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umka21 [38]
3 years ago
6

When 5.00 mol of benzene is vaporized at a constant pressure of 1.00 atm and at its normal boiling point of 80.1°C, 169.5 kJ are

absorbed and PΔV for the vaporization process is equal to 14.5 kJ.
What is the change of internal energy?


What is the change of enthalpy?


Is the change in entropy positive or negative for this process?


If the pressure would be only ½ atm, what are the changes of internal energy and enthalpy during vaporization (approximately)?


How much energy is absorbed?


Write the definition of the Gibbs Free Energy.


At the lower pressure, will the boiling point be higher or lower.

Chemistry
1 answer:
NNADVOKAT [17]3 years ago
3 0

Answer:

∆E=155kJ

Explanation:

Check attachment

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Explanation:

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Fe_2O_3+3CO(g)\rightarrow 2Fe+3CO_2

To calculate the moles, we use the equation:

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