Answer:
69.8%
Explanation:
k = 0.00813 sec-1
t = 44.1 s
What percentage of the compound has decomposed?
We can obtain this using the formular;
[A] = [A]o e^(-kt)
where;
Final Concentration = [A]
Initial concentration = [A]o
[A] / [A]o = e^(-kt)
[A] / [A]o = e^(-0.00813*44.1)
[A] / [A]o = e^(-0.359)
[A] / [A]o = 0.698
Percentage decomposed = Final Concentration / Initial concentration * 100
Percentage decomposed = [A] / [A]o * 100%
Percentage decomposed = 0.698 * 100 = 69.8%
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