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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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<u>Atomic Structure</u>

  • Reading a Periodic Table
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  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

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Explanation:

<u>Step 1: Define</u>

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[Solve] atoms Xe

<u>Step 2: Identify Conversions</u>

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Avogadro's Number

<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                       \displaystyle 57.5 \ L \ Xe(\frac{1 \ mol \ Xe}{22.4 \ L \ Xe})(\frac{6.022 \cdot 10^{23} \ atoms \ Xe}{1 \ mol \ Xe})
  2. [DA] Divide/Multiply [Cancel out units]:                                                           \displaystyle 1.54583 \cdot 10^{24} \ atoms \ Xe

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

1.54583 × 10²⁴ atoms Xe ≈ 1.55 × 10²⁴ atoms Xe

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