求高手帮忙初三化学,我在等……多谢啦各位,帮帮忙吧

2024-11-14 04:53:15
推荐回答(5个)
回答1:

500g*20%=100g

方案 25%的(g) 15%的(g) 固体食盐(g) 水(g)
1 200 100 35 165
2 100 200 45 155
3 100 400
4 100 75 325
5 200 70 230

回答2:

这个可以采用固体溶质+溶剂,浓溶液+水,稀溶液+溶质,浓溶液+稀溶液等几种方式,本质是溶液的稀释,抓住浓度计算公式:质量分数=溶质质量/溶液质量
1.固体溶质+溶剂,
溶质质量=500g*20%=100g 水的质量=500g-100g=400g 即100g硝酸钾,400g水
2.浓溶液+水+溶质,(情况很多,举一个例子)
250g质量分数25%的溶液中溶质质量=250*25%=62.5g,溶质还需100g-62.5g=37.5g,加水=500-250g-37.5g=212.5g
即250g质量分数25%的溶液,37.5g溶质,212.5g溶剂
3.稀溶液+溶质/溶剂,(情况很多,举一个例子)
200g溶质质量分数15%的溶液中溶质质量=200*15%=30g 溶质还需100g-30g=70g 溶剂质量=500g-200g-70g=230g
即200g15%的硝酸钾,70g硝酸钾固体,230g水
4.浓溶液+稀溶液
这个比较复杂,抓住溶质不变即可!
250g质量分数25%的溶液中溶质质量=250*25%=62.5g,还学要溶质 37.5g 需15%的溶液质量 37.5g/15%=250g,本题不能满足
希望对你有所帮助,满意请采纳,亲!

回答3:

500g*20%=100g
1,称量100g的食盐,再加水400g,
2,往250g质量分数为25%的溶液,加入37.5g的食盐,再加水212.5
3,两种溶液混合,再加7.5g食盐,再加42.5g水

回答4:

第一种:在250g溶质质量分数为25%的食盐溶液中加入212.5g水和37.5g食盐
第二种:在200g溶质质量分数为15%的食盐溶液中加入230g水和70g食盐
第三种:足量的固体食盐和水 把100g食盐溶于400g水中

回答5:

1,称量100g的食盐,再加水400g,
2,往250g质量分数为25%的溶液,加入37.5g的食盐,再加水212.5
3,两种溶液混合,再加7.5g食盐,再加42.5g水

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