Answer:
pH = 6.82
Explanation:
To solve this problem we can use the<em> Henderson-Hasselbach equation</em>:
- pH = pKa + log
![\frac{[NaOCl]}{[HOCl]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BNaOCl%5D%7D%7B%5BHOCl%5D%7D)
We're given all the required data to <u>calculate the original pH of the buffer before 0.341 mol of HCl are added</u>:
- pKa = -log(Ka) = -log(2.9x10⁻⁸) = 7.54
- [HOCl] = [NaOCl] = 0.500 mol / 0.125 L = 4 M
- pH = 7.54 + log
By adding HCl, w<em>e simultaneously </em><u><em>increase the number of HOCl</em></u><em> and </em><u><em>decrease NaOCl</em></u>:
- pH = 7.54 + log
![\frac{[NaOCl-HCl]}{[HOCl+HCl]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BNaOCl-HCl%5D%7D%7B%5BHOCl%2BHCl%5D%7D)
- pH = 7.54 + log
Moles of ammonium sulfide = 5.80 mol
The formula of ammonium sulfide is (NH₄)₂S
So each molecule of ammonium sulfide has (4 x 2) or 8 atoms of H
One mole of ammonium sulfide has 8 moles of H
5.80 mol of ammonium sulfide has (8 x 5.8) or 46.4 moles of H
As per the definition of Avogadro's number, 1 mole = 6.022 x 10²³ atoms
46.4 moles of H x (6.022 x 10²³ atoms/ 1 mole of H)
= 2.8 x 10²⁵ H atoms
Therefore, 2.8 x 10²⁵ H atoms are in 5.80 mol of ammonium sulfide.
Answer:
- <u>b. three half-lives</u>
Explanation:
The number of half-lives elapsed, n, is calculated dividing the time by the half-life time:
- n = time / half-life time
<u>a. A sample of Ce-141 with a half-life of 32.5 days after 32.5 days</u>
- n = 32.5 days / 32.5 days = 1 half-life
<u />
<u>b. A sample of F-18 with a half-life of 110 min after 330 min</u>
- n = 330 min / 110 min = 3 half-lives
<u />
<u>c. A sample of Au-198 with a half-life of 2.7 days after 5.4 days</u>
- n = 5.4 days / 2.7 days = 2 half-lives
As we know that
P.E. = mgh
where,
P.E. = Potential energy of the object =?
m= mass of object= 3kg
g= acceleration due to gravity = 9.8 ms^-2
h = height between object and animal = 0 m
Then
P.E. = 3× 9.8 × 0 = 0 Joules or 0J
<em>Have a luvely day!</em>
