生成的二氧化碳质量相等<br />（2）煤燃烧时排放出SO<SUB>2</SUB>等污染物，可能导致降雨的酸性增强，我们把&nbsp;pH<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>5.6（填“＞”、“＜”或“=”）的降雨称为酸雨．将煤燃烧后的烟气通入吸收塔，用石灰水吸收其中的SO<SUB>2</SUB>，石灰水需要“喷淋”的目的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“增大”或“减小”）与烟气的接触面积，充分吸收SO<SUB>2</SUB>．<br />（3）在汽车尾气系统中安装催化转化器，可将有毒的CO、NO转化为无毒的CO<SUB>2</SUB>、N<SUB>2</SUB>，该反应中化合价发生变化的元素是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填元素符号）．<br />（4）2016年年初，诺贝尔化学奖获得者乔治•欧拉教授率领团队，首次采用基于金属钌的催化剂，将从空气中捕获的二氧化碳直接转化为甲醇燃料，转化率高达79%．二氧化碳转化为甲醇的化学方程式可表示为CO<SUB>2</SUB>+3H<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;一定条件&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>CH<SUB>3</SUB>OH+<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（5）乙醇是实验室常用的燃料．如果92g乙醇完全燃烧，至少需要氧气的质量是多少？（写出计算过程）','','','','','','CD$###$＜$###$增大$###$C、N$###$H<SUB>2</SUB>O','【解答】解：（1）A、煤、石油、天然气都是化石燃料，故正确；<br />B、液化石油气是石油加工而来的，是一种化工产品，故正确；<br />C、可燃冰将成为未来新能源，其中主要含有甲烷水合物，故错误；<br />D、甲烷的分子式为CH<SUB>4</SUB>，丙烷的分子式为C<SUB>3</SUB>H<SUB>8</SUB>，碳元素含量不同，生成的二氧化碳质量不相等，故错误．<br />（2）SO<SUB>2</SUB>的大量排放能形成酸雨，其pH＜5.6；将煤燃烧后的烟气通入吸收塔，用石灰水吸收其中的SO<SUB>2</SUB>，石灰水需要“喷淋”的目的是增大与烟气的接触面积，充分吸收SO<SUB>2</SUB>．<br />（3）依据方程式2NO<SUB>2</SUB>+4CO<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;催化剂&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>N<SUB>2</SUB>+4CO<SUB>2</SUB>可以看出氮元素由+4价变成了0价，碳元素由+2价变成了+4价；该反应中化合价发生变化的元素是C、N元素；<br />（4）：由题意“化学方程式为CO<SUB>2</SUB>+3H<SUB>2</SUB>═CH<SUB>3</SUB>OH+X”；根据质量守恒定律的意义：则可知X中含有2个氢原子、一个氧原子；故可推测其中X的化学式为H<SUB>2</SUB>O．<br />（5）设至少需要氧气的质量为x．<br />C<SUB>2</SUB>H<SUB>5</SUB>OH+3O<SUB>2</SUB>&nbsp;<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2CO<SUB>2</SUB>+3H<SUB>2</SUB>O<br />46&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;96<br />92g&nbsp;&nbsp;&nbsp;&nbsp;x<br /><span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">46</td></tr><tr><td>92g</td></tr></table>=<table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">96</td></tr><tr><td>x</td></tr></table></span><br />答案：<br />（1）CD；&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<br />（2）＜；增大；&nbsp;&nbsp;&nbsp;&nbsp;<br />（3）C、N；&nbsp;&nbsp;&nbsp;&nbsp;<br />（4）H<SUB>2</SUB>O<br />（5）至少需要氧气的质量是192g．','【分析】（1）根据煤的成分以及燃烧产物解释．<br />（2）根据题意，产生的SO<SUB>2</SUB>用石灰水淋洗，外加氧气的作用，使之充分反应生成硫酸钙和水，写出反应的化学方程式即可；<br />（3）由题意：凡有元素化合价变化的化学反应就叫做氧化还原反应，据此进行分析解答；<br />（4）根据根据质量守恒定律，化学反应前后原子的种类和数目不变解答；<br />（5）根据乙醇的质量利用反应的方程式求出氧气的质量．','书写',3.00,'85e79c6751da7b24eabccac853e86ea5',9,400,'溶液的酸碱性与pH值的关系,酸雨的产生、危害及防治,书写化学方程式、文字表达式、电离方程式,常用燃料的使用与其对环境的影响,化石燃料及其综合利用,石油加工的产物,资源综合利用和新能源开发','',2016,'37','2016•南京校级一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1844902,'食物放置在潮湿的环境中易腐败变质，学习完金属的化学性质后，学习小组对月饼包装袋内一个细孔泡沫塑料袋产生了兴趣，下面是学习小组的探究过程：<br />（1）上网查阅包装袋内是纯度较高的铁粉，可以做“干燥剂”使食物保存较长的时间，主要吸收空气中的两种物质是食物保鲜所以也称“双吸剂”，吸收的物质指的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>和<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（写化学式）<br />（2）用铁粉而不用铁块作“双吸剂”的理由是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）小组设计了以下实验探究“双吸剂”是否失效，请你帮他们完成下列实验报告：<br /><table class=\"edittable\"><TBODY><TR><td width=143>问题与猜想</TD><td width=143>实验步骤</TD><td width=143>实验现象</TD><td width=143>结论、化学方程式</TD></TR><TR><td>双吸剂是否失效</TD><td>取样品少量于试管中加入适量的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD><td><u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD><td>结论：双吸剂没有失效，有关反应的化学方程式：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD></TR></TBODY></TABLE>','','','','','','O<SUB>2</SUB>$###$H<SUB>2</SUB>O$###$增大与氧气、水分的接触面积，加快反应速率$###$稀盐酸$###$产生大量气泡$###$Fe+2HCl═FeCl<SUB>2</SUB>+H<SUB>2</SUB>↑','【解答】解：（1）上网查阅包装袋内是纯度较高的铁粉，主要吸收空气中的两种物质是食物保鲜所以也称“双吸剂”，铁与氧气、水分接触容易生锈，则吸收的物质指的是氧气、水分，其化学式为O<SUB>2</SUB>、H<SUB>2</SUB>O．<br />（2）用铁粉而不用铁块作“双吸剂”，是因为能增大与氧气、水分的接触面积，加快反应速率．<br />（3）判断双吸剂是否失效，即检验“双吸剂”中是否含铁，取样品少量于试管中加入适量的稀盐酸，产生大量气泡，说明“双吸剂”中含有铁粉，双吸剂没有失效．铁能与酸反应生成氯化亚铁和氢气，反应的化学方程式为：Fe+2HCl═FeCl<SUB>2</SUB>+H<SUB>2</SUB>↑．<br />故答案为：（1）O<SUB>2</SUB>、H<SUB>2</SUB>O；（2）增大与氧气、水分的接触面积，加快反应速率；（3）<table class=\"edittable\"><TBODY><TR><td width=143>问题与猜想</TD><td width=143>实验步骤</TD><td width=143>实验现象</TD><td width=143>结论、化学方程式</TD></TR><TR><td>双吸剂是否失效</TD><td>取样品少量于试管中加入适量的 稀盐酸</TD><td>产生大量气泡</TD><td>结论：双吸剂没有失效，有关反应的化学方程式：Fe+2HCl═FeCl<SUB>2</SUB>+H<SUB>2</SUB>↑</TD></TR></TBODY></TABLE>','【分析】（1）根据题意，上网查阅包装袋内是纯度较高的铁粉，据此结合铁锈蚀的条件，进行分析解答．<br />（2）根据铁粉能增大与氧气、水分的接触面积，进行分析解答．<br />（3）实验探究“双吸剂”是否失效，即检验“双吸剂”中是否含铁，据此结合铁能与酸反应生成氯化亚铁和氢气，进行分析解答．','书写',3.00,'92671b087fce445017a71ce54a712a98',9,400,'食品干燥剂、保鲜剂和真空包装的成分探究,金属的化学性质,金属锈蚀的条件及其防护,书写化学方程式、文字表达式、电离方程式','',2015,'37','2015秋•临汾校级月考',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1844910,'下列分析正确的是（　　）','铁钉浸入硫酸铜溶液中，铁钉变红，铁没有铜活泼','氢气通入灼热的氧化铜粉末中，红色粉末变黑，氢气具有还原性','向汗液中滴入硝酸银溶液和稀硝酸，产生白色沉淀，汗液中含有氯离子','纯碱与盐酸混合，有气泡产生，纯碱与盐酸发生中和反应','','C','【解答】解：A、铁钉浸入硫酸铜溶液中，铁钉变红，说明铁的金属活动性比铜强，故选项说法错误．<br />B、氢气通入灼热的氧化铜粉末中，黑色粉末变红，而不是红色粉末变黑，故选项说法错误．<br />C、向汗液中滴入硝酸银溶液和稀硝酸，产生白色沉淀，说明生成了不溶于硝酸的氯化银沉淀，说明汗液中含有氯离子，故选项说法正确．<br />D、纯碱与盐酸反应生成氯化钠、水和二氧化碳，有气泡产生，但反应物是盐和酸，不属于置换反应，故选项说法错误．<br />故选：C．','【分析】A、根据金属的化学性质，进行分析判断．<br />B、根据氢气的化学性质，进行分析判断．<br />C、根据氯离子的检验方法进行分析判断．<br />D、根据纯碱与盐酸反应生成氯化钠、水和二氧化碳，进行分析判断．','选择题',3.00,'e1e79f591a077591a0adc19ef38597f4',9,400,'证明盐酸和可溶性盐酸盐,金属的化学性质,盐的化学性质,氢气的化学性质与燃烧实验','',2016,'32','2016•雅安校级模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1844949,'下列有关溶液的叙述正确的是（　　）','稀溶液一定是不饱和溶液','饱和溶液就是不能再溶解任何物质的溶液','饱和溶液一定比不饱和溶液浓','溶液是均一、稳定的混合物','','D','【解答】解：A、饱和溶液不一定是浓溶液，也可能是稀溶液，如氢氧化钙的饱和溶液为稀溶液，故选项说法错误．<br />B、饱和溶液是指在一定温度下、一定量的溶剂里，不能再继续溶解这种溶质的溶液，还能溶解其它溶质，故选项说法错误．<br />C、饱和溶液不一定比不饱和溶液浓，如较低温度下硝酸钾的饱和溶液不一定比较高温度时不饱和溶液浓，故选项说法错误．<br />D、溶液是均一、稳定的混合物；溶液的本质特征是均一性、稳定性，属于混合物，故选项说法正确．<br />故选：D．','【分析】A、浓稀溶液是溶液中所含溶质质量分数的大小，溶液是否饱和与溶液的浓稀没有必然联系．<br />B、饱和溶液是指在一定温度下、一定量的溶剂里，不能再继续溶解这种溶质的溶液，进行分析判断．<br />C、根据饱和溶液、不饱和溶液溶液与溶液浓度的关系，进行分析判断．<br />D、根据溶液的概念、特征进行分析判断．','选择题',3.00,'f36ec7ffbae2aae3431d6386b9d2c8a8',9,400,'溶液的概念、组成及其特点,饱和溶液和不饱和溶液,浓溶液、稀溶液跟饱和溶液、不饱和溶液的关系','',2016,'32','2016•渝北区模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1844965,'对下列实验中出现的异常情况的原因分析，合理的是（　　）','氢氧化钠溶液与盐酸混合时无明显现象，说明它们一定没有反应','氯化钙溶液中混有盐酸，加碳酸钙，无气泡产生，说明盐酸已除尽','用块状石灰石与稀硫酸制取CO<SUB>2</SUB>时，始终收集不满，说明装置一定漏气','检验某氢氧化钠溶液是否变质时滴加几滴稀盐酸，没有气泡，说明一定没有变质','','A','【解答】解：A、氢氧化钠与盐酸反应生成氯化钠和水，无明显现象，故A说法错误；<br />B、酸钙和盐酸反应生成二氧化碳，氯化钙溶液中混有盐酸，加碳酸钙，无气泡产生，说明盐酸已除尽，故B说法正确；<br />C、碳酸钙与稀硫酸反应生成的硫酸钙微溶，附着在石灰石上，阻碍了石灰石与稀硫酸的反应，故收集不满，不一定是装置漏气，故C说法错误；<br />D、盐酸首先与氢氧化钠反应，故滴加几滴稀盐酸，会没有气泡产生，故D说法错误；<br />故选B','【分析】A、根据氢氧化钠与稀盐酸反应无明显现象解答；<br />B、根据碳酸钙和盐酸反应生成二氧化碳解答；<br />C、根据碳酸钙与稀硫酸反应生成的硫酸钙微溶解答；<br />D、根据盐酸首先与氢氧化钠反应，再与碳酸钠反应解答；','选择题',3.00,'b4479eb86038e806c717388a11e3f453',9,400,'制取二氧化碳的操作步骤和注意点,酸的化学性质,碱的化学性质,中和反应及其应用','',2016,'32','2016•长春模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1844985,'下列化学家与其做出的贡献对应一致的是（　　）<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao96/53bddc40-94d4-11e9-93f9-b42e9921e93e_xkb97.png\" style=\"vertical-align:middle\" />','①②③','①③','②③④','②④','','C','【解答】解：①元素周期表是俄国的化学家门捷列夫发现的，故错误；<br />②法国化学家拉瓦锡用定量的方法研究了空气的成分，并得出空气是由氧气和氮气组成的结论，并通过对水的组成和分解实验确定了水不是一种元素，故正确；<br />③波义尔发现了酸碱指示剂，故正确；<br />④我国的化学家侯德榜发明了“侯氏制碱法”打破了西方国家对纯碱的制作的垄断，并生产了氮肥，故正确．<br />故选C．','【分析】由我们识记的化学发展简史即可完成对该题的判断．','选择题',3.00,'1cd6c2e1c77458007a8c7761cc070a82',9,400,'化学的历史发展过程','',2016,'32','2016•锦江区模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1845059,'某校研究性学习小组做用红磷测定空气中氧气含量的实验时，提出将红磷换成镁条是否可以测定空气中氧气含量的疑问，于是进行了下列探究，请你共同参与．<br />【提出问题】用镁条是否可以测定空气中氧气的含量？<br />【实验探究】该学习小组用图1所示装置按教材上用红磷测定空气中氧气含量的实验操作步骤进行了实验．<br /><img  src=\"/tikuimages/9/0/400/shoutiniao75/549c5561-94d4-11e9-823f-b42e9921e93e_xkb40.png\" style=\"vertical-align:middle\" /><br />（1）瓶底铺细沙的原因是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（2）如果镁只和空气中的氧气反应，则进入集气瓶中的水的体积最多不超过其容积的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>%．冷却后打开止水夹，发现进入集气瓶中的水的体积约占集气瓶容积的70%，推测原因可能是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />【结论】不能用镁条来测定空气中氧气的含量<br />【反思与交流】通过以上探究，你对燃烧的有关知识有了什么新的认识？<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（写一点即可）<br />【拓展实验】小明同学用木炭设计了如图2所示的实验，但他观察到的现象是水面基本没有上升，其原因是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，如果要用木炭测定空气中氧气的含量，可将水换成<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．由此我们总结出测定空气中氧气含量所选的药品应满足的条件是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','防止瓶底受热不均而破裂（防止高热的燃烧溅落物使瓶底炸裂）$###$21$###$瓶中的氮气与镁条发生了反应$###$燃烧不一定有氧气参加$###$木炭燃烧生成了二氧化碳气体$###$氢氧化钠溶液$###$能把氧气反应掉；不能与空气中的其它成分反应；不生成新的气体杂质，生成物是固体','【解答】解：（1）镁条燃烧放出大量的热，所以瓶底铺细沙的原因是防止瓶底受热不均而破裂（防止高热的燃烧溅落物使瓶底炸裂）；<br />（2）氧气约占空气体积的21%，所以若镁只和氧气反应，水进入集气瓶的体积不会超过21%，而题中的实验结果为进入了约70%，所以结合空气的成分可以知道镁条可能是与氮气也发生了反应．<br />【反思与交流】通过以上探究，对燃烧的新认识是：燃烧不一定有氧气参加；镁条失火，不能用氮气和二氧化碳灭火等；故答案为：燃烧不一定有氧气参加；镁条失火，不能用氮气和二氧化碳灭火等；<br />【拓展实验】小明同学由该实验得到启发，设计了如图2所示的测定实验：但是他看到的现象是水面并没有上升，原因是木炭燃烧生成了二氧化碳气体，如果要用木炭测定空气中氧气的含量，可将水换成氢氧化钠溶液；由此我们总结出测定空气中氧气含量的生成物所满足的条件是能把氧气反应掉；不能与空气中的其它成分反应；不生成新的气体杂质，生成物是固体．<br />故答案为：<br />（1）防止瓶底受热不均而破裂（防止高热的燃烧溅落物使瓶底炸裂）；<br />（2）21；瓶中的氮气与镁条发生了反应．<br />【反思与交流】燃烧不一定有氧气参加；<br />【拓展实验】木炭燃烧生成了二氧化碳气体；氢氧化钠溶液；能把氧气反应掉；不能与空气中的其它成分反应；不生成新的气体杂质，生成物是固体．','【分析】（1）根据镁条燃烧放出大量的热进行解答；<br />（2）根据氧气占空气的体积分数来解答，并分析可能的原因；<br />【反思与交流】根据探究结果分析解答；<br />【拓展实验】根据木炭燃烧生成了二氧化碳气体解答．','填空题',3.00,'95a100c5103930b2627456b533e44dea',9,400,'测定空气里氧气含量的探究,燃烧与燃烧的条件','',0,'37','',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1845063,'材料一：钓鱼岛是我国固有领土，其附近海底蕴藏着丰富的可燃冰．可燃冰主要含有甲烷水合物，还含有氮气、氧气、二氧化碳、硫化氢等．<br />材料二：钓鱼岛附近海域是我国东海渔场，蕴藏着丰富的海产品，这些可食用的海产品为人类提供丰富的蛋白质和磷、钙、铁、锌、碘等元素．<br />根据上述材料，请回答：<br />（1）写出可燃冰成分中任意一种单质的化学式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（2）目前，可燃冰开采技术还不成熟，如果开采过程中<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>大量泄漏于大气中，将比二氧化碳造成的温室效应更严重；<br />（3）我国渔民捕获的可食用的海产品中，能补充人体所需要的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>等微量元素（写一种元素名称或符号即可）．<br />（4）海水通常浑浊也含有很多杂质，要使海水成为饮用水，需要采取的操作方法是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />A．沉降&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;B．过滤&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;C．吸附&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;D．蒸馏．','','','','','','O<SUB>2</SUB>$###$甲烷$###$铁$###$ABCD','【解答】解：（1）由同一种元素组成的纯净物属于单质，故氮气、氧气均为单质，故填：O<SUB>2</SUB>；<br />（2）可燃冰具有能量高、燃烧值大等优点．但它的主要成分甲烷是一种温室气体，如果开采不当，甲烷气体将大量泄漏于空气中，其造成的温室效应将比二氧化碳更加严重，故填：甲烷；<br />（3）海产品中富含铁元素、锌元素和碘元素等；故填：铁；<br />（4）通过沉降将悬浮杂质沉降，然后过滤除去海水中的不溶性杂质，能使浑浊的海水变得澄清，通过吸附除去水中的污染物，若要使其变成较纯净的饮用水，需要的操作方法为蒸馏．<br />故填：ABCD．','【分析】（1）根据单质的概念以及化学式的写法来分析；<br />（2）根据温室效应的成因和可燃冰的成分进行分析；<br />（3）根据食物中含有的营养素来分析；<br />（4）根据净化水的措施分析回答．','书写',3.00,'68cb4cb60faf855919d20358ee93d4be',9,400,'过滤的原理、方法及其应用,二氧化碳对环境的影响,化学式的书写及意义,海洋中的资源,矿物质与微量元素','',2016,'32','2016•濠江区模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1845071,'工业上电解氧化铝制取金属铝．在这个反应中氧化铝、铝、氧气的质量比是（　　）','102：27：32','43：27：24','2：4：3','204：108：96','','D','【解答】解：工业上电解氧化铝制取金属铝，反应的化学方程式为：<br />2Al<SUB>2</SUB>O<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;通电&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>4Al+3O<SUB>2</SUB>↑<br />204&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;108&nbsp;&nbsp;96<br />在此反应中氧化铝、铝和氧气的质量比为204：108：96=51：27：24．<br />故选：D．','【分析】首先正确写出电解氧化铝反应的化学方程式，利用各物质之间的质量比等于相对分子质量和的比，进行分析解答即可．','选择题',3.00,'d859d55baa5dbb2bf43dd76d67a38a58',9,400,'常见化学反应中的质量关系','',2016,'35','2016春•沂源县期中',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1845120,'空气中含量最多的物质是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，地壳中含量最多的金属元素是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','N<SUB>2</SUB>$###$Al','【解答】解：空气中含量最多的气体是氮气，它约占空气体积的78%；地壳中含量最多的金属元素是铝元素．<br />故答案为：N<SUB>2</SUB>，Al．','【分析】根据空气中各气体的体积分数和地壳中各元素的含量顺序解答．','填空题',3.00,'21e9e451c82c44a03db942802825b08e',9,400,'空气的成分及各成分的体积分数,地壳中元素的分布与含量,元素的简单分类','',0,'37','',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1845149,'防止金属腐蚀，特别是钢铁的锈蚀是世界科学家研究和技术领域中的重大问题，铁生锈的条件为与<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>同时解除，铁锈成分复杂，主要成分为氧化铁（Fe<SUB>2</SUB>O<SUB>3</SUB>）<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>物质可加速铁的锈蚀．<br />用盐酸除去铁锈的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，为了防止铁制品生锈，应采取的措施是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（写出一种）．<br />【查阅资料】小明同学发现铁生锈是铁与空气中的物质发生的一系列复杂的化学反应，生成氢氧化亚铁【Fe（OH）<SUB>2</SUB>】，后再在空气中被氧化为氢氧化铁【Fe（OH）<SUB>3</SUB>】，Fe（OH）<SUB>3</SUB>不稳定，在阳光照射下发生缓慢的分解反应生成氧化铁（Fe<SUB>2</SUB>O<SUB>3</SUB>），铁锈的化学式可简单表示为Fe<SUB>2</SUB>O<SUB>3</SUB>•nH<SUB>2</SUB>O．<br />【提出问题】铁锈（Fe<SUB>2</SUB>O<SUB>3</SUB>•nH<SUB>2</SUB>O）中的n的值等于多少呢？<br />【问题探究】小明发现实验室中有一保管不善的铁粉，大部分已经结块成红褐色，为了探究铁锈（Fe<SUB>2</SUB>O<SUB>3</SUB>•nH<SUB>2</SUB>O）的组成，称取27.0g这种铁粉样品，按如图所示装置进行实验．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao12/55d6aa21-94d4-11e9-9927-b42e9921e93e_xkb64.png\" style=\"vertical-align:middle\" /><br />（1）为了保证实验安全，实验开始时应先<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，防止<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）A中的现象是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，B中浓硫酸的作用是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．D中碱石灰的作用是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />（3）请指出该装置中有一处明显不足<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />【数据处理】一下是加热时间和A中固体质量关系图象及B、C中质量补在变化时B中浓硫酸，C中碱石灰装置质量变化情况．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao8/55da0580-94d4-11e9-a568-b42e9921e93e_xkb72.png\" style=\"vertical-align:middle\" /><br />（1）n的值是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）写出T3-T4时间段发生反应的化学方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）计算原样品中单质铁的质量分数． <table class=\"edittable\"><TBODY><TR><td width=188>&nbsp;</TD><td width=188>反应前（g）&nbsp;</TD><td width=188>反应会（g）&nbsp;</TD></TR><TR><td>B</TD><td>100</TD><td>105.4</TD></TR><TR><td>C</TD><td>150</TD><td>163.2</TD></TR></TBODY></TABLE>','','','','','','水和氧气$###$酸、盐等$###$Fe<SUB>2</SUB>O<SUB>3</SUB>+6HCl═2FeCl<SUB>3</SUB>+3H<SUB>2</SUB>O$###$涂一层油漆$###$通入一会儿一氧化碳把玻璃管中的空气排尽$###$发生爆炸$###$红褐色固体变成黑色固体$###$吸收反应生成的水$###$防止空气中的水蒸气和二氧化碳进入C装置$###$没有处理尾气$###$3$###$Fe<SUB>2</SUB>O<SUB>3</SUB>+3CO<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2Fe+3CO<SUB>2</SUB>','【解答】解：铁生锈的条件为与水和氧气同时接触，酸、盐等物质可加速铁的锈蚀．<br />故填：水和氧气；酸、盐等．<br />用盐酸除去铁锈的化学方程式为：Fe<SUB>2</SUB>O<SUB>3</SUB>+6HCl═2FeCl<SUB>3</SUB>+3H<SUB>2</SUB>O，为了防止铁制品生锈，应采取的措施是涂一层油漆，或镀一层金属等．<br />故填：Fe<SUB>2</SUB>O<SUB>3</SUB>+6HCl═2FeCl<SUB>3</SUB>+3H<SUB>2</SUB>O；涂一层油漆．<br />【问题探究】<br />（1）为了保证实验安全，实验开始时应先通入一会儿一氧化碳把玻璃管中的空气排尽，防止发生爆炸．<br />故填：通入一会儿一氧化碳把玻璃管中的空气排尽；发生爆炸．<br />（2）A中的现象是红褐色固体变成黑色固体，B中浓硫酸的作用是吸收反应生成的水，D中碱石灰的作用是防止空气中的水蒸气和二氧化碳进入C装置．<br />故填：红褐色固体变成黑色固体；吸收反应生成的水；防止空气中的水蒸气和二氧化碳进入C装置．<br />（3）该装置中一处明显不足是没有处理尾气．<br />故填：没有处理尾气．<br />【数据处理】<br />（1）加热铁锈时，首先失去结晶水，结晶水的质量为：27g-21.6g=5.4g，<br />设氧化铁的质量为x，<br />反应生成水的质量为：105.4g-100g=5.4g，与左图中数据符合；<br />反应生成二氧化碳质量为：163.2g-150g=13.2g，<br />Fe<SUB>2</SUB>O<SUB>3</SUB>+3CO<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2Fe+3CO<SUB>2</SUB>，<br />160&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 132<br />x&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 13.2g<br /><span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">160</td></tr><tr><td>x</td></tr></table></span>=<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">132</td></tr><tr><td>13.2g</td></tr></table></span>，<br />x=16g，<br />根据题意有：<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">16g</td></tr><tr><td>5.4g</td></tr></table></span>=<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">160</td></tr><tr><td>18n</td></tr></table></span>，<br />n=3，<br />故填：3．<br />（2）T3-T4时间段时，氧化铁和一氧化碳反应生成铁和二氧化碳，发生反应的化学方程式为：Fe<SUB>2</SUB>O<SUB>3</SUB>+3CO<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2Fe+3CO<SUB>2</SUB>．<br />故填：Fe<SUB>2</SUB>O<SUB>3</SUB>+3CO<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2Fe+3CO<SUB>2</SUB>．<br />（3）样品中铁锈的质量为：16g+5.4g=21.4g，<br />原样品中单质铁的质量分数为：<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">27.0g-21.4g</td></tr><tr><td>27.0g</td></tr></table></span>×100%=20.7%，<br />答：原样品中单质铁的质量分数为20.7%．','【分析】铁与水和氧气同时接触时容易生锈，因此隔绝水和氧气可以防止铁制品生锈；<br />铁锈的主要成分是氧化铁，氧化铁能和稀盐酸反应生成氯化铁和水；<br />【问题探究】<br />一氧化碳和空气或氧气混合达到一定程度时，遇明火会发生爆炸；<br />高温条件下，氧化铁和一氧化碳反应生成铁和二氧化碳；<br />浓硫酸能够吸收水，碱石灰能够吸收水和二氧化碳；<br />一氧化碳有毒，扩散到空气中会污染环境；<br />【数据处理】<br />根据反应的化学方程式和提供的数据可以进行相关方密度计算和判断．','书写',3.00,'6861cc0d3b2346d0ba1433879ba91c66',9,400,'实验探究物质的组成成分以及含量,常见气体的检验与除杂方法,金属锈蚀的条件及其防护,铁锈的主要成分,酸的化学性质,书写化学方程式、文字表达式、电离方程式,根据化学反应方程式的计算','句容市',2016,'37','2016•句容市校级一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1845170,'某校综合实践活动小组为完成与红糖相关的研究课题，来到了制糖基地进行了实地调查．<br />（1）来到制糖间，只见屋内“白气”弥漫．在满屋的“白气”中可以看到九口大铁锅从大到小依次排开，铁锅里面翻滚着的浓稠液体颜色逐个加深．“白气”产生的原因是水蒸气遇冷<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>形成．<br />（2）在制糖的过程中，制糖师傅们会在锅里加入一种白色粉末后，液面上立即泛起白沫，此时制糖师傅便会用筛状勺子从上层液体中捞出许多杂质，同学们都很好奇．<br />老师对此作了如下解释：这种白色粉末叫小苏打，它的主要成分是NaHCO<SUB>3</SUB>，它受热分解的化学方程式是：2NaHCO<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>Na<SUB>2</SUB>CO<SUB>3</SUB>+<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>+CO<SUB>2</SUB>↑，产生的CO<SUB>2</SUB>会附着在杂质表面，从而使它们受到的浮力大于重力而上浮．','','','','','','液化$###$H<SUB>2</SUB>0','【解答】解：（1）“白气”的实质是液体的小液滴，是由于水蒸气遇冷液化形成的小液滴；<br />（2）碳酸氢钠在加热状态下发生分解反应，生成了水、二氧化碳和碳酸钠，化学方程式为：2NaHC0<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>Na<SUB>2</SUB>C0<SUB>3</SUB>+H<SUB>2</SUB>0+C0<SUB>2</SUB>↑；<br />答案：（1）液化；（2）H<SUB>2</SUB>0．','【分析】（1）在一定条件下，物体的三种状态--固态、液态、气态之间会发生相互转化，这就是物态变化；物质由气态变为液体叫液化；<br />（2）碳酸氢钠在加热状态下发生分解反应，生成了水、二氧化碳和碳酸钠，','填空题',3.00,'5433cad5f165eef674a8a908423e5a4d',9,400,'物质的三态及其转化,质量守恒定律及其应用','',2016,'32','2016•永嘉县模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1845193,'下列有关氧气的说法中，正确的是（　　）','氧气密度大于空气且易溶于水','氧气和臭氧（O<SUB>3</SUB>）组成元素相同，是同种物质','氧气性质活泼，能与所有物质反应','植物光合作用是空气中氧气的主要来源','','D','【解答】解：A、氧气密度略大于空气且不易溶于水，故A错误；<br />B、氧气和臭氧（O<SUB>3</SUB>）是由同种元素组成的不同单质，故B错误；<br />C、氧气性质比较活泼，在一定条件下能与很多种物质反应，故C错误；<br />D、光合作用的产物之一是氧气，植物光合作用是空气中氧气的主要来源，故D正确．<br />故选D．','【分析】根据氧气的组成、性质和空气中氧气的来源分析判断有关的说法．','选择题',2.00,'d7b280985a7198481c012bdab9b3c300',9,400,'氧气的物理性质,氧气的化学性质,自然界中的氧循环,氧元素组成的单质','',2016,'37','2016•延平区一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1845233,'下列说法中，符合实验室化学药品取用规则的是（　　）','为了节约药品，用剩的药品可放回原瓶','如果没有说明用量，应该按最少量取，液体取4-5mL','用剩的药品，只要不是危险药品就可以拿出实验室','闻气体药品时，应用手轻轻煽动气体，使少量气体飘入鼻孔','','D','【解答】解：A、对化学实验中的剩余药品，既不能放回原瓶，也不可随意丢弃，更不能带出实验室，应放入的指定的容器内，故选项说法错误．<br />B、为节约药品，根据实验室药品取用规则：取用化学药品时，如果没有说明用量，一般应按最少量（1～2mL）取用液体，故选项说法错误．<br />C、对化学实验中的剩余药品，既不能放回原瓶，也不可随意丢弃，更不能带出实验室，应放入的指定的容器内，故选项说法错误．<br />D、闻气体的气味时，应用手在瓶口轻轻的扇动，使极少量的气体飘进鼻子中，不能将鼻子凑到集气瓶口去闻气体的气味，故选项说法正确．<br />故选：D．','【分析】A、根据实验室剩余药品的处理原则（三不一要），进行分析判断．<br />B、根据实验室药品取用的用量原则，要遵循节约的原则，进行分析判断．<br />C、根据实验室剩余药品的处理原则（三不一要），进行分析判断．<br />D、根据闻气体的气味时的方法（招气入鼻法）进行分析判断．','选择题',3.00,'915ac67d65f2a80e79cef33e1a088bfc',9,400,'实验操作注意事项的探究','',2015,'35','2015秋•孝义市校级期中',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1845285,'小颖在地矿博物馆中看到了许多矿石标本，其中部分矿石信息如图所示：<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao43/574d448f-94d4-11e9-abbf-b42e9921e93e_xkb92.png\" style=\"vertical-align:middle\" /><br />（1）在钾长石的主要成分KAlSi<SUB>3</SUB>O<SUB>4</SUB>中，共含有<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>种金属元素．<br />（2）在上述四种矿石中，<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填序号）的主要成分属于氧化物．<br />（3）查阅博物馆资料，小颖了解到我们的祖先很早就掌握了炼铜工艺：他们将孔雀石和木炭一起加热就可得到红色的金属铜．（提示：孔雀石受热易分解，生成氧化铜、二氧化碳和水；高温灼烧时，用木炭作还原剂能把氧化铜中的铜还原出来．），请你从上述反应中，任选其一写出对应的化学方程式：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（4）在赤铁矿和黄铁矿中，你认为哪一种矿石更适合炼铁？请说明理由？<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','2$###$A$###$2CuO+C<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2Cu+CO<SUB>2</SUB>↑$###$赤铁矿，黄铁矿中含有硫元素，生成的二氧化硫会污染空气','【解答】解：（1）在钾长石的主要成分KAlSi<SUB>3</SUB>O<SUB>4</SUB>中，共含有钾、铝2种金属元素；<br />（2）氧化物是由两种元素组成的，其中一种是氧元素，所以A的主要成分属于氧化物；<br />（3）氧化铜和碳在高温的条件下生成铜和二氧化碳，化学方程式为：2CuO+C<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2Cu+CO<SUB>2</SUB>↑；<br />（4）在赤铁矿和黄铁矿中，赤铁矿矿石更适合炼铁，理由是：黄铁矿中含有硫元素，生成的二氧化硫会污染空气．<br />故答案为：（1）2；<br />（2）A；<br />（3）2CuO+C<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2Cu+CO<SUB>2</SUB>↑；<br />（4）赤铁矿，黄铁矿中含有硫元素，生成的二氧化硫会污染空气．','【分析】（1）根据物质的化学式分析物质的元素组成；<br />（2）根据氧化物的定义进行分析；<br />（3）根据氧化铜和碳在高温的条件下生成铜和二氧化碳进行分析；<br />（4）根据赤铁矿和黄铁矿反应的生成物进行分析．','书写',3.00,'fa97ed03cd3ecb465d9341f5c36fcf52',9,400,'金属元素的存在及常见的金属矿物,从组成上识别氧化物,元素的简单分类,书写化学方程式、文字表达式、电离方程式','',2016,'37','2016•历下区二模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1845348,'下列归纳和总结完全正确的一组是（　　）<br /><table class=\"edittable\"><TBODY><TR><td width=337>A　对除杂的认识</TD><td width=249>B&nbsp; 鉴别的方法</TD></TR><TR><td>①用化学方法除去食盐水中可溶性杂质<br />②用磁铁除去铁粉中木炭粉<br />③用碳酸钠除去CaCl<SUB>2</SUB>溶液中的HCl</TD><td>①用燃着的木条鉴别氧气和二氧化碳<br />②用熟石灰区别化肥硝酸钾和硝酸铵<br />③用酚酞试液鉴别稀盐酸和氯化钠溶液</TD></TR><TR><td>C　对现象的认识</TD><td>D&nbsp; 化学家与贡献</TD></TR><TR><td>①花香四溢，说明分子在不断运动<br />②镁条燃烧，有大量白雾产生<br />③电解水生成氢气和氧气的体积比为2：1</TD><td>①拉瓦锡-发现空气的主要成分<br />②门捷列夫-发现元素周期表<br />③侯德榜-发明联合制碱法</TD></TR></TBODY></TABLE>','A','B','C','D','','D','【解答】解：A、①食盐水中可溶性杂质如一些可溶性盐等，能发生化学反应，用化学方法除去食盐水中可溶性杂质；②磁铁能吸引铁粉，用磁铁除去铁粉中木炭粉；③HCl与碳酸钠反应生成氯化钠、水和二氧化碳，能除去杂质但引入了新的杂质氯化钠，不符合除杂原则；故选项归纳和总结有错误．<br />B、①氧气能支持燃烧，二氧化碳不能燃烧、不能支持燃烧，用燃着的木条进行检验，能使燃烧更旺的是氧气，熄灭的是二氧化碳，可以鉴别；②硝酸铵属于铵态氮肥，与熟石灰混合产生有刺激性气味的氨气，硝酸钾不能，可以鉴别；③稀盐酸和氯化钠溶液分别显酸性，中性，均不能使无色酚酞溶液变红色，不能鉴别；故选项归纳和总结有错误．<br />C、①花香四溢，是因为花香中含有的分子是在不断运动的，向四周扩散，使人们闻到花香；②镁条燃烧，发出耀眼的白光，产生大量的白烟；③电解水生成氢气和氧气的体积比为2：1，是实验结论而不是实验现象；故选项归纳和总结有错误．<br />D、①拉瓦锡发现空气的主要成分；②门捷列夫发现元素周期表；③侯德榜发明联合制碱法；故选项归纳和总结完全正确．<br />故选：D．','【分析】A、除杂质题至少要满足两个条件：①加入的试剂只能与杂质反应，不能与原物质反应；②反应后不能引入新的杂质．<br />B、根据两种物质与同种试剂反应产生的不同现象来鉴别它们，若两种物质与同种物质反应的现象相同，则无法鉴别它们．<br />C、根据分子的基本性质、镁条燃烧的现象、电解水的实验现象，进行分析判断．<br />D、根据科学家们各自在科学上做出的贡献，进行分析判断．','选择题',3.00,'263d3a3de98bbbbdf382497c4d57689c',9,400,'物质除杂或净化的探究,化学的历史发展过程,氧气与碳、磷、硫、铁等物质的反应现象,电解水实验,物质的鉴别、推断,分子的定义与分子的特性','',2016,'32','2016•泰安模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1845391,'下列事实和解释不相符的是（　　）','汽油洗油污--汽油能溶解油污','熟石灰调节酸性土壤--熟石灰显碱性','活性炭净水--活性炭能吸附水中的一些杂质','石墨可做电极材料--石墨质软','','D','【解答】解：A、汽油洗油污，是利用了汽油能溶解油污的原理，利用的溶解原理，故选项说法正确．<br />B、熟石灰调节酸性土壤，是利用了熟石灰显碱性，能与酸发生中和反应，故选项说法正确．<br />C、活性炭净水，是利用了活性炭具有吸附性，能吸附水中的一些杂质，故选项说法正确．<br />D、石墨可做电极材料，是利用了石墨具有优良的导电性，故选项说法错误．<br />故选：D．','【分析】A、根据汽油能溶解油污，进行分析判断．<br />B、根据熟石灰的用途，进行分析判断．<br />C、根据活性炭具有吸附性，进行分析判断．<br />D、根据石墨具有优良的导电性，进行分析判断．','选择题',3.00,'df31a30a7da65f05eb15158871849661',9,400,'溶解现象与溶解原理,中和反应及其应用,碳单质的物理性质及用途','',2016,'32','2016•宜昌模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1845418,'<img  src=\"/tikuimages/9/2016/400/shoutiniao41/58d5e05e-94d4-11e9-b73d-b42e9921e93e_xkb80.png\" style=\"vertical-align:middle;FLOAT:right\" />（2016•苏州一模）构建知识网络是化学学习中的常用方法．下图是某同学在复习“我们身边的化学物质”时构建的知识网络，其中A、B、C、D、E分别表示非金属氧化物、金属氧化物、酸、碱、盐中的某一种，“-”表示物质之间普遍能够发生的化学反应．<br />请按要求回答下列问题：<br />（1）从物质类别看，A物质属于<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；C物质属于<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）从反应基本类型看，属于复分解反应的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填图中编号）．<br />（3）写出任意一个符合序号③的化学方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（4）已知：盐酸与氢氧化钠反应的实质是盐酸、氢氧化钠溶液在水中解离出来的H<SUP>+</SUP>与OH<SUP>-</SUP>结合生成了H<SUB>2</SUB>O．据此分析：NH<SUB>4</SUB>NO<SUB>3</SUB>溶液与NaOH溶液反应的实质是它们在水中解离出来的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>结合而使反应发生；CaCl<SUB>2</SUB>溶液与Na<SUB>2</SUB>CO<SUB>3</SUB>溶液反应的实质是它们在水中解离出的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>结合而使反应发生．','','','','','','金属氧化物$###$碱$###$①④⑤⑥$###$Fe+CuSO<SUB>4</SUB>=FeSO<SUB>4</SUB>+Cu$###$NH<SUB>4</SUB><SUP>+</SUP>与OH<SUP>-</SUP>$###$Ca<SUP>2+</SUP>与CO<SUB>3</SUB><SUP>2-</SUP>','【解答】解：A、B、C、D、E分别表示非金属氧化物、金属氧化物、酸、碱、盐中的某一种，金属会与盐、酸反应，酸会与碱、金属氧化物反应，非金属氧化物呼吁碱反应，所以A是金属氧化物，B是酸，C是碱，D是非金属氧化物，E是盐，金属氧化物与酸、盐与酸、盐和碱、酸与碱都属于复分解反应，经过验证，推导正确，所以<br />（1）从物质类别看，A物质属于金属氧化物，C物质属于碱；<br />（2）从反应基本类型看，属于复分解反应的是①④⑤⑥；<br />（3）铁和硫酸铜反应生成硫酸亚铁和铜，化学方程式为：Fe+CuSO<SUB>4</SUB>=FeSO<SUB>4</SUB>+Cu；<br />（4）依据复分解反应的实质可知，盐酸与氢氧化钠反应的实质是盐酸、氢氧化钠溶液在水中解离出来的H<SUP>+</SUP>与OH<SUP>-</SUP>结合生成了H<SUB>2</SUB>O．据此分析：NH<SUB>4</SUB>NO<SUB>3</SUB>溶液与NaOH溶液反应的实质是它们在水中解离出来的NH<SUB>4</SUB><SUP>+</SUP>&nbsp;与&nbsp;OH<SUP>-</SUP>结合而使反应发生；CaCl<SUB>2</SUB>溶液与Na<SUB>2</SUB>CO<SUB>3</SUB>溶液反应的实质是它们在水中解离出的Ca<SUP>2+</SUP>&nbsp;与CO<SUB>3</SUB><SUP>2-</SUP>结合而使反应发生．<br />故答案为：（1）金属氧化物，碱；<br />（2）①④⑤⑥；<br />（3）Fe+CuSO<SUB>4</SUB>=FeSO<SUB>4</SUB>+Cu；<br />（4）NH<SUB>4</SUB><SUP>+</SUP>&nbsp;与&nbsp;OH<SUP>-</SUP>，Ca<SUP>2+</SUP>&nbsp;与CO<SUB>3</SUB><SUP>2-</SUP>．','【分析】根据A、B、C、D、E分别表示非金属氧化物、金属氧化物、酸、碱、盐中的某一种，金属会与盐、酸反应，酸会与碱、金属氧化物反应，非金属氧化物呼吁碱反应，所以A是金属氧化物，B是酸，C是碱，D是非金属氧化物，E是盐，金属氧化物与酸、盐与酸、盐和碱、酸与碱都属于复分解反应，然后将推出的物质类别选择适当的物质进行验证即可．','书写',3.00,'79bb985e7b074a5f90bc081f58775764',9,400,'物质的鉴别、推断,复分解反应及其应用,书写化学方程式、文字表达式、电离方程式','',2016,'37','2016•苏州一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1845448,'下列实验设计不正确的是（　　）','用肥皂水鉴别硬水与软水','用燃烧法区分羊毛和化纤布料','用润湿的pH试纸测白醋的pH','用淀粉溶液区分加碘食盐和无碘食盐','','C|D','【解答】解：A、取样品，加入肥皂水，产生泡沫较多的是软水，产生泡沫较少的是硬水，现象不同，可以鉴别，故A正确；<br />B、取样品，灼烧，产生烧焦羽毛气味的是羊毛，没有此气味的是化纤布料，现象不同，可以鉴别，故B正确；<br />C、pH试纸测定溶液的pH值时，不能湿润，故C错误；<br />D、淀粉只能与单质碘作用而不能与化合态碘反应显色，加碘盐中的碘为化合态的碘，故D错误．<br />故选：CD．','【分析】A、根据肥皂水在硬水和软水中的不同现象进行分析；<br />B、根据羊毛燃烧会产生烧焦羽毛气味的气体进行分析；<br />C、根据pH试纸测定溶液的pH值时，不能湿润进行分析；<br />D、根根据淀粉只能与单质碘作用而不能与化合态碘反应显色进行解答．','多选题',3.00,'010ad7f3fa093ab4bda4433e6899ee02',9,400,'化学实验方案设计与评价,溶液的酸碱度测定,硬水与软水,棉纤维、羊毛纤维和合成纤维的鉴别,加碘盐的检验','句容市',2016,'32','2016•句容市模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1845483,'以下说法正确的是（　　）','中和反应有水和盐生成，有水和盐生成的反应一定是中和反应','置换反应有单质生成，有单质生成的反应一定是置换反应','碳酸盐与酸反应放出气体，与酸反应放出气体的物质不一定是碳酸盐','盐能解离出酸根离子，能解离出酸根离子的化合物一定是盐','','C','【解答】解：A、中和反应生成盐和水，但生成盐和水的反应不一定是中和反应，如CO<SUB>2</SUB>+2NaOH═Na<SUB>2</SUB>CO<SUB>3</SUB>+H<SUB>2</SUB>O，故选项说法错误．