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Chap_13

Course: CHEM 102, Spring 2007
School: UMBC
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13 Chapter Structures, Properties, and Applications of Solids Crystalline solids have structures that are highly ordered, which has made their study much easier than liquids. We begin this chapter with discussions of the nature of this internal order, how it is described in simple ways, and how the properties of crystalline materials reflect the kinds of particles that make up a solid. The applications of solid...

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13 Chapter Structures, Properties, and Applications of Solids Crystalline solids have structures that are highly ordered, which has made their study much easier than liquids. We begin this chapter with discussions of the nature of this internal order, how it is described in simple ways, and how the properties of crystalline materials reflect the kinds of particles that make up a solid. The applications of solid materials, however, extend beyond those that are crystalline. Our goal in this chapter is to also introduce you to a variety of such materials and their uses, ranging from semiconductors to polymers to high tech ceramics found in modern electronic devices. We conclude by introducing you to a field of cuttingedge science called nanotechnology, which deals with structures having dimensions ranging from a few to several hundred atoms in size. Learning Objectives Throughout your study of this chapter, keep in mind the following objectives: 1 2 3 4 5 6 7 8 9 10 To learn how atoms, molecules, or ions are arranged in crystalline solids and how we are able to describe their structures in simple ways. To understand the concept of a lattice and how structures of crystalline solids can be described by giving the properties of a repeating unit called a unit cell. To learn the properties of the three kinds of cubic unit cells. To be able to count the number of atoms per unit cell given the arrangement of atoms in the cell. To learn what distinguishes cubic and hexagonal closest packed structures. To learn how the properties of amorphous solids differ from those of crystalline solids. To see how data obtained by X-ray diffraction experiments are used to obtain structural information about crystals. To learn how the physical properties of crystalline solids can be related to the kinds of particles at lattice positions and the kinds of attractive forces between the particles. To learn how some substances solidify without forming a crystalline solid. To learn how energy levels of atoms in a crystal merge to form energy bands. You should learn the distinction between valence and conduction bands, and how n- and ptype semiconductors differ. To learn how a polymer is formed by linking many monomer units. To learn how to write the formula for a polymer. To learn how addition polymers are formed through free radical polymerization. To learn the structures of polyethylene and polypropylene. To learn how condensation polymers are formed by elimination of water or other small molecule. 229 11 12 13 14 15 230 Structures, Properties and Applications of Solids 16 17 18 19 20 21 22 23 24 25 To learn the basic structures of polyesters and nylons. To learn what crosslinking means and how it affects the properties of a polymer. To learn how crystallinity affects the properties of a polymer. To learn the nature of the different kinds of liquid crystalline phases. To learn the structural properties possessed by molecules that form liquid crystals. To learn how cholesteric liquid crystals affect polarized light and how that property is used in liquid crystal displays. To learn the desirable properties of ceramics and the kinds of substances that form useful ceramic materials. To learn the essential features of the sol-gel process. To learn what nanotechnology hopes to achieve and the tools used to explore structures at the atomic level of detail. To learn the basic structures of graphite, diamond, fullerenes, and carbon nanotubes. 13.1 Review Crystalline solids have an ordered internal structure A crystalline solid is characterized by a highly organized, regular, repeating pattern of particles. To describe such structures, we use the concept of a lattice, which is a pattern of points that have the same repeat distances arranged at the same angles as the particles in the crystal. Study the discussion of two-dimensional lattices to be sure you understand the concept. When describing crystals, we often use the term crystal lattice for the three-dimensional lattice that describes their structures. Especially important is the concept that a single kind of lattice can be used to characterize many different structures (just as many different wallpaper patterns can be created using the same basic two-dimensional lattice.) In a lattice, the smallest repeating unit is called the unit cell. The number of kinds of lattices (or unit cells) is very small, and all substances can be described in