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A chemist has one solution that is 80 percent acid and another solution that is 30 percent acid how much of the second 30 percent solution is needed to make a 400L that is 62 percent acid?
Wiki User
∙ 14y ago
.80x+.30y=400*.62
.80x+.30y=248
This is equation one.
x+y=400
x=400-y
This is the second equation.
.80(400-y)+.30y=248
320-.80y+.30y=248
-.50y=-72
Y=144
X=256
CHECK
.80(256)+.30(144)=400(.62)
204.8+43.2=248
248=248
Wiki User
∙ 14y ago
To find out how much of the 30% acid solution is needed, we can set up a weighted average equation. Let x be the volume of the 30% solution needed. Since the final solution is 62% acid, we can write the equation as: 0.80*(400-x) + 0.30x = 0.62*400. Solving this equation, we find x to be 280 liters. Therefore, 280 liters of the 30% acid solution is needed to make a 400L solution that is 62% acid.
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