Welcome to Westonci.ca, where your questions are met with accurate answers from a community of experts and enthusiasts. Join our Q&A platform and get accurate answers to all your questions from professionals across multiple disciplines. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.

In a first order decomposition, the constant is 0.00586 sec-1. what percentage of the compound is left after 3.32 minutes?

Sagot :

The correct answer for percentage of the compound is left after 3.32 minutes is 55 % .

We use the formula A(t) = A0e-kt for decomposition, where A0 is the initial amount of decomposed substance at t = 0, k is the decay constant, t is the time elapsed, and A(t) is the amount after time t.

The percentage of decomposed compound left after a time t is, [A(t)], divided the the initial amount, [A0], time 100, or A/A0*100:

A = A0e-kt

A/A0 = e^- 0.00586t, [k = 0.00299, divide both sides by A0]

A/Ao = e^(-0.00299*199), [t = 3.32 min*(60 sec/ 1 min) =199 sec]

A/A0 = . 0.551= 55%, (* 100, rounded)

The Arrhenius equation is k = Ae(-Ea/kT), where A is the frequency or pre-exponential factor and e(-Ea/RT) is the fraction of collisions at temperature T that have enough energy to overcome the activation barrier (i.e., have energy greater than or equal to the activation energy Ea).

Learn more about percentage decomposition here :-

https://brainly.com/question/21438582

#SPJ4