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%

the positions which points up in upward carbon and down in downward carbon are axial

and vice versa holds true for equatorial positions

so the correct answer with labelled diagram will be