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Rina8888 [55]
2 years ago
15

Of ozone, molecular oxygen, and oxygen radicals, which is the most stable form of oxygen? How do you know?

Chemistry
1 answer:
PtichkaEL [24]2 years ago
5 0

Oxygen has a strong double bond which has more stability than the single co-ordinate bond in ozone, therefore more energy is required to break the O2 bonding than ozone, so the ozone molecule is more reactive than oxygen gas. ... The oxygen free radical contains two unpaired electrons in its valence shell.

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Ulfur dioxide gas can react in the presence of oxygen and vanadium(V) oxide to form sulfur trioxide. Sulfur trioxide is the only
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According to the illustration, the vanadium (V) oxide would be a catalyst.

<h3>What are catalysts?</h3>

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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

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P_{2} = 100 \ kPa

Putting the parameters into the formula, we get

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

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