13 Liquids and Solids

13 Liquids and Solids Chapter Goals 1. Kinetic-Molecular Description of Liquids and Solids 2. Intermolecular Attractions and Phase Changes The Liqu...
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Liquids and Solids

Chapter Goals 1. Kinetic-Molecular Description of Liquids and Solids 2. Intermolecular Attractions and Phase Changes The Liquid State 3. Viscosity 4. Surface Tension 5. Capillary Action 6. Evaporation 7. Vapor Pressure 8. Boiling Points and Distillation 9. Heat Transfer Involving Liquids 2

Chapter Goals 10. 11. 12. 13. 14. 15. 16. 17. 18.

The Solid State Melting Point Heat Transfer Involving Solids Sublimation and the Vapor Pressure of Solids Phase Diagrams (P versus T) Amorphous Solids and Crystalline Solids Structures of Crystals Bonding in Solids Band Theory of Metals Synthesis Question 3

Kinetic-Molecular Description of Liquids and Solids • Solids and liquids are condensed states. – The atoms, ions, or molecules in solids and liquids are much closer to one another than in gases. – Solids and liquids are highly incompressible.

• Liquids and gases are fluids. – They easily flow.

• The intermolecular attractions in liquids and solids are strong. 4

Kinetic-Molecular Description of Liquids and Solids • Schematic representation of the three common states of matter.

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Kinetic-Molecular Description of Liquids and Solids • If we compare the strengths of interactions among particles and the degree of ordering of particles, we see that Gases< Liquids < Solids • Miscible liquids are soluble in each other. – Examples of miscible liquids: • Water dissolves in alcohol. • Gasoline dissolves in motor oil. 6

Kinetic-Molecular Description of Liquids and Solids • Immiscible liquids are insoluble in each other. – Two examples of immiscible liquids: • Water does not dissolve in oil. • Water does not dissolve in cyclohexane.

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Intermolecular Attractions and Phase Changes •

There are four important intermolecular attractions. –

1.

This list is from strongest attraction to the weakest attraction.

Ion-ion interactions –

The force of attraction between two oppositely charged ions is governed by Coulomb’s law.

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Intermolecular Attractions and Phase Changes •

Coulomb’s law determines: 1. The melting and boiling points of ionic compounds. 2. The solubility of ionic compounds.



Example 13-1: Arrange the following ionic compounds in the expected order of increasing melting and boiling points. NaF, CaO, CaF2 You do it! What important points must you consider? 9

Intermolecular Attractions and Phase Changes 

Na F Ca F2 Ca O + -

2+

2+

2-

10

Intermolecular Attractions and Phase Changes 2. Dipole-dipole interactions – Consider BrF a polar molecule.

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Intermolecular Attractions and Phase Changes 3. Hydrogen bonding –

Consider H2O a very polar molecule.

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Intermolecular Attractions and Phase Changes 3. Hydrogen bonding –

Consider H2O a very polar molecule.

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Intermolecular Attractions and Phase Changes

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Intermolecular Attractions and Phase Changes 4. London Forces are very weak. – They are the weakest of the intermolecular forces. – This is the only attractive force in nonpolar molecules.



Consider Ar as an isolated atom.

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Intermolecular Attractions and Phase Changes • In a group of Ar atoms the temporary dipole in one atom induces other atomic dipoles.

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Intermolecular Attractions and Phase Changes • Similar effects occur in a group of I2 molecules.

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The Liquid State Viscosity • Viscosity is the resistance to flow. – For example, compare how water pours out of a glass compared to molasses, syrup or honey.

• Oil for your car is bought based on this property. – 10W30 or 5W30 describes the viscosity of the oil at high and low temperatures. 18

The Liquid State • An example of viscosity of two liquids.

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The Liquid State Surface Tension • Surface tension is a measure of the unequal attractions that occur at the surface of a liquid. • The molecules at the surface are attracted unevenly. 20

The Liquid State • Floating paper clip demonstration of surface tension.

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The Liquid State Capillary Action • Capillary action is the ability of a liquid to rise (or fall) in a glass tube or other container

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The Liquid State • Cohesive forces are the forces that hold liquids together. • Adhesive forces are the forces between a liquid and another surface. – Capillary rise implies that the: • Adhesive forces > cohesive forces

– Capillary fall implies that the: • Cohesive forces > adhesive forces 23

The Liquid State • Water exhibits a capillary rise.  Mercury exhibits a capillary fall.

