5 5. Calculate the change in enthalpy of formation when dihydrogen sulfide gas is bubbled through abasic solution, given that the heat of formation of each is: H2S = -20.6 kJ/mol, KOH = -482.37 kJ/mol,H2O = -285.8 kJ/mol, K2S = -406.2 kJ/mol.%3D%3D%3DH2S (g) +КОН (аq) —H2O (e) +K2S (aq)->

Answered: Image /qna-images/answer/d2e394bc-3b6e-4f05-b16d-c01e3632abbd.jpg

is bubbled through a 5. Calculate the change in enthalpy of formation when dihydrogen sulfide basic solution, given that the heat of formation of each is: H2S = -20.6 kJ/mol, KOH = -482.37 kJ/mol, H2O = -285.8 kJ/mol, K2S = -406.2 kJ/mol. gas H2S (g) КОН (аq) Н.О (е) + K2S (aq)

is bubbled through a 5. Calculate the change in enthalpy of formation when dihydrogen sulfide basic solution, given that the heat of formation of each is: H2S = -20.6 kJ/mol, KOH = -482.37 kJ/mol, H2O = -285.8 kJ/mol, K2S = -406.2 kJ/mol. gas H2S (g) КОН (аq) Н.О (е) + K2S (aq)

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Question

5. Calculate the change in enthalpy of formation when dihydrogen sulfide gas is bubbled through a
basic solution, given that the heat of formation of each is: H2S = -20.6 kJ/mol, KOH = -482.37 kJ/mol,
H2O = -285.8 kJ/mol, K2S = -406.2 kJ/mol.
%3D
%3D
%3D
H2S (g) +
КОН (аq) —
H2O (e) +
K2S (aq)
->

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