The spacing between the layers that is present in the potassium iodide crystal that gives rise to 14.7 degrees of diffraction has to be calculated.

Question

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Answer 1

Question

Chapter 11, Problem 11.88QE

(a)

Interpretation Introduction

Interpretation:

The spacing between the layers that is present in the potassium iodide crystal that gives rise to $14.7$ degrees of diffraction has to be calculated.

(a)

Expert Solution

Answer to Problem 11.88QE

The spacing between the layers in crystal of $KI$ is $3.526×10−10 m$ .

Explanation of Solution

Wavelength of X-ray is given as $1.790×10−10 m$ . The angle of diffraction is given as $14.7$ degrees.

Bragg equation can be used to determine the spacing between the layers. Bragg equation is given as shown below;

$d=nλ2sinθ$ (1)

Where,

$n$ is given as 1.

$d$ is the distance between layers.

$λ$ is the wavelength of X-ray.

$θ$ is the angle of diffraction.

Substituting the values in equation (1), the spacing between the layers can be calculated as shown below;

$d=1×1.790×10−10 m2(sin14.7°)=1.790×10−10 m2(0.2538)=3.526×10−10 m$

Therefore, the spacing between layers is $3.526×10−10 m$ .

(b)

Interpretation Introduction

Interpretation:

Density of $KI$ has to be calculated.

(b)

Expert Solution

Answer to Problem 11.88QE

Density of potassium iodide is $3.144 g/cm3$ .

Explanation of Solution

The crystal is of $KI$ . The spacing between the layers is $3.526×10−10 m$ . The edge length of the cube is given as twice of the length of spacing between the layers. Therefore, the edge length is calculated as follows;

$Edge length=2×3.526×10−10 m=7.052×10−10 m$

The unit cell of potassium iodide is face centered cubic cell. Therefore, there are eight corners and thus one atom is shared by eight other unit cells that are present. One atom is present at the center of each face of the cell. Therefore, the total number of atoms present is calculated as shown below;

$6 faces×1 atom2 faces+8 corners×1 atom8 corners = 4 atoms$

The volume of the cubic cell can be calculated using the edge length. This can be done as shown below;

$Volume=(edge length)3=(7.052×10−10 m)3=350.70×10−30 m3$

Volume of cubic cell can be converted into centimeters as shown below using conversion factors;

$350.70×10−30 m3=350.70×10−30 m3×100 cm31 m3=3.507×10−22 cm3$

Molar mass of $KI$ is $166.00 g/mol$ . Avogadro’s number is $6.022×1023 atoms$ . Therefore, the mass of $KI$ can be calculated as shown below;

$Mass of crystal=number of atoms×1Avogadro's number× molar mass= 4×16.022×1023 atoms×166.00 g/mol= 110.26×10-23 g$

Therefore, mass of $KI$ crystal is $110.26×10-23 g$ .

Density of the potassium iodide crystal can be calculated using the density formula as follows;

$Density=MassVolume=110.26×10-23 g3.507×10−22 cm3=31.44×10−1 g/cm3=3.144 g/cm3$

Thus, density of potassium iodide is calculated as $3.144 g/cm3$ .

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Answer 2

Question

Chapter 11, Problem 11.88QE

(a)

Interpretation Introduction

Interpretation:

The spacing between the layers that is present in the potassium iodide crystal that gives rise to $14.7$ degrees of diffraction has to be calculated.

(a)

Expert Solution

Answer to Problem 11.88QE

The spacing between the layers in crystal of $KI$ is $3.526×10−10 m$ .

Explanation of Solution

Wavelength of X-ray is given as $1.790×10−10 m$ . The angle of diffraction is given as $14.7$ degrees.

Bragg equation can be used to determine the spacing between the layers. Bragg equation is given as shown below;

$d=nλ2sinθ$ (1)

Where,

$n$ is given as 1.

$d$ is the distance between layers.

$λ$ is the wavelength of X-ray.

$θ$ is the angle of diffraction.

Substituting the values in equation (1), the spacing between the layers can be calculated as shown below;

$d=1×1.790×10−10 m2(sin14.7°)=1.790×10−10 m2(0.2538)=3.526×10−10 m$

Therefore, the spacing between layers is $3.526×10−10 m$ .

(b)

Interpretation Introduction

Interpretation:

Density of $KI$ has to be calculated.

(b)

Expert Solution

Answer to Problem 11.88QE

Density of potassium iodide is $3.144 g/cm3$ .

