anonymous
  • anonymous
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?
Chemistry
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SOLVED
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jamiebookeater
  • jamiebookeater
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Kainui
  • Kainui
A half-life is the amount of time it takes for half of the total isotope to decay. So if you know that 55g are left after 57 days and it decreases the total amount by half every 14.28 days, can you set up an equation that describes this?
anonymous
  • anonymous
thank you
Kainui
  • Kainui
You're welcome, if you need more help just ask =D

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More answers

anonymous
  • anonymous
wait is it 13.75 g
Kainui
  • Kainui
Nope, it can't be simply because you ended up with 55 g of the sample in the end. How could you have 13.75 g decaying away to become more than what you started with?
anonymous
  • anonymous
ok so i didnt understand how to work this out sorry
anonymous
  • anonymous
3.99g
Kainui
  • Kainui
You're on the right track, but not quite there. 3.99 is not the mass in grams, it is the number of half-lives that the sample has gone through! So now that you know how many times it has split in half, maybe that will help you out more?
anonymous
  • anonymous
I am not sure where to go from here is it mass to mole
Kainui
  • Kainui
Nope, no moles are involved in this problem. I'll type out a better explanation, give me a sec haha.
anonymous
  • anonymous
OK PLEASE, I HAVE STRUGGLED WITH THIS FOR 2 DAYS AND HAVE SO MUCH MORE TO DO.
Kainui
  • Kainui
So you know that the sample has been around for 57 days and the half life is 14.28 days. That must mean that 57/14.28 days. That gave you your 3.99 number of half-lives (not moles) So you have 1/2 to the 3.99 power because you have 1/2 times itself as many times as you have gone through a half-life. 55g = X.5^(57/14.28) Divide 55 by 1/2 to the 57/14.28 power and you should get something like 880g
Kainui
  • Kainui
I rounded 3.99 to 4 when I did my calculation, so your answer will be 873.92g if you use 3.99 instead. Hope that helps.
anonymous
  • anonymous
thank you your the best!!!!

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