某学生用如图所示装置制取氯气,并收集氯气进行实验.清回答下列问题:(1)装置A发生反应的离子方程式为

2025-04-06 14:18:50
推荐回答(1个)
回答1:

(1)实验室利用二氧化锰和浓盐酸加热反应制备氯气生成氯化锰、氯气和水,依据反应写出离子方程式为:MnO2+4H++2C1-

  △  
 
Mn2++C12↑+2H2O;
故答案为:MnO2+4H++2C1-
  △  
 
Mn2++C12↑+2H2O;
(2)氯气比空气重,收集气体用向上排气法,导气管长进短出,连接 A 的导管口是b;
故答案为:b;
(3)C装置用水吸收氯气发生反应,Cl2+H2O=HCl+HClO,盐酸使石蕊试液变红色,次氯酸具有漂白性使石蕊试液褪色;
故答案为:C中生成的氯化氢溶液使石蕊变红色,生成的次氯酸具有漂白性使红色褪去;
(4)进行氯气与铜的燃烧实验:首先进行的操作是用坩埚钳夹住一束铜丝灼烧,然后立刻放入充满氯气的集气瓶中,铜丝燃烧生成棕黄色烟,不是温度较高的固体熔融物,不会炸裂瓶底,集气瓶底不需要放少量的水或细沙;
故答案为:用坩埚钳夹住一束铜丝灼烧;不需要;实验过程中生成的是棕黄色烟,不是温度较高的固体熔融物,不会炸裂瓶底;
(5)一氯甲苯、甲苯都是有机物,互溶,氮沸点不同,可以利用控制沸点温度进行分离,分馏的装置选择玻璃仪器有蒸馏烧瓶、温度计、酒精灯、冷凝管、尾接管、锥形瓶;
故答案为:BDFHIK;

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