<br />B、置换反应有单质生成，但有单质生成的反应不一定是置换反应，如CO+CuO<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>Cu+CO<SUB>2</SUB>，故选项说法错误．<br />C、碳酸盐与酸反应放出气体，与酸反应放出气体的物质不一定是碳酸盐，也可能是活泼金属等，故选项说法正确．<br />D、盐能解离出酸根离子，能解离出酸根离子的化合物不一定是盐，也可能是酸，故选项说法错误．<br />故选：C．','【分析】A、中和反应是酸与碱作用生成盐和水的反应，反应物是酸和碱，生成物是盐和水．<br />B、置换反应是一种单质和一种化合物反应生成另一种单质和另一种化合物的反应．<br />C、根据酸能与活泼金属、碳酸盐等反应生成气体，进行分析判断．<br />D、盐是由金属离子（或铵根离子）和酸根离子组成的化合物．','选择题',3.00,'89b9719e05da5a7e0a7a2a34c67c8c93',9,400,'酸的化学性质,中和反应及其应用,氧化物、酸、碱和盐的概念,置换反应及其应用','',2016,'37','2016春•重庆校级月考',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1845501,'现有下列六种物质：①碳酸氢钠；②硝酸钾；③熟石灰；④甲醛；⑤塑料；⑥二氧化硫．请用上述物质的序号填空．<br />（1）用于改良酸性土壤的碱<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，<br />（2）可以用作发酵粉，也可用于治疗胃酸过多的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，<br />（3）在空气中会导致酸雨形成的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，<br />（4）使用方便，但废弃物会造成“白色污染”<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，<br />（5）其水溶液可用于浸泡动物标本的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，<br />（6）用作复合肥<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','③$###$①$###$⑥$###$⑤$###$④$###$②','【解答】解：<br />（1）用于改良酸性土壤的碱氢氧化钙；<br />（2）碳酸氢钠可以用作发酵粉，也可用于治疗胃酸过多；<br />（3）二氧化硫会导致酸雨；<br />（4）使用方便，但废弃物会造成“白色污染”是塑料；<br />（5）甲醛的水溶液可用于浸泡动物标本；<br />（6）硝酸钾中含有钾元素和氮元素，属于符合肥．<br />答案：（1）③；（2）①；（3）⑥；（4）⑤；（5）④；（6）②．','【分析】物质的性质决定物质的用途，根据物质的用途，分析选择物质．','填空题',3.00,'b333c9c8fee5f2f3675509511982cfb6',9,400,'酸雨的产生、危害及防治,酸碱盐的应用,常见化肥的种类和作用,白色污染与防治,亚硝酸钠、甲醛等化学品的性质与人体健康','',2016,'32','2016•射阳县模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1845582,'如图是工业上利用含SO<SUB>2</SUB>的烟气制备K<SUB>2</SUB>SO<SUB>4</SUB>的流程．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao65/5ae31580-94d4-11e9-a5bf-b42e9921e93e_xkb88.png\" style=\"vertical-align:middle\" /><br />相关物质的溶解度如下表所示：<br /><table class=\"edittable\"><TBODY><TR><td width=121>物质</TD><td width=77>KCl</TD><td width=93>K<SUB>2</SUB>SO<SUB>4</SUB></TD><td width=97>NH<SUB>4</SUB>Cl</TD><td width=97>（NH<SUB>4</SUB>）<SUB>2</SUB>SO<SUB>4</SUB></TD></TR><TR><td>溶解度/（&nbsp;25℃）</TD><td>34.0</TD><td>11.1</TD><td>37.2</TD><td>19.5</TD></TR></TBODY></TABLE>（1）在“脱硫”反应中，CaCO<SUB>3</SUB>粉碎后与水混合成<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“溶液”、“悬浊液”或“乳浊液”）后，与含SO<SUB>2</SUB>的烟气、空气中气体反生成石膏（&nbsp;CaSO<SUB>4</SUB>•2H<SUB>2</SUB>O&nbsp;），写出该反应的化学方程式：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）石膏与饱和（NH<SUB>4</SUB>）<SUB>2</SUB>CO<SUB>3</SUB>溶液相混合，发生反应I时需不断搅拌，其目的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）“操作I”的名称是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，实验室进行该操作时所用到的玻璃仪器有烧杯、玻璃棒、<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（4）你认为反应II在常温下能实现的原因是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（5）NH<SUB>4</SUB>C1在农业上可以用作化肥，从NH<SUB>4</SUB>C1溶液中得到NH<SUB>4</SUB>C1晶体，可以采用加热浓缩、<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>、过滤的方法．<br />（6）此工艺流程中可以循环利用的物质是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','悬浊液$###$2CaCO<SUB>3</SUB>+2SO<SUB>2</SUB>+O<SUB>2</SUB>═2CaSO<SUB>4</SUB>+2CO<SUB>2</SUB>$###$使原料充分反应，提高原料利用率$###$过滤$###$漏斗$###$常温下K<SUB>2</SUB>SO<SUB>4</SUB>的溶解度较小$###$降温结晶$###$CaCO<SUB>3</SUB>','【解答】解：（1）在“脱硫”反应中，CaCO<SUB>3</SUB>粉碎后与水混合成悬浊液后，与含SO<SUB>2</SUB>的烟气空气中气体反生成石膏（CaSO<SUB>4</SUB>），写出反应的化学方程式是：2CaCO<SUB>3</SUB>+2SO<SUB>2</SUB>+O<SUB>2</SUB>═2CaSO<SUB>4</SUB>+2CO<SUB>2</SUB>；（2）石膏与饱和（NH<SUB>4</SUB>）<SUB>3</SUB>CO<SUB>3</SUB>溶液相混合，发生反应Ⅰ时需不断搅拌，其目的是：使原料充分反应，提高原料利用率；<br />（3）“操作I”是过滤操作，其中需要用到的玻璃仪器有普通漏斗、烧杯和玻璃棒，玻璃棒的作用是：引流；<br />（4）从生成物的溶解性角度分析该反应常温下能够发生的原因可能是：常温下K<SUB>2</SUB>SO<SUB>4</SUB>的溶解度较小；<br />（5）由于NH<SUB>4</SUB>Cl溶解度受温度影响较大，从NH<SUB>4</SUB>Cl溶液中得到NH<SUB>4</SUB>Cl可以采用加热浓缩，降温结晶的方法；<br />（6）观察分析此工艺流程，其中可以循环利用的物质是CaCO<SUB>3</SUB>；此工艺的重要意义在于减少污染；<br />故答案为：<br />（1）悬浊液；2CaCO<SUB>3</SUB>+2SO<SUB>2</SUB>+O<SUB>2</SUB>═2CaSO<SUB>4</SUB>+2CO<SUB>2</SUB>；<br />（2）使原料充分反应，提高原料利用率；<br />（3）过滤；漏斗；<br />（4）常温下K<SUB>2</SUB>SO<SUB>4</SUB>的溶解度较小；<br />（5）降温结晶；<br />（6）CaCO<SUB>3</SUB>；','【分析】（1）根据“溶液”“悬浊液”或“乳浊液”的概念分析作答；根据CaCO<SUB>3</SUB>与SO<SUB>2</SUB>和空气中氧气反生成CaSO<SUB>4</SUB>和CO<SUB>2</SUB>的原理结合质量守恒定律分析作答；<br />（2）根据搅拌在化学反应中的作用分析作答；<br />（3）根据过滤的知识分析作答；<br />（4）根据反应Ⅱ中的反应物分析作答；<br />（5）根据NH<SUB>4</SUB>Cl在农业上可以用作化肥，及结合物质溶解性的特点分析作答；<br />（6）根据工艺流程中可以循环利用的物质情况分析作答．','书写',3.00,'0bcebe60e13233d8c0436ff4657a1a53',9,400,'过滤的原理、方法及其应用,悬浊液、乳浊液的概念及其与溶液的区别,物质的相互转化和制备,书写化学方程式、文字表达式、电离方程式','太仓市',2016,'32','2016•太仓市模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1845623,'<img  src=\"/tikuimages/9/2016/400/shoutiniao61/5b6b7f61-94d4-11e9-aa14-b42e9921e93e_xkb40.png\" style=\"vertical-align:middle;FLOAT:right\" />（2016•颍泉区模拟）如图图示中l与2是并列关系，3包含在2中，下列选项不正确的是（　　） <table class=\"edittable\"><TBODY><TR><td width=64>&nbsp;</TD><td width=142>&nbsp;A</TD><td width=123>&nbsp;B</TD><td width=123>&nbsp;C</TD><td width=139>&nbsp;D</TD></TR><TR><td>&nbsp;1</TD><td>&nbsp;饱和溶液</TD><td>&nbsp;纯净物</TD><td>&nbsp;单质</TD><td>&nbsp;微量元素</TD></TR><TR><td>&nbsp;2</TD><td>&nbsp;不饱和溶液</TD><td>&nbsp;混合物</TD><td>&nbsp;化合物</TD><td>&nbsp;常量元素</TD></TR><TR><td>&nbsp;3</TD><td>&nbsp;溶液</TD><td>&nbsp;合金</TD><td>&nbsp;酸</TD><td>&nbsp;钙</TD></TR></TBODY></TABLE>','A','B','C','D','','A','【解答】解：A、溶液分为饱和溶液和不饱和溶液，所以3包含着2，故A不正确；<br />B、物质分为纯净物和混合物，属于并列关系，合金中最少含有一种金属，属于混合物，故B正确；<br />C、纯净物分为单质和化合物，化合物分为酸、碱、盐，故C正确；<br />D、人体内元素分为微量元素和常量元素，钙属于常量元素，故D正确．<br />故选A．','【分析】A、溶液分为饱和溶液和不饱和溶液；B、物质分为纯净物和混合物，合金属于混合物；C、纯净物分为单质和化合物，化合物分为酸、碱、盐；D、人体内元素分为微量元素和常量元素，钙属于常量元素．','选择题',3.00,'a37438ff9d5ad7006f02849d24b1119e',9,400,'溶液的概念、组成及其特点,饱和溶液和不饱和溶液,合金与合金的性质,单质和化合物的概念,矿物质与微量元素','',2016,'32','2016•颍泉区模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1845672,'下列说法不正确的是（　　）','化学创造了众多的人造物质，使世界变得更加绚丽多彩','化学是在分子、原子层次上研究物质性质、组成、结构与变化规律的科学','道尔顿发现了元素周期律，使化学学习和研究变得有规律可循','原子论和分子学说的创立，奠定了近代化学的基础','','C','【解答】解：A、化学创造了众多的人造物质，使世界变得更加绚丽多彩，正确；<br />B、化学是在分子、原子层次上研究物质性质、组成、结构与变化规律的科学，正确；<br />C、1869年门捷列夫，编制了元素周期表，错误；<br />D、原子论和分子学说的创立，奠定了近代化学的基础，正确；<br />故选C','【分析】A、根据化学的用途知识判断．<br />B、根据化学的研究对象判断．<br />C、根据元素周期律是按照元素的内部结构特点和相似性质排列而成的，规律性强，是我们化学学习和研究的帮手判断．<br />D、根据原子论和分子学说判断．','选择题',3.00,'295cdc2cadbd809d3f21fea785ca29df',9,400,'化学的历史发展过程,化学的用途,空气组成的测定','峨眉山市',2016,'37','2016•峨眉山市二模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1845684,'对下列实验中出现的异常现象分析不合理的是（　　）','实验室取CO<SUB>2</SUB>气体时，很快不产生气泡了──使用了稀硫酸和石灰石反应','细铁丝在氧气中燃烧时，集气瓶底炸裂──集气瓶中未放入少量水','久置空气中的铝条放入稀盐酸中没有气泡冒出──铝的活动性在氢之后','验证空气中氧气含量时，进入集气瓶中的水少于五分之一体积──红磷过量','','C|D','【解答】解：A．稀硫酸与石灰石生成的硫酸钙是一种微溶物，会覆盖在石灰石的表面阻止反应的进一步进行，故分析合理；<br />B．细铁丝在氧气中燃烧时，为了防止高温生成物直接落在瓶底而使瓶底炸裂，需要在集气瓶底放少量水或铺一层细沙；故分析合理．<br />C．铝是一种活泼金属，与空气中的氧气反应生成的氧化铝具有致密结实的结构，对铝制品起到了保护作用，分析不合理；<br />D．测定空气中氧气的含量时，气体减小的体积小于1/5，可能是红磷的量不足，而非红磷过量，故分析不合理；<br />故选CD．','【分析】A．根据实验室制取二氧化碳的原理来分析；<br />B．根据细铁丝在氧气中燃烧实验时熔化的溅落物会炸裂瓶底进行分析判断．<br />C．根据金属的活动性来分析；<br />D．根据空气中氧气含量的测定方法来分析．','多选题',3.00,'c26dfe18ee07f02eaf3342d377ed6389',9,400,'空气组成的测定,氧气的化学性质,制取二氧化碳的操作步骤和注意点,金属的化学性质','',2015,'33','2015秋•滨州期末',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1845696,'水是生命之源，请回答下列有关水问题．<br />（1）水发生部分结冰变化后，形成的冰和水混合体系属于<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（选填“纯净物”或“混合物”）．<br />（2）生活中使硬水软化的一种最常用方法是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）若要测定某工业水样的酸碱度，最适宜使用<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（选填字母）．<br />A．无色酚酞度液　　　B．pH　试纸&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; C．紫色石蕊试液<br />（4）把下列物质分别加入纯净水中，用玻璃棒不断搅拌，能形成无色溶液的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（选填“字母”）<br />A．泥沙　　　　　　　　　B．无水硫酸铜　　　　　　　　　C．蔗糖　　　　　　　　D．花生油<br />（5）水草疯长，鱼虾几乎绝迹，主要原因是河水中N、P两种元素含量过高，水体富营养化，这种现象叫<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','纯净物$###$煮沸$###$B$###$C$###$水华','【解答】解：（1）水发生部分结冰变化后形成的冰和水混合体系中含有一种物质水，属于纯净物；<br />（2）生活中使硬水软化的一种最常用方法是煮沸；<br />（3）若要测定某工业水样的酸碱度，最适宜的是pH试纸，而酚酞试液和石蕊试液只能够测定溶液的酸碱性；<br />（4）把下列物质分别加入纯净水中，用玻璃棒不断搅拌，能形成无色溶液的是蔗糖，泥沙和花生油不溶于水，不能和水形成溶液，无水硫酸铜溶于水形成的硫酸铜溶液是蓝色的；<br />（5）水草疯长，鱼虾几乎绝迹，主要原因是河水中N、P两种元素含量过高，水体富营养化，这种现象叫水华；&nbsp;<br />故答案为：（1）纯净物；（2）煮沸；（3）B；（4）C；（5）水华．','【分析】（1）根据冰和水混合体系的组成分析；<br />（2）根据煮沸能使硬水软化进行分析；<br />（3）利用pH试纸可以测定溶液的酸碱度；<br />（4）蔗糖易溶于水，溶于水后形成无色溶液；<br />（5）水体富营养化，造成水华现象，是因为水中的氮元素、磷元素含量过高导致的．','填空题',3.00,'ef1a92fe1f43a0bcbafe7f633ee29415',9,400,'溶液的酸碱度测定,硬水与软水,溶液的概念、组成及其特点,纯净物和混合物的判别,富营养化污染与含磷洗衣粉的禁用','罗定市',2016,'35','2016春•罗定市期中',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1845823,'下列有关实验的评价正确的是（　　）','点燃某可燃物，在火焰上罩一个冷而干燥的烧杯，烧杯内壁有水雾出现，证明可燃物是H<SUB>2</SUB>','通过灼热的氧化铜后的气体不能使澄清石灰水变浑浊，证明原气体中没有一氧化碳','将活性炭放入紫色石蕊溶液中，紫色会褪去，这说明活性炭与石蕊发生反应的缘故','含二氧化碳、一氧化碳、氢气、水蒸气、氮气的混合气体，依次通过灼热的氧化铜、石灰水、干燥剂，最后只剩下氮气','','D','【解答】解：A、点燃某可燃物，在火焰上罩一个冷而干燥的烧杯，烧杯内壁有水雾出现，不能证明可燃物是氢气，因为甲烷燃烧也能够生成水蒸气，该选项说法不正确；<br />B、通过灼热的氧化铜后的气体不能使澄清石灰水变浑浊，不能证明原气体中没有一氧化碳，因为即使气体中含有一氧化碳，能和氧化铜反应生成二氧化碳，但是如果气体中含有氯化氢气体，则澄清石灰水不能变浑浊，该选项说法不正确；<br />C、将活性炭放入紫色石蕊溶液中，紫色会褪去，这说明活性炭具有吸附性，吸附了石蕊试液中的石蕊，该选项说法不正确；<br />D、氢气和氧化铜反应生成铜和水，石灰水能够吸收二氧化碳，干燥剂能够吸收水蒸气，含二氧化碳、一氧化碳、氢气、水蒸气、氮气的混合气体，依次通过灼热的氧化铜、石灰水、干燥剂，最后只剩下氮气，该选项说法正确．<br />故选：D．','【分析】氢气、甲烷等物质燃烧都能够生成水；<br />氧化铜能和一氧化碳反应生成铜和二氧化碳，二氧化碳能和氢氧化钙反应生成碳酸钙沉淀和水，稀盐酸能和碳酸钙反应生成氯化钙、水和二氧化碳；<br />活性炭具有吸附性，能够吸附色素和异味等；<br />氢气和氧化铜反应生成铜和水，石灰水能够吸收二氧化碳，干燥剂能够吸收水蒸气．','选择题',3.00,'b92daf7405f81054557291b90dad301f',9,400,'化学实验方案设计与评价,常见气体的检验与除杂方法,气体的净化（除杂）,碳单质的物理性质及用途','',0,'37','',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1845857,'下列实验操作、现象与结论对应关系正确的是（　　）<br /><table class=\"edittable\"><TBODY><TR><td width=45>选项</TD><td width=113>实验操作</TD><td width=170>现象</TD><td width=227>结论</TD></TR><TR><td>A</TD><td>涂有石灰水的烧杯罩在酒精灯火焰上方</TD><td>石灰水变浑浊</TD><td>酒精组成中一定含有碳元素和氧元素</TD></TR><TR><td>B</TD><td>向溶质质量分数为5%过氧化氢溶液中加入少量氧化铜</TD><td>有大量气泡产生</TD><td>氧化铜起催化作用</TD></TR><TR><td>C</TD><td>向某溶液中加入AgNO<SUB>3</SUB>溶液</TD><td>有白色沉淀生成</TD><td>该溶液中一定含有Cl<SUP>-</SUP></TD></TR><TR><td>D</TD><td>铝合金在铝片上刻划</TD><td>铝片上有明显划痕</TD><td>铝合金硬度比铝大</TD></TR></TBODY></TABLE>','A','B','C','D','','D','【解答】解：A、涂有石灰水的烧杯罩在酒精灯火焰上方，石灰水变浑浊，说明有二氧化碳生成，由质量守恒定律，反应前后元素种类不变，酒精组成中一定含有碳元素，可能含有氧元素，故选项实验操作、现象与结论对应关系错误．<br />B、向溶质质量分数为5%过氧化氢溶液中加入少量氧化铜，有大量气泡产生，可能是氧化铜加快了过氧化氢分解产生氧气的速率，氧化铜起催化作用；也可能是氧化铜与过氧化氢发生了化学反应，不能说明氧化铜起催化作用，故选项实验操作、现象与结论对应关系错误．<br />C、向某溶液中加入AgNO<SUB>3</SUB>溶液，有白色沉淀生成，该溶液中不一定含有Cl<SUP>-</SUP>，也可能含有碳酸根离子等，故选项实验操作、现象与结论对应关系错误．<br />D、铝合金在铝片上刻划，铝片上有明显划痕，说明铝合金硬度比铝大，故选项实验操作、现象与结论对应关系正确．<br />故选：D．','【分析】A、根据二氧化碳能使澄清的石灰水变浑浊，进行分析判断．<br />B、向溶质质量分数为5%过氧化氢溶液中加入少量氧化铜，有大量气泡产生，说明氧化铜加快了过氧化氢分解产生氧气的速率，进行分析判断．<br />C、根据AgNO<SUB>3</SUB>溶液能与盐酸盐、碳酸盐等反应生成白色沉淀，进行分析判断．<br />D、根据合金的性质，进行分析判断．','选择题',3.00,'a29d19386bd55e3861570ef814464596',9,400,'化学实验方案设计与评价,证明盐酸和可溶性盐酸盐,催化剂的特点与催化作用,合金与合金的性质,甲烷、乙醇等常见有机物的性质和用途','',0,'37','',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1845920,'如图是实验室用高锰酸钾制取氧气的装置．<br /><img  src=\"/tikuimages/9/0/400/shoutiniao26/5ecad700-94d4-11e9-9253-b42e9921e93e_xkb40.png\" style=\"vertical-align:middle\" /><br />（1）写出有标号的仪器名称：<br />①<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>②<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>③<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>④<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>⑤<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />（2）试管口塞一团棉花的作用<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）氧气可以用<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>收集，因为氧气不易溶于水；也可以用<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>法收集，因为氧气密度比空气略大，盛满氧气的集气瓶应<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>放在桌面上．检验氧气是否收集满的方法是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，证明已经收集满．<br />（4）实验室可以利用加热分解高锰酸钾的方法制氧气，该反应的文字表达式是：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>这种方法属于<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>变化．工业上采用<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>法制取氧气，这种方法属于<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>变化．<br />（5）在实验室制取氧气主要分为以下七步．<br />第①步：给试管加热<br />第②步：检查装置的气密性<br />第③步：用铁架台上的铁夹把试管固定在铁架台上<br />第④步：将高锰酸钾放入试管中，管口塞一团棉花，用带导管的塞子塞紧&nbsp;<br />第⑤步：用排水集气法收集一瓶氧气<br />第⑥步：熄灭酒精灯<br />第⑦步：将导管从水槽内拿出来<br />正确的操作顺序是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．第⑥、⑦步如果操作错误会造成什么后果<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','铁架台$###$集气瓶$###$水槽$###$酒精灯$###$试管$###$防止加热时高锰酸钾粉末进入导管$###$排水法$###$向上排空气$###$正$###$将带火星的木条放于集气瓶口，若木条复燃$###$高锰酸钾<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td style=\"font-size: 90%\">加热</td></tr><tr><td><div hassize=\"9\"><div style=\"background: url(\'/upimages/images/part/8594L.png\') repeat-x; width:1px; height:5px;float:left;overflow:hidden\" muststretch=\"h\"></div><div style=\"background: url(\'/upimages/images/part/8594.png\') no-repeat; width:9px; height:5px;float:left;overflow:hidden\"></div></div></td></tr></table></span>锰酸钾+二氧化锰+氧气$###$化学$###$分离液态空气$###$物理$###$②④③①⑤⑦⑥$###$水倒吸入热的试管，引起试管炸裂','【解答】解：（1）图中标号仪器分别是铁架台、集气瓶、水槽、酒精灯和试管；<br />（2）加热高锰酸钾试管口放棉花，是为了防止加热时高锰酸钾粉末进入导管；<br />（3）氧气不易溶于水，所以可用排水法收集，密度比空气略大，所以可用向上排空气法收集、且应正放在桌面上，防止气体逸出；验满氧气的方法是将带火星的木条放于集气瓶口，若木条复燃则满了；<br />（4）加热高锰酸钾生成锰酸钾、二氧化锰和氧气，反应的文字表达式是：高锰酸钾<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td style=\"font-size: 90%\">加热</td></tr><tr><td><div hassize=\"9\"><div style=\"background: url(\'/upimages/images/part/8594L.png\') repeat-x; width:1px; height:5px;float:left;overflow:hidden\" muststretch=\"h\"></div><div style=\"background: url(\'/upimages/images/part/8594.png\') no-repeat; width:9px; height:5px;float:left;overflow:hidden\"></div></div></td></tr></table></span>锰酸钾+二氧化锰+氧气，该反应有新物质生成，属于化学变化，工业上制取氧气是利用氧气和氮气的沸点不同，将液态空气分离，无新物质生成，属于物理变化；<br />（5）实验室制取氧气的操作步骤可简记为‘查装定点收移熄’，所以正确的操作顺序是②④③①⑤⑦⑥；实验完毕应先移导管后熄灯，若第⑥、⑦步颠倒，则会导致试管内温度降低、气压减小，水槽中的水倒吸入热的试管，引起试管炸裂；<br />故答案为：（1）铁架台；集气瓶；水槽；酒精灯；试管；<br />（2）防止加热时高锰酸钾粉末进入导管；<br />（3）排水法；向上排空气；正；将带火星的木条放于集气瓶口，若木条复燃；<br />（4）高锰酸钾<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td style=\"font-size: 90%\">加热</td></tr><tr><td><div hassize=\"9\"><div style=\"background: url(\'/upimages/images/part/8594L.png\') repeat-x; width:1px; height:5px;float:left;overflow:hidden\" muststretch=\"h\"></div><div style=\"background: url(\'/upimages/images/part/8594.png\') no-repeat; width:9px; height:5px;float:left;overflow:hidden\"></div></div></td></tr></table></span>锰酸钾+二氧化锰+氧气；化学；分离液态空气；物理；<br />（5）②④③①⑤⑦⑥；水倒吸入热的试管，引起试管炸裂．','【分析】（1）据常用仪器回答；<br />（2）加热高锰酸钾试管口放棉花，是为了防止加热时高锰酸钾粉末进入导管；<br />（3）据氧气的密度和溶解性确定收集方法，据氧气的助燃性验满；<br />（4）据反应原理书写文字表达式，并据有无新物质生成判断是物理变化，还是化学变化；<br />（5）实验室制取氧气的操作步骤可简记为‘查装定点收移熄’，并结合实验注意事项解答．','书写',3.00,'c24a4dc5fe7d2993097031535c93106d',9,400,'氧气的工业制法,氧气的制取装置,氧气的收集方法,氧气的检验和验满,制取氧气的操作步骤和注意点,书写化学方程式、文字表达式、电离方程式','',0,'37','',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1846164,'下列说法不正确的是（　　）','按质量计算空气中含氮气78%，氧气21%','因为氮气不支持燃烧，所以可以用作粮食的保护气','工业生产排放到空气中有害气体和烟尘会污染空气','从液态空气中可分离出氧气、氮气等气体','','A','【解答】解：A、按质量计算空气中含氮气78%，氧气21%错误，应该是按体积分数，错误符合题意，故选项正确；<br />B、氮气的用途是：制造氮肥、氮气充入食品包装袋内可以防腐做保护气、液态氮可以做制冷剂，正确但不符合题意，故选项错误；<br />C、工业生产排放到空气中有害气体和烟尘会污染空气正确，正确但不符合题意，故选项错误；<br />D、从液态空气中可分离出氧气、氮气等气体，是根据各成分沸点的不同而分离的，正确但不符合题意，故选项错误；<br />故选A','【分析】空气中各成分的体积分数分别是：氮气大约占78%、氧气大约占21%、稀有气体大约占0.94%、二氧化碳大约占0.03%、水蒸气和其它气体和杂质大约占0.03%；空气的成分主要以氮气和氧气为主，氧气约占五分之一，氮气约占五分之四．氮气的用途是：制造氮肥、氮气充入食品包装袋内可以防腐做保护气、液态氮可以做制冷剂；工业生产排放到空气中有害气体和烟尘会污染空气正确；从液态空气中可分离出氧气、氮气等气体，是根据各成分沸点的不同而分离的．','选择题',2.00,'d77f1d94134b954a7758f6092d6da325',9,400,'空气的成分及各成分的体积分数,空气的污染及其危害,氧气的工业制法,常见气体的用途','',0,'37','',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1846211,'下列实验不能达到目的是（　　）','<img  src=\"/tikuimages/9/2016/400/shoutiniao18/6186748f-94d4-11e9-8dcd-b42e9921e93e_xkb45.png\" style=\"vertical-align:middle\" /><br />&nbsp;检查装置气密性','<img  src=\"/tikuimages/9/2016/400/shoutiniao72/6188705e-94d4-11e9-bd47-b42e9921e93e_xkb83.png\" style=\"vertical-align:middle\" /><br />&nbsp; 比较物质着火点','<img  src=\"/tikuimages/9/2016/400/shoutiniao61/6189cff0-94d4-11e9-918c-b42e9921e93e_xkb46.png\" style=\"vertical-align:middle\" /><br />&nbsp;监控O<SUB>2</SUB>流速','<img  src=\"/tikuimages/9/2016/400/shoutiniao72/618c6800-94d4-11e9-879c-b42e9921e93e_xkb8.png\" style=\"vertical-align:middle\" /><br />&nbsp;&nbsp;&nbsp; 探究微粒运动','','C','【解答】解：A、推注射器使瓶内压强增大，长颈漏斗内会上升一段液柱，如果一段时间后液柱不再下降，说明装置不漏气，故A能达到实验目的；<br />B、煤块和塑料燃烧的难易程度不同，所以能验证不同物质的着火点不同，故B能达到实验目的；<br />C、氧气不易溶于水，进入水中会迅速溢出，所以通过气泡冒出的速度不同，来监控气体流速，所以要监控气体流速，应長进短出，故C不能达到实验目的．<br />D、浓氨水易挥发，氨气溶于水其溶液为碱性，则实验现象：A烧杯中的无色酞酚溶液变红；说明分子是在不断的运动的，故D能达到实验目的；<br />故选C．','【分析】A、根据推注射器使瓶内压强增大，长颈漏斗内会上升一段液柱考虑；<br />B、根据煤块和塑料的着火点不同考虑；<br />C、根据氧气不易溶于水考虑；<br />D、根据浓氨水易挥发，氨气溶于水其溶液为碱性，则根据实验现象分析．','选择题',3.00,'c4b9e4105799ab882de83ef3cd0b949c',9,400,'分离物质的仪器,检查装置的气密性,分子的定义与分子的特性,燃烧与燃烧的条件','丹阳市',2016,'32','2016•丹阳市模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1846220,'<img  src=\"/tikuimages/9/2011/400/shoutiniao64/61a98cee-94d4-11e9-8e04-b42e9921e93e_xkb84.png\" style=\"vertical-align:middle;FLOAT:right;\" />1768年拉瓦锡对当时人们认为水能变成土的观念发出挑战．他将一定量的蒸馏水加入蒸馏器中，反复加热101天，发现蒸馏器中产生少量沉淀，称得整个蒸馏装置的总质量没变，水的质量也没变，沉淀的质量等于蒸馏器减少的质量．下列的说法错误的是（　　）','水在长时间加热后能变成土','物质变化过程中总质量不变','精确称量是科学探究的重要方法','化学是通过实验探究成长起来的','','A','【解答】解：根据质量守衡定律可知由于水与土组成元素种类不同，故水变土的说法一定错误，从题干中得到的信息可知精确测量是研究问题很重要的方法，从质量关系上看整个变化过程质量未发生改变，且蒸馏器所减少的质量与生成的沉淀量相等，可知沉淀物来自于蒸馏器本身．<br />故选A．','【分析】精确称量是科学研究的要求之一，在整个过程中物质物质的质量没有增减，沉淀不是水产生的而是来自于蒸馏器本身，从而证明了水并没有变成土．','选择题',3.00,'f30522da1afa52c9dd6d9d82038056b8',9,400,'科学探究的基本方法,质量守恒定律及其应用','',2011,'37','2011秋•马鞍山月考',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1846242,'以下有关物质的说法中不正确的是（　　）','一氧化碳有毒','大理石的主要成分为碳酸钙','干冰可制作冷剂','植物光合作用吸收氧气，放出二氧化碳','','D','【解答】解：A、一氧化碳有毒，正确；<br />B、大理石的主要成分为碳酸钙，正确；<br />C、干冰升华吸收热量，可制作冷剂，正确；<br />D、植物的光合作用能吸收二氧化碳和释放氧气，从而维持大气中的氧气和二氧化碳的含量相对稳定，错误；<br />故选D','【分析】根据空气中各成分的性质和用途进行分析判断即可．','选择题',3.00,'49994a832b290669cd0057dc72f9f0f5',9,400,'二氧化碳的用途,一氧化碳的毒性,常用盐的用途,光合作用与呼吸作用','',2016,'37','2016•官渡区一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1846272,'下列说法不正确的是（　　）','拉瓦锡最先用试验的方法证明空气主要是由氮气和氧气组成的','因为人类活动加剧，空气中的氧气会越来越少、二氧化碳会越来越多','污染空气的主要物质是粉尘以及CO、SO<SUB>2</SUB>等有毒气体','露天焚烧稻草、桔杆、废旧塑料、橡胶等都会造成空气污染','','B','【解答】解：A、二百多年前，法国化学家拉瓦锡以天平作为研究工具，用定量的方法研究了空气的成分，首先提出了空气是由氮气和氧气组成的结论，故选项说法正确；<br />B、氧气和二氧化碳通过绿色植物，在自然界中进行循环，不会由于人口的增长，氧气会越来越少，二氧化碳会越来越多，故选项说法错误；<br />C、造成空气污染的主要物质是CO、SO<SUB>2</SUB>、NO<SUB>2</SUB>等有毒气体和粉尘，故选项说法正确；<br />D、露天焚烧稻草、桔杆、废旧塑料、橡胶等都会造成空气污染，故选项说法正确；<br />故选B．','【分析】A、根据拉瓦锡是最早提出空气由氮气和氧气的进行分析判断．<br />B、根据自然界的氧和二氧化碳循环进行分析判断．<br />C、根据空气污染物进行分析判断．<br />D、露天焚烧稻草、桔杆、废旧塑料、橡胶等的危害进行分析判断．','选择题',3.00,'c2c62c99722a25a2b8c905c67abeae17',9,400,'空气组成的测定,空气的污染及其危害,自然界中的氧循环','峨眉山市',2016,'37','2016•峨眉山市二模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1846299,'取用药品遵守“三不”原则：不能<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，不能<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，不要<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','用手接触药品$###$把鼻孔凑到容器口去闻药品的气味$###$尝任何药品的味道','【解答】解：实验室里所用的药品，很多是易燃、易爆、有腐蚀性或有毒的．在使用药品时为保证安全，要做到“三不”即：不能用手接触药品；不能把鼻孔凑到容器口去闻药品的气味；不能尝任何药品的味道．<br />故答案为：用手接触药品；把鼻孔凑到容器口去闻药品的气味；尝任何药品的味道．','【分析】根据实验室药品取用的“三不”原则，进行分析解答即可．','填空题',3.00,'deb58acf091c03722a8401ba5ff52bd2',9,400,'实验操作注意事项的探究','',0,'37','',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1846310,'<img  src=\"/tikuimages/9/2016/400/shoutiniao92/62a5520f-94d4-11e9-aa2c-b42e9921e93e_xkb88.png\" style=\"vertical-align:middle;FLOAT:right\" />（2016•鄞州区一模）如图所示，实验室模拟粉尘爆炸试验．连接好装置，在小塑料瓶中放入下列干燥的粉末，点燃蜡烛，快速鼓入大量的空气，肯定观察不到爆炸现象的是（　　）','镁粉','面粉','煤粉','大理石粉','','D','【解答】解：可燃性物质在有限的空间内与氧气接触发生剧烈燃烧，体积迅速膨胀，容易发生爆炸，面粉、煤粉、镁粉都能燃烧，都具有可燃性，所以都容易发生爆炸，大理石不能燃烧，不可能发生爆炸．<br />故选D．','【分析】可燃性物质在有限的空间内与氧气接触发生剧烈燃烧容易发生爆炸进行分析．','选择题',3.00,'0ff63c7d1f67f0eaf377c06493b88782',9,400,'燃烧和爆炸实验','',2016,'37','2016•鄞州区一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1846354,'实验室取用药品要注意节约，液体药品一般取<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>&nbsp;固体药品<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','1～2mL$###$只需盖满试管底部','【解答】解：为节约药品，根据实验室药品取用规则：取用化学药品时，如果没有说明用量，一般应按最少量（1～2mL）取用液体，固体只需盖满试管底部．<br />故答案为：1～2mL；只需盖满试管底部．','【分析】根据实验室药品取用的用量原则，要遵循节约的原则，所以取用时要尽量少用，在不影响实验结果的前提下，达到节约的目的．','填空题',3.00,'4b30505b55ac59e1992331066cfb5c3d',9,400,'实验操作注意事项的探究','',0,'37','',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1846507,'自2014年兰州市“4•11”自来水中有毒物质苯（C<SUB>6</SUB>H<SUB>6</SUB>）超标事件以来，人们开始更加科学的关注自来水饮用安全问题．自来水厂净水过程的主要操作流程如图：<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao78/64b6f400-94d4-11e9-9460-b42e9921e93e_xkb73.png\" style=\"vertical-align:middle\" /><br />友情提示：常用的絮凝剂是明矾[KAl（SO<SUB>4</SUB>）<SUB>2</SUB>•12H<SUB>2</SUB>O]，消毒剂中有液氯（Cl<SUB>2</SUB>）．试回答下列问题：<br />（1）上述过程中活性炭的作用X&nbsp;是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>作用．<br />（2）在氯气、苯（C<SUB>6</SUB>H<SUB>6</SUB>）和活性炭三种物质中，属于有机物的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）絮凝剂明矾[KAl（SO<SUB>4</SUB>）•12H<SUB>2</SUB>O]中，所含非金属元素共有<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>种．<br />（4）家用净水器能够利用膜法去除水中的重金属等杂质离子，吸附法吸收多余的氯，达到进一步净化饮用水的目的．&nbsp;若利用化学实验方法检测净化后的水中是否含有氯离子，下列试剂合理的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />A．烧碱&nbsp;&nbsp;B．酚酞试液&nbsp;&nbsp;C．硝酸银溶液&nbsp;&nbsp;&nbsp;D．生石灰．','','','','','','吸附$###$苯$###$3$###$C','【解答】解：（1）由于活性炭具有吸附性，在净水过程中活性炭的作用X是吸附作用．<br />（2）苯（C<SUB>6</SUB>H<SUB>6</SUB>）是含有碳元素化合物，属于有机物．<br />（3）絮凝剂明矾[KAl（SO<SUB>4</SUB>）•12H<SUB>2</SUB>O]中，所含非金属元素共有硫、氧、氢3种．<br />（4）由于硝酸银中的银离子能与氯离子结合生成了氯化银沉淀，该沉淀不溶于稀硝酸，常用硝酸银来检验净化后的水中是否含有氯离子．<br />故答为：（1）吸附；（2）苯；&nbsp;（3）3；（4）C．','【分析】（1）根据活性炭具有吸附性分析．<br />（2）根据物质的组成分析物质的类别．<br />（3）根据絮凝剂明矾[KAl（SO<SUB>4</SUB>）•12H<SUB>2</SUB>O]组成元素分析判断．<br />（4）根据氯离子的检验方法分析判断．','填空题',3.00,'7fcee9c87f400977e6c6882d22b952f7',9,400,'证明盐酸和可溶性盐酸盐,自来水的生产过程与净化方法,有机物与无机物的区别,元素的简单分类','',2016,'37','2016•槐荫区二模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1846528,'（1）图甲是镁原子和氧原子形成氧化镁的示意图<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao24/64e1d491-94d4-11e9-a09d-b42e9921e93e_xkb63.png\" style=\"vertical-align:middle\" /><br />①从得失氧的角度看，镁与氧气的反应属于<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>&nbsp;反应；从得失电子的角度看，反应中镁原子<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“得到”或“失去”）2个电子，形成相对稳定结构．<br />②由图甲可知，元素的原子得到电子后，其化合价将<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“升高”或“降低”）．<br />（2）图乙是NaCl与AgNO<SUB>3</SUB>两溶液反应的示意图．<br />①图中NO<SUB>3</SUB><SUP>-</SUP>的名称是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />②该反应的本质是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>结合生成沉淀．<br />③请再用AgNO<SUB>3</SUB>作反应物之一，写出另一个与上述反应本质相同的化学方程式：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','氧化$###$失去$###$降低$###$硝酸根离子$###$银离子与氯离子结合$###$HCl+AgNO<SUB>3</SUB>=AgCl↓+HNO<SUB>3</SUB>','【解答】解：（1）①从得失氧的角度看，镁与氧气的反应有氧的得失，属于氧化反应；从得失电子的角度看，反应中镁原子失去2个电子，形成相对稳定结构．<br />②由图甲可知，元素的原子得到电子后，其化合价将降低．<br />（2）①图中NO<SUB>3</SUB><SUP>-</SUP>的名称是硝酸根离子；<br />②通过分析微观图可知，该反应的实质是氯离子和银离子结合生成沉淀；<br />③盐酸和硝酸银反应生成氯化银沉淀和硝酸，化学方程式为：HCl+AgNO<SUB>3</SUB>=AgCl↓+HNO<SUB>3</SUB>．<br />故答案为：（1）①氧化，失去；②降低；<br />（2）①硝酸根离子；②氯离子和银离子；③HCl+AgNO<SUB>3</SUB>=AgCl↓+HNO<SUB>3</SUB>．','【分析】（1）①根据氧和电子的得失进行分析；<br />②根据化合价的变化分析；<br />（2）①根据常见离子的名称进行分析；<br />②根据复分解反应的实质进行分析；<br />③根据常见化学方程式的书写规则进行分析．','书写',3.00,'cb640c8961fc7fe8e471c32b78dab7d1',9,400,'原子结构示意图与离子结构示意图,复分解反应的条件与实质,书写化学方程式、文字表达式、电离方程式','',2016,'32','2016•锦江区模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1846542,'<img  src=\"/tikuimages/9/2016/400/shoutiniao65/650a9240-94d4-11e9-a70b-b42e9921e93e_xkb75.png\" style=\"vertical-align:middle;FLOAT:right\" />下列选项符合如图从属关系的是（　　）<br /><table class=\"edittable\"><TBODY><TR><td width=40></TD><td width=113>X</TD><td width=113>Y</TD><td width=113>Z</TD></TR><TR><td>A</TD><td>碱</TD><td>酸</TD><td>化合物</TD></TR><TR><td>B</TD><td>非金属单质</TD><td>金属元素</TD><td>元素</TD></TR><TR><td>C</TD><td>化合反应</TD><td>氧化反应</TD><td>化学反应</TD></TR><TR><td>D</TD><td>饱和溶液</TD><td>不饱和溶液</TD><td>浓溶液</TD></TR></TBODY></TABLE>','A','B','C','D','','A','【解答】解：A、酸是电离出的阳离子全部是氢离子的化合物；碱是电离出的阴离子全部是氢氧根离子的化合物，因此酸、碱都属于化合物，且酸和碱属于并列关系，故选项正确；<br />B、元素包括金属元素和非金属元素，而不是非金属单质，故选项错误；<br />C、化合反应和氧化反应都属于化学反应，它们是交叉关系，不是并列关系，故选项错误；<br />D、饱和溶液与不饱和溶液属于并列关系，但它们与浓溶液不是从属关系，故选项错误；<br />故选A．','【分析】A、根据酸和碱都是化合物进行解答；<br />B、根据元素包括金属元素和非金属元素进行解答；<br />C、根据化合反应和氧化反应都属于化学反应，它们是交叉关系进行解答；<br />D、根据饱和溶液与不饱和溶液属于并列关系，但它们与浓溶液不是从属关系进行解答．','选择题',3.00,'a302918593cdd8b04364e8bdfd4cc0e4',9,400,'饱和溶液和不饱和溶液,浓溶液、稀溶液跟饱和溶液、不饱和溶液的关系,氧化物、酸、碱和盐的概念,元素的简单分类,化合反应及其应用,氧化反应','',2016,'37','2016•高港区一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1846599,'有关水的说法正确的是（　　）','水变成水蒸气，水分子的体积变大','电解水实验中，原子没有发生改变','水分子是由氢分子和氧原子构成的','水具有分散性因此沿海城市的昼夜温差较小','','B','【解答】解：A．水转化为水蒸气，水分子间隔变大了，分子体积没有改变，故错误；<br />B．原子是化学变化中的最小粒子，所以在电解水的过程中，原子的种类没有改变，故正确；<br />C．水分子是由氢原子和氧原子构成的，故错误；<br />D．水具有分散性与温差大小无关，故错误．<br />故选B．','【分析】A．根据水的三态变化来分析；<br />B．根据化学变化的实质来分析；<br />C．根据分子结构来分析；<br />D．根据水具有分散性的原因来分析．','选择题',3.00,'ebcc5c672b96a4a62cee0311b6928d95',9,400,'电解水实验,物质的三态及其转化,分子的定义与分子的特性','',2016,'37','2016•浦东新区二模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1846708,'下列问题的研究中，未利用对比实验思想方法的是（　　）','<img  src=\"/tikuimages/9/2016/400/shoutiniao27/66bab340-94d4-11e9-85cb-b42e9921e93e_xkb67.png\" style=\"vertical-align:middle\" /><br />MnO<SUB>2</SUB>的催化作用','<img  src=\"/tikuimages/9/2016/400/shoutiniao22/66bc12cf-94d4-11e9-b335-b42e9921e93e_xkb19.png\" style=\"vertical-align:middle\" /><br />碘的溶解性','<img  src=\"/tikuimages/9/2016/400/shoutiniao64/66beaae1-94d4-11e9-ba4c-b42e9921e93e_xkb61.png\" style=\"vertical-align:middle\" /><br />白磷红磷着火点','<img  src=\"/tikuimages/9/2016/400/shoutiniao24/66c11bde-94d4-11e9-bf94-b42e9921e93e_xkb77.png\" style=\"vertical-align:middle\" /><br />水的组成','','D','【解答】解：A、研究催化剂影响化学反应速度，在双氧水浓度相同的情况下，利用有无催化剂的情况下，观察氧气放出速率的快慢，所以利用了对比实验的原理，故A不符合题意；<br />B、向水、汽油中分别加入适量的碘时，发现碘不溶于水，易溶于汽油，通过对比说明溶剂是影响溶解性的一个重要因素，该实验利用了对比实验的思想方法，故B不符合题意；<br />C、铜片上的白磷与铜片上的红磷属于对照实验，都与氧气接触，都属于可燃物，变量是着火点不同，白磷着火点低，红磷着火点高，热水能达到白磷的着火点，达不到红磷的着火点，从而证明可燃物燃烧温度必须达到可燃物的着火点，所以属于对比实验，故C不符合题意；<br />D、水通电分解正极生成氧气，负极生成氢气，从而得出水的组成元素，所以根本不存在对照实验，故D符合题意．