2NO2(8) 7 N204(g)Kc = 20.0 at 55 \deg CAt 55 \deg C a flask contains 0.040 mol-L-1 NO2(g) and 0.040 mol - L-\ N 204(g). When equilibrium is reached, (A) some of the NO2(g) will have been converted to N204(8). (B)some of the N 204(g) will have been converted to NO2(g). (C) the concentrations will remain at 0.040 mol-L-1 forN02(8) and 0.040 mol-L-1 for N204(8). (D) all of the NO2(g) will have been converted to N204(g).
2NO2(8) 7 N204(g)Kc = 20.0 at 55 \deg CAt 55 \deg C a flask contains 0.040 mol-L-1 NO2(g) and 0.040 mol - L-\ N 204(g). When equilibrium is reached, (A) some of the NO2(g) will have been converted to N204(8). (B)some of the N 204(g) will have been converted to NO2(g). (C) the concentrations will remain at 0.040 mol-L-1 forN02(8) and 0.040 mol-L-1 for N204(8). (D) all of the NO2(g) will have been converted to N204(g).
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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Transcribed Image Text:2NO2(8) 7 N204(g)Kc = 20.0 at 55 \deg CAt 55 \deg C a flask contains 0.040 mol-L-1 NO2(g) and 0.040 mol - L-\ N
204(g). When equilibrium is reached, (A) some of the NO2(g) will have been converted to N204(8). (B)some of the N
204(g) will have been converted to NO2(g). (C) the concentrations will remain at 0.040 mol-L-1 forN02(8) and 0.040
mol-L-1 for N204(8). (D) all of the NO2(g) will have been converted to N204(g).
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