Question

Part A how many grams will remain after Phosphorus 32 has a half-life of 14.0 days. Starting with 8.00 g of 840 days? Express your answer numerically in grams. View Available Hints) VALO Submit

Answer 1

Given:

Half life = 14 days

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

k = (ln 2) / k

= 0.693/(half life)

= 0.693/(14)

= 4.95*10^-2 days-1

we have:

[P]o = 8.00 g

t = 84.0 days

k = 4.95*10^-2 days-1

use integrated rate law for 1st order reaction

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

ln[P] = ln(8) - 4.95*10^-2*84

ln[P] = 2.079 - 4.95*10^-2*84

ln[P] = -2.079

[P] = e^(-2.079)

[P] = 0.125 g

Answer: 0.125 g