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Author Question: When 6.000 moles of H2(g) reacts with 3.000 mol of O2(g) to form 6.000 mol of H2O(l) at 25C and a ... (Read 134 times)

JMatthes

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on: Feb 15, 2020

Question 1

When 2.00 mol of benzene is vaporized at a constant pressure of 1.00 atm and at its normal boiling point of 80.1°C, 67.8 kJ are absorbed and PΔV for the vaporization process is equal to 5.80 kJ then
◦ ΔE = 62.0 kJ and ΔH = 67.8 kJ.
◦ ΔE = 67.8 kJ and ΔH = 62.0 kJ.
◦ ΔE = 67.8 kJ and ΔH = 73.6 kJ.
◦ ΔE = 73.6 kJ and ΔH = 67.8 kJ.

Question 2

When 6.000 moles of H2(g) reacts with 3.000 mol of O2(g) to form 6.000 mol of H2O(l) at 25°C and a constant pressure of 1.00 atm. If 409.8 kJ of heat are released during this reaction, and PΔV is equal to -22.20 kJ, then
◦ ΔH° = -409.8 kJ and ΔE° = -432.0 kJ.
◦ ΔH° = -409.8 kJ and ΔE° = -387.6 kJ.
◦ ΔH° = +409.8 kJ and ΔE° = +432.0 kJ.
◦ ΔH° = +409.8 kJ and ΔE° = +387.6 kJ.
Chemistry ¦ McMurry, Fay, Robinson ¦ 7th Edition
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Marked as best answer by JMatthes on Feb 15, 2020

nyrave

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Reply #1 on: Feb 15, 2020
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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JMatthes

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Reply 2 on: Feb 15, 2020
Excellent

adammoses97

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Reply 3 on: Yesterday
:D TYSM

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