Suppose a laboratory has a 38 g sample of polonium-210. the half-life of polonium-210 is about 138 days. a. how many half-lives of polonium-210 occur in 1104 days? b. how much polonium is in the sample 1104 days later?

Answer :

(a) The number of half-lives of polonium-210 are, 8

(b) The amount left in the sample after 1104 days will be, 0.391 grams.

Explanation :

First we have to determine the amount left in the sample after 1104 days.

This is a type of radioactive decay and all radioactive decays follow first order kinetics.

To calculate the rate constant, we use the formula :

Now we have to calculate the amount left.

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

where,

k = rate constant

t = time taken for decay process

a = initial amount or moles of the reactant

a - x = amount or moles left after decay process

Putting values in above equation, we get:

The amount left in the sample after 1104 days will be, 0.391 grams.

Now we have to calculate the number of half-lives of polonium-210.

where,

a = amount of reactant left after n-half lives = 0.391 g

= Initial amount of the reactant = 100 g

n = number of half lives  = ?

Putting values in above equation, we get:

Therefore, the number of half-lives of polonium-210 are, 8

in chemistry, a solution’s concentration is how much of a dissolvable substance, known as a solute, is mixed with another substance, called the solvent. the standard formula is c = m/v, where c is the concentration, m is the mass of the solute dissolved, and v is the total volume

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