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SVEN [57.7K]
4 years ago
14

Name the type(s) of intermolecular forces that exists between molecules (or basic units) in each of the following species. (Sele

ct all that apply.) (a) CH4 dipole-dipole dispersion ion-dipole ion-ion (b) NO2 dipole-dipole dispersion ion-dipole ion-ion (c) SOZ dipole-dipole dispersion ion-dipole O ion-ion
Chemistry
1 answer:
Sloan [31]4 years ago
5 0

Explanation:

(a) CH₄

The molecule of methane is non polar in nature , hence it will show only the dispersion forces .

(b) NO₂

The molecule of nitrogen dioxide will show both dipole-dipole interactions and dispersion forces as well .

Since , the molecule is has an electronegative atom , i.e. , oxygen , so will show dipole - dipole interactions .

(c) SO₂

The molecule of sulfur dioxide will show both  dipole-dipole interactions and dispersion forces as well .

Since , the molecule is has an electronegative atom , i.e. , oxygen , so will show dipole - dipole interactions .

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Compare the physical properties of two forms of carbon: diamond and graphite
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<em>Diamond and graphite are allotropes of carbon  and have various differences in their physical properties.</em>

Explanation:

<u>Diamond:</u>

  1. It is crystalline in nature
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3 years ago
How many molecules are there in 1024 grams of Na2SO4?
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3 years ago
Nitrogen gas in an expandable container is cooled from 45.0 C to 12.3 C with the pressure held constant at 2.58 ∗ 105 Pa. The to
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Answer:

The number of moles of the gas is: -27.14 mole

the charge in internal energy of the gas is -1.84 × 10⁴ J.

The work done by the gas is 4.42 × 10⁴ J

The heat liberated by the gas for the same temperature change while the volume was constant and is same as the change in internal energy.

As such ; Q = -1.84 × 10⁴ J.

Explanation:

The expression for the number of moles of a gas at constant pressure is as follows:

\mathbf{n = \frac{Q}{Cp \Delta T}}

\mathbf{n = \frac{Q}{Cp (T_2-T_1)}}

where ;

C_p is the specific heat at constant pressure of Nitrogen gas which is = 29.07 J/mol/K

Since heat is liberated from the gas ; then:

n = \dfrac{-2.58*10^4 }{29.07(45-12.3)}

n = -27.14 mole

The number of moles of the gas is: -27.14 mole

b) The expression to be used in order to determine the change internal energy is:

dU = nCv \Delta T

where ;

n= 27.14 mole

Cv = specific heat at constant volume of Nitrogen gas = 20.76 J/mol/K

ΔT = (12.3-45)

So;

dU = (27.14)(20.76)(12.3-45)

dU = 563.426(-32.7)

dU = -18424.04328

dU = -1.84 × 10⁴ J

Thus; the charge in internal energy of the gas is -1.84 × 10⁴ J.

c)  The workdone by the gas can be calculated as;

W = Q - ΔU

W = 2.58 × 10⁴ J - (-1.84 × 10⁴ J )

W = 2.58  × 10⁴ J + 1.84  × 10⁴ J

W = 4.42 × 10⁴ J

The work done by the gas is 4.42 × 10⁴ J

d) The expression to calculated the work done is given as:

W = pdV

since the volume is given as constant ; then dV = 0

so;

W = p(0)

W = 0

Replacing 0 for W in the equation W = Q - Δ U

0 = Q - ΔU

-Q = - ΔU

Q = ΔU

Thus , the heat liberated by the gas for the same temperature change while the volume was constant and is same as the change in internal energy.

As such ; Q = -1.84 × 10⁴ J.

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