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Phoenix [80]
3 years ago
11

Rationalize the difference in boiling points between the members of the following pairs of substances. Part APart complete HF (2

0 ∘C) and HCl (-85 ∘C) (20 ) and (-85 ) HF has the higher boiling point because HF molecules are more polar. HF has the higher boiling point because hydrogen bonding is weaker than dipole-dipole forces. HF has the higher boiling point because hydrogen bonding is stronger than dipole-dipole forces. HF has the higher boiling point because of ionic bonding. Previous Answers Correct HF has a much higher boiling point than HCl, indicating that it contains strong intermolecular forces. Specifically, it contains hydrogen bonding, which is a stronger force than dipole-dipole forces. Part B CHCl3 (61 ∘C) and CHBr3 (150 ∘C) (61 ) and (150 ) CHCl3 has the
Chemistry
1 answer:
Mrrafil [7]3 years ago
4 0

Answer:

HF has the higher boiling point because HF molecules are more polar. Part B: CHBr3 molecules possess stronger intermolecular interaction due to higher molar mass than CHCl3

Explanation:

Fluorine is more electronegative than chlorine. This implies that HF is more polar and possess stronger hydrogen bonds than HCl molecules.

In part B, the magnitude of dispersion forces depend on molar mass, the greater the molar mass, the greater the magnitude of dispersion forces between molecules, hence CHBr3 has a greater boiling point than CHCl3

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Answer:

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2 years ago
¿Cuál es la diferencia entre electrones internos y electrones externos? *
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True or false atoms with fewer then 4 outer elections lend elections?
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When two gases or two liquids form a solution, the substance that is present in the largest amount is the?
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3 years ago
How many moles of NH3 can be produced from 12.0 mol of H2 and excess N2? Express your answer numerically in moles. View Availabl
VladimirAG [237]

Answer:

A) 8.00 mol NH₃

B) 137 g NH₃

C) 2.30 g H₂

D) 1.53 x 10²⁰ molecules NH₃

Explanation:

Let us consider the balanced equation:

N₂(g) + 3 H₂(g) ⇄ 2 NH₃(g)

Part A

3 moles of H₂ form 2 moles of NH₃. So, for 12.0 moles of H₂:

12.0molH_{2}.\frac{2molNH_{3}}{3molH_{2}} =8.00molNH_{3}

Part B:

1 mole of N₂ forms 2 moles of NH₃. And each mole of NH₃ has a mass of 17.0 g (molar mass). So, for 4.04 moles of N₂:

4.04molN_{2}.\frac{2molNH_{3}}{1molN_{2}} .\frac{17.0gNH_{3}}{1molNH_{3}} =137gNH_{3}

Part C:

According to the <em>balanced equation</em> 6.00 g of H₂ form 34.0 g of NH₃. So, for 13.02g of NH₃:

13.02gNH_{3}.\frac{6.00gH_{2}}{34.0gNH_{3}} =2.30gH_{2}

Part D:

6.00 g of H₂ form 2 moles of NH₃. An each mole of NH₃ has 6.02 x 10²³ molecules of NH₃ (Avogadro number). So, for 7.62×10⁻⁴ g of H₂:

7.62 \times 10^{-4} gH_{2}.\frac{2molNH_{3}}{6.00gH_{2}} .\frac{6.02\times 10^{23}moleculesNH_{3}  }{1molNH_{3}}=1.53\times10^{20}moleculesNH_{3}

3 0
3 years ago
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