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laila [671]
3 years ago
13

The electron cloud of HF is smaller than that of F2, however, HF has a much higher boiling point than F2 has. Which of the follo

wing explains how the dispersion-force model of intermolecular attraction does not account for the unusually high boiling point of HF?
A. F2 is soluble in water, whereas HF is insoluble in water.
B. The F2 molecule has a greater mass than the HF molecule has.
C. Liquid F2 has weak dispersion force attractions between its molecules, whereas liquid HF has strong ionic interactions between H+ and F- ions.
D. Liquid F2 has weak dispersion force attractions between its molecules, whereas liquid HF has both weak dispersion force attractions and hydrogen bonding interactions between its molecules.
Chemistry
1 answer:
mafiozo [28]3 years ago
7 0

Answer:D

Explanation:

The high boiling point of HF is not attributable to the dispersion forces mentioned in the question. In HF, a stronger attraction is in operation, that is hydrogen bonding. This ultimately accounts for the high boiling point and not solely the dispersion model as in F2.

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Calculate the freezing point of a solution containing 5.0 grams of KCl and 550.0 grams of water. The molal-freezing-point-depres
yulyashka [42]

<u>Answer:</u> The freezing point of solution is -0.454°C

<u>Explanation:</u>

Depression in freezing point is defined as the difference in the freezing point of pure solution and freezing point of solution.

The equation used to calculate depression in freezing point follows:

\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}

To calculate the depression in freezing point, we use the equation:

\Delta T_f=iK_fm

Or,

\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

where,

Freezing point of pure solution = 0°C

i = Vant hoff factor = 2

K_f = molal freezing point elevation constant = 1.86°C/m

m_{solute} = Given mass of solute (KCl) = 5.0 g

M_{solute} = Molar mass of solute (KCl) = 74.55 g/mol

W_{solvent} = Mass of solvent (water) = 550.0 g

Putting values in above equation, we get:

0-\text{Freezing point of solution}=2\times 1.86^oC/m\times \frac{5\times 1000}{74.55g/mol\times 550}\\\\\text{Freezing point of solution}=-0.454^oC

Hence, the freezing point of solution is -0.454°C

3 0
3 years ago
Which of the following examples includes only chemical changes that occur in
Paul [167]
I don’t think you can get it for
6 0
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3. A measurement of the freezing temperature of a solution allowsyou to calculate the concentration of the solution. What else d
Harman [31]

Answer:

Osmotic pressure and boiling point elevation

Explanation:

In the the osmotic pressure one can determine the molar mass of a solid by calculating the number  of  moles from the Morality formula which involves the volume of the solution.

In the boiling point elevation you can determine the number of moles of the solute in the solution by using the Molality formula.

5 0
3 years ago
Typically if oxygen has a formal negative charge, it is participating in ____ covalent bond(s). Report your answer as a whole nu
ZanzabumX [31]

Oxygen is participating in non-polar covalent bond(s)

Oxygen forms 2 covalent bond.

This is because oxygen atoms have 6 valence electrons. This means that it has 2 lone pairs and 2 unpaired electrons that are shared in order to achieve octet configuration.

In this chemistry, the 2 lone pairs on the oxygen are not shared with any other atoms. Instead; they are assigned to the oxygen atom. The formal charge on the oxygen atom is zero. Oxygen's atomic number is 8 and is equal to the sum of the number of its valence and inner shell electrons.

<h3>What is non-polar covalent bonding?</h3>

Nonpolar covalent bonding is a kind of covalent bond in which the bonding electrons are shared equally between the two atoms.

Learn more about non-polar covalent bonding:

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8 0
3 years ago
Suppose you are titrating a sulfuric acid solution of unknown concentration with a sodium hydroxide solution according to the eq
pogonyaev

Answer:

Explanation:

H ₂ S O ₄ + 2 N a O H ⟶ 2 H ₂ O + N a ₂ S O ₄

29.09 mL of 0.639 M N a O H is mixed with 213.8 mL of H ₂ S O ₄

Let the concentration of H ₂ S O ₄ be S₂ .

In terms of normal or equivalent solution is will be 2 N  solution

From the formula S₁ V₁ = S₂ V₂

= 29.09 x .639 = 213.8 x  S₂

S₂ = .087 N solution

In terms of molar solution it will be .087 / 2 M

= .0435 M

5 0
3 years ago
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