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mihalych1998 [28]
3 years ago
15

1. A sample of polystyrene is found to have a number-average molar mass of 89,440 g mol−1 . Neglecting contributions from end gr

oups, calculate the number-average degree of polymerization of this sample. Assuming that the sample has a molar mass dispersity of 1.5, calculate its weight-average molar mass.
Chemistry
1 answer:
Snowcat [4.5K]3 years ago
8 0

Answer:

The weight-average molar mass of polystyrene is 134,160 g/mol.

Explanation:

Molar mass of the monomer styrene , C_8H_8, M=104 g/mol

Given , number average molar mass of the polymer , M'= 89,440 g/mol

Degree of polymerization = n

n=\frac{M'}{M}=\frac{89,440 g/mol}{104 g/mol}=860

The weight-average molar mass = M_{avg}=?

Molar mass dispersity is ratio of weight-average molar mass to the number average molar mass of the polymer.

\text{Molar mass dispersity}=\frac{M_{avg}}{M'}

1.5=\frac{M_{avg}}{89,440 g/mol}

M_{avg}=89,440 g/mol\times 1.5 = 134,160 g/mol

The weight-average molar mass of polystyrene is 134,160 g/mol.

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Answer:

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2 years ago
What mass of solid that has a molar mass of 46.0 g/mol should be added to 150.0 g of benzene to raise the boiling point of benze
enyata [817]

Answer : 17.12 g

Explanation:\Delta T =k_b\times m

\Delta T = elevation in boiling point

k_b = boiling point elevation constant

m= molality

molality=\frac{mass of solute}{molecular mass of solute\times weight of the solvent in kg}

given \Delta T=6.28^{\circ}C

Molar mass of solute = 46.0 gmol^{-1}

Weight of the solvent = 150.0 g = 0.15 kg

Putting in the values

molality=\frac{x}{46gmol^{-1}\times0.15kg}

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3 years ago
Cutting, melting, bending, or crushing are examples of what kind of change?
Ksivusya [100]

Answer:

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Explanation:

8 0
3 years ago
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What is the purpose of graphics in scientific articles
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Answer:

Graphics can sometimes convey more information in a brief amount of space than an author can explain in a paragraph.

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Red #40 has an acute oral LD50 of roughly 5000 mg dye/1 kg body weight. This means if you had a mass of 1 kg, ingesting 5000 mg
FrozenT [24]

Answer:

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3.5 × 10⁵ mg × (1 L / 12 mg) = 2.9 × 10⁴ L

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3 years ago
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