Asked by catalina
The molar enthalpy of vaporization for water
is 40.79 kJ/mol. Express this enthalpy of
vaporization in joules per gram.
b. The molar enthalpy of fusion for water is
6.009 kJ/mol.How much energy is absorbed when 3.03 g of liquid water boils.
is 40.79 kJ/mol. Express this enthalpy of
vaporization in joules per gram.
b. The molar enthalpy of fusion for water is
6.009 kJ/mol.How much energy is absorbed when 3.03 g of liquid water boils.
Answers
Answered by
DrBob222
a.
40.79 kJ/mol x (1000 J/kJ) x (1 mol/18.01 g) = ?
b.
Something is missing. I don't understand what the enthalpy of fusion has to do with boiling water. Are you starting with ice? liquid water? what's the temperature of the ice or liquid water. The amout of energy absorbed depends upon what you have initially.
40.79 kJ/mol x (1000 J/kJ) x (1 mol/18.01 g) = ?
b.
Something is missing. I don't understand what the enthalpy of fusion has to do with boiling water. Are you starting with ice? liquid water? what's the temperature of the ice or liquid water. The amout of energy absorbed depends upon what you have initially.
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.