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%

answer: the rate of diffusion increases with increase in temperature.

increase in temperature, increases the molecular/particles speed (kinetic energy. diffusion means movement of particles through the material.

generally, the heavier the particle the more it interacts with the environment and more the temperature matters. so higher temperature results in higher diffusion.

56 moles of   lieter in chemistry