Explanation:
what's li and na stand for ?
<span>I believe the correct 2nd reaction is:</span>
cof2(g)⇌1/2 co2(g)+1/2 cf4(g)
where we can see that it is exactly one-half of the
original
Therefore the new Kp is:
new Kp = (old Kp)^(1/2)
new Kp = (2.2 x 10^6)^(1/2)
<span>new Kp = 1,483.24 </span>
The atoms undergo chemical changes.
New substances may be a product of the reaction.
<u>Answer:</u> The equilibrium concentration of water is 0.597 M
<u>Explanation:</u>
Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as 
For a general chemical reaction:

The expression for
is written as:
![K_{c}=\frac{[C]^c[D]^d}{[A]^a[B]^b}](https://tex.z-dn.net/?f=K_%7Bc%7D%3D%5Cfrac%7B%5BC%5D%5Ec%5BD%5D%5Ed%7D%7B%5BA%5D%5Ea%5BB%5D%5Eb%7D)
The concentration of pure solids and pure liquids are taken as 1 in the expression.
For the given chemical reaction:

The expression of
for above equation is:
![K_c=\frac{[H_2O]^2}{[H_2S]^2\times [O_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BH_2O%5D%5E2%7D%7B%5BH_2S%5D%5E2%5Ctimes%20%5BO_2%5D%7D)
We are given:
![[H_2S]_{eq}=0.671M](https://tex.z-dn.net/?f=%5BH_2S%5D_%7Beq%7D%3D0.671M)
![[O_2]_{eq}=0.587M](https://tex.z-dn.net/?f=%5BO_2%5D_%7Beq%7D%3D0.587M)

Putting values in above expression, we get:
![1.35=\frac{[H_2O]^2}{(0.671)^2\times 0.587}](https://tex.z-dn.net/?f=1.35%3D%5Cfrac%7B%5BH_2O%5D%5E2%7D%7B%280.671%29%5E2%5Ctimes%200.587%7D)
![[H_2O]=\sqrt{(1.35\times 0.671\times 0.671\times 0.587)}=0.597M](https://tex.z-dn.net/?f=%5BH_2O%5D%3D%5Csqrt%7B%281.35%5Ctimes%200.671%5Ctimes%200.671%5Ctimes%200.587%29%7D%3D0.597M)
Hence, the equilibrium concentration of water is 0.597 M
3.
protium (A = 1), deuterium (A = 2), and tritium (A = 3).