terms of this limited set. The most symmetrical lattices (and unit cells) are cubic. There are three kinds of cubic unit cells--simple cubic, face-centered cubic, and body-centered cubic. Figure 13.6 illustrates how different substances can have the same kind of unit cell but with different cell dimensions. Counting atoms in a unit cell You should learn how to calculate the number of atoms in a unit cell. This requires that you realize that we have to assemble the parts actually enclosed within the cell into whole atoms, so we can count their number. Remember the following: An atom entirely with a unit cell counts as 1 atom. An atom at a corner contributes 1/8 of an atom to the unit cell. An atom in a face contributes 1/2 of an atom to the unit cell. An atom along an edge contributes 1/4 of an atom to the unit cell. Chapter 13 231 The rock salt structure, which is the structure of NaCl, is quite common among the alkali halides. You should learn how the atoms are arranged in the unit cell. Study Figures 13.8 and 13.11. Closest-packed structures There are two kinds of closest-packed structures--cubic closest packed (ccp) and hexagonal closest packed (hcp). Using letters A, B, and C to designate the relative orientations of layers of atoms, the ccp structure has the layer stacking arrangement A-B-C-A-B C ..., whereas the hcp structure has the layer stacking arrangement A-B-A-B-A-B .... Study Figures 13.12 through 13.14. Amorphous solids Unlike crystals, amorphous solids lack long-range order and resemble liquids in their arrangements of particles. They are often called supercooled liquids. Examples include glass and many plastics. Self-Test 1. Why do crystals have such regular surface features? 2. What is the difference between a lattice and a structure based on the lattice? 3. Below is an illustration of atoms arranged in a two-dimensional lattice. On the drawing, sketch the outline of a unit cell for the lattice. 4. How many atoms are there per unit cell in the structure in the preceding problem? 5. The compound ZnS forms a face centered cubic structure as shown in Figure 13.10. All four Zn2+ ions are located entirely within the unit cell. How many S2 ions are there per unit cell? 6. The unit cell for ZnS is shown in Figure 13.10. Could the compound AlCl3 crystallize with this same arrangement of ions in a cubic unit cell? Explain your answer. 232 Structures, Properties and Applications of Solids 7. 8. On a separate sheet of paper, make sketches of simple cubic, face-centered cubic and body-centered cubic unit cells. Compare your drawings to Figures 13.4, 13.5, and 13.7. How do the unit cells of KCl and NaCl compare? 9. Which kind of unit cell is depicted in each of the following? (a) (b) (c) (a) (b) (c) 10. The radius of a K+ ion is 133 pm and the radius of a Cl ion is 181 pm. The salt KCl has the same kind of unit cell that NaCl has (Figure 13.8). Calculate the length of an edge of this unit cell in units of picometers. 11. Below is a sketch of part of the first layer in a closest-packed structure. If the second layer is started by placing an atom over point B and the third layer is started by placing an atom over point A, is the result a ccp or hcp structure? A B 12. Why aren't the structures of amorphous solids described by lattices? New Terms Write the definitions of the following terms, which were introduced in this section. If necessary, refer to the Glossary at the end of the text. simple (primitive) cubic unit cell body-centered cubic (bcc) unit cell face-centered cubic (fcc) unit cell hexagonal closest-packed (hcp) lattice unit cell closest-packed structure amorphous solid crystal lattice rock salt structure cubic closest -packed (ccp) supercooled liquid Chapter 13 233 13.2 Review X-ray diffraction is used to study crystal structures When a crystal is bathed in X rays, it produces a diffraction pattern that can be recorded on film or by other detection devices. From the angles at which the diffracted beams emerge from the crystal and the wavelength of the X rays, the distances between planes of atoms can be computed using the Bragg equation, n = 2dsin where n is an integer, is the wavelength of the X rays, d is the distance between planes of atoms, and is the angle at which the X rays are diffracted. By a complex procedure, scientists can use these interplanar distances to figure out the structure of the crystal. X-ray diffraction data yields information about distances between atoms within a crystal, from which the sizes of unit cells and even atomic sizes can be measured. Study Example 13.2. Self-Test 13. Why do diffracted X-ray beams occur in only certain directions when a crystal is bathed in X rays? 14. Draw three additional sets of parallel lines through the points in Figure 13.19. Do all the sets of lines have the same spacing? 