Chapter 12: Problem 26
In terms of bonding, explain why silicate materials have relatively low densities.
Chapter 12: Problem 26
In terms of bonding, explain why silicate materials have relatively low densities.
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Get started for freeShow that the minimum cation-to-anion radius ratio for a coordination number of 6 is \(0.414 .[\) Hint: Use the NaCl crystal structure (Figure 12.2 ), and assume that anions and cations are just touching along cube edges and across face diagonals.
The unit cell for \(\mathrm{Fe}_{3} \mathrm{O}_{4}\left(\mathrm{FeO}-\mathrm{Fe}_{2} \mathrm{O}_{3}\right)\) has cubic symmetry with a unit cell edge length of \(0.839 \mathrm{nm} .\) If the density of this material is \(5.24 \mathrm{g} / \mathrm{cm}^{3},\) compute its atomic packing factor. For this computation, you will need to use ionic radii listed in Table 12.3.
(a) Using the ionic radii in Table \(12.3, \mathrm{com}-\) pute the theoretical density of CsCl. (Hint: Use a modification of the result of Prob\(\operatorname{lem} 3.3 .)\) (b) The measured density is \(3.99 \mathrm{g} / \mathrm{cm}^{3} .\) How do you explain the slight discrepancy between your calculated value and the measured one?
When kaolinite clay \(\left[\mathrm{Al}_{2}\left(\mathrm{Si}_{2} \mathrm{O}_{5}\right)(\mathrm{OH})_{4}\right]\) is heated to a sufficiently high temperature, chemical water is driven off. (a) Under these circumstances, what is the composition of the remaining product (in weight percent \(\mathrm{Al}_{2} \mathrm{O}_{3}\) )? (b) What are the liquidus and solidus temperatures of this material?
Using the Molecule Definition Utility found in both "Metallic Crystal Structures and Crystallography" and "Ceramic Crystal Structures" modules of VMSE, located on the book's web site [www.wiley.com/college/callister (Student Companion Site)], generate (and print out) a three-dimensional unit cell for lead oxide, \(\mathrm{PbO}\), given the following: (1) The unit cell is tetragonal with \(a=0.397 \mathrm{nm}\) and \(c=0.502 \mathrm{nm},(2)\) oxygen atoms are located at the following point coordinates: $$\begin{array}{ll} 000 & 001 \\ 100 & 101 \\ 010 & 011 \\ 110 & 111 \\ \frac{1}{2} \frac{1}{2} 0 & \frac{1}{2} \frac{1}{2} 1 \end{array}$$ and (3) \(\mathrm{Pb}\) atoms are located at the following point coordinates: $$\frac{1}{2} 00.763 \quad 0 \frac{1}{2} 0.237$$ $$\frac{1}{2} 10.763 \quad 1 \frac{1}{2} 0.237$$
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