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givi [52]
3 years ago
13

2KCIO3 -2KCL + 302

Chemistry
1 answer:
Firlakuza [10]3 years ago
5 0

Answer:

a. 6 moles of O₂

b. 4.06×10²⁴ molecules of O₂

c. 3.04 g of KCl

Explanation:

Reaction of decomposition is:

2KClO₃  →  2KCl  +  3O₂

a. See stoichiometry value

2 moles of potassium chlorate can decompose to 2 moles of potassium chloride and 3 moles of oxygen. Ratio is 2:3

If 2 moles of KClO₃ can decompose to 3 moles of O₂

Then 4 moles, may decompose to (4 . 3)/2 = 6 moles of O₂

b. In this case, the stoichiometry is the same.

Per 2 moles of KClO₃, I produce 2 moles of KCl

Then, 4.5 moles of KCl, were produced by 4.5 moles of KClO₃

We apply, the last relation:

(4.5 . 3) /2 = 6.75 moles of O₂ are also produced.

How many molecules are in 6.75 moles?

6.75 mol . 6.02×10²³ molecules/mol = 4.06×10²⁴ molecules of O₂

c. First of all, we convert the mass to moles:

5g . 1mol /122.55g = 0.0408 moles of salt

As ratio is 2:2, 0.0408 moles of salt, decompose to 0.0408 moles of KCl

We convert the moles to mass: 0.0408 mol . 74.55g /mol = 3.04g

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The solubility of a gas is 0.890 g/L at a pressure of 120 kPa. What is the solubility of the gas if the pressure is changed to 1
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The solubility of the gas if the pressure is changed to 100 kPa is 0.742 g/L

<h3>Effect of Pressure on Solubility </h3>

As the <em>pressure </em>of a gas increases, the <em>solubility </em>increases, and as the <em>pressure </em>of a gas decreases, the <em>solubility </em>decreases.

Thus, Solubility varies directly with Pressure

If S represents Solubility and P represents Pressure,

Then we can write that

S ∝ P

Introducing proportionality constant, k

S = kP

S/P = k

∴ We can write that

\frac{S_{1} }{P_{1} } = \frac{S_{2} }{P_{2} }

Where S_{1} is the initial solubility

P_{1} is the initial pressure

S_{2} is the final solubility

P_{2} is the final pressure

From the given information

S_{1} = 0.890 \ g/L

P_{1} = 120 \ kPa

P_{2} = 100 \ kPa

Putting the parameters into the formula, we get

\frac{0.890}{120}=\frac{S_{2}}{100}

S_{2}= \frac{0.890 \times 100}{120}

S_{2}= 0.742 \ g/L

Hence, the solubility of the gas if the pressure is changed to 100 kPa is 0.742 g/L

Learn more on Solubility here: brainly.com/question/4529762

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