The molarity of solution made by diluting 26.5ml of 6.0ml hno3 to a volume of 250ml is calculated using the following formula
M1V1 = M2V2, where
M1 = molality 1 (6.00m)
V1= volume 1 (26.5 ml)
M2 = molarity 2(?)
v2=volume 2 (250)
M2 = M1V1/V2
M2= 6 x26.5/250 = 0.636 M
Answer:
ΔU=-369.2 kJ/mol.
Explanation:
We start from the equation:
Δ(H)=ΔU+Δ(PV), which is an extension of the well known relation: H=U+PV.
If Δ(PV) were calculated by ideal gas law,
PV=nRT
Δ(PV)=RTΔn.
Where Δn is the change of moles due to the reaction; but, this reaction does not give a moles change (Four moles of HCl produced from 4 moles of reactants), so Δ(PV)=0.
So, for this case, ΔH=ΔU.
The enthalpy of reaction given is for one mole of reactant, so the enthalpy of reaction for the reaction of interest must be multiplied by two:

ΔU=-369.2 kJ/mol.
Explanation:
The given data is as follows.
Moles of propylene = 100 moles,
= 300 K
= 800 K,
,
of propylene = 100 J/mol
Now, we assume the following assumptions:
Since, it is a compression process therefore, work will be done on the system. And, work done will be equal to the heat energy liberating without any friction.
W = 

= 
= 5 MJ
Thus, we can conclude that a minimum of 5 MJ work is required without any friction.
When the concentration is expressed in molality, it is expressed in moles of solute per kilogram of solvent. Since we are given the mass of the solvent, which is water, we can compute for the moles of solute NaNO3.
0.5 m = x mol NaNO3/0.5 kg water
x = 0.25 mol NaNO3
Since the molar mass of NaNO3 is 85 g/mol, the mass is
0.25 mol * 85 g/mol = 21.25 grams NaNO3 needed
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.