A sample originally contained 1.28 g of a radioisotope. It now contains 1.12 g of its daughter isotope.
2 answers:
The answer is 3.
<span>The relation between number of half-lives (n) and decimal amount remaining (x) can be expressed as:
</span>
We need to calculate n, but we need x to do that. To calculate what p<span>ercentage of a radioactive species would be found as daughter material, we must calculate what amount remained:
1.28 -</span> 1.12 = 0.16
If 1.28 is 100%, how much percent is 0.16:
1.28 : 100% = 0.16 : x
x = 12.5%
Presented as decimal amount:
x = 0.125
Now, let's implement this in the equation:
<span>
</span>
Because of the exponent, we will log both sides of the equation:
<span>
</span>
Therefore, 3 half-lives have passed <span> since the sample originally formed.</span>
<span>The relation between number of half-lives (n) and decimal amount remaining (x) can be expressed as:
</span>
We need to calculate n, but we need x to do that. To calculate what p<span>ercentage of a radioactive species would be found as daughter material, we must calculate what amount remained:
1.28 -</span> 1.12 = 0.16
If 1.28 is 100%, how much percent is 0.16:
1.28 : 100% = 0.16 : x
x = 12.5%
Presented as decimal amount:
x = 0.125
Now, let's implement this in the equation:
<span>
</span>
Because of the exponent, we will log both sides of the equation:
<span>
</span>
Therefore, 3 half-lives have passed <span> since the sample originally formed.</span>
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Answer: a) The concentration after 8.8min is 0.17 M
b) Time taken for the concentration of cyclopropane to decrease from 0.25M to 0.15M is 687 seconds.
Explanation:
Expression for rate law for first order kinetics is given by:
where,
k = rate constant
t = age of sample
a = let initial amount of the reactant
a - x = amount left after decay process
a) concentration after 8.8 min:
b) for concentration to decrease from 0.25M to 0.15M