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Chemistry

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Published in: Chemistry
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Solids

Amiya B / Bhubaneswar

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Qualification: MSc

Teaches: Biology, Chemistry, Computer Science, C / C++, Python Programming

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  1. A Each of the millions of microhairs on a gecko's feet branches out to end in flattened tips called spatulae.
  2. TABLE 11.1 The Three States of Water Phase Gas (steam) Liquid (water) Solid (ice) Density Temperature (oc) (g/cm3, at 1 atm) Molar Volume 30.5 L 18.0 rnL 19.6 Molecular View 100 20 —4 5.90 x 10 0.998 0.917 TABLE 11.2 Properties of the States of Matter State Gas Liquid Solid Density Low High High Shape Indefinite Indefinite Definite Volume Indefinite Definite Definite Strength of Intermolecular Forces (Relative to Thermal Energy) Weak Moderate Strong A FIGURE 11.1 Liquids Assume the Shapes of Their Containers When we pour water into a flask, it assumes the shape of the flask because water molecules are free to flow.
  3. Molecules closely spaced not easily compressible Liquid Molecules widely spaced highly compressible Gas A FIGURE 11.2 Gases Are Compressible Molecules in a liquid are closely spaced and are not easily compressed. Molecules in a gas have a great deal of space between them, making gases compressible. Regular ordered structure Crystalline solid Solid No long-range order Amorphous solid Liquid < FIGURE 11.3 Crystalline and Amorphous Solids In a crystalline solid, the arrangement of the particles displays long-range order. In an amorphous solid, the arrangement of the particles has no long-range order. Gas
  4. C3H8 (g) C3H8 (1) A The propane in an LP gas tank is in the liquid state. When you open the tank, some propane vaporizes and escapes as a gas. This molecular diagram shows a sample of liquid water. Which of the diagrams below best depicts the vapor emitted from a pot of boiling water? (a) I CllC12 (b) (c) (When and are opposite in sign, Eis negative .) 4 neo Consider the interaction between two water molecules in liquid water:
  5. 96 pm Crystalline Solid Intermolecular Force 300 pm O O O o O O O o O Simple cubic I 2r A In the simple cubic lattice, the atoms touch along each edge so that the edge length is 2r.
  6. Cubic Cell Name Simple Cubic Body-Centered Cubic Face-Centered Cubic Simple Cubic Unit Cell Coordination number Atoms per Unit Cell 1 2 4 Structure Coordination Number 6 8 12 Edge Length in terms of r Packing Efficiency (fraction of volume occup 52% 68% Atoms per unit cell= 8 1 — atom at each 8 of 8 corners
  7. Body-centered cubic c2 12 b2 — 12 + b2 = 212 2 212 + 12 2 312 A In the body-centered cubic lattice, the atoms touch only along the cube diagonal. The edge length is 4r/Vfi. Face-centered cubic: extended structure Coordination number = 12 12 Face-Centered Cubic Unit Cell Face-centered cubic: unit cell Atoms/unit — Face-Centered Cubic Crystal Structure The different colors used on the atom: this figure are for clarity only. All atoms 'within the structure are identical,
  8. b2 = 12 + 12 = 212 2 = 212 (4r)2 Face-centered cubic A In the face-centered cubic lattice, the atoms touch along a face diagonal. The edge length is 2V5r.
  9. Relating Density to Crystal Structure Aluminum crystallizes with a face-centered cubic unit cell. The radius of an aluminum 3 atom is 143 pm. Calculate the density of solid crystalline aluminum in g/cm SORT You are given the radius of an aluminum atom and its crystal structure. You are asked to find the density of solid aluminum. STRATEGIZE The conceptual plan is based on the definition of density. Since the unit cell has the physical properties of the entire crystal, you can find the mass and volume of the unit cell and use these to calculate its density. SOLVE Begin by finding the mass of the unit cell. Determine the mass of an aluminum atom from its molar mass. Since the face-centered cubic unit cell contains four atoms per unit cell, multiply the mass of aluminum by 4 to get the mass of a unit cell. Next, calculate the edge length (l) of the unit cell (in m) from the atomic radius of aluminum. For the face-centered cubic structure, I 2V5r. Calculate the volume of the unit cell (in cm) by converting the edge length to cm and cubing the edge length. (Use cen- timeters because you will want to report the density in units of g/cm3.) Finally, calculate the density by dividing the mass of the unit cell by the volume of the unit cell. FOR PRACTICE GIVEN: r 143 pm, face-centered cubic FIND: d CONCEPTUAL PLAN m mass of unit cell — number of atoms in unit cell X mass of each at V volume of unit cell 3 (edge length) SOLUTION m(Al atom) m(unit cell) 1 2V2r I mol 26.98 mol 6.022 X 1023 atoms 4.480 X -10—23 g/ atom 4 ÄomS(4.480 X 10-23 g/ atom') 1.792 X 10-22 g 143 pm) 2VE(143 X 10-12m 4.045 X 10 10m 3 I cm 4.045 X 10 1 onf X 10-2 6.618 X 10 23 cm 3 m V 1.792 X 10-22 g 6.618 X 10 23 cm3 2.71 g/cm3 Chromium crystallizes with a body-centered cubic unit cell. The radius of a chromium 3 atom is 125 pm. Calculate the density of solid crystalline chromium in g/cm
  10. Orbital Blocks of the Periodic Table Groups IA Isl 11 3sl s-block elements p-block elements 2A 12 Mg 'I-block elements I-block elements 10 11 3s2 3B 4B 5B 6B 7B 58 59 Lanthanides Ce 6s24J15d1 6s24f3 Actinides Pa 60 6S24/4 7s25f36dl 61 Pm 6s24f5 7s25f46dl 8B 62 Sm Pu 7s25fG 1B 63 64 65 Eu Gd Tb Am Cm Bk 7s25f7 7s25f9 13 2s22p 66 6s24f10 Cf 14 2s22p2 Ho 6s24f11 7s25f11 115 2s22p3 68 6s24j12 Fm 7s25fl 16 2s22p4 69 Tm 6s24J13 Md 7s2sf13 17 2s22p 17 35 53 85 70 Yb 6s24fl 102 No 7s25fi The s, p, d, and f Blocks of the Periodic Table.

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