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Luden [163]
3 years ago
8

In a semi-crystalline polymer processed from solution, the presence of residual solvent will both decrease and broaden the melti

ng range of the crystals. Why? First, consider both the spherulites and the surrounding amorphous components. Next, consider only the structure found on the interior of the spherulites.
Chemistry
1 answer:
Rainbow [258]3 years ago
6 0

Explanation:

The presence of residual solvent is a factor that affects the glass temperature of the polymer. This is because a residual solvent decreases the free volume in a polymer. All types of polymers have total occupied volume, which is composed of an occupied volume and a free volume. The free volume is the volume needed for the polymer chains to move around. Thus, if there is a big free volume in the polymer it will decrease the glass transition temperature and will both decrease and broaden the melting range of the crystals.

In a polymer with crystalline spherulites and the surrounding amorphous, will have a bigger surrounding amorphous with the presence of residual solvent. Which will decrease the melting range of the crystals. The structure found on the interior of the spherulites is pure crystalline polymer, so probably the residual solvent will not be inside them.

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Answer:

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Explanation:

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4 0
3 years ago
PLEASE PLEASE HELP!!!!
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A supersaturated solution contains more solute at a given temperature than is needed to form a saturated solution.

Increased temperature usually increases the solubility of solids in liquids.

For example, the solubility of glucose at 25 °C is 91 g/100 mL of water. The solubility at 50 °C is 244 g/100 mL of water.

If we add 100 g of glucose to 100 mL water at 25 °C, 91 g dissolve. Nine grams of solid remain on the bottom. We have a saturated solution.

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3 years ago
A sample of gas initially occupies 3.35 L at a pressure of 0.950 atm at 13.0oC. What will the volume (in L) be if the temperatur
Marat540 [252]

Answer: V= 3.13 L

Explanation: solution attached:

Use combine gas law equation:

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Derive to find V2

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T1= 13.0°C + 273 = 286 K

T2= 22.5°C + 273 = 295.5 K

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4 years ago
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4 years ago
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