Consider the chemical equation below. Upper M g (upper O upper H) subscript 2 plus 2 upper H upper N upper O subscript 3 right a
2 answers:
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Answer:
All of the above.
Explanation:
In positive deviation from Raoult's Law occur when the vapour pressure of components is greater than what is expected value in Raoult's law.
When a solution is non ideal then it shows positive or negative deviation.
Let two solutions A and B to form non- ideal solutions.let the vapour pressure of component A is and vapour pressure of component B is
.
= Vapour pressure of component A in pure form
= Vapour pressure of component B in pure form
=Mole fraction of component A
=Mole fraction of component B
The interaction between A- B is less than the interaction A- A and B-B interaction.Therefore, the escaping tendency of liquid molecules in mixture is greater than the escaping tendency in pure form.Hence, the vapour pressure of a mixture is greater than the initial value of vapour pressure.
,
Therefore,
Therefore, the enthalpy of mixing is greater than zero and change in volume is greater than zero.
Hence, option a,b,c and d are true.
Answer:
See explanation and picture below
Explanation:
First, in the case of methyloxirane (Also known as propilene oxide) the mechanism that is taking place there is something similar to a Sn2 mechanism. Although a Sn2 mechanism is a bimolecular substitution taking place in only step, the mechanism followed here is pretty similar after the first step.
In both cases, the H atom of the HBr goes to the oxygen in the molecule. You'll have a OH⁺ in both. However, in the case of methyloxirane the next step is a Sn2 mechanism step, the bromide ion will go to the less substitued carbon, because the methyl group is exerting a steric hindrance. Not a big one but it has a little effect there, that's why the bromide will rather go to the carbon with more hydrogens. and the final product is formed.
In the case of phenyloxirane, once the OH⁺ is formed, the next step is a Sn1 mechanism. In this case, the bond C - OH⁺ is opened on the side of the phenyl to stabilize the OH. This is because that carbon is more stable than the carbon with no phenyl. (A 3° carbon is more stable than a 2° carbon). Therefore, when this bond opens, the bromide will go there in the next step, and the final product is formed. See picture below for mechanism and products.