Water

Mercury

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The Liquid State • Capillary action also affects the meniscus of liquids.

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The Liquid State •



Evaporation Evaporation is the process in which molecules escape from the surface of a liquid and become a gas. Evaporation is temperature dependent.

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The Liquid State

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The Liquid State • This is an animation of evaporation

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The Liquid State Vapor Pressure • Vapor pressure is the pressure exerted by a liquid’s vapor on its surface at equilibrium. • Vapor Pressure (torr) and boiling point for three liquids at different temperatures. 0oC

20oC

30oC

normal boiling point

diethyl ether 185 442 647 36oC ethanol 12 44 74 78oC water 5 18 32 100oC • What are the intermolecular forces in each of these compounds? You do it!

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The Liquid State Vapor Pressure as a function of temperature.

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The Liquid State Boiling Points and Distillation • The boiling point is the temperature at which the liquid’s vapor pressure is equal to the applied pressure. • The normal boiling point is the boiling point when the pressure is exactly 1 atm. • Distillation is a method we use to separate mixtures of liquids based on their differences in boiling points. 31

The Liquid State Distillation • Distillation is a process in which a mixture or solution is separated into its components on the basis of the differences in boiling points of the components. • Distillation is another vapor pressure phenomenon. 32

The Liquid State Heat Transfer Involving Liquids • From Chapter 1

q = m C T

•Example 13-2: How much heat is

released by 2.00 x 102 g of H2O as it cools from 85.0oC to 40.0oC? The specific heat of water is 4.184 J/goC. You do it! 33

The Liquid State ? J  2.00 102 g(4.184

J g

o

o )( 85 . 0  40 . 0 C) C

? J  3.76 104 J  37.6 kJ

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The Liquid State • Molar heat capacity is the amount of heat required to raise the temperature of one mole of a substance 1.00 oC. • Example 13-3: The molar heat capacity of ethyl alcohol, C2H5OH, is 113 J/moloC. How much heat is required to raise the T of 125 g of ethyl alcohol from 20.0oC to 30.0oC? 1 mol C2H5OH = 46.0 g You do it!

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The Liquid State 1 mol C 2 H 5OH ? mol = 125 g   2.72 mol C 2 H 5OH 46.0 g C 2 H 5OH





113 J   o ? J = 2.72 mol 30 . 0  20 . 0 C  3.07 kJ  o  mol C 

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The Liquid State • •

The calculations we have done up to now tell us the energy changes as long as the substance remains in a single phase. Next, we must address the energy associated with phase changes. – For example, solid to liquid or liquid to gas and the reverse.



Heat of Vaporization is the amount of heat required to change 1.00 g of a liquid substance to a gas at constant temperature. – Heat of vaporization has units of J/g.



Heat of Condensation is the reverse of heat of vaporization, phase change from gas to liquid. 2260 J

  1.00 g H O at 100.0o C 1.00 g H 2O( )at 100.0o C   2 (g) -2260 J 37

The Liquid State Molar heat of vaporization or Hvap • The Hvap is the amount of heat required to change 1.00 mole of a liquid to a gas at constant temperature. Hvap has units of J/mol.

Molar heat of condensation • The reverse of molar heat of vaporization is the heat of condensation. 40.7 kJ    1.00 mol H O at 100.0o C 1.00 mol H 2O(  ) at 100.0 C  2 (g) - 40.7 kJ o

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The Liquid State

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The Liquid State • Example 13-4: How many joules of energy must be absorbed by 5.00 x 102 g of H2O at 50.0oC to convert it to steam at 120oC? The molar heat of vaporization of water is 40.7 kJ/mol and the molar heat capacities of liquid water and steam are 75.3 J/mol oC and 36.4 J/mol oC, respectively. You do it! 40

The Liquid State 1 mol H 2 O ? mol = 500 g H 2 O   27.8 mol H 2 O 18 g H 2 O 1st let's calculate the heat required to warm water from 50 to 100 o C ?J =

27.8 mol





75.3 J  o 5 100 . 0  50 . 0 C  105 .  10 J  o  mol C

Next, let’s calculate the energy required to boil the water.  40.7  103 J  ? J = 27.8 mol .  105 J   1131 mol  

Finally, let’s calculate the heat required to heat steam from 100 to 120oC.





36.4 J   o 5 ? J = 27.8 mol 120.0 -100.0 C  0 . 20  10 J   mol o C  41

The Liquid State • The total amount of energy for this process is the sum of the 3 pieces we have calculated.

