document.write( "Question 553517: A sample of phosphorus-32 has a half-life of 14.28 days. If 55 g of this radioisotope remain unchanged after approximately 57 days, what was the mass of the original sample? could someone please help? \n" ); document.write( "

Algebra.Com's Answer #360929 by ankor@dixie-net.com(22740) $\"\"$ $\"About$

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A sample of phosphorus-32 has a half-life of 14.28 days.
\n" ); document.write( "If 55 g of this radioisotope remain unchanged after approximately 57 days, what was the mass of the original sample?
\n" ); document.write( ":
\n" ); document.write( "Using the radioactive decay formula: A = Ao*2^(-t/h), where
\n" ); document.write( "A = resulting amt after t time
\n" ); document.write( "Ao = initial amt (t=0)
\n" ); document.write( "t = time
\n" ); document.write( "h = half-life of substance
\n" ); document.write( ":
\n" ); document.write( "We want to find Ao:
\n" ); document.write( "Ao*2^(-57/14.28) = 55
\n" ); document.write( "using a calc to find 2^(-57/14.28)
\n" ); document.write( ".062865Ao = 55
\n" ); document.write( "Ao = $\"55%2F.062865\"$
\n" ); document.write( "Ao = 875 grams was the initial amt
\n" ); document.write( ":
\n" ); document.write( ":
\n" ); document.write( "We can check this solution on our calculator; enter: 875*2^(-57/14.28) results: 55.00 grams
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