The answer should be (4) they are negative subatomic particles and are found surrounding the nucleus. The nucleus is formed by neutrons and protons. Neutrons are neutral and protons are positive.
Mass of ammonium sulfate = 660.7 g
<h3>Further explanation</h3>
Given
3.01 x 10²⁴ molecules of ammonium sulfate
Required
mass
Solution
The mole is the number of particles(molecules, atoms, ions) contained in a substance
1 mol = 6.02.10²³ particles
Can be formulated
N=n x No
N = number of particles
n = mol
No = Avogadro's = 6.02.10²³
mol ammonium sulfate (NH₄)₂SO₄ :
n = N : No
n = 3.01 x 10²⁴ : 6.02 x 10²³
n = 5
mass ammonium sulfate :
= mol x MW
= 5 x 132,14 g/mol
= 660.7 g
Answer:
Amount of pyridine required = 0.0316 M
Explanation:
pH of a buffer solution is calculated by using Henderson - Hasselbalch equation.
![pH=pK_a+log\frac{[Conjugate\ base]}{[weak\ acid]}](https://tex.z-dn.net/?f=pH%3DpK_a%2Blog%5Cfrac%7B%5BConjugate%5C%20base%5D%7D%7B%5Bweak%5C%20acid%5D%7D)
Pyridinium is a weak acid and in the presence of its conjugate base, it acts as buffer.
Henderson - Hasselbalch equation for pyridine/pyridinium buffer is as follows:
![pH=pK_a+log\frac{[Py]}{PyH^+]}](https://tex.z-dn.net/?f=pH%3DpK_a%2Blog%5Cfrac%7B%5BPy%5D%7D%7BPyH%5E%2B%5D%7D)
pH = 4.7
(Pyridinium)=0.100 M
Substitute the values in the formula
![pH=pK_a+log\frac{[Py]}{PyH^+]}\\4.7=5.2 log\frac{[Py]}{0.100}](https://tex.z-dn.net/?f=pH%3DpK_a%2Blog%5Cfrac%7B%5BPy%5D%7D%7BPyH%5E%2B%5D%7D%5C%5C4.7%3D5.2%20log%5Cfrac%7B%5BPy%5D%7D%7B0.100%7D)
![4.7-5.2=log\frac{[Py]}{0.100} \\-0.5=log\frac{[Py]}{0.100}\\\frac{[Py]}{0.100}=antilog -0.5\\\frac{[Py]}{0.100}=0.316](https://tex.z-dn.net/?f=4.7-5.2%3Dlog%5Cfrac%7B%5BPy%5D%7D%7B0.100%7D%20%5C%5C-0.5%3Dlog%5Cfrac%7B%5BPy%5D%7D%7B0.100%7D%5C%5C%5Cfrac%7B%5BPy%5D%7D%7B0.100%7D%3Dantilog%20-0.5%5C%5C%5Cfrac%7B%5BPy%5D%7D%7B0.100%7D%3D0.316)
![\frac{[Py]}{0.100} =0.316](https://tex.z-dn.net/?f=%5Cfrac%7B%5BPy%5D%7D%7B0.100%7D%20%3D0.316)
[Py]=0.0316\ M
Amount of pyridine required = 0.0316 M
