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viktelen [127]
3 years ago
8

Why do compounds with strong intermolecular attractive forces have a higher boiling points than compounds with the weak intermol

ecular attractive forces? Provide an answer using 3-4 sentences.
Chemistry
2 answers:
mezya [45]3 years ago
4 0

Answer:

Higher the inter-molecular attractive force of a compound, higher the energy required to break those bonds. Therefore, their boiling points are higher.

Explanation:

The molecules of liquid are free to move in their lattice. But they cannot, leave their lattice or layer, because of the inter-molecular attractive forces. These forces form a bond among molecules. The boiling point is that temperature at which the molecules of liquid acquire enough energy that they overcome these inter-molecular attractive forces and leave the layers of liquid breaking the bonds. Hence, they are converted to gases. This is the reason that compounds with strong inter-molecular forces require more energy to break bond and hence, their boiling points are higher than compounds with weak inter-molecular attractive forces.

Brut [27]3 years ago
3 0

The boiling point is the temperature at which the vapour pressure of the solution or solvent becuase equal to the ambient pressure, and bubbles of vapour from directly in the liquid. The normal boiling point is specified when the ambient pressure ( And also thvapour pressure of the boiling liquid) is One atmosphere. Ans so where strong intermolecular forces of attraction operate for instance in Hydrogen-Bonded liquids, boiling point should be elevated. More energy has to be put into the system to disrupt the intermolecular force. And the best indicator or intermolecular or inter-particle force is boiling point.

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g Determine the empirical formula for a compound that contains C, H and O. It contains 40.92% C, 4.58% H, and 54.50% O by mass.
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Answer:

The empirical formula for the compound is C3H4O3

Explanation:

The following data were obtained from the question:

Carbon (C) = 40.92%

Hydrogen (H) = 4.58%

Oxygen (O) = 54.50%

The empirical formula for the compound can be obtained as follow:

C = 40.92%

H = 4.58%

O = 54.50%

Divide by their molar mass

C = 40.92/12 = 3.41

H = 4.58/1 = 4.58

O = 54.50/16 = 3.41

Divide by the smallest i.e 3.41

C = 3.41/3.41 = 1

H = 4.58/3.41 = 1.3

O = 3.41/3.41 = 1

Multiply through by 3 to express in whole number

C = 1 x 3 = 3

H = 1.3 x 3 = 4

O = 1 x 3 = 3

The empirical formula for the compound is C3H4O3

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