Answer:
25.89 × 10²³ molecules
Explanation:
Given data:
Mass of CoCl₂ = 560 g
Number of molecules present = ?
Solution:
Number of moles of CoCl₂:
Number of moles = mass/molar mass
Number of moles = 560 g/ 129.84 g/mol
Number of moles = 4.3 mol
Avogadro number:
The given problem will solve by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance. The number 6.022 × 10²³ is called Avogadro number.
1 mole = 6.022 × 10²³ molecules
4.3 mol × 6.022 × 10²³ molecules /1 mol
25.89 × 10²³ molecules
Explanation:
As it is given that solubility of water in diethyl ether is 1.468 %. This means that in 100 ml saturated solution water present is 1.468 ml.
Hence, amount of diethyl ether present will be calculated as follows.
(100ml - 1.468 ml)
= 98.532 ml
So, it means that 98.532 ml of diethyl ether can dissolve 1.468 ml of water.
Hence, 23 ml of diethyl ether can dissolve the amount of water will be calculated as follows.
Amount of water = 
= 0.3427 ml
Now, when magnesium dissolves in water then the reaction will be as follows.

Molar mass of Mg = 24.305 g
Molar mass of
= 18 g
Therefore, amount of magnesium present in 0.3427 ml of water is calculated as follows.
Amount of Mg =
= 0.462 g
<u>Answer:</u> The pH of the buffer is 5.25
<u>Explanation:</u>
Let the volume of buffer solution be V
We know that:

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:
![pH=pK_a+\log(\frac{[\text{conjugate base}]}{[acid]})](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%28%5Cfrac%7B%5B%5Ctext%7Bconjugate%20base%7D%5D%7D%7B%5Bacid%5D%7D%29)
We are given:
= negative logarithm of acid dissociation constant of weak acid = 4.90
![[\text{conjugate base}]=\frac{2.25}{V}](https://tex.z-dn.net/?f=%5B%5Ctext%7Bconjugate%20base%7D%5D%3D%5Cfrac%7B2.25%7D%7BV%7D)
![[acid]=\frac{1.00}{V}](https://tex.z-dn.net/?f=%5Bacid%5D%3D%5Cfrac%7B1.00%7D%7BV%7D)
pH = ?
Putting values in above equation, we get:

Hence, the pH of the buffer is 5.25
It would contain 0.105M moles of NaOH in each liter of solution