Portage learning CHEM 104 Week 1 Module 1 Problem Set QUESTIONS AND ANSWERS 2024.
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Course
Oncology Nursing
Institution
Oncology Nursing
Portage learning CHEM 104 Week 1 Module 1 Problem Set QUESTIONS AND ANSWERS 2024. Portage learning CHEM 104 Week 1 Module 1 Problem Set QUESTIONS AND ANSWERS 2024. Portage learning CHEM 104 Week 1 Module 1 Problem Set QUESTIONS AND ANSWERS 2024.
Portage learning CHEM 104 Week 1 Module 1
Problem Set QUESTIONS AND ANSWERS
2024
1. Show your work and be sure to give answer with the correct units.
In the reaction of 0.200 M gaseous N2O5 to yield NO2 gas and O2 gas
as shown below:
2 N2O5 (g) ’ 4 NO2 (g) + O2 (g)
the following data table is obtained:
Time (sec)
[N2O5]
[O2]
0
0.200 M
0
300
0.180 M
0.010 M
600
0.162 M
0.019 M
900
0.146 M
0.027 M
1200
0.132 M
0.034 M
1/8
, 1800
0.110 M
0.045 M
2400
0.096 M
0.052 M
3000
0.092 M
0.054 M
(a) Use the [O2] data from the table to calculate the average rate over
the measured time interval from 0 to 3000 secs.
(b) Use the [O2] data from the table to calculate the instantaneous rate
early in the reaction (0 secs to 300 sec).
(c) Use the [O2] data from the table to calculate the instantaneous rate
late in the reaction (2400 secs to 3000 secs).
(d) Explain the relative values of the average rate, early instantaneous
rate and late instantaneous rate.: (a) The average rate over the measured
time interval from 0 to 3000 secs is: rate = [O2] / t = (0.054 - 0) / 3000 - 0
= 1.80 x 10-5 mol/L•s
(b) The instantaneous rate early in the reaction from 0 to 300 secs is:
rate = [O2] / t = (0.010 - 0) / 300 - 0 = 3.33 x 10-5 mol/L•s
(c) The instantaneous rate late in the reaction from 2400 to 3000 secs
is: rate = [O2] / t = (0.054 - 0.052) / 3000 - 2400 = 3.33 x 10-6 mol/L•s
(d) It can be seen that the early instantaneous rate is the largest since
the concentrations of reactants is highest during the earliest stages of
the reaction and the late instantaneous rate is smallest since the
2/8
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