Asked by sammy
The standard molar heat of vaporiza-
tion for water is 40.79 kJ/mol. How
much energy would be required to vaporize
2.54 ×1010 molecules of water?
Answer in units of kJ
tion for water is 40.79 kJ/mol. How
much energy would be required to vaporize
2.54 ×1010 molecules of water?
Answer in units of kJ
Answers
Answered by
DrBob222
40.79 kJ/mol x # moles = ?
mols = 2.54E10/6.02E23
40.79 kJ/mol x (2.54E10/6.02E23) = ?
mols = 2.54E10/6.02E23
40.79 kJ/mol x (2.54E10/6.02E23) = ?
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