Answer:
= 913.84 mL
Explanation:
Using the combined gas laws
P1V1/T1 = P2V2/T2
At standard temperature and pressure. the pressure is 10 kPa, while the temperature is 273 K.
V1 = 80.0 mL
P1 = 109 kPa
T1 = -12.5 + 273 = 260.5 K
P2 = 10 kPa
V2 = ?
T2 = 273 K
Therefore;
V2 = P1V1T2/P2T1
= (109 kPa × 80 mL × 273 K)/(10 kPa× 260.5 K)
<u>= 913.84 mL</u>
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Answer:
1. The gas law used: Dalton's law of partial pressure.
2. Pressure of nitrogen = 331 mmHg
Explanation:
From the question given above, the following data were obtained:
Total pressure (Pₜ) = 592 mmHg
Pressure of Oxygen (Pₒ) = 261 mmHg
Pressure of nitrogen (Pₙ) =?
The pressure of nitrogen in the sample can be obtained by using the Dalton's law of partial pressure. This is illustrated below:
Pₜ = Pₒ + Pₙ
592 = 261 + Pₙ
Collect like terms
592 – 261 = Pₙ
331 = Pₙ
Pₙ = 331 mmHg
Therefore, the pressure of nitrogen in the sample is 331 mmHg
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