This question is asking for a method for the determination of the freezing point in a solution that does not have a noticeable transition in the cooling curve, which is basically based on a linear fit method.
The first step, would be to understand that when the transition is well-defined as the one on the attached file, we can just identify the temperature by just reading the value on the graph, at the time the slope has a pronounced change. For instance, on the attached, the transition occurs after about 43 seconds and the freezing point will be about 4 °C.
However, when we cannot identify a pronounced change in the slope, it will be necessary to use a linear fit method (such as minimum squares) to figure out the equation for each segmented line having a significantly different slope and then equal them so that we can numerically solve for the intercept.
As an example, imagine two of the segmented lines have the following equations after applying the linear fit method:
First of all, we equal them to find the x-value, in this case the time at which the freezing point takes place:
Next, we plug it in in any of the trendlines to obtain the freezing point as the y-value:
This means the freezing point takes place after 7.72 second of cooling and is about 1.84 °C. Now you can replicate it for any not well-defined cooling curve.
Learn more:
Maybe this example could help you to understand this problem.
https://image.slidesharecdn.com/121howmanyatoms-091201144624-phpapp02/95/12-1-how-many-atoms-17-728....
I found this....
Supraglacial Moraine
A supraglacial moraine is material on the surface of a glacier. Lateral and medial moraines can be supraglacial moraines. Supraglacial moraines are made up of rocks and earth that have fallen on the glacier from the surrounding landscape. Dust and dirt left by wind and rain become part of supraglacial moraines. Sometimes the supraglacial moraine is so heavy, it blocks the view of the ice river underneath.
If a glacier melts, supraglacial moraine is evenly distributed across a valley.
Ground Moraine
Ground moraines often show up as rolling, strangely shaped land covered in grass or other vegetation. They don’t have the sharp ridges of other moraines. A ground moraine is made of sediment that slowly builds up directly underneath a glacier by tiny streams, or as the result of a glacier meeting hills and valleys in the natural landscape. When a glacier melts, the ground moraine underneath is exposed.
Ground moraines are the most common type of moraine and can be found on every continent.
Terminal Moraine
A terminal moraine is also sometimes called an end moraine. It forms at the very end of a glacier, telling scientists today important information about the glacier and how it moved. At a terminal moraine, all the debris that was scooped up and pushed to the front of the glacier is deposited as a large clump of rocks, soil, and sediment.
Scientists study terminal moraines to see where the glacier flowed and how quickly it moved. Different rocks and minerals are located in specific places in the glacier’s path. If a mineral that is unique to one part of a landscape is present in a terminal moraine, geologists know the glacier must have flowed through that area.
Answer:
a): not necessarily due to London Dispersion Forces and dipole-dipole interactions.
b): not necessarily due to London Dispersion Forces.
Explanation:
There are three major types of intermolecular interaction:
- Hydrogen bonding between molecules with H-O, H-N, or H-F bonds and molecules with lone pairs.
- Dipole-dipole interactions between all molecules.
- London dispersion forces between all molecules.
The melting point of a substance is a result of all three forces, combined.
Note that the more electrons in each molecule, the stronger the London Dispersion Force. Generally, that means the more atoms in each molecule, the stronger the London dispersion force. The strength of London dispersion force between large molecules can be surprisingly strong.
For example, 
(water) molecules are capable of hydrogen bonding. The melting point of at 
is around 
. That's considerably high when compared to other three-atom molecules.
In comparison, the higher alkane hexadecane (
, straight-chain) isn't capable of hydrogen bonding. However, under a similar pressure, hexadecane melts at around 
above the melting point of water. The reason is that with such a large number of atoms (and hence electrons) per molecule, the London dispersion force between hexadecane molecules could well be stronger than that the hydrogen bonding between water molecules.
Similarly, the dipole moments in HCl (due to the highly-polar H-Cl bonds) are much stronger than those in hexadecane (due to the C-H bonds.) However, the boiling point of hexadecane under standard conditions is much higher (at around 
than that of HCl.
The isoelectric point<span> (</span>pI, pH(I),IEP<span>), is the pH at which a particular molecule carries no net electrical charge in the statistical mean.</span>
