/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 39 Aluminum metal crystallizes in a... [FREE SOLUTION] | 91影视

91影视

Aluminum metal crystallizes in a face-centered cubic unit cell. (a) How many aluminum atoms are in a unit cell? (b) What is the coordination number of each aluminum atom? (c) Estimate the length of the unit cell edge, \(a\) , from the atomic radius of aluminum \((1.43 \hat{\mathrm{A}}) .\) (d) Calculate the density of aluminum metal.

Short Answer

Expert verified
In a face-centered cubic unit cell, there are 4 aluminum atoms, and each atom has a coordination number of 12. The unit cell edge length, $a$, is 4.04 脜, and the density of aluminum metal is 2.70 g/cm鲁.

Step by step solution

01

Part (a): Number of aluminum atoms in a unit cell

In a face-centered cubic unit cell, there is one atom located at each corner and one atom at the center of each of the six faces. There are 8 corner atoms and 6 face atoms. Remember that each corner atom is shared by 8 unit cells, and each face atom is shared by 2, so we can calculate the total number of aluminum atoms in a unit cell like this: Number of aluminum atoms = (8 corner atoms 脳 1/8) + (6 face atoms 脳 1/2)
02

Calculate the Number of Aluminum Atoms

Now let's plug in the values and calculate the number of aluminum atoms in the FCC unit cell: Number of aluminum atoms = (8 脳 1/8) + (6 脳 1/2) Number of aluminum atoms = 1 + 3 = 4 In a face-centered cubic unit cell, there are 4 aluminum atoms.
03

Part (b): Coordination number of each aluminum atom

The coordination number is defined as the number of nearest neighbors that an atom has. In a face-centered cubic unit cell, each aluminum atom has 12 nearest neighbors. Therefore, the coordination number of each aluminum atom is 12.
04

Part (c): Estimate the length of the unit cell edge, a, from the atomic radius of aluminum

In a face-centered cubic unit cell, the atomic radius (r) is related to the edge length (a) by the following equation: \(a = 2\sqrt{2}r\) We are given the atomic radius of aluminum, r = 1.43 脜. Let's plug this value into the equation to find the edge length a.
05

Calculate the Edge Length a

By substituting the atomic radius in the equation, we get: \(a = 2\sqrt{2}(1.43\ 脜)\) \(a = 2(2.02)\ 脜\) \(a = 4.04\ 脜\) The unit cell edge length, a, for aluminum is 4.04 脜.
06

Part (d): Calculate the density of aluminum metal

To calculate the density (蟻) of aluminum, we will use the following equation: 蟻 = (mass of aluminum atoms in the unit cell) / (volume of the unit cell) The mass of each aluminum atom can be found using its molar mass, which is 26.98 g/mol, and Avogadro's number, N = 6.022 x 10^23 atoms/mol. The volume of the unit cell can be calculated as: Volume = a^3
07

