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Karolina [17]
3 years ago
7

Place the following covalent bonds in order from least to most polar: a. H-Cl b. H-Br c. H-S d. H-C

Chemistry
1 answer:
timofeeve [1]3 years ago
7 0

Answer:


The Order is as follow,

                                     C-H < S-H < H-Br < H-Cl


Explanation:

                      Polarity depends on the electronegativity difference between two atoms, greater the electronegativity difference, greater will be the polarity of bond and vice versa.


Electronegativity Difference between Hydrogen and other given elements are as follow,


1) C-H;

               E.N of Carbon     = 2.55

               E.N of Hydrogen = 2.20

                                           ------------

               Difference              0.35


2) S-H;

               E.N of Sulfur       = 2.58

               E.N of Hydrogen = 2.20

                                           ------------

               Difference               0.38


3) H-Br;

               E.N of Bromine   = 2.96

               E.N of Hydrogen = 2.20

                                          -------------

               Difference              0.76


4) H-Cl;

               E.N of Chlorine   = 3.16

               E.N of Hydrogen = 2.20

                                           -----------

               Difference               0.96


Hence it is proved that the greatest electronegativity difference is found between H and Chlorine in H-Cl, therefore it is highly polar bond and vice versa.

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Answer:

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

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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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Answer:

A and D

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