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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The solution would be like this for this specific problem:
<span>Given:
H2 = </span><span>2.6 atm
CL2 = 3.14 atm</span>
<span>
pressure H2 = 2.6 - x
pressure Cl2 = 3.14 - x
<span>pressure HBr = 2x = 1.13
x = 1.13 / 2 = 0.565
<span>pressure H2 = 2.6 - 0.565 = 2.035
pressure Br2 = 3.14 - 0.565 = 2.575
Kp = (1.13)^2 / 2.035 x 2.575</span></span></span>
= 1.2769 / (5.240125)
= 0.24367739319195629875241525726963
= 0.244
<span>Therefore, the Kp for the reaction at the given temperature
is 0.244.
To add, </span>the hypothetical pressure of a gas if
it alone occupied the whole volume of the original mixture at the same
temperature is called the partial pressure or Kp.
You can use P1V1/T1 = P2V2/T2 but since pressure is constant is becomes V1/T1=V2/T2
V1=0.5 L
T1=203 K
T2=273 K
V2=unknown
0.5L/203 = V2/273
V2= 0.67 L so C
Hope this helps :)
Equation: MgOH2 (s) --> MgO (s) + H2O (g)
And it's already balanced.
I had chemistry this semester too.
Answer:
cold
Warm air lifted over a moving cold air mass will produce a _____ front.