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IRISSAK [1]
3 years ago
11

Part a write an equation for the formation of nh3(g) from its elements in its standard states. express your answer as a chemical

equation. identify all of the phases in your answer. submitmy answersgive up part b find δh∘f for nh3(g) from appendix iib in the textbook. express your answer using three significant figures. δh∘f = kj/mol submitmy answersgive up part c write an equation for the formation of co2(g) from its elements in its standard states. express your answer as a chemical equation. identify all of the phases in your answer. submitmy answersgive up part d find δh∘f for co2(g) from appendix iib in the textbook. express your answer using four significant figures. δh∘f = kj/mol submitmy answersgive up part e write an equation for the formation of fe2o3(s) from its elements in its standard states. express your answer as a chemical equation. identify all of the phases in your answer. submitmy answersgive up part f find δh∘f for fe2o3(s) from appendix iib in the textbook. express your answer using four significant figures. δh∘f = kj/mol submitmy answersgive up part g write an equation for the formation of ch4(g) from its elements in its standard states. express your answer as a chemical equation. identify all of the phases in your answer. submitmy answersgive up part h find δh∘f for ch4(g) from appendix iib in the textbook. express your answer using three significant figures.
Chemistry
2 answers:
pochemuha3 years ago
8 0
Sourse 

https://quizlet.com/19584916/chemistry-chapter-6-and-7-flash-cards/
andrey2020 [161]3 years ago
3 0
A. N₂ (g) + 3 H₂ (g) --> 2 NH₃ (g)
B. The value for standard enthalpy of formation is empirical given that the reactants involved were pure elements. So, you can search this on the internet or in any textbook. The Hf for NH₃ is -46.0 kJ/mol.
C. C (s) + O₂ (g) --> CO₂ (g)
D. The Hf for CO₂ is <span>-393.5 kJ/mol
E. 4 Fe (s) + 3 O</span>₂ (g) --> 2 Fe₂O₃ (s)
F. The Hf for solid Fe₂O₃ is -826.0 kJ/mol.
G. C (s) + 2 H₂ (g) --> CH₄ (g)
H. The Hf for methane gas is -74.9 kJ/mol.
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g 2BrO3- + 5SnO22-+ H2O5SnO32- + Br2+ 2OH- In the above reaction, the oxidation state of tin changes from to . How many electron
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In the above reaction, the oxidation state of tin changes from 2+ to 4+.

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SnO₂²⁻ → SnO₃²⁻

Tin changes the oxidation state from +2 to +4. It has increased it so this is the oxidation from the redox (it released 2 e⁻). We are in basic medium, so we add water in the side of the reaction where we have the highest amount of oxygen. We have 2 O on left side and 3 O on right side so we add 1 water on the right and we complete with OH⁻ in the opposite side to balance the H.  

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First of all, we have unbalance the bromine, so we add 2 on the BrO₃⁻. We have 6 O in left side and there are no O on the right, so we add 6 H₂O on the left. To balance the H, we must complete with 12OH⁻. Bromate reduces to bromine at ground state, so it gained 5e⁻. We have 2 atoms of Br, so finally it gaines 10 e⁻.

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In order to balance the main reaction and balance the electrons we multiply  (x5) the oxidation and (x1) the reduciton

(SnO₂²⁻ + 2OH⁻ → SnO₃²⁻ + 2e⁻ + H₂O) . 5

(6H₂O + 10 e⁻ + 2BrO₃⁻ →  Br₂ + 12OH⁻) . 1

5SnO₂²⁻ + 10OH⁻ + 6H₂O + 10 e⁻ + 2BrO₃⁻ → Br₂ + 12OH⁻ + 5SnO₃²⁻ + 10e⁻ + 5H₂O

We can cancel the e⁻ and we substract:

12OH⁻ - 10OH⁻ = 2OH⁻ (on the right side)

6H₂O - 5H₂O = H₂O (on the left side)

2BrO₃⁻ + 5SnO₂²⁻ + H2O ⇄ 5SnO₃²⁻ + Br₂ + 2OH⁻

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