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The pressure of the gas when it's temperature reaches 928 °C is 3823,36 kPa
To solve that we need to apply
Gay-Lussac's Law. It states that the pressure of a gas when the volume is left constant (like in the case of a sealed container like an aerosol can) is proportional to temperature. This is the relationship derived from this law that we use to solve this problem:
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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
Answer:
sub-particle charge mass
protons +1 1
neutron 0 1
electron - 1 negligible
protons and neutrons are found in the nucleus
electrons in the shells orbiting the nucleus
Answer:
If the volume of the container is decreased by a factor of 2 the pressure is is increased by the same factor to 1664 torr.
Explanation:
Here we have Boyle's law which states that, at constant temperature, the volume of a given mass of gas is inversely proportional to its pressure
V ∝ 1/P or V₁·P₁ = V₂·P₂
Where:
V₁ = Initial volume
V₂ = Final volume = V₁/2
P₁ = Initial pressure = 832 torr
P₂ = Final pressure = Required
From V₁·P₁ = V₂·P₂ we have,
P₂ = V₁·P₁/V₂ = V₁·P₁/(V₁/2)
P₂ = 2·V₁·P₁/V₁ = 2·P₁ = 2× 832 torr = 1664 torr
Answer:
8.9 g / cm^3
Explanation:
Density = mass / volume
We are given volume which is 10.0cm^3 (cm^3 is just fancy way of saying ml), and mass of 89 gram. Just plug them in respective spots.
Density = 89 / 10.0 = 8.9 g / cm^3
