candkpe NOBR(g) = % N2(g) + O2(g) + % Br2(g) from the following information at 298 K: NO(9) = % N2jg) + % O2ig) % NO(g) + % Br2(g) = % NOBR(ga) Ke = 1.55 x 1015 Kp = 2.63 Type your answers below. Note: If answering in scientific notation, type your answer this way: 1.23 x 1045 → type 1.23e45 ANSWERS: Kc = Kp =
candkpe NOBR(g) = % N2(g) + O2(g) + % Br2(g) from the following information at 298 K: NO(9) = % N2jg) + % O2ig) % NO(g) + % Br2(g) = % NOBR(ga) Ke = 1.55 x 1015 Kp = 2.63 Type your answers below. Note: If answering in scientific notation, type your answer this way: 1.23 x 1045 → type 1.23e45 ANSWERS: Kc = Kp =
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
![Determine KeandK,at 298 K for the reaction:
NOBI(g) = % N2(g) + ½ O2(g) + % Br2(g)
from the following information at 298 K:
NO(g) = % N2(g) + ½ O2ig)
Ke = 1.55 x 1015
½ NO(9) + % Br2(g) = % NOBR(a)
Kp
= 2.63
Type your answers below.
Note: If answering in scientific notation, type your answer this way:
1.23 x 1045 → type 1.23e45
ANSWERS:
Kc =
Kp =](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F13fbfc8e-47fc-4cfb-baa3-f3f7c6c75828%2F9f379e84-ccb7-4201-9002-ed61b22ec0f8%2F48c2hr_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Determine KeandK,at 298 K for the reaction:
NOBI(g) = % N2(g) + ½ O2(g) + % Br2(g)
from the following information at 298 K:
NO(g) = % N2(g) + ½ O2ig)
Ke = 1.55 x 1015
½ NO(9) + % Br2(g) = % NOBR(a)
Kp
= 2.63
Type your answers below.
Note: If answering in scientific notation, type your answer this way:
1.23 x 1045 → type 1.23e45
ANSWERS:
Kc =
Kp =
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