MASC Midterm II page 3

MASC Midterm II page 3 - Contact # Atoms/Unit Cell Coord....

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Ionization Energy: energy to remove an electron from an element Increase right and up Electron Affinity: energy to add an electron to an element Increase right and up (more negative) R=.08206 L atm/mol/K =8.314 J/mol/K =62.37 L torr/mol/K 1 torr= 1 mm Hg 1 atm=760 torr 1 atm= 101,325 Pa Limiting Reactants: g1*mol/g*mol ratio*g/mol=g2 g1 requires g2, if g2>available, g2 is LR Gold has a very high density, d , of 19.3 g/mL, and it has the fcc structure. If the gold atoms were spheres as modeled by the fcc structure, what is the atomic radius? Solution Assume the radius of gold spheres to be r , and the edge of the face centered unit cell to be a . There are four gold (atomic mass 197.0, Avogadro's number = 6.023x10 23 ) atoms per unit cell, thus d = 4 * 197.0 / (6.023x10 23 * a 3 ) = 19.3 g / cm 3 a = 4.0774*10 -8 cm = 2 * 2 1/2 r r = 1.442 *10 -8 cm Please work out these formulas and numbers yourself. Unit Cell Direction of Closest
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Unformatted text preview: Contact # Atoms/Unit Cell Coord. No. Relation between r and a Simple cubic Edge 1 6 1 2 r a = Body-Centered Body Diagonal 2 8 3 4 r a = Face-Centered Face Diagonal 4 12 1 8 r a = solid mass of unit cell volume of unit cell d = . The mass of the unit cell will be equal to the number of atoms in the unit cell times the mass of a single atom. We saw earlier that simple cubic structures have 1 atom per unit cell; body-centered have 2 atoms per unit cell and face-centered have 4 atoms per unit cell. The mass of a single atom can be determined from the atomic weight of the atom. Recall that the atomic weight is the weight in grams of 1 mole of the substance and that 1 mole of a substance contains 6.022 10 23 atoms. Putting these ideas together, we get the following formula for the density of a solid: ( 29 23 solid 3 # atoms per unit cell AW/6.022 10 d a p = ....
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