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if the heat of combustion of glucose is -2808 kJ/mol, how many grams of glucose are required to supply 1650 kJ of heat?
Nov 12, 2012

Answers

2808 kJ/mol = 2808 kJ/180g.

So (2808 kJ/180g) x ?g = 1650 kJ.
Solve for ?g.
Nov 12, 2012

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