SOLUTION: How much pure water must be mixed with 10 liters of a 25% acid solution to reduce it to a 10% acid solution?

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Question 646081: How much pure water must be mixed with 10 liters of a 25% acid solution to reduce it to a 10% acid solution?
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
How much pure water must be mixed with 10 liters of a 25% acid solution to reduce it to a 10% acid solution?
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Equation:
water + water = water
0.75*10 + x = 0.90(10+x)
75*10 + 100x = 90*10 + 90x
10x = 15*10
x = 15 liters (amt. of water needed)
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Cheers,
Stan H.
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