作业帮there are some kilos of salt water which is 4% concentrated ,it becomes 10% concentratedafter evaporating ,then it becomes 6.4% concentrated after adding 300 grams of salty water which is 4% concentrated .how many kilos of slaty water are there origi
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there are some kilos of salt water which is 4% concentrated ,it becomes 10% concentratedafter evaporating ,then it becomes 6.4% concentrated after adding 300 grams of salty water which is 4% concentrated .how many kilos of slaty water are there origi
there are some kilos of salt water which is 4% concentrated ,it becomes 10% concentrated
after evaporating ,then it becomes 6.4% concentrated after adding 300 grams of salty water which is 4% concentrated .how many kilos of slaty water are there originally?
there are some kilos of salt water which is 4% concentrated ,it becomes 10% concentratedafter evaporating ,then it becomes 6.4% concentrated after adding 300 grams of salty water which is 4% concentrated .how many kilos of slaty water are there origi
假设有未知公斤的盐水,浓度为4%.蒸发后,浓度为10%.再加入浓度为4%的盐水300g后,盐水的浓度为6.4%.请问,最初的盐水是多少公斤?
设最初的盐水为X公斤
①4%·X 是最初盐水中盐的质量
② [（4%·X ）+ 300·4%] / 6.4% 加了300g 4%浓度的盐水之后,盐水的质量
③ [（4%·X ）+ 300·4%] / 6.4% - 300 蒸发后盐水的质量
④ { [（4%·X ）+ 300·4%] / 6.4% - 300}·10% 是蒸发后盐水中所含盐的质量
因为从最开始到蒸发后,盐水中盐的质量一直没有变化,所以①=④
4%·X = { [（4%·X ）+ 300·4%] / 6.4% - 300}·10%
备注：·是乘号,/是除号
这是我帮你推导出的算式啦 我喜欢化学,但是好久好久没做过数学题了,见谅啦!
一开始有一些质量浓度为4%的盐水；蒸发后，浓度成为10%；然后加入300克4%的盐水，其浓度成了6.4%。问一开始的盐水有多少千克？
一开始： 盐水质量X；其中盐质量 0.04X； 挥发后浓度10%，也就是说，盐水质量成了0.4X。
加入的盐水质量 0.3 kg ；其中盐质量 0.04*0.3=0.012kg。
此时的质量浓度为：全部的盐质量/全部的盐水质量，即（0.0...
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一开始有一些质量浓度为4%的盐水；蒸发后，浓度成为10%；然后加入300克4%的盐水，其浓度成了6.4%。问一开始的盐水有多少千克？
一开始： 盐水质量X；其中盐质量 0.04X； 挥发后浓度10%，也就是说，盐水质量成了0.4X。
加入的盐水质量 0.3 kg ；其中盐质量 0.04*0.3=0.012kg。
此时的质量浓度为：全部的盐质量/全部的盐水质量，即（0.04X+0.012）/（0.4X+0.3）=0.064；
X=0.5 千克
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