SOLUTION: A chemist has one solution that 34% acid and a second solution that is 58 % acid. How many liters(L) of the 58% acid should be mixed with the 34% to get 80L of a solution that is 4
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Question 361367: A chemist has one solution that 34% acid and a second solution that is 58 % acid. How many liters(L) of the 58% acid should be mixed with the 34% to get 80L of a solution that is 40% acid? Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! A chemist has one solution that 34% acid and a second solution that is 58 % acid. How many liters(L) of the 58% acid should be mixed with the 34% to get 80L of a solution that is 40% acid?
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Equation:
acid + acid = acid
0.34x + 0.58(80-x) = 0.40*80
Multiply thru by 100 to get:
34x + 58*80 - 58x = 40*80
-24x = -18*40
x = (3/4)40
x = 30 L (amt. of 34% solution needed in the mix)
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80-x = 50 L (amt of 58% solution needed in the mix)
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Cheers,
Stan H.
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