.  10 J   1131 .  10 J   0.20  10 J   105 5

5

5

12.56  10 J or 1.26  10 kJ 5

3

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The Liquid State • Example 13-5: If 45.0 g of steam at 140oC is slowly bubbled into 450 g of water at 50.0oC in an insulated container, can all the steam be condensed? You do it!

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The Liquid State  1 mol   1 mol    2.50 mol 450 g water    25.0 mol 45.0 g steam   18 g steam   18 g  (1) Calculate the amount of heat required to condense the steam. 2.50 mol 36.4

J

140.0 -100.0 C  2.50 mol 40.7 o

o

mol C

kJ

mol

  105. kJ

(2) Calculate the amount of heat available in the liquid water. 25.0 mol 75.3

J

o

mol C

 (100.0 - 50.0

o

C)  94.1 kJ

Amount of heat to condense all of the steam is 105 kJ. Amount of heat that the liquid water can absorb is 94.1 kJ. Thus all of the steam cannot be condensed. 44

The Liquid State • Clausius-Clapeyron equation – determine vapor pressure of a liquid at a new T – determine what T we must heat something to get a specified vapor pressure – way to determine Hvap if we know pressure at 2 T’s

 P2  H vap  1 1 ln       R  T1 T2   P1  45

The Liquid State • In Denver the normal atmospheric pressure is 630 torr. At what temperature does water boil in Denver?  P2  H vap  1 1 ln       R  T1 T2   P1  3 J 40 . 7  10 630 torr  1 1  mol  ln      J  760 torr  8.314 K mol  373 K T2 

 1 ln 0.829  4895 0.002681   T2   46

The Liquid State   0188 . 1   0.002681   4895  T2   383 .  10

5

 383 .  10

5

 1   0.002681   T2   1  0.002681   T2

1  0.00272   T2 T2  368 K or 95o C

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The Liquid State • Boiling Points of Various Kinds of Liquids Gas MW BP(oC) He

4

-269

Ne

20

-246

Ar

40

-186

Kr

84

-153

Xe

131

-107

Rn

222

-62 48

The Liquid State Noble Gases

Boiling Point (C)

0 -50

4

20

40

84

131

222

-100 -150 -200 -250 -300 Molar Mass 49

The Liquid State o

Compound MW(amu) B.P.( C) CH4

16

-161

C2H6

30

-88

C3H8

44

-42

n-C4H10

58

-0.6

n-C5H12

72

+36 50

The Liquid State Alkanes

Boiling Point (C)

50 0 -50

16

30

44

58

72

-100 -150 -200 Molar Mass

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The Liquid State o

Compound

MW(amu)

B.P.( C)

HF

20

19.5

HCl HBr

37 81

- 85.0 - 67.0

HI

128

- 34.0 52

The Liquid State Hydrogen Halides

Boiling Point (C)

40 20 0 -20 20

37

81

128

-40 -60 -80 -100 Molar Mass 53

The Liquid State Compound

MW(amu)

o

B.P.( C)

H 2O H 2S H 2Se

18 34 81

100 - 61 - 42

H 2 Te

130

-2 54

The Liquid State VIA Hydrides

Boiling Point (C)

150 100 50 0 -50

18

34

81

130

-100 Molar Mass 55

The Liquid State • At the molecular level what happens when a species boils?

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The Liquid State • Example 13-6: Arrange the following substances in order of increasing boiling points. C2H6, NH3, Ar, NaCl, AsH3 You do it! Ar < C2H6 < AsH3 < NH3 < NaCl nonpolar nonpolar polar very polar ionic London London dipole-dipole H-bonding ion-ion 57

The Solid State Normal Melting Point • The normal melting point is the temperature at which the solid melts (liquid and solid in equilibrium) at exactly 1.00 atm of pressure. • The melting point increases as the strength of the intermolecular attractions increase. 58

The Solid State • Which requires more energy?

 NaCl NaCls     or H O H 2 O s   2   

What experimental proof do you have?