15. In the Bragg equation (Equation 13.1): (a) (b) (c) What does the symbol stand for? What does the symbol stand for? What does the symbol d stand for? New Terms Write the definitions of the following terms, which were introduced in this section. If necessary, refer to the Glossary at the end of the text. diffraction pattern Bragg equation 13.3 Review Physical properties are related to crystal types In this section we assign crystalline solids to four crystalline types: ionic crystals, molecular crystals, covalent crystals, and metallic crystals. For each type you should learn the kinds of particles found at the lattice sites 234 Structures, Properties and Applications of Solids and the kinds of attractive forces that exist between them. You should be able to relate these forces to the physical properties of the solid. Study Table 13.1 and Example 13.3. Self-Test 16. What is the electron-sea model of a metallic crystal? 17. One of the elements was once named columbium, and that name is still used by some people. It is shiny, soft, ductile, and melts at 2468 C. It also conducts electricity. What type of crystal does columbium form? 18. Antimony pentachloride forms white crystals that do not conduct electricity and that melt at 2.8 C to give a nonconducting liquid. What kind of solid are SbCl5 crystals? 19. Strontium fluoride forms brittle, nonconducting crystals that melt at 1450 C and give an electrically conducting liquid. The solid itself is not an electrical conductor. What type of solid is SrF 2 ? New Terms Write the definitions of the following terms, which were introduced in this section. If necessary, refer to the Glossary at the end of the text. covalent crystal ionic crystal metallic crystal molecular crystal 13.4 Review Band theory explains the electronic structures of solids In a solid, the energy levels of the atoms merge to form sets of energy bands , each of which contains enormous numbers of closely spaced energy levels. The core electrons (electrons below the valence shell) of the atoms in a crystal are localized on the atoms. The valence shell energy levels combine to form a delocalized band called the valence band. This energy band contains the valence electrons of the atoms in the crystal. An energy band that is either filled or empty but which extends uninterrupted throughout the crystal is called a conduction band. (A conduction band is a delocalized band of energy levels.) In a metal, the valence band is either a partially filled band or overlaps an empty conduction band, so metals are good electrical conductors. In a nonmetal, the valence band is filled and there is a large energy gap (called the band gap) between the filled band and the lowest-energy conduction band. At normal temperatures, the conduction band is empty and there is no means for the substance to conduct electricity, so it is an insulator. Chapter 13 235 In a semiconductor, the band gap between the filled valence band and the conduction band is small. Thermal kinetic energy is able to raise some electrons into the conduction band, so semiconductors are weak electrical conductors. Raising the temperature increases the number of electrons populating the conduction band, so semiconductors become better electrical conductors as their temperatures are raised. p-type and n-type semiconductors If an element from Group IIIA is added as an impurity to a semiconductor from Group IVA, the valence band contains one electron vacancy for each atom of the Group IIIA element added. Electrical conduction begins when an electron from a neighboring atom fills this hole, leaving a "positive hole" elsewhere. Migration of such positive holes through the solid is the mode of electrical conduction in this p-type semiconductor. If an element from Group VA is added as an impurity to a semiconductor from Group IVA, the valence band contains one extra electron for each atom of the Group VA element added. Electrical conduction takes place by the migration of these extra electrons through the conduction band in this n-type semiconductor. Self-Test 20. Which kind of semiconductor is formed when (a) (b) antimony is added to silicon? boron is added to germanium? New Terms Write the definitions of the following terms, which were introduced in this section. If necessary, refer to the Glossary at the end of the text. band theory energy band valence band conduction band band gap p-type semiconductor n-type semiconductor solar battery 13.5 Review Polymers are composed of many repeating molecular units Polymers are macromolecules, which are large molecules containing hundreds or even thousands of atoms. They are formed by linking together many smaller molecules called monomers. The result is long chainlike molecules composed of large numbers of identical repeating units. Not all the molecules of a given polymer are identical in size, but they do contain the same repeating unit. Study the example of polypropylene shown on pages 564 and 565. Addition polymers are formed by just adding monomer units to form the chain. This is usually accomplished by a process involving free radicals (very reactive molecular fragments containing one or more unpaired electrons). Branching or linear chains are possible, depending on how the polymer is formed. Study the formation of polyethylene on page 566, as well the structure of polystyrene on page 567. Also, study Example 13.4 to be sure you know how to write the formula for an addition polymer. 