HCl - Red (strong acid)

- Blue (weak base)
NaOH - Purple (strong base)

Green (neutral)
Answer:
C₂H₂ + 3H₂ ⟶ 2CH₄
Explanation:
The initial concentrations are:
[CH₄] = 6.30 ÷ 6.00 = 1.05 mol·L⁻¹
[C₂H₂] = 4.20 ÷ 6.00 = 0.700 mol·L⁻¹
[H₂] = 11.15 ÷ 6.00 = 1.858 mol·L⁻¹
2CH₄ ⇌ C₂H₂ + 3H₂
I/mol·L⁻¹: 1.05 0.700 1.858
![Q = \dfrac{\text{[C$_{2}$H$_{2}$][H$_{2}$]}^{3}}{\text{[CH$_{4}$]}^{2}} = \dfrac{ 0.700\times 1.858^{3}}{1.05^{2}}= 4.07](https://tex.z-dn.net/?f=Q%20%3D%20%5Cdfrac%7B%5Ctext%7B%5BC%24_%7B2%7D%24H%24_%7B2%7D%24%5D%5BH%24_%7B2%7D%24%5D%7D%5E%7B3%7D%7D%7B%5Ctext%7B%5BCH%24_%7B4%7D%24%5D%7D%5E%7B2%7D%7D%20%3D%20%5Cdfrac%7B%200.700%5Ctimes%201.858%5E%7B3%7D%7D%7B1.05%5E%7B2%7D%7D%3D%204.07)
Q > K
That means we have too many products.
The reaction will go to the left to get rid of the excess products.
C₂H₂ + 3H₂ ⟶ 2CH₄
Answer:
Anode half reaction equation:
Ni(s)------> Ni^2+(aq) + 2e-
Explanation:
Looking at the values of reduction potential given in the question, Ni2+|Ni half cell has a more negative reduction potential than the Fe3+|Fe2+ half cell. The more negative the reduction potential of a half cell, the more its tendency to act as the anode. Hence based on the half cell reduction potentials presented in the question, Ni2+|Ni is the anode while the other is the cathode.
The lower the frequncy the longer the wavelength the longer the wavelength whoch would probably give you more energy.
if im incorrect im truly sorry
<span>N2H4 is a polar molecule </span>