For each of the following equilibria, write theequilibrium constant expression for KeYPart A2N₂Os (9) 4NO₂(g) + O₂(g)O Ke =O K₂=O K.OK-SubmitPart B[NO₂[0₂]N₂O₁Ke =[NO₂¹0₂]N₂0₂N₂O₁]NO₂0₁N₂O,NO₂] O₂Request Answer3F2(g) + Cl₂ (9) 2CIFs (9)(CIF₂CI, PICE,P Pearson Y▼Part B3F2 (g) + Cl2 (g) = 2C1F3 (9)[CIF 3][Cl₂][F2]O Kc =O KcO KO Kc = [C1₂] [F₂13[CIF3]2Submit-Part C[CIF3]2[Cl₂] [2]=[C1₂] [F2][CIF 3]Request Answer

The equilibrium constant expression for a general chemical equation: aA + bB ↔ cC + dD is: Kc =…

Answered: 2N₂Os (9) 4NO₂(g) + O₂(g) [NO₂0₂] N₂O₁…

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2N₂Os (9) 4NO₂(g) + O₂(g) [NO₂0₂] N₂O₁ O Ke OK, O K. = NO, No, N₂O₂ NO₂0₁

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

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Stoichiometry

Stoichiometry is a chemical division that evaluates the quantitative content of the chemical process with the help of the reacting molecules or products taking part in the reaction. The term “stoicheion” in “Stoichiometry”, is an old Greek term for "element" whereas “metron” means "measure".

Question

For each of the following equilibria, write the
equilibrium constant expression for Ke
Y
Part A
2N₂Os (9) 4NO₂(g) + O₂(g)
O Ke =
O K₂=
O K.
OK-
Submit
Part B
[NO₂[0₂]
N₂O₁
Ke =
[NO₂¹0₂]
N₂0₂
N₂O₁]
NO₂0₁
N₂O,
NO₂] O₂
Request Answer
3F2(g) + Cl₂ (9) 2CIFs (9)
(CIF₂
CI, P
ICE,
P Pearson

Y
▼
Part B
3F2 (g) + Cl2 (g) = 2C1F3 (9)
[CIF 3]
[Cl₂][F2]
O Kc =
O Kc
O K
O Kc = [C1₂] [F₂13
[CIF3]2
Submit
-
Part C
[CIF3]2
[Cl₂] [2]
=
[C1₂] [F2]
[CIF 3]
Request Answer

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