Answer:
[Acetic acid] = 0.07 M
[Acetate] = 0.13 M
Explanation:
pH of buffer = 5
pKa of acetic acid = 4.76
![pH=p_{Ka} + log\frac{[Salt]}{[Acid]}](https://tex.z-dn.net/?f=pH%3Dp_%7BKa%7D%20%2B%20log%5Cfrac%7B%5BSalt%5D%7D%7B%5BAcid%5D%7D)
Now using Henderson-Hasselbalch equation
![5=4.76 + log\frac{[Acetate]}{[Acetic\;acid]}](https://tex.z-dn.net/?f=5%3D4.76%20%2B%20log%5Cfrac%7B%5BAcetate%5D%7D%7B%5BAcetic%5C%3Bacid%5D%7D)
![log\frac{[Acetate]}{[Acetic\;acid]} = 0.24](https://tex.z-dn.net/?f=log%5Cfrac%7B%5BAcetate%5D%7D%7B%5BAcetic%5C%3Bacid%5D%7D%20%3D%200.24)
....... (1)
It is given that,
[Acetate] + [Acetic acid] = 0.2 M ....... (2)
Now solving both the above equations
[Acetate] = 1.74[Acetic acid]
Substitute the concentration of acetate ion in equation (2)
1.74[Acetic acid] + [Acetic acid] = 0.2 M
[Acetic acid] = 0.2/2.74 = 0.07 M
[Acetate] = 0.2 - 0.07 = 0.13 M
Explanation:
The given cell reaction is as follows.

Hence, reactions taking place at the cathode and anode are as follows.
At anode ; Oxidation-half reaction :
...... (1)
At cathode; Reduction-half reaction :
....... (2)
Hence, balance the half reactions by multiplying equation (1) by 2 and equation (2) by 3.
Therefore, net cell reaction is as follows.

Net reaction: 
Thus, we can conclude that the overall cell reaction is as follows.

The correct answer is the fourth option. The nervous system is the system in the body which can be affected by ionizing radiation. Exposure to this kind of radiation may lead to changes in motor function or neurological disorders.
Answer:
The difference between the energy of the reactants and the energy of the products is called the enthalpy change (∆H) of the reaction. For an exothermic reaction, the enthalpy change is always negative. In an endothermic reaction, the products are at a higher energy than the reactants.
Explanation:
<h3>
Answer:</h3>
812 kPa
<h3>
Explanation:</h3>
- According to Boyle's law pressure and volume of a fixed mass are inversely proportional at constant absolute temperature.
- Mathematically,

At varying pressure and volume;
P1V1=P2V2
In this case;
Initial volume, V1 = 2.0 L
Initial pressure, P1 = 101.5 kPa
Final volume, V1 = 0.25 L
We are required to determine the new pressure;

Replacing the known variables with the values;

= 812 kPa
Thus, the pressure of air inside the balloon after squeezing is 812 kPa