5. At a certain temperature, some X was used tocer-establish the equilibrium:X (s) 2 Y (g) + 22 (g). When the equilibriumwas established the partial pressure of Y was meas-ured to be 0.2 atm. The value of K, at that tempera-actsustture is :A) 0.01B) 0.02 C) 0.04D) 0.08 E) 0.032 3. Given : A (g) + 3B (g) 2 2C (g) + D (g)One mole of A and one mole of B are placed in a oneliter container. After the equilibrium has been esta-blished, 0.4 mol of C is present in the container. Kefor this reaction is:A) 0.250 B) 0.625 C) 1.560 D) 1.100 E) 2.785

We have to apply Kp and Kc formulas simply

Ai a certain temperature, some X was used to establish the equilibrium: X (s) 2 Y (g) + 22 (g). When the equilibrium was established the partial pressure of Y was meas- ured to be 0.2 atm. The value of K, at that tempera- ture is: A) 0.01 B) 0.02 C) 0.04 D) 0.08 E) 0.032

Ai a certain temperature, some X was used to establish the equilibrium: X (s) 2 Y (g) + 22 (g). When the equilibrium was established the partial pressure of Y was meas- ured to be 0.2 atm. The value of K, at that tempera- ture is: A) 0.01 B) 0.02 C) 0.04 D) 0.08 E) 0.032

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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Question

100%

5. At a certain temperature, some X was used to
cer-
establish the equilibrium:
X (s) 2 Y (g) + 22 (g). When the equilibrium
was established the partial pressure of Y was meas-
ured to be 0.2 atm. The value of K, at that tempera-
acts
ust
ture is :
A) 0.01
B) 0.02 C) 0.04
D) 0.08 E) 0.032

3. Given : A (g) + 3B (g) 2 2C (g) + D (g)
One mole of A and one mole of B are placed in a one
liter container. After the equilibrium has been esta-
blished, 0.4 mol of C is present in the container. Ke
for this reaction is:
A) 0.250 B) 0.625 C) 1.560 D) 1.100 E) 2.785

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