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bezimeni [28]
2 years ago
5

During the distillation nitrogen gas is obtained first, then argon and oxygen. What can u say about the boiling points of these

three gases?
Chemistry
2 answers:
svetlana [45]2 years ago
6 0

Explanation:

The boiling point of liquid nitrogen, liquid argon, and liquid oxygen are -196°C, -186°C, and -183°C respectively. So, the correct order is nitrogen, argon, oxygen.

marissa [1.9K]2 years ago
5 0

Nitrogen:

Since you don't want nitrogen to be condensing everywhere when it gets cold out, elemental nitrogen's boiling point is luckily below zero degrees Celsius. Thank goodness for nitrogen's extremely low boiling point. If it were at zero, boiling water would not produce a lot of heat, necessitating the use of a pressure cooker to prepare foods that typically call for boiling water. Additionally, since the air would be substantially enhanced with oxygen, items may catch fire more quickly (i.e. your home might burst in flame that much more easily). An ugly picture, to be sure.

Argon:

As nonpolar gases, Ar, F2, and Cl2 only display London Dispersion Forces (LDF). These largely negligible forces, which are the product of transient dipoles, must be resisted throughout the boiling process. Since bigger atoms and molecules have valence electrons further from their nucleus, LDF intensity is also directly correlated with atomic or molecular size. Compared to smaller atoms or molecules, these outermost electrons are less attracted to the positive nucleus or nuclei, making it simpler for them to create transient dipoles. As a result, the differing molar masses of each compound—Ar—39.95 g/mol, F—38.00 g/mol, and Cl—70.90 g/mol—can be used to explain the variation in boiling points.

Oxygen:

At 1 atm, oxygen would boil at -183 degrees Celsius if it were a liquid. Although the phase diagram for oxygen is more intricate than this, it is a general phase diagram. Imagine that the term "vaporization" is at the same height as 1 atm, which is roughly halfway up the y-axis. If the y value at that time was equal to 1 atm, the "vaporization" arrow to the right would be directly above -183 degrees Celsius. It's crucial to remember that pressure determines what temperature a substance will boil, melt, condense, or freeze, as you can probably guess. Additionally, the pressure in particular circumstances can affect what phase a material is in.

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swat32
<h3>Answer:</h3>

0.239 mol Sm(NO₃)₃

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
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<u>Chemistry</u>

<u>Atomic Structure</u>

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<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] 80.3 g Sm(NO₃)₃

[Solve] moles Sm(NO₃)₃

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

[PT] Molar Mass of Sm - 150.36 g/mol

[PT] Molar Mass of N - 14.01 g/mol

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Molar Mass of Sm(NO₃)₃ - 150.36 + 3[14.01 + 3(16.00)] = 336.39 g/mol

<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                 \displaystyle 80.3 \ g \ Sm(NO_3)_3(\frac{1  \ mol \ Sm(NO_3)_3}{336.39 \ g \ Sm(NO_3)_3})
  2. [DA] Divide [Cancel out units]:                                                                        \displaystyle 0.238711 \ mol \ Sm(NO_3)_3

<u>Step 4: Check</u>

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

0.238711 mol Sm(NO₃)₃ ≈ 0.239 mol Sm(NO₃)₃

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The rate change if(d) [Sucrose] and [H⁺] are both changed to 0.1 M will increase by a factor of 1.

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The rate law of reaction is R = k [H+] [sucorse]

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2 / 1

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