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svlad2 [7]
3 years ago
8

Explain how you determine the freezing point of a solution that does not have a well-defined transition in the cooling curve.

Chemistry
1 answer:
Sophie [7]3 years ago
7 0

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:

y=-3.5 x + 25\\\\y=-0.52 x + 2

First of all, we equal them to find the x-value, in this case the time at which the freezing point takes place:

-3.5 x + 25=-0.52 x + 2\\\\-3.5 x+0.52 x =2-25\\\\x=\frac{-23}{-2.98}=7.72

Next, we plug it in in any of the trendlines to obtain the freezing point as the y-value:

y=-3.5 (7.72) + 25\\\\y = 1.84

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:

  • brainly.com/question/22818252
  • brainly.com/question/9680530

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What is the relative atomic mass of a hypothetical element that consists isotopes in the indicated natural abundances
belka [17]

The given question is incomplete. The complete question is:What is the relative atomic mass of a hypothetical element that consists isotopes in the indicated natural abundances.

Isotope                    mass amu        Relative abundance

1                                77.9                     14.4

2                               81.9                     14.3

3                               85.9                      71.3

Express your answer to three significant figures and include the appropriate units.

Answer: 84.2 amu

Explanation:

Mass of isotope 1 = 77.9  

% abundance of isotope 1 = 14.4% = \frac{14.4}{100}=0.144

Mass of isotope 2 = 81.9

% abundance of isotope 2 = 14.3% = \frac{14.3}{100}=0.143

Mass of isotope 3 = 85.9

% abundance of isotope 2 = 71.3% = \frac{71.3}{100}=0.713

Formula used for average atomic mass of an element :

\text{ Average atomic mass of an element}=\sum(\text{atomic mass of an isotopes}\times {{\text { fractional abundance}})

A=\sum[(77.9\times 0.144)+(81.9\times 0.143)+(85.9\times 0.713)]

A=84.2amu

Therefore, the average atomic mass of a hypothetical element that consists isotopes in the indicated natural abundances is 84.2 amu

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