A chemist has one solution of 20% acid and another of 50% acid. how many millimeters of each are needed to obtain 30 ml of 30% acid?
.20x + .50(30-x) = .30(30)
x=20
So, 20ml of 20% and 10ml of 50%
Note that 30% is 1/3 of the way from 20% to 50%, so 1/3 of the acid is the higher concentration.
To solve this problem, we need to use a concentration equation. Let's assume the chemist needs x milliliters of the 20% acid solution and (30 - x) milliliters of the 50% acid solution to obtain 30 ml of a 30% acid solution.
First, let's calculate the amount of acid in each solution. The 20% acid solution has 20% of acid in each milliliter, so the amount of acid in x milliliters is (20/100) * x. Similarly, the 50% acid solution has (50/100) * (30 - x) amount of acid.
Now, we can set up an equation based on the concentration of acid in the final solution, which should be 30%:
(20/100) * x + (50/100) * (30 - x) = (30/100) * 30
Simplifying the equation:
(20/100) * x + (50/100) * (30 - x) = (3/10) * 30
Now, we can solve for x:
0.2x + 0.5(30 - x) = 0.3 * 30
0.2x + 15 - 0.5x = 9
-0.3x = 9 - 15
-0.3x = -6
x = -6 / -0.3
x = 20
So, the chemist needs 20 ml of the 20% acid solution and (30 - 20) = 10 ml of the 50% acid solution to obtain 30 ml of 30% acid solution.