Explanation:
The given reaction is as follows.

Hence, number of moles of NaOH are as follows.
n = 
= 0.005 mol
After the addition of 25 ml of base, the pH of a solution is 3.62. Hence, moles of NaOH is 25 ml base are as follows.
n = 
= 0.0025 mol
According to ICE table,

Initial: 0.005 mol 0.0025 mol 0 0
Change: -0.0025 mol -0.0025 mol +0.0025 mol
Equibm: 0.0025 mol 0 0.0025 mol
Hence, concentrations of HA and NaA are calculated as follows.
[HA] = 
[NaA] = 
![[A^{-}] = [NaA] = \frac{0.0025 mol}{V}](https://tex.z-dn.net/?f=%5BA%5E%7B-%7D%5D%20%3D%20%5BNaA%5D%20%3D%20%5Cfrac%7B0.0025%20mol%7D%7BV%7D)
Now, we will calculate the
value as follows.
pH = 
![pK_{a} = pH - log \frac{[A^{-}]}{[HA]}](https://tex.z-dn.net/?f=pK_%7Ba%7D%20%3D%20pH%20-%20log%20%5Cfrac%7B%5BA%5E%7B-%7D%5D%7D%7B%5BHA%5D%7D)
= 
= 3.42
Thus, we can conclude that
of the weak acid is 3.42.
Answer:
1.82 L
Explanation:
We are given the following information;
- Initial volume as 2.0 L
- Initial temperature as 60.0°C
- New volume as 30.0 °C
We are required to determine the new volume;
From Charles's law;

Where,
are initial and new volume respectively, while
are initial and new temperatures respectively;



Rearranging the formula;


Therefore, the new volume that would be occupied by the gas is 1.82 L
B) energy is absorbed by the reaction
is right answer.
Answer:
Impressionist paintings are a non-example of electric circuits.
Explanation:
hope this helped