将0.1molNa2O2投入到92.2g水（水过量）中，求：（1）所得溶液中Na_爱下问搜题

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问题问答题

将0.1mol Na₂O₂投入到92.2g水（水过量）中，求：（1）所得溶液中NaOH的物质的量是多少？（2）标准状况下得到O₂的体积为多少？（3）所得溶液中溶质的质量分数是多少？

答案

因过氧化钠完全反应，设生成NaOH的物质的量为x，O₂的物质的量为y，

2Na₂O₂+2H₂O═4NaOH+O₂↑

2 4 1

0.1mol x y

2

0.1mol

=

4

x

，

2

0.1mol

=

1

y

，

解得x=0.2mol、y=0.05mol，

（1）即NaOH的物质的量0.2mol，答：所得溶液中NaOH的物质的量是0.2mol；

（2）标准状况下得到O₂的体积为0.05mol×22.4L/mol=1.12L，答：标准状况下得到O₂的体积为1.12L；

（3）溶质的质量为0.2mol×40g/mol=8g，

溶液的质量为0.1mol×78g/mol+92.2g-0.05mol×32g/mol=98.4g，

则所得溶液中溶质的质量分数是

8g

98.4g

×100%=8.13%，

答：所得溶液中溶质的质量分数是8.13%．

完形填空

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He telephoned Johnny’s uncle, Peter, and asked him if he would let Johnny        with him.
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So Johnny, who was five,        to live with his uncle.
Two days later, Johnny’s parents       a telegram(电报) from Peter. It said, “I am returning the child. Please send an earthquake.”

D．interesting

单项选择题

目前，外商投资企业和外国企业暂不缴纳（）和教育费附加。

A．消费税

B．资源税

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D．土地增值税