Answer:
standard entropy of vaporization of ethanol = 142.105 J/K-mol
Explanation:
given data
enthalpy of vaporization of ethanol = 40.5 kJ/mol = 40.5 × J/mol
entropy of vaporization of ethanol boiling point = 285 K
to find out
standard entropy of vaporization of ethanol
solution
we get here standard entropy of vaporization of ethanol that is expess as
standard entropy of vaporization of ethanol ΔS = .............1
here ΔH is enthalpy of vaporization of ethanol and T is temperature
put value in equation 1
standard entropy of vaporization of ethanol ΔS =
standard entropy of vaporization of ethanol = 142.105 J/K-mol
The answer is D.1.950 g
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Answer:
i believe its A but im not sure
its the only one that makes any sense
plz mark brainliest
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Explanation:
Answer:
Explanation:
We know,
where, R = 0.0821 L.atm/(mol.K), T is temperature in kelvin and is difference in sum of stoichiometric coefficient of products and reactants
Here and T = 311 K
So,
Hence value of equilibrium constant in terms of partial pressure is