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

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Express the answer in numerically in grams.

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Answer 1 · 2018-05-17T06:55:44

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

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

And given that the initial quantity was $6.00*g$ ...we are left with $6.00*gxx1/16=??*g$ ....

Answer 2 · 2018-05-17T07:01:33

The mass remaining is $=0.375g$

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

The half life is $t_(1/2)=14 "days"$

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

$=t/t_(1/2)=56/14= 4 " half life "$

The initial mass of $" Phosphorus- 32 "$ is $m_0=6.00g$

After each half life, the mass will decrease by $xx1/2$

Therefore,

the mass remaining is $=6/16=3/8=0.375g$