The data in the table below were obtained for the reaction: 2H2(g) + O2(g) → 2H₂O(g) Expt. # [H₂] (M) 1 2 3 0.373 0.746 0.373 [0₂] (M) 0.539 1.078 1.078 Initial Rate (M/s) 0.147 0.589 0.294 Based on the data in the table above, when the [H₂] =1.3 M and [0₂] =2.2 M, then the initial rate is M/s.
The data in the table below were obtained for the reaction: 2H2(g) + O2(g) → 2H₂O(g) Expt. # [H₂] (M) 1 2 3 0.373 0.746 0.373 [0₂] (M) 0.539 1.078 1.078 Initial Rate (M/s) 0.147 0.589 0.294 Based on the data in the table above, when the [H₂] =1.3 M and [0₂] =2.2 M, then the initial rate is M/s.
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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![The data in the table below were obtained for the reaction:
2H2(g) + O2(g) → 2H₂O(g)
Expt. # [H₂] (M) [0₂] (M)
1
2
3
0.373
0.746
0.373
0.539
1.078
1.078
Initial Rate (M/s)
0.147
Please record your answer as in 1.2 M/s.
0.589
0.294
Based on the data in the table above, when the [H₂] =1.3 M and [0₂] =2.2 M, then the
initial rate is
M/s.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F69f1e81e-5408-438e-8a66-8963279f1243%2F2142053c-0e51-4519-b329-ce7d92fd0bbb%2Fdnxrc1_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The data in the table below were obtained for the reaction:
2H2(g) + O2(g) → 2H₂O(g)
Expt. # [H₂] (M) [0₂] (M)
1
2
3
0.373
0.746
0.373
0.539
1.078
1.078
Initial Rate (M/s)
0.147
Please record your answer as in 1.2 M/s.
0.589
0.294
Based on the data in the table above, when the [H₂] =1.3 M and [0₂] =2.2 M, then the
initial rate is
M/s.
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