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Author Question: For a particular process that is carried out at constant pressure, q = 145 kJ and w = -35 kJ. Therefore: (Read 167 times)

karateprodigy

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Question 1

Determine the enthalpy change in the following equation:

Cl2(g) + H2O(l) → 2 HCl(g) + 1/2 O2(g)
Δf H2O(l) = -285.8 kJ/mol
Δf HCl(g) = -92.31 kJ/mol

◦ 193.5 kJ/mol
◦ -470.4 kJ/mol
◦ 470.4 kJ/mol
◦ 101.2 kJ/mol
◦ -101.2 kJ/mol

Question 2

For a particular process that is carried out at constant pressure, q = 145 kJ and w = -35 kJ. Therefore:
◦ ΔU = 110 kJ and ΔH = 145 kJ
◦ ΔU = 145 kJ and ΔH = 110 kJ
◦ ΔU = 145 kJ and ΔH = 180 kJ
◦ ΔU = 180 kJ and ΔH = 145 kJ


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Marked as best answer by karateprodigy on Jul 8, 2021

irishcancer18

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karateprodigy

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Reply 2 on: Jul 8, 2021
Excellent


Dnite

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Reply 3 on: Yesterday
:D TYSM

 

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