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Dafna1 [17]
1 year ago
6

131i has a half-life of 8.04 days. assuming you start with a 1.53 mg sample of 131i, how many mg will remain after 13.0 days ___

_______?
a.0.835
b.0.268
c.0.422
d.0.440
e.0.499
Chemistry
1 answer:
inn [45]1 year ago
8 0
For this problem we can use half-life formula and radioactive decay formula.

Half-life formula,
t1/2 = ln 2 / λ

where, t1/2 is half-life and λ is radioactive decay constant.
t1/2 = 8.04 days

Hence,         
8.04 days    = ln 2 / λ                         
λ   = ln 2 / 8.04 days

Radioactive decay law,
Nt = No e∧(-λt)

where, Nt is amount of compound at t time, No is amount of compound at  t = 0 time, t is time taken to decay and λ is radioactive decay constant.

Nt = ?
No = 1.53 mg
λ   = ln 2 / 8.04 days = 0.693 / 8.04 days
t    = 13.0 days 

By substituting,
Nt = 1.53 mg e∧((-0.693/8.04 days) x 13.0 days))
Nt = 0.4989 mg = 0.0.499 mg

Hence, mass of remaining sample after 13.0 days = 0.499 mg

The answer is "e"

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The question is missing. Here is the complete question.

Which balanced redox reaction is ocurring in the voltaic cell represented by the notation of Al_{(s)}|Al^{3+}_{(aq)}||Pb^{2+}_{(aq)}|Pb_{(s)}?

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(c)Al^{3+}_{(aq)}+Pb_{(s)} ->Al_{(s)}+Pb^{2+}_{(aq)}

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For the cell notation: Al_{(s)}|Al^{3+}_{(aq)}||Pb^{2+}_{(aq)}|Pb_{(s)}

Aluminum's half-equation is oxidation:

Al_{(s)} -> Al^{3+}_{(aq)}+3e^{-}

For Lead, half-equation is reduction:

Pb^{2+}_{(aq)}+2e^{-} -> Pb_{(s)}

Multiply first half-equation for 2 and second half-equation by 3:

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Adding them:

2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}

The balanced redox reaction with cell notation Al_{(s)}|Al^{3+}_{(aq)}||Pb^{2+}_{(aq)}|Pb_{(s)} is

2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}

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