6. Calculate the amount of enengy relcased (QMe/mol) when 1 mole of \'H fuses wi

Question

6. Calculate the amount of enengy relcased (QMe/mol) when 1 mole of 'H fuses with 1 mole of 'H to make one mole of 'He (There is one more product. Figure out what it is.) The required masses are PH: 2014102 amu; 'H: 3.016029 amu; He: 4.002603 am For the following problems, be sure to use an appropriated formula ig half life or intergrade law), not any short cuts How old is a skeleton sample if the current mount of carbon in the bones is 3.125%? (Assume you started with 100%). The half life of carbon is 5730 years 7. 8 lodine-131 has a half life of 8 days. What fraction of the original sample would remain at the end of 32 days? 9. Actinium-226 have a half life of 29 hours. If 10.0 g of actinium-226 disintegrates over a period of 145 hours, how many mg of actinium-226 will remain? 10. The mass of cobalt-60 in a sample is found to have decreased to 0.200 g in a period of 1015.5 years. The half life of cobalt is 425.26 years, what was the initial amount of the sample of cobalt-60? Dang2

Explanation / Answer

The integrated rate law for first order radioactive decay is given as

ln Nt/N0 = -k*t where Nt = amount of the radioactive element remaining after time t, N0 = amount of the element present initially, k is the decay constant and t is the time. For a first order reaction,

k = 0.693/ where = half life of the radioactive element.

7) We have Nt = 3.125%, N0 = 100% and = 5730 y. Plug in values and get

ln (3.125%)/(100%) = -(0.693)/(5730 y)*t

=====> -3.4657 = -(0.00012094 y-1)*t

=====> t = 3.4657/(0.00012094 y-1) = 28656.35852 y 28656 y.

The age of the skeleton is 28656 years (ans).

8) Given = 8 d and t = 32 d, we are required to find out Nt/N0. Plug in values and get

ln Nt/N0 = -(0.693)/(8 d)*(32 d) = -(0.693)*4 = -2.772

====> Nt/N0 = exp(-2.772) = 0.0625

The fraction of I-131 left after 32 days is 0.0625 (ans).

9) The half life is given as = 29 h. We have N0 = 10.0 g and t = 145 h. Plug in values and get

ln Nt/N0 = -(0.693)/(29 h)*(145 h) = -(0.693)*5 = -3.465

====> Nt/N0 = exp(-3.465) = 0.03127

====> Nt/(10.0 g) = 0.03127

====> Nt = 0.03127*(10.0 g) = 0.3127 g = (0.3127 g)*(1000 mg/1 g) = 312.7 mg

The mass of Actinium-226 remaining after 145 hours is 312.7 mg (ans).

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