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Ganezh [65]
2 years ago
14

The triple point of nitrogen occurs at a temperature of 63.1 K and a pressure of 0.127 atm. Its normal boiling point is 77.4 K.

(a) What is the vapor pressure of solid nitrogen at 63.1 K
Chemistry
1 answer:
yulyashka [42]2 years ago
5 0

Nitrogen has a normal boiling point of 77.4 K and a melting point (at 1 atm) of 63.2 K. Its critical temperature is 126.2 K, and its critical pressure is 2.55 * 104 torr. It has a triple point at 63.1 K and 94.0 torr.

<h3>What is the triple point?</h3>

The temperature and pressure at which the solid, liquid, and vapour phases of a pure substance can coexist in equilibrium.

- Normal melting point: 63.2 K.

- Normal boiling point: 77.4 K.

- Triple point: 0.127 atm and 63.1 K.

- Critical point: 33.5 atm and 126.0 K.

In such a way:

- N2 does not exist as a liquid at pressures below 0.127 atm: that is because below this point, solid N2 exists only (triple point).

- N2 is a solid at 16.7 atm and 56.5 K: that is because it is above the triple point, below the critical point and below the normal melting point.

- N2 is a liquid at 1.00 atm and 73.9 K: that is because it is above the triple point, below the critical point and below the normal boiling point.

- N2 is a gas at 0.127 atm and 84.0 K: that is because it is above the triple point temperature at the triple point pressure.

Learn more about triple point here:

brainly.com/question/23307017

#SPJ1

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A sample of N2O effuses from a container in 47s .
malfutka [58]

Answer:

112.92 s.

Explanation:

Let M₁ be the molar mass of N₂O

Let t₁ be the time taken for N₂O to effuse.

Let M₂ be the molar mass of I₂

Let t₂ be the time taken for I₂ to effuse.

Molar mass (M₁) of N₂O = (14×2) + 16 = 28 + 16 = 44 g/mol

Time (t₁) of effusion of N₂O = 47 s

Molar mass (M₂) of I₂ = 127 × 2 = 254 g/mol

Time (t₂ ) of effusion of I₂ =?

The time take for the same amount of I₂ to effuse can be obtained as follow:

t₂/t₁ = √(M₂/M₁)

t₂/47 = √(254 / 44)

Cross multiply

t₂ = 47 × √(254 / 44)

t₂ = 112.92 s

Therefore, it will take 112.92 s for the same amount of I₂ to effuse.

7 0
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Bases on Graham's law of diffusion, how does the speed of diffusion of neon (M=20.2) compare to the Krpton (M=83.8)
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Answer:

2.037.

Explanation:

  • Thomas Graham found that, at a constant  temperature and pressure the rates of effusion  of various gases are inversely proportional to  the square root of their masses.

<em>∨ ∝ 1/√M.</em>

Where, ∨ is the rate of diffusion.

M is the molar mass of the gas.

  • For Ne and Kr:

<em>(∨)Ne/(∨)Kr = √(M of Kr/M of Ne)</em> = √(83.8 / 20.2) = <em>2.037.</em>

5 0
3 years ago
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