The heat of vaporization of alcohol is 879 J/g. What is the energy in J required to vaporize 4.50 g of C_2H_5OH?

May 27, 2017

This simply the product $\Delta = \Delta {H}_{\text{vaporization"xx"amount of alcohol.}}$ $\cong 4000 \cdot J$.

Explanation:

Practically, this process described the process:

$\text{Ethanol(l)"+Deltararr"Ethanol(g)}$

The ethanol would have to be at its normal boiling point.........

And so, we calculate..................

4.50*gxx879*J*g^-1=??*J