Dinitrogen pentoxide decomposes to form nitrogen dioxide oxygen, following the equation 2N2O3 - 4NO, + 0,. At
a certain timepoint, N2O5 is being consumed at a rate of 0.1 M/s. What's the rate of production of NO, and O, at the
same timepoint?
A. NO, is produced at 0.2 M/s, and O, is produced at 0.05 M/s.
B. NO, is produced at 0.2 M/s, and O, is produced at 0.2 M/s.
C. NO, is produced at 0.05 M/s, and O2 is produced at 0.2 M/s.
D. NO, is produced at 0.1 M/s, and O, is produced at 0.05 M/s.

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znk znk · Answer 1 · 2020-06-27T23:12:52+02:00

Answer:

A. N₂O₅ is produced at 0.2 mol·L⁻¹s⁻¹, and O₂ is produced at 0.05 mol·L⁻¹s⁻¹.

Explanation:

2N₂O₅ ⟶ 4NO₂ + O₂

For every two molecules of N₂O₅ that disappear, we get four molecules of NO₂ and one of O₂.

NO₂ is forming twice as fast, and O₂ is forming half as fast, as N₂O₅ is decomposing.

N₂O₅ is decomposing at a rate of 0.1 mol·L⁻¹s⁻¹.

So, N₂O₅ is produced at 0.2 mol·L⁻¹s⁻¹, and O₂ is produced at 0.05 mol·L⁻¹s⁻¹.

chase1posey410 chase1posey410 · Answer 2 · 2021-09-19T01:50:25+02:00

Answer:

NO2 is produced at 0.2 M/s, and O2 is produced at 0.05 M/s.

Explanation:

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