They give a double displacement reaction where the ions switch places and give sodium nitrate (NaNO3) and silver chloride (AgCl) as the products. Silver nitrate is also very soluble in water, but silver chloride is highly insoluble in water and will precipitate out of solution as a white solid.
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.
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Answer:
when mass is 1×10⁴ Kg then density is 5 g/cm³.
when mass is 104 Kg then density is 5.2 × 10⁻² g/ cm³.
Explanation:
Density:
Density is equal to the mass of substance divided by its volume.
Units:
SI unit of density is Kg/m3.
Other units are given below,
g/cm3, g/mL , kg/L
Formula:
D=m/v
D= density
m=mass
V=volume
Symbol:
The symbol used for density is called rho. It is represented by ρ. However letter D can also be used to represent the density.
Given data:
mass = 1×10⁴ Kg
volume= w ×l× h = 1×2× 1 = 2 m³
density = ?
first of all we will convert the given volume meter cube to cm³:
we know that
2×1000000 = 2 × 10⁶ cm³
Now we will convert the mass into gram.
1 Kg = 1000 g
1×10⁴ × 1000 = 1 ×10⁷ g
Now we will put the values in the formula,
d = m/v
d = 1 ×10⁷ g / 2×10⁶ cm³
d = 0.5 × 10¹ g/cm³
or
d = 5 g/cm³
If mas is 104 Kg:
104 × 1000 = 104000 g
d= m/v
d = 104000 g / 2×10⁶ cm³
d= 52000 ×10⁻⁶ g/ cm³
d= 5.2 × 10⁻² g/ cm³