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Neko [114]
3 years ago
6

Oxygen gas can be prepared by heating potassium chlorate according to the following equation:

Chemistry
1 answer:
Cerrena [4.2K]3 years ago
4 0

Moles KClO₃ = 0.239

<h3>Further explanation</h3>

Given

Reaction

2KClO₃(s) ⇒2KCl(s) + 3O₂(g)

P water = 23.8 mmHg

P tot = 758 mmHg

V = 9.07 L

T = 25 + 273 = 298 K

Required

moles of KClO₃

Solution

P tot = P O₂ + P water

P O₂ = P tot - P water

P O₂ = 758 - 23.8

P O₂ = 734.2 mmHg = 0.966 atm

moles O₂ :

n = PV/RT

n = 0.966 x 9.07 / 0.082 x 298

n = 0.358

From equation, mol ratio KClO₃ : O₂ = 2 : 3, so mol KClO₃ :

= 2/3 x mol O₂

= 2/3 x 0.358

= 0.239

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Which of the following solutes has the greatest effect on the colligative properties for a given mass of pure water? Explain.
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Option a.

0.01 mol of CaCl₂ will have the greatest effect on the colligative properties, because it has the biggest i

Explanation:

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Answer:

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CH₃NH₃⁺ is a weak acid, according to the following equation:

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