<br />故选D．','【分析】根据对比实验是指只允许一种变量，其它条件相同的实验进行解答．','选择题',3.00,'77f2be3747dc0e58fea0281d61a6551a',9,400,'科学探究的基本方法','',2016,'37','2016•陕西校级四模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1846720,'下列分离混合物的方法中，所依据的原理不正确的是（　　）','分离液态空气制取氧气--利用氮气和氧气的密度不同','氢氧化铝能中和过多的胃酸--利用了氢氧化铝的碱性','分离硝酸钾和氯化纳组成的混合物--利用两者溶解度受温度影响不同','分离氧气后剩余固体中的二氧化锰和氯化钾--利用二氧化锰和氯化钾的溶解性不同','','A','【解答】解：A、分离液态空气制取氧气，是利用了液态氮和液态氧沸点的不同，故选项说法错误．<br />B、氢氧化铝能中和过多的胃酸，是利用了氢氧化铝的碱性，能与酸发生中和反应，故选项说法正确．<br />C、分离硝酸钾和氧化纳组成的混合物，是利用了硝酸钾溶解度受温度影响较大，而氯化钠受温度影响较小，故选项说法正确．<br />D、氯化钾易溶于水，二氧化锰难溶于水，分离氧气后剩余固体中的二氧化锰和氯化钾，是利用了二氧化锰和氯化钾的溶解性不同，故选项说法正确．<br />故选：A．','【分析】A、根据分离液态空气制取氧气的原理，进行分析判断．<br />B、根据中和反应的应用，进行分析判断．<br />C、根据氯化钠与硝酸钾的溶解度受温度影响的不同，进行分析判断．<br />D、根据氯化钾易溶于水，二氧化锰难溶于水，进行分析判断．','选择题',3.00,'32d537400151e4b0c028316494ef05bc',9,400,'混合物的分离方法,氧气的工业制法,中和反应及其应用','',2016,'37','2016•呼和浩特一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1846725,'<img  src=\"/tikuimages/9/2016/400/shoutiniao95/66fc7730-94d4-11e9-894b-b42e9921e93e_xkb27.png\" style=\"vertical-align:middle;FLOAT:right\" />（2016春•龙口市期中）如图有多种功能，如手机气体，洗涤气体等．回答下列问题：<br />（1）用排空气法收集气体，若气体从b端进入，该气体必须具有的性质是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．（填“密度比空气大”或“密度比空气小”）<br />（2）装置中预先盛满水，将水排出，气体从<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>端进入．（填“a”或“b”）<br />（3）医院里给病人输氧，也可以在氧气瓶和病人之间连接该装置，在集气瓶装半瓶水，将<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>端接病人．（填“a”或“b”）','','','','','','密度比空气小$###$b$###$b','【解答】解：（1）收集气体从b端进入集气，相当于向下排空气收集气体，该气体必须具有的性质是密度比空气小；故填：密度比空气小；<br />（2）水的密度比气体的密度大，所以装置中预先盛满水，将水排出贮气，气体从b端进入；故填：b；<br />（3）在医院给病人输氧气时，也利用了类似的装置，并在装置中盛放大约半瓶蒸馏水．此时，该装置能起到的作用有观察是否有氧气输出．故填：b．','【分析】收集方法的选择是根据：难溶于水或不易溶于水用排水法收集，密度比空气大用向上排空气法收集，密度比空气小用向下排空气法收集；了解用此装置排空气法、排水集气法收集气体时，要收集的气体应从那端口通入．','填空题',3.00,'37a00813c70b02f231922057eeccb93b',9,400,'分离物质的仪器','龙口市',2016,'35','2016春•龙口市期中',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1846731,'阅读下面科普短文．<br />&nbsp;&nbsp;&nbsp; 食醋中一般含有3%～5%的醋酸，醋酸的化学名称叫乙酸（CH<SUB>3</SUB>COOH），是无色有刺激性气味的液体，能溶于水．<br />&nbsp;&nbsp; 食醋可以除去水壶内的水垢，水垢的主要成分是碳酸钙．除水垢时，可在水壶中加入水，倒入适量醋，浸泡一段时间，不溶于水的碳酸钙会转变成可溶于水的醋酸钙而被除掉，化学方程式为：CaCO<SUB>3</SUB>+2CH<SUB>3</SUB>COOH═Ca&nbsp;（CH<SUB>3</SUB>COO）<SUB>2</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑．在熬制骨头汤时，常常在汤中加入少量食醋，因为骨头中含有不溶于水的磷酸钙，当磷酸钙与醋酸作用时生成可溶于水的磷酸二氢钙，能够增加汤内的含钙量，促进人体对钙、磷的吸收和利用．<br />&nbsp;&nbsp; 食醋不仅在厨房中大显身手，还是一种杀菌剂，冬天在屋子里熬醋可以杀灭细菌，对抗感冒有很大作用．饮酒过量的人可以用食醋来解酒，因为乙酸能跟乙醇发生酯化反应生成乙酸乙酯（CH<SUB>3</SUB>COOC<SUB>2</SUB>H<SUB>5</SUB>）和水，从而达到解酒的目的．<br />&nbsp;&nbsp; 由于醋酸能与活泼的金属发生置换反应产生氢气，所以家用铝制品不能用来盛放食醋，以免被腐蚀．<br />根据文章内容，回答下列问题：<br />（1）乙酸属于<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“有机物”或“无机物”）．<br />（2）乙酸的物理性质<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）食醋能用来解酒的原因是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（4）用食醋除水垢，该反应属于基本反应类型中的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>反应．<br />（5）用铁锅炒菜时，放一点食醋能补铁的原因是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．（可用化学方程式表示）','','','','','','有机物$###$无色有刺激性气味液体，能溶于水$###$乙酸能跟乙醇发生酯化反应生成乙酸乙酯和水，从而达到解酒的目的$###$复分解$###$2CH<SUB>3</SUB>COOH+Fe═Fe（CH<SUB>3</SUB>COO）<SUB>2</SUB>+H<SUB>2</SUB>↑','【解答】解：（1）乙酸是含碳元素的化合物，属于有机物．<br />（2）由新信息可知，醋酸是无色有刺激性气味液体，能溶于水，不需要通过化学变化就能表现出来，属于物理性质．<br />（3）饮酒过量的人可以用食醋来解酒，因为乙酸能跟乙醇发生酯化反应生成乙酸乙酯（CH<SUB>3</SUB>COOC<SUB>2</SUB>H<SUB>5</SUB>）和水，从而达到解酒的目的．<br />（4）食醋的主要成分是醋酸，水垢的主要成分是碳酸钙．其反应的化学方程式为：化学方程式为：CaCO<SUB>3</SUB>+2CH<SUB>3</SUB>COOH═Ca（CH<SUB>3</SUB>COO）<SUB>2</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑；是由两种化合物相互交换成分，生成另外丙种化合物的反应，所以属于复分解反应；<br />（5）铁的金属活动性比氢强，能与醋酸反应生成醋酸亚铁和氢气，反应的化学方程式为：2CH<SUB>3</SUB>COOH+Fe=（CH<SUB>3</SUB>COO）<SUB>2</SUB>Fe+H<SUB>2</SUB>↑．<br />故答案为：（1）有机物<br />（2）无色有刺激性气味液体，能溶于水<br />（3）乙酸能跟乙醇发生酯化反应生成乙酸乙酯和水，从而达到解酒的目的<br />（4）复分解反应<br />（5）2CH<SUB>3</SUB>COOH+Fe═Fe&nbsp;（CH<SUB>3</SUB>COO）<SUB>2</SUB>+H<SUB>2</SUB>↑或铁能与食醋中的醋酸反应生成可被人体吸收的亚铁盐','【分析】（1）含有碳元素的化合物叫有机化合物，简称有机物．<br />（2）物理性质是不需要发生化学变化就能表现出来的性质，包括物质的颜色、状态、气味、熔点、沸点、密度、溶解性等．<br />（3）根据题意，饮酒过量的人可以用食醋来解酒，因为乙酸能跟乙醇发生酯化反应生成乙酸乙酯（CH<SUB>3</SUB>COOC<SUB>2</SUB>H<SUB>5</SUB>）和水，从而达到解酒的目的．<br />（4）反应类型的判定主要是指四大基本反应类型（即化合反应、分解反应、置换反应和复分解反应）的判定．判定的方法就是结合它们的概念，根据它们各自的特点来细心地判定．<br />（5）铁的金属活动性比氢强，能与醋酸反应生成醋酸亚铁和氢气，写出反应的化学方程式即可．','书写',3.00,'8721b1bef941394112cb58a1b8153a9d',9,400,'醋酸的性质及醋酸的含量测定,有机物与无机物的区别,反应类型的判定,书写化学方程式、文字表达式、电离方程式','',2016,'37','2016•门头沟区一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1846741,'已知碳12原子的质量为akg，A原子的质量为bkg，A原子的中子数为c，则A原子的核外电子数为（　　）','<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">ab</td></tr><tr><td>12</td></tr></table></span>-c','<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">12b</td></tr><tr><td>a</td></tr></table></span>+c','<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">12a</td></tr><tr><td>b</td></tr></table></span>+c','<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">12b</td></tr><tr><td>a</td></tr></table></span>-c','','D','【解答】解：已知碳12原子的质量为akg，A原子的质量为bkg，则A的相对原子质量为<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">bkg</td></tr><tr><td style=\"padding-top:1px;font-size:90%\">akg×<table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">1</td></tr><tr><td>12</td></tr></table></td></tr></table></span>=<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">12b</td></tr><tr><td>a</td></tr></table></span>；<br />相对原子质量=质子数+中子数，A原子的中子数为c，则该原子的质子数为<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">12b</td></tr><tr><td>a</td></tr></table></span>-c．<br />原子中：核电荷数=核内质子数=核外电子数，则A原子的核外电子数为<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">12b</td></tr><tr><td>a</td></tr></table></span>-c．<br />故选：D．','【分析】根据某原子的相对原子质量=<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">该元素的一个原子的质量</td></tr><tr><td style=\"padding-top:1px;font-size:90%\">一种碳原子质量×<table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">1</td></tr><tr><td>12</td></tr></table></td></tr></table></span>，相对原子质量=质子数+中子数，原子中：核电荷数=核内质子数=核外电子数，进行分析解答即可．','选择题',3.00,'fff68fa9f7cb62a4592e46e6105ddf04',9,400,'原子的有关数量计算,相对原子质量的概念及其计算方法','',2016,'32','2016•邻水县模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1846765,'下列实验现象描述正确的是（　　）','红磷在空气中燃烧产生大量白雾','铝片在纯氧中燃烧生成三氧化二铝','铁生锈后产生一种疏松多孔的红棕色固体','碱式碳酸铜粉末与稀盐酸反应，绿色溶液变成蓝色','','C','【解答】解：A、红磷在空气中燃烧，产生大量的白烟，而不是白色烟雾，故选项说法错误．<br />B、铝片在纯氧中燃烧生成三氧化二铝是实验结论，而非实验现象，故选项说法错误．<br />C、铁丝生锈后产生的铁锈是一种疏松多孔的红棕色物质，故选项说法正确．<br />D、碱式碳酸铜粉末与稀盐酸反应生成氯化铜、水和二氧化碳，会观察到产生大量气泡，绿色粉末逐渐溶解，溶液变成蓝色，故选项说法错误．<br />故选：C．','【分析】A、根据红磷在空气中燃烧的现象进行分析判断．<br />B、根据现象和结论的区别进行分析判断．<br />C、根据铁生锈的现象进行分析判断．<br />D、根据碱式碳酸铜粉末与稀盐酸反应的现象进行分析判断．','选择题',3.00,'9d394df7f93c9e938d044bc9e045fca6',9,400,'氧气与碳、磷、硫、铁等物质的反应现象,铁锈的主要成分,盐的化学性质','昆山市',2016,'37','2016•昆山市二模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1846807,'为了防止事故的发生，下列做法错误的是（　　）','炒菜时，锅内油着火用锅盖盖灭','霉变的大米可以煮熟后食用','燃放烟花爆竹时，要远离人群和可燃物','厨房煤气泄漏时，要关闭阀门并开窗通风','','B','【解答】解：A、屋里发生煤气泄露，立即开窗使气体尽快散失，且不可开关电器及使用明火，属于安全措施；<br />B、霉变的大米含有有毒物质，不可食用，煮熟后也不能食用，属于不安全措施；<br />C、烟花爆竹燃放时，会生成有毒气体、且产生火花，而且有巨大的响声，所以要远离人群和可燃物，属于安全措施；<br />D、厨房煤气泄漏时，要关闭阀门并开窗通风，属于安全措施．<br />故选B．','【分析】A、根据煤气是可燃性气体分析；<br />B、根据霉变的大米含有有毒物质，不可食用分析；<br />C、根据燃放烟花爆竹的危险性分析；<br />D、根据厨房煤气泄漏的处理方法分析．','选择题',3.00,'3c4cd555161f5d1a920324859ab05883',9,400,'灭火的原理和方法,防范爆炸的措施,易燃物和易爆物安全知识,亚硝酸钠、甲醛等化学品的性质与人体健康','',2015,'37','2015•长春二模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1846830,'化学有关的知识内容完全正确的一组是（　　）<br /><table class=\"edittable\"><TBODY><TR><td width=37 rowSpan=2>A</TD><td width=245>化学之最</TD><td width=48 rowSpan=2>B</TD><td width=249>符号中数字“2”的意义</TD></TR><TR><td>地壳中含量多的金属元素：Fe<br />人体中含量最多的金属元素：Ca<br />空气中性质最稳定的物质：N<SUB>2</SUB></TD><td>2O：两个氧分子<br />O<SUP>-2</SUP>：一个氧离子带两个单位负电荷<br />O<SUB>2</SUB>：两个氧原子</TD></TR><TR><td rowSpan=2>C</TD><td>化学记录</TD><td rowSpan=2>D</TD><td>物质及其应用</TD></TR><TR><td>托盘天平称得食盐质量为3.62g<br />量筒量取9.26mL水<br />pH试纸测纯碱溶液pH为12.3</TD><td>一氧化碳：工业上炼铁<br />食盐：医疗上配制0.9%的生理盐水<br />碳酸氢钠：面食发酵粉</TD></TR></TBODY></TABLE>','A','B','C','D','','D','【解答】解：A、地壳中含量多的金属元素是铝，人体中含量最多的金属元素是钙，空气中性质最稳定的气体是稀有气体，不完全正确；<br />B、2O：两个氧原子；离子所带的电荷标在元素符号的右上角，数字在前，符号在后，因此O<SUP>2-</SUP>表示一个氧离子带两个单位负电荷；O<SUB>2</SUB>：1个氧分子，全部错误；<br />C、根据天平、量筒、pH试纸的精确度可知：无法用托盘天平称得3.62g食盐，无法用量筒量取9.26mL水，用pH试纸测溶液的酸碱度只能测准到整数，全部错误；<br />D、一氧化碳具有还原性，所以可用于炼铁；食盐可用于医疗上配制0.9%的生理盐水；碳酸氢钠常用发酵粉，全部正确；<br />故选D','【分析】A、地壳中含量多的金属元素是铝，人体中含量最多的金属元素是钙，空气中性质最稳定的气体是稀有气体；<br />B、原子用元素符号表示；分子用化学式表示；离子所带的电荷标在元素符号的右上角，数字在前，符号在后；、<br />C、托盘天平能精确到0.1g，量筒精确度为0.1ml，pH试纸测定的溶液酸碱度应该是整数值；<br />D、一氧化碳具有还原性，食盐可用于医疗上配制0.9%的生理盐水，碳酸氢钠是发酵粉的主要成分．','选择题',3.00,'4d58619ced96b2baf0674423ef4edd05',9,400,'测量容器-量筒,称量器-托盘天平,溶液的酸碱度测定,有关化学之最,常用盐的用途,化学符号及其周围数字的意义','',2016,'37','2016•黑龙江一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1846875,'化学就在我们身边，请从A--F中选择适当的物质填空（填字母）：<br />A、食醋&nbsp;B、尿素&nbsp;C、氧气&nbsp;D、聚乙烯&nbsp;E、不锈钢&nbsp;F、氢气<br />（1）抢救危重病人常用的物质<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>&nbsp;&nbsp;（2）可用于制造手术刀的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />（3）用于农作物肥料的物质是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>&nbsp;（4）厨房调味品中pH＜7的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />（5）最清洁燃料的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;（6）可制成食品包装袋的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','C$###$E$###$B$###$A$###$F$###$D','【解答】解：（1）氧气能供给呼吸，可用于抢救危重病人；<br />（2）不锈钢的硬度大，常用作制造手术刀；<br />（3）尿素用于农作物肥料；<br />（4）食醋是厨房调味品，主要成分是醋酸，pH＜7；<br />（5）氢气燃烧的产物是水，是最清洁燃料；<br />（6）聚乙烯可制成食品包装袋；<br />故答案为：C；E；B；A；F；D．','【分析】物质的性质决定物质的用途，根据常见物质的性质和用途分析回答．','填空题',3.00,'d39c2e040512c85ced427fcdd1d3430e',9,400,'氧气的用途,合金与合金的性质,溶液的酸碱性与pH值的关系,常见化肥的种类和作用,氢气的用途和氢能的优缺点,塑料制品使用的安全','',2014,'32','2014•黄石校级模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1847112,'水是生命之源，下列有关水的说法错误的是（　　）','水通过三态变化，实现了自身的天然循环','实验室电解水时，负极端玻璃管内的气体是氢气','生活中可以煮沸的方法降低水的硬度','用滤纸过滤可以除去所有的杂质','','D','【解答】解：A．水通过三态变化实现天然循环，故正确；<br />B．实验室电解水时，正极端玻璃管内的气体是氧气，负极端玻璃管内的气体是氢气，故正确；<br />C．硬水是指含有较多钙、镁离子的水，煮沸能使钙、镁离子形成沉淀，析出，这是生活中最常用的最简单的降低水硬度的方法，故正确；<br />D．过滤可以除去水中不溶性的杂质，而不能滤纸过滤除去所有的杂质，故D错误．<br />故选：D．','【分析】A．水通过三态变化，实现了自身的天然循环；<br />B．根据实验室电解水时，正极端玻璃管内的气体是氧气，负极端玻璃管内的气体是氢气进行解答；<br />C．根据硬水软化的方法来分析；<br />D．根据过滤可以除去水中不溶性的杂质进行解答．','选择题',3.00,'74a90a46a131c63f2b54a92a5547111f',9,400,'过滤的原理、方法及其应用,电解水实验,硬水与软水,物质的三态及其转化','龙口市',2016,'35','2016春•龙口市期中',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1847255,'水和空气是人类赖以生存的自然资源．<br />①下列有关水和空气的叙述中正确的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />A．地球上水的总储量很大，但淡水资源并不充裕<br />B．充分利用太阳能、氢能等清洁能源可减少酸雨、温室效应等环境问题的发生<br />C．目前计入空气质量日报的主要污染物中已包括了二氧化碳<br />D．空气是重要的自然资源，利用其成分可以制造许多产品<br />②溶解了较多的可溶性钙和镁的化合物的水属于硬水，实验室中一般可用<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>的方法来降低水的硬度．<br />③实验室欲将200g质量分数为10%的氯化钠溶液稀释为5%的稀溶液，需加水的质量为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>g．','','','','','','ABD$###$蒸馏$###$200','【解答】解：<br />①A、地球上水的总储量很大，但淡水资源缺乏，故正确；<br />B、充分利用太阳能、氢能等清洁能源可减少酸雨、温室效应等环境问题的发生，故正确；<br />C、目前计入空气质量日报的主要污染物为：二氧化硫、二氧化氮、一氧化氮、可吸入颗粒物等，不是二氧化碳，故错误；<br />D、空气中含有多种物质，利用其成分可以制造许多产品，是一种重要的资源，故正确；<br />②解了较多的可溶性钙和镁的化合物的水属于硬水，实验室中一般可用蒸馏的方法来降低水的硬度；<br />③设要加水的质量为x，根据溶液稀释前后，溶质的质量不变，<br />则200g×10%=（200g+x）×5%&nbsp;&nbsp;&nbsp; <br />x=200g<br />答案：<br />①ABD；<br />②蒸馏；<br />③200．','【分析】①A、根据地球上水的总储量很大，但淡水资源缺乏解答；<br />B、根据清洁能源的好处解答；<br />C、根据二氧化碳不是空气污染物解答；<br />D、根据空气中各成分的用途解答．<br />②根据实验室中一般可用蒸馏的方法来降低水的硬度解答；<br />③根据溶液稀释前后，溶质的质量不变，结合题意进行分析解答．','填空题',3.00,'d28b7b5df0be643416188132de85049a',9,400,'空气对人类生活的重要作用,空气的污染及其危害,防治空气污染的措施,硬水与软水,水资源状况,用水稀释改变浓度的方法','',2016,'35','2016春•无锡期中',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1847395,'化学就在我们身边，它能改善我们的生活．<br />（1）用化学用语表示：<br />①天然气的主要成分<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；&nbsp;&nbsp;&nbsp;②缺<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>元素易患佝偻病；<br />（2）现有以下几种物质：<br />A．烧碱&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;B．熟石灰&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;C．纯碱&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;D．聚乙烯&nbsp;&nbsp;&nbsp;&nbsp;E．聚氯乙烯<br />请从A～E&nbsp;中选择适当的物质填空（填字母）：<br />①改良酸性土壤的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;②可以制成食品包装袋的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','CH<SUB>4</SUB>$###$Ca$###$B$###$D','【解答】解：（1）①天然气的主要成分是甲烷，其化学式为：CH<SUB>4</SUB>．<br />②人体缺乏钙元素易患佝偻病，其化学式为：Ca．<br />（2）①熟石灰是氢氧化钙的俗称，能与酸发生中和反应，可用于改良酸性土壤．<br />②聚氯乙烯塑料在使用时会分解出对健康有害的物质，不宜用于包装食品，可用于食品包装的塑料是聚乙烯．<br />故答案为：（1）①CH<SUB>4</SUB>；②Ca；（2）①B；②D．','【分析】（1）①天然气的主要成分是甲烷，写出其化学式即可．<br />②人体缺乏钙元素易患佝偻病，写出其元素符号即可．<br />（2）物质的性质决定物质的用途，熟石灰是氢氧化钙的俗称，能与酸发生中和反应，聚氯乙烯塑料在使用时会分解出对健康有害的物质，进行分析解答．','填空题',3.00,'d3b416a78ae33f6471720e25ed94455b',9,400,'常见碱的特性和用途,化学符号及其周围数字的意义,塑料制品使用的安全','',2016,'37','2016•淮安校级二模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1847407,'下列叙述正确的是（　　）','饱和溶液一定是浓溶液，不饱和溶液一定是稀溶液','铝比铁具有更好的抗腐蚀性能，是因为铁比铝更活泼','碱溶液能使酚酞溶液变红，那么能使酚酞溶液变红的一定是碱的溶液','用汽油和加了洗涤剂的水分别除去衣服上的油污，两者去污原理不同','','D','【解答】解：A、饱和溶液不一定是浓溶液，也可能是稀溶液，如氢氧化钙的饱和溶液为稀溶液；不饱和溶液不一定是稀溶液，也可能是浓溶液，故选项说法错误．<br />B、铝比铁具有更好的抗腐蚀性能，是因为铝的表面有一层致密的氧化铝薄膜，故选项说法错误．<br />C、碱溶液能使酚酞溶液变红，那么能使酚酞溶液变红的不一定是碱的溶液，也可能是碳酸钠等盐溶液，故选项说法错误．<br />D、用汽油和加了洗涤剂的水分别除去衣服上的油污，利用的是汽油能溶解油污、洗洁精具有乳化作用，两者去污原理不同，故选项说法正确．<br />故选：D．','【分析】A、浓稀溶液是溶液中所含溶质质量分数的大小，溶液是否饱和与溶液的浓稀没有必然联系．<br />B、根据金属的化学性质，进行分析判断．<br />C、无色酚酞溶液遇酸性溶液不变色，遇碱性溶液变红，进行分析判断．<br />D、根据汽油能溶解油污、洗洁精具有乳化作用，进行分析判断．','选择题',3.00,'58ffc77dc02d3fd6118fd73bf8c247b2',9,400,'溶解现象与溶解原理,乳化现象与乳化作用,浓溶液、稀溶液跟饱和溶液、不饱和溶液的关系,金属的化学性质,碱的化学性质','',2016,'37','2016春•石峰区月考',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1847412,'通过一年的学习，你认为下列对化学学科的认知有误的是（　　）','化学为人类研制了新材料','化学的发展导致了生态环境的恶化','化学为人类提供了新能源','化学已成为生命科学的重要基础','','B','【解答】解：A、化学是研究物质的组成、结构、性质及其变化规律的科学，不仅研究自然界中存在的物质，还研究和创造自然界中不存在的物质，故A认识正确．<br />B、化学为人类提供了丰富的生活和生产资料，同时化工生产也带来了一些环境污染问题，但这些问题是利用化学知识可控的，故B认识错误．<br />C、利用化学可以开发新能源和新材料，以改善人类生存的条件，故C认识正确．<br />D、化学分支有生物化学学科，化学在生命科学中发挥着越来越重要的作用，故D认识正确．<br />故选B．','【分析】解答时根据化学与生活、生产、交通、医疗保健、环境等方面联系密切的相关知识．也就是，化学中所学习到的有关物质在生活、生产、交通、医疗保健、环境等方面的具体用途等等来进行解答．','选择题',3.00,'b48949db3d3c5d1a24d6c16824cd1f82',9,400,'化学的用途','',2016,'32','2016•泗阳县模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1847449,'<img  src=\"/tikuimages/9/2016/400/shoutiniao69/6f62e300-94d4-11e9-ba36-b42e9921e93e_xkb53.png\" style=\"vertical-align:middle;FLOAT:right\" />（2016•房山区一模）钛（Ti）被称为继铁、铝之后的第三金属．如图所示，将钛厂、氯碱厂和甲醇（CH<SUB>4</SUB>O）厂组成产业链可以大大提高资源利用率，减少环境污染．<br />（1）FeTiO<SUB>3</SUB>中铁元素为+3价，则钛元素的化合价为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）“绿色化学”能实现“零排放”（即反应物中的原子利用率达到100%）．一氧化碳和氢气合成甲醇就符合“绿色化学”特点，此反应的基本反应类型为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）电解食盐水反应的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','+3$###$化合反应$###$2NaCl+2H<SUB>2</SUB>O<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;通电&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2NaOH+H<SUB>2</SUB>↑+Cl<SUB>2</SUB>↑','【解答】解：（1）FeTiO<SUB>3</SUB>中铁元素为+3价，氧元素显示-2价，所以钛元素的化合价为+3；<br />（2）一氧化碳和氢气合成甲醇就符合“绿色化学”特点，该反应是多变一的化合反应，符合）“绿色化学”能实现“零排放”（即反应物中的原子利用率达到100%）；<br />（3）氯化钠和水在通电的条件下生成氢氧化钠、氢气和氯气，化学方程式为：2NaCl+2H<SUB>2</SUB>O<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;通电&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2NaOH+H<SUB>2</SUB>↑+Cl<SUB>2</SUB>↑．<br />故答案为：（1）+3；<br />（2）化合反应；<br />（3）2NaCl+2H<SUB>2</SUB>O<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;通电&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2NaOH+H<SUB>2</SUB>↑+Cl<SUB>2</SUB>↑．','【分析】（1）根据化合价代数和为零进行分析；<br />（2）根据一氧化碳和氢气合成甲醇就符合“绿色化学”特点，该反应是多变一的化合反应进行分析；<br />（3）根据氯化钠和水在通电的条件下生成氢氧化钠、氢气和氯气进行分析．','书写',3.00,'a4fd04652a2a08c9b1399e8f246c71ad',9,400,'绿色化学,物质的相互转化和制备,有关元素化合价的计算,书写化学方程式、文字表达式、电离方程式','',2016,'37','2016•房山区一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1847499,'下列工农业生产、日常生活中，错误运用化学知识的是（　　）','干冰用于人工降雨','小苏打用于烘焙糕点','用氢氧化钠改良酸性土壤','用食醋除去水壶中的水垢[主要成分是CaCO<SUB>3</SUB>和Mg（OH）<SUB>2</SUB>]','','C','【解答】解：A、干冰升华吸收热量，可用于人工降雨，故说法正确；<br />B、小苏打是碳酸氢钠，用于烘焙糕点，故说法正确；<br />C、氢氧化钠具有极强的腐蚀性，不能用于改良酸性土壤，故说法错误；<br />D、食醋可以和碳酸钙和氢氧化镁反应生成可溶性的盐，因此用食醋除去水壶中的水垢[主要成分是CaCO<SUB>3</SUB>和Mg（OH）<SUB>2</SUB>]，故说法正确．<br />故选：C．','【分析】物质的性质决定物质的用途，根据已有的物质的性质进行分析解答即可．','选择题',3.00,'f7b97c5fa47d5cae9f5053f334fcc2d3',9,400,'二氧化碳的用途,常见碱的特性和用途,常用盐的用途,醋酸的性质及醋酸的含量测定','',2016,'32','2016•漳州模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1847522,'<img  src=\"/tikuimages/9/2016/400/shoutiniao19/704b6e40-94d4-11e9-a252-b42e9921e93e_xkb39.png\" style=\"vertical-align:middle;FLOAT:right\" />（2016•安庆一模）根据如图回答问题．<br />（1）仪器①的名称是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，A装置的优点是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）B装置在化学实验中有广泛的用途：<br />①集气：用排水法收集氧气，可在B装置装满水后，使氧气从<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“a”或“b”）口进入；<br />②储气：若要将储存在B中的氧气排出，水应从<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“a”或“b”）口进入；<br />③量气：若用B装置测量收集到的气体体积，还必须要增加到的仪器是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','长颈漏斗$###$可以控制反应随时发生或停止$###$b$###$a$###$量筒','【解答】解：（1）仪器①的名称是长颈漏斗；A装置有弹簧夹，其优点是可以控制反应随时发生或停止；<br />（2）①根据氧气密度比水小，难溶于水，用排水法收集氧气，可在B装置装满水后，使氧气从短进长出；<br />②若要将储存在B中的氧气排出，氧气的密度比水小，水应长进短出，便于氧气的排出；<br />③若用B装置测量收集到的气体体积，还必须要增加到的仪器是量筒，量取一定体积液体时，必须要用到的仪器是量筒．<br />故答案为：（1）（长颈）漏斗、可以控制反应随时发生或停止；&nbsp;&nbsp;（2）①b&nbsp;&nbsp;②a&nbsp;&nbsp;③量筒或注射器','【分析】（1）根据实验室常用仪器的名称和题中所指仪器的作用进行分析；<br />根据A装置制取气体可以控制反应随时发生或停止；<br />（2）①根据氧气密度比水小，难溶于水进行分析；<br />②根据氧气的密度比水小，水应长进短出，便于氧气的排出分析；<br />③根据量取一定体积液体时，必须要用到的仪器是量筒解答．','填空题',3.00,'c7a9e568b765b1516a1f8c499683684e',9,400,'分离物质的仪器,二氧化碳的实验室制法','',2016,'37','2016•安庆一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1847560,'除去下列各物质中的少量杂质，所用方法不可行的是（　　）<br /><table class=\"edittable\"><TBODY><TR><td width=142>&nbsp;选项</TD><td width=142>&nbsp;物质</TD><td width=142>&nbsp;杂质</TD><td width=142>&nbsp;除去杂质的方法</TD></TR><TR><td>&nbsp;A</TD><td>&nbsp;Al<SUB>2</SUB>（SO<SUB>4</SUB>）<SUB>3</SUB>溶液</TD><td>&nbsp;H<SUB>2</SUB>SO<SUB>4</SUB>溶液</TD><td>&nbsp;加入足量Al<SUB>2</SUB>O<SUB>3</SUB>粉末后过滤</TD></TR><TR><td>&nbsp;B</TD><td>&nbsp;Cu</TD><td>&nbsp;CuO</TD><td>&nbsp;通入氧气并加热</TD></TR><TR><td>&nbsp;C</TD><td>&nbsp;CaO</TD><td>&nbsp;CaCO<SUB>3</SUB></TD><td>&nbsp;高温煅烧</TD></TR><TR><td>&nbsp;D</TD><td>&nbsp;H<SUB>2</SUB>气体</TD><td>&nbsp;HCl气体</TD><td>&nbsp;先通过NaOH溶液，再通过浓硫酸</TD></TR></TBODY></TABLE>','A','B','C','D','','B','【解答】解：A、H<SUB>2</SUB>SO<SUB>4</SUB>溶液能与足量Al<SUB>2</SUB>O<SUB>3</SUB>粉末反应生成硫酸铝和水，再进行过滤，能除去杂质且没有引入新的杂质，符合除杂原则，故选项所采取的方法正确．<br />B、Cu通入氧气并加热生成氧化铜，反而会把原物质除去，不符合除杂原则，故选项所采取的方法错误．<br />C、CaCO<SUB>3</SUB>固体高温煅烧生成氧化钙和二氧化碳，能除去杂质且没有引入新的杂质，符合除杂原则，故选项所采取的方法正确．<br />D、HCl气体能与NaOH溶液反应生成氯化钠和水，氢气不氢氧化钠溶液反应，再通过浓硫酸进行干燥，能除去杂质且没有引入新的杂质，符合除杂原则，故选项所采取的方法正确．<br />故选：B．','【分析】根据原物质和杂质的性质选择适当的除杂剂和分离方法，所谓除杂（提纯），是指除去杂质，同时被提纯物质不得改变．除杂质题至少要满足两个条件：①加入的试剂只能与杂质反应，不能与原物质反应；②反应后不能引入新的杂质．','选择题',3.00,'c3045955db1221b24b8cf63bec213760',9,400,'物质除杂或净化的探究,常见气体的检验与除杂方法,铁锈的主要成分,酸的化学性质','',2016,'37','2016春•重庆校级月考',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1847564,'<img  src=\"/tikuimages/9/2016/400/shoutiniao9/70b46940-94d4-11e9-b3b9-b42e9921e93e_xkb10.png\" style=\"vertical-align:middle;FLOAT:right;\" />如图装置所示，经数小时后，U形管A、B两处的液面将会出现下列哪种情况（实验装置足以维持实验期间小白鼠的生命活动，瓶口密封）（　　）','A处上升，B处下降','A、B两处都下降','A处下降，B处上升','A、B两处都不变','','C','【解答】解：根据图示的装置可知：小白鼠进行呼吸作用时消耗氧气，呼出二氧化碳，呼出的二氧化碳能被氢氧化钠溶液吸收，导致集气瓶中的气体体积减小，瓶内压强减小，则A液面下降，B液面上升．<br />故选：C．','【分析】根据小白鼠进行呼吸作用时消耗氧气，呼出二氧化碳，呼出的二氧化碳能被氢氧化钠溶液吸收，导致集气瓶中的气体体积减小，瓶内压强减小，据此进行分析判断．','选择题',3.00,'6406a06d0c2cbc7eec4454a4962e11f5',9,400,'光合作用与呼吸作用','',2016,'37','2016春•灵台县校级月考',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1847582,'8•12天津滨海新区爆炸事故震惊全国<br />事故调查组查明，事故的直接原因是：瑞海公司危险品仓库运抵区南侧集装箱内硝化棉由于湿润剂散失出现局部干燥，在高温（天气）等因素的作用下加速分解放热，积热自燃，引起相邻集装箱内的硝化棉和其他危险化学品长时间大面积燃烧，导致堆放于运抵区的硝酸铵等危险化学品发生爆炸．<br />请你分析：<br />（1）从你所学的化学知识观点分析燃烧引起爆炸事故原因<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）“血的教训极其深刻，必须牢牢记取”，请你列举有关易燃物和易爆物防火灭火、防范爆炸知识．<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>至少一点．','','','','','','硝化棉等化学物质受热分解产生易燃物，温度达到它的着火点，发生自然，而硝酸铵属于易爆炸的危险品，遇到明火后有限的空间内急剧燃烧，体积迅速膨胀，从而发生爆炸$###$易燃物和易爆物的仓库，要有良好的通风设备','【解答】解：（1）根据燃烧和发生爆炸的条件：硝化棉等化学物质受热分解产生易燃物，温度达到它的着火点，发生自然，而硝酸铵属于易爆炸的危险品，遇到明火后有限的空间内急剧燃烧，体积迅速膨胀，从而发生爆炸．故填：硝化棉等化学物质受热分解产生易燃物，温度达到它的着火点，发生自然，而硝酸铵属于易爆炸的危险品，遇到明火后有限的空间内急剧燃烧，体积迅速膨胀，从而发生爆炸；<br />（2）根据贮存易燃物和易爆物的仓库，要有良好的通风设备，这样缓慢氧化等原因产生的热量就能及时扩散出去，防止发生自燃；<br />要扑灭一种物质大火时，灭火剂可能与另种物质反应，引起另种物质的燃烧或爆炸，要选择合适的灭火剂；<br />贮存易燃物和易爆物时，堆与堆、堆与墙之间要留有足够距离的通道利于热量扩散（通风）．<br />故答案为：易燃物和易爆物的仓库，要有良好的通风设备（答案合理即可）．','【分析】（1）根据爆炸的原理来分析；<br />（2）据易燃物和易爆物防火灭火、防范爆炸知识来分析解答．','填空题',3.00,'af71d346acfacfc459f7888182c5d3f6',9,400,'燃烧、爆炸、缓慢氧化与自燃,易燃物和易爆物安全知识','',2016,'32','2016•海南模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1847715,'回答下列问题：<br />（1）工厂大量燃烧煤时易产生大量二氧化硫气体，二氧化硫气体对人体有毒，会污染大气，可用氢氧化钠溶液吸收，二氧化硫的化学性质与二氧化碳相似，写出二氧化硫与氢氧化钠反应的化学方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）下列哪个科学家用实验获知空气的组成<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />A．法国化学家拉瓦锡&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;B．英国化学家普利斯特里<br />C．瑞典化学家舍勒&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;D．意大利科学家阿伏伽德罗<br />（3）使用硬水会给生活和生产带来许多麻烦．在日常生活中，人们常用<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>的方法来降低水的硬度，用此方法得到的软水是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“纯净物”或“混合物“）．我们还可以用<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>来检验水是软水还是硬水．<br />（4）“低碳”是一种生活理念，也是一种生活态度．下列做法中不符合“低碳”要求的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填序号）<br />A．少用一次性用品：纸杯、筷子、塑料袋等<br />B．多挖煤、多发电、加快经济发展<br />C．提倡骑自行车、乘公交车出行．<br />（5）钢铁工业是高能耗、高污染产业．我国冶炼钢铁的铁矿石大量从国外进口，钢铁年产量为11亿吨，约占世界年产量的45%．我国钢铁的年使用量约为8.4亿吨，根据以上信息，写出当前我国过剩钢铁生产的弊端．<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（写出其中一点）','','','','','','SO<SUB>2</SUB>+2NaOH=Na<SUB>2</SUB>SO<SUB>3</SUB>+H<SUB>2</SUB>O$###$A$###$煮沸$###$混合物$###$肥皂水$###$B$###$产品积压，企业利润下降或亏损或浪费资源和能源或增加有害气体排放，增加PM2.5排放','【解答】解：（1）二氧化硫与氢氧化钠反应生成了亚硫酸钠和水，反应的化学方程式是：SO<SUB>2</SUB>+2NaOH=Na<SUB>2</SUB>SO<SUB>3</SUB>+H<SUB>2</SUB>O．<br />（2）法国化学家拉瓦锡通过实验获知空气是由氮气和氧气组成的结论；<br />（3）使用硬水会给生活和生产带来许多麻烦，在日常生活中，人们常用煮沸的方法来降低水的硬度，用此方法得到的软水还含有可溶性的杂质，是混合物，我们还可以用肥皂水来检验水是软水还是硬水．<br />（4）A．少用一次性用品：纸杯、筷子、塑料袋等，能减少了二氧化碳的排放，符合“低碳”要求；<br />B．多挖煤、多发电、加快经济发展，产生了大量的二氧化碳．不符合符合“低碳”要求；<br />C．提倡骑自行车、乘公交车出行，减少了二氧化碳的排放，符合“低碳”要求．<br />（5）我国过剩钢铁生产的弊端有：产品积压，企业利润下降或亏损或浪费资源和能源或增加有害气体排放，增加PM2.5排放等．<br />故答为：（1）SO<SUB>2</SUB>+2NaOH=Na<SUB>2</SUB>SO<SUB>3</SUB>+H<SUB>2</SUB>O；（2）A；（3）煮沸，混合物，肥皂水，B；产品积压，企业利润下降或亏损或浪费资源和能源或增加有害气体排放，增加PM2.5排放．','【分析】（1）根据二氧化硫与氢氧化钠的反应，写出化学方程式；<br />（2）根据法国化学家拉瓦锡研究空气的组成分析回答；<br />（3）根据软水、硬水的区别、硬水的软化和组成分析回答；<br />（4）根据“低碳”要求分析判断；<br />（5）根据过剩钢铁生产的弊端．','填空题',3.00,'41712499da8ec020d955bc39e61021c7',9,400,'化学的历史发展过程,防治空气污染的措施,二氧化碳的化学性质,硬水与软水,纯净物和混合物的判别','中山市',2016,'35','2016春•中山市期中',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1847746,'选择合适操作或仪器的序号填空．<br />①坩埚钳&nbsp;&nbsp;&nbsp;&nbsp;②搅拌&nbsp;&nbsp;&nbsp;&nbsp;③10.0mL量筒&nbsp;&nbsp;&nbsp;&nbsp;④50.0mL量筒&nbsp;&nbsp;&nbsp;&nbsp;⑤加热<br />（1）将浓硫酸注入水中，为使热量及时散发应<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（2）转移热的蒸发皿到石棉网上用<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（3）除去Na<SUB>2</SUB>CO<SUB>3</SUB>固体中的NaHCO<SUB>3</SUB>，操作是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','②$###$①$###$⑤','【解答】解：（1）稀释浓硫酸时，要把浓硫酸缓缓地沿器壁注入水中，同时用玻璃棒不断搅拌，以使热量及时地扩散．<br />（2）转移热的蒸发皿到石棉网上用坩埚钳．<br />（3）除去Na<SUB>2</SUB>CO<SUB>3</SUB>固体中的NaHCO<SUB>3</SUB>，可采用加热的方法，碳酸氢钠在加热条件下生成碳酸钠、水和二氧化碳，能除去杂质且没有引入新的杂质，符合除杂原则．<br />故答案为：（1）②；（2）①；（3）⑤．','【分析】（1）根据浓硫酸的稀释方法（酸入水，沿器壁，慢慢倒，不断搅），进行分析解答．<br />（2）根据常用的夹持仪器，进行分析解答．<br />（3）除杂质题至少要满足两个条件：①加入的试剂只能与杂质反应，不能与原物质反应；②反应后不能引入新的杂质．','填空题',3.00,'1ef4506a0e059ca15f598bd56fd2ed1c',9,400,'挟持器-铁夹、试管夹、坩埚钳,浓硫酸的性质及浓硫酸的稀释,盐的化学性质','',2016,'32','2016•内江模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1847758,'<img  src=\"/tikuimages/9/0/400/shoutiniao92/72cdac51-94d4-11e9-ab7c-b42e9921e93e_xkb6.png\" style=\"vertical-align:middle;FLOAT:right\" />根据图所示的装置，回答下列问题．<br />（1）测定绿色植物光合作用时，能测定的变化有很多种，通常只测定其中之一的变化就可以了．从下列列举的变化里，选择其中最易操作的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．（本题只选最佳的一个答案）<br />A．生成的葡萄糖 B．生成的淀粉<br />C．生成的氧气 D．消耗的水<br />E．消耗的二氧化碳 F．消耗的氧<br />（2）用图中所示的装置，研究水草光合作用时，三角烧瓶里的水草中，加入<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>溶液效果最好（从以下溶液中选填序号：①葡萄糖溶液、②氢氧化钠溶液、③CO<SUB>2</SUB>释放溶液）<br />（3）加入上述溶液，水草在光照一定时间后，实验装置U形管两侧的液柱左侧要<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“高于”“相平”或“低于”）右侧．<br />（4）若不加入上述溶液，水草在光照一定时间后，实验装置U形管两侧的液柱高度变化不大，原因是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','C$###$③$###$低于$###$由于不进行光合作用，只进行呼吸作用，呼吸消耗的O<SUB>2</SUB>和产生的CO<SUB>2</SUB>体积相等，锥形瓶内气压不变，所以左侧液柱与右侧液柱相比，液面高度相等','【解答】解：（1）光合作用的原料是水和二氧化碳，产物是氧气和淀粉（葡萄糖）．在这些物质中氧气能够助燃，是最好测定的，用带火星的木条进行实验，如果木条复燃，说明有氧气生成，否则没有氧气生成．<br />（2）光合作用的原料是二氧化碳和水，所以用图中所示的装置，研究水草光合作用时，三角烧瓶里的水草中，加入二氧化碳溶液效果最好，加入该溶液的作用是不断补充CO<SUB>2</SUB>，供植物进行光合作用．<br />（3）光合作用产生的氧气在左侧装置内聚集，使装置内气体的压强大于大气压，所以水草在光照一定时间后，实验装置U形管两侧的液柱左侧要低于右侧．