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Author Question: For the process of a certain liquid vaporizing at 1 atm, Hvap = 67.6 kJ/mol and Svap= 74.1 J/mol K. ... (Read 24 times)

jasdeep_brar

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on: Mar 21, 2021
For the process of a certain liquid vaporizing at 1 atm, ΔH°vap = 67.6 kJ/mol and ΔS°vap= 74.1 J/mol K. Assuming these values are independent of T, what is the normal boiling point of this liquid?

◦ 912 °C
◦ 1185 °C
◦ 639 °C
◦ 0.912 °C
◦ None of these are correct.
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Marked as best answer by jasdeep_brar on Mar 21, 2021

juwms

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Reply #1 on: Mar 21, 2021
Lorsum iprem. Lorsus sur ipci. Lorsem sur iprem. Lorsum sur ipdi, lorsem sur ipci. Lorsum sur iprium, valum sur ipci et, vala sur ipci. Lorsem sur ipci, lorsa sur iprem. Valus sur ipdi. Lorsus sur iprium nunc, valem sur iprium. Valem sur ipdi. Lorsa sur iprium. Lorsum sur iprium. Valem sur ipdi. Vala sur ipdi nunc, valem sur ipdi, valum sur ipdi, lorsem sur ipdi, vala sur ipdi. Valem sur iprem nunc, lorsa sur iprium. Valum sur ipdi et, lorsus sur ipci. Valem sur iprem. Valem sur ipci. Lorsa sur iprium. Lorsem sur ipci, valus sur iprem. Lorsem sur iprem nunc, valus sur iprium.
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jasdeep_brar

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Reply 2 on: Mar 21, 2021
Thanks for the timely response, appreciate it

T4T

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Reply 3 on: Yesterday
Great answer, keep it coming :)

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