Asked by LaMetria
The following information is given for ether, C2H5OC2H5, at 1atm:
boiling point = 34.6 °C Hvap(34.6 °C) = 26.5 kJ/mol
specific heat liquid = 2.32 J/g°C
At a pressure of 1 atm,
kJ of heat are needed to vaporize a 41.0 g sample of liquid ether at its normal boiling point of 34.6 °C.
boiling point = 34.6 °C Hvap(34.6 °C) = 26.5 kJ/mol
specific heat liquid = 2.32 J/g°C
At a pressure of 1 atm,
kJ of heat are needed to vaporize a 41.0 g sample of liquid ether at its normal boiling point of 34.6 °C.
Answers
Answered by
DrBob222
I ASSUME the temperature of the liquid ether is at its boiling point of 34.6 C because you didn't list that; however, in problems of this type, especially since you gave the specific heat of the liquid, that the initial T is less than 34.6 C.
q needed to convert liquid ether at its boiling point to vapor at its boiling point is
q = mols ether x H vap = ?
q needed to convert liquid ether at its boiling point to vapor at its boiling point is
q = mols ether x H vap = ?
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