SOLUTION: A chemistry student wants to mix an 13% acid solution with a 51% acid solution to g 13L of a 29% acid solution. How many liters of the 13% solution and how many liters of the 51% s
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Question 224365: A chemistry student wants to mix an 13% acid solution with a 51% acid solution to g 13L of a 29% acid solution. How many liters of the 13% solution and how many liters of the 51% solution should be mixed
The mixture contains 7.52632 L of 13% acid solution and 5.47363 L of 51%acid solution
19 __ Answer by checkley77(12844) (Show Source):
You can put this solution on YOUR website! .51X+.13(13-X)=.29*13
.51X+1.69-.13X=3.77
.38X=3.77-1.69
.38X=2.08
X=2.08/.38
X=5.47368 LITERS OF 51% ACID IS USED.
13-5.47368=7.52632 LITERS OF 13% ACID IS USED.
PROOF:
.51*5.47368+.13*7.52632=3.77
2.79158+.97842=3.77
3.77=3.77
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