236 Structures, Properties and Applications of Solids Condensation polymers are formed by the elimination of a small molecule and the joining of monomer units. The monomer units being joined do not have to be identical; when they are different, the polymer formed is called a copolymer. Nylon is an example; learn the formula for the repeating unit of nylon 6,6. Crosslinking occurs by the formation of bridges between adjacent polymer strands and increases the rigidity of the polymer. Rubber is formed by crosslinking the latex polymer called polyisoprene by heating it with sulfur. Polymer strands are linked by sulfursulfur bonds. Crystallinity affects the properties of polymers, which become stronger as the polymer more becomes crystalline. Learn the differences between the different polyethylene polymers (LDPE, HDPE, UHMWPE). Inorganic polymers Most polymers have a carbon backbone and are called organic polymers. Polymers in which the backbone chain consists of atoms other than carbon are called inorganic polymers. Silicone polymers are formed by reaction of water with compounds in which silicon is bonded to organic groups (e.g., --CH 3 ) and chlorine. Study the formation of the methylsilicone polymer described on page 574. Self-Test 21. What is the difference between an addition polymer and a condensation polymer? 22. What is the function of an initiator in the creation of an addition polymer? 23. Write the structural formula for the addition polymer formed from CH 3 CH2 CH=CH2. 24. Write the structural formula for the condensation polymer formed by elimination of CH3 OH from the following two monomer units. O CH3 O C CH2 CH2 CH2 O C O CH3 and HO CH2 CH2 OH 25. What would be the repeating unit in nylon 4,4? (Sketch its structure.) Chapter 13 237 26. What is the difference between LDPE and HDPE? 27. How does crosslinking affect the physical properties of a polymer? 28. Write the structural formula of the compound formed by the reaction of water with Si(CH3)3Cl. New Terms Write the definitions of the following terms, which were introduced in this section. If necessary, refer to the Glossary at the end of the text. addition polymers branching copolymer macromolecules polymer vulcanized rubber amide bond condensation polymer crosslinks monomer polymerization backbone of a polymer condensation polymerization initiator nylon 6,6 polystyrene 13.6 Review Liquid crystals have properties of both liquids and crystals Liquid crystals are substances with properties of both crystals and liquids. For example, they possess a relatively high degree of internal order, like a crystal, but they also have the ability to flow like a liquid. Different kinds of liquid crystals have different degrees of order within them. Here's a summary. Nematic phase: Molecules are aligned parallel to each other but are able to move in any direction. Smectic phase: Molecules are aligned parallel to each other in layers that can slide over each other. Smectic C phase: Like the smectic phase, but the axes of the molecules tilt relative to a line perpendicular to the plane of a layer. Cholesteric phase: Molecules are aligned in thin layers. The axes of the molecule rotate from one layer to the next, imparting a twist going down through a stack of layers. The cholesteric phase is able to rotate polarized light and is the kind of liquid crystal used in LCD displays. Study Figure 13.36 to be sure you understand how polarized light differs from unpolarized light. Self-Test 29. (a) How is a liquid crystal like a liquid? 238 Structures, Properties and Applications of Solids (b) How is a liquid crystal like a crystal? 30. What property characterizes molecules that are able to form a liquid crystalline phase? 31. Which molecule below would be more likely to form a liquid crystalline phase? (Note: At each intersection of bonds there is a carbon atom plus enough hydrogen atoms to give a total of four bonds to carbon.) Justify your choice. (a) HO H C HO OH OH (b) O HO OH O HO OH O O C H O New Terms Write the definitions of the following terms, which were introduced in this section. If necessary, refer to the Glossary at the end of the text. liquid crystal cholesteric phase nematic phase smectic C phase smectic phase plane polarized light 13.7 Review Modern ceramics have applications far beyond porcelain and pottery This section deals mainly with advanced ceramics, which are specialized ceramic materials that have exceptional properties. All ceramics have high melting points and are very hard. Some general uses include: Chapter 13 239 Abrasives: Materials such as silicon carbide (SiC) and alumina (Al2O3) are used in sandpaper and grinding wheels. Refractories: Because of their high melting points, ceramics are used in special bricks that line high temperature furnaces. The surface of the Space Shuttle is fitted with special light-weight ceramic