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Mixture Problems
Asked Oct 31, 2006, 01:54 PM
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1 Answer How many ounces of pure water must be added to 80oz of a 20% acid solution to make a solution that is 12% acid? (hint: Pure water is 0% Acid.) |
1 Answer
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| Let x = # of ounces of pure water You have 80 oz. Of 20% acid (before adding the water) You have x + 80 oz. Of 12% acid after adding the water The amount of acid is the same before the water is added and after the water is added So: .2(80) = .12(x + 80) Solve the above equation for x | ||||
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