<br />（4）由于不进行光合作用，只进行呼吸作用，呼吸消耗的O<SUB>2</SUB>和产生的CO<SUB>2</SUB>体积相等，锥形瓶内气压不变，所以左侧液柱与右侧液柱相比，液面高度相等．<br />答案：<br />（1）C；<br />（2）③<br />（3）低于；<br />（4）由于不进行光合作用，只进行呼吸作用，呼吸消耗的O<SUB>2</SUB>和产生的CO<SUB>2</SUB>体积相等，锥形瓶内气压不变，所以左侧液柱与右侧液柱相比，液面高度相等．','【分析】（1）根据绿色植物光合作用时的反应物和产物分析；<br />（2）根据光合作用的原料分析；<br />（3）根据光合作用可生成气体，以及气体的压强分析；<br />（4）根据由于不进行光合作用，只进行呼吸作用，呼吸消耗的O<SUB>2</SUB>和产生的CO<SUB>2</SUB>体积相等，锥形瓶内气压不变，所以左侧液柱与右侧液柱相比，液面高度相等解答．','简答题',3.00,'4dfaafafcd34fd3b29ec321d9da94df3',9,400,'光合作用与呼吸作用','',0,'37','',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1847761,'工业上冶炼金属一般用热分解法、热还原法和电解法．不同的金属应选用不同的冶炼方法，你认为选择的原则主要是依据（　　）','金属在自然界的存在形式','金属元素在地壳中的含量','金属的活动性','金属熔点的高低','','C','【解答】解：A、不同的金属应选用不同的冶炼方法，与金属在自然界的存在形式关系不大，故选项错误．<br />B、不同的金属应选用不同的冶炼方法，与金属元素在地壳中的含量无关，故选项错误．<br />C、不同的金属应选用不同的冶炼方法，本质上是由金属活动性决定的，故选项正确．<br />D、不同的金属应选用不同的冶炼方法，与金属熔点的高低无关，故选项错误．<br />故选：C．','【分析】工业上冶炼金属一般用热分解法、热还原法和电解法．不同的金属应选用不同的冶炼方法，本质上是由金属活动性决定的，进行分析判断．','选择题',3.00,'722e25084d824d073132b5e0392133f7',9,400,'常见金属的冶炼方法','',0,'37','',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1847776,'下列关于化合价的说法中，正确的是（　　）','金属元素只能显正价','在氧气中，氧元素显-2价','一种元素在同一种化合物中，可能有几种不同的化合价','非金属元素在化合物中总是显负价','','C','【解答】解：A、金属元素不一定只显正价，单质中金属的化合价为0，故选项说法错误．<br />B、单质中元素的化合价为0，氧气属于单质，故氧元素的化合价为0，故选项说法错误．<br />C、一种元素在同一种化合物中，可能有几种不同的化合价，如在NH<SUB>4</SUB>NO<SUB>3</SUB>中含有铵根离子和硝酸根离子；铵根离子显+1价，而氢元素显+1价，故氮元素为-3价；硝酸根离子显-1价，而氧元素显-2价，故氮元素显+5价；所以硝酸铵中氮元素分别显-3价和+5价；一种元素在同一种化合物里，可能显不同的化合价，故选项说法正确．<br />D、非金属元素在化合物中不一定总显负价，如H<SUB>2</SUB>O中氢元素显+1价，故选项说法错误．<br />故选：C．','【分析】A、根据化合价的规律，进行分析判断．<br />B、根据单质中元素的化合价为0，进行分析判断．<br />C、根据常见元素的化合价、化合价的规律进行分析判断．<br />D、非金属元素在化合物中不一定总显负价．','选择题',3.00,'4565a19709f392b0ee869b3e09bc49bd',9,400,'常见元素与常见原子团的化合价,化合价规律和原则','',0,'37','',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1847896,'消防安全人人有责，生活中的下列做法正确的是（　　）','<img  src=\"/tikuimages/9/2016/400/shoutiniao27/7446699e-94d4-11e9-a299-b42e9921e93e_xkb45.png\" style=\"vertical-align:middle\" /><br />液化石油气泄漏，点火检查看得清','<img  src=\"/tikuimages/9/2016/400/shoutiniao94/74486570-94d4-11e9-aacd-b42e9921e93e_xkb43.png\" style=\"vertical-align:middle\" /><br />躺在床上抽口烟，自在逍遥真舒服','<img  src=\"/tikuimages/9/2016/400/shoutiniao96/7449ec0f-94d4-11e9-be5e-b42e9921e93e_xkb5.png\" style=\"vertical-align:middle\" /><br />炒菜得油锅着火，立即盖严锅盖儿','<img  src=\"/tikuimages/9/2016/400/shoutiniao91/744a8851-94d4-11e9-ad6c-b42e9921e93e_xkb72.png\" style=\"vertical-align:middle\" /><br />随心所欲拉电线，省事方便又好用','','C','【解答】解：A、液化气具有可燃性，泄露遇明火可能发生爆炸，为防止发生爆炸，不能用打火机点火检查液化气罐是否漏气，故选项说法错误．<br />B、咽产生大量的有害气体，危害人体健康，故选项说法错误．<br />C、炒菜得油锅着火，立即盖严锅盖儿，可隔绝空气灭火，故选项说法正确．<br />D、随心所欲拉电线，会造成触电事故，故选项说法错误．<br />故选：C．','【分析】根据可燃性气体与空气混合后点燃可能发生爆炸；灭火原理：①清除或隔离可燃物，②隔绝氧气或空气，③使温度降到可燃物的着火点以下，据此结合灭火方法进行分析判断．','选择题',3.00,'22c4046a8e08208f071aadca2251eef0',9,400,'灭火的原理和方法,防范爆炸的措施,烟的危害性及防治','',2016,'37','2016•吉林一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1848197,'物质由元素组成，相同元素可以组成不同的物质．<br />（1）碳元素可以组成金刚石或石墨．<br />①金刚石和石墨硬度不同的原因是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />②在一定条件下，石墨可转变为金刚石，这一变化属于<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“物理变化”或“化学变化”）．<br />③由同种元素组成的不同单质，互为同素异形体．下列属于同素异形体的有<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填标号）．<br />A．C<SUB>60</SUB>和C<SUB>70</SUB>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;B．CO和CO<SUB>2</SUB>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;C．O<SUB>2</SUB>和O<SUB>3</SUB>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;D．生铁和钢<br />（2）CO与CO<SUB>2</SUB>都是由碳、氧元素组成的不同物质．<br />①用化学方法区别两种物质，可选择的试剂是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，写出反应的化学方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />②CO与CO<SUB>2</SUB>相互可以转化．写出CO<SUB>2</SUB>转变为CO的化学方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />（3）C、H、O三种元素可组成碳酸、乙醇（C<SUB>2</SUB>H<SUB>5</SUB>OH）和乙酸（CH<SUB>3</SUB>COOH）等．<br />①以上三种物质中不属于有机物的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />②在汽油中添加一定量的乙醇可制得“乙醇汽油”．与普通汽油比较，使用“乙醇汽油”的优点是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>&nbsp;（写一点）．','','','','','','构成它们的碳原子的排列方式不同（结构不同）$###$化学变化$###$AC$###$澄清石灰水$###$Ca（OH）<SUB>2</SUB>+CO<SUB>2</SUB>═CaCO<SUB>3</SUB>↓+H<SUB>2</SUB>O$###$CO<SUB>2</SUB>+C<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2CO$###$碳酸$###$燃烧产生的废气对环境污染较少','【解答】解：（1）①金刚石和石墨硬度不同的原因是：构成它们的碳原子的排列方式不同．<br />②在一定条件下，石墨可转变为金刚石，有新物质生成，这一变化属于化学变化；<br />③由同种元素组成的不同单质，互称同素异形体．C<SUB>60</SUB>和C<SUB>70</SUB>、O<SUB>2</SUB>和O<SUB>3</SUB>属于同素异形体．<br />（2）①由于二氧化碳能与氢氧化钙反应生成了碳酸钙沉淀，一氧化碳不与氢氧化钙反应，如用化学方法区别两种物质，可选择的试剂是：澄清石灰水，反应的化学方程式为：Ca（OH）<SUB>2</SUB>+CO<SUB>2</SUB>═CaCO<SUB>3</SUB>↓+H<SUB>2</SUB>O．<br />②CO与CO<SUB>2</SUB>相互间可以转化，在该高温条件下，二氧化碳与碳反应生成了一氧化碳碳，CO<SUB>2</SUB>转变为CO的化学方程式是：CO<SUB>2</SUB>+C<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2CO；<br />（3）①碳酸虽然含有碳元素，但性质同无机物相似，常归到无机物类．<br />②在汽油中添加一定量的乙醇可制得“乙醇汽油”．与普通汽油相比较，使用“乙醇汽油”的优点是燃烧产生的废气对环境污染较少等．<br />故答为：（1）①构成它们的碳原子的排列方式不同，②化学变化，③AC；（2）①澄清石灰水，Ca（OH）<SUB>2</SUB>+CO<SUB>2</SUB>═CaCO<SUB>3</SUB>↓+H<SUB>2</SUB>O，②CO<SUB>2</SUB>+C<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2CO；（3）①碳酸；②燃烧产生的废气对环境污染较少．','【分析】（1）根据物质的构成变化的特征分析判断；<br />（2）根据CO与CO<SUB>2</SUB>的性质分析回答；<br />（3）根据物质的组成、“乙醇汽油”的优点分析回答．','书写',3.00,'46e57d0eaf585c8835a8fb9de087bab4',9,400,'二氧化碳的化学性质,一氧化碳的化学性质,有机物与无机物的区别,碳元素组成的单质,同素异形体和同素异形现象,化学变化和物理变化的判别,书写化学方程式、文字表达式、电离方程式,常用燃料的使用与其对环境的影响','',2015,'37','2015•福州二模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1848232,'现在有许多家庭都在使用管道煤气，其主要成分是甲烷（CH<SUB>4</SUB>），甲烷在足量的氧气中完全燃烧时生成二氧化碳和水，此时，反应物的质量比是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；如果氧气不足时，甲烷燃烧除生成上述物质外还有一氧化碳生成，其反应的化学方程式是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','1：4$###$4CH<SUB>4</SUB>+7O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2CO+2CO<SUB>2</SUB>+8H<SUB>2</SUB>O','【解答】解：甲烷在足量的氧气中完全燃烧时生成二氧化碳和水，反应的化学方程式为：<br />CH<SUB>4</SUB>+2O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>CO<SUB>2</SUB>+2H<SUB>2</SUB>O<br />16&nbsp;&nbsp;64&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <br />在此反应中甲烷、氧气的质量比为16：64=1：4．<br />如果氧气不足时，甲烷燃烧除生成上述物质外还有一氧化碳生成，反应的化学方程式为：4CH<SUB>4</SUB>+7O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2CO+2CO<SUB>2</SUB>+8H<SUB>2</SUB>O．<br />故答案为：1：4；4CH<SUB>4</SUB>+7O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2CO+2CO<SUB>2</SUB>+8H<SUB>2</SUB>O．','【分析】根据题意，甲烷在足量的氧气中完全燃烧时生成二氧化碳和水，利用各物质之间的质量比等于相对分子质量和的比，进行分析解答即可．<br />如果氧气不足时，甲烷燃烧除生成上述物质外还有一氧化碳生成，写出反应的化学方程式即可．','书写',3.00,'d972ab6f8cc2d503b740f7774e995719',9,400,'常见化学反应中的质量关系,书写化学方程式、文字表达式、电离方程式,常用燃料的使用与其对环境的影响','',2012,'33','2012春•杭州期末',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1848281,'为了研究问题的方便，科学家创造性地引入了一系列科学概念，譬如：<br />①在研究光的传播时引入了“光线”；&nbsp;&nbsp;&nbsp;&nbsp;②在研究物质结构时引入了“原子”；&nbsp;&nbsp;③在研究物<br />质导电性时引入了“绝缘体”；④在研究磁场时引入了“磁感线”；&nbsp;&nbsp;⑤在研究人类听觉范围时<br />引入了“超声”．其中属于科学假想而实际并不存在的是（　　）','①②','①④','③④','②⑤','','B','【解答】解：五个选项中的“原子”、“绝缘体”、“超声”都是真实存在的物质，只有①“光线”和④“磁感线”是利用建模法，物理学家创造性地引入的一系列科学概念．<br />故选B．','【分析】首先确定题干的要求是：找出科学家创造性地引入的一系列科学概念，这是建模法，然后从五个例子中找出用建模法的即可．','选择题',3.00,'5a2d52d47db39f6e650bfb7d77f1b57b',9,400,'科学探究的基本方法','',2015,'35','2015春•台州校级期中',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1848409,'如图甲是氢气和氧化铜反应的实验，图乙是木炭和氧化铁反应的实验．<br /><img  src=\"/tikuimages/9/2011/400/shoutiniao23/7a84c551-94d4-11e9-a340-b42e9921e93e_xkb3.png\" style=\"vertical-align:middle\" /><br />（1）上述两个实验中，氢气和木炭表现出相同的化学性质是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>性．<br />（2）实验步骤的先后次序非常重要，比如甲实验中，反应前必须先通氢气后点燃酒精灯，否则易发生爆炸；反应结束后必须先<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>后<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，否则生成的红色的铜会变成黑色．<br />（3）某同学正确操作做完乙实验后，发现澄清石灰水变浑浊，试管中粉末全部变为黑色，取少量黑色粉末，加入足量稀硫酸充分振荡，但她惊讶地发现黑色粉末没有溶解，试管中也未产生预期的气泡，这说明该反应并没有生成<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','还原$###$停止加热$###$停止通入氢气$###$铁','【解答】解：（1）氢气和木炭都得到氧，都能夺取氧化物中的氧，是还原剂，具有还原性．<br />故答案为：还原；<br />（2）氢气还原氧化铜的实验开始要先通氢气，再加热，反应之加热氢气与空气的混合气体，可能会发生爆炸．<br />故答案为：停止加热，停止通入氢气；<br />（3）乙实验完成后发现黑色粉末没有溶解，试管中也未产生预期的气泡，说明反应没有铁生成（如有铁生成，铁会和稀硫酸反应生成氢气，图中酒精灯没加网罩，没有达到反应所需的温度）．<br />故答案为：铁．','【分析】（1）在化学反应中得到氧的物质叫还原剂，具有还原性；<br />（2）氢气还原氧化铜的实验中，氢气要依据“早出晚归”的原则；<br />（3）根据“黑色粉末没有溶解，试管中 也未产生预期的气泡，”做出判断；','填空题',3.00,'e9394fb19a08c63436253b663fc6af5f',9,400,'碳、一氧化碳、氢气还原氧化铜实验','桐乡市',2011,'35','2011秋•桐乡市期中',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1848517,'化学概念在逻辑上存在如图所示的关系，对下列概念间的关系说法正确的是（　　）<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao39/7c11cdf0-94d4-11e9-89db-b42e9921e93e_xkb25.png\" style=\"vertical-align:middle\" /><br />A．纯净物与混合物属于包含关系<br />B．复分解反应与中和反应属于包含关系<br />C．单质与化合物属于交叉关系<br />D．氧化反应与化合反应属于<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>关系．','','','','','','交叉','【解答】解：A、物质按含有物质种类的多少可分为纯净物与混合物，是并列关系，故此选项错误．<br />B、中和反应一定是复分解反应，复分解反应不一定是中和反应，复分解反应与中和反应属于包含关系，故此选项正确．<br />C、纯净物按元素的含有情况可分为化合物与单质，故是并列关系，故此选项错误．&nbsp;&nbsp;&nbsp;<br />D、氧化反应不一定是化合反应，化合反应不一定是氧化反应，氧化反应和化合反应属于交叉关系．<br />故答案为：B；交叉','【分析】应用各知识点的概念，理解概念间相关的关系，结合图示所提供的关系意义，分析相关的选项从而判断正确与否，从物质分类的知识可知物质可分为纯净物与混合物，纯净物又可分为化合物与单质，化合物中又可分为酸碱盐及氧化物等．','填空题',3.00,'86eb54a3a1b9b3afc34115c2da3bb86f',9,400,'中和反应及其应用,复分解反应及其发生的条件,纯净物和混合物的概念,单质和化合物的概念,化合反应及其应用,氧化反应','',2016,'37','2016•会昌县一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1848542,'097型核潜艇是传闻中中国人民解放军海军正在研制的核潜艇，为攻击型潜艇或巡航导弹潜艇．潜水艇里要配备氧气的再生装置，以保证长时间潜航．有以下几种种制氧的方法：①加热高锰酸钾；②电解水；&nbsp;③在常温下过氧化钠固体（Na<SUB>2</SUB>O<SUB>2</SUB>）与CO<SUB>2</SUB>反应生碳酸钠和氧气．<br />（1）写出方法①和③的化学方程式：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>、<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）你认为最适合在潜水艇里制氧气的方法是（填序号）<SUB><u>&nbsp;&nbsp;&nbsp;&nbsp;</u></SUB>，与其他两种方法相比，这种方法的优点是<SUB><u>&nbsp;&nbsp;&nbsp;&nbsp;</u></SUB>（写出一点）．','','','','','','2KMnO<SUB>4</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>K<SUB>2</SUB>MnO<SUB>4</SUB>+MnO<SUB>2</SUB>+O<SUB>2</SUB>↑$###$2Na<SUB>2</SUB>O<SUB>2</SUB>+2CO<SUB>2</SUB>=2Na<SUB>2</SUB>CO<SUB>3</SUB>+O<SUB>2</SUB>$###$③$###$常温下就能进行（或操作简单，还能将人呼吸产生的二氧化碳转变为呼吸所需的氧气等，合理即可）','【解答】解：①加热高锰酸钾，需要在加热的条件下才能放出氧气，而加热时也会消耗氧气，不适宜在潜水器里供给氧气，反应的化学方程式为：2KMnO<SUB>4</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>K<SUB>2</SUB>MnO<SUB>4</SUB>+MnO<SUB>2</SUB>+O<SUB>2</SUB>↑．<br />②电解水，需要通电，消耗电能，不适宜在潜水器里供给氧气．<br />③在常温下过氧化钠固体（Na<SUB>2</SUB>O<SUB>2</SUB>）与CO<SUB>2</SUB>反应生碳酸钠和氧气，常温下即可发生反应，且可把空气中的二氧化碳转化为氧气，供给氧气的同时吸收了人呼出的二氧化碳，适宜在潜水器里供给氧气．反应的化学方程式为：2Na<SUB>2</SUB>O<SUB>2</SUB>+2CO<SUB>2</SUB>=2Na<SUB>2</SUB>CO<SUB>3</SUB>+O<SUB>2</SUB>．<br />与其它方法相比，这种方法常温下就能进行，不用加热、不用通电，操作简单，还能将人呼吸产生的二氧化碳转变为呼吸所需的氧气．<br />故答案为：（1）2KMnO<SUB>4</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>K<SUB>2</SUB>MnO<SUB>4</SUB>+MnO<SUB>2</SUB>+O<SUB>2</SUB>↑；2Na<SUB>2</SUB>O<SUB>2</SUB>+2CO<SUB>2</SUB>=2Na<SUB>2</SUB>CO<SUB>3</SUB>+O<SUB>2</SUB>；<br />（2）③；常温下就能进行（或操作简单，还能将人呼吸产生的二氧化碳转变为呼吸所需的氧气等，合理即可）．','【分析】根据潜水艇里的环境及对供给氧气反应的要求，分析所涉及反应的特点，判断出最适合潜水艇里供给氧气的反应原理即可．','书写',3.00,'7e514a8c0b320df0b219598db8bcd6f3',9,400,'制取气体的反应原理的探究,书写化学方程式、文字表达式、电离方程式','',2014,'35','2014秋•余杭区期中',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1848674,'化学概念在逻辑上存在如图所示关系，对下列概念间的关系说法正确的是（　　）<br /><img  src=\"/tikuimages/9/2015/400/shoutiniao20/7e15b440-94d4-11e9-8bbf-b42e9921e93e_xkb11.png\" style=\"vertical-align:middle\" /><br />包含关系并列关系交叉关系<br />①纯净物与混合物属于包含关系②化合物与氧化物属于包含关系<br />③单质与化合物属于交叉关系<br />④金属元素与非金属元素属于并列关系<br />⑤中和反应与复分解反应属于并列关系<br />⑥氧化反应与化合反应属于交叉关系．','2个','3个','4个','5个','','B','【解答】解：①物质按含有物质种类的多少可分为纯净物与混合物，是并列关系，故此选项错误．<br />②化合物有多种元素组成，其中氧化物是含有氧元素和另外一种元素的化合物，是包含关系，故此选项正确．<br />③纯净物按元素的含有情况可分为化合物与单质，故是并列关系，故此选项错误．<br />④元素分为金属元素和非金属元素，因此金属元素与非金属元素属于并列关系，故此选项正确．<br />⑤中和反应属于复分解反应，故中和反应与复分解反应属于包含关系，故此选项错误．<br />⑥氧化反应与化合反应属于交叉关系，正确；<br />故选B．','【分析】应用各知识点的概念，理解概念间相关的关系，结合图示所提供的关系意义，分析相关的选项从而判断正确与否，从物质分类的知识可知物质可分为纯净物与混合物，纯净物又可分为化合物与单质，化合物中又可分为酸碱盐及氧化物等；从元素的分类看，元素分为金属元素和非金属元素．','选择题',3.00,'67f889a4938c8572408a18789841e28e',9,400,'中和反应及其应用,复分解反应及其发生的条件,纯净物和混合物的概念,单质和化合物的判别,元素的简单分类,化合反应及其应用,氧化反应','',2015,'37','2015秋•武汉校级月考',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1848715,'<img  src=\"/tikuimages/9/2016/400/shoutiniao66/7ea215c0-94d4-11e9-8830-b42e9921e93e_xkb12.png\" style=\"vertical-align:middle;FLOAT:right\" />（2016•长春模拟）根据图示完成有关气体制备的相关问题：<br />（1）写出标号为①的仪器名称<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（2）制取气体前图C的操作名称是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（3）实验室用图A装置制取4.4g二氧化碳至少需要碳酸钙的质量为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，将产生的气体通入图B装置中，无明显现象，可能的原因是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','铁架台$###$检查装置的气密性$###$10g$###$二氧化碳中混有氯化氢气体','【解答】解：（1）标号为①的仪器名称为铁架台；<br />（2）制取气体前图C的操作是检查装置的气密性；<br />（3）设需CaCO<SUB>3</SUB>的质量为x．<br />CaCO<SUB>3</SUB>+2HCl=CaCl<SUB>2</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑<br />100&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;44 <br />&nbsp;x&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;4.4g<br />则<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">100</td></tr><tr><td>44</td></tr></table>=<table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">x</td></tr><tr><td>4.4g</td></tr></table></span>，解得x=10g．<br />答：需CaCO<SUB>3</SUB>的质量为10g．<br />因为制取二氧化碳中混有氯化氢气体，B中的澄清石灰水无明显现象．<br />故答案为：<br />（1）铁架台；（2）检查装置的气密性；（3）10g，二氧化碳中混有氯化氢气体．','【分析】（1）根据仪器名称回答；<br />（2）制取气体前图C的操作是检查装置的气密性；<br />（3）根据碳酸钙与盐酸反应的方程式，由二氧化碳的质量可以求出碳酸钙的质量；考虑制取二氧化碳中混有氯化氢气体，B中的澄清石灰水无明显现象．','简答题',3.00,'b4155ca0183812e727a4c7fe572d5ff4',9,400,'二氧化碳的实验室制法,制取二氧化碳的操作步骤和注意点,根据化学反应方程式的计算','',2016,'32','2016•长春模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1848732,'<img  src=\"/tikuimages/9/2014/400/shoutiniao0/7ec30b40-94d4-11e9-91ab-b42e9921e93e_xkb15.png\" style=\"vertical-align:middle;FLOAT:right\" />（2014秋•武侯区期末）水是生命之源，“珍惜水、节约水、保护水”是每个公民的义务和责任．<br />①用如图甲装置进行电解水的实验，b中收集到的气体是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，该反应的基本反应类型是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />②有些地方的村民用地下水作为生活用水，人们常用<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>检验地下水是硬水还是软水；生活中可用<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>的方法降低水的硬度；某同学自制如图乙所示简易净水器，图中所标注的材料中属于可燃物的有<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>、<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>、<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；该简易净水器主要利用了活性炭的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>性．','','','','','','氢气$###$分解反应$###$肥皂水$###$煮沸$###$纱布$###$活性炭$###$膨松棉$###$吸附','【解答】解：①由电解水的装置可知，b中收集到的气体较多是氢气，则a中收集的是氧气，该反应的基本类型是分解反应；<br />②人们常用肥皂水检验地下水是硬水还是软水，遇肥皂水产生的泡沫少的是硬水，遇肥皂水产生的泡沫多的是软水；生活中可用煮沸的方法降低水的硬度；图中所标注的材料中属于可燃物的有：纱布、活性炭、膨松棉，由于活性炭有吸水性，在简易净水器中，活性炭的主要作用是吸附．<br />故答案为：①氢气，分解反应；<br />②肥皂水，煮沸；纱布、活性炭、膨松棉，吸附．','【分析】①根据电解水实验的现象、原理分析回答；<br />②根据硬水、软水的鉴别、硬水的软化，活性炭的吸附性分析回答．','填空题',3.00,'003ea52621ac707148ffb89f1d4bd8a7',9,400,'电解水实验,硬水与软水,反应类型的判定,易燃物和易爆物安全知识','',2014,'33','2014秋•武侯区期末',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1848884,'在生活中，下列做法不正确的是（　　）','通常铝制品很耐腐蚀，清洗时不宜用钢刷、沙等来擦洗','为了保护金属资源，目前已经广泛用塑料来代替钢和其他合金制造管道、齿轮和汽车零部件等','粉刷墙壁用的乳胶漆是悬浊液','把水喷向空中，可以增加养鱼池水中氧气的溶解量','','C','【解答】解：A、通常铝制品很耐腐蚀，是因为铝能和空气中的氧气反应生成氧化铝，用钢刷、沙等来擦洗时能够把氧化铝擦去，因此清洗时不宜用钢刷、沙等来擦洗，该选项说法正确；<br />B、为了保护金属资源，目前已经广泛用塑料来代替钢和其他合金制造管道、齿轮和汽车零部件等，该选项说法正确；<br />C、粉刷墙壁用的乳胶漆是乳浊液，该选项说法不正确；<br />D、把水喷向空中时，能够增大水和空气的接触面积，从而可以增加养鱼池水中氧气的溶解量，该选项说法正确．<br />故选：C．','【分析】铝能和空气中的氧气反应生成氧化铝；<br />寻找金属材料的替代品可以保护金属资源；<br />乳胶漆属于乳浊液；<br />水和空气的接触面积增大时，能够增加水的溶氧量．','选择题',3.00,'eb80b38d72944a24d655aa31954524fb',9,400,'悬浊液、乳浊液的概念及其与溶液的区别,气体溶解度的影响因素,金属的化学性质,金属资源的保护','',2015,'37','2015秋•香坊区校级月考',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1848892,'为保证长时间潜航，在潜水艇里要配备氧气再生装置，有以下几种制氧气的方法：①加热高锰酸钾，②电解水，③在常温下过氧化钠固体与二氧化碳反应生成碳酸钠和氧气．<br />（1）写出方法③的文字表达式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（2）你认为最适合在潜水艇里制氧气的方法是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填序号），与其他两种方法相比，这种方法的优点是（写一条即可）<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','2Na<SUB>2</SUB>O<SUB>2</SUB>+2CO<SUB>2</SUB>=2Na<SUB>2</SUB>CO<SUB>3</SUB>+O<SUB>2</SUB>$###$③$###$常温下就能进行（或操作简单，还能将人呼吸产生的二氧化碳转变为呼吸所需的氧气等，合理即可）','【解答】解：（1）在常温下过氧化钠固体与二氧化碳反应生成碳酸钠和氧气，反应的化学方程式为：2Na<SUB>2</SUB>O<SUB>2</SUB>+2CO<SUB>2</SUB>=2Na<SUB>2</SUB>CO<SUB>3</SUB>+O<SUB>2</SUB>．<br />（2）①加热高锰酸钾，需要在加热的条件下才能放出氧气，而加热时也会消耗氧气，不适宜在潜水器里供给氧气．<br />②电解水，需要通电，消耗电能，不适宜在潜水器里供给氧气．<br />③在常温下过氧化钠固体（Na<SUB>2</SUB>O<SUB>2</SUB>）与CO<SUB>2</SUB>反应生碳酸钠和氧气，常温下即可发生反应，且可把空气中的二氧化碳转化为氧气，供给氧气的同时吸收了人呼出的二氧化碳，适宜在潜水器里供给氧气．<br />与其它方法相比，这种方法常温下就能进行，不用加热、不用通电，操作简单，还能将人呼吸产生的二氧化碳转变为呼吸所需的氧气．<br />故答案为：（1）2Na<SUB>2</SUB>O<SUB>2</SUB>+2CO<SUB>2</SUB>=2Na<SUB>2</SUB>CO<SUB>3</SUB>+O<SUB>2</SUB>；（2）③；常温下就能进行（或操作简单，还能将人呼吸产生的二氧化碳转变为呼吸所需的氧气等，合理即可）．','【分析】根据潜水艇里的环境及对供给氧气反应的要求，分析所涉及反应的特点，判断出最适合潜水艇里供给氧气的反应原理即可．','书写',3.00,'bf83cdf843afee682e84f60efd216308',9,400,'制取气体的反应原理的探究,书写化学方程式、文字表达式、电离方程式','',2015,'35','2015秋•广元校级期中',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1848935,'请根据如图所示的实验室中实验装置图填空．<br /><img  src=\"/tikuimages/9/0/400/shoutiniao80/812453cf-94d4-11e9-9c48-b42e9921e93e_xkb99.png\" style=\"vertical-align:middle\" /><br />（1）完成图中标有数字的仪器的名称：①<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>&nbsp;②<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />（2）用高锰酸钾制备氧气选择<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>装置，写出反应化学方程式：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）在实验室中，制取并收集二氧化碳选用装置为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>和<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填字母编号），反应的化学方程式为：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（4）实验室制得的CO<SUB>2</SUB>气体中常含有水蒸气．为了得到纯净、干燥的CO<SUB>2</SUB>气体，除杂装置的导管气流方向连接顺序是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填选项&nbsp;A．a→b&nbsp;&nbsp;&nbsp;B．b→a）．<br />（5）有一名同学欲用F装置收集H<SUB>2</SUB>，则H<SUB>2</SUB>应从导管口<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>&nbsp;&nbsp;通入；若用G来收集一瓶O<SUB>2</SUB>，则O<SUB>2</SUB>应从导管口<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>通入．<br />（6）已知MnO<SUB>2</SUB>固体和浓盐酸混合共热可制得氯气（Cl<SUB>2</SUB>），则应选用<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>做气体发生装置．','','','','','','长颈漏斗$###$集气瓶$###$D$###$2KMnO<SUB>4</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>K<SUB>2</SUB>MnO<SUB>4</SUB>+MnO<SUB>2</SUB>+O<SUB>2</SUB>↑$###$B$###$F$###$CaCO<SUB>3</SUB>+2HCl═CaCl<SUB>2</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑$###$A$###$b$###$a$###$A','【解答】解：（1）①是长颈漏斗，②是集气瓶．<br />故填：长颈漏斗；集气瓶．<br />（2）用高锰酸钾制备氧气需要加入，应该选择D装置，反应化学方程式为：2KMnO<SUB>4</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>K<SUB>2</SUB>MnO<SUB>4</SUB>+MnO<SUB>2</SUB>+O<SUB>2</SUB>↑．<br />故填：D；2KMnO<SUB>4</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>K<SUB>2</SUB>MnO<SUB>4</SUB>+MnO<SUB>2</SUB>+O<SUB>2</SUB>↑．<br />（3）在实验室中，制取二氧化碳不需要加入，应该用B装置作为发生装置；<br />二氧化碳能够溶于水，不能用排水法收集，密度比空气大，可以用向上排空气法收集，即用F装置收集；<br />反应的化学方程式为：CaCO<SUB>3</SUB>+2HCl═CaCl<SUB>2</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑．<br />故填：B；F；CaCO<SUB>3</SUB>+2HCl═CaCl<SUB>2</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑．<br />（4）实验室制得的CO2气体中常含有水蒸气，为了得到纯净、干燥的二氧化碳气体，除杂装置的导管气流方向连接顺序是a→b．<br />故填：A．<br />（5）欲用F装置收集氢气，由于氢气密度比空气小，则氢气应从导管口b通入；<br />若用G来收集一瓶氧气，则氧气应从导管口a通入．<br />故填：b；a．<br />（6）已知二氧化锰固体和浓盐酸混合共热可制得氯气，则应选用A做气体发生装置．<br />故填：A．','【分析】（1）要熟悉各种仪器的名称、用途和使用方法；<br />（2）高锰酸钾受热时能够分解生成锰酸钾、二氧化锰和氧气；&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <br />为了防止高锰酸钾进入导管，通常在试管口塞一团棉花；<br />（3）实验室通常用大理石或石灰石和稀盐酸反应制取二氧化碳，反应不需要加热，大理石和石灰石的主要成分是碳酸钙，能和稀盐酸反应生成氯化钙、水和二氧化碳；&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<br />二氧化碳能够溶于水，密度比空气大；<br />（4）浓硫酸可以用来干燥二氧化碳；<br />（5）氢气密度比空气小，氧气不易溶于水；<br />（6）根据制取气体的反应物状态、反应条件、气体的性质可以选择发生装置和收集装置．','书写',3.00,'42dca37282f7214c89e0ebfc7df9cd44',9,400,'常用气体的发生装置和收集装置与选取方法,气体的净化（除杂）,实验室制取氧气的反应原理,二氧化碳的实验室制法,书写化学方程式、文字表达式、电离方程式','',0,'37','',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1848962,'下列叙述中正确的是（　　）','中和反应时四大基本类型反应之一','正常雨水的pH为7，酸雨的pH小于7','农作物一般适应在pH为7或接近7的土壤中生长','溶液碱性越强越易洗去油污，因此洗发液的pH应大于13','','C','【解答】解：A、中和反应属于复分解反应，不是四大基本类型反应之一，故选项说法错误．<br />B、空气中的二氧化碳气体溶于水生成碳酸，故正常雨水的pH约为5.6，酸雨的pH小于5.6，故选项说法错误．<br />C、一般作物在5.5＜pH＜8.5的土壤上，都能生长良好，过酸、过碱才对其生长不利，也才有改良的必要．5.5＜pH＜8.5的土壤被称为中性或接近中性，说法正确．<br />D、pH大于13的溶液碱性太强，对皮肤有害，故选项说法错误．<br />故选C．','【分析】A、根据四大基本类型反应进行分析．<br />B、正常雨水的pH≈5.6．<br />C、农作物一般适宜在pH=7或接近7的土壤中生长．<br />D、pH大于13的溶液碱性太强．','选择题',3.00,'cf56ba381267645ef81b607647035227',9,400,'中和反应及其应用,溶液的酸碱性与pH值的关系,酸碱性对生命活动和农作物生长的影响,酸雨的产生、危害及防治','',2016,'32','2016•重庆校级模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1849018,'下列实验方案的设计中，没有正确体现对比这种科学思想的是（　　）','<img  src=\"/tikuimages/9/2015/400/shoutiniao13/822262e1-94d4-11e9-8718-b42e9921e93e_xkb73.png\" style=\"vertical-align:middle\" /><br />探究二氧化碳的含量','B、<img  src=\"/tikuimages/9/2015/400/shoutiniao75/82252200-94d4-11e9-a16b-b42e9921e93e_xkb29.png\" style=\"vertical-align:middle\" /><br />食盐溶液的配制','C、<img  src=\"/tikuimages/9/2015/400/shoutiniao77/8227ba11-94d4-11e9-8ac5-b42e9921e93e_xkb30.png\" style=\"vertical-align:middle\" /><br />探究溶剂种类对物质溶解性的影响','<img  src=\"/tikuimages/9/2015/400/shoutiniao64/82280830-94d4-11e9-96d1-b42e9921e93e_xkb96.png\" style=\"vertical-align:middle\" /><br />探究二氧化锰的催化作用','','B','【解答】解：A、探究空气和人体呼吸产生的气体中二氧化碳的含量，加入等量的澄清石灰水起到了对比的作用；<br />B、10g食盐和5g食盐分别溶解在50g水和80g水中没有起对比的作用，错误；<br />C、2g碘在10ml酒精和10mL水中溶解，是对比的实验方法；<br />D、在过氧化氢溶液和加入二氧化锰的过氧化氢溶液的试管中分别伸进带火星的木条，起到了对比的作用；<br />故选B．','【分析】根据已有的实验的过程以及对比的方法进行分析解答即可．','选择题',3.00,'e9750a222f66e733c9a331a220bab9d5',9,400,'科学探究的基本方法','',2015,'35','2015秋•李沧区期中',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1849037,'下列化肥通过外观即可与其他三种化肥区分开来的是（　　）','磷酸二氢钾','尿素','氯化钾','磷矿粉','','D','【解答】解：A、磷酸二氢钾为白色晶体．<br />B、尿素为白色晶体．<br />C、氯化钾为白色晶体．<br />D、磷矿粉是灰白色的固体．<br />磷酸二氢钾、磷矿粉、尿素、磷矿粉从外观看均为白色晶体，只有磷矿粉是灰白色粉末，故从外观看与磷矿粉可与其他化肥相区别．<br />故选：D．','【分析】根据磷矿粉是灰白色的，碳酸铵、氯化钾和硝酸钾都是白色的晶体，进行分析判断．','选择题',3.00,'c565f605e320ec83d1ef864a51393dec',9,400,'化肥的简易鉴别','',2016,'32','2016•谷城县模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1849157,'小明和小红对市场上销售的一种真空充气包装的形如小枕头“蛋黄派”产生了兴趣，他们查阅资料获知：这种真空充气包装技术，即将食品装入包装袋，抽出包装袋内空气，再充入某种气体，然后封口．它能使食品保持原有的色、香、味及营养价值，防止食品受压而破碎变形．那么，这是什么气体呢？小红猜想是氮气，小明猜想是二氧化碳．<br />（1）请你帮助他们设计一个简单的实验方案，来判断小明的猜想是正确的，简要写出操作步骤（方法、现象和结论）：<table class=\"edittable\"><TBODY><TR><td width=187>实验步骤</TD><td width=187>现象及结论</TD></TR><TR><td><u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD><td><u>&nbsp;&nbsp;&nbsp;&nbsp;</u></TD></TR></TBODY></TABLE>（2）你认为食品充气包装，对所充气体的要求是：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．（至少写两点）','','','','','','用注射器抽取袋内气体，然后通入澄清的石灰水$###$石灰水变浑浊，则气体是二氧化碳$###$无毒；廉价；不与食品反应','【解答】解：（1）用注射器抽取袋内气体，然后通入澄清的石灰水中，若石灰水变浑浊，则是二氧化碳，否则不是二氧化碳；<br />故答案为：<br /><table class=\"edittable\"><TBODY><TR><td width=189>实验步骤</TD><td width=189>现象</TD><td width=189>结论</TD></TR><TR><td>用注射器抽取袋内气体，然后通入澄清的石灰水</TD><td>石灰水变浑浊</TD><td>气体是二氧化碳</TD></TR></TBODY></TABLE>（2）由于包装内充入的气体用于保护食品的变质，因此所充气体应无毒、不能与食品发生反应；为不增加食品售价，所充气体还就价格低廉；<br />故答案为：无毒；廉价；不与食品反应．','【分析】根据二氧化碳能使澄清的石灰水变浑浊，而氮气不能使石灰水变浑浊，故用澄清的石灰水去检验气体是否为二氧化碳．','填空题',3.00,'ea66eb292904dcbc5f179a836251e620',9,400,'食品干燥剂、保鲜剂和真空包装的成分探究,常见气体的检验与除杂方法','',2015,'35','2015秋•临汾校级期中',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1849159,'下列说法中，错误的是（　　）','原子一定都是由原子核和核外电子构成的','化学变化中元素种类一定不变化','最外层电子数为8的粒子一定是稀有气体的原子','物理变化中分子种类一定不变化','','C','【解答】解：A、原子都是由居于原子中心的原子核和核外电子构成的，故说法正确；<br />B、根据质量守恒定律，在化学变化中，元素种类不变，故说法正确；<br />C、最外层电子数为8的粒子是稀有气体的原子或某些离子；故说法错误；<br />D、在物理变化中，没新物质生成，分子种类不变，故说法正确．<br />故选C．','【分析】A、根据原子的构成分析，原子都是由原子核和核外电子构成的；<br />B、根据质量守恒定律分析，在化学变化中，元素种类不变；<br />C、最外层电子数为8的粒子是稀有气体的原子或某些离子；<br />D、根据物理变化的概念分析，在物理变化中，没新物质生成，分子种类不变．','选择题',3.00,'1e1a1113cc0490dfafbea83f309f5ce7',9,400,'原子的定义与构成,核外电子在化学反应中的作用,分子的定义与分子的特性,元素在化学变化过程中的特点','',2014,'35','2014春•成都校级期中',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1849384,'厨房中有一瓶白色固体，小明认为它可能是食盐，小刚认为它可能是白糖，就两个人的看法而言，应属于科学探究环节中的（　　）','猜想与假设','设计实验','获得结论','收集证据','','A','【解答】解：A、小明认为它可能是食盐，小刚认为它可能是白糖，这一过程属于科学探究环节中的猜想与假设，故选项正确．<br />B、小明认为它可能是食盐，小刚认为它可能是白糖，这一过程属于科学探究环节中的猜想与假设，不是设计实验，故选项错误．<br />C、小明认为它可能是食盐，小刚认为它可能是白糖，这一过程属于科学探究环节中的猜想与假设，不是获得结论，故选项错误．<br />D、小明认为它可能是食盐，小刚认为它可能是白糖，这一过程属于科学探究环节中的猜想与假设，不是收集证据，故选项错误．<br />故选：A．','【分析】科学探究的主要环节有提出问题→猜想与假设→制定计划（或设计方案）→进行实验→收集证据→解释与结论→反思与评价→拓展与迁移，据此结合题意进行分析判断．','选择题',3.00,'b51ff349a23daea6ce35870b821bda09',9,400,'科学探究的基本环节','',2015,'32','2015•黄陵县校级模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1849573,'下列实验设计，不能达到实验目的是（　　）<br /><img  src=\"/tikuimages/9/2015/400/shoutiniao1/88e1d56e-94d4-11e9-af02-b42e9921e93e_xkb32.png\" style=\"vertical-align:middle\" />','甲装置验证空气中有水分','乙装置验证CaO和水反应的热效应','丙装置验证质量守恒定律','丁装置验证X溶液中是否含有Cl<SUP>-</SUP>','','B','【解答】解：A、甲装置中，无水硫酸铜能和空气中的水蒸气结合成蓝色五水硫酸铜，可以验证空气中有水分；<br />B、装置中是水和钠，没有氧化钙，因此不能验证CaO和水反应的热效应；<br />C、硫酸铜和氢氧化钠反应生成氢氧化铜沉淀和硫酸钠，反应前后质量不变，能够验证质量守恒定律；<br />D、银离子能够氯离子结合成白色沉淀氯化银，丁装置能够验证X溶液中是否含有银离子．<br />故选：B．','【分析】无水硫酸铜能和水反应生成蓝色五水硫酸铜；<br />氧化钙和水反应生成氢氧化钙，同时放出热量；<br />化学反应遵循质量守恒定律；<br />银离子能够氯离子结合成白色沉淀氯化银．','选择题',3.00,'4d3999333f9ff9ad62cf9a47b7b498e8',9,400,'化学实验方案设计与评价,常见气体的检验与除杂方法,证明盐酸和可溶性盐酸盐,生石灰的性质与用途,质量守恒定律及其应用','余姚市',2015,'32','2015•余姚市模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1849612,'小明发现月饼盒里的脱氧剂部分呈红褐色，查阅资料得知脱氧剂中含有铁粉和活性炭．他猜想这包脱氧剂中可能含有①Cu和C；②Fe<SUB>2</SUB>O<SUB>3</SUB>和C；③Fe<SUB>3</SUB>O<SUB>4</SUB>和C；④Fe<SUB>3</SUB>O<SUB>4</SUB>、C和Fe；⑤Fe<SUB>2</SUB>O<SUB>3</SUB>、C和Fe其中猜想合理的是（　　）','①③⑤','②④⑤','④⑤','②⑤','','D','【解答】解：由题意可知，脱氧剂中含有铁粉和活性炭．