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SVEN [57.7K]
3 years ago
15

Determine the time required for an original 24.0-gram sample of sr-94 to decay until only 1.5 grams of the sample remains unchan

ged.

Chemistry
2 answers:
postnew [5]3 years ago
8 0
For the answer to the question above,
<span>24 --> 12 --> 6 --> 3 --> 1.5 
.....1 .......2 ......3 ......4 ............. four half-lives 

The time is 4(t½) ...... where t½ is the half-life. 

There are two types of problems dealing with half-life. One is the "super-simple" which is what you have here and the solution is also simple. It involves counting half-lives. 

A more realistic problem wouldn't work out so nicely and would involve this equation or a variation since radioactive decay is a first-order process. 

A = Ao e^(-kt) ...... where A is the activity (or amount) after some time (t), Ao is the original activity (or amount) at t=0, k is the decay constant and t is the elapsed time. The decay constant is related to the half-life. t½=ln2/k or k=ln2/(t½). 

A = Ao e^(-kt) 
ln A = -kt + lnAo 
ln(A/Ao) = -kt 
t = (ln(A/Ao)) / -k 
t = ln(1.5/24) / -k 
t = ln(0.0625) / -k 
t = -2.773 / -k 
t = -2.773 / -ln2 / (t½) 
t = -2.773 / (-0.693 / (t½)) 
t = 4(t½)</span>
Bess [88]3 years ago
7 0

Answer: Check explanation.

Explanation:

In this question, we are to calculate for the time required for an original mass of a sample to to decay to another mass of the sample.

We can alive this by using the equation below;

N(t) = N(o) × (1/2)^t/t-half life------------------------------------------------------------(1).

Where N(t/) is the amount left after time t, which is given as 1.5 gram from the question.

N(o) is the initial amount of the substance, this is given as 24 gram.

t-half life= half life if the decaying substance. We make an assumption that the Half life is 10 years.

t= is the time required for an original mass of a sample to to decay to another mass of the sample.

Therefore, 1.5 g/24 g= (1/2)^t/10 years.

=> 0.625 = (1/2)^t/10.

==> t/10 log 0.5 = log 0.625.

==> t = 6.781 years.

Check attached picture for the exclusive solution.

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<h3>Further explanation</h3>

1.

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