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NikAS [45]
3 years ago
7

Arrange the molecules H2O, NH3, Ar, NaCl in order of expected increasing boiling points. 1. None of these 2. NaCl, H2O, NH3, Ar

3. Ar, NH3, H2O, NaCl 4. NH3, Ar, H2O, NaCl
Chemistry
1 answer:
Aleks04 [339]3 years ago
4 0

Answer:

Option (4) is correct

Explanation:

  • NaCl is an ionic solid. Na^{+} and Cl^{-} ions are held together by strong coloumbic attractive force. Hence NaCl has highest boiling point'
  • Ar is a monoatomic molecule. Hence only weak london dispersion force exists between Ar atom/molecules. Hence it requires lowest amount of energy to boil and thereby possess lowest boiling point.
  • NH_{3} and H_{2}O are polar protic molecules. Hence they possess london dispersion force, dipole-dipole force and hydrogen bonding as intermolecular forces.
  • Dipole-dipole force is stronger in H_{2}O than NH_{3} due to more polar O-H bond than N-H bond. Also molecular weight of H_{2}O is higher than NH_{3}. So, more energy is required to boil H_{2}O. So, H_{2}O has higher boiling point than NH_{3}

Hence order of boiling point : NaCl>H_{2}O>NH_{3}>Ar

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Solid sodium iodide is slowly added to a solution that is 0.0050 M Pb 2+ and 0.0050 M Ag +. [K sp (PbI 2) = 1.4 × 10 –8; K sp (A
UkoKoshka [18]

Answer:

[Ag⁺] = 5.0x10⁻¹⁴M

Explanation:

The product solubility constant, Ksp, of the insoluble salts PbI₂ and AgI is defined as follows:

Ksp(PbI₂) = [Pb²⁺] [I⁻]² = 1.4x10⁻⁸

Ksp(AgI) = [Ag⁺] [I⁻] = 8.3x10⁻¹⁷

The PbI₂ <em>just begin to precipitate when the product  [Pb²⁺] [I⁻]² = 1.4x10⁻⁸</em>

<em />

As the initial [Pb²⁺] = 0.0050M:

[Pb²⁺] [I⁻]² = 1.4x10⁻⁸

[0.0050] [I⁻]² = 1.4x10⁻⁸

[I⁻]² = 1.4x10⁻⁸ / 0.0050

[I⁻]² = 2.8x10⁻⁶

<h3>[I⁻] = 1.67x10⁻³</h3><h3 />

So, as the [I⁻] concentration is also in the expression of Ksp of AgI and you know concentration in solution of I⁻ = 1.67x10⁻³M:

[Ag⁺] [I⁻] = 8.3x10⁻¹⁷

[Ag⁺] [1.67x10⁻³] = 8.3x10⁻¹⁷

<h3>[Ag⁺] = 5.0x10⁻¹⁴M</h3>

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Answer:

i can relate

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