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?

Question

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#

chemwiki.ucdavis.edu

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#