/*! 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 71 Gold is a face-centered cubic st... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Gold is a face-centered cubic structure that has a unit cell edge length of \(4.08 \AA\) (Figure 11.34). How many gold atoms are there in a sphere that is \(20 \mathrm{~nm}\) in diameter? Recall that the volume of a sphere is \(\frac{4}{3} \pi r^{3}\).

Short Answer

Expert verified
There are approximately 247111 gold atoms in a sphere with a diameter of 20 nm, given that gold has a face-centered cubic structure with a unit cell edge length of 4.08 Ã….

Step by step solution

01

Convert Edge Length to Nanometers

First, we need to convert the unit cell edge length from Ã… (angstroms) to nm (nanometers), as we need to work with the same unit of measurement. The conversion factor is 1 Ã… = 0.1 nm. Thus, we have: Unit cell edge length in nm = \(4.08 Ã… \times 0.1 nm/Ã… = 0.408 nm\)
02

Calculate the Volume of the Unit Cell

The volume of a cube with edge length 'a' is given by the formula \(V = a^{3}\). Using the edge length in nm, we can find the volume of the unit cell: Volume of unit cell = \((0.408 nm)^{3} = 0.0678 nm^{3}\)
03

Calculate the Volume of the Sphere

The volume of a sphere with radius 'r' is given by the formula \(V = \frac{4}{3}\pi r^{3}\). To find the volume of the sphere with a diameter of 20 nm, we need to divide the diameter by 2 to determine the radius: Radius of sphere = \(20 nm / 2 = 10 nm\) Now we can calculate the volume of the sphere: Volume of sphere = \(\frac{4}{3}\pi (10 nm)^{3} = \frac{4}{3}\pi (1000 nm^{3}) = 4188.79 nm^{3}\)
04

Calculate the Number of Unit Cells in the Sphere

To find out how many unit cells fit into the sphere, divide the volume of the sphere by the volume of the unit cell: Number of unit cells = \(\frac{Volume\ of\ sphere}{Volume\ of\ unit\ cell} = \frac{4188.79 nm^{3}}{0.0678 nm^{3}} = 61777.7\)
05

Calculate the Number of Gold Atoms in the Sphere

Since there are 4 gold atoms in each unit cell, we can calculate the total number of gold atoms in the sphere by multiplying the number of unit cells by 4: Number of gold atoms = \(Number\ of\ unit\ cells \times 4 = 61777.7 \times 4 = 247110.8\) Because the number of atoms must be an integer, we will round this result to the nearest whole number: Number of gold atoms in the sphere ≈ 247111

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 91Ó°ÊÓ!

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

Explain why "bands" may not be the most accurate description of bonding in a solid when the solid has nanoscale dimensions.

(a) What is the superconducting transition temperature, \(T_{c} ?\) (b) The discovery by Müller and Bednorz of superconductivity in a copper oxide ceramic at \(35 \mathrm{~K}\) set off a frantic scramble among physicists and chemists to find materials that exhibit superconductivity at higher temperatures. What is the significance of achieving \(T_{c}\) values above \(77 \mathrm{~K}\) ?

In contrast to ordinary liquids, liquid crystals are said to possess "order." What does this mean?

Classify each of the following materials as metal, semiconductor, or insulator: (a) InAs (b) \(\mathrm{MgO}\) (c) \(\mathrm{HgS}\) (d) \(\mathrm{Sn}\)

The group \(4 \mathrm{~B}\) metal nitrides (TiN, ZrN, and HfN) as well as the group \(5 \mathrm{~B}\) metal nitrides \((\mathrm{V} \mathrm{N}, \mathrm{Nb} \mathrm{N}\), and \(\mathrm{TaN})\) are all superconductors at low temperature. Niobium(III) nitride, which has the highest \(T_{c}\), superconducts below \(16.1 \mathrm{~K}\). All of these compounds have crystal structures that are analogous to the sodium chloride structure. Scandium nitride also adopts the sodium chloride structure, but it is not a superconductor. (a) At room temperature will NbN be a metallic conductor or an insulator? (b) At room temperature will ScN be a metallic conductor or an insulator? (c) Why do you think the properties of \(\mathrm{ScN}\) are so different than the group \(4 \mathrm{~B}\) and \(5 \mathrm{~B}\) metal nitrides? (Hint: Consider the electron configuration of the metal cation.)

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.