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Author Question: The second-order decomposition of NO2 has a rate constant of 0.255 L mol-1 s-1. How much NO2 ... (Read 118 times)

robinn137

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on: Nov 18, 2019

Question 1

The first-order decomposition of N2O5 at 328 K has a rate constant of 1.70 × 10-3 s-1. If the initial concentration of N2O5 is 2.88 mol L-1, what is the concentration of N2O5 after 12.5 minutes?
◦ 2.82 mol L-1
◦ 0.355 mol L-1
◦ 0.174 mol L-1
◦ 0.124 mol L-1
◦ 0.805 mol L-1

Question 2

The second-order decomposition of NO2 has a rate constant of 0.255 L mol-1 s-1. How much NO2 decomposes in 4.00 s if the initial concentration of NO2 (1.00 L volume) is 1.33 mol L-1?
◦ 0.56 mol
◦ 1.8 mol
◦ 0.77 mol
◦ 0.85 mol
◦ 0.48 mol
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Marked as best answer by robinn137 on Nov 18, 2019

dyrone

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Reply #1 on: Nov 18, 2019
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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robinn137

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Reply 2 on: Nov 18, 2019
Excellent

alexanderhamilton

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Reply 3 on: Yesterday
Great answer, keep it coming :)

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