Sample of acid are brought to a laboratory for analysis. several titrations are performed and it is determined that a 20.0-mililiter sample of acid rain is neutralized with 6.50 milliliters of 0.010 m naoh. what is the molarity of the h+ ions in the acid rain?

Since NaOH is a strong base, we know that it completely dissociates in water and will react with any strong or weak acid to completion.  That means that you can multiply 0.0065L by 0.010M NaOH which will give you the moles of OH⁻ in solution which is also the moles of H⁺ since all of the OH⁻ reacted with the acid in the rain.
0.0065Lx0.01M NaOH=6.5x10^-5mol OH⁻=6.5x10^-5 mol H⁺
To find the concetration of H⁺ is the original solution you divide the moles of H⁺ by the volume (in liters) of the original solution.
6.5x10^-5mol H⁺/0.020L=0.00325M H⁺

A short cut way of doing this is using the equation M₁V₁=M₂V₂
M₂ is the concentration of H⁺, V₁ is the volume of acidic solution, V₂ is the volume of basic solution, and M₂ is the concentration of the OH⁻.
M₁=M₂V₂/V₁
M₁=(0.01M OH⁻ x 0.0065L)/0.02L
M₁=0.00325M H⁺

I hope this helps.  Let me know in the comments if you have anything doesn't make sense to you so that I can try to fix it.

Forgot the question hahaha