Explanation of Solution

The crystal is of $KI$ . The spacing between the layers is $3.526×10−10 m$ . The edge length of the cube is given as twice of the length of spacing between the layers. Therefore, the edge length is calculated as follows;

$Edge length=2×3.526×10−10 m=7.052×10−10 m$

The unit cell of potassium iodide is face centered cubic cell. Therefore, there are eight corners and thus one atom is shared by eight other unit cells that are present. One atom is present at the center of each face of the cell. Therefore, the total number of atoms present is calculated as shown below;

$6 faces×1 atom2 faces+8 corners×1 atom8 corners = 4 atoms$

The volume of the cubic cell can be calculated using the edge length. This can be done as shown below;

$Volume=(edge length)3=(7.052×10−10 m)3=350.70×10−30 m3$

Volume of cubic cell can be converted into centimeters as shown below using conversion factors;

$350.70×10−30 m3=350.70×10−30 m3×100 cm31 m3=3.507×10−22 cm3$

Molar mass of $KI$ is $166.00 g/mol$ . Avogadro’s number is $6.022×1023 atoms$ . Therefore, the mass of $KI$ can be calculated as shown below;

$Mass of crystal=number of atoms×1Avogadro's number× molar mass= 4×16.022×1023 atoms×166.00 g/mol= 110.26×10-23 g$

Therefore, mass of $KI$ crystal is $110.26×10-23 g$ .

Density of the potassium iodide crystal can be calculated using the density formula as follows;

$Density=MassVolume=110.26×10-23 g3.507×10−22 cm3=31.44×10−1 g/cm3=3.144 g/cm3$

Thus, density of potassium iodide is calculated as $3.144 g/cm3$ .

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Answer 3

Question

Chapter 11, Problem 11.88QE

(a)

Interpretation Introduction

Interpretation:

The spacing between the layers that is present in the potassium iodide crystal that gives rise to $14.7$ degrees of diffraction has to be calculated.

(a)

Expert Solution

Answer to Problem 11.88QE

The spacing between the layers in crystal of $KI$ is $3.526×10−10 m$ .

Explanation of Solution

Wavelength of X-ray is given as $1.790×10−10 m$ . The angle of diffraction is given as $14.7$ degrees.

Bragg equation can be used to determine the spacing between the layers. Bragg equation is given as shown below;

$d=nλ2sinθ$ (1)

Where,

$n$ is given as 1.

$d$ is the distance between layers.

$λ$ is the wavelength of X-ray.

$θ$ is the angle of diffraction.

Substituting the values in equation (1), the spacing between the layers can be calculated as shown below;

$d=1×1.790×10−10 m2(sin14.7°)=1.790×10−10 m2(0.2538)=3.526×10−10 m$

Therefore, the spacing between layers is $3.526×10−10 m$ .

(b)

Interpretation Introduction

Interpretation:

Density of $KI$ has to be calculated.

(b)

Expert Solution

Answer to Problem 11.88QE

Density of potassium iodide is $3.144 g/cm3$ .

Explanation of Solution

The crystal is of $KI$ . The spacing between the layers is $3.526×10−10 m$ . The edge length of the cube is given as twice of the length of spacing between the layers. Therefore, the edge length is calculated as follows;

$Edge length=2×3.526×10−10 m=7.052×10−10 m$

The unit cell of potassium iodide is face centered cubic cell. Therefore, there are eight corners and thus one atom is shared by eight other unit cells that are present. One atom is present at the center of each face of the cell. Therefore, the total number of atoms present is calculated as shown below;

$6 faces×1 atom2 faces+8 corners×1 atom8 corners = 4 atoms$

The volume of the cubic cell can be calculated using the edge length. This can be done as shown below;

$Volume=(edge length)3=(7.052×10−10 m)3=350.70×10−30 m3$

Volume of cubic cell can be converted into centimeters as shown below using conversion factors;

$350.70×10−30 m3=350.70×10−30 m3×100 cm31 m3=3.507×10−22 cm3$

Molar mass of $KI$ is $166.00 g/mol$ . Avogadro’s number is $6.022×1023 atoms$ . Therefore, the mass of $KI$ can be calculated as shown below;

$Mass of crystal=number of atoms×1Avogadro's number× molar mass= 4×16.022×1023 atoms×166.00 g/mol= 110.26×10-23 g$

Therefore, mass of $KI$ crystal is $110.26×10-23 g$ .

Density of the potassium iodide crystal can be calculated using the density formula as follows;

$Density=MassVolume=110.26×10-23 g3.507×10−22 cm3=31.44×10−1 g/cm3=3.144 g/cm3$

Thus, density of potassium iodide is calculated as $3.144 g/cm3$ .

