How much heat would be required to convert 5.31 mol of a pure substance from a liquid at 40.0°C to a gas at 113.0°C?

Q = 7.7 KJ

Explanation:

The complete question carries the following data:

Specific Heat Capacity of Liquid State = C(l) = 1.45 J/mol.°C

Specific Heat Capacity of Gaseous State = C(g) = 0.65 J/mol.°C

Boiling Temperature = Tb = 88.5°C

Heat of Vaporization = ΔH(vap) = 1.23 KJ/mol

Now, first we calculate the heat required (Q₁) to raise he temperature to boiling point:

Q₁ = n*C(l)*ΔT₁

where,

n = no. of moles = 5.31 mol

ΔT₁ = Temperature difference from 40°C to Tb = 88.5°C - 40°C = 48.5°C

Therefore,

Q₁ = (5.31 mol)(1.45 J/mol.°C)(48.5°C)

Q₁ = 373.4 J = 0.37 KJ

Now, we calculate the heat required (Q₂) to change its phase:

Q₂ = nΔH(vap)

Q₂ = (5.31 mol)(1.23 KJ/mol)

Q₂ = 6.53 KJ

Now, we calculate the heat required (Q₃) to raise he temperature to boiling point:

Q₃ = n*C(g)*ΔT₃

where,

n = no. of moles = 5.31 mol

ΔT₃ = Temperature difference from Tb to 113°C = 113°C - 88.5°C = 24.5°C

Therefore,

Q₃ = (5.31 mol)(0.65 J/mol.°C)(24.5°C)

Q₃ = 84.5 J = 0.08 KJ

So, the total heat required to convert 5.31 mol of this pure substance from a liquid at 40°C to a gas at 113°C is:

Q = Q₁ + Q₂ + Q₃

Q = 0.37 KJ + 6.53 KJ + 0.08 KJ

Q = 7.7 KJ

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metals are good conductor of electricity. it has free electrons on their surface. according to electron pool theory, each atom present in a metal crystal losses all of its valence electron and as a result a pool of free is form. these pool of electrons are free to move and   conduct electricity easily   while making metal as good conductor.

the correct answer is a. the ability of electrons to flow throughout the metal.

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