SOLUTION: A science teacher has a supply of 5% hydrochloric acid and a supply of 65% hydrochloric acid (HCl). How much of each solution should the teacher mix together to get 42 mL of 45% HC

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Question 933473: A science teacher has a supply of 5% hydrochloric acid and a supply of 65% hydrochloric acid (HCl). How much of each solution should the teacher mix together to get 42 mL of 45% HCl for an experiment?
I tried to solve this but I got crazy decimals. I do not understand. I'm thinking your're going to have to turn the percents into decimals but I'm not sure.
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39618)   (Show Source): You can put this solution on YOUR website!
The description maybe is forcing you to expect percents as mass per volume; otherwise, you may need to know the densities also. (65% HCl seems very high.)

You can handle the arithmetic in decimal numbers if you want. You do not really need to turn the percents into decimals. Here, I WILL.

x for volume of 5%, and y for volume of the "65%".
to account for mililiters of volume, and is the quantity of PURE HCl. The concentration as a DECIMAL fraction is . That accounts for the concentration as a decimal fraction.

You know what to do from that?

A strong suggestion is to solve the system for x and y, but do not do any computations for simplification until LAST.
Answer by MathTherapy(10552)   (Show Source): You can put this solution on YOUR website!

I tried to solve this but I got crazy decimals. I do not understand. I'm thinking your're going to have to turn the percents into decimals but I'm not sure.
What type of crazy decimals did you get? Why didn't you share them?



Let amount of 5% acid to be mixed, be F

Then amount of 65% acid to be mixed = 42 – F

Therefore, we get: .05F + .65(42 – F) = .45(42)

.05F + 27.3 - .65F = 18.9

.05F - .65F = 18.9 – 27.3 

- .6F = - 8.4

F, or amount of 5% acid to be mixed = , or  mL

Amount of 65% acid to be mixed: 42 – 14, or  mL

You can do the check!! 

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