22 You have the following acids and their conjugate bases available: Acetic Acid: CH 3COOH/CH3COO-; Ka = 1.8 × 10-5 Carbonic Aci
1 answer:
You might be interested in
The answer is <span>(3) 3 × 12.4 hours
</span>
To calculate this, we will use two equations:
where:
<span>n - number of half-lives
</span>x - remained amount of the sample, in decimals
<span>
- half-life length
</span>t - total time elapsed.
First, we have to calculate x and n. x is <span>remained amount of the sample, so if at the beginning were 16 grams of potassium-42, and now it remained 2 grams, then x is:
2 grams : x % = 16 grams : 100 %
x = 2 grams </span>× 100 percent ÷ 16 grams
x = 12.5% = 0.125
Thus:
<span>
</span>
It is known that the half-life of potassium-42 is 12.36 ≈ 12.4 hours.
Thus:
<span>
</span><span>
</span>
Therefore, it must elapse 3 × 12.4 hours <span>before 16 grams of potassium-42 decays, leaving 2 grams of the original isotope</span>
</span>
To calculate this, we will use two equations:
where:
<span>n - number of half-lives
</span>x - remained amount of the sample, in decimals
<span>
</span>t - total time elapsed.
First, we have to calculate x and n. x is <span>remained amount of the sample, so if at the beginning were 16 grams of potassium-42, and now it remained 2 grams, then x is:
2 grams : x % = 16 grams : 100 %
x = 2 grams </span>× 100 percent ÷ 16 grams
x = 12.5% = 0.125
Thus:
<span>
</span>
It is known that the half-life of potassium-42 is 12.36 ≈ 12.4 hours.
Thus:
<span>
</span><span>
</span>
Therefore, it must elapse 3 × 12.4 hours <span>before 16 grams of potassium-42 decays, leaving 2 grams of the original isotope</span>