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Answer 4

Question

Chapter 11, Problem 11.88QE

(a)

Interpretation Introduction

Interpretation:

The spacing between the layers that is present in the potassium iodide crystal that gives rise to $14.7$ degrees of diffraction has to be calculated.

(a)

Expert Solution

Answer to Problem 11.88QE

The spacing between the layers in crystal of $KI$ is $3.526×10−10 m$ .

Explanation of Solution

Wavelength of X-ray is given as $1.790×10−10 m$ . The angle of diffraction is given as $14.7$ degrees.

Bragg equation can be used to determine the spacing between the layers. Bragg equation is given as shown below;

$d=nλ2sinθ$ (1)

Where,

$n$ is given as 1.

$d$ is the distance between layers.

$λ$ is the wavelength of X-ray.

$θ$ is the angle of diffraction.

Substituting the values in equation (1), the spacing between the layers can be calculated as shown below;

$d=1×1.790×10−10 m2(sin14.7°)=1.790×10−10 m2(0.2538)=3.526×10−10 m$

Therefore, the spacing between layers is $3.526×10−10 m$ .

(b)

Interpretation Introduction

Interpretation:

Density of $KI$ has to be calculated.

(b)

Expert Solution

Answer to Problem 11.88QE

Density of potassium iodide is $3.144 g/cm3$ .

Explanation of Solution

The crystal is of $KI$ . The spacing between the layers is $3.526×10−10 m$ . The edge length of the cube is given as twice of the length of spacing between the layers. Therefore, the edge length is calculated as follows;

$Edge length=2×3.526×10−10 m=7.052×10−10 m$

The unit cell of potassium iodide is face centered cubic cell. Therefore, there are eight corners and thus one atom is shared by eight other unit cells that are present. One atom is present at the center of each face of the cell. Therefore, the total number of atoms present is calculated as shown below;

$6 faces×1 atom2 faces+8 corners×1 atom8 corners = 4 atoms$

The volume of the cubic cell can be calculated using the edge length. This can be done as shown below;

$Volume=(edge length)3=(7.052×10−10 m)3=350.70×10−30 m3$

Volume of cubic cell can be converted into centimeters as shown below using conversion factors;

$350.70×10−30 m3=350.70×10−30 m3×100 cm31 m3=3.507×10−22 cm3$

Molar mass of $KI$ is $166.00 g/mol$ . Avogadro’s number is $6.022×1023 atoms$ . Therefore, the mass of $KI$ can be calculated as shown below;

$Mass of crystal=number of atoms×1Avogadro's number× molar mass= 4×16.022×1023 atoms×166.00 g/mol= 110.26×10-23 g$

Therefore, mass of $KI$ crystal is $110.26×10-23 g$ .

Density of the potassium iodide crystal can be calculated using the density formula as follows;

$Density=MassVolume=110.26×10-23 g3.507×10−22 cm3=31.44×10−1 g/cm3=3.144 g/cm3$

Thus, density of potassium iodide is calculated as $3.144 g/cm3$ .

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Answer 5

Chapter 11, Problem 11.88QE

(a)

Interpretation Introduction

Interpretation:

The spacing between the layers that is present in the potassium iodide crystal that gives rise to $14.7$ degrees of diffraction has to be calculated.

(a)

Expert Solution

Answer to Problem 11.88QE

The spacing between the layers in crystal of $KI$ is $3.526×10−10 m$ .

Explanation of Solution

Wavelength of X-ray is given as $1.790×10−10 m$ . The angle of diffraction is given as $14.7$ degrees.

Bragg equation can be used to determine the spacing between the layers. Bragg equation is given as shown below;

$d=nλ2sinθ$ (1)

Where,

$n$ is given as 1.

$d$ is the distance between layers.

$λ$ is the wavelength of X-ray.

$θ$ is the angle of diffraction.

Substituting the values in equation (1), the spacing between the layers can be calculated as shown below;

$d=1×1.790×10−10 m2(sin14.7°)=1.790×10−10 m2(0.2538)=3.526×10−10 m$

Therefore, the spacing between layers is $3.526×10−10 m$ .

(b)

Interpretation Introduction

Interpretation:

Density of $KI$ has to be calculated.

(b)

Expert Solution

Answer to Problem 11.88QE

Density of potassium iodide is $3.144 g/cm3$ .

