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Sagot :
To determine how long it takes for the amount of argon-44 to decrease from 28.0 mg to 3.50 mg, we will follow these steps:
1. Identify the half-life of argon-44:
The half-life of argon-44 is given as 12 minutes.
2. Determine the initial and final amounts:
The initial amount of argon-44 is 28.0 mg.
The final amount is 3.50 mg.
3. Calculate the number of half-lives required:
We need to find out how many half-lives are required to reduce the amount of argon-44 from 28.0 mg to 3.50 mg.
The relationship between the initial amount (A[tex]\(_0\)[/tex]), the final amount (A), and the number of half-lives (n) is given by the formula:
[tex]\[ A = A_0 \times \left( \frac{1}{2} \right)^n \][/tex]
Solving for [tex]\( n \)[/tex]:
[tex]\[ \frac{A}{A_0} = \left( \frac{1}{2} \right)^n \][/tex]
[tex]\[ \frac{3.5}{28.0} = \left( \frac{1}{2} \right)^n \][/tex]
[tex]\[ \frac{1}{8} = \left( \frac{1}{2} \right)^n \][/tex]
[tex]\(\frac{1}{8}\)[/tex] can be expressed as [tex]\( \left( \frac{1}{2} \right)^3\)[/tex], so:
[tex]\[ \left( \frac{1}{2} \right)^n = \left( \frac{1}{2} \right)^3 \][/tex]
Hence, [tex]\( n = 3 \)[/tex].
4. Calculate the total time:
If one half-life is 12 minutes and it takes 3 half-lives to reach the final amount, the total time is:
[tex]\[ \text{Total time} = n \times \text{half-life} \][/tex]
[tex]\[ \text{Total time} = 3 \times 12 = 36 \text{ minutes} \][/tex]
Therefore, it will take 36 minutes for the amount of argon-44 to decrease from 28.0 mg to 3.50 mg.
1. Identify the half-life of argon-44:
The half-life of argon-44 is given as 12 minutes.
2. Determine the initial and final amounts:
The initial amount of argon-44 is 28.0 mg.
The final amount is 3.50 mg.
3. Calculate the number of half-lives required:
We need to find out how many half-lives are required to reduce the amount of argon-44 from 28.0 mg to 3.50 mg.
The relationship between the initial amount (A[tex]\(_0\)[/tex]), the final amount (A), and the number of half-lives (n) is given by the formula:
[tex]\[ A = A_0 \times \left( \frac{1}{2} \right)^n \][/tex]
Solving for [tex]\( n \)[/tex]:
[tex]\[ \frac{A}{A_0} = \left( \frac{1}{2} \right)^n \][/tex]
[tex]\[ \frac{3.5}{28.0} = \left( \frac{1}{2} \right)^n \][/tex]
[tex]\[ \frac{1}{8} = \left( \frac{1}{2} \right)^n \][/tex]
[tex]\(\frac{1}{8}\)[/tex] can be expressed as [tex]\( \left( \frac{1}{2} \right)^3\)[/tex], so:
[tex]\[ \left( \frac{1}{2} \right)^n = \left( \frac{1}{2} \right)^3 \][/tex]
Hence, [tex]\( n = 3 \)[/tex].
4. Calculate the total time:
If one half-life is 12 minutes and it takes 3 half-lives to reach the final amount, the total time is:
[tex]\[ \text{Total time} = n \times \text{half-life} \][/tex]
[tex]\[ \text{Total time} = 3 \times 12 = 36 \text{ minutes} \][/tex]
Therefore, it will take 36 minutes for the amount of argon-44 to decrease from 28.0 mg to 3.50 mg.
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