SOLUTION: a chemist has two large containers of sulfuric acid solution. each has a different concentration of acid. Blending 300mls of the first solution (25% acid solution) with a current
Algebra ->
Customizable Word Problem Solvers
-> Mixtures
-> SOLUTION: a chemist has two large containers of sulfuric acid solution. each has a different concentration of acid. Blending 300mls of the first solution (25% acid solution) with a current
Log On
Question 148233: a chemist has two large containers of sulfuric acid solution. each has a different concentration of acid. Blending 300mls of the first solution (25% acid solution) with a currently unknown amount of the second solution (10% acid solution) gives a mixture that is 15% acid. How much of the second solution was mixed in and how much was in the final mixture? You must find the solutions by substitution and the process of elimination. Demonstrate both approaches in your work. Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! a chemist has two large containers of sulfuric acid solution. each has a different concentration of acid. Blending 300mls of the first solution (25% acid solution) with a currently unknown amount of the second solution (10% acid solution) gives a mixture that is 15% acid. How much of the second solution was mixed in and how much was in the final mixture?
:
Let x = amt of the 2nd solution
then
(x+300) = final amt
:
.25(300) + .10x = .15(x+300)
:
75 + .10x = .15x + 45
:
75 - 45 = .15x - .10x
:
30 = .05x
:
x =
x = 600 mls of the 2nd solution (10%)
:
Find the final mixture amt: (x+300), substitute 600 for x
600 + 300 = 900 mls final amt