How much energy is needed to vaporize 75.0 g of diethyl ether (c4h10o) at its boiling point (34.6°c), given that δhvap of diethyl ether = 26.5 kj/mol?

26.8 kJ of energy is needed to vaporize 75.0 g of diethyl ether

Explanation:

First we have to calculate the moles of diethyl ether

As, 1 mole of diethyl ether require heat = 26.5 kJ

So, 1.01  moles of diethyl ether require heat =

Thus 26.8 kJ of energy is needed to vaporize 75.0 g of diethyl ether

the atomic number will stay the same no matter how many electrons it has, so it doesn’t decrease nor increase. it’ll only change with a different number of protons