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THERMODYNAMICS: The combustion of 1.00 mol of glucose (CGH1206) liberates 2820 kj of heat. If 1.38 g of glucose is burned in a calorimeter containing 801 g of water, and the temperature of the assembly increases from 20.10°C to 24.21°C, what is the heat capacity of the calorimeter in kj/°C? MW of glucose is 180 g/mol. (Round off the final answer to TWo decimal places). Round your answer to 2 decimal places.

THERMODYNAMICS: The combustion of 1.00 mol of glucose (CGH1206) liberates 2820 kj of heat. If 1.38 g of glucose is burned in a calorimeter containing 801 g of water, and the temperature of the assembly increases from 20.10°C to 24.21°C, what is the heat capacity of the calorimeter in kj/°C? MW of glucose is 180 g/mol. (Round off the final answer to TWo decimal places). Round your answer to 2 decimal places.

Chemistry: An Atoms First Approach
Chemistry: An Atoms First Approach
2nd Edition
ISBN:9781305079243
Author:Steven S. Zumdahl, Susan A. Zumdahl
Publisher:Steven S. Zumdahl, Susan A. Zumdahl
Chapter7: Chemical Energy
Section: Chapter Questions
Problem 59E: In a coffee-cup calorimeter, 50.0 mL of 0.100 M AgNO3 and 50.0 mL of 0.100 M HCl are mixed to yield...
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THERMODYNAMICS: The combustion of 1.00 mol of glucose (CGH1206) liberates 2820 kJ of heat. If 1.38 g of glucose is burned in a calorimeter containing 801 g of water, and the temperature of the assembly increases from 20.10°C to 24.21°C, what is the heat capacity of the calorimeter in kJ/°C? MW of glucose is 180 g/mol. (Round off the final answer to TWO decimal places). Round your answer to 2 decimal places. Add your answer
Transcribed Image Text:THERMODYNAMICS: The combustion of 1.00 mol of glucose (CGH1206) liberates 2820 kJ of heat. If 1.38 g of glucose is burned in a calorimeter containing 801 g of water, and the temperature of the assembly increases from 20.10°C to 24.21°C, what is the heat capacity of the calorimeter in kJ/°C? MW of glucose is 180 g/mol. (Round off the final answer to TWO decimal places). Round your answer to 2 decimal places. Add your answer
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