# what equation can you use to solve for x based on your answer to part a?

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a chemist mixed x milliliters of 25% acid solution with some 15% acid solution to produce 100 milliliters of a 19% acid solution.

Nov 15th, 2014

We need to set up a table to solve this mixture problem.

mL of acid            % acid              total mL acid

25% sol'n               x                     0.25                    0.25x

15% sol'n               y                     0.15                    0.15y

Mixture            x+y = 100               0.19               0.19(100) = 19

Since we know that x + y = 100, we can solve for x = 100 - y.  Now substitute this into the table:

mL of acid               % acid             total mL acid

25% sol'n            100-y                    0.25                   0.25(100-y)

15% sol'n                y                       0.15                         0.15y

Mixture                 100                      0.19                   0.19(100) = 19

Now we can take the last column and solve:

0.25(100-y) + 0.15y = 19

25 - 0.25y + 0.15y = 19

-0.25y + 0.15y = -6

-0.10y = -6

y = 60 mL

Since we know that x + y = 100:

x+60 = 100

x = 40 mL

Therefore, you need 40 mL of 25% acid solution and 60 mL of 15% solution.

Nov 15th, 2014

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Nov 15th, 2014
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Nov 15th, 2014
Nov 24th, 2017
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