anonymous
  • anonymous
a mummy is found to be 3940 years old by ^14C dating. Using half life of 5730 years for ^14C demonstrate that 62.1% of ^14C at the death of the person remains the mummy
Chemistry
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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aaronq
  • aaronq
For 1st order reactions you can use these formulas: find the decay constant \[t _{1/2}=\frac{ \ln2 }{ k }\] t1/2 = half-life then use the general exponential decay/growth equation: \[A _{t}=A _{0}*e ^{-kt}\] Ao=initial amount At= amount after time elapsed t=time elapsed k=decay constant start off by assuming you have 100% 14C, i.e. Ao=100
anonymous
  • anonymous
ok then what
anonymous
  • anonymous
.125=e^-k(5730)

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aaronq
  • aaronq
you set it up wrong
aaronq
  • aaronq
find k first
anonymous
  • anonymous
how?
aaronq
  • aaronq
with the first equation i wrote
anonymous
  • anonymous
what would At be though
aaronq
  • aaronq
okay rewind a little bit, its asking you to prove that 62.1% remains of the mummy. now you can do this several ways, find what At is assuming all other variables OR use 62.1% for At and find the initial amount Ao.
anonymous
  • anonymous
im sorry I still don't understand if Ao equals 100 then what would At and t be
aaronq
  • aaronq
well if you're using Ao as 100% then you're solving for At if you're using At as 62.1% then you're solving for Ao
aaronq
  • aaronq
t is time elapsed so 3940 years
anonymous
  • anonymous
what is the k value we are using
aaronq
  • aaronq
you found it in the first equation using the half life
anonymous
  • anonymous
ok then I got 1.21x10^-4 for k
aaronq
  • aaronq
use it in the other equation
anonymous
  • anonymous
i did and i got .621 which is 62.1% right?
aaronq
  • aaronq
yep, in decimal, 0.621 is 62.1%
anonymous
  • anonymous
ok thanks so much do you have time for one more?
aaronq
  • aaronq
no problem, yeah, what is it?
anonymous
  • anonymous
nevermind i figured it out thanks so much i was so screwed because i have test tomorrow
aaronq
  • aaronq
okay, good luck on your test !

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