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Heat Transfer Involving Solids Heat of Fusion • Heat of fusion is the amount of heat required to melt one gram of a solid at its melting point at constant temperature.  334 J o     1.00 g H O at 0o C 1.00 g H O at 0 C   2

(s)

-334 J

2

()

• Heat of crystallization is the reverse of the heat of fusion. 60

Heat Transfer Involving Solids Molar heat of fusion or Hfusion • The molar heat of fusion is the amount of heat required to melt a mole of a substance at its melting point. • The molar heat of crystallization is the reverse of molar heat of fusion 6012 J     1.00 mole H O at 0o C 1.00 mole H 2O(s) at 0 C  2 () -6012 J o

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Heat Transfer Involving Solids • Here is a summary of the heats of transformation for water. 40.7 kJ    1.00 mol H O at 100.0o C 1.00 mol H 2O(  ) at 100.0 C  2 (g) - 40.7 kJ o

6012 J     1.00 mole H O at 0o C 1.00 mole H 2O(s) at 0 C  2 () -6012 J o

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Heat Transfer Involving Solids • Example 13-7: Calculate the amount of heat required to convert 150.0 g of ice at 10.0oC to water at 40.0oC. specific heat of ice is 2.09 J/goC you do it

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Heat Transfer Involving Solids J ? J = (150.0 g)(2.09 o )(10 o C) = 3.14  103 J g C J 4 ? J = (150.0 g)(334 ) = 5.01  10 J g J ? J = (150.0 g)(4.18 o )(40 o C) = 2.51  10 4 J g C 7.83  10 J 4

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Sublimation and the Vapor Pressure of Solids Sublimation • In the sublimation process the solid transforms directly to the vapor phase without passing through the liquid phase. • Solid CO2 or “dry” ice does this well.

     solid     gas condensation sublimation

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Phase Diagrams (P versus T) • Phase diagrams are a convenient way to display all of the different phase transitions of a substance. • This is the phase diagram for water.

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Phase Diagrams (P versus T) • Compare water’s phase diagram to carbon dioxide’s phase diagram.

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Amorphous Solids and Crystalline Solids • Amorphous solids do not have a well ordered molecular structure. – Examples of amorphous solids include waxes, glasses, asphalt.

• Crystalline solids have well defined structures that consist of extended array of repeating units called unit cells. – Crystalline solids display X-ray diffraction patterns which reflect the molecular structure. – The Bragg equation, detailed in the textbook, describes how an X-ray diffraction pattern can be used to determine the interatomic distances in crystals. 68

Structure of Crystals • Unit cells are the smallest repeating unit of a crystal. – As an analogy, bricks are repeating units for buildings.

• There are seven basic crystal systems.

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Structure of Crystals • We shall look at the three variations of the cubic crystal system. • Simple cubic unit cells. – The balls represent the positions of atoms, ions, or molecules in a simple cubic unit cell.

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Structure of Crystals • In a simple cubic unit cell each atom, ion, or molecule at a corner is shared by 8 unit cells – Thus 1 unit cell contains 8(1/8) = 1 atom, ion, or molecule.

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Structure of Crystals • Body centered cubic (bcc) has an additional atom, ion, or molecule in the center of the unit cell. • On a body centered cubic unit cell there are 8 corners + 1 particle in center of cell. – 1 bcc unit cell • contains 8(1/8) + 1 = 2 particles.

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Structure of Crystals • A face centered cubic (fcc) unit cell has a cubic unit cell structure with an extra atom, ion, or molecule in each face.

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Structure of Crystals • A face centered cubic unit cell has 8 corners and 6 faces. – 1 fcc unit cell contains • 8(1/8) + 6(1/2) = 4 particles.

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Bonding in Solids • Molecular Solids have molecules in each of the positions of the unit cell. – Molecular solids have low melting points, are volatile, and are electrical insulators.

• Examples of molecular solids include: – water, sugar, carbon dioxide, benzene

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Bonding in Solids • Covalent Solids have atoms that are covalently bonded to one another • Some examples of covalent solids are: • Diamond, graphite, SiO2 (sand), SiC

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Bonding in Solids • Ionic Solids have ions that occupy the positions in the unit cell. • Examples of ionic solids include: – CsCl, NaCl, ZnS

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Bonding in Solids • Metallic Solids may be thought of as positively charged nuclei surrounded by a sea of electrons. • The positive ions occupy the crystal lattice positions. • Examples of metallic solids include: – Na, Li, Au, Ag, ……..

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Bonding in Solids • Variations in Melting Points for Molecular Solids • What are the intermolecular forces in each solid?

