"Sodium-24 has a half-life of 15 hours. How much of a 30g sample will remain undecayed after one day?"
I know it's really easy but have completely forgotten how to do these, working out would be appreciated :D cheers
I know it's really easy but have completely forgotten how to do these, working out would be appreciated :D cheers
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Simplest is to graph since you are only dealing with 2 significant figures.
0.30 15,15 30, 7.5 45, 3.75 and find the vale for 24. Remeber to subtract this from the original 30 g.
0.30 15,15 30, 7.5 45, 3.75 and find the vale for 24. Remeber to subtract this from the original 30 g.
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ln(A0/A)=kt, where A0 is the initial amount and A is the current amount remaining at time t
Now t(1/2)=0.693/k
or k=0.693/t(1/2)=0.693/15=0.0462
now no. of moles in 30g sample=30/24=5/4=1.25 moles
So
ln(1.25/A)=0.0462*24 (because 1 day=24 hours)
so ln(1.25/A)=1.1088
hence 1.25/A=3.031
and A=1.25/3.031=0.412 moles
so amount remaining=24*0.412=9.89g
Now t(1/2)=0.693/k
or k=0.693/t(1/2)=0.693/15=0.0462
now no. of moles in 30g sample=30/24=5/4=1.25 moles
So
ln(1.25/A)=0.0462*24 (because 1 day=24 hours)
so ln(1.25/A)=1.1088
hence 1.25/A=3.031
and A=1.25/3.031=0.412 moles
so amount remaining=24*0.412=9.89g
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The relevant time equation is
Amount = Amount at start x exp( - k * time) where k is a specific constant.
To find k we need the fact that
15 = 30*exp(-15k) which gives a k of ln(1/2)/-15
Thus one day = 24 hours ==> A = 30*exp( -( ln(1/2)/-15 )*24)
==> A = 9.896g
Amount = Amount at start x exp( - k * time) where k is a specific constant.
To find k we need the fact that
15 = 30*exp(-15k) which gives a k of ln(1/2)/-15
Thus one day = 24 hours ==> A = 30*exp( -( ln(1/2)/-15 )*24)
==> A = 9.896g