在脱氧时，活性炭具有吸附性，化学性质稳定；铁粉与氧气、水反应生成锈，呈红褐色，铁锈的主要成分是Fe<SUB>2</SUB>O<SUB>3</SUB>．如果铁全部生成了铁锈，这包脱氧剂中只有Fe<SUB>2</SUB>O<SUB>3</SUB>和C；若果铁粉部分生成铁锈，这包脱氧剂中Fe<SUB>2</SUB>O<SUB>3</SUB>、C和Fe．<br />A、脱氧剂中含有铁粉和活性炭，由质量守恒定律可知，脱氧后的物质中不可能含有铜，故错误；<br />B、由铁生锈可知，锈的主要成分是Fe<SUB>2</SUB>O<SUB>3</SUB>、不是Fe<SUB>3</SUB>O<SUB>4</SUB>，故错误；<br />C、由锈的主要成分是Fe<SUB>2</SUB>O<SUB>3</SUB>可知，这包脱氧剂中不可能含有Fe<SUB>3</SUB>O<SUB>4</SUB>，故错误．<br />D、由脱氧剂的脱氧情况可知，②⑤猜想合理．故正确．<br />故选：D．','【分析】根据质量守恒定律进行分析猜想，铁粉生锈成红褐色，活性炭具有吸附性，化学性质稳定．','选择题',3.00,'d86da75b08760ade5416818af2cb50f7',9,400,'食品干燥剂、保鲜剂和真空包装的成分探究','',2013,'37','2013秋•萧山区月考',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1849660,'将少量的（1）面粉（2）味精（3）胡椒粉（4）醋（5）菜油，分别加入水中，振荡后，其中<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>形成悬浊液，<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>形成乳浊液，<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>形成溶液．','','','','','','①③$###$⑤$###$②④','【解答】解：①面粉不溶于水，和水混合后形成悬浊液；②味精能溶于水形成溶液；③胡椒粉不溶于水，本身是固体，它和水混合后形成悬浊液；④醋能溶于水形成溶液；⑤菜油是液体，不溶于水，能形成乳浊液，故答案：①③；⑤；②④．','【分析】根据溶液、悬浊液和乳浊液的概念和物质的溶解性分析解答．','填空题',3.00,'30b9a635288437709f776e502e62094e',9,400,'悬浊液、乳浊液的概念及其与溶液的区别','',0,'37','',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1850162,'<img  src=\"/tikuimages/9/2015/400/shoutiniao13/8fb3be8f-94d4-11e9-bedb-b42e9921e93e_xkb28.png\" style=\"vertical-align:middle;FLOAT:right\" />（2015秋•金华校级期末）二氧化碳是一种奇妙的气体，某校化学兴趣小组在老师指导下，进行了《实验室二氧化碳的制取及其性质》实验，请你帮助回答下列问题．<br />（1）在大试管中加入5粒石灰石、倾倒约试管体积三分之一的稀盐酸（1：1），取一支试管尝试用排水法收集CO<SUB>2</SUB>．该操作过程中合理的排列顺序是（选填序号）<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />①检验装置的气密性<br />②待有连续稳定气泡再排水集气<br />③将水槽中待收集气体的小试管注满水<br />④旋紧连有导管的单孔胶塞<br />⑤向大试管内5粒石灰石再倒入三分之一的盐酸（1：1）<br />（2）小刚在做完实验后，准备将反应后的废液倒进废液缸时，发现实验桌上有一瓶未知质量分数的Na<SUB>2</SUB>CO<SUB>3</SUB>溶液，他决定利用该废液，测定Na<SUB>2</SUB>CO<SUB>3</SUB>溶液中溶质的质量分数．他将废液过滤，然后向废液中慢慢滴加Na<SUB>2</SUB>CO<SUB>3</SUB>溶液，加入Na<SUB>2</SUB>CO<SUB>3</SUB>溶液的质量与生成沉淀质量的关系如图所示．<br />①在加入Na<SUB>2</SUB>CO<SUB>3</SUB>溶液的过程中，开始时没有发现沉淀生成，说明滤液中的溶质除含有CaCl<SUB>2</SUB>外，还含有的微粒是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />②所滴加的Na<SUB>2</SUB>CO<SUB>3</SUB>溶液中溶质的质量分数是多少？（计算结果精确到0.1%）','','','','','','①③⑤④②（或③①⑤④②）$###$H<SUP>+</SUP>、Cl<SUP>-</SUP>、H<SUB>2</SUB>O','【解答】解：（1）根据实验室制取二氧化碳的具体操作步骤和注意事项可得，操作过程中合理的排列顺序是：①③⑤④②（或③①⑤④②）；故填：①③⑤④②（或③①⑤④②）；<br />（2）①在加入Na<SUB>2</SUB>CO<SUB>3</SUB>溶液的过程中，开始时没有发现沉淀生成，说明碳酸钠并没有和氯化钙反应，滤液中的溶质除含有CaCl<SUB>2</SUB>外，还含有盐酸，含有的微粒是：H<SUP>+</SUP>、Cl<SUP>-</SUP>、H<SUB>2</SUB>O；<br />②分析图意，Na<SUB>2</SUB>CO<SUB>3</SUB>溶液与氯化钙溶液全部反应，生成5g碳酸钙沉淀．<br />&nbsp; 设：20gNa<SUB>2</SUB>CO<SUB>3</SUB>溶液中溶质的质量为x<br />Na<SUB>2</SUB>CO<SUB>3</SUB>+CaCl<SUB>2</SUB>=CaCO<SUB>3</SUB>↓+2NaCl&nbsp;<br />106&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 100<br />x&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 5g<br />&nbsp;<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">106</td></tr><tr><td>x</td></tr></table>=<table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">100</td></tr><tr><td>5g</td></tr></table></span><br />x═5.3g&nbsp;&nbsp; <br />该Na<SUB>2</SUB>CO<SUB>3</SUB>溶液中溶质的质量分数为<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">5.3g</td></tr><tr><td>25g-5g</td></tr></table></span>×100%=26.5%<br />答：该Na<SUB>2</SUB>CO<SUB>3</SUB>溶液中溶质的质量分数为26.5%．','【分析】（1）根据实验室制取二氧化碳的具体操作步骤和注意事项分别是：连→查（检查装置的气密性）→加（加入固体药品）→倒（倒入液体药品；注意长颈漏斗的下端要浸在液面下）→收集，进行解答；<br />（2）在加入Na<SUB>2</SUB>CO<SUB>3</SUB>溶液的过程中，开始时没有发现沉淀生成，说明碳酸钠并没有和氯化钙反应，滤液中的溶质除含有CaCl<SUB>2</SUB>外，还含有盐酸，由图意，Na<SUB>2</SUB>CO<SUB>3</SUB>溶液与氯化钙溶液全部反应，生成5 g碳酸钙沉淀，要计算碳酸钠溶液中溶质的质量分数，可根据生成沉淀的质量为5g，求出碳酸钠的质量，而这些碳酸钠是20g碳酸钠溶液中的碳酸钠的质量，就不难求出质量分数了．','填空题',3.00,'f3bda6fe69f26007c59e8a88e608cb5c',9,400,'制取二氧化碳的操作步骤和注意点,有关溶质质量分数的简单计算,根据化学反应方程式的计算','',2015,'33','2015秋•金华校级期末',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1850249,'小明同学发现某的包装袋内有一个小纸袋，上面写着“食品保鲜剂”，他很好奇，想和同学们对“食品保鲜剂”的成分进行探究．<br />【发现问题】“食品保鲜剂”的成分是什么？<br />【查阅资料】a．食品保鲜剂是利用其中的有效成分吸收空气中的氧气和水蒸气，防止食品变质．<br />b．初中阶段学习过的能吸收氧气或水蒸气的物质有：<br />①氢氧化钠&nbsp;&nbsp;②生石灰&nbsp;&nbsp;③铁粉&nbsp;&nbsp;④浓硫酸<br />【初步探究】同学们经过讨论，一致认为不可能是浓硫酸和氢氧化钠，理由是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />同学们打开小纸袋，发现袋内固体为白色，则该保鲜剂的有效成分一定不是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />【继续探究】同学们对袋内白色固体的成分继续探究：<br /><table class=\"edittable\"><TBODY><TR><td width=189>实验操作</TD><td width=189>实验现象</TD><td width=189>结论与解释</TD></TR><TR><td>1．取样品放入盛有水的试管中，用温度计测量水温先后的情况</TD><td>温度升高</TD><td>白色固体中含有<br /><u>&nbsp;&nbsp;&nbsp;&nbsp;</u>&nbsp;</TD></TR><TR><td>2．取样品放入研钵，与硝酸铵一起研磨</TD><td>&nbsp;<br /><u>&nbsp;&nbsp;&nbsp;&nbsp;</u>&nbsp;</TD><td>白色固体中含有氢氧化钙</TD></TR><TR><td>3．取样滴加稀盐酸</TD><td>有气体产生</TD><td>白色固体中含有<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，产生该气体的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．</TD></TR></TBODY></TABLE>【拓展应用】食品保鲜剂除了能够吸收氧气或水蒸气外，还应具备一些要求，如：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（写一点）等．','','','','','','氢氧化钠和浓硫酸具有强烈的腐蚀性且浓硫酸是液体$###$铁粉$###$氧化钙$###$有刺激性气味的气体产生$###$碳酸钙$###$CaCO<SUB>3</SUB>+2HCl═CaCl<SUB>2</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑$###$无毒','【解答】解：【初步探究】氢氧化钠和浓硫酸具有强烈的腐蚀性且浓硫酸是液体，所以不可能是浓硫酸和氢氧化钠，铁是黑色的固体，而袋内固体为白色，则该保鲜剂的有效成分一定不是铁；故填：氢氧化钠和浓硫酸具有强烈的腐蚀性且浓硫酸是液体；铁粉；<br />【继续探究】氧化钙溶于水放热、氢氧化钙和硝酸铵研磨会放出有刺激性气味的氨气，碳酸钙和盐酸反应放出二氧化碳气体，所以实验方案：<br /><table class=\"edittable\"><TBODY><TR><td width=189>实验操作</TD><td width=189>实验现象</TD><td width=189>结论与解释</TD></TR><TR><td>1．取样品放入盛有水的试管中，用温度计测量水温先后的情况</TD><td>温度升高</TD><td>白色固体中含有氧化钙</TD></TR><TR><td>2．取样品放入研钵，与硝酸铵一起研磨</TD><td>有刺激性气味的气体产生</TD><td>白色固体中含有氢氧化钙</TD></TR><TR><td>3．取样滴加稀盐酸</TD><td>有气体产生</TD><td>白色固体中含有碳酸钙，产生该气体的化学方程式为CaCO<SUB>3</SUB>+2HCl═CaCl<SUB>2</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑．</TD></TR></TBODY></TABLE>；故填：氧化钙；有刺激性气味的气体产生；碳酸钙；CaCO<SUB>3</SUB>+2HCl═CaCl<SUB>2</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑；<br />【拓展应用】食品保鲜剂除了能够吸收氧气或水蒸气外，还应该无毒．故填：无毒．','【分析】【初步探究】根据氢氧化钠和浓硫酸具有强烈的腐蚀性且浓硫酸是液体以及铁是黑色的固体进行解答；<br />【继续探究】根据氧化钙溶于水放热、氢氧化钙和硝酸铵研磨会放出有刺激性气味的氨气，碳酸钙和盐酸反应放出二氧化碳气体进行解答；<br />【拓展应用】根据食品保鲜剂除了能够吸收氧气或水蒸气外，还应该无毒进行解答．','书写',3.00,'97578c97368f23eb72518fd26df01ab2',9,400,'食品干燥剂、保鲜剂和真空包装的成分探究,浓硫酸的性质及浓硫酸的稀释,证明碳酸盐,生石灰的性质与用途,常见碱的特性和用途,碱的化学性质,书写化学方程式、文字表达式、电离方程式','',2016,'32','2016•临沂模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1850397,'普通锌锰电池由锌、碳棒和黑然糊状五物组成，其中黑色糊状物的主要成分是淀粉、氯化铵、氯化锌和二氧化锰等．某研究性学习小组设计如下流程，变废为宝．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao20/92bc43a1-94d4-11e9-8a02-b42e9921e93e_xkb41.png\" style=\"vertical-align:middle\" /><br />（1）“溶解”操作中，为提高溶解速率，可采取的措施是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>等；“操作I”的名称是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．请写出实验室进行该操作用到的一种仪器名称<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）废旧电池处理后得到的碳棒，具有导电性，还有<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>等（填一种即可）化学性质，请写出具有这种性质的一个反应方程式：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）黑色滤渣中的二氧化锰与浓盐酸共热可制氯气（Cl<SUB>2</SUB>）．同时生成氯化锰（MnCl<SUB>2</SUB>）和一种化合物，反应的化学方程式表示如下：MnO<SUB>2</SUB>+4HCl（浓）<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;加热&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>MnCl<SUB>2</SUB>+Cl<SUB>2</SUB>+2<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','用玻璃棒搅拌$###$过滤$###$漏斗$###$可燃$###$C+O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>CO<SUB>2</SUB>$###$H<SUB>2</SUB>O','【解答】解：（1）“溶解”操作中，为提高溶解速率，可采取的措施是用玻璃棒搅拌；过滤可用于难溶性固体和可溶性固体的分离，所以“操作Ⅰ”的名称是过滤，实验室进行该操作用到的一种仪器是漏斗；<br />（2）碳具有可燃性，可以用作燃料，碳和氧气在点燃的条件下生成二氧化碳，化学方程式为：C+O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>CO<SUB>2</SUB>；<br />（3）二氧化锰与浓盐酸共热反应生成氯气（Cl<SUB>2</SUB>），同时生成氯化锰（MnCl<SUB>2</SUB>）和水，化学方程式为MnO<SUB>2</SUB>+4HCl（浓）<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>MnCl<SUB>2</SUB>+Cl<SUB>2</SUB>↑+2H<SUB>2</SUB>O，所以需要填的物质是：H<SUB>2</SUB>O．<br />故答案为：（1）用玻璃棒搅拌，过滤，漏斗；<br />（2）可燃，C+O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>CO<SUB>2</SUB>；<br />（3）H<SUB>2</SUB>O．','【分析】（1）根据“溶解”操作中，为提高溶解速率，可采取的措施是用玻璃棒搅拌以及过滤可用于难溶性固体和可溶性固体的分离，以及过滤操作中需要的仪器进行解答；<br />（2）根据碳具有可燃性或还原性进行解答；<br />（3）根据二氧化锰与浓盐酸共热反应生成氯气（Cl<SUB>2</SUB>），同时生成氯化锰（MnCl<SUB>2</SUB>）和水进行解答．','书写',3.00,'c9830672a1fa70f179288ec9b669aa3d',9,400,'混合物的分离方法,过滤的原理、方法及其应用,影响溶解快慢的因素,碳的化学性质,质量守恒定律及其应用,书写化学方程式、文字表达式、电离方程式','',2016,'37','2016春•太和县校级月考',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1850405,'俗话说“桂花香，蟹黄肥”．温州是盛产螃蟹的地方，青色的螃蟹煮熟后颜色会变成红色．一些同学认为这种红色物质可能就象酸碱指示剂一样，遇到酸或碱颜色会发生改变．就这些同学的看法而言应属于科学探究中的（　　）','实验','做结论','观察','假设','','D','【解答】解：A、一些同学认为这种红色物质可能就象酸碱指示剂一样，遇到酸或碱颜色会发生改变，属于科学探究中的假设，而不是实验，故选项错误．<br />B、一些同学认为这种红色物质可能就象酸碱指示剂一样，遇到酸或碱颜色会发生改变，属于科学探究中的假设，而不是做结论，故选项错误．<br />C、一些同学认为这种红色物质可能就象酸碱指示剂一样，遇到酸或碱颜色会发生改变，属于科学探究中的假设，而不是观察，故选项错误．<br />D、一些同学认为这种红色物质可能就象酸碱指示剂一样，遇到酸或碱颜色会发生改变，属于科学探究中的假设，故选项正确．<br />故选：D．','【分析】科学探究的主要环节有提出问题→猜想与假设→制定计划（或设计方案）→进行实验→收集证据→解释与结论→反思与评价→拓展与迁移，据此结合题意进行分析判断．','选择题',3.00,'32d8d87ca7f45eb303f56cd51dad6a91',9,400,'科学探究的基本环节','嵊州市',2011,'37','2011秋•嵊州市校级月考',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1850430,'有下列装置，其中A是气体发生装置，B、C、D、E是气体收集装置，E还可作气体的净化装置（E装置有多个备用）．<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao34/93247b51-94d4-11e9-9d2b-b42e9921e93e_xkb3.png\" style=\"vertical-align:middle\" /><br />（1）当用大理石和盐酸制取CO<SUB>2</SUB>时，选取的收集装置是（填B或C或D）<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，发生反应的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；为了得到纯净、干燥的CO<SUB>2</SUB>，小睿同学在A装置后连续接了两个E装置，她想利用第一个E装置注入足量的NaOH溶液来除去HCl气体，你认为她的方法是否恰当并说明理由<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，她在第二个E装置中注入了浓硫酸，你认为其作用是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）当用过氧化氢溶液和二氧化锰来制取氧气时，A中发生反应的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，若导管口a与导管口c相连来收集氧气，还需将装置E<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','C$###$CaCO<SUB>3</SUB>+2HCl=CaCl<SUB>2</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑$###$不恰当，因CO<SUB>2</SUB>也会被除去$###$除去水蒸气$###$2H<SUB>2</SUB>O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span>&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O+O<SUB>2</SUB>↑$###$注满水或倒立','【解答】解：（1）二氧化碳密度比空气大，易溶于水，所以选取的收集装置是C，碳酸钙和盐酸反应生成氯化钙、水和二氧化碳，化学方程式为：CaCO<SUB>3</SUB>+2HCl=CaCl<SUB>2</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑，氢氧化钠会与氯化氢、二氧化碳反应，所以利用第一个E装置注入足量的NaOH溶液来除去HCl气体的做法不恰当，因CO<SUB>2</SUB>也会被除去，浓硫酸有吸水性，所以第二个E装置中注入了浓硫酸的作用是除去水蒸气；<br />（2）过氧化氢在二氧化锰的催化作用下生成水和氧气，化学方程式为：2H<SUB>2</SUB>O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span>&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O+O<SUB>2</SUB>↑，氧气密度大于空气，不易溶于水，可用向上排空气法和排水法收集．当a与c相连收集氧气时，若集气瓶内是空气，则需将集气瓶倒放，若集气瓶正放，则需装满水用排水法收集．<br />故答案为：（1）C，CaCO<SUB>3</SUB>+2HCl=CaCl<SUB>2</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑，不恰当，因CO<SUB>2</SUB>也会被除去，除去水蒸气；<br />（2）2H<SUB>2</SUB>O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span>&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O+O<SUB>2</SUB>↑，注满水或倒立．','【分析】（1）根据二氧化碳密度比空气大，易溶于水，碳酸钙和盐酸反应生成氯化钙、水和二氧化碳，氢氧化钠会与氯化氢、二氧化碳反应，浓硫酸有吸水性进行分析；<br />（2）根据过氧化氢在二氧化锰的催化作用下生成水和氧气，多功能装置用排水法收集时要注满水，气体从短管进，用排空气法收集要根据密度大小及瓶子的正倒放确定进气管．','书写',3.00,'4cffe065084fed2dd2fe5edfc2e31163',9,400,'气体的净化（除杂）,实验室制取氧气的反应原理,二氧化碳的实验室制法,书写化学方程式、文字表达式、电离方程式','靖江市',2016,'37','2016春•靖江市校级月考',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1850450,'研究和学习化学，有许多方法．下列方法中所举例错误的是（　　）<br /><table class=\"edittable\"><TBODY><TR><td width=54>选项</TD><td width=66>方法</TD><td width=416>示例</TD></TR><TR><td>A</TD><td>实验法</TD><td>用磷做“测定空气中氧气含量”的实验</TD></TR><TR><td>B</TD><td>分类法</TD><td>根据组成物质的元素种类，将纯净物分为单质和化合物</TD></TR><TR><td>C</TD><td>归纳法</TD><td>根据稀盐酸、稀硫酸等物质的化学性质，归纳出酸的通性</TD></TR><TR><td>D</TD><td>类推法</TD><td>根据金属铝能与稀盐酸反应，推测金属铜也能与稀盐酸反应</TD></TR></TBODY></TABLE>','A','B','C','D','','D','【解答】解：A．足量的磷在集气瓶中完全燃烧，冷却至室温后，进入集气瓶内的水的体积即为集气瓶内空气中氧气的体积，故A正确；<br />B．根据组成纯净物的元素的种类，有同一种元素组成的纯净物，属于单质，由两种或两种以上的元素组成的纯净物，属于化合物，故B正确；<br />C．因为稀盐酸和稀硫酸都能电离出氢离子，故二者具有相似的化学性质--酸的通性，故C正确；<br />D．在金属活动顺序中，铝排在氢的前面，能与稀盐酸反应生成氢气，而铜排在氢的后面，不能与稀盐酸反应产生氢气，故D错误．<br />故选D．','【分析】A．根据测定空气中氧气含量的方法来分析；<br />B．根据纯净物的分类来分析；<br />C．根据酸的通性来分析解答；<br />D．根据金属的活动性顺序来分析．','选择题',3.00,'66a1a2d1badf3db710fe74d07f011363',9,400,'科学探究的基本方法','',2016,'32','2016•洛阳模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1850481,'盐酸是一种重要的化工产品，也是实验室中重要的化学试剂，用途广泛．<br />（1）增大压强，HCl由气态为液态，从微观的角度分析该变化过程中改变的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）蘸浓盐酸的玻璃棒和蘸浓氨水的玻璃棒接近但不接触，发生了如图1的现象，反应的化学方程式是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，浓盐酸和浓氨水都具有挥发性，气体有刺激性和腐蚀性，实验时要注意<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）观察图2并回答问题：<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao22/93bfa9e1-94d4-11e9-80c3-b42e9921e93e_xkb43.png\" style=\"vertical-align:middle\" />&nbsp;<br />由图2可知，HCl和NaOH的反应实质是H<SUP>+</SUP>和OH<SUP>-</SUP>之间的反应，此反应可表示为：H<SUP>+</SUP>+0H<SUP>-</SUP>=H<SUB>2</SUB>O．像这种用实际参与反应的离子来表示反应的式子称为离子方程式．任何复分解反应都可用离子方程式来表示．<br />【练习】按照书写化学方程式的要求写出下列反应的离子方程式<br />HCl溶液与AgNO溶液反应：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />HNO<SUB>3</SUB>溶液与Na<SUB>2</SUB>CO<SUB>3</SUB>溶液反应：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />【归纳】复分解反应的实质是：阴阳离子结合生成<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>的反应．<br />【应用】判断在水溶液中一定能大量共存的离子组是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />A．Na<SUP>+</SUP>、H<SUP>+</SUP>、Cl<SUP>-</SUP>、CO<SUB>3</SUB><SUP>2-</SUP>　　　　　B．H<SUP>+</SUP>、Ba<SUP>2+</SUP>、Cl<SUP>-</SUP>、SO<SUB>4</SUB><SUP>2-</SUP><br />C．Cu<SUP>2+</SUP>、Na<SUP>+</SUP>、NO<SUB>3</SUB><SUP>-</SUP>、OH<SUP>-</SUP>　　　　 D．H<SUP>+</SUP>、K<SUP>+</SUP>、Cl<SUP>-</SUP>、SO<SUB>4</SUB><SUP>2-</SUP>．','','','','','','氯化氢分子之间的间隔变小$###$NH<SUB>3</SUB>+HCl=NH<SUB>4</SUB>Cl$###$在通风厨中进行$###$Ag<SUP>+</SUP>+Cl<SUP>-</SUP>=AgCl$###$2H<SUP>+</SUP>+CO<SUB>3</SUB><SUP>2-</SUP>=H<SUB>2</SUB>O+CO<SUB>2</SUB>↑$###$水、沉淀、气体$###$D','【解答】解：（1）增大压强，HCl由气态为液态，从微观的角度分析该变化过程中改变的是氯化氢分子之间的间隔变小．<br />故填：氯化氢分子之间的间隔变小．<br />（2）氨气和氯化氢反应的化学方程式是：NH<SUB>3</SUB>+HCl=NH<SUB>4</SUB>Cl，浓盐酸和浓氨水都具有挥发性，气体有刺激性和腐蚀性，实验时要注意在通风厨中进行．<br />故填：NH<SUB>3</SUB>+HCl=NH<SUB>4</SUB>Cl；在通风厨中进行．<br />（3）HCl溶液与AgNO溶液反应的离子方程式为：Ag<SUP>+</SUP>+Cl<SUP>-</SUP>=AgCl．<br />故填：Ag<SUP>+</SUP>+Cl<SUP>-</SUP>=AgCl．<br />HNO<SUB>3</SUB>溶液与Na<SUB>2</SUB>CO<SUB>3</SUB>溶液反应的离子方程式为：2H<SUP>+</SUP>+CO<SUB>3</SUB><SUP>2-</SUP>=H<SUB>2</SUB>O+CO<SUB>2</SUB>↑．<br />故填：2H<SUP>+</SUP>+CO<SUB>3</SUB><SUP>2-</SUP>=H<SUB>2</SUB>O+CO<SUB>2</SUB>↑．<br />复分解反应的实质是：阴阳离子结合生成 水、沉淀、气体的反应．<br />故填：水、沉淀、气体．<br />A、氢离子和碳酸根离子结合生成水和二氧化碳，因此该选项离子不能共存；<br />B、钡离子能和硫酸根离子结合成硫酸钡沉淀，因此该选项不能共存；<br />C、铜离子和氢氧根离子结合成氢氧化铜沉淀，因此该选项不能共存；<br />D、离子之间不能反应，因此该选项能够共存．<br />故填：D．','【分析】（1）微观粒子之间有间隔；<br />（2）氨气和氯化氢反应生成氯化铵；<br />（3）书写离子方程式要注意遵循电荷守恒和质量守恒．','书写',3.00,'6785bbdeaec903a61e57435bc1503a9b',9,400,'分子的定义与分子的特性,复分解反应的条件与实质,书写化学方程式、文字表达式、电离方程式','诸城市',2016,'37','2016•诸城市一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1850498,'为了净化和收集由盐酸和大理石制得的CO<SUB>2</SUB>气体，从如图中选择合适的装置并连接．合理的是（　　）<br /><img  src=\"/tikuimages/9/2013/400/shoutiniao0/93e903cf-94d4-11e9-b413-b42e9921e93e_xkb51.png\" style=\"vertical-align:middle\" />','a-a′→d-d′→e','b-b′→d-d′→g','c-c′→d-d′→g','d-d′→c-c′→f','','C','【解答】解：实验室用稀盐酸和固体大理石在常温下制取二氧化碳，其反应化学式为：CaCO3+2HCl=CaCl2+H2O+CO2↑，因氯化氢易挥发，所以制得的二氧化碳中含有氯化氢和水蒸气，为了净化二氧化碳，图中提供了四组洗气装置和三组收集装置．<br />洗气装置：用水吸收：水无法吸收氯化氢、二氧化碳混合气体中的氯化氢，该法不能选；用氢氧化钠溶液吸收：氯化氢能和氢氧化钠发生中和反应，但二氧化碳和氢氧化钠反应生成碳酸钠和水，该法不能选；用碳酸氢钠溶液吸收：碳酸氢钠和氯化氢反应生成氯化钠水和二氧化碳、将杂质氯化氢转化为二氧化碳，该法能选；用浓硫酸吸收：浓硫酸和二氧化碳不反应且能吸收其中的水蒸气，故选；<br />收集装置：二氧化碳能溶于水，不能用排水法收集；二氧化碳密度比空气大，需用向上排空气法收集．<br />因此仪器的连接顺序为：c-c′→d-d′→g；<br />故选C．','【分析】实验室用稀盐酸和固体大理石在常温下制取二氧化碳，因氯化氢易挥发，所以为了得到干燥的二氧化碳，需除去氯化氢和水蒸气，饱和碳酸氢钠溶液可以吸收HCl而不吸收CO2，用浓硫酸可以干燥二氧化碳，收集装置主要由气体的密度和溶水性决定，二氧化碳密度比空气大，应用向上排空气法收集，据此分析解答．','选择题',3.00,'e48c7ce3fb602c0ff59f8805d4220bad',9,400,'气体的净化（除杂）','',2013,'32','2013•宁波模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1850662,'<img  src=\"/tikuimages/9/0/400/shoutiniao79/95cdf070-94d4-11e9-9638-b42e9921e93e_xkb69.png\" style=\"vertical-align:middle;FLOAT:right;\" />卤代烷烃曾用作冰箱制冷剂，但在高空中与某些气体发生反应，破坏臭氧层，而被逐渐淘汰．如氟氯甲烷（CF<SUB>2</SUB>Cl<SUB>2</SUB>）在高空中会发生如图所示的变化，根据图示，判断下列说法中不正确的是（　　）','②过程发生的变化是：Cl+O<SUB>2</SUB>→ClO+O<SUB>2</SUB>','③过程发生的变化是：ClO+O→Cl+O<SUB>2</SUB>','Cl反复作用使O<SUB>3</SUB>不断转变为O<SUB>2</SUB>','氟氯甲烷是上述总反应的催化剂','','A','【解答】解：A．由图分析，②过程发生的变化是：Cl+O<SUB>3</SUB>→ClO+O<SUB>2</SUB>，故错误；<br />B．由图分析，③过程发生的变化是：ClO+O→Cl+O<SUB>2</SUB>，故正确；<br />C．根据总的反应过程可以看出，Cl没有发生变化，只是使O<SUB>3</SUB>不断转变为O<SUB>2</SUB>，故正确；<br />D．氟氯甲烷是上述反应中，加快了反应的速率，而其质量和化学性质没有发生变化，属于催化剂，故正确．<br />故选A．','【分析】A．根据反应②的过程来分析；<br />B．根据反应③的过程来分析；<br />C．根据反应中Cl的作用来分析．<br />D．根据氟氯甲烷在整个反应中的作用来分析．','选择题',3.00,'9976ef469d981d5dcb898f2290460698',9,400,'催化剂的特点与催化作用,臭氧空洞和臭氧层保护','',0,'37','',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1850682,'<img  src=\"/tikuimages/9/0/400/shoutiniao14/9613d30f-94d4-11e9-a0d4-b42e9921e93e_xkb20.png\" style=\"vertical-align:middle;FLOAT:right;\" />背景1：市场上销售的豆腐干、盐水鸭等食品，常采用真空包装．真空包装的目的是除去空气使大多数微生物因缺少<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>气而受到抑制，停止繁殖，从而防止食品变质．<br />背景2：“可比克”等膨化食品、茶叶等常采用真空充气包装，即将食品装入包装袋，抽出包装袋内的空气，再充入防止食品变质的气体，然后封口．真空充气包装能使食品保持原有的色、香、味及营养价值，防止食品受压而破碎变形．<br />【提出问题】充入食品包装袋中的气体是什么？<br />【猜想假设】经过同学们的激烈讨论，最后同学们提出了气体可能三种情况：①N<SUB>2</SUB>、②CO<SUB>2</SUB>、③H<SUB>2</SUB>．<br />【提出方案】<br />（1）小明同学认为气体不可能是H<SUB>2</SUB>，因为它具有<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>性．<br />（2）为了继续探究气体的成分，小华和小红两位同学各提出了一种方案．<br />①小华方案：收集一瓶该气体，把一根燃着的木条伸入瓶中，发现木条熄灭．由此得出气体是CO<SUB>2</SUB>．该方案是否可行？<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（是或否）．理由是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />②小红的方案被同学们认可．走进食品生产工厂，在装有该气体的容器（容器内压强接近大气压强）上连接了如图一个装置．<br />（a）为了使容器内的气体通过澄清石灰水，活塞应向<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（左或右）移动．<br />（b）不管气体是CO<SUB>2</SUB>还是N<SUB>2</SUB>，试管中都有的实验现象是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（c）经过实验后，发现<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，证明气体是CO<SUB>2</SUB>．','','','','','','氧$###$可燃$###$否$###$氮气也能使木条熄灭$###$有$###$有气泡冒出$###$稀有气体','【解答】解：背景1、大多数微生物在缺少氧气的环境中生命活动受到抑制，有机物在氧气存在的情况下容易发生缓慢氧化．<br />（2）因为氢气有可燃性，用做保护气会不安全；①因为氮气和二氧化碳都不支持燃烧，都能使木条熄灭，故小华的方案不可行．<br />（a）为了使容器内的气体通过澄清石灰水，活塞应向右移动．<br />（b）不管气体是CO<SUB>2</SUB>还是N<SUB>2</SUB>，试管中都有的实验现象是有气泡冒出．<br />（c）根据对食品中充入的气体应该无毒并且不和食物反应，同学们经过讨论后，认为该气体除了是CO<SUB>2</SUB>，还可能是稀有气体．<br />故答案为：（1）氧气；（2）可燃，①否，氮气也能使木条熄灭②（a）右（b）有气泡冒出；（c）稀有气体．','【分析】背景1、大多数的食品因为氧气的存在而发生氧化，并且大多数的微生物的生存环境中氧气是必不可少的一种气体；<br />（2）①从氢气具有可燃性，氮气、二氧化碳都不支持燃烧判断．<br />②结合物理知识中压强的改变和会出现的实验现象分析．','填空题',3.00,'004a0f3591e4472cd07b8c4854e0db1a',9,400,'食品干燥剂、保鲜剂和真空包装的成分探究,常见气体的检验与除杂方法,燃烧、爆炸、缓慢氧化与自燃','',0,'37','',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1850698,'某种大理石除主要成分为CaCO<SUB>3</SUB>外，还有少量的硫化物．某同学用这种大理石和稀盐酸反应，分别开展以下探究：<br />查阅资料一：已知复分解反应CaCO<SUB>3</SUB>+2HCI═CO<SUB>2</SUB>↑+H<SUB>2</SUB>O+CaCl<SUB>2</SUB>可自发进行．在常温下，测得浓度均为a%的下列四种溶液的pH大小情况：HCl＜H<SUB>2</SUB>SO<SUB>4</SUB>＜H<SUB>2</SUB>S＜H<SUB>2</SUB>CO<SUB>3</SUB><br />资料二：常见干燥剂有①五氧化二磷②无水氯化钙③碱石灰④生石灰<br />请你参与探究并回答相关问题．<br />（1）pH大小情况揭示出复分解反应的一条规律：较强酸发生类似反应可以生成较弱酸．下列反应均能发生，其中不符合该规律的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填字母）<br />A．HCl+NaHCO<SUB>3</SUB>═NaCl+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑<br />B．2HCl+CaS═CaCl<SUB>2</SUB>+H<SUB>2</SUB>S↑<br />C．H<SUB>2</SUB>S+CuSO<SUB>4</SUB>═H<SUB>2</SUB>SO<SUB>4</SUB>+CuS↓<br />（2）为了得到纯净的二氧化碳，设计了如图装置，请你分析：<br /><img  src=\"/tikuimages/9/2013/400/shoutiniao0/9653c240-94d4-11e9-ab4a-b42e9921e93e_xkb90.png\" style=\"vertical-align:middle\" /><br />a．制备的CO<SUB>2</SUB>气体中，可能含有的杂质是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />b．上述装置中，A是CuSO<SUB>4</SUB>溶液，此溶液主要可以吸收<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />c．上述装置中，B物质的名称不可能是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（用标号表示）．如果B物质失效，测定CO<SUB>2</SUB>结果偏高．','','','','','','C$###$HCl、H<SUB>2</SUB>S、和水蒸气$###$HCl$###$碱石灰、生石灰','【解答】解：（1）由信息可知酸性强的酸能制取酸性弱的酸，因H<SUB>2</SUB>S的酸性比硫酸酸性弱，故C错误；<br />（2）a、因大理石中含有少量硫化物，加入盐酸后可能与盐酸反应生成硫化氢气体，所以生成的二氧化碳气体中可能含有H<SUB>2</SUB>S、HCl和水蒸气；<br />b、根据除杂的要求及题目信息，CuSO<SUB>4</SUB>溶液主要可以吸收HCl；<br />c、B物质用于干燥制得的CO<SUB>2</SUB>，可选用无水CaCl<SUB>2</SUB>或五氧化二磷，碱石灰、生石灰也能作干燥剂，但能与二氧化碳反应，故不可用；<br />若氯化钙失效会使水分干燥不彻底，使二氧化碳中混有水蒸气，使质量偏大，质量越大，相对分子质量越大．<br />故答案为：（1）C；<br />（2）a、HCl、H<SUB>2</SUB>S、和水蒸气；<br />b、HCl；<br />c、碱石灰、生石灰．','【分析】（1）根据题意：较强酸发生类似反应可以生成较弱酸分析解答；<br />（2）a、根据实验所用药品，制得的气体中可能含有H<SUB>2</SUB>S、HCl和水蒸气进行分析；<br />b、根据除杂的要求，碳酸氢钠溶液用于吸收HCl气体或吸收酸性气体进行分析；<br />c、根据B物质用于干燥制得的CO<SUB>2</SUB>，可选用无水CaCl<SUB>2</SUB>；根据氯化钙的作用分析解答．','填空题',3.00,'fd56262b76b19551b82ed7d4bdbdf34e',9,400,'制取气体的反应原理的探究,常见气体的检验与除杂方法,复分解反应的条件与实质','东阳市',2013,'37','2013春•东阳市月考',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1850762,'下列有关说法正确的是（　　）','含氧元素质量分数最高的氧化物是水','空气是一种重要的自然资源，它是由氧气和氮气两种气体组成的','地球上的水资源是丰富的，但可利用的淡水资源只约占全球水储量的2.53%','海洋中含有80多种元素，其中含量最多的金属元素是钠元素','','D','【解答】解：A．含氧元素质量分数最高的氧化物是过氧化氢，故错误；<br />B．空气主要是由氮气和氧气组成的，此外还包括稀有气体、二氧化碳等多种物质，故错误；<br />C．地球上的淡水只约占全球水储量的2.53%，其中可利用的淡水不足1%，故错误；<br />D．海洋中含有80多种元素，其中最多的金属元素是钠元素，故正确．<br />故填：D．','【分析】A．根据氧化物中氧元素的含量来分析；<br />B．根据空气的组成来分析；<br />C．根据水资源的分布进行解答；<br />D．根据海洋中含有80多种元素，其中最多的金属元素是钠元素解答．','选择题',3.00,'f19fd640af1752024a5bce33d1c1472b',9,400,'有关化学之最,空气的成分及各成分的体积分数,水资源状况,海洋中的资源','',2016,'37','2016春•哈尔滨校级月考',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1850775,'下列操作正确的是（　　）','<img  src=\"/tikuimages/9/2015/400/shoutiniao75/972c6f00-94d4-11e9-ab06-b42e9921e93e_xkb95.png\" style=\"vertical-align:middle\" />','<img  src=\"/tikuimages/9/2015/400/shoutiniao50/972f0711-94d4-11e9-b183-b42e9921e93e_xkb24.png\" style=\"vertical-align:middle\" />','<img  src=\"/tikuimages/9/2015/400/shoutiniao38/97301880-94d4-11e9-adb6-b42e9921e93e_xkb90.png\" style=\"vertical-align:middle\" />','<img  src=\"/tikuimages/9/2015/400/shoutiniao8/973102e1-94d4-11e9-a118-b42e9921e93e_xkb8.png\" style=\"vertical-align:middle\" />','','C','【解答】解：A、取用液体后胶头滴管不能平放或倒置，以防止液体倒流，腐蚀胶头，图中所示操作错误．<br />B、用试管夹夹持试管时，应从试管底部往上套，夹在距管口约三分之一处，图中所示操作错误．<br />C、使用酒精灯时要注意“两查、两禁、一不可”，熄灭酒精灯应用灯帽盖灭，图中所示操作正确．<br />D、取用液体药品时，要先拿下瓶塞，倒放在桌面上，标签对准手心，瓶口与试管口紧挨，图中所示操作错误．<br />故选C．','【分析】A、根据胶头滴管的使用方法进行分析判断．<br />B、根据用试管夹夹持试管的方法进行分析判断．<br />C、根据使用酒精灯时的注意事项分析判断．<br />D、根据液体药品的取用方法进行分析判断．','选择题',3.00,'58f66fb5ce1183ad5a320a56371483c6',9,400,'加热器皿-酒精灯,挟持器-铁夹、试管夹、坩埚钳,液体药品的取用','',2015,'35','2015秋•漯河校级期中',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1850781,'甲酸（HCOOH）通常是一种无色易挥发的液体，它在浓硫酸作用下易分解，反应方程式为：<br />HCOOH&nbsp;<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td style=\"font-size: 90%\">浓硫酸</td></tr><tr><td><div hassize=\"9\"><div style=\"background: url(\'/upimages/images/part/8594L.png\') repeat-x; width:1px; height:5px;float:left;overflow:hidden\" muststretch=\"h\"></div><div style=\"background: url(\'/upimages/images/part/8594.png\') no-repeat; width:9px; height:5px;float:left;overflow:hidden\"></div></div></td></tr></table></span>&nbsp;CO↑+H<SUB>2</SUB>O，某课外活动小组的同学欲用该反应来制取CO，并还原红棕色的氧化铁粉末．<br />现有下列仪器或装置供选择：<br /><img  src=\"/tikuimages/9/2010/400/shoutiniao17/9743525e-94d4-11e9-aaa6-b42e9921e93e_xkb21.png\" style=\"vertical-align:middle\" /><br />（1）用甲酸滴入浓硫酸的方法制取CO，应选<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填序号）装置；如果要除去CO中混有少量甲酸气体，最好选择上图中<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填序号）进行洗气．<br />（2）用上述方法制取的CO还原氧化铁，并检验气体产物，则所选仪器的接口连接顺序：G→H→<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>→<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填字母）．<br />（3）在对氧化铁粉末加热前，为安全起见，应进行的一项重要操作是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（4）从开始加热到实验结束，氧化铁粉末的颜色变化为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（5）上述实验的尾气不能直接排放到空气中，请你说出一种处理尾气的方法：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（6）在该制取CO的反应中，浓硫酸所起的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>作用．','','','','','','②$###$③$###$F$###$E$###$加热前要先通CO排出空气$###$由红棕色变为黑色$###$将尾气点燃（或用气球、塑料袋收集尾气）$###$催化的','【解答】解：（1）因甲酸、浓硫酸都为液态，液液混合，应该选择②装置．<br />故填：②．<br />氢氧化钠溶液、氢氧化钙溶液都能够吸收CO中混有的少量甲酸气体，但是氢氧化钠溶液的吸收能力比氢氧化钙溶液强．<br />故填：③．<br />（2）生成的一氧化碳气体从B中排出，从D进入氢氧化钠溶液除去甲酸气体，从C中排出，再从G（或H）进入，再从H（或G）排出，再从F进入氢氧化钙溶液中．<br />故填：F；E．<br />（3）为安全起见，应进行的一项重要操作是加热前要先通CO排出空气．<br />故填：加热前要先通CO排出空气．<br />（4）从开始加热到实验结束，氧化铁粉末的颜色变化为由红棕色变为黑色．<br />故填：由红棕色变为黑色．<br />（5）将尾气点燃或用气球、塑料袋收集尾气等措施可以防止一氧化碳污染环境．<br />故填：将尾气点燃（或用气球、塑料袋收集尾气）．<br />（6）在该制取CO的反应中，浓硫酸作催化剂．<br />故填：催化的．','【分析】根据甲酸为液态，在浓硫酸作用下反应生成一氧化碳，利用生成的一氧化碳还原氧化铁．根据高炉炼铁的原理及实验中的注意点解决问题．还要掌握催化剂的概念．','书写',3.00,'57eddd7b9687d7a82ee8367cb9595145',9,400,'气体的净化（除杂）,一氧化碳还原氧化铁,书写化学方程式、文字表达式、电离方程式','',2010,'35','2010秋•深圳校级期中',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1851153,'科学家计划用质子数为20的钙离子，轰击核电荷数为98的铜元素靶，使两元素的原子核合并在一起，合成核电荷数为118的新元素，下列对此理解正确的是（　　）','新元素的原子核外电子数为116','新元素的原子序数为118','变化过程中元素种类没有发生改变','新元素的相对原子质量为118','','B','【解答】解：A、原子中：原子序数=核电荷数=核内质子数=核外电子数，由题意，新元素的核电荷数为118，则新元素原子的核外电子数为118，故选项说法错误．<br />B、原子中：原子序数=核电荷数，由题意，新元素的核电荷数为118，则新元素的原子序数为118，故选项说法正确．<br />C、根据题意，用质子数为20的钙离子，轰击核电荷数为98的元素靶，使两元素的原子核合并在一起，合成核电荷数为118的新元素，则变化过程中元素种类发生了改变，故选项说法错误．<br />D、相对原子质量=质子数+中子数，新元素原子的质子数为118，则新元素的相对原子质量应大于118，故选项说法错误．<br />故选：B．','【分析】根据原子中：原子序数=核电荷数=核内质子数=核外电子数、相对原子质量=质子数+中子数，结合题意进行分析解答．','选择题',3.00,'4f4dbada0cbff8ec4b5d328e94b95f3a',9,400,'原子的有关数量计算,元素在化学变化过程中的特点','',2015,'35','2015秋•河南校级期中',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1851168,'下列古代发明不属于化学工艺的是（　　）','制火药','烧瓷器','指南针','钻木取火','','C','【解答】解：A、火药的主要成分有硫磺、木炭、硝酸钾等物质，其中的木炭是由木材隔绝空气加热制成，有新物质生成，属于化学工艺．