polymer chemistry 2nd edition hiemenz and lodge solutions manual pdf

1. for download complete version of solutions (all chapter 1 to 13 ) click here. SOLUTIONS MANUAL FOR Polymer Chemistry, Second Edition by Paul C. Hiemenz Timothy P. Lodge https://gioumeh.com/product/polymer-chemistry-2nd-edition-hiemenz-and-lodge-solutions-manual-pdf/ DK7140.indd 1 1/23/08 10:20:15 AM

2. for download complete version of solutions (all chapter 1 to 13 ) click here. https://gioumeh.com/product/polymer-chemistry-2nd-edition-hiemenz-and-lodge-solutions-manual-pdf/ DK7140.indd 2 11/6/07 1:54:13 PM

3. for download complete version of solutions (all chapter 1 to 13 ) click here. SOLUTIONS MANUAL FOR Polymer Chemistry, Second Edition by Paul C. Hiemenz Timothy P. Lodge https://gioumeh.com/product/polymer-chemistry-2nd-edition-hiemenz-and-lodge-solutions-manual-pdf/ DK7140.indd 3 1/23/08 10:20:15 AM

4. for download complete version of solutions (all chapter 1 to 13 ) click here. CRC Press Taylor & Francis Group 6000 Broken Sound Parkway NW, Suite 300 Boca Raton, FL 33487-2742 © 2007 by Taylor & Francis Group, LLC CRC Press is an imprint of Taylor & Francis Group, an Informa business No claim to original U.S. Government works Printed in the United States of America on acid-free paper 10 9 8 7 6 5 4 3 2 1 International Standard Book Number-13: 978-1-4200-7140-5 (Softcover) This book contains information obtained from authentic and highly regarded sources. Reprinted material is quoted with permission, and sources are indicated. A wide variety of references are listed. Reasonable efforts have been made to publish reliable data and information, but the author and the publisher cannot assume responsibility for the validity of all materials or for the consequences of their use. Except as permitted under U.S. Copyright Law, no part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information storage or retrieval system, without written permission from the publishers. For permission to photocopy or use material electronically from this work, please access www.copyright.com (http://www.copyright.com/) or contact the Copyright Clearance Center, Inc. (CCC) 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400. CCC is a not-for-profit organization that provides licenses and registration for a variety of users. For organizations that have been granted a photocopy license by the CCC, a separate system of payment has been arranged. Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation with- out intent to infringe. Visit the Taylor & Francis Web site at http://www.taylorandfrancis.com and the CRC Press Web site at http://www.crcpress.com https://gioumeh.com/product/polymer-chemistry-2nd-edition-hiemenz-and-lodge-solutions-manual-pdf/ T&F_LOC_C_Master.indd 1 DK7140.indd 4 11/6/07 12:49:26 PM 11/6/07 1:54:14 PM

5. for download complete version of solutions (all chapter 1 to 13 ) click here. Hiemenz and Lodge, Polymer Chemistry, 2nd Edition, Solutions Manual 1.1 B’s: 5 cis, 5 trans and 1 vinyl. The 5 I’s may all be shown as cis. B’s and I’s are in blocks; the isomers of B are random. The ratio of B:1 is 2:1; therefore, roughly 10 B’s to 5 I’s. To illustrate %’s of B, show 11 cis cis trans trans vinyl trans cis cis trans cis cis trans cis cis cis cis 1.2 Consider 100 repeat units: 47.2 + 44.9 = 92.1 units with 4 backbone C’s and 7.9 units with 2 backbone C’s: 92.1 units × 4 C’s/unit = 368 C’s 7.9 units × 2 C’s/unit = 16 C’s 368+16 ( ) C's/100 units 7.9 ethyl groups/100 units= 48.6C atoms ? ethyl 1.3 ? propylene glycol O O O O O O O O O ethylene glycol adipic acid O O O N O crosslink O N N O N O butanediol N O O O naphthalene 1 https://gioumeh.com/product/polymer-chemistry-2nd-edition-hiemenz-and-lodge-solutions-manual-pdf/

6. for download complete version of solutions (all chapter 1 to 13 ) click here. Hiemenz and Lodge, Polymer Chemistry, 2nd Edition, Solutions Manual 1.4 (a) polyethylene CH2 CH2n (b) polyoxymethylene O CH2n (c) poly(oxy-1,4-phenylene) O n (d) poly(vinyl butyral) n O O (e) poly(vinyl acetate) n O O (f) poly[(1-methoxycarbonyl)ethylene] n O O 1.5 (a) R = cyclohexanone for b = 4 and bicyclohexanone for b = 8. b = number of arms in star y = degree of polymerization of each linear chain (b) 1 – L = number of equivalents of reacted groups per mole monomer Q + L = total number of equivalents of “ends” per mole monomer ? ratio = number of reacted groups number of ends (c) For a molecule with b arms = y ? 2 https://gioumeh.com/product/polymer-chemistry-2nd-edition-hiemenz-and-lodge-solutions-manual-pdf/

