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2 years ago

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Which of the five type of equilibrium problems best applies to this question: Consider the following reaction at equilibrium. Wh

at effect will reducing the volume of the reaction mixture have on the system

1 answer:

andreev551 [17]2 years ago

3 0

Answer:

d

Explanation:

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Which one of the following best represents the predicted approximate chemical shift and coupling for the hydrogen(s) indicated w

melomori [17]

Answer:

The answer is quartet 2.40 ppm.

Note: Kindly find an attached image below for the part of the solution to this question

Sources: The image was researched from Course hero platform

Explanation:

Solution

Multiplicity or (n+1) rule:

It helps in determination of multiplicity of an individual proton or individual types of proton which are available in the molecule.

Multiplicity =(n+1)

Thus

The non equivalent protons which are attached from adjacent atom is denoted by n.

Now because there are three non-equivalent protons are present at adjacent carbon of methylene group, hence the multiplicity of methylene hydrogen is given as follows:

The multiplicity will be the same for the two hydrogen's. thus we compute multiplicity only for one hydrogen atom stated below:

Non- equivalent = 3

Multiplicity = (3 +1)

= 4

= Quartet for 2H

A quartet for 2H indicates that the hydrogen atoms attached from the carbon, which is attached one side from a methyl group and the other side form an atom that have no any hydrogens.

Now due +I effect of carbonyl group, chemical shift value is high for these two hydrogens which is exactly at 2.40 ppm or 2.40 Quartet.

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2 years ago

Assume that each atom is a sphere, and that the surface of each atom is in contact with its nearest neighbor. Determine the perc

tatiyna

Answer:

The percentage of unit cell volume that is occupied by atoms in a face- centered cubic lattice is 74.05%
The percentage of unit cell volume that is occupied by atoms in a body-centered cubic lattice is 68.03%
The percentage of unit cell volume that is occupied by atoms in a diamond lattice is 34.01%

Explanation:

The percentage of unit cell volume = Volume of atoms/Volume of unit cell

Volume of sphere = $\frac{4 }{3} \pi r^2$

a) Percentage of unit cell volume occupied by atoms in face- centered cubic lattice:

let the side of each cube = a

Volume of unit cell = Volume of cube = a³

Radius of atoms = $\frac{a\sqrt{2} }{4}$

Volume of each atom = $\frac{4 }{3} \pi (\frac{a\sqrt{2}}{4})^3$ = $\frac{\pi *a^3\sqrt{2}}{24}$

Number of atoms/unit cell = 4

Total volume of the atoms = $4 X \frac{\pi *a^3\sqrt{2}}{24} = \frac{\pi *a^3\sqrt{2}}{6}$

The percentage of unit cell volume = $\frac{\frac{\pi *a^3\sqrt{2}}{6}}{a^3} =\frac{\pi *a^3\sqrt{2}}{6a^3} = \frac{\pi \sqrt{2}}{6}$ = 0.7405

= 0.7405 X 100% = 74.05%

b) Percentage of unit cell volume occupied by atoms in a body-centered cubic lattice

Radius of atoms = $\frac{a\sqrt{3} }{4}$

Volume of each atom = $\frac{4 }{3} \pi (\frac{a\sqrt{3}}{4})^3$ = $\frac{\pi *a^3\sqrt{3}}{16}$

Number of atoms/unit cell = 2

Total volume of the atoms = $2X \frac{\pi *a^3\sqrt{3}}{16} = \frac{\pi *a^3\sqrt{3}}{8}$

The percentage of unit cell volume = $\frac{\frac{\pi *a^3\sqrt{3}}{8}}{a^3} =\frac{\pi *a^3\sqrt{3}}{8a^3} = \frac{\pi \sqrt{3}}{8}$ = 0.6803

= 0.6803 X 100% = 68.03%

c) Percentage of unit cell volume occupied by atoms in a diamond lattice

Radius of atoms = $\frac{a\sqrt{3} }{8}$

Volume of each atom = $\frac{4 }{3} \pi (\frac{a\sqrt{3}}{8})^3$ = $\frac{\pi *a^3\sqrt{3}}{128}$

Number of atoms/unit cell = 8

Total volume of the atoms = $8X \frac{\pi *a^3\sqrt{3}}{128} = \frac{\pi *a^3\sqrt{3}}{16}$

The percentage of unit cell volume = $\frac{\frac{\pi *a^3\sqrt{3}}{16}}{a^3} =\frac{\pi *a^3\sqrt{3}}{16a^3} = \frac{\pi \sqrt{3}}{16}$ = 0.3401

= 0.3401 X 100% = 34.01%

4 0

2 years ago