Oxidation reaction
In ---> In³⁺ + 3e ---1)
reduction reaction
Cd²⁺ + 2e ---> Cd ---2)
when balancing the reactions, electrons have to be balanced. to balance the electrons multiple 1st reaction by 2 and 2nd reaction by 3
1) x 2
2) x 3
2In ---> 2In³⁺ + 6e
3Cd²⁺ + 6e ---> 3Cd
add the 2 equations to obtain the overall reaction
2In + 3Cd²⁺ ---> 2In³⁺ + 3Cd
the answer would be winter because the north would be facing away from the son therefor making the northern states cold.
sorry but I don't know so sorry
Answer:
[H₂] = 1.61x10⁻³ M
Explanation:
2H₂S(g) ⇋ 2H₂(g) + S₂(g)
Kc = 9.30x10⁻⁸ = ![\frac{[H_{2}]^2[S_{2}]}{[H_{2}S]^2}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BH_%7B2%7D%5D%5E2%5BS_%7B2%7D%5D%7D%7B%5BH_%7B2%7DS%5D%5E2%7D)
First we <u>calculate the initial concentration</u>:
0.45 molH₂S / 3.0L = 0.15 M
The concentrations at equilibrium would be:
[H₂S] = 0.15 - 2x
[H₂] = 2x
[S₂] = x
We <u>put the data in the Kc expression and solve for x</u>:
![\frac{(2x^2) * x}{(0.15-2x)^2}=9.30x10^{-8}](https://tex.z-dn.net/?f=%5Cfrac%7B%282x%5E2%29%20%2A%20x%7D%7B%280.15-2x%29%5E2%7D%3D9.30x10%5E%7B-8%7D)
![\frac{4x^3}{0.0225-4x^2}=9.30*10^{-8}](https://tex.z-dn.net/?f=%5Cfrac%7B4x%5E3%7D%7B0.0225-4x%5E2%7D%3D9.30%2A10%5E%7B-8%7D)
We make a simplification because x<<< 0.0225:
![\frac{4x^3}{0.0225} =9.30*10^{-8}](https://tex.z-dn.net/?f=%5Cfrac%7B4x%5E3%7D%7B0.0225%7D%20%3D9.30%2A10%5E%7B-8%7D)
x = 8.058x10⁻⁴
[H₂] = 2*x = 1.61x10⁻³ M