Question

Calcium-47, used to evaluate bone metabolism, has a half-life of 4.5 days.

Answer 1

Given:

Half life = 4.5 days

use relation between rate constant and half life of 1st order reaction

k = (ln 2) / k

= 0.693/(half life)

= 0.693/(4.5)

= 0.154 days-1

we have:

[A]o = 49 mg

t = 54.0 days

k = 0.154 days-1

use integrated rate law for 1st order reaction

ln[A] = ln[A]o - k*t

ln[A] = ln(49) - 0.154*54

ln[A] = 3.892 - 0.154*54

ln[A] = -4.424

[A] = e^(-4.424)

[A] = 1.198*10^-2 mg

Answer: 1.2*10^-2 mg