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Chapter 24 Metals and Metallurgy 511 (25 g/mol) = (25 + (25 (55.85 + X 100% = 26.6% Fe by mass and percent by mass V = 100.0% - (24.8% + 26.6%) = 48.6% V. 24.33 Cr and Fe both form body-centered cubic structures. In addition, they are very close together in the periodic table (atomic numbers 24 and 26, respectively); so their respective atomic radii are probably close enough to form an alloy. 24.35 This phase diagram indicates that the solid and liquid phases are completely miscible. The composition is found by determining the x-axis value at the point. A: solid with 20% Cr and 100% - 20% = 80% Fe B: liquid with 50% Cr and 100% - 50% = 50% Fe 24.37 This phase diagram indicates that the solid and liquid phases are not completely miscible. Single phases only exist between the pure component and the red line. The composition will be a mixture of the two structures. The composi- tion of the two phases is determined by moving to the left and right until reaching the red lines. The x-axis value is the composition of that structure. According to the lever rule, the phase that is closer to the point is the dominant phase. A: solid with 20% Co and 100% 20% = 80% Cu overall composition. One phase will be the copper structure with ~4% Co, and the other phase will be the Co structure with ~7% Cu. According to the lever rule, there will be more of the Cu structure because point A is closer to the Cu structure phase boundary line. B: solid single phase with Co structure and composition of 90% Co and 100% 90% = 10% Cu. 24.39 Because C (77 pm) is much smaller than Fe (126 pm), it will fill interstitial holes. Because Mn (130 pm) and Si (118 pm) are close to the same size as Fe, they will substitute for Fe in the lattice. 24.41 (a) Because there are the same number of octahedral holes as there are metal atoms in a closest-packed structure and half are filled with N, the formula is Mo₂N. (b) Because there are twice as many tetrahedral holes as there are metal atoms and all of the tetrahedral holes are occupied, the formula is CrH₂. Sources, Properties, and Products of Some of the 3d Transition Metals 24.43 (a) Zn because sphalerite is ZnS (b) Cu because malachite is Cu₂(OH)₂CO₃ (c) Mn because hausmannite is Mn₃O₄ 24.45 Given: Calcination of rhodochrosite (MnCO₃) Find: heat of reaction Conceptual Plan: Write the reaction. Then = Solution: MnCO₃(s) + Heat MnO₂(s) + Reactant/Product from Appendix IIB) MnCO₃(s) -894.1 0.0 MnO₂(s) -520.0 -393.5 Be sure to pull data for the correct formula and phase. = = + - + = [1(-520.0kJ) + [1(-894.1kJ) + = [-913.5kJ] [-894.1kJ] = -19.4kJ Check: The units (kJ) are correct. The answer is negative, which means that the reactions are exothermic. Copyright © 2017 Pearson Education, Inc.