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labwork [276]
3 years ago
7

The solubility of a gas in water is 0.16g/L at 104 kPa. What is the solubility when the pressure of the gas is increased to 288

kPa? Assume the temperature remains constant.
Chemistry
1 answer:
stepladder [879]3 years ago
7 0
Answer:
new solubility = 0.443 g/l

Explanation:
To solve this question, we will use Henry's law.
Henry's law states that, at constant temperature:
S1 / P1 = S2 / P2

We are given that:
S1 = 0.16 g/l
P1 = 104 kPa
S2 is the solubility we want to calculate
P2 = 288 kPa

Substitute with the givens in the above relation to get the new solubility (S2) as follows:
0.16 / 104 = S2 / 288
S2 = (0.16/104) * 288
S2 = 0.443 g/l

Hope this helps :)

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Answer:

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Explanation:

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2 NaN₃(s) ⇒ 2 Na(s) + 3 N₂(g)

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N₂ occupies a 80.0 L bag at 1.3 atm and 27 °C (300 K). We will calculate the moles of N₂ using the ideal gas equation.

P × V = n × R × T

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n = 1.3 atm × 80.0 L / (0.0821 atm.L/mol.K) × 300 K = 4.2 mol

We can also calculate the mass of nitrogen using the molar mass (M) 28.01 g/mol.

4.2 mol × 28.01 g/mol = 1.2 × 10² g

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The mass ratio of NaN₃ to N₂ is 130.02:84.03.

1.2 × 10² g N₂ × 130.02 g NaN₃/84.03 g N₂ = 1.9 × 10² g NaN₃

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We will use the following expression.

ρ = P × M / R × T

ρ = 1.3 atm × 28.01 g/mol / (0.0821 atm.L/mol.K) × 300 K = 1.5 g/L

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