Half - Life Chemistry Questions?

julianbato

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11 years ago

3. A meteorite strikes the earth in Western Wyoming. Chemical analysis shows it contains 44.62 kg of radioactive iron-59. How much of this isotope will remain in the meteorite after 310 days? The half - life is 44.3 day.

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Tokii

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11 years ago

So a half life means the mass of the sample will decrease by 50% per every half life. in 310 days there are (310/44.3) 7 half lifes.

44.62/2= 22.31 kg after one half life, the sample goes through 7 half lives so after halving the initial mass 7 times we end up with : 0.349 kg of Iron-59 remains.

EDIT : The other answers are too complicated for such a simple question. Yes they are correct but just use some common sense here.

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tola2001

wrote...

11 years ago

We make use of this formula.
N = Noe^(-kt)

N = final amount
No = original amount
k = first-order rate constant of radioactive isotope
t = time

However, we do not know k yet. We make use of this formula then.
-kt(1/2) = ln(1/2) (Where t(1/2) is the half-life)

Solving for k:
k = [ln(1/2)] / -t(1/2)
k = [ln(1/2)] / -44.3 days
k = 1.56 x 10^-2 / day

We can now solve for the final amount of the radioactive substance.
N = Noe^(-kt)
N = (44.62 kg)(e^[(-1.56 x 10^-2 / day)(310 days)])
N = 0.35 kg (Answer)

Have a good day.

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juliar33

wrote...

11 years ago

Use the first order rate equation:

t = 2.303/k log[a0]/[a]
310 = [(2.303 x t1/2) / 0.693] log[44.62]/[a]

310 = [(2.303 x 44.3)/0.693] log [44.62]/[a]
log[44.62]/[a] = 2.1
[44.62] / [a] = 125.89
[a] = 44.62/125.89 = 0.3544 kgs or 354.4g

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