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3. In a 2.0 L container an equilibrium exists between 0.400 mol of N202, 0.600 mol of H2, 1.000 mol of N20 and 1.000 mol of H₂O. N₂O2 (g) + H₂ (g) = N₂O(g) + H₂O(g) If 0.200 mol of N₂02 is added, what will the new equilibrium concentrations be of N₂O(g) + H₂O(g)?

3. In a 2.0 L container an equilibrium exists between 0.400 mol of N202, 0.600 mol of H2, 1.000 mol of N20 and 1.000 mol of H₂O. N₂O2 (g) + H₂ (g) = N₂O(g) + H₂O(g) If 0.200 mol of N₂02 is added, what will the new equilibrium concentrations be of N₂O(g) + H₂O(g)?

Chemistry
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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3 [H₂0] = [N₂0] = 0.537M
Transcribed Image Text:3 [H₂0] = [N₂0] = 0.537M
3. In a 2.0 L container an equilibrium exists between 0.400 mol of N202, 0.600 mol of H₂, 1.000 mol of N₂0 and 1.000 mol of H₂O. N₂O2 (g) + H₂ (g) = N₂O(g) + H₂O(g) If 0.200 mol of N₂02 is added, what will the new equilibrium concentrations be of N2O(g) + H₂O(g)?
Transcribed Image Text:3. In a 2.0 L container an equilibrium exists between 0.400 mol of N202, 0.600 mol of H₂, 1.000 mol of N₂0 and 1.000 mol of H₂O. N₂O2 (g) + H₂ (g) = N₂O(g) + H₂O(g) If 0.200 mol of N₂02 is added, what will the new equilibrium concentrations be of N2O(g) + H₂O(g)?
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