Pentaborane B5H9(s) burns vigorously in 02 to give B203(s) and H20(1). Calculate AHrxn for the combustion of 5.00 mol of B5Hg. AH°F[B203(s)] =-1,273.5 kJ/mol AH°f[B5H9(s)] = 73.2 kJ/mol AH°f[H2O(1)] = -285.8 kJ/mol Multiple Choice - 8,790 kJ - 45,400 kJ 45,400 kJ - 22,700 kJ
Pentaborane B5H9(s) burns vigorously in 02 to give B203(s) and H20(1). Calculate AHrxn for the combustion of 5.00 mol of B5Hg. AH°F[B203(s)] =-1,273.5 kJ/mol AH°f[B5H9(s)] = 73.2 kJ/mol AH°f[H2O(1)] = -285.8 kJ/mol Multiple Choice - 8,790 kJ - 45,400 kJ 45,400 kJ - 22,700 kJ
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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![Multiple Choice
- 8,790 kJ
- 45,400 kJ
45,400 kJ
- 22,700 kJ
- 9,090 kJ](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fea29189a-484d-4c07-a526-248f9a98b501%2F108dcaf2-b49d-4d69-bf9d-4abdff5f6de3%2F1pvun58_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Multiple Choice
- 8,790 kJ
- 45,400 kJ
45,400 kJ
- 22,700 kJ
- 9,090 kJ
![Pentaborane B5H9(s) burns vigorously in 02 to give B203(s) and H20(1). Calculate AHrxn for the combustion of 5.00 mol of B5H9.
AH°F[B2O3(s)] =-1,273.5 kJ/mol
AH°f[B5H9(s)] = 73.2 kJ/mol
AH°f[H2O(1)] = -285.8 kJ/mol
Multiple Choice
- 8,790 kJ
- 45,400 kJ
45,400 kJ
- 22,700 kJ](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fea29189a-484d-4c07-a526-248f9a98b501%2F108dcaf2-b49d-4d69-bf9d-4abdff5f6de3%2F3ncpss_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Pentaborane B5H9(s) burns vigorously in 02 to give B203(s) and H20(1). Calculate AHrxn for the combustion of 5.00 mol of B5H9.
AH°F[B2O3(s)] =-1,273.5 kJ/mol
AH°f[B5H9(s)] = 73.2 kJ/mol
AH°f[H2O(1)] = -285.8 kJ/mol
Multiple Choice
- 8,790 kJ
- 45,400 kJ
45,400 kJ
- 22,700 kJ
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