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svetlana [45]
3 years ago
7

Which molecule will undergo only London dispersion forces when interacting with other molecules of the same kind? a.HCl b.C2H5OH

c.NH3 d.CH4
Chemistry
1 answer:
ohaa [14]3 years ago
5 0

Answer:

CH4

Explanation:

London dispersion forces are a type of force acting between atoms and molecules that are normally electrically symmetric; that is, the electrons are symmetrically distributed with respect to the nucleus. They are part of the van der Waals forces(Wikipedia).  It is a temporary attractive force that results when the electrons in two adjacent atoms occupy positions that make the atoms form temporary dipoles. This force is sometimes called an induced dipole-induced dipole attraction(Chem Purdue).

If we consider HCl, C2H5OH  and NH3, we will discover that they all have polar bonds. Consequently, dipole - dipole interaction as well as hydrogen bonds exist between their molecules.

CH4 is nonpolar. Being a nonpolar molecule, the only possible intermolecular interaction forces between its molecules is the London dispersion forces.

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2.5 % butane, 42.2 % pentane and 55.3 % hexane

Explanation:

Hello,

In this case, the mass balance for each substance is given by:

Butane:z_bF=y_bD+x_bB\\\\Pentane: z_pF=y_pD+x_pB\\\\Hexane: z_hF=y_hD+x_hB

Whereas y accounts for the fractions at the outlet distillate and x for the fractions at the outlet bottoms. Moreover, with the 90 % recovery of butane, we can write:

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So we can compute the product of the molar fraction of butane at the distillate by total distillate flow by assuming a 100-mol feed:

y_bD=0.9*z_bF=0.9*0.2*100mol=18mol

The total distillate flow:

y_bD=18mol\\\\D=\frac{18mol}{0.95} =18.95mol

And the total bottoms flow:

F=D+B\\\\B=F-D=100mol-18.95mol=81.05mol

Next, by using the mass balance of butane, we compute the molar fraction of butane at the bottoms:

x_b=\frac{z_bF-y_bD}{B} =\frac{0.2*100mol-18mol}{81.05} =0.025

Then, the molar fraction of pentane and hexane:

x_p=\frac{z_pF-y_pD}{B} =\frac{0.35*100mol-0.04*18.95mol}{81.05} =0.422

x_h=\frac{z_hF-y_hD}{B} =\frac{(1-0.2-0.35)*100mol-(1-0.95-0.04)*18.95mol}{81.05} =0.553

Therefore, the molar composition of the bottom product is 2.5 % butane, 42.2 % pentane and 55.3 % hexane.

NOTE: notice the result is independent of the value of the assumed feed, it means that no matter the basis, the compositions will be the same for the same recovery of butane at the feed, only the flows will change.

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