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Anarel [89]
3 years ago
15

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
Chemistry
1 answer:
grigory [225]3 years ago
3 0

Answer:

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 = \frac{7.35 g}{78 g/mol}=0.09423 mol

Heat fusion of benzene,\Delta H_{fus} = 9.92 kJ/mol

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

Q=\Delta H_{fus}\times 0.09423 mol

=9.92 kJ/mol\times 0.09423 mol=0.9348 kJ=934.8 J

(1 kJ = 1000 J)

2) Heat vaporization of benzene,\Delta H_{vap} = 30.7 kJ/mol

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

Q=\Delta H_{Vap}\times 0.09423 mol

=30.7 kJ/mol\times 0.09423 mol=2.893 kJ=2,993 J

(1 kJ = 1000 J)

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