2,3 please 2. Initially, pO₂ = 1.0 atm3.IсEICET2-2x)2PbS(s) + 302(g)2H₂SSet up the expression for Kp. You do not need to solve for x.(g)pCO2 = PSO2=0+ C(s)[H₂S] = 1M, [CH4] =1.0M+ CH4(g)(2-2x1²2Pb(s) + CO2(g) +[CS2] = [H₂] = 0CS2(g) + 4H2(g)Set up the expression for Kc. You do not need to solve for x.Equilibrium PART 22SO2(g)

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2PbS (s) + 302(g) the expression for Kp. Y [H₂S] = 1M, [CH4] =1 2H₂S (g) + CH4(8) The expression for Kc. Y-

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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2,3 please

2. Initially, pO₂ = 1.0 atm
3.
I
с
E
I
C
E
T2-2x)
2PbS(s) + 302(g)
2H₂S
Set up the expression for Kp. You do not need to solve for x.
(g)
pCO2 = PSO2=0
+ C(s)
[H₂S] = 1M, [CH4] =1.0M
+ CH4(g)
(2-2x1²
2Pb(s) + CO2(g) +
[CS2] = [H₂] = 0
CS2(g) + 4H2(g)
Set up the expression for Kc. You do not need to solve for x.
Equilibrium PART 2
2SO2(g)

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