{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

e45_hw_9 - E45 HW09 Due November 5 2010 Resources J.F...

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

View Full Document Right Arrow Icon
E45 HW09 Due November 5, 2010 Resources: J.F. Shackelford, Introduction to Materials Science for Engineers, 6th Edition, Prentice Hall, New Jersey (2005) Bond lengths and energies. University of Waterloo. http://www.science.uwaterloo.ca/~cchieh/cact/c120/bondel.html Problem 1 13.1 What is the average molecular weight of a polypropylene with a degree of polymerization of 500? (Note Table 13.1) SOLUTION Polypropylene (Source: Shackelford, Table 13.1, p. 478) 1 “mer” of polypropylene = C 3 H 6 Mol. weight of 1 “mer” of polypropylene = 3 (12 g) + 6 (1 g) = 42 g Mol. weight of polymer = Degree of Polymerization * Mol. weight of “mer” Mol. weight of polymer = (500) (42 g/mol) = 21000 g/mol Problem 2 13.10 If the polymer evaluated in Problem 13.9 is polypropylene, what would be the a. coiled length and b. extended length of the average molecule? SOLUTION Ans 13.9: 0.01(250) + 0.1(350) + 0.21(450) + 0.22(550) + 0.18(650) + 0.12(750) + 0.07(850) + 0.05(950) + 0.02(1050) + 0.01(1150)+ 0.01(1250) = 612 ≈ 600 a. L = l √[2n] (Source: Shackelford, p. 471, eq. 13.4)
Background image of page 1

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

View Full Document Right Arrow Icon
= [134 pm] √[2(600)] (Source: Bond lengths and energies, C=C bond) = 4641.9 pm = 4.6419 * 19 -9 m b. L ext = ml sin (109.5/2) (Source: Shackelford, p. 472, eq. 13.5) = [2(600)] [154 pm] [sin(109.5/2)]
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}