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Bess [88]
3 years ago
7

Which intermolecular force is characteristic of compounds with low molar mass, which are liquids at room temperature and have re

latively high boiling points? Which intermolecular force is characteristic of compounds with low molar mass, which are liquids at room temperature and have relatively high boiling points? covalent bonds London forces dipole-dipole forces hydrogen bonds ionic bonds
Chemistry
1 answer:
eduard3 years ago
6 0

Answer:

Hydrogen bonds.

Explanation:

The intermolecular forces are the forces that are presented between molecules in a substance. As higher is the force, as closer are the molecules, and more difficult will be to separate them.

Because of that, the solids have stronger forces than the liquids, which have stronger forces than the gases. Also, as strong the force, as higher will be the boiling point.

The ionic bond is the strongest and it's presented at ionic compounds, which are solids at room temperature and have high boiling points.

The covalent bonds are presented in the molecules and are formed when atoms share pairs of electrons. Between these molecules, the forces can be London forces, dipole-dipole forces, or hydrogen bonds.

The London forces are the weaker, they are presented at the nonpolar molecules, so it's easy to boil it. For a compound with low molar mass and London forces, it'll probably be at the gas phase at room temperature.

The dipole-dipole forces are presented at the polar molecules, and it's strong than the London forces. And the hydrogen bonds are a specific type of dipole-dipole forces, which is stronger and is formed between hydrogen and F, O, or N.

Because hydrogen has a low molar mass (1 g/mol), the compounds formed by it intends to have a low molar mass. So, to be liquid at room temperature, low mass, and high boiling point, it's more probable that the compound has hydrogen bonds.

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Answer:

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Explanation:

<u>1) Convert absolute zero to celsius:</u>

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<u>2) Convert - 273.15°C to Fahrenheit:</u>

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  • T (°F) = - 273.15 × 1.8 + 32 = - 459.67 °F ← answer

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Suppose that coal of density 1.5 g/cm^3 is pure carbon. (It is, in fact, much more complicated, but this is a reasonable first a
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Answer:

q = -6464.9 kJ

Explanation:

We are given that the heat of combustion is  ∆H° = −394 kJ per mol of carbon.Therefore what we need to do is calculate how many moles of C are in the lump of coal by finding its mass since the density is given.

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q =  −394 kJ /mol C x 16.41 mol C = -6464.9 kJ

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Answer:

Explanation: I think its 4.91 x 10^25. Im not very sure, i just multipled 1.15 mol by the molar mass of Cl 2, which was 70.9 g. Then I multiplied that by avogadro's number. sorry if im wrong

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