Chemistry II question....
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Chemistry II question....

[From: ] [author: ] [Date: 12-06-28] [Hit: ]
.please explain the steps thanks :)-The decay equation is m(t) = m0*e^-t/τWhen t = th,½*m0 = m0*e^-th/τ;½ = e^- th/τ -ln(2) = -th/τ so τ = th/ln(2) = 20.4/ln(2) = 29.m(t) = m0*e^-t/29.m0 = 500 µg th = 20.......
Carbon-11 has a half life of 20.4 minutes.Assuming you start with a 500 ug sample of carbon-11 , how much will remain after 24 hours?

The answer is supposed to be 2.82 x10^-19 ug

But I'm not getting it....please explain the steps thanks :)

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The decay equation is m(t) = m0*e^-t/τ When t = th, m(th) = ½*m0

½*m0 = m0*e^-th/τ; ½ = e^- th/τ -ln(2) = -th/τ so τ = th/ln(2) = 20.4/ln(2) = 29.43

m(t) = m0*e^-t/29.43

m0 = 500 µg th = 20.4 min 24 hrs = 1440 min

m(1440) = 500*e^-(1440/29.43) = 2.82*10^-19 µg

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well start with this idea.
half life = divided by 2.
24 hours = 1440 mins.
there are to find how many times carbon-11 goes through half life you'd divide the total(1440mins) by the period(20.4mins). this equals 70.58
so take your beginning amount, 500 and you're going to divide it by 2, 70.58 times. so you'd raise 2 to the 70.58th power, which equals 1.764e+21. then take 500 and divide it by 1.764e+21 which equals 2.83e-19ug of Carbon-11. which whoever gave you the answer didn't use proper sig fig rules. all my numbers are correct
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