Which one of the following statements about solid Cu (face-centered cubic unit cell) is incorrect? O a. The length of a face diagonal is four times the Cu radius. O b. There are two atoms per unit cell. O c. The solid has a cubic closest-packed structure. O d. The number of atoms surrounding each Cu atom is 12. O e. It will conduct electricity.

Assume X has a body-centered cubic lattice with all atoms at the lattice points. The edge length of the unit cell is 379.0 pm. The atomic mass of X is 195.0 amu. Calculate the density of X.

Iridium metal, Ir, crystallizes in a face-centered cubic (close-packed) structure. The edge length of the unit cell was found by x-ray diffraction to be 383.9 pm. The density of iridium is 22.42 g/cm3. Calculate the mass of an iridium atom. Use Avogadros number to calculate the atomic mass of iridium.

Calculate the percent of volume that is actually occupied by spheres in a body-centered cubic lattice of identical spheres You can do this by first relating the radius of a sphere, r, to the length of an edge of a unit cell, l. (Note that the spheres do not touch along an edge but do touch along a diagonal passing through the body-centered sphere.) Then calculate the volume of a unit cell in terms of r. The volume occupied by spheres equals the number of spheres per unit cell times the volume of a sphere (4r3/3).

(a) Determining an Atom Radius from Lattice Dimensions: Gold has a face-centered unit cell, and its density is 19.32 g/cm3. Calculate the radius of a gold atom. (b) The Structure of Solid Iron: Iron has a density of 7.8740 g/cm3, and the radius of an iron atom is 126 pm. Verify that solid iron has a body-centered cubic unit cell. (Be sure to note that the atoms in a body-centered cubic unit cell touch along the diagonal across the cell. They do not touch along the edges of the cell.) (Hint: The diagonal distance across the unit cell = edge 3.)

Phase diagrams for materials that have allotropes can be more complicated than those shown in the chapter. Use the phase diagram for carbon given here to answer the following questions. (a) How many triple points are present and what phases are in equilibrium for each? (b) Is there a single point where all four phases are in equilibrium? (c) Which is more stable at high pressures, diamond or graphite? (d) Which is the stable phase of carbon at room temperature and 1 atmosphere pressure?

Aluminum metal crystallizes with a face-centered cubic unit cell. The volume of the cell is 0.0662 nm3. (a) What is the atomic radius of aluminum in cm? (b) What is the volume of a single aluminum atom? (c) What is the density of a single aluminum atom? (d) In face-centered cubic cell packing, the fraction of empty space is 26.0%. When this is factored in, what is the calculated density of aluminum?

The CsCl structure is a simple cubic array of chloride ions with a cesium ion at the center of each cubic array (see Exercise 69). Given that the density of cesium chloride is 3.97 g/cm3, and assuming that the chloride and cesium ions touch along the body diagonal of the cubic unit cell, calculate the distance between the centers of adjacent Cs+ and Cl ions in the solid. Compare this value with the expected distance based on the sizes of the ions. The ionic radius of Cs+ is 169 pm, and the ionic radius of Cl is 181 pm.

The memory metal, nitinol, is an alloy of nickel and titanium. It is called a memory metal because after being deformed, a piece of nitinol wire will return to its original shape. The structure of nitinol consists of a simple cubic array of Ni atoms and an inner penetrating simple cubic array of Ti atoms. In the extended lattice, a Ti atom is found at the center of a cube of Ni atoms; the reverse is also true. a. Describe the unit cell for nitinol. b. What is the empirical formula of nitinol? c. What are the coordination numbers (number of nearest neighbors) of Ni and Ti in nitinol?

Calculate the percent of volume that is actually occupied by spheres in a face-centered cubic lattice of identical spheres. You can do this by first relating the radius of a sphere, r, to the length of an edge of a unit cell, l. (Note that the spheres do not touch along an edge but do touch along the diagonal of a face.) Then calculate the volume of a unit cell in terms of r. The volume occupied by spheres equals the number of spheres per unit cell times the volume of a sphere (4r3/3).