Will Upvote!!! Determine the correct IUPAC name for this branched chain hydrocarbon.
1 answer:
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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>
Explanation:
It is known that of propionic acid = 4.87
And, initial concentration of propionic acid =
= 0.158 M
Concentration of sodium propionate = \frac{0.26}{1.20}[/tex]
= 0.216 M
Now, in the given situation only propionic acid and sodium propionate are present .
Hence, pH =
=
= 4.87 + log (1.36)
= 5.00
- Therefore, when 0.02 mol NaOH is added then,
Moles of propionic acid = 0.19 - 0.02
= 0.17 mol
Hence, concentration of propionic acid =
= 0.14 M
and, moles of sodium propionic acid = (0.26 + 0.02) mol
= 0.28 mol
Hence, concentration of sodium propionic acid will be calculated as follows.
= 0.23 M
Therefore, calculate the pH upon addition of 0.02 mol of NaOH as follows.
pH =
=
= 4.87 + log (1.64)
= 5.08
Hence, the pH of the buffer after the addition of 0.02 mol of NaOH is 5.08.
- Therefore, when 0.02 mol HI is added then,
Moles of propionic acid = 0.19 + 0.02
= 0.21 mol
Hence, concentration of propionic acid =
= 0.175 M
and, moles of sodium propionic acid = (0.26 - 0.02) mol
= 0.24 mol
Hence, concentration of sodium propionic acid will be calculated as follows.
= 0.2 M
Therefore, calculate pH upon addition of 0.02 mol of HI as follows.
pH =
=
= 4.87 + log (0.114)
= 4.98
Hence, the pH of the buffer after the addition of 0.02 mol of HI is 4.98.