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Bezzdna [24]
3 years ago
12

which of the following substances have polar interactions (dipole-dipole forces) between molecules? F2, Cl2, ClF, NF3

Chemistry
2 answers:
Alex787 [66]3 years ago
5 0
F2 and Cl2 are non-polar because both contains same atom and they have same electronegativity.

ClF and NF3 have polar interactions because there is electronegativity difference between atoms.

Hope this helps!
serious [3.7K]3 years ago
3 0

Answer:

Explanation:

The covalent bond is the chemical bond between atoms where electrons are shared, forming a molecule. Covalent bonds are established between non-metallic elements, such as hydrogen H, oxygen O and chlorine Cl. These elements have many electrons in their outermost level (valence electrons) and have a tendency to gain electrons to acquire the stability of the electronic structure of noble gas. The shared electron pair is common to the two atoms and holds them together.

Electronegativity is defined as the ability of an element to attract the electrons that link it to another element.

The covalent bond between two atoms can be polar or nonpolar.

When two atoms have different electronegativities, the one with the highest electronegativity will attract the electrons towards each other, giving rise to two opposite charges on the bond. That is, this generates that in a polar molecule there is separation between positive and negative charges. The bonds will be all the more polar the greater the difference in electronegativity between the bound atoms.

On the other hand, the non-polar covalent bond occurs between atoms of the same element or between atoms with very little electronegativity difference. It is thus characterized to molecules or bonds that do not exhibit any polarity.

A molecule is a dipole when there is an asymmetric distribution of electrons because the molecule is made up of atoms of different electronegativity. That is, dipole is formed when the molecule is polar.

So, dipole-dipole forces are forces of attraction between polar molecules. These molecules attract when the positive end of one of them is close to the negative of the other.

Given the above, F₂ and Cl₂ are non-polar molecules and ClF and NF₃ are polar molecules. Then ClF and NF₃ have polar interactions (dipole-dipole forces).

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The equilibrium constant in terms of pressure. Kp for the following process is 0.179 at 50 °C. It increases to 0.669 at 86 °C. C
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Answer:

a) ΔHvap=35.3395 kJ/mol

b) Tb=98.62 °C

Explanation:

Given the reaction:

C₇H₁₆ (l) ⇔ C₇H₁₆ (g)

Kp=P(C₇H₁₆) since the concentration ratio for a pure liquid is equal to 1.

When

T₁=50°C=323.15K ⇒P₁=0.179

T₂=86°C=359.15K ⇒P₂=0.669

The Clasius-Clapeyron equation is:

ln(\frac{P_2}{P_1}) =-\frac{AH_{vap}}{R} (\frac{1}{T_2}-\frac{1}{T_1})

ln(\frac{0.669}{0.179}) =-\frac{AH_{vap}}{8.3145 J.mol^{-1}K^{-1}} (\frac{1}{359.15K}-\frac{1}{323.15K})

1.3184 =-\frac{AH_{vap}}{8.3145 J.mol^{-1}K^{-1}} (-3.10186*10^{-4}K{^-1})

ΔHvap=35339.5 J/mol=35.3395 KJ/mol

Normal boiling point ⇒ P=1 atm

Hence, we find the normal boiling point where:

T₁=323.15K

P₁=0.179 atm

P₂=1 atm

ln(\frac{P_2}{P_1}) =-\frac{AH_{vap}}{R} (\frac{1}{T_2}-\frac{1}{T_1})

ln(\frac{1atm}{0.179atm}) =-\frac{35339.5 J/mol}{8.3145 J.mol^{-1}K^{-1}} (\frac{1}{T_2}-\frac{1}{323.15K})

1.7203=-4250.34 (\frac{1}{T_2}-\frac{1}{323.15K})

T₂=371.77 K= 98.62 °C

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Answer:

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Explanation:

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