Hi, you have not provided structure of the aldehyde and alkoxide ion.
Therefore i'll show a mechanism corresponding to the proton transfer by considering a simple example.
Explanation: For an example, let's consider that proton transfer is taking place between a simple aldehyde e.g. acetaldehyde and a simple alkoxide base e.g. methoxide.
The hydrogen atom attached to the carbon atom adjacent to aldehyde group are most acidic. Hence they are removed by alkoxide preferably.
After removal of proton from aldehyde, a carbanion is generated. As it is a conjugated carbanion therefore the negative charge on carbon atom can conjugate through the carbonyl group to form an enolate which is another canonical form of the carbanion.
All the structures are shown below.
Answer:
pH = 12.52
Explanation:
Given that,
The [H+] concentration is 
.
We need to find its pH.
We know that, the definition of pH is as follows :
![pH=-log[H^+]](https://tex.z-dn.net/?f=pH%3D-log%5BH%5E%2B%5D)
Put all the values,
![pH=-log[3\times 10^{-13}]\\\\pH=12.52](https://tex.z-dn.net/?f=pH%3D-log%5B3%5Ctimes%2010%5E%7B-13%7D%5D%5C%5C%5C%5CpH%3D12.52)
So, the pH is 12.52.
Answer:
1.62
Explanation:
From the given information:
number of moles of benzamide 
= 0.58 mole
The molality = 
= 0.6837
Using the formula:
where;
dT = freezing point = 27
l = Van't Hoff factor = 1
kf = freezing constant of the solvent
∴
2.7 °C = 1 × kf × 0.6837 m
kf = 2.7 °C/ 0.6837m
kf = 3.949 °C/m
number of moles of NH4Cl = 
= 1.316 mol
The molality = 
= 1.5484
Thus;
the above kf value is used in determining the Van't Hoff factor for NH4Cl
i.e.
9.9 = l × 3.949 × 1.5484 m
l = 1.62
By direct heating of an element with oxygen : many metals and non-. metals burn rapidly when heated in oxygen or air producing their oxides e.g.