7. for download complete version of solutions (all chapter 1 to 13 ) click here. Hiemenz and Lodge, Polymer Chemistry, 2nd Edition, Solutions Manual M n= bMn,arm+ Mcentral and Mn,arm= yMo=1?L Q+ LMo ? Mb=Mcentral b 1?L Q+ L+ Mb ? ?? ? ?? ? ? M n= b Mo ? 1.6 utilize the equations By dividing mi by Mi, we obtain the number of moles for each fraction, ni. We then ? ? i ? i ? i 2 i 3 i n M n M n M i i i i = = = M M M , and ? i ? i ? i n w z 2 i n n M n M i i i i to obtain Mn=29.1 kDa, Mw=46.1 kDa and Mz=62.5 kDa. 1.7 We consider ? ? i the ? equation for the viscosity average molecular weight: ? ? i 1 + ? 1 i n M ? ? i = M ? ? . ? n M ? ? ? ? i i Mw=671.2 kDa. So Mn<Mν,acetone< Mν,chloroform<Mw. If we consider the meaning of α, it quantifies the dependence of the intrinsic viscosity on molecular weight. The higher value means that the volume occupied by a chain, and consequently the solution viscosity, is more sensitive to the larger polymers in the distribution. We obtain Mν=661.7 kDa for acetone, Mν=665.1 kDa for chloroform and Mn=596.9 kDa, 1.8 Example solution (inspired by the characters of Marvel Comics): Character Mass (kg) wi wiMi Jean Gray 52 0.11 5.7 Rogue 54 0.11 6.2 Bruce Banner 58 0.12 7.1 Elektra 59 0.12 7.3 Spider-Man 75 0.16 11.8 Magneto 86 0.18 15.7 Wolverine 88 0.19 16.5 3 https://gioumeh.com/product/polymer-chemistry-2nd-edition-hiemenz-and-lodge-solutions-manual-pdf/

8. for download complete version of solutions (all chapter 1 to 13 ) click here. Hiemenz and Lodge, Polymer Chemistry, 2nd Edition, Solutions Manual For the listed characters, ? Mi Mn= 7= 67.5 kg ? Mw= wiMi= 70.6 kg PDI =Mw Mn=1.04 Based on PDI, the selected distribution is quite narrow. ? 1.9 We’ll present two approaches to solve this problem. First, ?? + ? n n M M 1 z ? + z 1 z i z m zm ? i ?? ? ?? ? = = M w m w exp i and . ( ) w i i i + z 1 To obtain an analytical expression for Mw that does not include a summation, we convert the sum to an integral according to the rule ' k ' k ? ? = k n = dn : k ? ? ?? ? + z 1 z 1 zm ( ) m ? 0 ? 0 ?? ? ?? ? + = = z 1 M w mdm m exp dm so . ( ) w + ? + z n 1 z 1 M M n With the aid of an integral table, we find that ( a ) ? ? ? + n 1 ( ) ? = n x exp ax dx . + n 1 0 Applying this to our expression of Mw, we get z M + ? ( M ) + + ? + z 1 1 z 2 = ( ) ( ) w + z 2 z n 1 z 1 M z n If z only assumes integer values, then Γ(z+1)=z! and Γ(z+2)=(z+1)!=(z+1)z!. Then the expression for Mw simplifies to 1 z M = + M w n z Another option is to consider moments of distributions, as presented in the book: ? () di i ? ? µ = = k k x i i P . k i i 0 Mw is proportional to the ratio of the second to the first moment of the distribution. 4 https://gioumeh.com/product/polymer-chemistry-2nd-edition-hiemenz-and-lodge-solutions-manual-pdf/

