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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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You are instructed to create 900. mL of a 0.29 M phosphate buffer with a pH of 7.8. You have phosphoric acid and the sodium salt
Aleksandr [31]

Answer:

B and C

Explanation:

When we have to do a buffer solution we always have to choose the reaction that has the <u>pKa closer to the desired pH value</u>. When we find the pKa values we will obtain:

pKa_1=-Log[6.9x10^-^3]=2.16

pKa_2=-Log[6.2x10^-^8]=7.20

pKa_3=-Log[4.8x10^-^13]=12.31

The closer value is pKa2 with a value of 7.2. Therefore we have to use the second reaction. In which  H_2PO_4^-^1 is the <u>acid</u> and HPO_4^-^2 is the <u>base</u>. Therefore the answer for the first question is B and the answer for the second question is C.

8 0
3 years ago
Ethanol (CH3CH2OH), the intoxicant in alcoholic beverages, is also used to make other organic compounds. In concentrated sulfuri
Fed [463]

Answer:

a) 88.48%

b) 0.05625 mol

Explanation:

2CH₃CH₂OH(l) → CH₃CH₂OCH₂CH₃(l) + H₂O(g)         Reaction 1

CH₃CH₂OH(l) → CH₂═CH₂(g) + H₂O(g)                        Reaction 2

a) CH₃CH₂OH = 46.0684 g/mol

   CH₃CH₂OCH₂CH₃ = 74.12 g/mol

1 mol CH₃CH₂OH ______  46.0684 g

x                            ______   50.0 g

x = 1.085 mol  CH₃CH₂OH

1 mol  CH₃CH₂OCH₂CH₃ ______  74.12 g g

y                           ______   35.9 g

y = 0.48 mol   CH₃CH₂OCH₂CH₃

100% yield _____ 0.5425 mol CH₃CH₂OCH₂CH₃

w                _____  0.48 mol CH₃CH₂OCH₂CH₃

w = 88.48%

b) Only 0.96 mol of ethanol reacted to form diethyl ether. This means that 0.125 mol of ethanol did not react. 45% of 0.125 mol reacted to form ethylene. Therefore, 0.05625 mol of ethanol reacted by the side reaction (reaction 2). Since 1 mol of ethanol leads to 1 mol of ethylene, 0.05625 mol of ethanol produces 0.05625 mol of ethylene.

4 0
3 years ago
Group I elements are metals with ................. density
kherson [118]

group 1 elements are metals with<u> low</u> density

8 0
3 years ago
How many moles of xenon gas, Xe, would occopy 37.8L at stp
Charra [1.4K]

Answer:

The number of moles of  xenon are 1.69 mol.

Explanation:

Given data:

Number of moles of xenon = ?

Volume of gas = 37.8 L

Temperature = 273 K

Pressure = 1 atm

Solution:

The given problem will be solve by using general gas equation,

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K  

T = temperature in kelvin

Now we will put the values in formula.

1 atm × 37.8 L = n ×  0.0821 atm.L/ mol.K   ×273 K

37.8 atm.L =  n × 22.413 atm.L/ mol.

n = 37.8 atm.L /  22.413 atm.L/ mol.

n = 1.69 mol

The number of moles of  xenon are 1.69.

8 0
3 years ago
What does a nubula turn into
LekaFEV [45]

Answer:

a white dwarf

Explanation:

6 0
3 years ago
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