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lions [1.4K]
3 years ago
6

Poly(ethylene terephthalate) (PET), which has glass transition (Tg) and crystalline melting (Tm) temperature of 69 and 267 °C, r

espectively, can exist in a number of different states depending upon temperature and thermal history. Thus, it is possible to prepare materials that are semicrystalline with amorphous regions that are either glassy or rubbery and amorphous materials that are glassy, rubbery or melts. Consider a sample of PET cooled rapidly from 300 °C (state A) to room temperature. The resulting material is rigid and perfectly transparent (state B). The sample is then heated to 100 °C and maintained at this temperature, during which time is gradually becomes translucent (state C). It is then cooled to room temperature, where it is again observed to be translucent (state D).
Chemistry
1 answer:
anastassius [24]3 years ago
8 0

Answer:

Following are the solution to the given points:

Explanation:

For point A:

  1. The sample cooking (PET) is between 300°C and room temperature.
  2. Now in nature, the substance is exceedingly stiff.
  3. Samples of PET up to 100°C were heated and stayed on equal footing.
  4. Now it has cooled off the same sample below 100° C and we may see how it is again TRASNEPARENT in nature.

For point B:

In point 3, the mixture was added to 100°C, which implies that the granular material flows and deforms, enabling it to become elongated. This is termed solid-state crystalline such that grains are flexible, but this material contaminates numerous little crystalline that has spheres when we cool down in point  4 polymers. It forms therefore an unstructured solid, which then in point 4 is higher in particles and less pliable in orderly atoms.

For point C:

In point 2, the specimen gets forced at room temperature to organize a huge molecule in an ordinary and crystal fashion and therefore is transparent due to highly crystalline atoms in point 2 of the PET sample.

In point 4, however, we notice how amorphous, firm but not crystalline develops. It's why light tends to disperse over many cereal limits, since many microscopic crystallines, therefore dispersion, PET in point 4 is translucent.

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Rank the following measurements in order from least to greatest
Harman [31]
It’s
1.A
2.C
3.B

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4 0
3 years ago
Question 23 (3 points)
Mice21 [21]

Answer:

<h2>The answer is 3.0 mL</h2>

Explanation:

The volume of a substance when given the density and mass can be found by using the formula

volume =  \frac{mass}{density}  \\

From the question

mass of aluminum = 8.1 g

density = 2.7 g/mL

It's volume is

volume =  \frac{8.1}{2.7}  \\

We have the final answer as

<h3>3.0 mL</h3>

Hope this helps you

3 0
3 years ago
A Calorie unit used in food is equal to the amount of energy necessary to raise the temperature of 1 kilogram of water ________
Kitty [74]

A Calorie unit used in food is equal to the amount of energy necessary to raise the temperature of 1 kilogram of water by <u>1</u> degrees Celsius.

<h3>What is One Calorie ?</h3>

The amount of heat energy required to raise the temperature by 1 gram of water through 1°C is known as One Calorie.

1 Calorie = 4.18 J

Thus from the above conclusion we can say that A Calorie unit used in food is equal to the amount of energy necessary to raise the temperature of 1 kilogram of water by <u>1</u> degrees Celsius.

Learn more about the One calorie here: brainly.com/question/1061571

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7 0
2 years ago
The process in which an organic acid and an alcohol react to form an ester and water is known as esterification. Ethyl butanoate
egoroff_w [7]

Answer:

697 g

Explanation:

Ethanol (C₂H₅OH) and butanoic acid (C₃H₇COOH) react to form ethyl butanoate (C₃H₇COOC₂H₅) and water (H₂O).

C₂H₅OH + C₃H₇COOH → C₃H₇COOC₂H₅ + H₂O

The molar ratio of C₂H₅OH to C₃H₇COOC₂H₅ is 1:1. The moles of C₃H₇COOC₂H₅ produced from 6.00 moles of C₂H₅OH are:

6.00 mol C₂H₅OH × (1 mol C₃H₇COOC₂H₅/1 mol C₂H₅OH) = 6.00 mol C₃H₇COOC₂H₅

The molar mass of C₃H₇COOC₂H₅ is 116.16 g/mol. The mass corresponding to 6.00 mol is:

6.00 mol × (116.16 g/mol) = 697 g

7 0
3 years ago
calcium reacts with fluorine to produce calcium fluoride. how does oxidation and reduction take place in this reaction?
Assoli18 [71]

The oxidation is occurring on Calcium ions as it release one electron and reduction will be occurring on fluorine ion as it accepts one electron.

<u>Explanation:</u>

An element will undergo oxidation and form a positive ion on releasing one or more electrons from its valence shell. While reduction is occurred in a chemical reaction, then the element will be forming a negative ion with the acceptance of one or more electrons in its valence shell.

So in the given process of calcium fluoride, the one electron from the valence shell of calcium will be released making it as c a^{+} ions and this is termed as oxidation process. This one electron will be getting accepted by the fluorine ion and thus it will convert to F^{-} ions. This process of acceptance of electrons is termed as reduction.

3 0
3 years ago
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