Asked by Nancy
A 1.50 liter gas sample at 745mm HG and 25 celsius contains 3.55% radon by volume radon-220 is an alpha emitter with a half life of 55.6 seconds. how many alpha particles are emitted in 5 minutes by this sample?
I got an answer 1.59x10^19
I started at 745/760=.980263
then
.980263(1.50) =.0821n (298.15)
n =.0600 moles
.06006986x.0355=.002132 x 6.022 x10^23
=1.28x10^21
K=.012464
1.28x10^21x.012464= 1.59 x10^19 then I multiplied by 300 seconds but my book is giving me the answer 1.24x10^21 which is not what I have. Please help and explain why this answer
I got an answer 1.59x10^19
I started at 745/760=.980263
then
.980263(1.50) =.0821n (298.15)
n =.0600 moles
.06006986x.0355=.002132 x 6.022 x10^23
=1.28x10^21
K=.012464
1.28x10^21x.012464= 1.59 x10^19 then I multiplied by 300 seconds but my book is giving me the answer 1.24x10^21 which is not what I have. Please help and explain why this answer
Answers
Answered by
DrBob222
If I do this I found
n = 0.06010 and that times 0.0355 = 0.002134 mols Rn-220.
ln(0.002134/N) = 0.01246*5min x (60s/min)
N = mols Rn after 5 min =
N = 5.08E-5 mols
How much has been used in the 5 min?
That's 0.002134-5.08E-5 = 0.00208
0.00208E x 6.02E23 = 1.25E21
You need to go back through my work and watch the significant figures. Probably the subtraction step will clear up the small difference between the book answer and my answer.
n = 0.06010 and that times 0.0355 = 0.002134 mols Rn-220.
ln(0.002134/N) = 0.01246*5min x (60s/min)
N = mols Rn after 5 min =
N = 5.08E-5 mols
How much has been used in the 5 min?
That's 0.002134-5.08E-5 = 0.00208
0.00208E x 6.02E23 = 1.25E21
You need to go back through my work and watch the significant figures. Probably the subtraction step will clear up the small difference between the book answer and my answer.
Answered by
Nancy
Yes this does help! Thanks!
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