Explanation of Solution

The crystal is of $KI$ . The spacing between the layers is $3.526×10−10 m$ . The edge length of the cube is given as twice of the length of spacing between the layers. Therefore, the edge length is calculated as follows;

$Edge length=2×3.526×10−10 m=7.052×10−10 m$

The unit cell of potassium iodide is face centered cubic cell. Therefore, there are eight corners and thus one atom is shared by eight other unit cells that are present. One atom is present at the center of each face of the cell. Therefore, the total number of atoms present is calculated as shown below;

$6 faces×1 atom2 faces+8 corners×1 atom8 corners = 4 atoms$

The volume of the cubic cell can be calculated using the edge length. This can be done as shown below;

$Volume=(edge length)3=(7.052×10−10 m)3=350.70×10−30 m3$

Volume of cubic cell can be converted into centimeters as shown below using conversion factors;

$350.70×10−30 m3=350.70×10−30 m3×100 cm31 m3=3.507×10−22 cm3$

Molar mass of $KI$ is $166.00 g/mol$ . Avogadro’s number is $6.022×1023 atoms$ . Therefore, the mass of $KI$ can be calculated as shown below;

$Mass of crystal=number of atoms×1Avogadro's number× molar mass= 4×16.022×1023 atoms×166.00 g/mol= 110.26×10-23 g$

Therefore, mass of $KI$ crystal is $110.26×10-23 g$ .

Density of the potassium iodide crystal can be calculated using the density formula as follows;

$Density=MassVolume=110.26×10-23 g3.507×10−22 cm3=31.44×10−1 g/cm3=3.144 g/cm3$

Thus, density of potassium iodide is calculated as $3.144 g/cm3$ .

Answer 6

Chapter 11, Problem 11.88QE

(a)

Interpretation Introduction

Interpretation:

The spacing between the layers that is present in the potassium iodide crystal that gives rise to $14.7$ degrees of diffraction has to be calculated.

(a)

Expert Solution

Answer to Problem 11.88QE

The spacing between the layers in crystal of $KI$ is $3.526×10−10 m$ .

Explanation of Solution

Wavelength of X-ray is given as $1.790×10−10 m$ . The angle of diffraction is given as $14.7$ degrees.

Bragg equation can be used to determine the spacing between the layers. Bragg equation is given as shown below;

$d=nλ2sinθ$ (1)

Where,

$n$ is given as 1.

$d$ is the distance between layers.

$λ$ is the wavelength of X-ray.

$θ$ is the angle of diffraction.

Substituting the values in equation (1), the spacing between the layers can be calculated as shown below;

$d=1×1.790×10−10 m2(sin14.7°)=1.790×10−10 m2(0.2538)=3.526×10−10 m$

Therefore, the spacing between layers is $3.526×10−10 m$ .

Answer 7

Chapter 11, Problem 11.88QE

(a)

Interpretation Introduction

Interpretation:

The spacing between the layers that is present in the potassium iodide crystal that gives rise to $14.7$ degrees of diffraction has to be calculated.

Answer 8

(a)

Expert Solution

Answer to Problem 11.88QE

The spacing between the layers in crystal of $KI$ is $3.526×10−10 m$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Wavelength of X-ray is given as $1.790×10−10 m$ . The angle of diffraction is given as $14.7$ degrees.

Bragg equation can be used to determine the spacing between the layers. Bragg equation is given as shown below;

$d=nλ2sinθ$ (1)

Where,

$n$ is given as 1.

$d$ is the distance between layers.

$λ$ is the wavelength of X-ray.

$θ$ is the angle of diffraction.

Substituting the values in equation (1), the spacing between the layers can be calculated as shown below;

$d=1×1.790×10−10 m2(sin14.7°)=1.790×10−10 m2(0.2538)=3.526×10−10 m$

Therefore, the spacing between layers is $3.526×10−10 m$ .

Answer 13

(b)

Interpretation Introduction

Interpretation:

Density of $KI$ has to be calculated.

(b)

Expert Solution

Answer to Problem 11.88QE

Density of potassium iodide is $3.144 g/cm3$ .

Explanation of Solution

The crystal is of $KI$ . The spacing between the layers is $3.526×10−10 m$ . The edge length of the cube is given as twice of the length of spacing between the layers. Therefore, the edge length is calculated as follows;

$Edge length=2×3.526×10−10 m=7.052×10−10 m$

The unit cell of potassium iodide is face centered cubic cell. Therefore, there are eight corners and thus one atom is shared by eight other unit cells that are present. One atom is present at the center of each face of the cell. Therefore, the total number of atoms present is calculated as shown below;

$6 faces×1 atom2 faces+8 corners×1 atom8 corners = 4 atoms$

The volume of the cubic cell can be calculated using the edge length. This can be done as shown below;

$Volume=(edge length)3=(7.052×10−10 m)3=350.70×10−30 m3$

Volume of cubic cell can be converted into centimeters as shown below using conversion factors;

$350.70×10−30 m3=350.70×10−30 m3×100 cm31 m3=3.507×10−22 cm3$

Molar mass of $KI$ is $166.00 g/mol$ . Avogadro’s number is $6.022×1023 atoms$ . Therefore, the mass of $KI$ can be calculated as shown below;

$Mass of crystal=number of atoms×1Avogadro's number× molar mass= 4×16.022×1023 atoms×166.00 g/mol= 110.26×10-23 g$

Therefore, mass of $KI$ crystal is $110.26×10-23 g$ .

