<u>Answer:</u> The concentration of nitrogen dioxide at equilibrium is 0.063 M
<u>Explanation:</u>
We are given:
Initial concentration of nitrogen dioxide = 0.0250 M
Initial concentration of dinitrogen tetraoxide = 0.0250 M
For the given chemical equation:

<u>Initial:</u> 0.025 0.025
<u>At eqllm:</u> 0.025-x 0.025+2x
The expression of
for above equation follows:
![K_c=\frac{[NO_2]^2}{[N_2O_4]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BNO_2%5D%5E2%7D%7B%5BN_2O_4%5D%7D)
We are given:

Putting values in above expression, we get:

Neglecting the negative value of 'x', because concentration cannot be negative
So, equilibrium concentration of nitrogen dioxide = (0.025 + 2x) = [0.025 + 2(0.019)] = 0.063 M
Hence, the concentration of nitrogen dioxide at equilibrium is 0.063 M
There it is, I hope it gets to be helpful
Agree definitely :P poop dance
Answer:
C) Q < K, reaction will make more products
Explanation:
- 1/8 S8(s) + 3 F2(g) ↔ SF6(g)
∴ Kc = 0.425 = [ SF6 ] / [ F2 ]³
∴ Q = [ SF6 ] / [ F2 ]³
∴ [ SF6 ] = 2 mol/L
∴ [ F2 ] = 2 mol/L
⇒ Q = ( 2 ) / ( 2³)
⇒ Q = 0.25
⇒ Q < K, reaction will make more products
Answer:
They are similarly charged, which is why they repel each other.