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Answer:
The enthalpy of vaporization of water at 273 K and 1 bar = 44.9 KJ/mol
Explanation:
Enthalpy of vaporization of water at 273 K, ΔHvap(T₂) is given as;
ΔHvap(T₂) = ΔHvap(T₁) + ΔCp * (T₂ - T₁)
where ΔCp = molar heat capacity of gas - molar heat capacity of liquid
Therefore, ΔCp = (33.6 - 75.3) = -41.70 J/(mol K) = 0.0417 kJ/(molK)
substituting ΔCp = 0.0417 kJ/(mol K) in the initial formula
;
ΔHvap(T) = ΔHvap(T1) + ΔCp * (T₂ - T₁)
ΔHvap(T₂)= 40.7 kJ/mol + {-0.0417 kJ/(mol K) * (272 - 373 K)}
ΔHvap(T₂) = 44.9 kJ/mol
Therefore, enthalpy of vaporization of water at 273 K and 1 bar = 44.9kJ/mol
Polar Covalent.
This is because both Calcium (Ca) and Carbon (C) are non metals, thus eliminating the ionic option.
It is polar covalent rather than nonpolar covalent due to the fact that the electrons in a polar covalent bond are shared unequally while those in a nonpolar bond are shared equally.
This can be determined by looking at the amount of valence electrons; if the electrons cancel each other out, it is shared equally. If not, it is shared unequally.
Hope this helps!
