Phosphorus-32 has a half-life of 14.0 days. Starting with 6.00 g of $^{32} P$ $""^32P$ , how many grams will remain after 56.0 days?

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Phosphorus-32 has a half-life of 14.0 days. Starting with 6.00 g of $^{32} P$ $""^32P$ , how many grams will remain after 56.0 days?

Express the answer in numerically in grams.

2 Answers

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Answer 1 · 2018-05-17T06:55:44

$Well what is a half of a half of a half of a half?$ $"Well what is a half of a half of a half of a half?"$

Explanation:

And $56 \cdot days$ $56*"days"$ is FOUR half-lives....and so we got $\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{2^{4}} = \frac{1}{16}$ $1/2xx1/2xx1/2xx1/2=1/2^4=1/16$ of the original mass of $^{32} P$ $""^32P$ ...

And given that the initial quantity was $6.00 \cdot g$ $6.00*g$ ...we are left with $6.00 \cdot g \times \frac{1}{16} = ? ? \cdot g$ $6.00*gxx1/16=??*g$ ....

Answer 2 · 2018-05-17T07:01:33

The mass remaining is $= 0.375 g$ $=0.375g$

Explanation:

![http://chemwiki.ucdavis.edu](https://useruploads.socratic.org/i0nzxwQiR1uX7T9ILCxu_Radioactive_Decay.jpg)

The half life is $t_{\frac{1}{2}} = 14 days$ $t_(1/2)=14 "days"$

The number of half life of $Phosphorus- 32$ $" Phosphorus- 32 "$ in $56$ $56$ days is

$= \frac{t}{t_{\frac{1}{2}}} = \frac{56}{14} = 4 half life$ $=t/t_(1/2)=56/14= 4 " half life "$

The initial mass of $Phosphorus- 32$ $" Phosphorus- 32 "$ is $m_{0} = 6.00 g$ $m_0=6.00g$

After each half life, the mass will decrease by $\times \frac{1}{2}$ $xx1/2$

Therefore,

the mass remaining is $= \frac{6}{16} = \frac{3}{8} = 0.375 g$ $=6/16=3/8=0.375g$

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Phosphorus-32 has a half-life of 14.0 days. Starting with 6.00 g of $^{32} P$ $""^32P$ , how many grams will remain after 56.0 days?

Express the answer in numerically in grams.

2 Answers

Explanation:

Explanation:

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Science

Math

Humanities

... and beyond

Phosphorus-32 has a half-life of 14.0 days. Starting with 6.00 g of 32P""^32P, how many grams will remain after 56.0 days?

Express the answer in numerically in grams.

2 Answers

Explanation:

Explanation:

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Phosphorus-32 has a half-life of 14.0 days. Starting with 6.00 g of $^{32} P$ $""^32P$ , how many grams will remain after 56.0 days?