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问题 计算题

向50mL1mol/L的Al2（SO4）3溶液中加入100mLKOH溶液，充分反应得到3.9g沉淀，求KOH溶液的物质的量浓度。（原子量Al27 O16 S32 K39 H1 ）

Al2(SO4)3+ 6KOH="2" Al(OH)3↓+3K2 SO4

答案

3.5mol/L

考查根据方程式进行的有关计算。

硫酸铝是0.05L×1mol/L＝0.05mol，含有铝离子是0.1mol。

由于生成的氢氧化铝沉淀是3.9g÷78g/mol＝0.05mol

这说明铝离子没有完全转化为氢氧化铝沉淀

如果加入的氢氧化钾不足，铝离子有剩余，则

Al2(SO4)3+ 6KOH="2" Al(OH)3↓+3K2 SO4

6mol    2mol

0.15mol  0.05mol

所以氢氧化钾的浓度是0.15mol÷0.1L＝1.5mol/L

如果氢氧化钾过量，过量的氢氧化钾将部分氢氧化铝溶解，则

Al2(SO4)3+ 6KOH="2" Al(OH)3↓+3K2 SO4

1mol    6mol    2mol

0.05mol  0.3mol  0.1mol

Al(OH)3＋NaOH=NaAlO2＋2H2O

1mol     1mol

0.05mol   0.05mol

即实际消耗氢氧化钾是0.3mol＋0.05mol＝0.35mol

所以浓度所以氢氧化钾的浓度是0.35mol÷0.1L＝3.5mol/L

单项选择题

Observe that for the programmer，as for the chef,the urgency of the patron(顾客) may govern the scheduled completion of the task，but it cannot govern the actual completion．An omelette(煎鸡蛋)，promised in two minutes，may appear to be progressing nicely．But when it has not set in two minutes，the customer has two choices—waits or eats it raw．Software customers have had (71) choices．
Now I do not think software (72) have less inherent courage and firmness than chefs，nor than other engineering managers．But false (73) to match the patron’s desired date is much more common in our discipline than elsewhere in engineering．It is very (74) to make a vigorous，plausible，and job risking defense of an estimate that is derived by no quantitative method，supported by little data，and certified chiefly by the hunches of the managers．
Clearly two solutions are needed．We need to develop and publicize productivity figures，bug-incidence figures，estimating rules，and so on．The whole profession can only profit from (75) such data．Until estimating is on a sounder basis，individual managers will need to stiffen their backbones and defend their estimates with the assurance that their poor hunches are better than wish derived estimates．

A．tasks

B．jobs

C．works

D．scheduling
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判断题

在文学创作方面，与西方浪漫主义相比，东方浪漫主义持续时间都比较短。
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