In a first order decomposition, the constant is 0.00813 sec-1. What percentage of the compound has decomposed after 44.1 seconds

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%

your answer is a. the enthalpy of formation for a pure element in its standard state is always positive.