<br />B、烧制瓷器的原理：瓷器是以粘土为主要原料以及各种天然矿物经过粉碎混炼、成型和煅烧制得的材料以及各种制品，有新物质生成，属于化学工艺．<br />C、指南针的应用原理：指南针在地球的磁场中受磁场力的作用，所以会一端指南一端指北，不属于化学工艺．<br />D、钻木取火时，有新物质生成，属于化学工艺．<br />故选C．','【分析】我国古代的一些发明中，蕴含着化学知识，属于化学工艺的，在变化过程中都有其他物质生成．本题的实质是对物理变化和化学变化的判断，在熟知物理变化和化学变化的概念的基础上，关键是知道四大发明的制作原理．','选择题',3.00,'b28f183dcb4a99192e3fe0f0ee5bab2d',9,400,'化学的研究领域','',2015,'32','2015•元阳县校级模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1851218,'保护环境，提倡“低碳生活“，是我们应共同关注的社会问题．<br />（1）近几十年来大气中二氧化碳含量在不断上升．目前有些城市的出租车已改用压缩天然气（CH<SUB>4</SUB>）作燃料，以减少对空气污染．天然气完全燃烧的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（2）下列做法不符合“低碳经济”理念的是（填序号）<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />A．改造或淘汰高能耗、高污染产业<br />B．研制和开发新能源替代传统能源<br />C．大力发展火力发电<br />D．优化建筑设计，增强室内自然采光，减少照明用电<br />（3）最近有科学家提出“绿色自由”构想：把空气吹入碳酸钾溶液，然后再把CO<SUB>2</SUB>从生成物中提取出来，再经过化学反应转化为有机物有效实施碳循环．“绿色自由”构想技术流程如下：<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao15/9c202240-94d4-11e9-9ef0-b42e9921e93e_xkb79.png\" style=\"vertical-align:middle\" /><br />①上述流程中吸收池中发生的主要化学反应为：K<SUB>2</SUB>CO<SUB>3</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>═2KHCO<SUB>3</SUB>．则分解池<br />中反应的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，该流程中可循环利用的物质是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />②在合成塔中若气体X是H<SUB>2</SUB>，反应生成甲醇（CH<SUB>3</SUB>OH）和水，该反应化学反应方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；若气体X是CH<SUB>4</SUB>，它与CO<SUB>2</SUB>的质量比是4：11，反应分子个数比最简，原子利用率最高时，生成的有机物化学式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','CH<SUB>4</SUB>+2O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span> CO<SUB>2</SUB>+2H<SUB>2</SUB>O$###$C$###$2KHCO<SUB>3</SUB>═K<SUB>2</SUB>CO<SUB>3</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑$###$碳酸钾$###$CO<SUB>2</SUB>+3H<SUB>2</SUB>═CH<SUB>3</SUB>OH+H<SUB>2</SUB>O$###$C<SUB>2</SUB>H<SUB>4</SUB>O<SUB>2</SUB>','【解答】解：（1）反应物是甲烷和氧气，生成物是水和二氧化碳，所以方程式是：CH<SUB>4</SUB>+2O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>CO<SUB>2</SUB>+2H<SUB>2</SUB>O．<br />（2）火力发电是利用燃烧化石燃料煤矿进行，产生大量的空气污染物，故答案为：C<br />（3）①从图示看出二氧化碳与碳酸钾能反应生成碳酸氢钾，它分解应该是此反应的可逆过程，可循环物质为碳酸钾<br />故答案为：2KHCO<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>K<SUB>2</SUB>CO<SUB>3</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑ K<SUB>2</SUB>CO<SUB>3</SUB><br />②从叙述中可以看出反应物是氢气与二氧化碳；反应物是CH<SUB>4</SUB>和CO<SUB>2</SUB>，质量比是4：11时其分子个数比是1：1 所以生成物是C<SUB>2</SUB>H<SUB>4</SUB>O<SUB>2</SUB>故答案为：3H<SUB>2</SUB>+CO<SUB>2</SUB>=CH<SUB>3</SUB>OH+H<SUB>2</SUB>O C<SUB>2</SUB>H<SUB>4</SUB>O<SUB>2</SUB><br />故答案为：（1）CH<SUB>4</SUB>+2O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>&nbsp;CO<SUB>2</SUB>+2H<SUB>2</SUB>O；（2）C；<br />（3）①2KHCO<SUB>3</SUB>═K<SUB>2</SUB>CO<SUB>3</SUB>+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑&nbsp;碳酸钾<br />②CO<SUB>2</SUB>+3H<SUB>2</SUB>═CH<SUB>3</SUB>OH+H<SUB>2</SUB>O&nbsp;&nbsp;&nbsp;C<SUB>2</SUB>H<SUB>4</SUB>O<SUB>2</SUB>','【分析】（1）根据化学方程式的写法写出化学方程式．<br />（2）火力发电燃烧的是矿物燃料<br />（3）利用题目所给的信息解决．','书写',3.00,'f2f65d168db54267e77275e46e57bc7e',9,400,'绿色化学,书写化学方程式、文字表达式、电离方程式,常用燃料的使用与其对环境的影响','兴化市',2016,'37','2016•兴化市校级一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1851271,'某化学小组Ⅰ探究用大理石和稀盐酸反应制取二氧化碳气体，并进行性质实验．如图是有关实验的部分装置，请根据要求回答问题：<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao45/9ccbcb8f-94d4-11e9-bb90-b42e9921e93e_xkb66.png\" style=\"vertical-align:middle\" /><br />（1）a仪器的名称为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；甲同学用A和C组合制取二氧化碳，乙同学用B和C组合制取二氧化碳，你认为那个组合更便于控制反应<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填写甲或乙）；<br />（2）甲同学利用装置D进行性质实验时，观察到紫色石蕊试液变红色，将红色液体充分加热未能重新变为紫色，你认为可能的原因是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；为使上述红色液体加热后能重新变为紫色，在气体通入装置D之前可接入上图装置<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填字母序号）；在该装置中发生的主要化学反应方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（3）乙同学将CO<SUB>2</SUB>通入到氢氧化钠溶液中，无明显现象，经过思考讨论后，设计了如图G装置，使该反应有了明显现象，则该装置G中的现象为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','长颈漏斗$###$乙$###$通入的二氧化碳气体中含有少量氯化氢气体$###$F$###$NaHCO<SUB>3</SUB>+HCl=NaCl+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑$###$小气球逐渐膨胀起来','【解答】解：（1）通过分析题中所指仪器的作用可知，a是长颈漏斗，B装置可以利用气体的压强将固体和液体分离，所以乙组合更便于控制反应；<br />（2）盐酸具有挥发性，所以甲同学利用装置D进行性质实验时，观察到紫色石蕊试液变红色，将红色液体充分加热未能重新变为紫色，原因是：通入的二氧化碳气体中含有少量氯化氢气体，为使上述红色液体加热后能重新变为紫色，在气体通入装置D之前可接入如图装置F，碳酸氢钠和盐酸反应会生成二氧化碳气体、氯化钠和水，化学方程式为：NaHCO<SUB>3</SUB>+HCl=NaCl+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑；<br />（3）二氧化碳会与氢氧化钠反应，导致锥形瓶内的压强减小，所以装置G中的现象为小气球逐渐膨胀起来．<br />故答案为：<br />（1）长颈漏斗，乙；<br />（2）通入的二氧化碳气体中含有少量氯化氢气体，F，NaHCO<SUB>3</SUB>+HCl=NaCl+H<SUB>2</SUB>O+CO<SUB>2</SUB>↑；<br />（3）小气球逐渐膨胀起来．','【分析】（1）根据实验室常用仪器的名称和题中所指仪器的作用进行分析；<br />根据B装置可以利用气体的压强将固体和液体分离进行分析；<br />（2）根据盐酸具有挥发性进行分析；<br />根据碳酸氢钠和盐酸反应会生成二氧化碳气体进行分析；<br />（3）根据二氧化碳会与氢氧化钠反应，导致锥形瓶内的压强减小进行分析．','书写',3.00,'6a6c8e0e6c5151c9520340b7351371ec',9,400,'制取气体的反应原理的探究,二氧化碳的实验室制法,二氧化碳的化学性质,书写化学方程式、文字表达式、电离方程式','',2016,'37','2016•东平县一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1851386,'下列变化中，后者一定包括前者的是（　　）<br />①化学变化、物理变化&nbsp;&nbsp;&nbsp;&nbsp;<br />②氧化反应、化合反应&nbsp;&nbsp;<br />③分解反应、化合反应&nbsp;&nbsp;<br />④状态变化、物理变化．','①和②','③和④','只有③','只有④','','D','【解答】解：A、化学变化是指有新物质生成的变化，物理变化是指没有新物质生成的变化，物理变化与化学变化是并列关系，而不是包含关系，后者不一定包括前者，故错误；<br />B、化合反应是两种或两种以上物质反应后生成一种物质的反应，物质与氧发生的化学反应是氧化反应，化合反应并不都是氧化反应，如氧化钙+水→氢氧化钙，则后者不一定包括前者，故错误，<br />C、分解反应是一种物质反应后生成两种或两种以上的物质的反应，化合反应是两种或两种以上物质反应后生成一种物质的反应，则后者不一定包括前者，故错误，<br />D、物理变化是没有新物质生成的变化，只是物质的形状、状态等发生改变，根据定义可以看出，物理变化包括状态变化，故正确．<br />答案：D．','【分析】A、根据物理变化与化学变化的关系进行分析判断；<br />B、根据氧化反应和化合反应的定义进行分析；<br />C、根据分解反应、化合反应的特点进行分析；<br />D、根据物理变化的定义分析．','选择题',3.00,'7cdeace56e3700ab51aab1da2ef94687',9,400,'化学变化的基本特征,物理变化的特点,化合反应及其应用,分解反应及其应用,氧化反应','',2014,'37','2014春•尧都区校级月考',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1851471,'如图甲是氢气和氧化铜反应的实验，如图乙是木炭和氧化铁反应的实验．<br /><img  src=\"/tikuimages/9/2011/400/shoutiniao61/9f4cd11e-94d4-11e9-a33f-b42e9921e93e_xkb31.png\" style=\"vertical-align:middle\" /><br />（1）在上述两个实验中，氢气和木炭表现出相同的化学性质是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>性．<br />（2）某同学正确操作做完乙实验后，发现澄清石灰水变浑浊，试管中粉末全部变为黑色，取少量黑色粉末，加入足量稀硫酸充分振荡，但她惊讶地发现黑色粉末没有溶解，试管中&nbsp;也未产生预期的气泡，这说明该反应并没有生成<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）该同学查阅资料，得到关于铁的氧化物如下信息：<br /><table class=\"edittable\"><TBODY><TR><td width=138>铁的氧化物化学式</TD><td width=96>Fe<SUB>2</SUB>O<SUB>3</SUB></TD><td width=146>Fe<SUB>3</SUB>O<SUB>4</SUB></TD><td width=79>FeO</TD></TR><TR><td>颜色</TD><td>红</TD><td>黑</TD><td>黑</TD></TR><TR><td>化学性质</TD><td>可溶于酸</TD><td>常温下不溶于稀酸</TD><td>可溶于酸</TD></TR></TBODY></TABLE>根据以上信息，试写出乙实验试管中反应的化学方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />经分析后，该同学试着提高反应温度后，实验出现了预期的现象．','','','','','','还原$###$铁$###$C+6Fe<SUB>2</SUB>O<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>4Fe<SUB>3</SUB>O<SUB>4</SUB>+CO<SUB>2</SUB>↑','【解答】解：（1）在两个实验中，氢气和木炭都得到氧是还原剂，具有还原性．<br />（2）乙实验完成后，发现澄清石灰水变浑浊，生成部分黑色粉末，黑色粉末加入足量稀硫酸没有溶解，说明不是FeO，试管中也未产生预期的气泡，说明无铁单质生成，因为粉末是黑色的，则不是Fe<SUB>2</SUB>O<SUB>3</SUB>，可知黑色粉末是Fe<SUB>3</SUB>O<SUB>4</SUB>．<br />（3）根据上述分析可知发生的反应是：碳与氧化铁加热反应生成四氧化三铁和二氧化碳气体，化学方程式是：C+6Fe<SUB>2</SUB>O<SUB>3</SUB> <span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>4Fe<SUB>3</SUB>O<SUB>4</SUB>+CO<SUB>2</SUB>↑．<br />答案：（1）还原；（2）铁；（3）C+6Fe<SUB>2</SUB>O<SUB>3</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>4Fe<SUB>3</SUB>O<SUB>4</SUB>+CO<SUB>2</SUB>↑','【分析】（1）在化学反应中得到氧的物质叫还原剂，具有还原性；<br />（2）根据“黑色粉末没有溶解，试管中&nbsp;也未产生预期的气泡，”做出判断；<br />（3）根据反应物、反应条件、生成物写出化学方程式．','书写',3.00,'59f4c4edd29c1d0859207471d589c5ff',9,400,'碳、一氧化碳、氢气还原氧化铜实验,碳的化学性质,书写化学方程式、文字表达式、电离方程式','',2011,'37','2011秋•绍兴校级月考',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1851514,'二氧化碳若合理利用，是一种与水一样重要的原料．<br />I&nbsp;最近，美国研究人员利用镍和钯作催化剂，将二氧化碳转化为具有多种用途的一氧化碳或甲醇．该转化属于化学变化，其催化剂是两种<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“金属”或“非金属”）．<br />II&nbsp;日本核灾难后，清洁煤电受到追捧．我国已成功运行燃煤电厂二氧化碳捕集技术，使液态二氧化碳成为最终产品．<br />（1）以二氧化碳为原料可生产更环保的碳基钾肥，如碳酸钾等．碳酸钾中的阴离子是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（用离子符号表示）；<br />（2）该技术的重要一步是在低温条件下用化学溶剂吸收烟气中的二氧化碳．所用化学溶剂可能是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填一种物质的化学式）．','','','','','','金属$###$CO<SUB>3</SUB><SUP>2-</SUP>$###$NaOH（或KOH或NH<SUB>3</SUB>．H<SUB>2</SUB>O）','【解答】解：Ⅰ、汉字中带有钅字旁的元素属于金属元素（金和汞除外），因为．“镍”、“钯”中都含有钅字旁，所以都属于金属元素；<br />Ⅱ、（1）碳酸钾是一种盐，由钾离子和碳酸根离子构成，其中阴离子为CO<SUB>3</SUB><SUP>2-</SUP>．<br />（2）二氧化碳气体的溶解度随温度的降低而增大，低温时二氧化碳溶解度大，利于吸收；二氧化碳能与碱反应生成盐和水，所以所用化学溶剂可能是可溶性的碱，如NaOH或KOH或NH<SUB>3</SUB>．H<SUB>2</SUB>O．<br />故答案为：Ⅰ．金属；<br />Ⅱ．（1）CO<SUB>3</SUB><SUP>2-</SUP>；（2）NaOH（或KOH或NH<SUB>3</SUB>．H<SUB>2</SUB>O）．','【分析】Ⅰ、根据金属元素的判断方法考虑；<br />Ⅱ、（1）根据碳酸钾的结构回答．<br />（2）根据二氧化碳的溶解度受温度影响的情况和二氧化碳的化学性质回答．','填空题',3.00,'d11bfd8bd7aa36788ed699fd2fd4414d',9,400,'二氧化碳的用途,常见的溶剂,常见的金属和非金属的区分,化学符号及其周围数字的意义','',2011,'37','2011秋•浙江校级月考',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1851589,'实验室里有一瓶标签残缺的酸，为能立即确定它是否为盐酸，你认为下列做法合理的是（　　）','猜想假设','查阅资料','交流讨论','进行试验','','D','【解答】解：A、猜想假设只是对问题的可能性进行合理的推测，不能通过猜想假设确定该无色气体是否为盐酸，故该做法不合理．<br />B、查阅资料只是为实验探究过程找寻理论支持，不能通过猜想假设确定该无色气体是否为盐酸，故该做法不合理．<br />C、交流讨论可以使实验探究更合理、更完善，不能代替实验做出实验证明，不能通过猜想假设确定该无色气体是否为盐酸，故该做法不合理．<br />D、通过进行实验以及对实验现象的观察、记录和分析等可以确定该无色气体是否为盐酸，故该做法合理．<br />故选：D．','【分析】科学探究的主要环节有提出问题→猜想与假设→制定计划（或设计方案）→进行实验→收集证据→解释与结论→反思与评价→拓展与迁移，据此结合题意进行分析判断．','选择题',3.00,'b4ca6e106770e7f9c2cabf7ea087010b',9,400,'科学探究的基本环节','',2012,'33','2012秋•锦江区期末',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1851607,'下列说法正确的是（　　）','pH小于5.6的雨水是酸雨','所有作物都适宜在pH为6.5～7.5的土壤中生长','稀盐酸遇到紫色石蕊试液会变红','能使无色酚酞试液变红的一定是碱溶液','','A','【解答】解：<br />A、pH小于5.6的雨水是酸雨，故正确；<br />B、不是所有作物都适宜在pH在6.5--7.5的土壤中生长，故B错误；<br />C、紫色石蕊试液遇到盐酸变红色，故C错误；<br />D、有些盐溶液呈碱性，如碳酸钠溶液呈碱性，也能使无色酚酞试液变红，故D错误．<br />故选：A．','【分析】A、根据pH小于5.6的雨水是酸雨解答；<br />B、根据不是所有作物都适宜在pH在6.5--7.5的土壤中生长进行解答；<br />C、根据紫色石蕊试液遇到盐酸变红色进行解答；<br />D、根据有些盐溶液呈碱性，也能使无色酚酞试液变红进行解答．','选择题',3.00,'febdf208209b66f44628605c81bac000',9,400,'酸碱指示剂及其性质,酸的化学性质,酸碱性对生命活动和农作物生长的影响,酸雨的产生、危害及防治','',2016,'37','2016•苏州校级一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1851711,'同学们为证明影响物质溶解快慢的因素，设计了如下部分实验：<br /><img  src=\"/tikuimages/9/2016/400/shoutiniao97/a1e7a040-94d4-11e9-8e21-b42e9921e93e_xkb17.png\" style=\"vertical-align:middle\" /><br />（1）各实验中都可观察到溶液颜色为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）通过实验①②的对比，可得出的实验结论是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）若要证明物质溶解的快慢与固体颗粒大小有关，需增加实验④与实验③进行对比，实验④的烧杯中需加入<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>和<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','蓝色$###$温度越高，溶解速率越快$###$50mL冷水$###$0.5g硫酸铜粉末','【解答】解：（1）硫酸铜在溶液中显蓝色，所以各实验中都可观察到溶液颜色为蓝色；<br />（2）通过对比实验①和②，可得出的实验结论是温度越高，溶解速率越快；<br />（3）要证明物质溶解的快慢与固体颗粒大小有关，需增加实验④与实验③进行对比，实验④的烧杯中需加入50mL冷水和0.5g硫酸铜粉末．<br />故答案为：（1）蓝色；<br />（2）温度越高，溶解速率越快；<br />（3）50mL冷水，0.5g硫酸铜粉末．','【分析】根据影响固体在水中的溶解的因素进行分析解答，温度高、颗粒小，则溶解的快，依据控制变量法的具体操作进行分析．','填空题',3.00,'b2831dd393781abad829485582d53bb7',9,400,'影响溶解快慢的因素','',2016,'37','2016•桂林一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1851765,'将H<SUB>2</SUB>、CO、CO<SUB>2</SUB>、N<SUB>2</SUB>依次通过足量下列物质：澄清石灰水、灼热的氧化铜、浓H<SUB>2</SUB>SO<SUB>4</SUB>（注：有吸水性），最终剩余的气体为（　　）','N<SUB>2</SUB>','N<SUB>2</SUB>　CO<SUB>2</SUB>','H<SUB>2</SUB>　CO<SUB>2</SUB>　N<SUB>2</SUB>','N<SUB>2</SUB>　H<SUB>2</SUB>','','B','【解答】解：将H<SUB>2</SUB>、CO、CO<SUB>2</SUB>、N<SUB>2</SUB>通过澄清石灰水后，二氧化碳会与石灰水中的氢氧化钙反应生成碳酸钙和水，CO<SUB>2</SUB>被吸收；再通过灼热的氧化铜后，一氧化碳和氢气会与灼热的氧化铜反应分别生成铜和二氧化碳、铜和水，一氧化碳和氢气被吸收，但生成了二氧化碳和水蒸气；再通过浓硫酸除去水，最后剩余的气体为二氧化碳和氮气．<br />故选：B．','【分析】根据二氧化碳能与澄清石灰水反应，氢气和一氧化碳能与灼热的氧化铜反应，浓硫酸具有吸水性，据此进行分析判断即可．','选择题',3.00,'545696bae41f03aaa82f9528b8a25928',9,400,'气体的净化（除杂）','',2016,'37','2016春•曲靖校级月考',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1851852,'善于归纳知识，有利于培养素质．下列对部分化学知识点的归纳完全正确的一组是（　　）<br />①碱的水溶液显碱性，但显碱性的不一定是碱的溶液；②用PH试纸测得苹果汁的PH为3.2；③酸和碱中都一定有氢元素；④稀盐酸、稀硫酸化学性质相似是因为它们的溶液中都有氢离子；⑤打开浓盐酸瓶盖有白烟；⑥浓硫酸不慎沾在皮肤上，要用大量水冲洗，再涂上硼酸溶液．','①③④','①③⑤⑥','①③④⑥','①②③④','','A','【解答】解：①碱的水溶液显碱性，但显碱性的不一定是碱的溶液，例如碳酸钠属于盐，水溶液显碱性，该选项说法正确；<br />②用PH试纸可以测得苹果汁的PH，但是不能是3.2，因为利用pH试纸测定的酸碱度是整数，该选项说法不正确；<br />③酸和碱中都一定有氢元素，该选项说法正确；<br />④稀盐酸、稀硫酸化学性质相似是因为它们的溶液中都有氢离子，该选项说法正确；<br />⑤浓盐酸易挥发，打开浓盐酸瓶盖有白雾，该选项说法不正确；<br />⑥浓硫酸不慎沾在皮肤上，要用大量水冲洗，再涂上硼酸钠溶液或碳酸氢钠溶液，硼酸显酸性，因此不能涂上硼酸溶液，该选项说法不正确．<br />故选：A．','【分析】一些盐的水溶液显碱性，例如碳酸钠属于盐，溶液显碱性；<br />利用pH试纸可以测定溶液的酸碱度；<br />酸和碱中都含有氢元素；<br />酸溶液的化学性质相似，是因为溶液中都含有自由移动的氢离子；<br />浓盐酸易挥发；<br />浓硫酸具有腐蚀性．','选择题',3.00,'ae2b0b718093a36d06caf24f17a0b237',9,400,'溶液的酸碱度测定,酸的物理性质及用途,酸的化学性质,碱的化学性质,氧化物、酸、碱和盐的概念','江阴市',2016,'37','2016春•江阴市月考',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1851890,'化学实验在化学学习中具有重要作用，请据图回答下列问题：<br /><img  src=\"/tikuimages/9/2015/400/shoutiniao48/a44a9680-94d4-11e9-b0bd-b42e9921e93e_xkb97.png\" style=\"vertical-align:middle\" /><br />（1）若利用图1所示器材组装成氢气还原氧化铜的实验装置，请你指出一点不足之处<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）如图2为某同学用甲酸与热的浓硫酸反应制取一氧化碳的微型实验装置（夹持仪器略），此装置的主要优点是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（只需写一点）．','','','','','','有橡胶塞或导管进入试管太短$###$环保','【解答】解：（1）若利用图1所示器材组装成氢气还原氧化铜的实验装置，其中一点不足之处有橡胶塞或导管进入试管太短；<br />（4）此实验装置的主要优点是环保，其理由是一氧化碳被点燃生成二氧化碳，不污染环境，并且节约能源；<br />故答案为：<br />（1）有橡胶塞或导管进入试管太短；（2）环保（合理即可）','【分析】（1）考虑图中有橡胶塞或导管进入试管太短；<br />（2）考虑此实验装置的主要优点是环保，其理由是一氧化碳被点燃生成二氧化碳，不污染环境，并且节约能源．','简答题',3.00,'fcca9682a699f1b91c6afd185c04bbfe',9,400,'碳、一氧化碳、氢气还原氧化铜实验','',2015,'33','2015秋•金华校级期末',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1851982,'煤是社会生产、生活中最重要的能源，工业中常把煤进行气化和液化处理，使煤变成清洁能源，煤气化和液化流程示意图如下：<br /><img  src=\"/tikuimages/9/2013/400/shoutiniao77/a52e8dde-94d4-11e9-903f-b42e9921e93e_xkb25.png\" style=\"vertical-align:middle\" /><br />（1）第①步操作发生的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“物理”或“化学”）变化．<br />（2）第②步是精炼煤与水蒸气的反应，基本类型属于<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）第③步反应的化学方程式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（4）含硫化合物洗液经过提炼后，可用来制硫酸，过程是：含硫化合物氧化得到SO<SUB>2</SUB>，SO<SUB>2</SUB>进一步氧化得到X，X与水反应得到H<SUB>2</SUB>SO<SUB>4</SUB>，则X的化学式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（5）从“绿色化学”的角度分析，“煤的汽化和煤的液化”生产流程的优点<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；结合能源利用问题，请你谈谈对改善成都雾霭天气的看法<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．（谈一点即可）','','','','','','物理$###$置换反应$###$CO+2H<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;催化剂&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>CH<SUB>3</SUB>OH$###$SO<SUB>3</SUB>$###$原料全部转化为产品或实现“零排放”，不对环境造成污染$###$减少化石燃料的利用，使用清洁能源','【解答】解：（1）第①步操作中没有新物质生成，发生的是物理变化．<br />（2）第②步是精炼煤与水蒸气的反应，在高温条件下生成了氢气和一氧化碳，反应化学方程式为：H<SUB>2</SUB>O+C<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;高温&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>&nbsp;H<SUB>2</SUB>+CO；其反应物和生成物均为一种单质和一种化合物，所以该反应属于置换反应；<br />（3）第③步反应是氢气和一氧化碳在催化剂的作用下生成了甲醇，反应化学方程式为：CO+2H<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;催化剂&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>CH<SUB>3</SUB>OH．<br />（4）SO<SUB>2</SUB>进一步氧化得到X，X与水反应得到H<SUB>2</SUB>SO<SUB>4</SUB>，则X是三氧化硫，三氧化硫的化学式为SO<SUB>3</SUB>．<br />故填：SO<SUB>3</SUB>．<br />（5）从“绿色化学”的角度分析，“煤的气化和煤的液化”生产流程的优点是：原料全部转化为产品或实现“零排放”，不对环境造成污染；雾霾形成的主要原因还是各类工业、农业、生活的烟尘等颗粒物的排放造成的．因此可采取的改善措施有：低碳生活、节能减排、提高热机效率、植树造林，爱护花草树木，多乘坐公共交通工具等．<br />故答案为：（1）物理；（2）置换反应；（3）CO+2H<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;催化剂&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>CH<SUB>3</SUB>OH；（4）SO<SUB>3</SUB>；（5）原料全部转化为产品或实现“零排放”，不对环境造成污染；减少化石燃料的利用，使用清洁能源．','【分析】（1）根据变化的特征分析变化的类型；<br />（2）根据基本反应类型的特点分析；<br />（3）根据氢气和一氧化碳在催化剂的作用下生成了甲醇进行分析；<br />（4）根据二氧化硫能被氧化成三氧化硫分析回答；<br />（5）根据原料的利用和对环境的影响分析回答．','书写',3.00,'55bfff615af99eaabb15de3648cbb9c5',9,400,'绿色化学,化学变化和物理变化的判别,反应类型的判定,质量守恒定律及其应用,书写化学方程式、文字表达式、电离方程式,化石燃料及其综合利用','',2013,'32','2013•青羊区模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1852034,'通过实验可以获取证据，得出结论．下列做法中不能获得明显证据的是（　　）','为了证明汗水中含有Cl<SUP>-</SUP>，加入含稀硝酸的硝酸银溶液','为了证明铁粉与硫磺加热后生成新物质，用磁铁靠近生成物','为了证明某些金属氧化物可以和水发生反应，把氧化铁粉末放入水中','为了证明氢氧化钠固体容易潮解，在表面皿上放置少量氢氧化钠固体暴露在空气中','','C','【解答】解：A、在检验氯离子时，是加入硝酸银出现白色沉淀，证明氯离子的存在，故A正确；<br />B、明铁粉与硫磺加热后生成新物质，用磁铁靠近生成物，不能吸引，故B正确；<br />C、证明某些金属氧化物可以和水发生反应，把氧化铁粉末放入水中，不会反应，不能证明，故C错误；<br />D、证明氢氧化钠固体容易潮解，在表面皿上放置少量氢氧化钠固体暴露在空气中，一段时间，表面潮湿，故D正确．<br />故选：C．','【分析】A、根据氯离子和银离子会生成氯化银沉淀进行分析；<br />B、根据铁能被磁铁吸引进行分析；<br />C、根据氧化铁不会与水反应进行分析；<br />D、根据氢氧化钠具有吸水性进行分析．','选择题',3.00,'6f21a88ebf53c0412d7b64409f5ddae8',9,400,'化学实验方案设计与评价,证明盐酸和可溶性盐酸盐,常见金属的特性及其应用,常见碱的特性和用途','嵊州市',2013,'33','2013秋•嵊州市期末',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1852068,'淀粉遇碘酒变蓝紫色的实验，可以检验含淀粉多的食物．我们可以先对一种食物进行猜测，再进行检验，看我们是不是了解了淀粉类食物的特点．<br />我的猜测：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />实验方法：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />实验现象：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />结论及解释：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />通过验证多种食物，发现淀粉类食物的特点是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','淀粉遇碘变蓝色$###$取一片马铃薯片，滴加几滴碘酒$###$马铃薯片变蓝色$###$马铃薯的主要成分是淀粉，淀粉遇碘变蓝色$###$遇到碘会变蓝色','【解答】解：根据淀粉的性质可知，淀粉遇碘变蓝色．<br />我的猜测是：淀粉遇碘变蓝色；<br />实验方法：取一片马铃薯片，滴加几滴碘酒；<br />实验现象：马铃薯片变蓝色；<br />结论及解释：马铃薯的主要成分是淀粉，淀粉遇碘变蓝色．<br />可以验证馒头、面包等其他富含淀粉的物质，滴加碘酒观察现象，均变成蓝色．<br />故答案为：淀粉遇碘变蓝色；取一片马铃薯片，滴加几滴碘酒；马铃薯片变蓝色；马铃薯的主要成分是淀粉，淀粉遇碘变蓝色；遇到碘会变蓝色．','【分析】根据科学探究的基本环节以及淀粉的检验方法来分析．','填空题',3.00,'40e5ef1a96dcf2ca3d1e9aa3718be4d0',9,400,'猜想与事实验证,鉴别淀粉、葡萄糖的方法与蛋白质的性质','',0,'37','',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1852240,'<img  src=\"/tikuimages/9/2015/400/shoutiniao78/a85151ae-94d4-11e9-8a7e-b42e9921e93e_xkb57.png\" style=\"vertical-align:middle;FLOAT:right\" />（2015秋•甘肃月考）如图是某同学进行一氧化碳还原氧化铜实验的装置图，请回答：<br />（1）在此实验过程中，可以观察到哪些现象？<br />玻璃管内的黑色粉末逐渐变成<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>色，生成的气体使澄清石灰水变浑浊．<br />（2）写出玻璃管中发生反应的化学方程式：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）此反应中被还原的物质是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，说明一氧化碳具有<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>性，可用来冶炼金属．<br />（4）该实验装置有何不足之处？会造成什么后果？应如何改进？<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','红$###$CO+CuO<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>Cu+CO<SUB>2</SUB>$###$氧化铜$###$还原$###$缺少尾气处理装置；会污染空气；可在装置末端出气口添加一燃着的酒精灯，把未反应的一氧化碳燃烧掉','【解答】解：（1）CO具有还原性，能与氧化铜反应生成铜和二氧化碳，氧化铜和木炭是黑色粉末，铜是红色的粉末，故会观察到：黑色粉末变红色；二氧化碳能使澄清的石灰水变浑浊．<br />故答案是：红；<br />（2）一氧化碳具有还原性，还原氧化铜生成铜和二氧化碳，反应的化学方程式为：CO+CuO<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>Cu+CO<SUB>2</SUB>．<br />故答案为：CO+CuO<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>Cu+CO<SUB>2</SUB>．<br />（3）用一氧化碳还原氧化铜，主要是利用CO的还原性，反应的化学方程式为：CO+CuO<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>Cu+CO<SUB>2</SUB>，在此反应中一氧化碳夺取了氧化铜中的氧，发生了氧化反应，具有还原性，作还原剂．氧化铜失去了氧，发生了还原反应，被还原．<br />故答案是：氧化铜；还原；<br />（4）通过观察可知该装置中缺少尾气的处理装置，而把一氧化碳直接排放到空气中了，会造成空气污染．我们可以在出气口处添加一燃着的酒精灯，把未反应的一氧化碳燃烧掉；<br />故答案是：缺少尾气处理装置；会污染空气；可在装置末端出气口添加一燃着的酒精灯，把未反应的一氧化碳燃烧掉．','【分析】（1）根据在高温条件下碳能还原氧化铜，二氧化碳能与氢氧化钙反应分析在实验过程中看到的实验现象．<br />（2）根据一氧化碳和氧化铜高温条件下反应生成铜和二氧化碳解答；<br />（3）用一氧化碳还原氧化铜，主要是利用CO的还原性；<br />（4）根据一氧化碳有毒，扩散到空气中能够污染环境，并依据一氧化碳的可燃性分析解答．','书写',3.00,'441e6dfc036563db075b3d0517e697ba',9,400,'碳、一氧化碳、氢气还原氧化铜实验,书写化学方程式、文字表达式、电离方程式','',2015,'37','2015秋•甘肃月考',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1852502,'下列化肥中，从外观即可与其他化肥相区别的是（　　）','氯化钾','硝酸铵','磷矿粉','硫酸钾','','C','【解答】解：A、氯化钾为白色晶体．<br />B、硝酸铵为白色晶体．<br />C、磷矿粉是灰白色的固体．<br />D、硫酸钾为白色晶体．<br />硫酸钾、磷矿粉、硝酸铵、氯化钾从外观看均为白色晶体，只有磷矿粉是灰白色粉末，故从外观看与磷矿粉可与其他化肥相区别．<br />故选：C．','【分析】根据磷矿粉是灰白色的，碳酸铵、氯化钾和硝酸钾都是白色的晶体，进行分析判断．','选择题',3.00,'3fe329cef6c1272fbde4cde29ae7e448',9,400,'化肥的简易鉴别','',2016,'32','2016•河西区模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1852769,'二百多年前，哪位法国科学家利用天平进行定量实验，研究了空气的成分（　　）','<img  src=\"/tikuimages/9/2015/400/shoutiniao45/ae4fe540-94d4-11e9-aad3-b42e9921e93e_xkb99.png\" style=\"vertical-align:middle\" />','<img  src=\"/tikuimages/9/2015/400/shoutiniao16/ae564de1-94d4-11e9-b8ef-b42e9921e93e_xkb71.png\" style=\"vertical-align:middle\" />','<img  src=\"/tikuimages/9/2015/400/shoutiniao94/ae59a940-94d4-11e9-b48c-b42e9921e93e_xkb22.png\" style=\"vertical-align:middle\" />','<img  src=\"/tikuimages/9/2015/400/shoutiniao18/ae5bf330-94d4-11e9-8948-b42e9921e93e_xkb39.png\" style=\"vertical-align:middle\" />','','D','【解答】解：二百多年前，法国化学家拉瓦锡以天平作为研究工具，用定量的方法研究了空气的成分，首先提出了空气是由氮气和氧气组成的结论．舍勒、普利斯特里在空气成分的研究中也做出了一定的贡献，但并没有得出空气的成分．<br />故选：D','【分析】根据对空气组成研究史的了解、根据各位科学家的突出贡献判断，选择最早提出空气组成的科学家即可．','选择题',3.00,'9ce9d9a0d0e6d1303be896ebcacb3c9b',9,400,'化学的历史发展过程','',2015,'35','2015秋•官渡区校级期中',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1853097,'归纳总结是学习化学的重要方法，下面是某同学对有关知识的总结，请你完成填空．<br />（1）逻辑关系：物质类属间存在着如下关系，其中酸和盐属于<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>关系．<br /><img  src=\"/tikuimages/9/2015/400/shoutiniao30/b2281661-94d4-11e9-b977-b42e9921e93e_xkb67.png\" style=\"vertical-align:middle\" /><br />（2）转化关系：下图表示几种化合物能通过一步反应转化为含镁化合物M．<br /><img  src=\"/tikuimages/9/2015/400/shoutiniao98/b22b23a1-94d4-11e9-a22f-b42e9921e93e_xkb83.png\" style=\"vertical-align:middle\" /><br />上图中物质M属于<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（选填“单质”、“氧化物”、“酸”、“碱”或“盐”）；<br />从①～③表示的化学反应中任选一个，写出化学方程式：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','B$###$盐$###$MgO+H<SUB>2</SUB>SO<SUB>4</SUB>═MgSO<SUB>4</SUB>+H<SUB>2</SUB>O','【解答】解：（1）酸和盐属于化合物，为并列关系，故选B；<br />（2）硫酸锌、氧化镁和硫酸都能转化生成M，M是含镁化合物，故M是硫酸镁，属于盐类物质，氧化镁能与硫酸反应生成硫酸镁和水，<br />故填：盐，MgO+H<SUB>2</SUB>SO<SUB>4</SUB>═MgSO<SUB>4</SUB>+H<SUB>2</SUB>O','【分析】根据图示进行分析，酸和盐属于化合物，为并列关系；根据图示可以看出硫酸锌、氧化镁和硫酸都能转化生成M，故M是硫酸镁，属于盐类物质，氧化镁能与硫酸反应生成硫酸镁．','书写',3.00,'f8737c8bc7ec67c132eb9984b1032877',9,400,'氧化物、酸、碱和盐的概念,常见的氧化物、酸、碱和盐的判别,物质的相互转化和制备,书写化学方程式、文字表达式、电离方程式','',2015,'33','2015秋•镇海区期末',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1853115,'<img  src=\"/tikuimages/9/2016/400/shoutiniao33/b242f15e-94d4-11e9-b984-b42e9921e93e_xkb48.png\" style=\"vertical-align:middle;FLOAT:right\" />（2016春•萝岗区校级月考）小明和小红对市场上销售的一种真空充气包装的形如小枕头“蛋黄派”发生了兴趣，如图所示．他们查阅资料获知：这种真空充气包装技术，即将食品装入包装袋，抽出包装袋内空气，再充入某种气体，然后封口．它能使食品保持原有的色、香、味及营养价值，防止食品受压而破碎变形．那么，这是什么气体呢？小红猜想是氮气，小明猜想是二氧化碳．<br />（1）现有一个注射器和少量澄清的石灰水，请你用上述仪器和试剂，帮他们设计一个简单的实验方案，来判断小明的猜想是否正确，简要写出操作步骤（方法、现象和结论）：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>、<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>、<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）你认为食品充气包装，对所充气体的要求是：①<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；②<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；③<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','用注射器抽取袋内气体，然后通入澄清的石灰水$###$澄清的石灰水变浑浊$###$气体是二氧化碳，小明的猜想正确$###$无毒$###$不与食品反应$###$价廉或来源广泛','【解答】解：（1）根据二氧化碳的性质，可以用澄清石灰水来检验小明的猜想，具体的步骤为：用注射器抽取包装袋内的气体，然后通入适量的澄清石灰水中，<br />澄清石灰水会变浑浊，证明小明的猜想正确；<br />（2）由于气体是充入食品包装中所以不能影响身体健康，要考虑到无毒、不与食品反应、价廉或来源广泛，经济便宜，<br />故答案为：<br />（1）用注射器抽取袋内气体，然后通入澄清的石灰水；澄清的石灰水变浑浊；气体是二氧化碳，小明的猜想正确；<br />（2）无毒、不与食品反应，价廉或来源广泛．','【分析】（1）根据二氧化碳的性质可以选择澄清石灰水来进行验证；<br />（2）由于气体是充入食品包装中所以考虑不能影响身体健康考虑．','填空题',3.00,'e518649969a88aa8e2c81fee43d8ed87',9,400,'食品干燥剂、保鲜剂和真空包装的成分探究,常见气体的检验与除杂方法','',2016,'37','2016春•萝岗区校级月考',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1853393,'氢气是一种密度<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>溶于水的气体，因此可用<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>方法来收集它．纯净的氢气在空气里能够安静的燃烧，发出<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>火焰，发生反应的文字表达式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．由于氢气是一种具有<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>性的气体，因此点燃前必须<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','比空气小$###$难$###$向下排空气法$###$排水法$###$淡蓝色$###$氢气+氧气<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td style=\"font-size: 90%\">点燃</td></tr><tr><td><div hassize=\"9\"><div style=\"background: url(\'/upimages/images/part/8594L.png\') repeat-x; width:1px; height:5px;float:left;overflow:hidden\" muststretch=\"h\"></div><div style=\"background: url(\'/upimages/images/part/8594.png\') no-repeat; width:9px; height:5px;float:left;overflow:hidden\"></div></div></td></tr></table></span>水$###$可燃$###$检验氢气的纯度','【解答】解：氢气是一种无色、无味、密度比空气小、难溶于水的气体，可用排水法或向下排空气法来收集；纯净的氢气在空气里能够安静的燃烧，发出淡蓝色火焰，发生反应的文字表达式为氢气+氧气<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td style=\"font-size: 90%\">点燃</td></tr><tr><td><div hassize=\"9\"><div style=\"background: url(\'/upimages/images/part/8594L.png\') repeat-x; width:1px; height:5px;float:left;overflow:hidden\" muststretch=\"h\"></div><div style=\"background: url(\'/upimages/images/part/8594.png\') no-repeat; width:9px; height:5px;float:left;overflow:hidden\"></div></div></td></tr></table></span>水．由于氢气是一种具有可燃性的气体，因此点燃前必须检验氢气的纯度，以防发生爆炸．<br />故答案为：比空气小；难；向下排空气法；排水法；淡蓝色；氢气+氧气<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td style=\"font-size: 90%\">点燃</td></tr><tr><td><div hassize=\"9\"><div style=\"background: url(\'/upimages/images/part/8594L.png\') repeat-x; width:1px; height:5px;float:left;overflow:hidden\" muststretch=\"h\"></div><div style=\"background: url(\'/upimages/images/part/8594.png\') no-repeat; width:9px; height:5px;float:left;overflow:hidden\"></div></div></td></tr></table></span>水；可燃；检验氢气的纯度．','【分析】根据氢气的物理性质、化学性质、文字表达式的写法来分析．','书写',3.00,'bd362e991f9225dfac38af23c687ebb8',9,400,'常用气体的收集方法,书写化学方程式、文字表达式、电离方程式,氢气的物理性质,氢气的化学性质与燃烧实验,氢气、一氧化碳、甲烷等可燃气体的验纯','',0,'37','',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1853411,'学习化学的一个重要途径是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，通过实验以及对<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>的观察、记录和分析等，可以<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>和验证化学原理，学习<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>的方法并获得化学知识．','','','','','','实验$###$实验现象$###$发现$###$科学探究','【解答】解：学习化学的一个重要途径是实验，通过实验以及对实验现象的认真观察、准确记录和分析讨论，可以发现和验证化学原理，学习科学探究的方法并获得化学知识．