Question

How much cesium (half-life = 2 years) would remain from a 10 g sample after

4 years?
O A. Og
OB. 8g
O c. 59
O D. 29
SUBMIT

Answer 1

Answer: 2.5 grams

Step-by-step explanation:

Expression for rate law for first order kinetics is given by:

`t=(2.303)/(k)\log(a)/(a-x)`

where,

k = rate constant

t = age of sample

a = let initial amount of the reactant

a - x = amount left after decay process

a) to calculate the rate constant:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

`t_{(1)/(2)}=(0.693)/(k)`

`k=(0.693)/(2years)=0.346years^(-1)`

b) to find amount left after 4 years

`4=(2.303)/(0.346)\log(10)/(a-x)`

`(a-x)=2.5g`

Thus 2.5 g of cesium would remain from a 10 g sample after 4 years