Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 10: Problem 86

Superalloys have been made of nickel and aluminum. The alloy owes its strength to the formation of an ordered phase, called the gamma-prime phase, in which Al atoms are at the corners of a cubic unit cell and Ni atoms are at the face centers. What is the composition (relative numbers of atoms) for this phase of the nickel–aluminum superalloy?

Short Answer

Expert verified
The composition of the gamma-prime phase of the nickel-aluminum superalloy in terms of the relative numbers of atoms is 1 Al atom and 3 Ni atoms, which can be represented as Al : Ni = 1:3.

Step by step solution

01

1. Collect the information

We know that aluminum atoms are at the corners of the unit cell (8 corners) and nickel atoms are at the face centers (6 faces) of the cubic unit cell.
02

2. Calculate the number of Al atoms in the unit cell

Each corner atom of the cubic unit cell is shared by 8 different cells. Since there are 8 corner positions in the unit cell, the total number of Al atoms contributed to the unit cell is (8 corners x (1/8)) = 1 Al atom.
03

3. Calculate the number of Ni atoms in the unit cell

Each face-centered atom of the cubic unit cell is shared by 2 different cells. Since there are 6 face center positions in the unit cell, the total number of Ni atoms contributed to the unit cell is (6 faces x (1/2)) = 3 Ni atoms.
04

4. Determine the composition

The composition of the gamma-prime phase of the nickel-aluminum superalloy in terms of the relative numbers of atoms can be given as: 1 Al atom and 3 Ni atoms. This can be simplified by removing the units and stating the composition as Al : Ni, which is 1:3.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Which of the following compound(s) exhibit only London dispersion intermolecular forces? Which compound(s) exhibit hydrogen-bonding forces? Considering only the compounds without hydrogen-bonding interactions, which compounds have dipole–dipole intermolecular forces? a. \(\mathrm{SF}_{4}\) b. \(\mathrm{CO}_{2}\) c. \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) d. \(\mathrm{HF}\) e. \(\mathrm{ICl}_{5}\) f. \(\mathrm{XeF}_{4}\)

Barium has a body-centered cubic structure. If the atomic radius of barium is 222 \(\mathrm{pm}\) , calculate the density of solid barium.

Predict which substance in each of the following pairs would have the greater intermolecular forces. a. \(\mathrm{CO}_{2}\) or OCS b. \(\mathrm{SeO}_{2}\) or \(\mathrm{SO}_{2}\) c. \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\) or \(\mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\) d. \(\mathrm{CH}_{3} \mathrm{CH}_{3}\) or \(\mathrm{H}_{2} \mathrm{CO}\) e. \(\mathrm{CH}_{3} \mathrm{OH}\) or \(\mathrm{H}_{2} \mathrm{CO}\)

Carbon tetrachloride, \(\mathrm{CCl}_{4},\) has a vapor pressure of 213 torr at \(40 .^{\circ} \mathrm{C}\) and 836 torr at \(80 .^{\circ} \mathrm{C}\) . What is the normal boiling point of \(\mathrm{CCl}_{4} ?\)

Iron has a density of 7.86 \(\mathrm{g} / \mathrm{cm}^{3}\) and crystallizes in a body-centered cubic lattice. Show that only 68\(\%\) of a body-centered lattice is actually occupied by atoms, and determine the atomic radius of iron.
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

Nuclear Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Inorganic Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Ionic and Molecular Compounds

Read Explanation
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.

Sign-up for free