Textbook Question
Explain why silicon carbide–reinforced alumina is strongerand tougher than pure alumina.
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Ch.12 - Solids and Solid-State Materials
McMurry8th EditionChemistryISBN: 9781292336145
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Table of contents
- Ch.1 - Chemical Tools: Experimentation & Measurement143
- Ch.2 - Atoms, Molecules & Ions134
- Ch.3 - Mass Relationships in Chemical Reactions149
- Ch.4 - Reactions in Aqueous Solution157
- Ch.5 - Periodicity & Electronic Structure of Atoms92
- Ch.6 - Ionic Compounds: Periodic Trends and Bonding Theory73
- Ch.7 - Covalent Bonding and Electron-Dot Structures47
- Ch.8 - Covalent Compounds: Bonding Theories and Molecular Structure51
- Ch.9 - Thermochemistry: Chemical Energy104
- Ch.10 - Gases: Their Properties & Behavior138
- Ch.11 - Liquids & Phase Changes43
- Ch.12 - Solids and Solid-State Materials56
- Ch.13 - Solutions & Their Properties98
- Ch.14 - Chemical Kinetics79
- Ch.15 - Chemical Equilibrium107
- Ch.16 - Aqueous Equilibria: Acids & Bases94
- Ch.17 - Applications of Aqueous Equilibria125
- Ch.18 - Thermodynamics: Entropy, Free Energy & Equilibrium66
- Ch.19 - Electrochemistry130
- Ch.20 - Nuclear Chemistry73
- Ch.21 - Transition Elements and Coordination Chemistry122
- Ch.22 - The Main Group Elements155
- Ch.23 - Organic and Biological Chemistry43
Chapter 12, Problem 136a
A group 3A metal has a density of 2.70 g/cm3 and a cubic unit cell with an edge length of 404 pm. Reaction of A 1.07 cm3 chunk of the metal with an excess of hydrochloric acid gives a colorless gas that occupies 4.00 L at 23.0 °C and a pressure of 740 mm Hg. (a) Identify the metal.
Verified step by step guidance1
Calculate the molar mass of the metal using the given density and unit cell edge length. First, convert the edge length from picometers to centimeters. Then, use the formula for the volume of a cube (V = edge length^3) to find the volume of the unit cell. Multiply the volume by the density to find the mass of the unit cell. Since the metal is in group 3A, assume there are four atoms per unit cell (based on its cubic structure). Divide the mass of the unit cell by four to find the mass of one atom, and then convert this mass to molar mass (g/mol).
Determine the amount of gas produced by using the ideal gas law, PV = nRT. Convert the pressure from mm Hg to atm, and use the temperature in Kelvin. Solve for n, the number of moles of gas produced.
Assume the gas produced is hydrogen gas (H2), which is typical when metals react with hydrochloric acid. The stoichiometry of the reaction between the metal and hydrochloric acid is typically M + 2HCl -> MCl2 + H2, where M represents the metal. This indicates that the moles of metal reacted will be half the moles of hydrogen gas produced.
Use the moles of metal calculated in the previous step and the molar mass from the first step to find the mass of the metal that reacted. Compare this mass to the initial mass of the metal chunk to ensure the calculation is consistent.
Identify the metal by comparing the calculated molar mass with known molar masses of group 3A metals (Aluminum, Gallium, Indium, etc.). The closest match will likely be the identity of the metal.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Density and Molar Mass
Density is defined as mass per unit volume and is crucial for identifying substances. In this question, the density of the metal (2.70 g/cm³) can be used to calculate its molar mass by determining the mass of the given volume of the metal. This relationship helps in identifying the metal by comparing the calculated molar mass with known values of group 3A metals.
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Molar Mass Concept
Ideal Gas Law
The Ideal Gas Law (PV = nRT) relates the pressure, volume, temperature, and number of moles of a gas. In this scenario, the gas produced from the reaction with hydrochloric acid can be analyzed using this law to find the number of moles of gas generated. This information is essential for determining the stoichiometry of the reaction and ultimately identifying the metal.
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Unit Cell and Atomic Radius
The unit cell is the smallest repeating unit in a crystal lattice, and its edge length can be used to calculate the atomic radius of the metal. Given the edge length of 404 pm, one can derive the volume of the unit cell and relate it to the number of atoms per unit cell. This information, combined with the density, aids in identifying the metal by comparing its calculated atomic radius with known values.
Related Practice
Textbook Question
A cube-shaped crystal of an alkali metal, 1.62 mm on an edge, was vaporized in a 500.0 mL evacuated flask. The resulting vapor pressure was 12.5 mm Hg at 802 °C. The structure of the solid metal is known to be body-centered cubic. (c) What are the densities of the solid and the vapor in g>cm3?
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Textbook Question
Europium(II) oxide is a semiconductor with a band gap of 108 kJ/mol. Below 69 K, it is also ferromagnetic, meaning all the unpaired electrons on europium are aligned in the same direction. How many f electrons are present on each europium ion in EuO? (In lanthanide ions the 4f orbitals are lower in energy than the 6s orbitals.)
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Textbook Question
Assume that 1.588 g of an alkali metal undergoes complete reaction with the amount of gaseous halogen contained in a 0.500 L flask at 298 K and 755 mm Hg pressure. In the reaction, 22.83 kJ is released 1ΔH = -22.83 kJ2. The product, a binary ionic compound, crystallizes in a unit cell with anions in a face-centered cubic arrangement and with cations centered along each edge between anions. In addition, there is a cation in the center of the cube. (c) Sketch a space-filling, head-on view of the unit cell, labeling the ions. Are the anions in contact with one another?
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Textbook Question
Explain why graphite/epoxy composites are good materials for making tennis rackets and golf clubs.
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