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A.) ΔHr =prod-react1/2 H2(g) + 1/2(O2)(g) —>OH(g) ΔHr=38.45 so, OH(g) + O(g) = -38.95J/K b.) prod-react(435.994)+1/2(498.34)-(-241.814)435.994+249...

Question

A.) ΔHr =prod-react1/2 H2(g) + 1/2(O2)(g) —>OH(g) ΔHr=38.45 so, OH(g) + O(g) = -38.95J/K b.) prod-react(435.994)+1/2(498.34)-(-241.814)435.994+249.17685.164 +241.814ΔHr= 926.98J/Kc.) H2O(g) —> (-1(g)+HO(g)

a.) ΔHr =prod-react1/2 H2(g) + 1/2(O2)(g) —>OH(g) ΔHr=38.45 so, OH(g) + O(g) = -38.95J/K b.) prod-react(435.994)+1/2(498.34)-(-241.814)435.994+249.17685.164 +241.814ΔHr= 926.98J/Kc.) H2O(g) —> (-1(g)+HO(g)



Answers

Balance the following equations in basic solution:
a) $\mathrm{PbO}_{2}+\mathrm{KCl} \longrightarrow \mathrm{KClO}+\mathrm{KPb}(\mathrm{OH})_{3}$
b) $\mathrm{KMnO}_{4}+\mathrm{KIO}_{3} \longrightarrow \mathrm{MnO}_{2}+\mathrm{KIO}_{4}$
c) $\mathrm{K}_{2} \mathrm{MnO}_{4} \longrightarrow \mathrm{MnO}_{2}+\mathrm{KMnO}_{4}$

In this problem, I can write very extended Channel two. This is A plus To Ash to S 03. This is B plus Ash Tour will give the product ID NH 20 H. This is C plus two H two S 04 This is the therefore according to the given option in this problem, only option C Only option CH correct answer. Only option C. Correct answer for this problem.

In this problem, I can write DDX and Edge CO two plus and S T D plus as to will react to form and at four Edge TCU three. This is compound x and NH food at Ceo today plus any cl will react to form any Edge coterie plus, NH four Cl and two. Any Edge Co three will react to form And I do see you three tiered Plus Ash two Plus Co two. So option it, correct here.

A Z. We know that the component will must be from its constituent in their most favorable elemental form. So balanced reactions should consist off anto and 02 and this has given in tow Blessed 0.5 or two that forms and to and the second one is and took this half or two forms and to on the third is to ch three ch two c w age in liquid form plus seven oxygen molecules Forms six. CEO to plastics H two off.

Who are looking at some violence reactions. So in this first brought me up to R B, the solid state, uh, two h 20 in the liquid state. What we get is to be oh h and the request form at H two in the gaseous form, so that reaction goes ahead. This is an alkali metal that gives an alkali metal height dockside upon hydraulic sis. So in the next one they're given reaction will not occur because bromine is more electro negative than iodine. So in this reduction reaction, the iodine is not able to oxidize the BR minus iron because of the small electro negativity value. Next, we have an alkaline earth metal that oxidizes to give the corresponding hydroxide upon hydraulic assist. So we have S r. O in the solid state at H 20 in the liquid state that generates S R. O H. Two in the acquis statement. Lastly, we have s 03 in the gaseous state at H 20 in the liquid state that generates H two s 04 That is our sulfuric acid


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