"what volume of .50 m h2so4 must be added to 65 ml of .20 m h2so4 to give a final solution of .35 M? assume volumes are additive"

Use M1V2=M2V2

Plug in your given molarity and volume and solve for V1.

Note that Alison's post should read

M1V1 = M2V2

Thanks, but there are 3 Molarity amounts given, I am not sure how to do this step by step. I know the formula, but don't know how to fit all of them in....?

To find the volume of 0.50 M H2SO4 needed to be added to 65 ml of 0.20 M H2SO4 to obtain a final solution of 0.35 M, we can use the dilution formula.

The formula for dilution is:

C1*V1 = C2*V2

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

First, let's assign the given values to the corresponding variables:

C1 = 0.20 M (initial concentration of 65 ml of 0.20 M H2SO4)
V1 = 65 ml (initial volume of 0.20 M H2SO4)
C2 = 0.35 M (final concentration)
V2 = unknown (final volume, which we need to find)

Now, we can plug the values into the dilution formula:

(0.20 M) * (65 ml) = (0.35 M) * (65 ml + V2)

By rearranging the formula, we can solve for V2:

(0.20 M * 65 ml) / (0.35 M) = 65 ml + V2

0.20 M * 65 ml = 65 ml + V2

13 ml = V2

Therefore, 13 ml of 0.50 M H2SO4 must be added to 65 ml of 0.20 M H2SO4 to obtain a final solution of 0.35 M.