Chemistry, 18.07.2019 14:30 demi9646

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?

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Answer from: adam15200031

A. 1104 / 138 = 8 half lives

b after 1104 days amount of polonium in sample = 38 (1/2)^8 = 0.184 g to nearest thousandth of a gram

Answer from: Michcardwell8570

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 :

$k=\frac{0.693}{t_{1/2}}$

$k=\frac{0.693}{138\text{ days}}$

$k=5.022\times 10^{-3}\text{ days}^{-1}$

Now we have to calculate the amount left.

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

$k=\frac{2.303}{t}\log\frac{a}{a-x}$

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:

$5.022\times 10^{-3}=\frac{2.303}{138\text{ days}}\log\frac{100g}{a-x}$

a-x=0.391g

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.

$a=\frac{a_o}{2^n}$

where,

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

a_o = Initial amount of the reactant = 100 g

n = number of half lives = ?

Putting values in above equation, we get:

$0.391=\frac{100}{2^n}$

256=2^n

2^8=2^n

n=8

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

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Suppose a laboratory has a 38 g sample of polonium-210. the half-life of polonium-210 is about 138 d...

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