Question

32P is a radioactive isotope with a half-life of 14.3 days. If you currently have 82.7 g of 32P, how much 32P was present 5.00 days ago?

Please show detailed steps

Answer #1

K = 0.693/t1/2

= 0.693/14.3

= 0.04846days^-1

K = 2.303/t log[A0]/[A]

0.04846 = 2.303/5 log82.7/[A]

log82.7/[A] = 0.04846*5/2.303

log82.7/[A] = 0.1052

82.7/[A] = 10^0.1052

82.7/[A] = 1.274

[A] = 82.7/1.274 = 65g >>>>answer

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in grams

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1. Given that a freshly prepared radioactive isotope has a
half-life of 10 days, the percentage of it remaining after 30 days
is
A
30.0 %.
B
10.0 %.
C
12.5 %.
D
72.5 %.
2. During a second half-life, the original
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A
25%.
B
50%.
C
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sample containing this isotope has an initial activity at (t=0) of
1.50e-12 Bq. Calculate the number of nuclei that will decay in the
time interval between t1=10 hours and t2=20 hours
Answer is 4.60e16 but I'm not sure how. Thanks and please show
work

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Please show and explain work, and do not use calculus to solve
it.

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