tiles to protect the Shuttle from the high temperatures generated when the space vehicle reenters Earth's atmosphere. (Failure of the tiles has been blamed for the tragic loss of the Columbia, February 1, 2003.) Electrical insulators: Most ceramics have very low electrical conductivities and are used as insulators in high-voltage electrical systems. Hardness and high melting points reflect the large lattice energies of the ceramic materials, which arise because the metal ions are small and have large positive charges. Ceramics can be made by mixing the raw materials and then heating the mixture to the point where it almost melts. At the high temperature, the fine particles begin to join, producing a solid. The process is called sintering. The sol-gel process enables the synthesis of some ceramics that can't be made by other methods. The process involves hydrolysis (reaction with water) of metal alkoxides. These are metal compounds formed with alcohol molecules that have lost a proton to yield an anion. Using ethanol as an example, CH3 CH2 ethanol OH H+ CH3 CH2 O - ethoxide (an alkoxide ion) The hydrolysis takes place stepwise, with polymerization occurring during the hydrolysis. For example, the first step for the hydrolysis of a metal ethoxide and subsequent polymerization is C 2H 5O C 2H 5O M OC2H5 + H2O C 2H 5O C 2H 5O M OH C 2H 5O C 2H 5O C 2H 5O C 2H 5O M O H H OC2H5 O M OC2H5 C 2H 5O C 2H 5O M OC2H5 O M OC2H5 C 2H 5O H2O OC2H5 C 2H 5O OC2H5 Continued hydrolysis yields further polymerization. The ultimate product is a gel-like substance that can be treated in a variety of ways, as described in Figure 13.39. Study the applications described on page 581. Silicon carbide fibers can be made from poly(methylsilane). See the structure on page 582. Superconducting ceramics lose resistance to the flow of electricity at significantly higher temperatures than metals. The temperature at and below which superconductivity occurs is given the symbol T c. Among the most well characterized materials are mixed oxides of yttrium, barium, and copper. An example is 240 Structures, Properties and Applications of Solids YBa2Cu 3 O7, which has T c = 93 K. This means inexpensive liquid nitrogen, with a boiling point of 77K, is cold enough to bring the solid into a superconducting condition. Superconducting solids are repelled by a magnetic field. If problems related to transforming superconducting materials into wires can be overcome, it may someday be possible to have mag-lev trains suspended in air and traveling at high speed with no more resistance than experienced by an airplane. Self-Test 32. Which oxide is more likely to be used to form a ceramic, SnO or ZrO2 ? Why? 33. Which oxide has the larger lattice energy, Na2 O or Al 2 O3? 34. What is the formula for the methoxide ion? 35. What is a xerogel? 36. What is an aerogel? 37. What electrical property does a piezoelectric ceramic possess? 38. Write an equation for the thermal decomposition of poly(methylsilane), (CH3 SiH)n . 39. Which ceramic is used to coat the surfaces of tools to make them ore wear resistant? 40. Why is boron nitride used in cosmetics? New Terms Write the definitions of the following terms, which were introduced in this section. If necessary, refer to the Glossary at the end of the text. ceramic alkoxide aerogel sintering hydrolysis superconductor sol-gel process xerogel 13.8 Review Nanotechnology deals with controlling structure at the molecular level Nanotechnology (also called molecular nanotechnology) deals with extremely small-scale structures that have dimensions on the order of tens to hundreds of atoms. The goal is ultimately to be able to build structures atom by atom so that properties can be controlled in precise ways. At present, scientists are not capable of Chapter 13 241 doing this, but the tools are being developed that permit manipulation of very small structures and sometimes, individual atoms themselves. Like the scanning tunneling microscope (STM) discussed in Chapter 1, the atomic force microscope (AFM) enables the imaging of surface features down to the atomic scale. A description of how the instrument works is given in Figure 13.46. Unlike the STM, the AFM is able to study surface features of nonconducting samples. This section describes different forms of the element carbon. In diamond, carbon atoms are joined by single bonds only, so each carbon is at the center of a tetrahedron of other carbon atoms. The most stable form is graphite, which consists of layers of fused (joined) benzene-like rings, as illustrated on page 587 and in Figure 13.48. A more recently discovered form of carbon consists of tiny balls of carbon atoms arranged in five- and six-member rings. The simplest is C60, called buckminsterfullerene (buckyball for short). Much interest by scientists working in the field of nanotechnology has recently been focused on still another form of carbon referred to as carbon nanotubes. These consist of long tubes of carbon atoms that can be viewed as rolled up sheets of graphite, capped at each end by part of a buckyball. The computer-generated image on the cover of the textbook shows the structure of a carbon nanotube looking down its axis. See also Figure 13.48d. Study some of the properties of carbon nanotubes described on page 588 that make them unique. Self-Test 41. Why does the STM require that the sample be electrically conducting? 