在酸性溶液中,碘酸钾(KIO3)和亚硫酸钠可发生如下反应: 2I+5S+2H+I2+5S+H2O

在酸性溶液中,碘酸钾(KIO3)和亚硫酸钠可发生如下反应:

2I+5S+2H+I2+5S+H2O

生成的I2可以用淀粉溶液检验,根据反应溶液出现蓝色所需的时间来衡量该反应的速率。某同学设计实验如下表所示:

0.01mol/L

KIO3酸性溶液(含淀粉)的体积/mL

0.01mol/L

Na2SO3溶液的体积/mL

H2O的体积/mL

实验温度/

溶液出现蓝色时所需时间/s

实验1

5

V1

35

25

实验2

5

5

40

25

实验3

5

5

V2

0

该实验的目的是            ;表中V2=    

答案

解析:由实验中改变的反应条件有温度、水的体积(引起反应物浓度的变化),本实验探究的是反应速率与温度、浓度之间的关系,由实验2、实验3中的反应条件知,二者温度不同,故其他条件应该相同,V2应该是40

答案:探究该反应的反应速率与温度、浓度的关系(或其他合理答案) 40

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