Suppose that there is an air sample and a water sample, each of the same mass and each with an initial temperature, Tinitial, of
2 answers:
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Answer : The correct option is, (b) +115 J/mol.K
Explanation :
Formula used :
where,
= change in entropy
= change in enthalpy of vaporization = 40.5 kJ/mol
= boiling point temperature = 352 K
Now put all the given values in the above formula, we get:
Therefore, the standard entropy of vaporization of ethanol at its boiling point is +115 J/mol.K
Answer:
0.2425 M
Explanation:
Given mass = 9.36 g
Molar mass of NaOH = 39.997 g/mol
The formula for the calculation of moles is shown below:
Thus,
Given that:- Volume = 965 mL = 0.965 L ( 1 mL = 0.001 L )
So,
Considering: