Calculate the heat required to melt 7.35 g of benzene at its normal melting point. heat of fusion (benzene) = 9.92 kj/mol heat = kj

calculate the heat required to vaporize 7.35 g of benzene at its normal boiling point. heat of vaporization (benzene) = 30.7 kj/mol heat = kj

The heat required to melt 7.35 g of benzene at its normal melting point is 934.8 Joules.

The heat required to vaporize 7.35 g of benzene at its normal melting point is 2,893 Joules.

Explanation:

Mass of benzene = 7.35 g

Moles of benzene =

Heat fusion of benzene,

1) Heat required to melt 7.35 g of benzene at its normal melting point = Q

(1 kJ = 1000 J)

2) Heat vaporization of benzene,

Heat required to vaporize 7.35 g of benzene at its normal melting point = Q

(1 kJ = 1000 J)

i have a feeling this is a gay person trolling to make straights look bad.

ill try to . no promises tho.

explanation: