1) A chemist mixes an 8% nitric acid solution with a 5% nitric acid solution. How many milliliters of the 5% solution should the chemist use to make a 500-millileter solution that is 6.8% nitric acid? a. 167 milliliters c. 333 milliliters b. 200 milliliters d. 300 milliliters

Suppose the chemist uses x mL of the 8% acid solution, and y mL of the 5% solution.

The chemist wants to end up with a 500 mL solution, so

x + y = 500

and wants this solution to have a concentration of 6.8% acid. This means 6.8% of this volume should be acid. By volume, the 8% solution contributes 0.08x mL of acid, and the 5% solution contributes 0.05y mL, so

0.08x + 0.05y = 0.068(x + y) = 0.068 * 500 = 34

Now,

y = 500 - x

0.08x + 0.05(500 - x) = 34

0.08x + 25 - 0.05x = 34

0.03x = 9

x = 300

The chemist must use 300 mL of the 5% solution.

if you mean

then it's c. 8, -7/3

but if you mean

then it is

x = 25

step-by-step explanation:

a^2+b^2=c^2

7^2+24^2=c^2

49+576=625

sqrt625=25

x=25