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Author Question: Consider the following processes:2A (1/2)B + CH1 = 5 kJ/mol(3/2)B + 4C 2A + C + 3DH2 = 15 kJ/molE ... (Read 42 times)

NClaborn

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on: Mar 21, 2021

Question 1

​
Consider the following processes:
 
ΔH (kJ/mol)
3B → 2C + D    
–125.
(1/2)A → B
150
E + A → D
350
Calculate ΔH for: B → E + 2C

​


◦ 325 kJ/mol
◦ 525 kJ/mol
◦ –175 kJ/mol
◦ –325 kJ/mol
◦ none of these are correct

Question 2

Consider the following processes:
2A → (1/2)B + C
ΔH1 = 5 kJ/mol
(3/2)B + 4C → 2A + C + 3D
ΔH2 = –15 kJ/mol
E + 4A → C
ΔH3 = 10 kJ/mol
Calculate ΔH for:     C → E + 3D


◦ 0 kJ/mol
◦ 10 kJ/mol
◦ –10 kJ/mol
◦ –20 kJ/mol
◦ 20 kJ/mol
Chemistry: An Atoms First Approach ¦ Zumdahl, DeCoste ¦ 3rd Edition
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Marked as best answer by NClaborn on Mar 21, 2021

shewald78

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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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NClaborn

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Reply 2 on: Mar 21, 2021
Excellent

parker125

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Reply 3 on: Yesterday
Gracias!

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