A chemist wants to mix a 22% acid solution with a 36% acid solution to get 28 L of a 26% acid solution. How many liters of the 22% solution and how many liters of the 36% solution should be mixed?

20 L of 22% solution and 8 L of 36% solution

Step-by-step explanation:

Volume of 22% solution + volume of 36% solution = volume of 26% solution

x + y = 28

Acid in 22% solution + acid in 36% solution = acid in 26% solution

0.22x + 0.36y = 0.26 (28)

0.22x + 0.36y = 7.28

Solve the system of equations using either elimination or substitution. I'll use substitution:

x = 28 - y

0.22 (28 - y) + 0.36y = 7.28

6.16 - 0.22y + 0.36y = 7.28

0.14y = 1.12

y = 8

x = 28 - y

x = 20

The chemist should use 20 L of 22% solution and 8 L of 36% solution.

There should be mixed 20 L of the 22% acid solution with 8L of the 36% acid solution

Step-by-step explanation:

We are mixing two acids.

x = liters of 22% acid solution

y = liters of 36% acid solution

x + y = 28 (total liters)

0.22x + 0.36y = 0.26 * 28

Since x+y=28 means y = 28-x

Now we will use substitution to find x

0.22x + 0.36 (28-x) = 0.26 * 28

0.22x + 10.08 - 0.36x = 7.28

0.14x = 2.8

x = 20

y = 28 - 20 = 8

⇒ We use 20 liters of the 22% solution to be mixed with 8 liters of the 36% solution to form 28l of a 26% acid solution.