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Gnom [1K]
3 years ago
14

The stronger the intermolecular forces, the _______the energy needed to separate the molecules, the ____________the boiling poin

t.
Please only answer if you actually know it!
Chemistry
1 answer:
slamgirl [31]3 years ago
7 0

Answer:

higher, higher

Explanation:

It takes more energy to rip apart stronger bonds (that's mostly just common sense there). The boiling point increases because it would take more energy to get the molecules to go from a stuck together liquid, to separating in a gaseous form.

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What is the ratio of effusion rates for the lightest gas, h2, to the heaviest known gas, uf6?
andriy [413]

The ratio of effusion rates for the lightest gas H₂ to the heaviest known gas UF₆ is 13.21 to 1

<h3>What is effusion?</h3>

Effusion is a process by which a gas escapes from its container through a tiny hole into evacuated space.

Rate of effusion ∝ 1/√Ц, (where Ц is molar mass)

Rate H₂ = 1/√ЦH₂

Rate UF₆ = 1/√ЦUF₆

Therefore, Rate H₂/ Rate UF₆ = √ЦH₂/√ЦUF₆

ЦH₂= 2.016 g/mol

ЦUF₆= 352.04 g/mol

Rate H₂ / Rate UF₆ = √352.04/√2.016 = 18.76/1.42

Rate H₂ / Rate UF₆ = 13.21

Therefore, H₂ is lower mass than UF₆. Thus H₂ gas will effuse 13 times more faster than UF₆ because the most probable speed of H₂ molecule is higher; therefore, more molecules escapes per unit time.

learn more about effusion rate: brainly.com/question/28371955

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8 0
1 year ago
Identify the oxidizing agent and the reducing agent for 4Li(s) + O_2 (g) to 2Li_2O(s).
wolverine [178]

Answer: Li is the reducing agentg and O is the oxidizing agent.

Explanation:

1) The oxidizing agent is the one that is reduced and the reducing agent is the one that is oxidized.

2) The given reaction is:

4Li(s) + O₂ (g) → 2 Li₂O(s)

3) Determine the oxidation states of each atom:

Li(s): oxidation state = 0 (since it is alone)

O₂ (g): oxidation state = 0 (since it is alone)

Li in Li₂O (s) +1

O in Li₂O -2

That because 2× (+1) - 2 = 0.

4) Determine the changes:

Li went from 0 to + 1, therefore it got oxidized and it is the reducing agent.

O went from 0 to - 2, therefore it got reduced and it is the oxidizing agent.

7 0
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Answer:

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Explanation:

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Answer:An iodide ion is the ion I−. Compounds with iodine in formal oxidation state −1 are called iodides. This page is for the iodide ion and its salts, not organo iodine compounds. In everyday life, iodide is most commonly encountered as a component of iodized salt, which many governments mandate.

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