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iris [78.8K]
3 years ago
8

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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Answer:

A = -213.09°C

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Explanation:

Given data:

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According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

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2)

V₁/T₁ = V₂/T₂

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3)

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T₂ = 0.0875 L × 273 K / 5.00 L

T₂ = 23.8875 L.K / 5.00 K

T₂ = 4.78 K

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