SOLUTION: a chemist must mix 8 L of a 40% acid solution with some 70% solution to get a 50% solution. how much of the 70% solution should be used?

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Question 399310: a chemist must mix 8 L of a 40% acid solution with some 70% solution to get a 50% solution. how much of the 70% solution should be used?
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
(liters of acid in final solution)/(liters of final solution) = 50%
Acid in 40% solution = .4%2A8+=+3.2
Let x = liters of 70% solution to use
.7x = liters of acid in 70% solution
--------------------------------------------
%28.7x+%2B+3.2%29%2F%288+%2B+x%29+=+.5
.7x+%2B+3.2+=+.5%2A%288+%2B+x%29
.7x+%2B+3.2+=+4+%2B+.5x
.2x+=+.8
x+=+4
4 liters of 70% solution are needed
check answer:
%28.7x+%2B+3.2%29%2F%288+%2B+x%29+=+.5
%28.7%2A4+%2B+3.2%29%2F%288+%2B+4%29+=+.5
%282.8+%2B+3.2%29+%2F+12+=+.5
6%2F12+=+.5
.5+=+.5
OK