Asked by Andrew
The combustion of 1.00 mol of sucrose,C12H22,O11, evolves 5650 kJ of heat. A bomb calorimeter has a calibrated heat capacity of 1.23 kj/degree Celsius. How many grams of sucrose should be burned to raise the temperature of the calorimeter and its contents from 23 degrees C to 80 degrees C?
Answers
Answered by
DrBob222
How much heat is need to raise the T of the calorimeter and its contents from 23 to 80 C?
1.23 kJ/oC x (80-23) = 70.11 kJ.
So burning 1 mole sucrose (342 grams is it--check me out on that) produces 5650 kJ.
342 g x (70.11/5650) = ?? g sucrose. Check my thinking.
1.23 kJ/oC x (80-23) = 70.11 kJ.
So burning 1 mole sucrose (342 grams is it--check me out on that) produces 5650 kJ.
342 g x (70.11/5650) = ?? g sucrose. Check my thinking.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.