1. F2N2(g) + 202(g)2 NO2(g)Initially, 0.45 mol/L of nitrogen and 0.68 mol/L of oxygen are introduced into acontainer. What are the equilibrium values of all the species?O[N₂] = 0.44 M[0₂] = 0.66 M#3O[N₂] = 0.66 M[NO₂] = 0.020 MED[NO₂] = 0.44 M[2] U. UZU IVE80F3[N₂] = 0.44 M[NO₂] = 0.44 M[0₂] = 0.020 MO[N₂] = 0.020 MC[NO2] = 0.020 M[0₂] = 0.020 M$4888F4RF67 do%5VF5TG^6MacBook AirBF6YIv lo7F7UNK = 1.96 x 10-3* 008JDIIF81M(9KDDF9O)01LF10P>.F11

Answered: Image /qna-images/answer/e8e21650-f576-4dc1-b6da-10558e60aaa0.jpg

N2(g) +202(g) 2 NO2(g) Initially, 0.45 mol/L of nitrogen and 0.68 mol/L of oxygen are introduced into a container. What are the equilibrium values of all the species? O[N₂] = 0.44 M [02] = 0.66 M #3 3 [NO₂] = 0.020 M O[N₂] = 0.66 M [NO₂] = 0.44 M [2] - O[N₂] = 0.44 M [NO₂] = 0.44 M [0₂] = 0.020 M 80 F3 O[N₂] = 0.020 M - C. U2U IVE [NO₂] = 0.020 M [0₂] = 0.020 M 54 $ 4 000 000 F4 % 5 6 MacBook Air & 7 K = 1.96 x 10-3 F7 00 * 8 DII FB 9 DD F9 0 C F10 11 d F11 + 11

N2(g) +202(g) 2 NO2(g) Initially, 0.45 mol/L of nitrogen and 0.68 mol/L of oxygen are introduced into a container. What are the equilibrium values of all the species? O[N₂] = 0.44 M [02] = 0.66 M #3 3 [NO₂] = 0.020 M O[N₂] = 0.66 M [NO₂] = 0.44 M [2] - O[N₂] = 0.44 M [NO₂] = 0.44 M [0₂] = 0.020 M 80 F3 O[N₂] = 0.020 M - C. U2U IVE [NO₂] = 0.020 M [0₂] = 0.020 M 54 $ 4 000 000 F4 % 5 6 MacBook Air & 7 K = 1.96 x 10-3 F7 00 * 8 DII FB 9 DD F9 0 C F10 11 d F11 + 11

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

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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