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Author Question: Given that fH [Ag2S(s)] = -32.6 kJ/mole, what is the enthalpy change for the precipitation of ... (Read 11 times)

rlane42

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on: Jul 8, 2021
Given that ΔfH° [Ag2S(s)] = -32.6 kJ/mole, what is the enthalpy change for the precipitation of argentite, Ag2S(s)?

ΔfH° [Ag+(aq)] = +105.6 kJ/mol
ΔfH° [S2-(aq)] = +33.05 kJ/mol

◦ -276.9 kJ/mol
◦ -106.1 kJ/mol
◦ +171.2 kJ/mol
◦ +106.1 kJ/mol
◦ +138.7 kJ/mol
General Chemistry: Principles and Modern Applications ¦ Petrucci, Herring, Madura, Bissonnette ¦ 11th Edition
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Marked as best answer by rlane42 on Jul 8, 2021

jazzlynnnnn

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Reply #1 on: Jul 8, 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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rlane42

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Reply 2 on: Jul 8, 2021
Wow, this really help

raili21

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Reply 3 on: Yesterday
Thanks for the timely response, appreciate it

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