42. In what major way is the structure of diamond different from the other forms of carbon? 43. Why was the C60 molecule named buckminsterfullerene? 44. How is the structure of a carbon nanotube related to the structure of graphite? How is it related to the structure of a buckyball? 45. Which other substances besides carbon are known to form nanotubes? 46. How does the strength of a carbon nanotube compare with that of stainless steel? 47. Why are nanosized rods of BaTiO3 and SrTiO3 of interest to makers of computer storage devices? 48. The textbook describes potential uses for nanosized fibers of alumina. What is alumina? (Hint: you may have to look elsewhere in this chapter to find the answer.) 242 Structures, Properties and Applications of Solids New Terms Write the definitions of the following terms, which were introduced in this section. If necessary, refer to the Glossary at the end of the text. nanotechnology atomic force microscope fullerene buckyball molecular nanotechnology graphite buckminsterfullerene carbon nanotube Answers to Self-Test Questions 1. Because they have such ordered internal structures. 2. A lattice is a symmetrical array of points; a structure based on a lattice has chemical units associated with the lattice points. 3. Unit Cell 4. One. 5. Four. 6. No. AlCl 3 would require three times as many anion sites as cation sites, but the ratio on the ZnS structure is one to one. 7. See Figures 13.4, 13.5, and 13.7. 8. Both are fcc, but the edge length is greater for KCl than for NaCl because K+ is larger than Na+. 9. (a) face centered cubic, (b) body centered cubic, (c) simple cubic 10. Edge length = 2(133 pm) + 2(181 pm) = 628 pm 11. hcp 12. Because amorphous solids lack long range order. 13. Constructive and destructive interference 14. More sets can be drawn. Their spacings differ. 15. (a) The angle at which X rays are reflected from a set of planes of atoms in a crystal. (b) The wavelength of the X rays. (c) The distance between planes of atoms that are giving the reflection. 16. Positive ions of the metal at lattice sites surrounded by a sea of mobile electrons. 17. Metallic (the element is niobium, Nb) 18. Molecular 19. Ionic 20. (a) n-type, (b) p-type 21. In an addition polymer, monomer units are simply joined end to end. In a condensation polymer, a small molecule is eliminated when the monomer units become joined. 22. To start a free radical polymerization of monomer units. (See pages 565 and 566). Chapter 13 243 23. H C H C n O CH2 CH2 CH2 C O CH2 CH2 n H CH2 CH2 CH2 CH2 N CH2 H CH3 24. O 25. O C CH2 CH2 O C O H C N n 26. LDPE has shorter chains and more branching than HDPE. 27. Crosslinking makes a polymer stiffer, stronger, and raises its melting point. CH3 CH3 28. CH3 Si CH3 O Si CH3 CH3 29. (a) It is able to flow like a liquid. (b) It has some internal order similar to a crystal. 30. The molecules tend to be long, thin, and have a rigid central portion. 31. Molecule (a), because molecule (b) has many branches and single bonds, which makes it more flexible and more difficult to align with other molecules of the same type. 32. ZrO 2 , because the more highly charged Zr 4+ ion would give the oxide a high lattice energy. 33. Al2 O3 (because of the large charge on the Al3+ cation.) 34. CH3 O. 35. A very porous solid formed by removal of solvent from the product of the sol-gel process. 36. An extremely low density ceramic solid formed by rapid removal of solvent at a temperature above which the solvent cannot exist as a liquid. 37. They produce an electric potential when their shapes are deformed, as when made to bend or twist. heat 38. (CH3SiH)n (s) (SiC) n (s) + 2nH2(g) 39. Titanium nitride, TiN. 40. It consists of flat platelike crystals that slide over each other easily and thereby give the cosmetic a smooth and silky texture. 41. It relies on a flow of electricity. 42. In diamond, carbon uses only single bonds. In the other forms, a degree of multiple bonding is involved between carbon atoms. 43. The arrangement of carbon rings is similar to the structural members of the geodesic dome, a type of architectural structure designed by R. Buckminster Fuller. 44. The tube is structured like a rolled-up sheet of graphite. The tubes are capped by parts of buckyballs. 45. Molybdenum disulfide, MoS2 , and boron nitride, BN. 46. A carbon nanotube is about 100 times stronger than the same weight of stainless steel. 47. The rods can be electrically polarized in either of two directions. 