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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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Hoochie [10]
First we need to know that the boiling point of water in C is 100 and we just need to solve for x in the equation:

-33.75-(-77.75) / 100 = 100-(-77.75) / x
44.4/100 = 177.75 / x
 x = 177.75*100/44.4 = 400.33

The boiling point of water in ∘a would be 400.33∘a.

7 0
2 years ago
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For the reaction A +B+ C D E, the initial reaction rate was measured for various initial concentrations of reactants. The follow
lora16 [44]

Answer:

Rate constant of the reaction is 3.3\times 10^{-3} M^{-2} s^{-1}.

Explanation:

A + B + C → D + E

Let the balanced reaction be ;

aA + bB + cC → dD + eE

Expression of rate law of the reaction will be written as:

R=k[A]^a[B]^b[C]^c

Rate(R) of the reaction in trail 1 ,when :

[A]=0.30 M,[B]=0.30 M,[C]=0.30 M

R=9.0\times 10^{-5} M/s

9.0\times 10^{-5} M/s=k[0.30 M]^a[0.30 M]^b[0.30 M]^c...[1]

Rate(R) of the reaction in trail 2 ,when :

[A]=0.30 M,[B]=0.30 M,[C]=0.90 M

R=2.7\times 10^{-4} M/s

2.7\times 10^{-4} M/s=k[0.30 M]^a[0.30 M]^b[0.90 M]^c...[2]

Rate(R) of the reaction in trail 3 ,when :

[A]=0.60 M,[B]=0.30 M,[C]=0.30 M

R=3.6\times 10^{-4} M/s

3.6\times 10^{-4} M/s=k[0.60 M]^a[0.30 M]^b[0.30 M]^c...[3]

Rate(R) of the reaction in trail 4 ,when :

[A]=0.60 M,[B]=0.60 M,[C]=0.30 M

R=3.6\times 10^{-4} M/s

3.6\times 10^{-4} M/s=k[0.60 M]^a[0.60 M]^b[0.30 M]^c...[4]

By [1] ÷ [2], we get value of c ;

c = 1

By [3] ÷ [4], we get value of b ;

b = 0

By [2] ÷ [3], we get value of a ;

a = 2

Rate law of reaction is :

R=k[A]^2[B]^0[C]^1

Rate constant of the reaction = k

9.0\times 10^{-5} M/s=k[0.30 M]^2[0.30 M]^0[0.30 M]^1

k=\frac{9.0\times 10^{-5} M/s}{[0.30 M]^2[0.30 M]^0[0.30 M]^1}

k=3.3\times 10^{-3} M^{-2} s^{-1}

7 0
3 years ago
What is the concentration of an unknown Mg(OH)2 solution if it took an average of 15.4mL of
vova2212 [387]

Answer:

0.077M

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

2HCl + Mg(OH)2 —> MgCl2 + 2H2O

From the balanced equation above,

The mole ratio of the acid (nA) = 2

The mole ratio of the base (nB) = 1

Step 2:

Data obtained from the question.

Concentration of base Cb =...?

Volume of base (Vb) = 10mL

Concentration of acid (Ca) = 0.1M

Volume of acid (Va) = 15.4mL

Step 3:

Determination of the concentration of the base, Mg(OH)2.

The concentration of the base can be obtained as follow:

CaVa/CbVb = nA/nB

0.1 x 15.4 /Cb x 10 = 2/1

Cross multiply to express in linear form

Cb x 10 x 2 = 0.1 x 15.4

Divide both side by 10 x 2

Cb = (0.1 x 15.4) /(10 x 2)

Cb = 0.077M

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7 0
3 years ago
During the phase change from solid ice to liquid water, which bonds/forces are being weakened?
vampirchik [111]
I think Intramolecular forces are being weakened
5 0
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Calculate the entropy change for the reaction: Fe2O3(s) +3C(s) -> 2Fe(s) + 3CO(g)
Gelneren [198K]

Answer:

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Explanation:

Given parameters:

Standard Entropy of Fe₂O₃ = 90Jmol⁻¹K⁻¹

Standard Entropy of C = 5.7Jmol⁻¹K⁻¹

Standard Entropy of Fe = 27.2Jmol⁻¹K⁻¹

Standard Entropy of CO = 198Jmol⁻¹K⁻¹

To find the entropy change of the reaction, we first write a balanced reaction equation:

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To calculate the entropy change of the reaction we simply use the equation below:

      ΔS = ∑S_{products} - ∑S_{reactants}

Therefore:

     ΔS = [(2x27.2) + (3x198)] - [(90) + (3x5.7)] = 648.4 - 107.1

                          ΔS = +541.3Jmol⁻¹K⁻¹

5 0
2 years ago
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