Nitric acid is abailable commercially at a concentration of 16M. What volume would you use to prepare 750 ml of a 0.20 M solution?

I am sure that I have answered this one recently.

Anyway you need
750 x 0.20 mmoles of HCl

if you need v ml of 16M HCl, then

v x 16 = 750 x 0.20

I got 9.4 ml

(only 2 sig figs as both 16 M and 0.20 M are to 2 sig figs)

To find the volume of 16M nitric acid required to prepare a 0.20M solution, we can use the formula for dilution:

C1V1 = C2V2

Where:
C1 = initial concentration
V1 = initial volume
C2 = final concentration
V2 = final volume

In this case, we are given:
C1 = 16M
C2 = 0.20M
V2 = 750 ml

Rearranging the formula to solve for V1:

V1 = (C2 * V2) / C1

Let's substitute the given values:

V1 = (0.20M * 750 ml) / 16M

V1 ≈ 9.375 ml

To prepare a 0.20M solution, you would need to use approximately 9.375 ml of 16M nitric acid.