SOLUTION: How much of a 50% acid solution must be mixed with 8 gallons of a 22% acid solution to obtain a solution that is 40% acid?
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Question 255487: How much of a 50% acid solution must be mixed with 8 gallons of a 22% acid solution to obtain a solution that is 40% acid?
Answer by checkley77(12844) (Show Source): You can put this solution on YOUR website!
.50x+.22*8=.40(8+x)
.50x+1.76=3.2+.40x
.50x-.40x=3.2-1.76
.1x=1.44
x=1.44/.1
x=14.4 gals. of 50% acid is used.
Proof:
.50*14.4+.22*8=.40(8+14.4)
7.2+1.76=.40*22.4
8.96=8.96
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