The half-life for the following first order reaction is 688 hours at 10000C. Calculate the specific rate constant, k, at 10000C and the amount of a 3.0 g sample of CS2 that remains after 48 hours. g

2.858g

Explanation:

First order reaction.

Half life (t1/2) = 688

Specific rate constant, K = ?

The relationship between t1/2 and K id given as;

t1/2 = 0.693 / k

k = 0.693 / t1/2

k = 0.693 / 688 = 0.0010073 hr-1

Amount of 3.0 g sample of CS2 that remains after 48 hours.

Integrated rate law for a first order reaction is given as;

ln[A] = ln[A]o - kt

ln[A] = ln[3] - 0.0010073(48)

ln[A] = 1.05026

[A] = 2.858g

your answers are a,b,,e and f

i hope this

answer; i believe that the correct answer is (a) potassium(k)