48. Al2 O3 244 Structures, Properties and Applications of Solids Tools you have learned Consider removing this chart from the Study Guide so you can have it handy when tackling homework problems. Tool Bragg equation How it Works Using the wavelength of X rays and angles at which X rays are diffracted from a crystal, the distances between planes of atoms can be calculated. By knowing the arrangements of atoms in these unit cells, we can use the dimensions of the unit cell to calculate atomic radii and other properties. By examining certain physical properties of a solid (hardness, melting point, electrical conductivity in the solid and liquid state), we can often predict the nature of the particles that occupy lattice sites in the solid and the kinds of attractive forces between them. From the structure of a monomer, we can predict the structure of the polymer. From the structures of the monomers, we can predict the structure of the polymer. We can use structural formulas to write reactions for the formation of oxygen bridges between metals and nonmetals as part of the creation of ceramic materials. Unit cell structures for simple cubic, facecentered cubic, and body-centered cubic lattices Properties of crystal types Polymerization reactions forming addition polymers Polymerization reactions forming condensation polymers Hydrolysis reactions in the sol-gel process
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UC Irvine - PHYS - 47020
UC Irvine - PHYS - 47020
Bucknell - PHYS - 212
Solutions for Hand-In Set #1 PHYS 212 - Spring 2007Sketch Force Diagram for Left Ballon (Blue)Problem A2 - Repulsive BalloonsBlew up the balloons (i.e., inflated them - I didn't use any explosives or anything) and tied them to the ends of a piec
Bucknell - PHYS - 212
Solutions for Hand-In Set #2 PHYS 212 - Spring 2007Problem A5 (Problem 2 on E-fields program)The signs are (1) positive; (2) negative; (3) negative; (4) negative; (5) positive. The way to figure this out: bring the positive test charge very close
Bucknell - PHYS - 212
Solutions for Hand-In Set #3 PHYS 212 - Spring 2007Problem A15 - Mapping the field of a permanent magnetor vf = 2IHBL . m(a) Sketch of a bar magnet with direction of magnetic field around the magnet:Comment: The solution given above makes it
Bucknell - PHYS - 212
Solutions for Hand-In Set #4 PHYS 212 - Spring 2007Problem A21- An electromagnetTipler 27-54Tried this out with two batteries in series - it worked very well. I first tried touching the two nails together before applying the current - there was
Bucknell - PHYS - 212
Review Sheet for Hour-Test I - PHYS 212 NOTE: This review sheet is NOT a substitute for doing the assigned and hand-in problems and drills. You CAN'T expect to do well on the test unless you successfully complete all of the problems, including the EP
Bucknell - PHYS - 212
Solutions for Hand-In Set #6 PHYS 212 Spring 2007A32 - Traveling Waves on a Magic SpringA35 - Annoying your roommateFor the longitudinal wave, I found that the pulse did about 5 full round trips in about 4.4 seconds. (It was a bit difficult to
Bucknell - PHYS - 212
Review Sheet for Hour-Test II - PHYS 212, Spring 2007 NOTE: This review sheet is NOT a substitute for doing the assigned problems, hand-in problems, reading quizzes and drills. You CAN'T expect to do well on the test unless you successfully complete
Bucknell - PHYS - 212
Bucknell - PHYS - 212
Bucknell - PHYS - 212
Solutions for Hand-In Set #9 PHYS 212 Spring 2007A62 - Uncertainty(c) For the second trial solution we have 2 (x) = B sin kx d2 = Bk cos kx dx d 2 1 = -Bk 2 sin kx. dx2 Plugging these into the rearranged Schrdinger equation o gives -Bk 2 sin kx
Bucknell - PHYS - 212
Solutions for Hand-In Set #10 PHYS 212 Spring 2007Supplementary Reading 1.4Supplementary Reading 1.10When you plot the square of the wavefunction for the ground state you see a broad "bump" between 0 and L. From this graph you see that the par
Bucknell - PHYS - 212
Solutions for Hand-In Set #12 PHYS 212 Spring 2007Supplemental Reading - 5.4Supplementary Reading - 5.12To complete this problem, use the same steps as in 5.3, but in reverse: use "distance = speed time" (with c as a speed) to get an estimate
Bucknell - PHYS - 212
Solutions for Hand-In Set #12 PHYS 212 Spring 2007A81 - Flipping inside atomsThe energy of the photon is equal to the energy difference between the two energy levels of the antiparallel spin state and the parallel spin state. The energy of a pho
Bucknell - PHYS - 212
Solutions for Hand-In Set #12 PHYS 212 Spring 2007A81 - Flipping inside atomsThe energy of the photon is equal to the energy difference between the two energy levels of the antiparallel spin state and the parallel spin state. The energy of a pho
Pittsburgh - ENGR - 0031
ProblemsSection 1-2 Electric Circuits and Current Flow P1.2-1i (t ) =P1.2-2d 4 1 - e-5t = 20 e-5t A dt()t t t t 4 4 q ( t ) = i ( ) d + q ( 0 ) = 4 1 - e -5 d + 0 = 4 d - 4 e -5 d = 4 t + e -5t - C 0 0 0 0 5 5()P1.2-3 q ( t ) =
Pittsburgh - ENGR - 0031
Section 1-3 Systems of Units P1.3-1q = i t =( 4000 A )( 0.001 s )= 4CP1.3-2 i= q 45 10-9 = = 9 10-6 = 9 A t 5 10-3P1.3-3 electron C -19 i = 10 billion 1.602 10 electron = s C 9 electron -19 1010 1.602 10 electron s elect
Pittsburgh - ENGR - 0031
