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:
The description including its given problem is outlined in the following section on the clarification.
Explanation:
The given values are:
RBCC = 0.12584 nm
RFCC = 0.12894 nm
The unit cell edge length (ABCC) as well as the atomic radius (RBcc) respectively connected as measures for BCC (α-phase) structure:
√3 ABCC = 4RBCC
⇒ ABCC = 
⇒ = 
⇒ = 
Likewise AFCC as well as RFCC are interconnected by
√2AFCC = 4RFCC
⇒ AFCC = 
⇒ = 
⇒ = 
Now,
The Change in Percent Volume,
= 
= 
= 
= 
= 
Note: percent = %
Answer:
No, it is not.
Explanation:
Most solutions do not behave ideally. Designating two volatile substances as A and B, we can consider the following two cases:
Case 1: If the intermolecular forces between A and B molecules are weaker than those between A molecules and between B molecules, then there is a greater tendency for these molecules to leave the solution than in the case of an ideal solution. Consequently, the vapor pressure of the solution is greater than the sum of the vapor pressures as predicted by Raoult’s law for the same concentration. This behavior gives rise to the positive deviation.
Case 2: If A molecules attract B molecules more strongly than they do their own kind, the vapor pressure of the solution is less than the sum of the vapor pressures as predicted by Raoult’s law. Here we have a negative deviation.
The benzene/toluene system is an exception, since that solution behaves ideally.
To determine the Keq, we need the chemical reaction in the system. In this case it would be:
CO + 2H2 = CH3OH
The Keq is the ration of the amount of the product and the reactant. We use the ICE table for this. We do as follows:
CO H2 CH3OH
I .42 .42 0
C -0.13 -2(0.13) 0.13
-----------------------------------------------
E = .29 0.16 0.13
Therefore,
Keq = [CH3OH] / [CO2] [H2]^2 = 0.13 / 0.29 (0.16^2)
Keq = 17.51
Answer:
the enthalpy of the second intermediate equation is halved and has its sign changed.
Explanation:
Let us take a look at the first and second intermediate reactions as well as the overall reaction equation for the process under review;
First reaction;
Ca (s) + CO₂ (g) + ½O₂ (g) → CaCO₃ (s) ΔH₁ = -812.8 kJ
Second reaction;
2Ca (s) + O₂ (g) → 2CaO (s) ΔH₂ = -1269 kJ
Hence the overall equation is now;
CaO (s) + CO₂ (g) → CaCO₃ (s) ΔH = ?
According to the Hess law of constant heat summation, the enthalpy of the overall reaction is supposed to be obtained as a sum of the enthalpy of both reactions but this will not give the enthalpy of the overall reaction in this case. The enthalpy of the overall reaction is rather obtained by halving the enthalpy of the second intermediate reaction and reversing its sign before taking the sum as shown below;
Enthalpy of Intermediate reaction 1 + ½(- Enthalpy of Intermediate reaction 2) = Enthalpy of Overall reaction