How does the 10.9 M^-1 turn into 0.0940 M? + kt[N2O],t = 1250 seconds%3D%3D20], = 0.108 Mk = 1.10×10-3 M-'s%3D1+ (1.10×10-3 M's)(1250 s1%3DN20],0.108 M110.6 M-1%3D[N2O],[N2O],0.0940 MUse Integrated Rate Law to Calculate ReactantConcentration at Time tsition of nitrous oxide at 565 °CN2(g) + 2 O2(g)er in N20 with a rate constant of 1.10 x 10-3 M-1s1.ent is performed where the initial concentration ofM, what is the N,0 concentration after 1250

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How does the 10.9 M^-1 turn into 0.0940 M?

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

How does the 10.9 M^-1 turn into 0.0940 M?

+ kt
[N2O],
t = 1250 seconds
%3D
%3D
20], = 0.108 M
k = 1.10×10-3 M-'s
%3D
1
+ (1.10×10-3 M's)(1250 s
1
%3D
N20],
0.108 M
1
10.6 M-1
%3D
[N2O],
[N2O],
0.0940 M
Use Integrated Rate Law to Calculate Reactant
Concentration at Time t
sition of nitrous oxide at 565 °C
N2(g) + 2 O2(g)
er in N20 with a rate constant of 1.10 x 10-3 M-1s1.
ent is performed where the initial concentration of
M, what is the N,0 concentration after 1250

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