<br />故答案为：实验；实验现象；发现；科学探究．','【分析】根据化学是一门以实验为基础的科学，化学的许多重大发现和研究成果都是通过实验得到的解答．','填空题',3.00,'dd64c98c9b2c8dcdce316b1206219644',9,400,'学习化学的重要途径及学习方法','',0,'37','',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1853968,'下列说法不正确的是（　　）','人体中血浆的PH为7.35～7.45，则身体健康','正常雨水的PH≈5.6','厕所清洁剂的PH小于7','农作物一般适宜在PH=7或接近7的土壤中生长','','A','【解答】解：A、人体内血浆的正常pH为7.35-7.45，血浆的pH值正常，只能说明血液的一项指标，人体是否健康有好多因素，故选项说法错误．<br />B、空气中的二氧化碳气体溶于水生成碳酸，故正常雨水的pH约为5.6，故选项说法正确．<br />C、厕所清洁剂显酸性，pH小于7，故选项说法正确．<br />D、一般作物在5.5＜pH＜8.5的土壤上，都能生长良好，过酸、过碱才对其生长不利，也才有改良的必要．5.5＜pH＜8.5的土壤被称为中性或接近中性，说法正确．<br />故选A．','【分析】A、根据人体血液的pH值分析．<br />B、正常雨水的PH≈5.6．<br />C、厕所清洁剂显酸性，pH小于7；<br />D、农作物一般适宜在PH=7或接近7的土壤中生长．','选择题',3.00,'ae10e3692fea010bbdea415fb14afec4',9,400,'溶液的酸碱性与pH值的关系,酸碱性对生命活动和农作物生长的影响','',2015,'37','2015秋•哈尔滨校级月考',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1854084,'无土栽培是利用营养液栽培作物的一种方法．<br />表中是栽培绿叶菜营养液配方．<br /><table class=\"edittable\"><TBODY><TR><td width=101>肥料成份名称</TD><td width=129>用量（毫克∕升）</TD></TR><TR><td>硝酸钙</TD><td>1260</TD></TR><TR><td>硫酸钾</TD><td>250</TD></TR><TR><td>磷酸二氢钾</TD><td>350</TD></TR><TR><td>硫酸镁</TD><td>537</TD></TR><TR><td>硫酸铵</TD><td>237</TD></TR></TBODY></TABLE>①化学肥料成份中属于复合肥的肥料名称是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />②该营养液中存在的一种金属离子符号<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />③硫酸镁的化学式是：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />④若在实验里配制2升该营养液，需要称取的硫酸钾的质量是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>毫克．','','','','','','磷酸二氢钾$###$K<SUP>+</SUP> （ Ca<SUP>2+</SUP>，Mg<SUP>2+</SUP>）$###$MgSO<SUB>4</SUB>$###$500','【解答】解：①磷酸二氢钾中含有磷元素和钾元素，属于复合肥料，故填：磷酸二氢钾．<br />②该溶液中含有钾离子、钙离子、镁离子，故填：K<SUP>+</SUP> （ Ca<SUP>2+</SUP>，Mg<SUP>2+</SUP>）．<br />③硫酸镁的化学式为MgSO<SUB>4</SUB>，故填：MgSO<SUB>4</SUB>．<br />④根据表格提供的数据可以看出，每升营养液中含有硫酸钾250mg，故2升营养液中含有硫酸钾500mg，故填：500．','【分析】根据复合肥料的概念解答；根据离子符号的书写方法解答；根据物质化学式的书写方法解答；根据溶质质量的计算解答即可．','书写',3.00,'1eeb1ec1791415342becd553ce66dbfb',9,400,'有关溶质质量分数的简单计算,常见化肥的种类和作用,化学式的书写及意义,化学符号及其周围数字的意义,无土栽培','',2016,'37','2016•潍坊一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1854274,'通过对已学知识的对比和归纳，我们往往可以得出一些十分有趣的规律，这些规律可以帮助我们掌握学习化学的方法．请你仔细阅读下表中的内容，并回答相应的问题．<br />（1）由前两行内容对照可得出的规律是：元素或原子团的化合价数值往往与相应离子所带的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>数相等．<br />（2）根据氯化亚铁化学式FeCl<SUB>2</SUB>，可推出该物质所含阳离子符号为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．写出硫酸钠的化学式<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','电荷数$###$Fe<SUP>2+</SUP>$###$Na<SUB>2</SUB>SO<SUB>4</SUB>','【解答】解：（1）对比离子符号与其化合价可知：化合价数值与离子数值相同，其符号即正负也相同．故填：电荷数；<br />（2）在FeCl<SUB>2</SUB>中，氯元素显-1，故铁元素的化合价为+2，故阳离子所带的电荷数为2个单位的正电荷，故填：Fe<SUP>2+</SUP>．<br />硫酸钠中，钠元素显+1价，硫酸根显-2价，故硫酸钠的化学式为Na<SUB>2</SUB>SO<SUB>4</SUB>，故填：Na<SUB>2</SUB>SO<SUB>4</SUB>．','【分析】（1）对比离子数与其化合价可知规律．<br />（2）根据物质的结构以及化学式的写法来分析．','书写',3.00,'af13fe76cf8be9ae6b18750ec0b9b415',9,400,'化学式的书写及意义,化合价与离子表示方法上的异同点,化学符号及其周围数字的意义','',2013,'37','2013秋•永嘉县校级月考',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1854579,'<img  src=\"/tikuimages/9/2016/400/shoutiniao94/c29568e1-94d4-11e9-9fee-b42e9921e93e_xkb70.png\" style=\"vertical-align:middle;FLOAT:right\" />（2016•抚州校级模拟）甲同学设计了如下实验装置验证一氧化碳的部分性质并验证产物．实验时，在点燃B处酒精灯之前先通入一氧化碳排出装置中的空气，然后继续实验．<br />（1）B中反应的化学方程式是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，该反应的还原剂是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）实验过程中，C中的现象是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，D处点燃的目的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）对该实验的分析正确的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（选填编号）<br />A．实验结束时应先熄灭B处酒精灯<br />B．C中增加的质量与B中固体减少的质量相等<br />C．实验过程中，共通入0.56g一氧化碳可生成1.28g铜<br />D．反应结束后继续通入一氧化碳的目的是防止铜被氧化<br />（4）甲同学认为A装置用于证明一氧化碳不能和石灰水反应，乙同学认为省略A可达到同样的目的，理由是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','CO+CuO<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>Cu+CO<SUB>2</SUB>$###$CO$###$澄清的石灰水变浑浊$###$防止有毒的一氧化碳逸散到空气中污染环境$###$AD$###$为了排尽装置内的空气，反应前已经通入一段时间的CO','【解答】解：（1）一氧化碳和氧化铜在加热是生成铜和二氧化碳，故B中的反应是CO+CuO<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>Cu+CO<SUB>2</SUB>，该反应中一氧化碳是还原剂，<br />故填：CO+CuO<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;△&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>Cu+CO<SUB>2</SUB>，CO；<br />（2）一氧化碳还原氧化铜生成二氧化碳，二氧化碳能使澄清的石灰水变浑浊，一氧化碳是有毒的气体，点燃能防止有毒的一氧化碳排放到空气中污染环境，<br />故填：澄清的石灰水变浑浊；防止有毒的一氧化碳逸散到空气中污染环境；<br />（3）A．实验结束时应先熄灭B处酒精灯，正确；<br />B．C中增加的质量是二氧化碳的质量，B中固体减小的质量是氧元素的质量，故质量不相等，错误；<br />C．该反应前后都要通入一氧化碳，故反应开始后通入0.56g一氧化碳不可生成1.28g铜，错误；<br />D．反应结束后继续通入一氧化碳的目的是防止铜被氧化，正确；<br />故填：AD；<br />（4）省略A可达到同样的目的，因为为了排尽装置内的空气，反应前已经通入一段时间的CO，<br />故填：为了排尽装置内的空气，反应前已经通入一段时间的CO．','【分析】根据已有的一氧化碳还原氧化铜的知识进行分析解答，一氧化碳能与氧化铜反应生成铜和二氧化碳，其中一氧化碳具有还原性；一氧化碳是有毒的气体，需要进行尾气处理，根据一氧化碳和氧化铜反应的注意事项以及物质的量的关系进行解答即可．','书写',3.00,'c6c2d0b758c1be0d3c6aa30ea64bdabe',9,400,'碳、一氧化碳、氢气还原氧化铜实验,书写化学方程式、文字表达式、电离方程式','',2016,'32','2016•抚州校级模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1854749,'在CO、CO<SUB>2</SUB>和N<SUB>2</SUB>的混合气体50g，碳元素的质量分数是36%，将该混合气体与足量的灼热氧化铁完全反应，再将气体通入过量的澄清石灰水中，充分反应后得到白色沉淀的质量为（　　）','50g','100g','150g','200g','','C','【解答】解：在CO、CO<SUB>2</SUB>和N<SUB>2</SUB>的混合气体50g，碳元素的质量分数是36%，混合气体中碳元素的质量为50g×36%=18g．<br />将该混合气体与足量的灼热氧化铁完全反应，一氧化碳与氧化铁反应生成铁和二氧化碳，再将气体通入过量的澄清石灰水中，二氧化碳与石灰水反应生成碳酸钙沉淀和水，由质量守恒定律，碳元素最终转化为碳酸钙沉淀，则充分反应后得到白色沉淀的质量为18g÷（<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">12</td></tr><tr><td>40+12+16×3</td></tr></table></span>×100%）=150g．<br />故选：C．','【分析】根据题意，在CO、CO<SUB>2</SUB>和N<SUB>2</SUB>的混合气体50g，碳元素的质量分数是36%，据此可计算出混合气体中碳元素的质量；将该混合气体与足量的灼热氧化铁完全反应，再将气体通入过量的澄清石灰水中，由质量守恒定律，碳元素最终转化为碳酸钙沉淀，据此进行分析解答．','选择题',3.00,'dae9be9c590b03268c880e629c567892',9,400,'化合物中某元素的质量计算,混合物中某元素的质量计算,质量守恒定律及其应用','',0,'37','',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1854779,'<img  src=\"/tikuimages/9/2015/400/shoutiniao64/c4ec030f-94d4-11e9-9536-b42e9921e93e_xkb73.png\" style=\"vertical-align:middle;FLOAT:right\" />（2015•烟台模拟）多角度认识物质，能帮助我们更全面了解物质世界．以氧气和二氧化碳为例，回答下列问题：<br />（1）认识物质的组成和构成<br />①从宏观上看，氧气和二氧化碳都由<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“元素”、“原子”或“分子”，下同）组成．<br />②从微观上看，氧气和二氧化碳都由<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>构成．<br />（2）认识物质的性质<br />①氧气的化学性质比较活泼．纳米铁粉在氧气中可自燃生成氧化铁，反应的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />②将蘸有酒精的棉芯点燃后放入烧杯中，向烧杯中缓缓倾倒二氧化碳，观察到烧杯中的棉芯自下而上熄灭（如图）．说明二氧化碳具有的性质有<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．若使棉芯下半段恢复燃烧，操作方法是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）认识物质的制法<br />①某同学用软塑料瓶自制气体发生装置，通过捏放瓶身可随时控制反应发生和停止（如图2）．若利用该装置制氧气，反应的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．若利用该装置制二氧化碳，无纺布包内药品为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（写名称）．<br />②工业上常用液化空气制氧气，该过程发生<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填“物理变化”或“化学变化”）．<br />（4）辩证地认识物质<br />①量变引起质变．例如：碳在充足氧气中燃烧生成二氧化碳，在不充足的氧气中燃烧生成<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（写化学式）．<br />②功过相对论．从“二氧化碳导致温室效应”的事实分析，“过”：使全球气候变暖导致海平面上升等；“功”：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>　（举一例）．','','','','','','元素$###$分子$###$3Fe+2O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>Fe<SUB>3</SUB>O<SUB>4</SUB>$###$二氧化碳密度比空气大，不能燃烧也不支持燃烧$###$将棉芯从烧杯中取出（合理均可）$###$2H<SUB>2</SUB>O<SUB>2</SUB> <span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span>&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O+O<SUB>2</SUB>↑$###$大理石$###$物理变化$###$CO$###$使地球气温保持在适宜生物生存的温度范围（合理均可）','【解答】解：（1）①氧气和一氧化碳都由元素组成的，故填：元素；<br />②氧气和二氧化碳都由分子构成的，故填：分子；<br />（2）①纳米铁粉在氧气中可自燃生成氧化铁，故填：3Fe+2O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>Fe<SUB>3</SUB>O<SUB>4</SUB>；<br />②烧杯中的棉芯自下而上熄灭说明二氧化碳的密度比空气大、既不燃烧也不支持燃烧；隔绝氧气可以灭火；若使棉芯下半段恢复燃烧，可以将棉芯从烧杯中取出，使其与氧气接触；<br />（3）①过氧化氢在二氧化锰的催化作用下能生成水和氧气，若是制取二氧化碳，则使用的固体药品是石灰石，故填：2H<SUB>2</SUB>O<SUB>2</SUB> <span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span>&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O+O<SUB>2</SUB>↑；大理石；<br />②工业上常用液化空气制氧气，是利用氮气和氧气沸点的不同，没有产生新的物质，是物理变化，故填：物理变化；<br />（4）①碳在充足氧气中燃烧生成二氧化碳，在不充足的氧气中燃烧生成一氧化碳，化学式是：CO；<br />②二氧化碳的温室效应，能使地球的温度维持在一定的正常范围内．<br />答案：<br />答案：<br />（1）①元素&nbsp;&nbsp;②分子&nbsp;&nbsp;<br />（2）①3Fe+2O<SUB>2</SUB>&nbsp;<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>&nbsp;Fe<SUB>3</SUB>O<SUB>4 </SUB>②二氧化碳密度比空气大，不能燃烧也不支持燃烧；将棉芯从烧杯中取出（合理均可）；<br />（3）①2H<SUB>2</SUB>O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;Mn<span><span>O</span><span style=\"vertical-align:sub;font-size:90%\">2</span></span>&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O+O<SUB>2</SUB>↑；&nbsp;大理石；&nbsp;&nbsp;②物理变化；<br />（4）①CO；&nbsp;&nbsp;&nbsp;&nbsp;②使地球气温保持在适宜生物生存的温度范围（合理均可）．','【分析】（1）根据物质的组成进行分析；<br />（2）纳米铁粉在氧气中燃烧生成氧化铁；根据二氧化碳的性质进行解答；<br />（3）根据制取氧气的原理分析解答；<br />（4）碳不完全燃烧生成一氧化碳，根据二氧化碳的功与过进行解答即可．碳不完全燃烧生成一氧化碳，根据二氧化碳的功与过进行解答即可．','书写',3.00,'43917a6ccadd284330ffc45974776525',9,400,'空气对人类生活的重要作用,氧气的化学性质,实验室制取氧气的反应原理,二氧化碳的实验室制法,二氧化碳的化学性质,分子、原子、离子、元素与物质之间的关系,碳的化学性质,书写化学方程式、文字表达式、电离方程式','',2015,'32','2015•烟台模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1855454,'俗话说“桂花香，蟹黄肥”．青色的螃蟹煮熟后颜色会变成红色．一些同学认为这种红色物质可能就象酸碱指示剂一样，遇到酸或碱颜色会发生改变．就这些同学的看法而言应属于科学探究中的（　　）','实验','推理','观察','假设','','D','【解答】解：A、一些同学认为这种红色物质可能就象酸碱指示剂一样，遇到酸或碱颜色会发生改变，属于科学探究中的假设，而不是实验，故选项错误．<br />B、一些同学认为这种红色物质可能就象酸碱指示剂一样，遇到酸或碱颜色会发生改变，属于科学探究中的假设，而不是做推理，故选项错误．<br />C、一些同学认为这种红色物质可能就象酸碱指示剂一样，遇到酸或碱颜色会发生改变，属于科学探究中的假设，而不是观察，故选项错误．<br />D、一些同学认为这种红色物质可能就象酸碱指示剂一样，遇到酸或碱颜色会发生改变，属于科学探究中的假设，故选项正确．<br />故选：D．','【分析】科学探究的主要环节有提出问题→猜想与假设→制定计划（或设计方案）→进行实验→收集证据→解释与结论→反思与评价→拓展与迁移，据此结合题意进行分析判断．','选择题',3.00,'37f76525b16abdf4f0a7b743e86ef1b3',9,400,'科学探究的基本环节','桐乡市',2011,'35','2011秋•桐乡市期中',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1855615,'下列图片中，不属于模型的是（　　）','<img  src=\"/tikuimages/9/2013/400/shoutiniao0/ce0962cf-94d4-11e9-88a8-b42e9921e93e_xkb82.png\" style=\"vertical-align:middle\" /><br />细胞结构图','<img  src=\"/tikuimages/9/2013/400/shoutiniao63/ce0b85ae-94d4-11e9-b70b-b42e9921e93e_xkb47.png\" style=\"vertical-align:middle\" /><br />伦敦奥运会会徽','<img  src=\"/tikuimages/9/2013/400/shoutiniao9/ce0e1dc0-94d4-11e9-ac7a-b42e9921e93e_xkb25.png\" style=\"vertical-align:middle\" /><br />地球仪','<img  src=\"/tikuimages/9/2013/400/shoutiniao42/ce0fa461-94d4-11e9-9b22-b42e9921e93e_xkb62.png\" style=\"vertical-align:middle\" /><br />足球烯分子','','B','【解答】解：建立模型可以帮助人们认识和理解一些不能直接观察到的事物，因为细胞太小，难以观察，所以人们画出了细胞模式图；因为地球太大，难以认识，所以人们做出了地球仪；因为分子太小，难以观察，人们建造了分子模型．伦敦奥运会会徽只是一个图案标志，不属于模型．<br />故答案为：B','【分析】根据模型思想进行分析解答．','选择题',3.00,'3895482c33024ca04e553641e4a31114',9,400,'化学常识','',2013,'37','2013春•台州校级月考',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1855741,'下列四瓶气体存放方法正确的是（　　）','<img  src=\"/tikuimages/9/2010/400/shoutiniao78/cf90ed30-94d4-11e9-912d-b42e9921e93e_xkb70.png\" style=\"vertical-align:middle\" />','<img  src=\"/tikuimages/9/2010/400/shoutiniao8/cf931011-94d4-11e9-8f97-b42e9921e93e_xkb10.png\" style=\"vertical-align:middle\" />','<img  src=\"/tikuimages/9/2010/400/shoutiniao92/cf990380-94d4-11e9-8a34-b42e9921e93e_xkb51.png\" style=\"vertical-align:middle\" />','<img  src=\"/tikuimages/9/2010/400/shoutiniao23/cf9e0c8f-94d4-11e9-8868-b42e9921e93e_xkb44.png\" style=\"vertical-align:middle\" />','','A','【解答】解：毛玻璃的使用时粗糙面要和与之接触的仪器相接触，与之接触仪器面也是粗糙面，这样才利于密封和使用．<br />A、氧气密度比空气大，气体下降，瓶口朝上，故A正确；<br />B．氢气密度比空气小，气体上升，瓶口应朝下，故B错误；<br />C．二氧化碳密度比空气大，气体下降，瓶口朝上，光面在下，故C错误；<br />D．甲烷密度比空气小，气体上升，瓶口朝下，故D错误．<br />故选A．','【分析】根据密度的大小判断瓶口朝上朝下，磨砂面始终是要和接触的一起接触．','选择题',3.00,'6033f36e4c79b4ae3072dcbc3010f6a2',9,400,'氧气的物理性质,二氧化碳的物理性质,甲烷、乙醇等常见有机物的性质和用途,氢气的物理性质','',2010,'37','2010•太原一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1855843,'下列说法错误的是（　　）','原子的质量几乎都集中在原子核上','在化学变化中元素的种类和质量都不会改变','原子、分子、离子都是构成物质的微粒','由分子构成的物质发生化学变化时，分子本身不变','','D','【解答】解：A、原子的质量几乎都集中在原子核上，正确；<br />B、在化学变化中元素的种类和质量都不会改变，正确；<br />C、原子、分子、离子都是构成物质的微粒，正确；<br />D、由分子构成的物质发生化学变化时，分子本身一定改变，不正确；<br />故选D．','【分析】根据已有的物质的微观构成粒子之间的关系进行分析解答即可．','选择题',3.00,'7b81885e68efa4aa8455f9b8c1a88bc4',9,400,'分子、原子、离子、元素与物质之间的关系,原子的定义与构成,分子的定义与分子的特性,元素在化学变化过程中的特点','',2015,'33','2015秋•桐梓县期末',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1856407,'下列物质的鉴别方法不正确的是（　　）','用水鉴别蔗糖和硝酸铵','用看颜色的方法鉴别氧化铁和氧化铜','用肥皂水鉴别硬水和软水','用燃烧的木条鉴别二氧化碳和氮气','','D','【解答】解：A、蔗糖和硝酸铵都易溶于水，但硝酸铵溶于水吸热会使温度降低；故方法正确；<br />B、因为氧化铁为红色，氧化铜为黑色，故可以用看颜色的方法鉴别氧化铁和氧化铜；故方法正确；<br />C、区分硬水和软水的方法是：用肥皂水，加入肥皂水，泡沫多的是软水，泡沫少的是硬水，故方法正确；<br />D、二氧化碳和氮气都不能支持燃烧，都能使燃着的木条熄灭，现象相同，不能鉴别，故方法错误；<br />故选：D．','【分析】根据物质的性质差异进行分析，蔗糖和硝酸铵都易溶于水，但硝酸铵溶于水吸热会使温度降低；区分硬水和软水的方法是：用肥皂水，加入肥皂水，泡沫多的是软水，泡沫少的是硬水；但氮气和二氧化碳都不能够支持燃烧等；出现不同的实验现象，据此解答即可．','选择题',3.00,'dd6e35954431e490ea6dc231606e6644',9,400,'常见气体的检验与除杂方法,硬水与软水,碳酸钠、碳酸氢钠与碳酸钙,物质的鉴别、推断','',2015,'33','2015秋•扬州期末',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1856744,'把下列符合题意要求的物质名称写在横线上：二氧化碳、二氧化硫、高锰酸钾、四氧化三铁、五氧化二磷、稀有气体、过氧化氢溶液．<br />（1）通常用作保护气和通电时发出有色光的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />（2）铁在氧气中燃烧的产物是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />（3）白色固体物质有<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />（4）能使澄清石灰水变浑浊且参与植物光合作用的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />（5）有刺激性气味的气体是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u><br />（6）绿色化学又称环境友好化学，其核心就是要利用化学反应原理从源头消除污染．上述物质中能用于制取氧气，且其反应具备绿色化学的某些特点的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','稀有气体$###$四氧化三铁$###$五氧化二磷$###$二氧化碳$###$二氧化硫$###$过氧化氢溶液','【解答】解：（1）稀有气体化学性质比较稳定，通常用作保护气，通电时发出有色光；<br />（2）铁能在氧气中燃烧，生成四氧化三铁；<br />（3）白色固体物质有五氧化二磷；<br />（4）二氧化碳能和氢氧化钙反应生成碳酸钙沉淀和水，故能使澄清的石灰水变浑浊；光合作用是指植物利用光能，通过叶绿体，把二氧化碳和水合成贮存能量的有机物，同时释放氧气的过程，故二氧化碳可作光合作用的原料；<br />（5）有刺激性气味的气体是二氧化硫；<br />（6）过氧化氢溶液能用于制取氧气，且生成的另一种产物是水，对环境无污染，其反应具备绿色化学的某些特点．<br />故答案为：（1）稀有气体；（2）四氧化三铁；（3）五氧化二磷；（4）二氧化碳；（5）二氧化硫；（6）过氧化氢溶液．','【分析】（1）根据稀有气体的性质分析．<br />（2）根据铁在氧气中燃烧的产物分析．<br />（3）根据物质的颜色分析．<br />（4）根据二氧化碳的化学性质分析．<br />（5）根据二氧化硫的状态、气味、毒性分析．<br />（6）根据实验室制氧气的原理分析．','填空题',3.00,'2c34ec75a4a3ec4d44cfcd2be5ebefeb',9,400,'绿色化学,氧气的化学性质,常见气体的用途','',2015,'37','2015秋•孝昌县校级月考',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1856752,'电焊时会闻到一股异味，这是因为在放电条件下部分氧气转化成了臭氧（O<SUB>3</SUB>）．这一过程可用化学方程式表示为：3O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;放电&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2O<SUB>3</SUB>．下列有关说法中正确的是（　　）','发生的是物理变化','O<SUB>3</SUB>与O<SUB>2</SUB>的性质相同','O<SUB>3</SUB>的相对分子质量为48','臭氧比氧气多一个氧原子','','C','【解答】解：A、氧气由氧分子构成，臭氧是由臭氧分子构成，氧气与臭氧是两种不同的物质，所以该变化属于化学变化，故A说法错误；<br />B、O<SUB>2</SUB>与O<SUB>3</SUB>是氧元素的同素异形体，物理性质不同，化学性质相似，故B错误；<br />C、O<SUB>3</SUB>的相对分子质量=16×3═48，故C说法正确；<br />D、应该是臭氧分子比氧气分子多一个氧原子，故D说法错误．<br />故选C','【分析】根据物理变化与化学变化的本质区别：是否有新物质生成判断属于什么变化，氧气和臭氧化学性质相似，物理性质不同；相对分子质量为各个原子的相对原子质量之和；','选择题',3.00,'e27891e3797191e786ede657c2320407',9,400,'氧元素组成的单质,相对分子质量的概念及其计算,化学变化和物理变化的判别','',0,'37','',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1856775,'阅读下列信息，回答有关问题：<br />1909年哈伯在实验室将氮气和氢气在600℃、2.02×10<SUP>4</SUP>kPa和铁作催化剂的条件下首次合成了氨（NH<SUB>3</SUB>）．常温下，氨是一种无色有刺激性气味的气体，能经过下列反应制得生产炸药的原料--硝酸．<br />A．氨和氧气在铂催化剂和一定温度下反应&nbsp;生成一氧化氮和水；<br />B．一氧化氮和氧气反应生成二氧化氮；<br />C．二氧化氮和水反应生成硝酸和一氧化氮．<br />工业合成氨的原料来自空气、煤和水，这是一种经济的固氮方法．这一成果生产的化肥给农业带来了丰收，也获得了代替智利硝石生产炸药的原料．1914年第一次世界大战爆发时，由于德国垄断了合成氨技术，能快速生产氨和硝酸，使粮食和炸药的供应有了保障，这也促成了德皇威康二世开战的决心，给世界人民带来了灾难．<br />（1）请从以上信息中总结有关氨的知识：<br />①氨的物理性质：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />②氨的化学性质：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />③氨的制法（写化学方程式）：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />④氨的用途：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）根据氨生产硝酸的三个反应，回答下列问题：<br />①B反应的反应类型是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />②对于C反应中产生的一氧化氮（一种大气污染物）尾气，你认为最好的处理方法是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（3）你认为合成氨中氢元素主要来自原料中的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（4）从合成氨技术给人类带来的好处与灾难，有同学提出如下看法：化学新技术给人类进步带来了贡献，也带来了灾难．可见发明化学新技术对人类并没有实际意义．你是否同意此看法，请谈谈你的观点？<br /><u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','常温下，无色有刺激性气味的气体$###$氨和氧气在铂催化剂和一定温度下反应生成一氧化氮和水$###$N<SUB>2</SUB>+3H<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;催化剂&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>高温高压</td></tr></table></span>2NH<SUB>3</SUB>$###$制取化肥、炸药$###$化合反应$###$将尾气返回第B步继续反应$###$水$###$不同意；只要人类正确使用新技术，就能给人类的发展带来贡献，如炸药可用于开山修路','【解答】解：（1）①常温下，氨气是无色有刺激性气味的气体，属于氨气的物理性质．故填：常温下，无色有刺激性气味的气体．<br />②氨和氧气在铂催化剂和一定温度下反应能生成一氧化氮和水，属于氨气的化学性质．故填：氨和氧气在铂催化剂和一定温度下反应能生成一氧化氮和水．<br />③氮气和氢气在一定条件下能生成氨气，反应的化学方程式为：N<SUB>2</SUB>+3H<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;&nbsp;&nbsp;催化剂&nbsp;&nbsp;&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>高温高压</td></tr></table></span>2NH<SUB>3</SUB>．<br />④氨气能够用来制取化肥、炸药．故填：制取化肥、炸药．<br />（2）①一氧化氮和氧气反应生成二氧化氮，属于化合反应．故填：化合反应．<br />②最好的处理方法是将尾气返回第B步继续反应．故填：将尾气返回第B步继续反应．<br />（3）合成氨中氢元素主要来自原料中的水．故填：水．<br />（4）化学是一把双刃剑，既能给人类带来好处，也能给人类带来害处；但是只要利用得当，趋利避害，化学对人类的贡献还是相当大的．故填：不同意；只要人类正确使用新技术，就能给人类的发展带来贡献，如炸药可用于开山修路．','【分析】（1）颜色、气味、状态等性质不需要通过化学变化表现出来，属于物质的物理性质；<br />氨气和氧气在一定条件下能生成一氧化氮和水，属于物质的化学性质；<br />根据反应物和生成物及其质量守恒定律可以书写化学方程式；<br />氨气能够用来制取化肥、炸药．<br />（2）由两种或两种物质反应生成一种物质的反应属于化合反应；<br />对于C反应中产生的一氧化氮可以循环利用；<br />（3）合成氨中氢元素主要来自原料中的水；<br />（4）化学是一把双刃剑，既能给人类带来好处，也能给人类带来害处；但是只要利用得当，趋利避害，化学对人类的贡献还是相当大的．','书写',3.00,'7fc617eb4e79062d771176ec518ebe87',9,400,'化学的用途,化学性质与物理性质的差别及应用,反应类型的判定,书写化学方程式、文字表达式、电离方程式','',0,'37','',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1857338,'化学中存在很多“化学之最”，请完成下列填空：<br />空气成分中体积分数最大的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；天然存在的最硬的物质是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；形成化合物种类最多的元素是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','氮气/N<SUB>2</SUB>$###$金刚石$###$碳元素/C','【解答】解：空气的成分中体积分数最大的是氮气；天然存在的最硬的物质是金刚石；形成化合物种类最多的元素是碳．<br />故答案为：氮气/N<SUB>2</SUB>&nbsp;&nbsp;&nbsp;&nbsp;金刚石&nbsp;&nbsp;&nbsp;&nbsp;碳元素/C','【分析】空气中各成分的体积分数分别是：氮气78%、氧气21%、稀有气体0.94%、二氧化碳0.03%、其它气体和杂质0.03%；天然存在的最硬的物质是金刚石；形成化合物种类最多的元素是碳．','填空题',3.00,'9da366ac6c9dbc6de9091853a43797b1',9,400,'有关化学之最,空气的成分及各成分的体积分数,碳单质的物理性质及用途,物质的元素组成','',2016,'37','2016•滑县一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1857848,'燃料的合理使用是解决环境污染重要途径．<br />①氢气被誉为“最理想的燃料”，请用化学方程式解释原因<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />②今年各地雾霾频发导致呼吸系统疾病病人增多，是因为空气中新增了大量的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填字母序号）．<br />A、二氧化碳&nbsp;&nbsp; B、氮氧化物&nbsp;&nbsp; C、二氧化硫&nbsp;&nbsp; D、可吸入颗粒物<br />③我国已经开发和推广使用乙醇汽油，其中含乙醇（C<SUB>2</SUB>H<SUB>5</SUB>OH）10%，乙醇的摩尔质量是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，每摩尔乙醇分子中含<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>个氢原子．<br />④乙醇和甲烷都是常用的燃料，完全燃烧后产物相同，若燃烧2mol乙醇释放出的二氧化碳与<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>g甲烷燃烧释放出的二氧化碳的质量相等．','','','','','','2H<SUB>2</SUB>+O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O$###$C D$###$46g/mol$###$6×6.02×10<SUP>23</SUP>$###$64','【解答】解：①氢气燃烧产物是水，无污染，被誉为最理想的清洁能源，反应的化学方程式是为：2H<SUB>2</SUB>+O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O；<br />②PM2.5是首要污染物，PM2.5是指大气中直径小于或等于2.5微米的颗粒物，雾霾可使空气中增加大量的可吸入颗粒物；二氧化硫气体大量排放到空气中，会形成酸雨；<br />③乙醇的相对分子质量是46，摩尔质量是46g/mol；每摩尔乙醇中含有6×6.02×10<SUP>23</SUP>个氢原子；<br />④由C<SUB>2</SUB>H<SUB>5</SUB>OH+3O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2CO<SUB>2</SUB>+3H<SUB>2</SUB>O；&nbsp;可知，1摩尔乙醇反应生成2摩尔二氧化碳；据CH<SUB>4</SUB>+2O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>CO<SUB>2</SUB>+2H<SUB>2</SUB>O．可知，1摩尔甲烷反应生成1摩尔二氧化碳，故燃烧2mol乙醇释放出的二氧化碳与燃烧4mol甲烷放出的二氧化碳质量相等．甲烷的相对分子质量是16，摩尔质量是16g/mol，故4mol甲烷的质量是：4mol×16g/mol=64g．<br />故答案为：①2H<SUB>2</SUB>+O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O；&nbsp; ②D C；&nbsp; ③46g/mol、6×6.02×10<SUP>23</SUP>； ④64．','【分析】①氢气燃烧产物是水，无污染，被誉为最理想的清洁能源；<br />②空气污染的途径主要有两个：有害气体和粉尘．有害气体主要有一氧化碳、二氧化硫、二氧化氮等气体；粉尘主要指一些固体小颗粒．PM2.5是首要污染物，PM2.5是指大气中直径小于或等于2.5微米的颗粒物，雾霾可使空气中增加大量的可吸入颗粒物；<br />③乙醇的相对分子质量是46，摩尔质量是46g/mol；每摩尔乙醇中含有6×6.02×10<SUP>23</SUP>个氢原子；<br />④','书写',3.00,'60e9fae79b8e601423ab9e6bebdacc12',9,400,'空气的污染及其危害,有关化学式的计算和推断,书写化学方程式、文字表达式、电离方程式,常用燃料的使用与其对环境的影响','',2016,'37','2016•黄浦区二模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1858068,'阅读材料，回答问题：<br />材料1．臭氧是淡蓝色气体，大气中的臭氧层能有效阻挡紫外线，保护地球的生存环境，但目前南极出现了臭氧层空洞，并有继续扩大的趋势．<br />材料2．复印机在工作时，会因高压放电产生一定浓度的臭氧．长期吸入大量臭氧会引起口干舌燥，咳嗽等不适症状，还可能诱发中毒性肺气肿．<br />材料3．臭氧发生器是在高压电极的作用下将空气中的氧气转化为臭氧（化学式为O<SUB>3</SUB>）的装置．利用臭氧的强氧化性，可将其应用于游泳池、生活用水、污水的杀菌和消毒．<br />（1）请总结臭氧的有关知识：<br />①物理性质：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />②化学性质：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />③用途：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（2）请从分子构成的角度，指出氧气和臭氧的不同点：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（3）从上述材料中可见臭氧对人类有利有弊．请再举出一种物质，并说出其利弊：<br /><u>&nbsp;&nbsp;&nbsp;&nbsp;</u>：利：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>、弊：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','淡蓝色气体$###$强氧化性$###$游泳池、生活用水、污水的杀菌和消毒$###$每个臭氧分子中有3个氧原子，而每个氧气分子中只有2个氧原子$###$二氧化碳$###$空气中含有二氧化碳，有利于植物的光合作用$###$过多会引起温室效应','【解答】解：（1）物理性质主要包括：色态味、密度、溶解性等等；常见的化学性质有：氧化性、还原性、毒性、酸碱性、稳定性等等，故物理组性质：①淡蓝色气体；化学性质为：②强氧化性；③游泳池、生活用水、污水的杀菌和消毒．<br />（2）氧气、臭氧的化学式分别为：O<SUB>2</SUB>、O<SUB>3</SUB>，可以看出两种物质的分子构成中原子个数不同．<br />故答：每个臭氧分子中有3个氧原子，而每个氧气分子中只有2个氧原子．<br />（3）任何一种物质都存在利弊两个方面，所谓利即为人所利用的用途，弊则是指使用中所带来的问题．回答本题时尽可能选择自己熟悉的物质．<br />故可答：二氧化碳；空气中含有二氧化碳，有利于植物的光合作用，但过多会引起温室效应','【分析】（1）物理性质主要包括：色态味、密度、溶解性等等；常见的化学性质有：氧化性、还原性、毒性、酸碱性、稳定性等等．<br />（2）分子由原子构成，包括原子的种类、原子的个数．<br />（3）化学方程书写基础是确定反应物和生成物，然后用化学式表示出来．<br />（4）物质的使用往往利弊共存，这是由物质性质所决定的','填空题',3.00,'184d490e98fad6d30c2495f69179666b',9,400,'分子的定义与分子的特性,氧元素组成的单质,化学性质与物理性质的差别及应用','',2014,'37','2014秋•明水县校级月考',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1858223,'下列有关资源、能源的叙述不正确的是&nbsp;&nbsp;（　　）','节约水资源就要提高水的利用效益','空气是一种空贵的资源，广泛用于生产化肥、化工产品、石油加工等领域','天然气是一种较为清洁的新能源','稀土是不可再生的重要战略资源','','C','【解答】解：A、节约水资源就要提高水的利用效益，故A正确；<br />B、空气是一种空贵的资源，广泛用于生产化肥、化工产品、石油加工等领域，故B正确；<br />C、天然气是一种较为清洁的能源，但不是新能源，故C错误；<br />D、稀土是不可再生的重要战略资源，故D正确．<br />故选：C．','【分析】A、根据节约水资源就要提高水的利用效益进行解答；<br />B、根据空气是一种空贵的资源进行解答；<br />C、根据天然气是一种较为清洁的能源，但不是新能源进行解答；<br />D、根据稀土是不可再生的重要战略资源进行解答．','选择题',3.00,'ddb615ede1c23cfd92f49e2352db7ab3',9,400,'空气对人类生活的重要作用,化石燃料及其综合利用,保护水资源和节约用水','',2016,'37','2016•香坊区一模',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1858511,'<img  src=\"/tikuimages/9/2012/400/shoutiniao10/ecb1e591-94d4-11e9-940a-b42e9921e93e_xkb76.png\" style=\"vertical-align:middle;FLOAT:right\" />某学生设计了一个“黑笔写红字”的趣味实验．如图所示，滤纸先用氯化钠、无色酚酞的混合液浸湿，然后平铺在一块铂片上，接通电源后，用铅笔在滤纸上写字，会出现红色字迹．据此，下列猜想肯定不正确的是（　　）','接通电源后铅笔写字时溶液中发生了化学反应','出现红色字迹可能是因为生成了NaOH','出现红色字迹可能是因为生成了HCl','断开电源，写字时字迹可能不会变红','','C','【解答】解：出现红色的字迹，说明有碱性物质生成，在化学反应前后，元素的种类不变；<br />A、接通电源后铅笔写字时发生了化学反应，正确；<br />B、出现红色字迹可能是因为生成了NaOH，氢氧化钠呈碱性，正确；<br />C、盐酸不能使酚酞变色，错误；<br />D、断开电源写字字迹不会变红，正确；<br />故选C．','【分析】根据已有的知识进行分析解答，酚酞试液在碱性溶液中会变红，据此解答．','选择题',3.00,'2c93037f611df4822144f7342d4668b3',9,400,'猜想与事实验证,酸碱指示剂及其性质','温岭市',2012,'32','2012•温岭市模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1858514,'下列有关说法中错误的是（　　）<br />①原子一定比分子小<br />②最外层电子数为8的粒子一定是稀有气体元素的原子<br />③浓溶液一定是饱和溶液；稀溶液一定是不饱和溶液．<br />④同种元素在同一化合物中不一定显示一种化合价．<br />⑤单质也可以发生分解反应．<br />⑥凡金属与酸发生的置换反应，反应后溶液的质量一定增加．','①②④⑤','①③④⑥','①②③⑤','①②③④⑤⑥','','C','【解答】解：①不能笼统比较分子与原子的大小，只能说是构成某分子的原子比该分子小，所以错误．<br />②很多原子的离子最外层电子数是8，但不是稀有气体元素，例如氯离子等，故凡是最外层排8个电子的粒子一定是稀有气体元素错误；<br />③溶液的浓稀与饱和不饱和无关；浓溶液不一定是饱和溶液；稀溶液不一定是不饱和溶液．错误；<br />④同种元素在同一化合物中不一定显示一种化合价．如硝酸铵；正确；<br />⑤单质不可以发生分解反应．错误<br />⑥金属与酸发生的置换反应，反应后溶液的质量一定增加，正确．<br />答案：C．','【分析】①根据分子的结构判断．②很多原子的离子最外层电子数是8，但不是稀有气体元素；③溶液的浓稀与饱和不饱和无关；④同种元素在同一化合物中不一定显示一种化合价．<br />⑤单质不可以发生分解反应．⑥金属与酸发生的置换反应，反应后溶液的质量一定增加．','选择题',4.00,'1449d19ed1f65750b928c30dac76f469',9,400,'浓溶液、稀溶液跟饱和溶液、不饱和溶液的关系,金属的化学性质,分子和原子的区别和联系,核外电子在化学反应中的作用,常见元素与常见原子团的化合价,分解反应及其应用','',2012,'33','2012秋•黄梅县期末',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1858598,'本学期，同学们探究了氧气和二氧化碳的性质，请回忆相关探究过程，回答下列问题．<br /><img  src=\"/tikuimages/9/2015/400/shoutiniao83/eda71b00-94d4-11e9-95bc-b42e9921e93e_xkb13.png\" style=\"vertical-align:middle\" /><br />（1）甲图实验中观察到硫在空气中燃烧发出微弱的淡蓝色火焰，而在氧气中燃烧得更旺，发出<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>火焰．说明氧气的化学性质比较活泼，具有<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>性．实验中集气瓶中水的作用是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）乙图A中观察到的现象是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，B中发生反应的化学方程式是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，通过C实验能证明二氧化碳具有的性质是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','蓝紫色$###$氧化$###$吸收燃烧生成的二氧化硫气体，防止污染空气$###$紫色石蕊试液变红$###$CO<SUB>2</SUB>+Ca（OH）<SUB>2</SUB>=CaCO<SUB>3</SUB>↓+H<SUB>2</SUB>O$###$二氧化碳不能燃烧，也不能支持燃烧，二氧化碳的密度比空气的密度大','【解答】解：（1）甲图实验中观察到硫在空气中燃烧发出微弱的淡蓝色火焰，而在氧气中燃烧得更旺，发出蓝紫色火焰．说明氧气的化学性质比较活泼，具有氧化性．由于二氧化硫有毒能污染空气，所以实验中集气瓶中水的作用是：吸收燃烧生成的二氧化硫气体，防止污染空气．<br />（2）由于二氧化碳能与水化合生成了碳酸，能与氢氧化钙反应生成了碳酸钙和水，所以，乙图A中观察到的现象是紫色石蕊试液变红，B中发生反应的化学方程式是：CO<SUB>2</SUB>+Ca（OH）<SUB>2</SUB>=CaCO<SUB>3</SUB>↓+H<SUB>2</SUB>O，通过C实验可知，下面的蜡烛先熄灭，上面的蜡烛后熄灭，能证明二氧化碳具有的性质是：二氧化碳不能燃烧，也不能支持燃烧，二氧化碳的密度比空气的密度大．<br />故答为：（1）蓝紫色，氧化；吸收燃烧生成的二氧化硫气体，防止污染空气．