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fiasKO [112]
3 years ago
13

Explain the difference in the boiling point of 2-methylpropane and 2-iodo-2-methylpropane in terms of both molecular polarity an

d imf
Chemistry
1 answer:
Len [333]3 years ago
6 0
The Boiling Point of 2-methylpropane is approximately -11.7 °C, while, Boiling Point of <span>2-iodo-2-methylpropane is approximately 100 </span>°C.

As both compounds are Non-polar in nature, So there will be no dipole-dipole interactions between the molecules of said compounds.

The Interactions found in these compounds are London Dispersion Forces.

And among several factors at which London Dispersion Forces depends, one is the size of molecule.

Size of Molecule:
                          There is direct relation between size of molecule and London Dispersion forces. So, 2-iodo-2-methylpropane containing large atom (i.e. Iodine) experience greater interactions. So, due to greater interactions 2-iodo-2-methylpropane need more energy to separate from its partner molecules, Hence, high temperature is required to boil them.
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What is the empirical formula of a compound with 35.94% aluminum and 64.06% sulfur?
Marat540 [252]

<u>Given:</u>

% Al = 35.94

% S = 64.06

<u>To determine:</u>

Empirical formula of a compound with the above composition

<u>Explanation:</u>

Atomic wt of Al = 27 g/mol

Atomic wt of S = 32 g/mol

Based on the given data, for 100 g of the compound: Mass of Al = 35.94 g and mass of S = 64.06 g

# moles of Al = 35.94/27 = 1.331

# moles of S = 64.06/32 = 2.002

Divide by the smallest # moles:

Al = 1.331/1.331 = 1

S = 2.002/1,331 = 1.5 ≅ 2

Empirical formula = AlS₂

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In a heat engine, 500 J of heat enters the system, and the piston does 300 J of work. What is the final internal (thermal) energ
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