D-examples of state functions only occur in chemistry and physics and not in real life.
Answer:
As2O3 → H3AsO4
As2O3 → 2H3AsO4 balance As
5H2O + As2O3 → 2H3AsO4 balance O by adding H2O to one side
5H2O + As2O3 → 2H3AsO4 + 4H+ balance H by adding H+ to one side
5H2O + As2O3 → 2H3AsO4 + 4 H+ + 4e- balance charge by adding electrons to one side
Now do the same for the other part of the reaction
NO3- → NO
NO3- → NO + 2H2O
4H+ + NO3- → NO + 2H2O
3e- + 4H+ + NO3- → NO + 2H2O
Now cancel the electrons by multiplying the first equation by 3 and the second equation by 4, then add them together
.
3As2O3 + 4NO3- + 7H2O + 4H+ → 6H3AsO4 + 4NO
Answer: Concentration of N₂ is 4.8.
M.
Explanation: 
is a constant of equilibrium and it is dependent of the concentrations of the reactants and the products of a balanced reaction. For
N2(g) + 2 O2(g) ⇄ 2 NO2(g)
= ![\frac{[NO2]^{2} }{[N2][O2]^{2} }](https://tex.z-dn.net/?f=%5Cfrac%7B%5BNO2%5D%5E%7B2%7D%20%7D%7B%5BN2%5D%5BO2%5D%5E%7B2%7D%20%7D)
From the question concentration of NO2 is twice of O2:
[NO2] = 2[O2]
Substituting this into :
= ![\frac{[2O2]^{2} }{[N2][O2]^{2} }](https://tex.z-dn.net/?f=%5Cfrac%7B%5B2O2%5D%5E%7B2%7D%20%7D%7B%5BN2%5D%5BO2%5D%5E%7B2%7D%20%7D)
8.3.
= ![\frac{4O2^{2} }{[N2].O2^{2} }](https://tex.z-dn.net/?f=%5Cfrac%7B4O2%5E%7B2%7D%20%7D%7B%5BN2%5D.O2%5E%7B2%7D%20%7D)
[N2] = 
[N2] = 
[N2] = 4.8.
The concentration of N2 in the equilibrium is [N2] = 4.8.M.
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