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Mariana [72]
3 years ago
13

Which of the following are dipole-dipole interactions that occur only between molecules containing N-H, O-H or F-H bonds?

Chemistry
2 answers:
Lelu [443]3 years ago
8 0

Answer:

Hydrogen bonding

Explanation:

Hydrogen bonding is a type of dipole dipole interaction. It exist between H atom bonded to electronegative atom; O, N, F and other O, N, F of another molecule.

Some of the examples of compounds having hydrogen bonding are:

NH_3,\ H_2O,\ HF,\ CH_3OH,\ etc.

Hydrogen bonding exists because of development of dipole dipole interaction between H and electronegative atom.

As H is less electronegative as compared to O, N and F. So partial positive charge develop on H and partial negative chagre develop develop on the attached electronegative atom. Because of development of partial positive and partial negative charge, dipole dipole interaction occurs which leads to the development of hydrogen bonding.

ivolga24 [154]3 years ago
5 0

Answer:

Hydrogen bonding

Explanation:

Hydrogen bonds are special dipole-dipole attraction between polar molecules in which a hydrogen molecule atom is directly joined to a highly electronegative atom.

This type of bond is an intermolecular bond caused by an electrostatic attraction between the hydrogen atom of one molecule and the electronegative atom(O or N or F) of a neighbouring molecule.

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Express the van der Waals equation of state as a virial expansion in powers of 1/Vm and obtain expressions for B and C in terms
nirvana33 [79]

Answer:

PV_{m} = RT[1 + (b-\frac{a}{RT})\frac{1}{V_{m} } + \frac{b^{2} }{V^{2} _{m} } + ...]

B = b -a/RT

C = b^2

a = 1.263 atm*L^2/mol^2

b = 0.03464 L/mol

Explanation:

In the given question, we need to express the van der Waals equation of state as a virial expansion in powers of 1/Vm and obtain expressions for B and C in terms of the parameters a and b. Therefore:

Using the van deer Waals equation of state:

P = \frac{RT}{V_{m}-b } - \frac{a}{V_{m} ^{2} }

With further simplification, we have:

P = RT[\frac{1}{V_{m}-b } - \frac{a}{RTV_{m} ^{2} }]

Then, we have:

P = \frac{RT}{V_{m} } [\frac{1}{1-\frac{b}{V_{m} } } - \frac{a}{RTV_{m} }]

Therefore,

PV_{m} = RT[(1-\frac{b}{V_{m} }) ^{-1} - \frac{a}{RTV_{m} }]

Using the expansion:

(1-x)^{-1} = 1 + x + x^{2} + ....

Therefore,

PV_{m} = RT[1+\frac{b}{V_{m} }+\frac{b^{2} }{V_{m} ^{2} } + ... -\frac{a}{RTV_{m} }]

Thus:

PV_{m} = RT[1 + (b-\frac{a}{RT})\frac{1}{V_{m} } + \frac{b^{2} }{V^{2} _{m} } + ...]           equation (1)

Using the virial equation of state:

P = RT[\frac{1}{V_{m} }+ \frac{B}{V_{m} ^{2}}+\frac{C}{V_{m} ^{3} }+ ...]

Thus:

PV_{m} = RT[1+ \frac{B}{V_{m} }+ \frac{C}{V_{m} ^{2} } + ...]     equation (2)

Comparing equations (1) and (2), we have:

B = b -a/RT

C = b^2

Using the measurements on argon gave B = −21.7 cm3 mol−1 and C = 1200 cm6 mol−2 for the virial coefficients at 273 K.

b = \sqrt{C} = \sqrt{1200} = 34.64[tex]cm^{3}/mol[/tex] = 0.03464 L/mol

a = (b-B)*RT = (34.64+21.7)*(1L/1000cm^3)*(0.0821)*(273) = 1.263 atm*L^2/mol^2

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