Section 1-5 Power and Energy P1.5-1 a.) q =b.) P = v i = (110 V )(10 A ) = 1100 W c.) Cost = P1.5-2 i dt = it = (10 A )( 2 hrs )( 3600s/hr ) = 7.2100.06$ 1.1kW 2 hrs = 0.132 $ kWhr4CP = ( 6 V )(10 mA ) = 0.06 W t = 200 Ws w = = 3.33103 s
Pittsburgh - ENGR - 0031
Section 1.8 How Can We Check.? P 1.8-1 Notice that the element voltage and current of each branch adhere to the passive convention. The sum of the powers absorbed by each branch are: (-2 V)(2 A)+(5 V)(2 A)+(3 V)(3 A)+(4 V)(-5 A)+(1 V)(5 A) = -4 W + 1
Pittsburgh - ENGR - 0031
ProblemsSection 2-2 Engineering and Linear Models P2.2-1 The element is not linear. For example, doubling the current from 2 A to 4 A does not double the voltage. Hence, the property of homogeneity is not satisfied. P2.2-2 (a) The data points do ind
Oregon State - ECON - 201
chapter1> FirstPrinciplesSection 1: Individual Choice: The Core of EconomicsIndividual choice is the decision by an individual of what to do, which necessarily involves a decision of what not to do.Every economic issue involves, on its most
Oregon State - ECON - 201
chapter1> FirstPrinciplesSection 2: Interaction: How Economies WorkAs we learned in the Introduction, an economy is a system for coordinating the productive activities of many people. In a market economy, such as the one we live in, that coor
Oregon State - PHL - 280
Writing Philosophy Papers: A Student GuideWelcome toWriting Philosophy Papers: A Student GuideDepartment of Philosophy Oregon State Universitycopyright 1997 by Department of Philosophy, Oregon State University. Reprinted with permission.file
Clarkson - ES - 220
Stevens - MA - 115
MA115 Mathematical Analysis I Test 1Instructions The test consists of two sections, Part 1: Questions 1 6 Part 2: Questions 7 10 You are to do all of Part 1. For Part 2, select and complete 2 of the 4 problems given. Please clearly indicate which
Stevens - MA - 115
MA115 Mathematical Analysis I Test 2Instructions The test consists of three sections, Part 1: Questions 1 6 Part 2: Questions 7 10 Part 3: Bonus Question You are to do all of Part 1. For Part 2, select and complete 2 of the 4 problems given. Plea
Stevens - MA - 115
MA115 Mathematical Analysis I Test 3 Answer KeyInstructions The test consists of three sections, Part 1: Questions 1 6 Part 2: Questions 7 10 You are to do all of Part 1. For Part 2, select and complete 2 of the 4 problems given. Please clearly i
Stevens - MA - 115
Ma 115 Homework Solutions for Week 1 1.1 #23f (x) = 4 + 3x - x2 f (3 + h) = 4 + 3(3 + h) - (3 + h)2 = 4 + 9 + 3h - (9 + 6h + h2 ) = 4 - 3h - h2 (4-3h-h2 -4) f (3+h)-f (3) = = h(-3-h) = -3 - h h h h 1.1 f (x) = #285x+4 x2 +3x+2is defined for al
Johns Hopkins - MATH - 110.109
Johns Hopkins - MATH - 110.109
Johns Hopkins - MATH - 110.109
Stevens - MA - 115
Ma 115 Homework Solutions for Week 2 2.2 #4x0(a) lim f (x) = 3 (c) lim f (x) = 2 +x3(b) lim f (x) = 4 -x3(d) lim f (x) does not exist because the limits in (b) and (c) are not equal.x3(e) f (3) = 32.2xa#6lim f (x) exists for all a
Stevens - CS - 181
Programmer Defined Types - Classes and Objects1Programmer Defined Types - Classes and ObjectsObject oriented programming (OOP) revolves around the notion of an object. Objects are created using programmer defined types, called classes. In C+, a
Stevens - MA - 115
MA 115 Homework Solutions for Week 3 2.7 #4(a) Since g(5) = -3, the point (5, -3) is on the graph of g. Since g (5) = 4, the slope of the tangent line at x = 5 is 4. Using the point-slope form of a line gives us y - (-3) = 4(x - 5), or y = 4x - 23
Stevens - MA - 115
MA 115 Homework Solutions for Week 43.5#2 u = u1/2 .Let u = g(x) = 4 + 3x and y = f (u) = Thendy dx=dy du du dx1 = 2 u-1/2 (3) =3 2 u=3 . 2 4+3x3.5#6Let u = g(x) = ex and y = f (u) = sin u. Thendy dx=dy du du dx= (co
Stevens - MA - 115
MA 115 Homework Solutions for Week 5 4.2 #6Absolute maximum value is f (8) = 5; absolute minimum value is f (2) = 0; local maximum values are f (1) = 2, f (4) = 4, and f (6) = 3; local minimum values are f (2) = 0, f (5) = 2, and f (7) = 1.4.2#
Stevens - MA - 115
MA 115 Homework Solutions for Week 6 4.9 #12 x +x+1 1 = x + 1 + F (x) = f (x) = x x 4.921 2 2x 1 2 2x+ x + ln |x| + C1 + x + ln |x| + C2if x < 0 if x > 0#19 f (x) = x(6 + 5x) = 6x1/2 + 5x3/2 f (x) = 4x3/2 + 2x5/2 + C f (1) = 6 + C
Stevens - MA - 115
MA 115 Homework Solutions for Week 7 5.5 #10 xex dx =2Let u = x2 . Then du = 2x, so 5.51 21 eu ( 2 du) = 1 eu + C = 1 ex + C 2 22#201 2 du, soLet u = 2. Then du = 2d and d = sec 2 + C #22sec 2 tan 2d =sec u tan u( 1 du) = 21 2sec
Stevens - MA - 115
MA 115 Homework Solutions for Week 8 5.7 #18 A B x-1 = + . Multiply both sides by (x + 1)(x + 2) to get x - 1 = A(x + 2) + B(x + 1). x2 + 3x + 2 x+1 x+2 Substituting -2 for x gives -3 = -B B = 3. Substituting -1 for x gives -2 = A. Thus,1 0x-1 dx
Stevens - MA - 115
C+ Basics - Part 11C+ BasicsInteger data types In our first program the data type of the variables we have used is int. Thus we can store whole numbers in these variables. In this part, we will discuss: two other integer types long and short fo