Asked by Anonymous

a lab assistant wants to make 5 liters of 27.8% acid solutions. if solutions of 50% and 13% are in stock, how many liters of each must be mixed to prepare the solution?
Answered by Steve
just work with the amount of acid in each part. If there's x L of 50% acid,

.50x + .13(5-x) = .278(5)
x=2
Answered by jack
yup
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