（2）紫色石蕊试液变红&nbsp; CO<SUB>2</SUB>+Ca（OH）<SUB>2</SUB>=CaCO<SUB>3</SUB>↓+H<SUB>2</SUB>O；&nbsp;二氧化碳不能燃烧，也不能支持燃烧，二氧化碳的密度比空气的密度大．','【分析】（1）根据硫在氧气中燃烧的现象、生成物的性质等分析回答；<br />（2）根据二氧化碳的性质分析回答．','书写',4.00,'0c3058b5025e1fd49c72998f1c8bd078',9,400,'探究氧气的性质,探究二氧化碳的性质,书写化学方程式、文字表达式、电离方程式','',2015,'33','2015秋•孝义市期末',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1858767,'下列实验基本操作或叙述中，不正确的是（　　）','酒精流到实验台上起火燃烧应用湿抹布扑灭','试管夹夹持试管应从试管底部往上套','称量腐蚀性强的药品不要直接放在托盘上称量，应该放在纸上称量','给试管内液体加热时，液体不能超过试管容积的1/3，不能将管口对着人','','C','【解答】解：A、酒精洒在实验桌上燃烧起来，可用湿抹布盖灭，隔绝氧气灭火，故A操作正确；<br />B、试管夹是一种专门用来夹持试管的挟持器．需要注意的是在夹持时应该从试管底部往上套，夹持在试管的中上部．故B操作正确；<br />C、无腐蚀性的药品也不能直接放在天平托盘上称量，腐蚀性强的药品要放在玻璃容器中称量，故C错误；<br />D、给试管里的液体加热，试管口应向上倾斜，与桌面约成45°角，且试管内液体不能超过试管容积的<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">1</td></tr><tr><td>3</td></tr></table></span>，管口不能朝着有人的地方；故D操作正确；<br />故选：C','【分析】A、酒精洒在实验桌上燃烧起来，可用湿抹布盖灭；<br />B、根据试管夹的使用方法及注意事项进行分析解答；<br />C、无腐蚀性的药品也不能直接放在托盘上称量；<br />D、依据给试管内药品加热的注意事项分析解答．','选择题',3.00,'273ef75656db4bcf6b04c5904d66fa90',9,400,'称量器-托盘天平,加热器皿-酒精灯,挟持器-铁夹、试管夹、坩埚钳,给试管里的液体加热','丹阳市',2014,'37','2014秋•丹阳市校级月考',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1858887,'<img  src=\"/tikuimages/9/2015/400/shoutiniao63/f17783f0-94d4-11e9-85aa-b42e9921e93e_xkb53.png\" style=\"vertical-align:middle;FLOAT:right\" />（2015秋•成都期末）下列是金属或合金的性质或变化或应用，请根据题意填空：<br />（1）自行车作为常用的代步工具，既轻便灵活，又符合环保要求，如图是一款自行车的示意图．<br />①所标物质中，属于合金的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，②车架表面刷漆主要是为了防锈，其原理是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；③当今自行车外形美观，材质轻便、牢固，除了代步，还可以作为健身工具．由此你对化学与人类生活的关系有何感想？<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）下列是金属铜的有关描述：①铜在潮湿的空气中会生锈②铜绿是一种绿色粉末③加热铜绿生成氧化铜、二氧化碳和水④加热时产生的水蒸气在试管口冷凝成水珠其中属于物理变化的是（填序号，下同）；属于化学变化的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；属于物理性质的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；属于化学性质的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','不锈钢、钢$###$使铁与水和氧气隔绝$###$化学的发展促进了人类社会的进步$###$③$###$②$###$①','【解答】解：（1）不锈钢属于合金材料，钢中含有的金属是铁；车架表面刷油漆能使铁与水和氧气隔绝，化学的发展促进了人类社会的进步，故填：不锈钢、钢；使铁与水和氧气隔绝；化学的发展促进了人类社会的进步．<br />（2）：①铜生锈生成了碱式碳酸铜，有新物质生成，属于化学变化，所以铜在潮湿的空气中会生锈，属于铜的化学性质；<br />②铜绿是一种绿色粉末，描述的物质的颜色、状态，不需要通过化学变化表现出来属于物理性质；<br />③加热铜绿生成氧化铜、二氧化碳与水，有新物质生成，属于化学变化；<br />④加热时产生的水蒸气在试管口冷凝成水珠，是水的状态的改变，没有新物质生成属于物理变化；<br />故答案为：（1）④；（2）③；（3）②；（4）①．','【分析】（1）根据已有的材料的类别解答，铁在与水和氧气并存时易生锈，防锈就是破坏铁生锈的条件，根据化学与生活的关系解答即可．<br />（2）分别根据物理性质，化学性质以及化学变化和物理变化的定义以及特点来判断即可．注意：有没有新物质生成，是物理变化与化学变化的根本区别；物理性质：物质不需要经过化学变化就表现出来的性质叫做物理性质；化学性质：物质在发生化学变化中才能表现出来的性质叫做化学性质．','填空题',3.00,'ffccde6235dedbda86caccc6da71d333',9,400,'化学的用途,合金与合金的性质,金属锈蚀的条件及其防护,化学变化和物理变化的判别,化学性质与物理性质的差别及应用','',2015,'33','2015秋•成都期末',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1859384,'西瓜适合在弱酸性土壤中生长，某校化学课外小组的同学测定我市某地土壤溶液的pH为8，为了适合西瓜种植，需对该土壤的酸碱性进行调节，则下列各项中，可采用的是（　　）','施以氢氧化钙粉末','施以碳酸钙粉末','用氨水进行喷灌','适量施用酸性肥料','','D','【解答】解：某地土壤溶液的pH为8，呈碱性，而西瓜适合在弱酸性土壤中生长，故需要施加酸性肥料，故选D．','【分析】根据已有的酸碱中和反应的知识进行分析解答即可．','选择题',3.00,'f139c45bb6e641261007f3186189ad33',9,400,'溶液的酸碱性与pH值的关系,酸碱性对生命活动和农作物生长的影响','肥城市',2015,'33','2015秋•肥城市期末',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1859401,'<img  src=\"/tikuimages/9/2015/400/shoutiniao55/f73d061e-94d4-11e9-9311-b42e9921e93e_xkb38.png\" style=\"vertical-align:middle;FLOAT:right\" />如图表示的是纯净物、化合物、氧化物之间的包含与不包含关系，若整个大圆圈代表纯净物，则在下列选项中，则其中能正确表示化合物的是（　　）','①','②','③','无法确定','','B','【解答】解：由图中信息可知，①代表纯净物，②代表化合物，③代表氧化物．<br />故选：B','【分析】物质包括混合物和纯净物，纯净物包括单质和化合物，化合物包括氧化物、酸碱盐等．','选择题',3.00,'1fbc45da1da298fb7e4f123883c0a2dd',9,400,'物质的简单分类,单质和化合物的概念,氧化物、酸、碱和盐的概念','',2015,'37','2015秋•温州校级月考',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1859405,'pH与人类的生产、生活关系密切，人的血液pH正确范围是（　　）','7.35～7.45','8～10','3～4','4.55～5.55','','A','【解答】解：人的血液pH正确范围是7.35～7.45，故选A．','【分析】根据已有的血液的正常pH范围进行分析解答即可．','选择题',3.00,'a4c494d6395ab4e7d7910d568871ccb5',9,400,'溶液的酸碱性与pH值的关系,酸碱性对生命活动和农作物生长的影响','',2013,'32','2013•武平县模拟',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1859570,'物理变化与化学变化的本质区别<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，下列能判断镁燃烧是化学变化的依据是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />A．发出耀眼白光&nbsp;&nbsp;&nbsp;B．放出热量&nbsp;&nbsp;&nbsp;C．生成白色粉末&nbsp;&nbsp;D．颜色发生改变．','','','','','','是否有新物质生成$###$C','【解答】解：物理变化与化学变化的本质区别：是否有新物质生成．<br />判断镁在空气中燃烧是否发生了化学变化，关键是看是否有新物质的生成，选项A、B、D均是描述了镁在空气中燃烧的现象，不能作为判断化学变化的依据；而选项C说明有新物质生成，是判断发生化学变化的依据．<br />故填：是否有新物质生成；C．','【分析】根据生成其他物质的变化叫化学变化，它的本质特征是有其他（新的）物质生成，分析过程要注意不要被变化过程中的现象所迷惑，要找到本质．据此即可进行分析解答本题．','填空题',3.00,'951d8afefc46a0cd7c9fc565a3057a4e',9,400,'化学变化的基本特征,物理变化的特点','',0,'37','',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1859756,'水与人类的生活和生产密切相关．请回答下列问题：<br />（1）保持水的化学性质的最小粒子是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（2）人们对水组成的认识是从氢气在氧气中燃烧实验开始的，该反应的化学方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（3）去年五月份我省部分地区天气干旱，严重缺水，因地制宜利用水资源迫在眉睫．<br />①打井取用地下水，检验地下水是否为硬水可用<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，生活中降低水的硬度的方法是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />②除去河水中的异味和色素，可用活性炭，这是利用了活性炭的<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>性；<br />③若你生活在干旱地区，请提出一条节水措施：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','水分子$###$2H<SUB>2</SUB>+O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O；$###$肥皂水$###$煮沸$###$吸附$###$洗菜水浇花等','【解答】解：（1）水是由水分子构成，保持水的化学性质的最小粒子是水分子；<br />（2）氢气在氧气中燃烧生成了水，该反应的化学方程式为2H<SUB>2</SUB>+O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O；<br />（3）①打井取用地下水，检验地下水是否为硬水可用肥皂水，硬水遇肥皂水产生的泡沫少，生活中降低水的硬度的方法是煮沸；<br />②除去河水中的异味和色素，可用活性炭，这是利用了活性炭的吸附性；<br />③节水措施有很多，例如：洗菜水浇花、使用节水龙头等．<br />故答为：（1）水分子；（2）2H<SUB>2</SUB>+O<SUB>2</SUB><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<SUB>2</SUB>O；（3）①肥皂水，煮沸；②吸附；③洗菜水浇花等．','【分析】（1）根据水的构成及分子的定义分析；<br />（2）根据氢气在氧气中燃烧生成了水，写出反应的化学方程式；<br />（3）①根据硬水的检验和硬水的软化措施分析回答；<br />②根据活性炭的具有吸附性分析回答；<br />③根据常见的节水措施分析回答．','填空题',3.00,'89000cde3bdd4d5e1415dcd1b104bbb9',9,400,'水的合成与氢气的燃烧,硬水与软水,分子的定义与分子的特性,碳单质的物理性质及用途','',2016,'37','2016•崇仁县一模',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1859778,'下列做法不合理的是（　　）','无水硫酸铜可用于鉴别水和酒精两种物质','医院用生理盐水实际上就是0.9% NaCl溶液','误食重金属盐中毒后，可立即服用鸡蛋清解毒','可用氢氧化钠溶液和湿润的蓝色石蕊试纸来检验铵盐','','D','【解答】解：A、无水硫酸铜遇到水变蓝色，所以可用来检验水和酒精，故A说法正确；<br />B、生理盐水的浓度是0.9%，故B说法正确；<br />C、重金属盐能使蛋白质变性，所以误食重金属盐中毒后，可立即服用鸡蛋清解毒，故C说法正确；<br />D、铵根离子的检验方法：可用氢氧化钠溶液和湿润的红色石蕊试纸来检验铵盐，故D说法错误．<br />故选D．','【分析】A、根据无水硫酸铜遇到水变蓝色，考虑本题；B、根据生理盐水的浓度考虑；C、根据重金属盐能使蛋白质变性考虑；D、根据铵根离子的检验方法考虑．','选择题',3.00,'f1b576cbc030fd6dac89962537b852bd',9,400,'证明铵盐,氯化钠与粗盐提纯,盐的化学性质,常见中毒途径及预防方法','',2014,'35','2014秋•台州期中',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1859828,'逻辑推理是一种重要的化学思维方法，下列推理合理的是（　　）','分子可以保持物质的化学性质，所以保持化学性质的一定是分子','因为红磷和白磷的组成元素相同，所以它们是同一种物质','将CO<SUB>2</SUB>通入紫色石蕊溶液中，溶液变红，所以使紫色石蕊变红的是CO<SUB>2</SUB>','实验室用KClO<SUB>3</SUB>制取氧气，所以制取氧气的物质中一定含有氧元素','','D','【解答】解：A、由原子、离子直接构成的物质化学性质由原子、离子保持，错误；<br />B、红磷和白磷是不同的物质，错误；<br />C、将CO<SUB>2</SUB>通入紫色石蕊溶液中，溶液变红，使紫色石蕊变红的是二氧化碳与水反应生成的碳酸，不是二氧化碳，错误；<br />D、实验室用KClO<SUB>3</SUB>制取氧气，根据质量守恒定律可知，制取氧气的物质中一定含有氧元素，故正确．<br />故选D．','【分析】A、根据由原子、离子直接构成的物质化学性质由原子、离子保持进行分析；<br />B、根据红磷和白磷是不同的物质进行分析；<br />C、根据二氧化碳的化学性质进行分析；<br />D、根据质量守恒定律进行分析．','选择题',4.00,'57c38e730abfca1db13cf4fe7148128a',9,400,'二氧化碳的化学性质,分子的定义与分子的特性,同素异形体和同素异形现象,质量守恒定律及其应用','',2015,'33','2015秋•合肥期末',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1859835,'有关化学变化和物理变化的关系叙述中，不正确的是（　　）','化学变化和物理变化常常同时发生','化学变化时一定同时发生物理变化','物理变化时一定发生化学变化','物理变化时不一定发生化学变化','','C','【解答】解：A、物理变化和化学变化的联系是：化学变化过程中一定有物理变化，物理变化中不一定有化学变化，因此化学变化和物理变化常常同时发生，正确但不符合题意，故选项错误；<br />B、物理变化和化学变化的联系是：化学变化过程中一定有物理变化，物理变化中不一定有化学变化，因此化学变化时一定同时发生物理变化，正确但不符合题意，故选项错误；<br />C、物理变化和化学变化的联系是：化学变化过程中一定有物理变化，物理变化中不一定有化学变化，因此物理变化时一定发生化学变化是错误的，错误符合题意，故选项正确；<br />D、物理变化和化学变化的联系是：化学变化过程中一定有物理变化，物理变化中不一定有化学变化，正确但不符合题意，故选项错误；<br />故选C','【分析】有新物质生成的变化叫化学变化，没有新物质生成的变化叫物理变化．化学变化的特征是：有新物质生成．判断物理变化和化学变化的依据是：是否有新物质生成．化学变化中伴随的现象有：发光、发热、颜色的改变、放出气体、生成沉淀等．物理变化和化学变化的联系是：化学变化过程中一定有物理变化，物理变化中不一定有化学变化．','选择题',2.00,'897896a96955daaef4cbb181458d36f1',9,400,'化学变化的基本特征,物理变化的特点','',2015,'37','2015秋•青岛月考',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1859896,'下列有关“化学之最”的叙述中正确的是（　　）','空气中含量最多的气体是氧气','地壳中含量最多的元素是铝元素','熔点最低的金属是汞','人体中含量最多的金属元素是钠元素','','C','【解答】解：A、空气中含量最多的气体是氮气，故选项说法错误．<br />B、地壳中含量最多的元素是氧元素，故选项错误．<br />C、熔点最低的金属是汞，故选项说法正确．<br />D、人体中含量最多的金属元素是钙，故选项说法错误．<br />故选：C．','【分析】根据常见的化学之最的知识（空气中含量最多的气体、地壳中含量最多的元素、熔点最低的金属、人体中含量最多的金属元素等）进行分析判断即可．','选择题',2.00,'c5ac6334ee736487fbbd77ac706e4a3c',9,400,'有关化学之最,空气的成分及各成分的体积分数,金属的物理性质及用途,地壳中元素的分布与含量,人体的元素组成与元素对人体健康的重要作用','',2015,'37','2015•曲靖',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1859897,'化学就在我们身边，它与我们的生活密切相关．请从下列物质中选择适当的物质填空（填字母）：<br />A．盐酸&nbsp;&nbsp;&nbsp;&nbsp;B．石墨&nbsp;&nbsp;&nbsp;&nbsp;C．聚乙烯　&nbsp;D．碳酸氢钠　&nbsp;&nbsp;E．大理石&nbsp;&nbsp;　F．肥皂水<br />（1）可用于建筑材料的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（2）可用于生产食品包装袋的有机高分子材料的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（3）可用于制铅笔芯的是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<br />（4）发酵粉的主要成分之一是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','E$###$C$###$B$###$D','【解答】解：（1）大理石可用于建筑材料；<br />（2）聚乙烯是可用于生产食品包装袋的有机高分子材料；<br />（3）石墨质软，可用于制铅笔芯； <br />（4）发酵粉的主要成分之一是小苏打碳酸氢钠；<br />故填：（1）E&nbsp;&nbsp;（2）C&nbsp;&nbsp;（3）B&nbsp;&nbsp;&nbsp;（4）D．','【分析】物质的性质决定物质的用途，根据已有的物质的性质进行分析解答即可．','填空题',3.00,'7f721ed714c249c3f62eedfd9d27c995',9,400,'常用盐的用途,碳单质的物理性质及用途,有机高分子材料的分类及鉴别','',2016,'32','2016•无锡模拟',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1860027,'在“蜡烛燃烧”的探究实验中，“对问题的回答”属于“实验方案设计”的是（　　）<br /><table class=\"edittable\"><TBODY><TR><td width=46>选项</TD><td width=303>问题</TD><td width=218>对问题的回答</TD></TR><TR><td>A</TD><td>吹灭蜡烛时产生的白烟是什么？</TD><td>可能是石蜡蒸气</TD></TR><TR><td>B</TD><td>用干燥的烧杯罩在火焰上方，观察到什么？</TD><td>烧杯内壁有水雾</TD></TR><TR><td>C</TD><td>蜡烛燃烧后的产物是什么？</TD><td>燃烧后生成二氧化碳</TD></TR><TR><td>D</TD><td>蜡烛火焰温度哪层最高？</TD><td>将火柴梗平放在火焰中1s后取出，观察其炭化程度</TD></TR></TBODY></TABLE>','A','B','C','D','','D','【解答】解：A、可能是石蜡蒸气，属于科学探究环节中的猜想与假设，故选项错误．<br />B、烧杯内壁有水雾，属于科学探究环节中的收集证据，故选项错误．<br />C、燃烧后生成二氧化碳，属于科学探究环节中的解释与结论，故选项错误．<br />D、将火柴梗平放在火焰中1s后取出，观察其炭化程度，属于科学探究环节中的实验方案设计，故选项正确．<br />故选：D．','【分析】科学探究的主要环节有提出问题→猜想与假设→制定计划（或设计方案）→进行实验→收集证据→解释与结论→反思与评价→拓展与迁移，据此结合题意进行分析判断．','选择题',4.00,'e3d2fdb44aa0ff53c21501b4ebc37c43',9,400,'科学探究的基本环节','',2015,'33','2015秋•贵阳校级期末',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1860532,'现有下列仪器：<br /><img  src=\"/tikuimages/9/2015/400/shoutiniao46/051de161-94d5-11e9-bcdf-b42e9921e93e_xkb18.png\" style=\"vertical-align:middle\" /><br />（1）吸取和滴加少量液体时<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，洗涤试管应使用<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>（填仪器名称）；<br />（2）加热前用试管夹夹持试管的具体操作是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；<br />（3）写出造成下列现象的原因：①某同学在量筒中溶解固体烧碱（溶于水放热），发现量筒炸裂：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>；②给试管内液体加热时液体溅出烫伤了人，请你分析造成此后果的原因：<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','胶头滴管$###$试管刷$###$试管夹从底部往上套，夹在离管口约三分之一处$###$量筒不能受热（或受热炸裂），而烧碱溶于水放出大量的热$###$液体量超过了试管容积的三分之一，并且加热时试管口对着人','【解答】解：（1）胶头滴管 用于吸取和滴加少量液体．清洗试管要用到试管刷．<br />故答案为：胶头滴管、试管刷；<br />（2）加热前用试管夹夹持试管的具体操作是：试管夹从底部往上套，夹在离管口约三分之一处；<br />故答案为：试管夹从底部往上套，夹在离管口约三分之一处<br />（3）①固体烧碱溶于水时要释放大量的热．量筒在突然受热的情况下会炸裂．②给试管内的液体加热时，若液体量超过了试管容积的三分之一，并且加热时试管口对着人，则液体沸腾时会溅出烫伤人；<br />故答案为：量筒不能受热（或受热炸裂），而烧碱溶于水放出大量的热；②液体量超过了试管容积的三分之一，并且加热时试管口对着人．','【分析】（1）胶头滴管用于少量液体的吸取和滴加；试管刷可用于洗涤试管；<br />（2）用试管夹夹试管时，应从下向上套，夹在据试管口的三分之一；<br />（3）①考虑烧碱与水反应放出热量，可能会炸裂量筒；给试管内的液体加热时，若液体量超过了试管容积的三分之一，并且加热时试管口对着人，则液体沸腾时会溅出烫伤人．','填空题',2.00,'0eade7448d155682c2895f5de30ea338',9,400,'挟持器-铁夹、试管夹、坩埚钳,常用仪器的名称和选用','',2015,'37','2015秋•莒南县月考',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1860664,'你认为下列选项不属于化学研究范畴的是（　　）','物质的组成','物质的性质','物质的运动','物质的用途','','C','【解答】解：A、物质的组成，属于化学研究范畴，故选项错误．<br />B、物质的性质，属于化学研究范畴，故选项错误．<br />C、物质的运动，属于物理学研究的范畴，故选项正确．<br />D、物质的用途，属于化学研究范畴，故选项错误．<br />故选：C．','【分析】根据化学的定义和研究内容进行分析判断，化学是一门在分子、原子的层次上研究物质的性质、组成、结构及其变化规律的科学，研究对象是物质，研究内容有组成、结构、性质、变化、用途等．','选择题',2.00,'ec8447449a7d9481d2829470d8ddea07',9,400,'化学的研究领域','',2015,'37','2015秋•安徽校级月考',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1860683,'<img  src=\"/tikuimages/9/0/400/shoutiniao63/06f364b0-94d5-11e9-9e35-b42e9921e93e_xkb91.png\" style=\"vertical-align:middle;FLOAT:right;\" />如图是电解水的实验装置图，试回答下列问题：<br />（1）Ⅰ中的气体是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，Ⅱ中的气体是<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，Ⅰ和Ⅱ中气体的体积比约为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．<br />（2）反应的化学反应方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，该反应属于<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>反应．<br />（3）若将两只试管中的气体充分混合后点燃，会发生<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>，反应的反应方程式为<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>．','','','','','','氧气$###$氢气$###$1：2$###$2H<sub>2</sub>O<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;通电&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<sub>2</sub>↑+O<sub>2</sub>↑$###$分解$###$爆炸$###$2H<sub>2</sub>+O<sub>2</sub><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<sub>2</sub>O','【解答】解：（1）Ⅰ中的气体是氧气，Ⅱ中的气体是氢气，Ⅰ和Ⅱ中气体的体积比约为1：2．<br />故填：氧气；氢气；1：2．<br />（2）电解水生成氢气和氧气，反应的化学反应方程式为：2H<sub>2</sub>O<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;通电&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<sub>2</sub>↑+O<sub>2</sub>↑，属于分解反应．<br />故填：2H<sub>2</sub>O<span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;通电&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<sub>2</sub>↑+O<sub>2</sub>↑；分解．<br />（3）若将两只试管中的气体充分混合后点燃，会发生爆炸，反应的反应方程式为：2H<sub>2</sub>+O<sub>2</sub><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<sub>2</sub>O．<br />故填：爆炸；2H<sub>2</sub>+O<sub>2</sub><span dealflag=\"1\" class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black;padding-bottom:1px;font-size:90%\"><table style=\"margin-right: 1px\" cellspacing=\"-1\" cellpadding=\"-1\"><tr><td>&nbsp;点燃&nbsp;</td></tr><tr><td style=\"font-size: 90%\"><div style=\"border-top:1px solid black;line-height:1px\">.</div></td></tr></table></td></tr><tr><td>&nbsp;</td></tr></table></span>2H<sub>2</sub>O．','【分析】电解水时，正极产生的是氧气，负极产生的是氢气，氧气和氢气的体积比约为1：2；<br />氢气在氧气中燃烧生成水．','书写',3.00,'48e6b39066b5fdf1678587dacce54978',9,400,'电解水实验,水的合成与氢气的燃烧,反应类型的判定,书写化学方程式、文字表达式、电离方程式','',0,'37','',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1860800,'下列说法不正确的是（　　）','实验剩余的药品不能放回原瓶，也不要随意丢弃','称量易潮解的药品时，需把药品放在纸上称量','硫粉在氧气中燃烧时，集气瓶中放少量水可吸收二氧化硫','在实验室，不得尝任何药品的味道','','B','【解答】解：A、对化学实验中的剩余药品，既不能放回原瓶，也不可随意丢弃，更不能带出实验室，应放入的指定的容器内，故选项说法正确．<br />B、称量易潮解的药品时，应放在玻璃器皿中称量，故选项说法错误．<br />C、二氧化硫能溶于水，硫粉在氧气中燃烧时，集气瓶中放少量水可吸收二氧化硫，故选项说法正确．<br />D、实验室里所用的药品，很多是易燃、易爆、有腐蚀性或有毒的．在使用药品时为保证安全，要做到“三不”，其中之一是不能尝任何药品的味道，故选项说法正确．<br />故选：B．','【分析】A、根据实验室剩余药品的处理原则（三不一要），进行分析判断．<br />B、根据托盘天平的使用注意事项，进行分析判断．<br />C、根据二氧化硫能溶于水，进行分析判断．<br />D、根据实验室药品取用的“三不”原则，进行分析判断．','选择题',3.00,'6450edfec5f7de984c2ef6318ef47c5b',9,400,'实验操作注意事项的探究,称量器-托盘天平,氧气的化学性质','',2015,'37','2015秋•开封月考',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1860846,'物理变化和化学变化的关系（　　）','物质发生物理变化时一定伴随化学变化','物质发生化学变化是一定伴随物理变化','物质发生化学变化时可能伴随物理变化','二者之间没有任何关系','','B','【解答】解：物质发生化学变化是一定伴随物理变化，但发生物理变化时不一定伴随化学变化，观察选项，故选B．','【分析】根据化学变化和物理变化的概念、关系以及伴随的现象等进行分析．','选择题',4.00,'2fd9098f94b48b42610088004dbddb2f',9,400,'化学变化的基本特征,物理变化的特点','东莞市',2014,'37','2014秋•东莞市校级月考',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1861253,'下列有关操作的叙述说法正确的是（　　）','实验室使用后的剩余药品直接倒入废液缸中，防止污染','实验室的易燃物品、有毒物品等可以与其它药品混合存放','量取7.5ml水时需要选择10ml的量筒和胶头滴管','检查装置气密性时先用手紧握试管，再将导管一端浸入水中观察现象','','C','【解答】解：A、对化学实验中的剩余药品，既不能放回原瓶，也不可随意丢弃，更不能带出实验室，应放入的指定的容器内，故选项说法错误．<br />B、实验室的易燃物品、有毒物品等要分类存放，不能与与其它药品混合存放，故选项说法错误．<br />C、要量取7.5ml水，应选择10mL规格的量筒；向量筒内加液时，先用倾倒法加液到近刻度线，再改用胶头滴管加液到刻度线，用到的玻璃仪器是10mL量筒和胶头滴管，故选项说法正确．<br />D、检查装置气密性时，先把导管的一端浸没在水里，双手紧贴容器外壁，若导管口有气泡冒出，装置不漏气，故选项说法错误．<br />故选：C．','【分析】A、根据实验室剩余药品的处理原则（三不一要），进行分析判断．<br />B、根据实验室的易燃物品、有毒物品的存放方法进行分析判断．<br />C、从减小实验误差的角度去选择量筒的量程，量筒量程选择的依据有两点：一是保证测量一次，二是量程要与液体的取用量最接近．<br />D、根据检查装置气密性的方法，进行分析判断．','选择题',3.00,'e3fdb9b599cd29f5f8e8232cecfc0bcb',9,400,'实验操作注意事项的探究,测量容器-量筒,检查装置的气密性','',0,'37','',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1861311,'在温度和体积相同的条件下，密闭容器中含有的气体分子数目越多压强就越大．在体积相同的两个密闭容器中分别充满O<SUB>2</SUB>、O<SUB>3</SUB>气体，当这两个容器内温度和气体密度相等时，下列说法正确的是（　　）','两种气体的压强相等','O<SUB>2</SUB>比O<SUB>3</SUB>的质量小','两种气体的分子数目相等','两种气体的氧原子数目相等','','D','【解答】解：A、氧原子个数相等时，气体的物质的量之比为：n（O<SUB>2</SUB>）：n（O<SUB>3</SUB>）=3：2，根据PV=nRT，可以得出其压强之比为3：2，错误；<br />B、相同体积相同密度时，量容器中的气体的质量相等，错误；<br />C、相同体积相同密度时，量容器中的气体的质量相等，都是由氧元素组成，故氧原子的个数相同，故分子个数不相等，错误；<br />D、相同体积相同密度时，量容器中的气体的质量相等，都是由氧元素组成，故氧原子的个数相同，正确；<br />故选D．','【分析】相同体积相同密度时，量容器中的气体的质量相等，都是由氧元素组成，故氧原子的质量、个数以及物质的量相同，结合PV=nRT判断即可．','选择题',3.00,'03ee33a7f1ef80a0170da478cc2ef3a0',9,400,'有关化学式的计算和推断','',2016,'37','2016春•苍南县校级月考',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1861597,'金属材料在我们的生活中随处可见，如做饭用的铁锅、自行车的铁支架还有切菜用的菜刀等．下列关于金属材料的说法正确的是（　　）','金属都是坚硬的固体','夏商时期的青铜器属于合金','金属的颜色都是银白色的','合金不属于金属材料','','B','【解答】解：A、金属不一定都是坚硬的固体，如汞在常温下是液体，故选项说法错误．<br />B、夏商时期的青铜器是铜和锡的合金，属于合金，故选项说法正确．<br />C、金属的颜色不一定都是银白色的，铜呈紫红色，故选项说法错误．<br />D、金属材料包括纯金属以及它们的合金，合金属于金属材料，故选项说法错误．<br />故选：B．','【分析】A、根据常见金属的状态、硬度，进行分析判断．<br />B、根据青铜器是铜和锡的合金，进行分析判断．<br />C、根据常见金属的颜色，进行分析判断．<br />D、根据金属材料包括纯金属以及它们的合金，进行分析判断．','选择题',3.00,'83c8523e67398d84d63317223bcf5a28',9,400,'金属的物理性质及用途,常见的金属和非金属的区分,金属材料及其应用','',2015,'33','2015秋•赵县期末',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1862549,'在学习化学的过程中，及时对所学知识进行整理，是一种好的学习方法．以下归纳中，其中存在错误的一组是（　　）','化学知识中有许多的“相等”：<br />原子核中质子数与中子数相等<br />稀释前后溶液中溶质的质量相等<br />化合物中元素化合价的正价总数与负价总数的数值相等','相似物质（或微粒）之间的区别：<img  src=\"/tikuimages/9/2015/400/shoutiniao73/1c9baca1-94d5-11e9-9e2f-b42e9921e93e_xkb24.png\" style=\"vertical-align:middle\" />','选择实验室制取气体装置与净化方法的依据：<img  src=\"/tikuimages/9/2015/400/shoutiniao94/1c9f0800-94d5-11e9-83c5-b42e9921e93e_xkb27.png\" style=\"vertical-align:middle\" />','环境问题----主要原因-----解决对策：<img  src=\"/tikuimages/9/2015/400/shoutiniao33/1ca0678f-94d5-11e9-8095-b42e9921e93e_xkb59.png\" style=\"vertical-align:middle\" />','','A','【解答】解：A、原子核中质子数与中子数不一定相等，如：氢原子中有一个质子没有中子．故选项错误；<br />B、氯原子和氯离子最外层电子数不同，二氧化硫和三氧化硫分子构成不同，生铁和钢含碳量不同，故选项正确；<br />C、制取气体选择发生装置时要根据反应物的状态和反应条件选择；收集气体的装置要根据气体的密度和溶解性选择；气体净化时要考虑气体及所含杂质的化学性质，故正确；<br />D、造成温室效应的气体是二氧化碳，解决这一问题的措施是植树造林，禁止滥砍滥伐；水质恶化的主要原因是污水的任意排放，应该先处理后排放；酸雨形成的原因是排放在空气中的二氧化硫和二氧化氮等气体造成的，应该减少化石燃料的使用；故正确；<br />故答案为A．','【分析】根据原子的构成，微粒和物质之间的联系可判断A错误，B正确；根据实验室制取气体装置与净化方法可判断C正确；根据环境污染的主要原因和防止的办法可判断D正确．','选择题',4.00,'51a55d24987e9c77bf737fb834b497d0',9,400,'实验室制取气体的思路,目前环境污染问题,用水稀释改变浓度的方法,生铁和钢,原子的定义与构成,原子和离子的相互转化,分子的定义与分子的特性,化合价规律和原则','',2015,'33','2015秋•沧州期末',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1862745,'水是我们日常生活中必不可少的物质之一．实验中常用<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>法来制取纯净程度较高的水；同时我们还可以通过点燃<u>&nbsp;&nbsp;&nbsp;&nbsp;</u>和氧气来获得水．','','','','','','蒸馏$###$氢气','【解答】解：实验室所用的净化程度较高的水，可以通过蒸馏自来水制取；氢气燃烧生成水，同时我们还可以通过点燃氢气和氧气来获得水．<br />答案：蒸馏；氢气．','【分析】根据实验室所用的净化程度较高的水，可以通过蒸馏自来水制取、以及氢气燃烧生成水进行解答．','填空题',3.00,'117c1148dd9f67e4e3861ac4b011728b',9,400,'水的合成与氢气的燃烧,水的净化','沙河市',2015,'33','2015秋•沙河市期末',0,0,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1862885,'下列关于元素化合价的有关说法中，正确的是（　　）','铁单质中铁元素的化合价为+2价和+3价','一种元素只能表现出一种化合价','在任何化合物中元素正负化合价的代数和为零','一种物质中同一种元素不可能表现出不同的化合价','','C','【解答】解：A、单质铁中铁元素的化合价为0，不是+2或+3，错误；<br />B、一种元素可以表现多种化合价，例如在化合物中，铁元素的化合价为+2价或+3价，故错误<br />C、在任何化合物中，各元素正负化合价的代数和为零，正确；<br />D、在化合物中，同一种元素可能表现出不同的化合价，比如硝酸铵中氮元素为-3、+5价，错误；<br />故选C．','【分析】A、单质中元素的化合价为0．<br />B、一种元素可以表现多种化合价．<br />C、根据在化合物中正负化合价代数和为零，进行分析判断．<br />D、根据常见元素的化合价、化合价的规律进行分析判断．','选择题',2.00,'7f0c42ae997d1857456bd3664efdee01',9,400,'常见元素与常见原子团的化合价,化合价规律和原则','',0,'37','',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1863109,'“元素观”是化学学科的重要观念，自然界中的氧循环和碳循环正是这一观念的体现，各循环又是由许许多多物质间的转化组成的．下列自然过程中的物质转化只涉及一种元素的是（　　）','植物的光合作用','动物的呼吸作用','化石燃料的燃烧','石墨在地壳深处特殊条件下转化成金刚石','','D','【解答】解：A．植物的光合作用是由二氧化碳和水转化为葡萄糖和氧气，此过程中涉及到碳、氢、氧三种元素，故错误；<br />B．呼吸作用是将体内的有机物与氧气反应生成水和二氧化碳，此过程中涉及到碳、氢、氧三种元素，故错误；<br />C．化石燃料与氧气在点燃的条件下转化为水、二氧化碳等其他物质，此过程中涉及到碳、氢、氧等元素，故错误；<br />D．石墨与金刚石都是由碳元素组成的单质，由石墨转化为金刚石只涉及到碳元素，故正确．<br />故选D．','【分析】根据四个转化过程中物质的转化、以及元素的组成来分析解答．','选择题',4.00,'711674f29c7a7fe01cbe33ecdd0a2197',9,400,'物质的元素组成,元素在化学变化过程中的特点','',2015,'33','2015秋•黄山期末',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1872987,'在甘油（分子式为C<SUB>3</SUB>H<SUB>8</SUB>O<SUB>3</SUB>）和甲苯（分子式为C<SUB>7</SUB>H<SUB>8</SUB>）混合物中，测得氧的质量分数为19%，则混合物中碳元素的质量分数是（　　）','72.3%','68.4%','57.1%','无法计算','','A','【解答】解：甲苯的分子式为C<SUB>7</SUB>H<SUB>8</SUB>，相对分子质量为92，甘油的分子式为C<SUB>3</SUB>H<SUB>8</SUB>O<SUB>3</SUB>，相对分子质量为92，<br />由于两种有机物中，每个分子中氢原子数目相等，则氢元素的质量分数是定值，氢元素质量分数为：<span class=\"MathJye\" mathtag=\"math\" style=\"whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal\"><table cellpadding=\"-1\" cellspacing=\"-1\" style=\"margin-right:1px\"><tr><td style=\"border-bottom:1px solid black\">8</td></tr><tr><td>92</td></tr></table></span>×100%=8.7%，<br />又因为氧元素质量分数是19%，则碳元素的质量分数是：1-8.7%-19%=72.3%，<br />故选：A．','【分析】根据甘油和甲苯的化学式及其相对原子质量和氧元素质量分数可以计算碳元素的质量分数．','选择题',3.00,'e24496f461a129557d7cc80f05cc420f',9,400,'有关化学式的计算和推断,石油加工的产物','平湖市',2013,'37','2013•平湖市校级自主招生',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1873165,'科学的假设与猜想是科学探究的先导和价值所在．在下列假设（猜想）引导下的探究肯定没有意义的选项是（　　）','探究二氧化硫和水反应可能有硫酸生成','探究钠与水的反应产生的气体可能是氧气','探究Mg在CO<SUB>2</SUB>气体中燃烧生成的白色固体可能Mg（OH）<SUB>2</SUB>','探究Cu（OH）<SUB>2</SUB>粉末加热后生成的黑色物质可能是CuO','','C','【解答】解：A．二氧化硫为酸性氧化物，可以与水反应生成酸，故可探究该反应可能生成硫酸，故正确；<br />B．根据质量守恒定律，化学反应前后元素的种类不变，钠和水中含有钠元素、氢元素和氧元素，可猜测产生的气体可能为氧气，故正确；<br />C．根据质量守恒定律，化学反应前后元素的种类不变，Mg在CO<SUB>2</SUB>气体中燃烧生成的白色固体不可能Mg（OH）<SUB>2</SUB>，因为反应物中没有氢元素，生成物中就不能出现氢元素，故错误；<br />D．不溶性碱受热能分解，氢氧化铜受热能分解生成黑色的氧化铜，符合质量守恒定律和反应规律，故正确．<br />故选：C．','【分析】进行科学探究时，要有合理的理论依据，不能凭空猜测．二氧化硫为酸性氧化物，可以与水反应生成酸；化学反应前后元素的种类不变，钠和水中含有钠元素、氢元素和氧元素，故生成物中含钠元素、氢元素和氧元素；镁和二氧化碳中含有镁、碳、氧三种元素，生成物不会存在氢元素；根据氢氧化铜的元素组成，分解后可以生成氧化铜．','选择题',3.00,'6c611ed39494d4e3946987d7e3dcd11e',9,400,'猜想与事实验证','',2015,'33','2015秋•咸丰县校级期末',0,1,1);
insert  into `questions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`option_e`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`md5`,`subjectId`,`gradeId`,`knowledges`,`area`,`year`,`paperTpye`,`source`,`isSub`,`isNormal`,`isKonw`) values (1874082,'化学实验成功的关键是（　　）<br />①严谨的科学态度；②合理的实验步骤；③正确的操作方法．','①','①②','①③','①②③','','D','【解答】解：做化学实验要具有严谨的科学实验态度，实验方案要科学、合理，实验步骤和操作方法要可行，不能仅靠一次实验结果就下结论，防止偶然的发生，而造成错误的结论，因此上述①、②、③都是成功的关键．<br />故选D．','【分析】化学实验要想成功，必须具有严谨的科学实验态度，不能凭着侥幸心理、偶然的巧合就得出结论．','选择题',2.00,'d8d5e53fa76c6e5a6b3e357cef0eb9e9',9,400,'学习化学的重要途径及学习方法','',2012,'37','2012秋•安康月考',0,1,1);

/*Table structure for table `subject` */

DROP TABLE IF EXISTS `subject`;

CREATE TABLE `subject` (
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  `subjectName` varchar(255) DEFAULT NULL,
  `pinyin` varchar(255) DEFAULT NULL,
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) ENGINE=MyISAM DEFAULT CHARSET=utf8;

/*Data for the table `subject` */

insert  into `subject`(`subjectId`,`subjectName`,`pinyin`) values (1,'语文',NULL);
insert  into `subject`(`subjectId`,`subjectName`,`pinyin`) values (2,'数学',NULL);
insert  into `subject`(`subjectId`,`subjectName`,`pinyin`) values (3,'英语',NULL);
insert  into `subject`(`subjectId`,`subjectName`,`pinyin`) values (4,'历史',NULL);
insert  into `subject`(`subjectId`,`subjectName`,`pinyin`) values (5,'地理',NULL);
insert  into `subject`(`subjectId`,`subjectName`,`pinyin`) values (6,'政治',NULL);
insert  into `subject`(`subjectId`,`subjectName`,`pinyin`) values (7,'生物',NULL);
insert  into `subject`(`subjectId`,`subjectName`,`pinyin`) values (8,'物理',NULL);
insert  into `subject`(`subjectId`,`subjectName`,`pinyin`) values (9,'化学',NULL);
insert  into `subject`(`subjectId`,`subjectName`,`pinyin`) values (10,'科学',NULL);
insert  into `subject`(`subjectId`,`subjectName`,`pinyin`) values (11,'道德与法治',NULL);

/*Table structure for table `subquestions` */

DROP TABLE IF EXISTS `subquestions`;

CREATE TABLE `subquestions` (
  `id` int(10) NOT NULL AUTO_INCREMENT,
  `title` text COMMENT '试题-题干',
  `option_a` text COMMENT '选项A',
  `option_b` text COMMENT '选项B',
  `option_c` text COMMENT '选项C',
  `option_d` text COMMENT '选项D',
  `pid` int(11) DEFAULT NULL COMMENT '父ID',
  `answer1` text,
  `answer2` text COMMENT '非标准格式答案或含部分过程说明的答案',
  `parse` text COMMENT '试题解析',
  `qtpye` varchar(80) DEFAULT NULL COMMENT '试题题型',
  `diff` float(3,2) DEFAULT NULL,
  `subjectId` tinyint(2) DEFAULT NULL COMMENT '学科Id',
  `gradeId` int(5) DEFAULT NULL COMMENT '年级ID',
  `knowledges` varchar(200) DEFAULT NULL COMMENT '知识点名称',
  `source` varchar(200) DEFAULT NULL COMMENT '试题来源(试卷)',
  PRIMARY KEY (`id`),
  KEY `index_subject_fromsite` (`subjectId`),
  KEY `index_qtypes` (`qtpye`),
  KEY `index_knowedges` (`knowledges`),
  KEY `index_pid` (`pid`)
) ENGINE=MyISAM AUTO_INCREMENT=559295 DEFAULT CHARSET=utf8;

/*Data for the table `subquestions` */

insert  into `subquestions`(`id`,`title`,`option_a`,`option_b`,`option_c`,`option_d`,`pid`,`answer1`,`answer2`,`parse`,`qtpye`,`diff`,`subjectId`,`gradeId`,`knowledges`,`source`) values (407778,'下列混合气体点燃后可能发生爆炸的是（　　）','氢气、氧气','一氧化碳、氢气','一氧化碳、氮气','氢气、氩气',1842859,'A','【解答】解：根据爆炸是由于急速的燃烧在有限的空间而引起的，因此要满足燃烧的条件，气体必须是可燃性：<br />A、氢气具有可燃性，因此氢气与氧气混合后遇明火，可能发生爆炸；故A符合题意；<br />B、一氧化碳和氢气都具有可燃性，但没有助燃物不能燃烧；故B不符合题意；<br />C、一氧化碳具有可燃性，氮气不具有助燃性，没有助燃物不能燃烧；故C不符合题意．<br />D、氢气具有可燃性，氩气不具有助燃性，不可能发生爆炸，故D不符合题意．<br />故选A．','【分析】点燃可燃性气体与空气（或与氧气）的混合物都容易发生爆炸，一旦达到爆炸极限就能发生爆炸<br/>【点评】通过回答本题知道了点燃可燃性气体与空气（或与氧气）的混合物都容易发生爆炸，所以在点燃它们前要进行验纯．','选择题',3.00,9,0,'【考点】燃烧、爆炸、缓慢氧化与自燃．','');

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