• • • • • •

Compound Melting Point (oC) ice 0.0 ammonia -77.7 benzene, C6H6 5.5 napthalene, C10H8 80.6 benzoic acid, C6H5CO2H 122.4 79

Bonding in Solids • Variations in Melting Points for Covalent Solids

• • • • •

Substance Melting Point (oC) sand, SiO2 1713 carborundum, SiC ~2700 diamond >3550 graphite 3652-3697

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Bonding in Solids • •

Variations in Melting Points for Ionic Solids Compound Melting Point (oC)

• • • • • • • •

LiF LiCl LiBr LiI CaF2 CaCl2 CaBr2 CaI2

842 614 547 450 1360 772 730 740 81

Bonding in Solids • Variations in Melting Points for Metallic Solids

• Metal • • • • • •

Na Pb Al Cu Fe W

Melting Point (oC) 98 328 660 1083 1535 3410 82

Bonding in Solids • Example 13-8. A group IVA element with a density of 11.35 g/cm3 crystallizes in a face-centered cubic lattice whose unit cell edge length is 4.95 Å. Calculate the element’s atomic weight. What is the atomic radius of this element?

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Bonding in Solids • •

Face centered cubic unit cells have 4 atoms, ions, or molecules per unit cell. Problem solution pathway: 1. Determine the volume of a single unit cell. 2. Use the density to determine the mass of a single unit cell. 3. Determine the mass of one atom in a unit cell. 4. Determine the mass of 1 mole of these atoms

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Bonding in Solids 1. Determine the volume of a single unit cell. 0

0

1 A  10 cm thus 4.95 A  4.95 10-8 cm -8

Face centered cubic unit cells are cubic so V   3

4.95 10

2.

-8

cm



3

 1.2110 -22 cm 3

Use the density to determine the mass of a unit cell.

1.2110

- 22



g  11.35 g  - 21 cm 3   1.38  10 3  one unit cell  cm 

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Bonding in Solids 3. Determine the mass of one atom in the unit cell. Because face centered cubic has 4 atoms per unit cell the mass of one atom can be determined in this fashion. 1.38 10-21 g unit cell  3.44 10  22 g atom 4 atoms unit cell

4. Determine the mass of one mole of these atoms.

3.44 1022 g

6.022 10 atom

23

atoms

mole

  207 g/mole 86

Bonding in Solids • To determine an atomic radius requires some geometry. • For simple cubic unit cells: – The edge length = 2 radii

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Bonding in Solids • For face-centered cubic unit cells: – The face diagonal = 2 x edge length. – The diagonal length = 4 radii.

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Bonding in Solids • For body-centered cubic unit cells: – The body diagonal = 3 x edge length. – The diagonal length = 4 radii.

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Bonding in Solids • Determine the diagonal length then divide by 4 to get the atomic radius.

diagonal =



2 4.95  10 cm -8



 7.00  10 cm -8

-8 7 . 00  10 cm radius =  175 .  10 cm 4 -8

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Band Theory of Metals • Sodium’s 3s orbitals can interact to produce overlapping orbitals

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Band Theory of Metals • The 3s orbitals can also overlap with unfilled 3p orbitals

92

Band Theory of Metals • Insulators have a large gap between the s and p bands. – Gap is called the forbidden zone.

• Semiconductors have a small gap between the bands.

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Synthesis Question • Maxwell House Coffee Company decaffeinates its coffee beans using an extractor that is 7.0 feet in diameter and 70.0 feet long. Supercritical carbon dioxide at a pressure of 300.0 atm and temperature of 100.0oC is passed through the stainless steel extractor. The extraction vessel contains 100,000 pounds of coffee beans soaked in water until they have a water content of 50%. 94

Synthesis Question • This process removes 90% of the caffeine in a single pass of the beans through the extractor. Carbon dioxide that has passed over the coffee is then directed into a water column that washes the caffeine from the supercritical CO2. How many moles of carbon dioxide are present in the extractor?

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Synthesis Question Diameter of vessel  (7.0 ft)(30.48 cm/ft)  213.4 cm Radius of vessel  213.4 cm/2  106.7 cm Length of vessel  (70.0 ft)(30.48 cm/ft)  2134 cm Volume of vessel   r 2 h  (3.1416)(1 06.7cm) 2 (2134cm)  (7.633 107 cm 3 )(1 mL/cm 3 )(1 L/1000 mL)  7.633 10 4 L

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Synthesis Question PV  nRT n  PV

 300 atm 7.633 10 L   RT 0.08206 L atm mol K 373 K  4

n  748,000 mol of CO 2

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Group Question • How many CO2 molecules are there in 1.0 cm3 of the Maxwell House Coffee Company extractor? How many more CO2 molecules are there in a cm3 of the supercritical fluid in the Maxwell House extractor than in a mole of CO2 at STP?

98

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Liquids and Solids