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Author Question: Given: 4 NO2(g) + O2(g) 2 N2O5(g)H = -110.2 kJfind H for 8 N2O5(g) 16 NO2(g) + 4 O2(g). (Read 53 times)

CORALGRILL2014

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Question 1

10.0 g of a metal, initially at 25°C, are placed into 10.0 g of water, initially at 100°C. Which metal will have the highest final temperature? Shown after each metal is its specific heat in J/(g∙°C).
◦ calcium (0.650)
◦ sodium (1.23)
◦ nickel (0.440)
◦ lead (0.160)

Question 2

Given: 4 NO2(g) + O2(g) → 2 N2O5(g)ΔH° = -110.2 kJ
find ΔH° for 8 N2O5(g) → 16 NO2(g) + 4 O2(g).
◦ 440.8 kJ
◦ -220.4 kJ
◦ 220.4 kJ
◦ -440.8 kJ


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Marked as best answer by CORALGRILL2014 on Feb 15, 2020

al

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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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CORALGRILL2014

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Reply 2 on: Feb 15, 2020
Gracias!


komodo7

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

 

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