MidTrm-PrepQs - CHAPTER 2 Q1- How many grams are there in a...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: CHAPTER 2 Q1- How many grams are there in a one amu of a material? A1- In order to determine the number of grams in one amu of material, appropriate manipulation of the amu/atom, g/mol, and atom/mol relationships is all that is necessary, as # g/amu = 1 mol 6.023 x 10 23 atoms 1 g / mol 1 amu / atom = 1.66 x 10-24 g/amu Q2- Show the electron configuration for Fe 3+ , Cu + , and S 2- A2- Fe 3+- 1s 2 2s 2 2p 6 3s 2 3p 6 3d 5 Cu +- 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 S 2-- 1s 2 2s 2 2p 6 3s 2 3p 6 Q3. Why water expands on freezing? A3. The geometry of the H 2 O molecules, which are hydrogen bonded to one another, is more restricted in the solid phase than for the liquid. This results in a more open molecular structure in the solid, and a less dense solid phase. CHAPTER 3 Q1- What is the difference between a crystal structure and a crystal system? A1- A crystal structure is described by both the geometry of, and atomic arrangements within, the unit cell, whereas a crystal system is described only in terms of the unit cell geometry. For example, face-centered cubic and body-centered cubic are crystal structures that belong to the cubic crystal system. Q2- Show that the ideal c / a ratio for HCP is 1.633. A2- A sketch of one-third of an HCP unit cell is shown below. Consider the tetrahedron labeled as JKLM , which is reconstructed as The atom at point M is midway between the top and bottom faces of the unit cell--that is MH = c / 2. And, since atoms at points J , K , and M , all touch one another, JM = JK = 2R = a where R is the atomic radius. Furthermore, from triangle JHM , (JM ) 2 = (JH ) 2 + (MH ) 2 , or a 2 = (JH ) 2 + c 2 2 Now, we can determine the JH length by consideration of triangle JKL , which is an equilateral triangle, cos 30 = a / 2 JH = 3 2 , and JH = a 3 Substituting this value for JH in the above expression yields a 2 = a 3 2 + c 2 2 = a 2 3 + c 2 4 and, solving for c / a c a = 8 3 = 1.633 Q3- Calculate the unit cell volume for Co which has an HCP crystal structure....
View Full Document

This note was uploaded on 09/26/2010 for the course MATRL 101 taught by Professor Fredlange during the Spring '09 term at UCSB.

Page1 / 6

MidTrm-PrepQs - CHAPTER 2 Q1- How many grams are there in a...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online