<h2>Acetic Acid + Sodium ethoxide ⇄ Butyric Acid + Sodium Hydroxide</h2>
Explanation:
An ionic equation for the reaction of acetic acid with sodium ethoxide is as follows -
Acetic Acid and Sodium ethanolate (sodium ethoxide)
Butyric Acid and Sodium hydroxide
Hence,
Acetic Acid + Sodium ethoxide ⇄ Butyric Acid + Sodium Hydroxide
⇄ 
- Weak acids and bases have low energy than strong acids and bases.
- The chemical equilibria shift the reaction side with the species having lower energy.
- Given reaction is an acid-base reaction in which the equilibrium favors the starting material that means it will go to the side of the weakest acid that is acetic acid is weaker than butyric acid.
Answer:
a) [A⁻]/[HA] = 0.227
b) [A⁻]/[HA] = 0.991
c) [A⁻]/[HA] = 2.667
Explanation:
In the Henderson-Hasselbalch equation, HA stands from an acid an A⁻ stands from its conjugate base, as follows:
pH = pka + Log [A⁻]/[HA]
pH = 4.874 + Log[CH₃CH₂CO₂⁻]/[CH₃CH₂CO₂H]
4.23 = 4.874 + Log [A⁻]/[HA]
-0.644 = Log [A⁻]/[HA]
= [A⁻]/[HA]
0.227 = [A⁻]/[HA]
4.87 = 4.874 + Log [A⁻]/[HA]
-0.004 = Log [A⁻]/[HA]
= [A⁻]/[HA]
0.991 = [A⁻]/[HA]
5.30 = 4.874 + Log [A⁻]/[HA]
0.426 = Log [A⁻]/[HA]
= [A⁻]/[HA]
2.667 = [A⁻]/[HA]
Solution :
Comparing the solubility of silver chromate for the solutions :
----- Less soluble than in pure water.
----- Less soluble than in pure water.
----- Similar solubility as in the pure water
----- Similar solubility as in the pure water
The silver chromate dissociates to form :
When 0.1 M of 
is added, the equilibrium shifts towards the reverse direction due to the common ion effect of 
, so the solubility of 
decreases.
Both 
and 
are neutral mediums, so they do not affect the solubility.
C. Accelerating should be the correct onw
3 L will be the final volume for the gas as per Charle's law.
Answer:
Explanation:
The kinetic theory of gases has two significant law which forms the backdrop of motion of gases. They are Charle's law and Boyle's law. As per Charle's law, the volume of any gas molecule at constant pressure is directly proportional to the temperature of the molecule.
V∝ T
Since, here two volumes are given and at two different temperatures with constant pressure. Then as per Charle's law, the relation between the volumes of air at different temperature will be
So in this case, V1 = 6 L and T1 = 80° C. Similarly, T2 = 40° C. So we have to determine the V2.
So, 3 L will be the final volume for the gas as per Charle's law.
