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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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In the reaction 2H2O (1)+ 2Cl^- (aq)= H2(g)+Cl2 (g)+ 2OH^-(aq), which substance is reduced?
Oksanka [162]

Answer:- C. H

Explanations:- Reduction is gain of electron. In other words we could say that decrease in oxidation number is reduction.

As per the rules, oxidation number of hydrogen in its compounds is +1(except metal hydrides) and the oxidation number of oxygen in its compounds is -2.

The oxidation number in elemental form is zero.

In H_2O , the oxidation number of H is +1 and oxidation number of O is -2. Oxidation number of Cl in Cl^- is -1. On product side, the oxidation number of hydrogen in H_2 is zero and in OH^- the oxidation number of H is +1 and that of O is -2. Oxidation number of Cl in Cl_2 is 0.

From above data, Oxidation number of O is -2 on both sides so it is not reduced.

Oxidation number of Cl is changing from -1 to 0 which is oxidation.

Oxidation number of H is changing from +1 to 0 which is reduction.

So, the right choice is C.H

8 0
2 years ago
A mouse is placed in a sealed chamber with air at 769.0 torr. This chamber is equipped with enough solid KOH to absorb any CO2 a
svlad2 [7]

<u>Answer:</u> The amount of oxygen gas consumed by mouse is 0.202 grams.

<u>Explanation:</u>

We are given:

Initial pressure of air = 769.0 torr

Final pressure of air = 717.1 torr

Pressure of oxygen = Pressure decreased = Initial pressure - Final pressure = (769.0 - 717.1) torr = 51.9 torr

To calculate the amount of oxygen gas consumed, we use the equation given by ideal gas which follows:

PV=nRT

where,

P = pressure of the gas = 51.9 torr

V = Volume of the gas = 2.20 L

T = Temperature of the gas = 292 K

R = Gas constant = 62.364\text{ L. Torr }mol^{-1}K^{-1}

n = number of moles of oxygen gas = ?

Putting values in above equation, we get:

51.9torr\times 2.20L=n\times 62.364\text{ L. Torr }mol^{-1}K^{-1}\times 292K\\\\n=\frac{51.9\times 2.20}{62.364\times 292}=0.0063mol

To calculate the mass from given number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Moles of oxygen gas = 0.0063 moles

Molar mass of oxygen gas = 32 g/mol

Putting values in above equation, we get:

0.0063mol=\frac{\text{Mass of oxygen gas}}{32g/mol}\\\\\text{Mass of oxygen gas}=(0.0063mol\times 32g/mol)=0.202g

Hence, the amount of oxygen gas consumed by mouse is 0.202 grams.

5 0
2 years ago
Ethanol (CH3CH2OH), the intoxicant in alcoholic beverages, is also used to make other organic compounds. In concentrated sulfuri
Fed [463]

Answer:

a) 88.48%

b) 0.05625 mol

Explanation:

2CH₃CH₂OH(l) → CH₃CH₂OCH₂CH₃(l) + H₂O(g)         Reaction 1

CH₃CH₂OH(l) → CH₂═CH₂(g) + H₂O(g)                        Reaction 2

a) CH₃CH₂OH = 46.0684 g/mol

   CH₃CH₂OCH₂CH₃ = 74.12 g/mol

1 mol CH₃CH₂OH ______  46.0684 g

x                            ______   50.0 g

x = 1.085 mol  CH₃CH₂OH

1 mol  CH₃CH₂OCH₂CH₃ ______  74.12 g g

y                           ______   35.9 g

y = 0.48 mol   CH₃CH₂OCH₂CH₃

100% yield _____ 0.5425 mol CH₃CH₂OCH₂CH₃

w                _____  0.48 mol CH₃CH₂OCH₂CH₃

w = 88.48%

b) Only 0.96 mol of ethanol reacted to form diethyl ether. This means that 0.125 mol of ethanol did not react. 45% of 0.125 mol reacted to form ethylene. Therefore, 0.05625 mol of ethanol reacted by the side reaction (reaction 2). Since 1 mol of ethanol leads to 1 mol of ethylene, 0.05625 mol of ethanol produces 0.05625 mol of ethylene.

4 0
3 years ago
Sean drew this diagram and labeled an unknown gas as gas A. Which gas does A represent?
ANEK [815]
Gas A is Carbon Dioxide
3 0
3 years ago
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The diagram shows the box for an element in the periodic
mr_godi [17]

Answer: A 13

Explanation: hope this helps!

6 0
3 years ago
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