School Work

20 pages
769 views

Science and Engineering of Materials 7th Edition Askeland Solutions Manual

of 20
All materials on our website are shared by users. If you have any questions about copyright issues, please report us to resolve them. We are always happy to assist you.
Share
Description
Science and Engineering of Materials 7th Edition Askeland Solutions Manual Download at: https://goo.gl/HHsWNE People also search: the science and engineering of materials 7th edition pdf the science and engineering of materials donald r. askeland pdf free download the science and engineering of materials 7th edition solutions the science and engineering of materials 7th edition pdf download the science and engineering of materials 6th edition solution manual the science and engineering of materials 7th edition askeland solutions the science and engineering of materials 7th edition solution manual the science and engineering of materials 6th edition solution manual askeland pdf
Transcript
    © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Science and Engineering of Materials 7th Edition Askeland Solutions Manual  Full clear download (no formatting errors) at: http://testbanklive.com/download/science-and-engineering-of-materials-7th-edition-askeland-solutions-manual/  Chapter 2: Atomic Structure   2-1 What is meant by the term composition of a material? Solution:  The chemical make-up of the material. 2-2 What is meant by the term structure of a material? Solution:  The spatial arrangement of atoms or ions in the material. 2-3 What are the different levels of structure of a material? Solution:  Atomic structure, short- and long-range atomic arrangements, nanostructure, microstructure, and macrostructure. 2-4 Why is it important to consider the structure of a material when designing and fabricating engineering components? Solution:  The structure of the material at all levels will affect the physical and mechanical properties of the final product. 2-5 What is the difference between the microstructure and macrostructure of a material? Solution:  A length scale of about 100,000 nm (100 μ m) separates microstructure (less than 100,000 nm) from macrostructure (greater than 100,000 nm). 2-6 (a) Aluminum foil used for storing food weighs about 0.3 grams per square inch. How many atoms of aluminum contained in one square inch of the foil? (b) Using the densities and atomic weights given in Appendix A, calculate and compare the number of atoms per cubic centimeter in (i) lead and (ii) lithium.    © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Solution:  (a) = 1 in.   0.3 g Al 1 mol Al 6.022 × 10   atoms #1 in.   26.982 g Al 1 mol    © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. cm   = 6.7 × 10   % Al atoms (b) 11.36 g Pb 1 mol Pb 6.022 × 10   atoms   = 1 cm   207.19 g Pb 1 mol #   = 33.0 × 10   %   Pb atoms cm   0.534 g Li 1 mol Li 6.022 × 10   atoms   = 1 6.94 g Li 1 mol #   = 46.3 × 10   %   Li atoms cm   2-7 (a) Using data in Appendix A, calculate the number of iron atoms in one ton (2000 pounds). (b) Using data in Appendix A, calculate the volume in cubic centimeters occupied by one mole of boron. Solution:   2000 lb 454 g 1 mol Fe 6.022 × 10   atoms   = 1 ton 1 lb 55.847 g Fe 1 mol #  = 9.8 × 10   .   Fe atoms ton   10.81  /  = 1 mol B   g B 1 cm   B #1 mol B 2.36 g B   / = 4.58 cm   B  2-8 In order to plate a steel part having a surface area of 200 in. 2 with a 0.002 in.-thick layer of nickel: (a) how many atoms of nickel are required? (b) How many moles of nickel are required? Solution:  (a) We start with the volume required: 2.54 cm /  = 200 in.   0.002 in. 1 in. = 6.55 cm      © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. = 6.55 cm   8.902 g Ni 1 mol Ni 6.022 × 10   atoms  # 1 cm   58.71 g Ni 1 mol   = 598 × 10   % atoms Ni  (b) 8.902 g Ni  2 3   = 6.55 cm   1 mol Ni 1 cm   58.71 g Ni   2 3   = 0.99 mol Ni  2-9 Define electronegativity  . Solution:  Electronegativity is the tendency of an atom to accept an electron (which has a negative charge) and become an anion. 2-10 Write the electronic configuration of the following elements (a) tungsten, (b) cobalt, (c) zirconium, (d) uranium, and (e) aluminum. Solution:  (a) W: [Xe] 4  f  14 5 d  4 6 s 2  (b) Co: [Ar] 3 d  7 4 s 2  (c) Zr: [Kr] 4 d  2 4 s 2  (d) U: [Rn] 5  f  3 6 d  1 7 s 2  (e) Al: [Ne] 3 s 2 3  p 1   2-11 Write the electron configuration for the element Tc. Solution:  Since Technetium is element 43: [Tc] = 1 s 2 2 s 2 2  p 6 3 s 2 3  p 6 4 s 2 3 d  10 4  p 6 5 s 2 4 d  5  2-12 Assuming that the Aufbau Principle is followed, what is the expected electronic configuration of the element with atomic number  Z = 116? Solution:  Using the Aufbau diagram produces: [116] = 1 s 2 2 s 2 2  p 6 3 s 2 3  p 6 4 s 2 3 d  10 4  p 6 5 s 2 4 d  10 5  p 6 6 s 2 4  f  14 5 d  10 6  p 6 7 s 2 5  f  14 6 d  10 7  p 4  Or in shorthand:
We Need Your Support
Thank you for visiting our website and your interest in our free products and services. We are nonprofit website to share and download documents. To the running of this website, we need your help to support us.

Thanks to everyone for your continued support.

No, Thanks
SAVE OUR EARTH

We need your sign to support Project to invent "SMART AND CONTROLLABLE REFLECTIVE BALLOONS" to cover the Sun and Save Our Earth.

More details...

Sign Now!

We are very appreciated for your Prompt Action!

x