9. for download complete version of solutions (all chapter 1 to 13 ) click here. Hiemenz and Lodge, Polymer Chemistry, 2nd Edition, Solutions Manual ? ? ?? ? + z 1 z z m zm ( ) m ? 0 ? 0 ?? ? ?? ? µ = = = exp dm M w dm M . ( ) 1 n n ? + z n z 1 M M n Substituting this into the moment ratio for Mw, we get: ? 2 z M ?? ? + ? z 1 z 1 z m zm ? 0 ?? ? ?? ? m exp dm ( ) ? ? + ?? ? z n + µ z 1 1 M M z 1 zm ? 0 ?? ? ?? ? + = = = z 1 n m exp dm 2 . ( ) w + µ ? + z n 1 M z 1 M M 1 n n 1.10 Find the maximum of wi w.r.t. Mi: zz+1 ?(z+1) ? ? ? ? dwi dMi 1 z Mnz+1zMiz?1e?zMi/Mn? ? Mize?zMi/Mn = ? = 0 Mn Dividing through all the common terms we reach ? ? ? ? 1 Miz?1? ? Miz ? = 0 ? Mn and therefore at the maximum of wi Mi= Mn The half-width at half-height is about 3,500, so σ≈ 3000. From eqn 1.7.16 ? 1.11 ? 1/2 ? ? ? ? Mw Mn ? = 48,916 1.008?1 ( )1/2= 4,400 ? = Mn ?1 ? This is in the ballpark, but not exact. ? 1.12 (a) poly(methacrylic acid) (T) or poly[1-(carboxy)-1-methylethylene] (I) Me OH n Me n H2C O O OH A chain growth process (free radical) through the carbon-carbon double bond; initiation and termination steps are not specified). (b) poly(tetramethylene adipamide) (T) or poly(iminoadipoyliminobutane-1,4-diyl) (I) O O O H N H2N NH2 4 + n + n 2n HCl N H Cl Cl n 4 O A step growth (polycondensation) process bewteen a diamine and a diacid chloride. 5 https://gioumeh.com/product/polymer-chemistry-2nd-edition-hiemenz-and-lodge-solutions-manual-pdf/

10. for download complete version of solutions (all chapter 1 to 13 ) click here. Hiemenz and Lodge, Polymer Chemistry, 2nd Edition, Solutions Manual (c) poly(6-hydroxyhexanoic acid) (T) or poly(oxycarbonyl pentamethylene) (I) O O O + n H2O n HO n OH 5 A step growth (self-condensation) of an α-hydroxy, ω-carboxy alkane. (d) same as (c) by mistake! It was supposed to be poly(acrylonitrile) (T) or poly(1-cyanoethylene) (I) N n H2C n CN A chain growth process (free radical) through the carbon-carbon double bond; initiation and termination steps are not specified). (e) 1,4-phenyleneiminocarbonyl) (I) O C N poly(tetramethylene phenylurethane) (T) or poly(oxytetramethyleneoxycarbonylimino- O n + n HO OH 4 HN NH O O n N C O O A step growth polymerization (strictly, not a condensation) between a diisocyanate and a diol. Note that in reality the NH groups can react with the NCO groups to produce some crosslinking. (f) poly(alanine) (T) or poly[imino(1-methyl-2-oxo-ethylene)] (I) Me OH n N O H + n H2O H2N n O Me A step growth (self-condensation) of an amino acid. 1.13 The molecular ion will include one silver ion (108 g/mol), leaving 1098 for the polymer. As Mo = 104, it is tempting to infer a degree of polymerization of 10, leaving 58 for the mass of the endgroups. This corresponds exactly to butyl group and a hydrogen atom, suggesting an anionic polymerization. configuration of the side groups. There are two possibilities for each repeat unit, leading to 210=1024 possible structures. Note that the two ends of the chain are different, so we do not need to worry about overcounting. Imagine this polystyrene in the all trans backbone conformation. We are interested in the 1.14 We can write the molecular structure as 6 https://gioumeh.com/product/polymer-chemistry-2nd-edition-hiemenz-and-lodge-solutions-manual-pdf/

11. for download complete version of solutions (all chapter 1 to 13 ) click here. Hiemenz and Lodge, Polymer Chemistry, 2nd Edition, Solutions Manual H2 COCH3 H3COC H2 There are 6 methyl protons from the endgroups, and 4 methylene protons per repeat unit. As there are 20 times as many methylene protons, that means there are 120 methylene protons per molecule, or 30 repeat units. The number average molecular weight is therefore Mn= 30 12+12+16+ 4 ( )+12+12+16+ 6 =1366 1.15 The mole fractions can be obtained from the sample masses and Mn values: 10/70,000 10/70,000+ 20/20,000=1 Therefore for the mixture Mn=70,000 8 8 The weight fractions are easily seen to be 1/3 and 2/3, giving Mw=1 3?60,000 = 73,333 ? 8; x2=7 x1= 8 +7?20,000 = 26,250 ? 3?100,000+2 ? 1.16 respectively. For an equimolar mixture, xi = 0.5, and therefore Mn=1 2198+142 The weight fractions are given by the proportion of mass, and therefore 198 198+142?198+ The polydispersity is 1.027. Tetradecane (C14H30) and decane (C10H22) have molecular weights of 198 and 142, ? ( )=170 142 Mw= 198+142?142 =174.6 ? ? 7 https://gioumeh.com/product/polymer-chemistry-2nd-edition-hiemenz-and-lodge-solutions-manual-pdf/