Chemistry, 29.11.2019 00:31 chant9

Nitric oxide (no) can be formed from nitrogen, hydrogen and oxygen in two steps. in the first step, nitrogen and hydrogen react to form ammonia: n2(g) + 3h2(g) →2nh3(g)δh=−92.kjin the second step, ammonia and oxygen react to form nitric oxide and water: 4nh3(g) + 5o2(g) → 4no(g) +6h2o(g)δh=−905.kjcalculate the net change in enthalpy for the formation of one mole of nitric oxide from nitrogen, hydrogen and oxygen from these reactions. round your answer to the nearest kj.

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Answer from: melanieambrosy

ΔH = - 272 kJ

Explanation:

We are going to use the fact that Hess law allows us to calculate the enthalpy change of a reaction no matter if the reaction takes place in one step or in several steps. To do this problem we wll add two times the first step to second step as follows:

N2(g) + 3H2(g) → 2NH3(g) ΔH=−92.kJ Multiplying by 2:

2N2(g) + 6H2(g) → 4NH3(g) ΔH=− 184 kK

plus

4NH3(g) + 5O2(g) → 4NO(g) +6H2O(g) ΔH=−905.kJ

2N2(g) + 6H2(g) + 5O2(g)→ 4NO(g) + 6H2O(g) ΔH = (-184 +(-905 )) kJ

ΔH = -1089 kJ

Notice how the intermediate NH3 cancels out.

As we can see this equation is for the formation of 4 mol NO, and we are asked to calculate the ΔH for the formation of one mol NO:

-1089 kJ/4 mol NO x 1 mol NO = -272 kJ (rounded to nearest kJ)

Answer from: Quest

answer: any goods being imported from another country may be subject to issued by the federal government are tariffs

explanation:

tariffs are a tax on anything that are imported from abroad. government use them to protect and boost their local industries and the thought is that they encourage consumers to buy local instead of importing and they will hope will protect jobs and economy in the country that implements them

Answer from: Quest

Answerthe half-reaction having the greatest tendency to occur at the anode: i₂(aq) + 2e⁻ ⇄ 2i⁻ ; e = +0.53 v====== $\begin{aligned} \text{i}_2\text{\footnotesize (aq)}+2\,\text{e}^{-} & \rightleftharpoons 2\,\text{i}^{-}\text{\footnotesize (aq)}& \text{e} & = +0.53\text{ v} \\ \text{co}^{3+}\text{\footnotesize (aq)} + \text{e}^{-} & \rightleftharpoons \text{co}^{2+}\text{\footnotesize (aq)} & \text{e} & = +1.80\text{ v} \\ \text{o}_2\text{\footnotesize (g)}+4\,\text{h}^{+}\text{\footnotesize (aq)} + 4\,\text{e}^{-} & \rightleftharpoons 2\,\text{h}_2\text{o}\text{\footnotesize(l)}& \text{e} & = +1.23\text{ v} \\ \end{aligned}$ oxidation occurs at the anode. we must therefore find the half-reaction with the greatest tendency to oxidize.the given values of e are reduction potentials (the form of the half-reactions are of reduction).the lower the reduction potential, the higher the potential to oxidize. the half-cell with the lowest reduction potential will have the greatest tendency to oxidize at the anode.the half-reaction having the greatest tendency must be i₂(aq) + 2e⁻ ⇄ 2i⁻ e = +0.53 vbecause its reduction potential is the lowest out of the three.

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Nitric oxide (no) can be formed from nitrogen, hydrogen and oxygen in two steps. in the first step,...