Answer: 0.35 m 
> 0.20 m 
>0.10 m 
> 0.15 m 
> 0.15 m 
Explanation:
Depression in freezing point:
where,
= freezing point of solution
= freezing point of solvent
= freezing point constant
m = molality
1. For 0.10 m
, i= 4 as it is a electrolyte and dissociate to give 4 ions. and concentration of ions will be 
2. For 0.35 m
, i= 2 as it is a electrolyte and dissociate to give 2 ions. and concentration of ions will be 
3. For 0.20 m
, i= 3 as it is a electrolyte and dissociate to give 3 ions. and concentration of ions will be 
4. For 0.15 m
, i= 1 as it is a non electrolyte and does not dissociate to give ions.
5. For 0.15 m
, i= 2 as it is a electrolyte and dissociate to give 2 ions and thus have 
As concentration is highest for 0.35 m , freezing point depression will be highest and thus has lowest freezing point. As concentration is lowest for 0.15 m , freezing point depression will be lowest and thus has highest freezing point
The answer is <span>Plastics, Medicine, Clothing, Paper. </span>Coal is used in the making of Plastics, Medicine, Clothing, Paper. Some important users of coal include alumina refineries, paper manufacturers, and the chemical and pharmaceutical industries. Thousands of different products have coal or coal by-products as components: soap, aspirins, solvents, dyes, plastics and fibres, such as rayon and nylon.
the answer would be winter because the north would be facing away from the son therefor making the northern states cold.
Answer:
See explanation below
Explanation:
The question is incomplete, cause you are not providing the structure. However, I found the question and it's attached in picture 1.
Now, according to this reaction and the product given, we can see that we have sustitution reaction. In the absence of sodium methoxide, the reaction it's no longer in basic medium, so the sustitution reaction that it's promoted here it's not an Sn2 reaction as part a), but instead a Sn1 reaction, and in this we can have the presence of carbocation. What happen here then?, well, the bromine leaves the molecule leaving a secondary carbocation there, but the neighbour carbon (The one in the cycle) has a more stable carbocation, so one atom of hydrogen from that carbon migrates to the carbon with the carbocation to stabilize that carbon, and the result is a tertiary carbocation. When this happens, the methanol can easily go there and form the product.
For question 6a, as it was stated before, the mechanism in that reaction is a Sn2, however, we can have conditions for an E2 reaction and form an alkene. This can be done, cause the extoxide can substract the atoms of hydrogens from either the carbon of the cycle or the terminal methyl of the molecule and will form two different products of elimination. The product formed in greater quantities will be the one where the negative charge is more stable, in this case, in the primary carbon of the methyl it's more stable there, so product 1 will be formed more (See picture 2)
For question 6b, same principle of 6a, when the hydrogen migrates to the 2nd carbocation to form a tertiary carbocation the methanol will promove an E1 reaction with the vecinal carbons and form two eliminations products. See picture 2 for mechanism of reaction.
<span>In the electron cloud model, the denser areas represent that there is a great probability that a good number of electrons are ganged up or crowded in that area. The electrons affect the density of some parts of the electron cloud when they condense in those locations.</span>
