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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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Explanation:

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The mass percentage of hydrochloric acid within a solution is 28.00%28.00% . Given that the density of this solution is 1.1411.1
sashaice [31]

Answer:

8.76M

Explanation:

Given that

Mass from the density = 1141g

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So,

Mass of HCl in 1-litre solution is

= \frac{28}{100} \times 1141

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8 0
3 years ago
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Answer:

52. The alcohol USED => methanol, CH3OH

The carboxylic acid USED => propanoic acid, CH3CH2COOH.

53. The alcohol USED => Ethanol, CH2CH3OH

The carboxylic acid USED => Formic acid, HCOOH.

Explanation:

52. To obtain Methyl propanoate, CH3CH2COOCH3, we simply react propanoic, CH3CH2COOH and methanol, CH3OH together as shown below:

CH3CH2COOH + CH3OH —> CH3CH2COOCH3 + H2O

The alcohol used: methanol, CH3OH

The carboxylic acid used: propanoic acid, CH3CH2COOH.

53. To obtain Ethyl methanoate, HCOOCH2CH3, we simply react

Formic acid, HCOOH and ethanol, CH3CH2OH together as show below:

HCOOH + CH3CH2OH —> HCOOCH2CH3 + H2O

The alcohol USED => Ethanol, CH2CH3OH

The carboxylic acid USED => Formic acid, HCOOH.

7 0
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Also keep in mind that atomic radii decreases from left to right as you move along a periodic table. It also decreases from bottom up. 
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As bromine is higher up in the periodic table than Iodine, it would have a smaller radius. Iodine would have a larger radius.

The correct answer is B. Br
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3 years ago
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