<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
Answer:
Sodium
(Na)
Just count the electrons and search which atom it is.
False. The number of neutrons can be more or less than the number of protons.
Answer:
1. KNO3
2. Ca(NO3)2
3. CaCl2
4. KCl
Explanation:
In each of the neutralization reactions, the H from one of the reactant(acid) will combine with the OH from the other reactant (base) to form water while the other elements combine to give the salt as shown below:
1. HNO3 + KOH → H2O + KNO3
The salt produced is KNO3
2. 2HNO3 + Ca(OH)2 → 2H2O + Ca(NO3)2
The salt produced is Ca(NO3)2
3. 2HCl +Ca(OH)2 → 2H2O + CaCl2
The salt produced is CaCl2
4. HCl +KOH → H2O + KCl
The salt produced is KCl
