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Answer:
0.296 mg
Explanation:
8.00/2.90 half lives have passed, so this means the remaining mass is
[tex]2.00 {(0.5)}^{8.00 \div 2.90} [/tex]
which is about 0.296 mg.
The amount of polonium remaining after 8.00 years is 0.296 mg. This is obtained by using rate law of first order kinetics.
The expression for the rate law for first order kinetics is
k = [tex]\frac{2.303}{t}log\frac{a}{a-x}[/tex] years⁻¹
Where,
k - rate constant
t - time passed by the sample
a - initial amount of the reactant
a-x - final amount of the reactant
The expression for the half-life of the compound is
[tex]t_{\frac{1}{2}}=\frac{0.693}{k}[/tex] years
Given that,
A Polonium-208 is an alpha emitter with a half-life of 2.90 years.
I.e., [tex]t_{\frac{1}{2}}[/tex] = 2.90 years
Then,
2.90 = (0.693)/k
⇒ k = (0.693)/(2.90) = 0.2389 years⁻¹
Then,
for t = 8.00 years,
Since we have rate law of first order kinetics as
k = [tex]\frac{2.303}{t}log\frac{a}{a-x}[/tex]
⇒ k = (2.303/8.00) log(a/a-x)
Where the initial amount of the reactants is a = 2.00 mg
So,
k = 0.287 log (2.00/a-x) (a-x is the final amount of the reactants)
⇒ 0.238 = 0.287 log(2.00/a-x)
⇒ (0.238/0.287) = log(2.00/a-x)
⇒ 0.829 = log(2.00/a-x)
⇒ (2.00/a-x) = [tex]10^{0.829}[/tex] = 6.745
⇒ (a-x) = 2.00/6.745 = 0.296 mg
Therefore, the remaining polonium is 0.296 mg after 8.00 years.
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