For the reaction 4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g), explain how a reduction in pressure or the use of a more efficient catalyst would affect the yield of NO.

let's tackle question 34.

34)
The reaction is: $4\text{NH}_3(g) + 5\text{O}_2(g) \rightleftharpoons 4\text{NO}(g) + 6\text{H}_2\text{O}(g)$

a) Reduction in pressure
According to Le Chatelier's principle, a reduction in pressure will shift the equilibrium to the side with a greater number of moles of gas.
Reactant moles of gas: $4 (\text{NH}_3) + 5 (\text{O}_2) = 9$ moles.
Product moles of gas: $4 (\text{NO}) + 6 (\text{H}_2\text{O}) = 10$ moles.
Since the product side has more moles of gas (10) than the reactant side (9), reducing the pressure will shift the equilibrium to the right. This will increase the yield of Nitrogen (II) oxide gas ($\text{NO}$).

b) Using a more efficient catalyst
A catalyst increases the rate of both the forward and reverse reactions equally. It helps the reaction reach equilibrium faster but does not change the position of the equilibrium or the final yield of the products. Therefore, using a more efficient catalyst would not affect the yield of Nitrogen (II) oxide gas.

For question 35, the table showing the standard reduction potentials is not visible in the image. Please provide the table so I can help you with that question.