The answers that are correct are a, b, and d
Explanation:
1)  + 7 H_2(g)](https://tex.z-dn.net/?f=%202%20Al%28s%29%20%2B%202%20NaOH%28aq%29%20%2B%206%20H_2O%28l%29%20%5Clongleftrightarrow%202%20Na%5BAl%28OH%29_4%5D%28aq%29%20%2B%207%20H_2%28g%29)
![Kc=\frac{[Na[Al(OH)_4]]^2*[H_2]^7}{[NaOH]^2}](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BNa%5BAl%28OH%29_4%5D%5D%5E2%2A%5BH_2%5D%5E7%7D%7B%5BNaOH%5D%5E2%7D)
The Kc for the reverse reaction is the inverse of the Kc of the reaction:

2) 
![Kc=\frac{[H_2SO_4]}{[SO_3]^2}](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BH_2SO_4%5D%7D%7B%5BSO_3%5D%5E2%7D)
The Kc for the reverse reaction is the inverse of the Kc of the reaction:

3)
![Kc=\frac{1}{[O_2]^3}](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B1%7D%7B%5BO_2%5D%5E3%7D)
The Kc for the reverse reaction is the inverse of the Kc of the reaction:

55.9 kPa; Variables given = volume (V), moles (n), temperature (T)
We must calculate <em>p</em> from <em>V, n</em>, and <em>T</em>, so we use <em>the Ideal Gas Law</em>:
<em>pV = nRT</em>
Solve for <em>p</em>: <em>p = nRT/V</em>
R = 8.314 kPa.L.K^(-1).mol^(-1)
<em>T</em> = (265 + 273.15) K = 538.15 K
<em>V</em> = 500.0 mL = 0.5000 L
∴ <em>p</em> = [6.25 x 10^(-3) mol x 8.314 kPa·L·K^(-1)·mol^(-1) x 538.15 K]/(0.5000 L) = 55.9 kPa
It is a completely true statement that a <span>base increases the OH- ion concentration of water. The correct option among the two options that are given in the question is the first option. I hope that this is the answer that you were looking for and the answer has actually come to your desired help.</span>
Answer:
Partial pressure N₂ . (Partial pressure H₂O)² / (Partial pressure H₂)² . (Partial pressure NO)² = Kp
Explanation:
The reaction is:
2NO + 2H₂ → N₂ + 2H₂O
The expression for Kp (pressure equilibrium constant) would be:
Partial pressure N₂ . (Partial pressure H₂O)² / (Partial pressure H₂)² . (Partial pressure NO)²
There is another expression for Kp, where you work with Kc (equilibrium constant)
Kp = Kc (R.T)^Δn
where R is the Ideal Gases constant
T° is absolute temperature
Δn = moles of gases formed - moles of gases, I had initially