A chemist has two solutions of sulfuric acid. One is a 25% solution, and the other is a 50% solution. How many liters of each does the chemist mix to get 10 liters of a 40% solution?
25% solution L:
50% solution L:
If x is 25%, then the rest (10-x) is 50%. So,
.25x + .50(10-x) = .40(10)
To solve this problem, we can use a technique called the method of mixtures.
Let's assume that the chemist mixes L liters of the 25% solution and (10 - L) liters of the 50% solution to get a 10-liter solution.
The amount of acid in the 25% solution is 25% of L liters of solution. This can be calculated as 0.25L liters of acid.
Similarly, the amount of acid in the 50% solution is 50% of (10 - L) liters of solution. This can be calculated as 0.5(10 - L) liters of acid.
So, the total amount of acid in the resulting 10-liter solution is 0.25L + 0.5(10 - L) liters.
Since the resulting 10-liter solution is a 40% solution, we know that the amount of acid in the resulting solution is 40% of 10 liters. This can be calculated as 0.4 * 10 = 4 liters of acid.
Now, we can set up an equation and solve for L:
0.25L + 0.5(10 - L) = 4
Simplifying the equation:
0.25L + 5 - 0.5L = 4
-0.25L = -1
L = -1 / -0.25
L = 4
So, the chemist needs to mix 4 liters of the 25% solution and (10 - 4) = 6 liters of the 50% solution to get 10 liters of a 40% solution.
Therefore:
25% solution L: 4 liters
50% solution L: 6 liters