问题 填空题

某环保小组监测到一工厂向河水中排放的酸性废液中含有CuSO4

(1)CuSO4是重金属盐,能与蛋白质发生_________变化(选填“物理”或“化学”),使蛋白质失去原来的生理功能,所以CuSO4有毒。

(2)为了测定废液中CuSO4的质量分数,该小组取了100g废液,逐滴加入NaOH溶液至过量.测得生成Cu(OH)2沉淀(无其它沉淀)的质量与所加NaOH溶液的质量关系如上图所示:

①说明图中的OA段未生成Cu(OH)2沉淀的原因。

②计算废液中CuSO4的质量分数(写出计算过程,结果保留1位小数)。

③如果仍取l00g废液,向其中逐滴加入过量的BaCl2溶液,充分反应后,过滤,将所得沉淀洗涤、干燥称得质量为m g,能否用含m的代数式表达废液中CuSO4的质量分数?_____ (填“能’,或“不能”)并说明你的理由。                                                                

                                                                                    

答案

(1)化学(1分)

(2)(1)NaOH先与废液中的酸性物质反应(1分)

②解:设废液中CuSO4的质量为x

(3)不能 (1分)

若废液中含有H2SO4(SO42)会与BaCl2溶液反应生成沉淀而影响测定结果

考查知识点:根据化学反应方程式的计算;有关溶质质量分数的简单计算;化学变化和物理变化的判别.

思路分析:(1)重金属离子可以改变蛋白质的结构,使蛋白质失去原来的生理功能.

(2)①根据题中信息可知废液呈酸性,酸首先要和氢氧化钠反应,当酸反应完后,硫酸铜才和氢氧化钠反应.

②可根据图中所给氢氧化铜的质量,依据化学方程式求出硫酸铜的质量,进而根据废水的质量求出硫酸铜的质量分数.

废液中主要成分有硫酸和硫酸铜,考虑沉淀的来源.

具体解答过程:解:(1)重金属离子可以改变蛋白质的结构,使蛋白质失去原来的生理功能,所以发生的是化学变化.

故答案为:化学

(2)①因为废液还有酸,氢氧化钠溶液先与废液中的酸性物质发生反应,没有沉淀产生.

故答案为:氢氧化钠溶液先与废液中的酸性物质发生反应,没有沉淀产生.

②解:根据图可以看出生成氢氧化铜的质量为3.92g

设废液中CuSO4的质量为x

CuSO4+2NaOH=Na2SO4+Cu(OH)2

160                  98

X                   3.92g

160/x=98/3.92g

解之得:x=6.4g

废液中CuSO4的质量分数=硫酸铜的质量/废液的质量×100%=×100%=6.4%

答:废液中CuSO4的质量分数6.4%

③因为废液中含有硫酸,硫酸也可以和氯化钡生成硫酸钡沉淀,不能用沉淀的质量来求出硫酸铜的质量.

故答案为:不能;废液中的硫酸也能与氯化钡溶液反应生成硫酸钡

试题点评:本题是一道综合性计算题,首先要从题中整理信息(废液的成分),其次要分析图表,获取信息,最后根据所学知识根据化学方程式进行计算.

单项选择题
完形填空
完型填空。
     We arrived in Spain for the first time a few weeks ago. I decided to   1   a car because we had sold the
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