Density of the potassium iodide crystal can be calculated using the density formula as follows;

$Density=MassVolume=110.26×10-23 g3.507×10−22 cm3=31.44×10−1 g/cm3=3.144 g/cm3$

Thus, density of potassium iodide is calculated as $3.144 g/cm3$ .

Answer 14

(b)

Interpretation Introduction

Interpretation:

Density of $KI$ has to be calculated.

Answer 15

(b)

Expert Solution

Answer to Problem 11.88QE

Density of potassium iodide is $3.144 g/cm3$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

The crystal is of $KI$ . The spacing between the layers is $3.526×10−10 m$ . The edge length of the cube is given as twice of the length of spacing between the layers. Therefore, the edge length is calculated as follows;

$Edge length=2×3.526×10−10 m=7.052×10−10 m$

The unit cell of potassium iodide is face centered cubic cell. Therefore, there are eight corners and thus one atom is shared by eight other unit cells that are present. One atom is present at the center of each face of the cell. Therefore, the total number of atoms present is calculated as shown below;

$6 faces×1 atom2 faces+8 corners×1 atom8 corners = 4 atoms$

The volume of the cubic cell can be calculated using the edge length. This can be done as shown below;

$Volume=(edge length)3=(7.052×10−10 m)3=350.70×10−30 m3$

Volume of cubic cell can be converted into centimeters as shown below using conversion factors;

$350.70×10−30 m3=350.70×10−30 m3×100 cm31 m3=3.507×10−22 cm3$

Molar mass of $KI$ is $166.00 g/mol$ . Avogadro’s number is $6.022×1023 atoms$ . Therefore, the mass of $KI$ can be calculated as shown below;

$Mass of crystal=number of atoms×1Avogadro's number× molar mass= 4×16.022×1023 atoms×166.00 g/mol= 110.26×10-23 g$

Therefore, mass of $KI$ crystal is $110.26×10-23 g$ .

Density of the potassium iodide crystal can be calculated using the density formula as follows;

$Density=MassVolume=110.26×10-23 g3.507×10−22 cm3=31.44×10−1 g/cm3=3.144 g/cm3$

Thus, density of potassium iodide is calculated as $3.144 g/cm3$ .

Answer 20

Ch. 11 - Prob. 11.1QE Ch. 11 - Prob. 11.2QE Ch. 11 - Prob. 11.3QE Ch. 11 - Prob. 11.4QE Ch. 11 - Prob. 11.5QE Ch. 11 - Why does a perspiring body achieve greater cooling...Ch. 11 - Prob. 11.7QE Ch. 11 - Prob. 11.8QE Ch. 11 - Prob. 11.9QE Ch. 11 - Prob. 11.10QE

The spacing between the layers that is present in the potassium iodide crystal that gives rise to 14.7 degrees of diffraction has to be calculated.

Concept explainers

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Interpretation:

The spacing between the layers that is present in the potassium iodide crystal that gives rise to $14.7$ degrees of diffraction has to be calculated.

Answer to Problem 11.88QE

Explanation of Solution

Interpretation:

Density of $KI$ has to be calculated.

Answer to Problem 11.88QE

Explanation of Solution

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Chapter 11 Solutions

The spacing between the layers that is present in the potassium iodide crystal that gives rise to 14.7 degrees of diffraction has to be calculated.

Concept explainers

Videos

Interpretation: The spacing between the layers that is present in the potassium iodide crystal that gives rise to 14.7<m:math><m:mrow><m:mn>14.7</m:mn></m:mrow></m:math> degrees of diffraction has to be calculated.

Answer to Problem 11.88QE

Explanation of Solution

Interpretation: Density of KI<m:math><m:mrow><m:mtext>KI</m:mtext></m:mrow></m:math> has to be calculated.

Answer to Problem 11.88QE

Explanation of Solution

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Chapter 11 Solutions

Interpretation:

The spacing between the layers that is present in the potassium iodide crystal that gives rise to $14.7$ degrees of diffraction has to be calculated.

Interpretation:

Density of $KI$ has to be calculated.