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Answer:
a. 5.9 × 10⁻³ M/s
b. 0.012 M/s
Explanation:
Let's consider the following reaction.
2 N₂O(g) → 2 N₂(g) + O₂(g)
a.
Time (t): 12.0 s
Δn(O₂): 1.7 × 10⁻² mol
Volume (V): 0.240 L
We can find the average rate of the reaction over this time interval using the following expression.
r = Δn(O₂) / V × t
r = 1.7 × 10⁻² mol / 0.240 L × 12.0 s
r = 5.9 × 10⁻³ M/s
b. The molar ratio of N₂O to O₂ is 2:1. The rate of change of N₂O is:
5.9 × 10⁻³ mol O₂/L.s × (2 mol N₂O/1 mol O₂) = 0.012 M/s
Answer:
Explanation — This page looks at the oxidation of alcohols using acidified sodium or ... of sodium or potassium dichromate(VI) acidified with dilute sulfuric acid. ... The electron-half-equation for this reaction is as follows: ... To do that, oxygen from an oxidizing agent is represented as [O]. ... Article type: Section or Page.