How many joules of energy are required to vaporize 13.1 kg of lead at its normal boiling point?

Chemistry

1 answer:

Korolek [52]3 years ago

7 0

Answer: 1123000 Joules of energy are required to vaporize 13.1 kg of lead at its normal boiling point

Explanation:

Latent heat of vaporization is the amount of heat required to convert 1 mole of liquid to gas at atmospheric pressure.

Amount of heat required to vaporize 1 mole of lead = 177.7 kJ

Molar mass of lead = 207.2 g

Mass of lead given = 1.31 kg = 1310 g (1kg=1000g)

Heat required to vaporize 207.2 of lead = 177.7 kJ

Thus Heat required to vaporize 1310 g of lead = $\frac{177.7}{207.2}\times 1310=1123kJ=1123000J$

Thus 1123000 Joules of energy are required to vaporize 13.1 kg of lead at its normal boiling point

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