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iren2701 [21]
3 years ago
9

A sample of phosphorus-32 has a half-life of 14.28 days. If 55 g of this radioisotope remain unchanged after approximately 57 da

ys, what was the mass of the original sample?
Chemistry
2 answers:
Flura [38]3 years ago
7 0
55= No (1/2)^55/57
55= No (1/2)^3.9
55= No (1/2)^4
55= No (1/16)
No= 880 g
Julli [10]3 years ago
6 0

Answer: The mass of the original sample is 12.82\times 10^{7} grams.

Explanation:

Half-life of sample of phosphorus-32 = 14.28 days

\lambda =\frac{0.693}{t_{\frac{1}{2}}}=\frac{0.693}{14.28}=0.0485 day^{-1}

N=N_o\times e^{-\lambda t}

lnN=lnN_o-\lambda t

N = 55 g, t= 57 days

\ln[55 g]=-0.0485 day^{-1}\times 57 days+ln[N_o]

\log\frac{55 g}{N_o}=2.303\times (-0.0485 day^{-1})\times 57 days

\frac{55 g}{N_o}=antilog[-6.3666]

N_o=\frac{55 g}{4.29\times 10^{-7}}=12.82\times 10^{7} grams

The mass of the original sample is 12.82\times 10^{7} grams.


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