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lions [1.4K]
2 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]2 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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You are trying to determine the specific heat of a metal. You heat the 97 g piece of metal to 100 °C and place it in a calorimet
Brut [27]

Answer : The specific heat of the metal is, 0.658J/g^oC

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of metal = ?

c_2 = specific heat of water = 4.18J/g^oC

m_1 = mass of metal = 97 g

m_2 = mass of water  = 122 g

T_f = final temperature of mixture = 28.9^oC

T_1 = initial temperature of metal = 100^oC

T_2 = initial temperature of water = 20.0^oC

Now put all the given values in the above formula, we get

97g\times c_1\times (28.9-100)^oC=-122g\times 4.18J/g^oC\times (28.9-20.0)^oC

c_1=0.658J/g^oC

Therefore, the specific heat of the metal is, 0.658J/g^oC

8 0
3 years ago
An unknown material has a mass of 5.75 g and a volume of 7.5 cm3.
Shtirlitz [24]

Hey there!:

Mass = 5.75 g

Volume = 7.5 cm³

Therefore:

Density = mass / volume

D = 5.75 / 7.5

D = 0.7 g/cm³

Answer C

Hope that helps!

3 0
3 years ago
Elements within a period have ___.
Vlada [557]

Ans: Have the same number of electron shells

3 0
3 years ago
Three blocks are shown here pls HELPPPP:
White raven [17]

Answer: I am confident the answer is B

Explanation:

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5 0
2 years ago
What is the empirical formula for a substance that is composed of 40.66% carbon, 8.53% Hydrogen,23.72% Nitrogen, and 27.09% Oxyg
prohojiy [21]

Answer:

THE EMPIRICAL FORMULA OF THE SUBSTANCE IS C2H5NO

Explanation:

The steps involved in calculating the empirical formula of this substance in shown in the table below:

Element                            Carbon           Hydrogen          Nitrogen         Oxygen

1. % Composition              40.66              8.53                 23.72                27.09

2. Mole ratio =

%mass/ atomic mass       40.66/12         8.53/1          23.72/14            27.09/16

                                       =  3.3883            8.53              1,6943             1.6931

3. Divide by smallest

value (0.6931)          3.3883/1.6931    8.53/1.6931    1.6943/1.6931   1.6931/1.6931

                               =      2.001                  5.038           1.0007                      1

4. Whole number ratio        2                       5                   1                               1

The empirical formula = C2H5NO

7 0
3 years ago
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