Calculate the Density of Aluminum Metal

First, let's calculate the mass of 4 aluminum atoms in the unit cell: Mass of 4 Al atoms = (4 atoms x 26.98 g/mol) / (6.022 x 10^23 atoms/mol) Mass of 4 Al atoms = 1.79 x 10^-22 g Next, let's calculate the volume of the unit cell: Volume = a^3 Volume = (4.04 脜)^3 Volume = 65.78 脜^3 To convert 脜^3 to cm^3, multiply by 10^-24: Volume = 65.78 x 10^-24 cm^3 Now we can calculate the density of aluminum metal: 蟻 = (1.79 x 10^-22 g) / (65.78 x 10^-24 cm^3) 蟻 = 2.70 g/cm^3 The density of aluminum metal is 2.70 g/cm^3.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Aluminum Unit Cell
Aluminum, a key component for many industrial and household applications, forms a distinctive face-centered cubic (FCC) lattice in its crystalline state. Understanding this structure helps in grasping aluminum's physical properties, such as its strength and malleability.
In an FCC unit cell, each cell is made up of atoms located not just at the corners, but also at the center of each face of the cube. To break this down:
  • There are 8 atoms at the corners.
  • Additionally, each face of the cube has an atom in its center.
This means 6 face-centered atoms. Importantly, while each corner atom is shared among 8 surrounding cells, each face-centered atom is shared with one adjacent unit cell. This arrangement results in a total of 4 atoms per unit cell. This counting method is crucial for understanding the atomic packing in metals and influences characteristics like density and stability.
By counting these shared atoms accurately, we find the full complement of atoms per cell, displaying aluminum鈥檚 efficiency of packing within its unit cell.
Coordination Number
The coordination number provides insight into an atom's immediate environment within a structure. It indicates how many neighboring atoms a single atom is directly in contact with. For aluminum in a face-centered cubic structure, this is particularly pertinent.
In an FCC lattice, each aluminum atom is surrounded by 12 equidistant neighboring atoms. This high coordination number of 12 signifies a tightly packed structure. The significance of this feature dovetails with aluminum's strengths, contributing to its reputation as both lightweight and robust. A high coordination number, common to metals like copper and gold as well, is a hallmark of such a structure.
This concept is instrumental when considering metal alloys and compounds, as it impacts their mechanical properties.
Atomic Radius
The atomic radius is a fundamental measure used to describe the size of an atom from its nucleus to the outer boundary of the surrounding cloud of electrons. It plays a pivotal role in understanding the dimensions of materials, particularly when calculating the sizes of unit cells.
With aluminum, known for its face-centered cubic structure, its atomic radius is given as 1.43 脜. This specific measurement allows us to estimate the edge length of the unit cell. The relationship between the atomic radius and the edge length in FCC lattices is given by the formula:\[ a = 2\sqrt{2}r \]Where \( a \) is the edge length and \( r \) is the atomic radius. Plugging aluminum's radius into the equation, we find an edge length \( a \) of 4.04 脜. This method of calculation showcases how atomic dimensions influence structural dimensions, a key concept in any materials science study.
Understanding the atomic radius not only aids in visualizing how atoms are arranged spatially but also impacts how materials interact under various conditions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Amorphous silica, \(\mathrm{SiO}_{2},\) has a density of about 2.2 \(\mathrm{g} / \mathrm{cm}^{3}\) , whereas the density of crystalline quartz, another form of \(\mathrm{SiO}_{2},\) is 2.65 \(\mathrm{g} / \mathrm{cm}^{3} .\) Which of the following statements is the best explanation for the difference in density? $$ \begin{array}{l}{\text { (a) Amorphous silica is a network-covalent solid, but quartz }} \\ {\text { is metallic. }} \\ {\text { (b) Amorphous silica crystallizes in a primitive cubic lattice. }} \\ {\text { (c) Quartz is harder than amorphous silica. }}\end{array} $$ $$ \begin{array}{l}{\text { (d) Quartz must have a larger unit cell than amorphous }} \\ {\text { silica. }}\end{array} $$ $$ \begin{array}{l}{\text { (e) The atoms in amorphous silica do not pack as efficiently }} \\ {\text { in three dimensions as compared to the atoms in quartz. }}\end{array} $$

For each of these solids, state whether you would expect it to possess metallic properties: (a) TiCl_ \(_{4},(\mathbf{b})\) NiCo alloy, \((\mathbf{c}) \mathrm{W}\) \((\mathbf{d}) \mathrm{Ge},(\mathbf{e}) \mathrm{ScN}\)

Red light-emitting diodes are made from GaAs and GaP solid solutions, GaP \(_{x} A s_{1-x}(\) see Exercise 12.79\() .\) The original red LEDs emitted light with a wavelength of 660 nm. If we assume that the band gap varies linearly with composition between \(x=0\) and \(x=1,\) estimate the composition (the value of \(x\) ) that is used in these LEDs.

An ester is a compound formed by a condensation reaction between a carboxylic acid and an alcohol that eliminates a water molecule. Read the discussion of esters in Section 24.4 and then give. an example of a reaction forming an ester. How might this kind of reaction be extended to form a polymer (a polyester)?

(a) What are the \(\mathrm{C}-\mathrm{C}-\mathrm{C}-\mathrm{C}\) bond angles in diamond? (b) What are they in graphite (in one sheet)? (c) What atomic orbitals are involved in the stacking of graphite sheets with each other?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.