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Hess's Law Given the following data: (CH3)2CHOH(1) + H2(g)-(CH3)2CH2(g) + H2O(1) AH°=-73.7 kJ (CH3)2CHOH(1) + 1/202(g) 2H2(g) + N2(g) → H4Nz(!) H2(g) + 1/202-H2O(1) calculate AH° for the reaction: (CH3)2CO(1) + H2O(1) AH°--218.4 kJ AH°=50.6 kJ AH-286.0 kJ (CH3)2CO(1) + H4N2(1)(CH3)2CH2(g) + N2(g) + H2O(1) AH= Submit Answer Incorrect. Tries 3/5 Previous Tries Copyright:

Hess's Law Given the following data: (CH3)2CHOH(1) + H2(g)-(CH3)2CH2(g) + H2O(1) AH°=-73.7 kJ (CH3)2CHOH(1) + 1/202(g) 2H2(g) + N2(g) → H4Nz(!) H2(g) + 1/202-H2O(1) calculate AH° for the reaction: (CH3)2CO(1) + H2O(1) AH°--218.4 kJ AH°=50.6 kJ AH-286.0 kJ (CH3)2CO(1) + H4N2(1)(CH3)2CH2(g) + N2(g) + H2O(1) AH= Submit Answer Incorrect. Tries 3/5 Previous Tries Copyright:

Chemistry: Principles and Practice
Chemistry: Principles and Practice
3rd Edition
ISBN:9780534420123
Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Chapter5: Thermochemistry
Section: Chapter Questions
Problem 5.51QE: The enthalpy change when 1 mol methane (CH4) is burned is 890 kJ. It takes 44.0 kJ to vaporize 1 mol...
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Hess's Law Given the following data: (CH3)2CHOH(1) + H2(g)-(CH3)2CH2(g) + H2O(1) AH°=-73.7 kJ (CH3)2CHOH(1) + 1/202(g) 2H2(g) + N2(g) → H4Nz(!) H2(g) + 1/202-H2O(1) calculate AH° for the reaction: (CH3)2CO(1) + H2O(1) AH°--218.4 kJ AH°=50.6 kJ AH-286.0 kJ (CH3)2CO(1) + H4N2(1)(CH3)2CH2(g) + N2(g) + H2O(1) AH= Submit Answer Incorrect. Tries 3/5 Previous Tries Copyright:
Transcribed Image Text:Hess's Law Given the following data: (CH3)2CHOH(1) + H2(g)-(CH3)2CH2(g) + H2O(1) AH°=-73.7 kJ (CH3)2CHOH(1) + 1/202(g) 2H2(g) + N2(g) → H4Nz(!) H2(g) + 1/202-H2O(1) calculate AH° for the reaction: (CH3)2CO(1) + H2O(1) AH°--218.4 kJ AH°=50.6 kJ AH-286.0 kJ (CH3)2CO(1) + H4N2(1)(CH3)2CH2(g) + N2(g) + H2O(1) AH= Submit Answer Incorrect. Tries 3/5 Previous Tries Copyright:
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