A chemist mixes a 200 L of a solution that is 60% acid with a 300 L of a solution that is 20% acid. What is the percentage of the mixture?

Question

↳Redirected from "How do I solve #y'' + y = -2 sin(x)# with the initial conditions #y(0) = 0# and #y'(0) = 1#?"

3655 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-08-11T14:18:54

The resulting mixture is #36%# acid.

The resulting mixture's acid proportion is

#"volume of acid in the solution"/"volume of the solution"#

The volume of the mixture is #200+300=500L#

The volume of acid in the first solution is #0.6*200=120L#

The volume of acid in the second solution is #0.2*300=60L#

Therefore, the resulting mixture's acid proportion is:

#(120+60)/500=180/500=36%#