molar mass of methane CH4
= C + 4 H
= 12.0 + 4 x 1.008
= 12.0 + 4.032
= 16.042g/mol
7.31 x 10^25 molecules x <u> 1 mole CH4 </u> = 121.43 moles
6.02 x 10^23 CH4 molecules
121.43 moles CH4 are present.
2 and 2 and 3 hope that helped
Let the acid be HA.
The chemical formula for this acid will be the following:

The formula for the <span>acid dissociation constant will be the following:
</span>
![K_a= \dfrac{[H^+][A^-]}{[HA]}](https://tex.z-dn.net/?f=K_a%3D%20%5Cdfrac%7B%5BH%5E%2B%5D%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
<span>
We know [H+]=0.0001 (it's given).
However, we must find [A-] and [HA] in order to solve for the constant.
We find that [A-]=[H+] by using a electroneutrality equation.
Also, we can create a concentration equation to find [HA].
</span>
![0.5M=[A^-]+[HA]](https://tex.z-dn.net/?f=0.5M%3D%5BA%5E-%5D%2B%5BHA%5D)
![[HA]=0.5M-[A^-]](https://tex.z-dn.net/?f=%5BHA%5D%3D0.5M-%5BA%5E-%5D)
<span>
Now, we can find the acid dissociation constant.
</span>
![K_a= \dfrac{[H^+][A^-]}{0.5M-[A^-]}](https://tex.z-dn.net/?f=K_a%3D%20%5Cdfrac%7B%5BH%5E%2B%5D%5BA%5E-%5D%7D%7B0.5M-%5BA%5E-%5D%7D)