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Fiesta28 [93]
3 years ago
6

Consider the substances hydrogen (H2), fluorine (F2), and hydrogen fluoride (HF). Based on their molecular structures, how does

the boiling point of HF compare with the boiling points of H2 and F2?
The boiling point of HF is (higher than, lower than, similar to) the boiling point of H2, and it is (higher than, lower than, similar to) the boiling point of F2.
Chemistry
1 answer:
katovenus [111]3 years ago
3 0

Answer:

The boiling point of HF is <u><em>higher than</em></u> the boiling point of H2, and it is <u><em>higher than</em></u> the boiling point of F2.

Explanation:

In HF, inter- molecule forces will be present between the hydrogen and fluorine atoms. There will be hydrogen bonding present among the hydrogen and fluorine atoms. Hydrogen bonds are strong bonds and hence the boiling point for HF would be high as much energy will be required to break these bonds.

H2 and F2 will only have intra-molecular attractions and there will be no hydrogen bonds present in them. As a result, their boiling point will be lower.

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<h2>Answer:  6 moles</h2>

<h3>Explanation:</h3>

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   ↓            ↓

4 mol       3 mol

Since the moles of N₂ is the smaller of the two reactants, then N₂ is the limiting factor (the reactant that will decide how much ammonia is produced since it has the smaller amount of moles). ∴ we have to use it in calculating the number of moles of ammonia

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