.50 M KCl because 5% is the same as .05, which makes the .50M more concentrated.
<u>Answer:</u> The equilibrium concentration of
is 0.332 M
<u>Explanation:</u>
We are given:
Initial concentration of
= 2.00 M
The given chemical equation follows:

<u>Initial:</u> 2.00
<u>At eqllm:</u> 2.00-2x x x
The expression of
for above equation follows:
![K_c=\frac{[CO_2][CF_4]}{[COF_2]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCO_2%5D%5BCF_4%5D%7D%7B%5BCOF_2%5D%5E2%7D)
We are given:

Putting values in above expression, we get:

Neglecting the value of x = 1.25 because equilibrium concentration of the reactant will becomes negative, which is not possible
So, equilibrium concentration of ![COF_2=(2.00-2x)=[2.00-(2\times 0.834)]=0.332M](https://tex.z-dn.net/?f=COF_2%3D%282.00-2x%29%3D%5B2.00-%282%5Ctimes%200.834%29%5D%3D0.332M)
Hence, the equilibrium concentration of
is 0.332 M
They both turn something on, and the way they are different is the way they turn it off<span>
</span>
Here we are talking about ammonia which is a molecule. Now 1 mole of ammonia also contains 6.022<span> * 10^</span>23 molecules<span> of ammonia. In one molecule of ammonia,we have 1 nitrogen atom and 3 hydrogen atom.</span>