ORGANIC AND PHYSICAL CHEMISTRY OF POLYMERS phần 2 ppt

... the enthalpy of mixing is established as follows:Hmix= zn 2 1ε11 2( n1+n 2 X)+n 2 2X 2 ε 22 2( n1+n 2 X)+n1n 2 Xε 12 (n1+n 2 X)−n1ε11 2 +n 2 Xε 22 2 orHmix= ... expressions deliversu = (2/ Na)1 2 −χ 12 V 2 2M 2 2V01F(Y) (4.49a) and Y = [2/ Na]1 2 −χ 12 V 2 2M 2 2V0134πs 2 3 /2 (4.51a)where V10is the molar volume of the solvent in ... +un=un−u 2 2n 2 +u33n3−··· (5 .25 )which givesln w(n, u) = n ln 2 +n +1 2 ln n −1 2 ln 2 −1 2 (n +u +1)ln n −ln 2 +un−u 2 2n 2 −1 2 (n −u +1)ln n −ln 2 +un−u 2 2n 2 (5 .26 )Finally,ln...

ORGANIC AND PHYSICAL CHEMISTRY OF POLYMERS phần 4 pptx

... expression and since K1 and α1are knownfor the standard polymer 1 and K 2 and α 2 for the polymer analyzed, one can easilydeduce M 2 with the following expression:ln M 2 =11 +α 2 lnK1K 2 +1 ... variation of gwith the number of branches (f )have been proposed. The Fixman–Stockmayer model givesg= f 2/ 3 [2 −f +2 1 /2 (f −1)]−3 and for the Zimm and Kilb model we haveg= (2/ f )3 /2 [0.39(f ... is written as 2 2 ˜n 2 λ40Na(d ˜ndc) 2 In X-ray scattering, the scattering cross section corresponds todσd= r 2 e(Z) 2 204 DETERMINATION OF MOLAR MASSES AND STUDY OF CONFORMATIONS...

ORGANIC AND PHYSICAL CHEMISTRY OF POLYMERS phần 9 ppt

... HOOC–CH 2 –CH(NH 2 )–COOHGlutamic acid HOOC–CH 2 –CH 2 –CH(NH 2 )–COOHAlanine H 2 N–CH(CH3)–COOHArginine H 2 N–C(NH)–NH–(CH 2 )3–CH(NH 2 )–COOHAsparagine H 2 N–CO–CH 2 –CH(NH 2 )–COOHCysteine ... HS–CH 2 –CH(NH 2 )–COOHGlutamine H 2 N–CO–CH 2 –CH 2 –CH(NH 2 )–COOHGlycine H 2 N–CH 2 –COOHHistidine Imidazolyl–CH(NH 2 )–COOHIsoleucine CH3–CH 2 –CH(CH3)–CH(NH 2 )–COOHLeucine (CH3) 2 CH–CH 2 –CH(NH 2 )–COOHLysine ... (CH3) 2 CH–CH 2 –CH(NH 2 )–COOHLysine NH 2 –(CH 2 )4–CH(NH 2 )–COOHMethionine CH3–S–(CH 2 ) 2 –CH(NH 2 )–COOHPhenylalanine C6H5–CH 2 –CH(NH 2 )–COOHProline Pyrrolidyl–COOHSerine HO–CH 2 –CH(NH 2 )–COOHThreonine...

ORGANIC AND PHYSICAL CHEMISTRY OF POLYMERS phần 10 ppt

... (of polymerizations): 24 9, 25 0, 25 6 25 8, 26 0, 26 2, 26 3, 26 5, 26 6, 26 8 27 4, 27 8 28 0, 28 2 28 6, 28 8 29 4, 3 02 307,309–317, 322 – 327 , 330–337,340–345, 348–350, 379, 381–386,389–3 92, 397, 521 , 523 , ... 457–460, 464.Dyad: 22 , 26 , 29 , 34, 35, 38, 1 12, 131, 25 5, 25 6, 27 5, 29 5, 29 6, 359, 361,3 62, 524 , 526 , 546, 548.Efﬁciency (of initiators): 25 7, 26 3, 27 1 27 4, 27 9, 28 2, 28 5, 28 8, 324 ,330, 340, 347, ... of PA producedby DuPont de Nemours Company.INDEXAbrasion: 4 82, 517, 674.ABS: 134, 624 , 6 32, 661.Absolute modulus: 429 , 430.Abstraction (reaction): 26 1, 26 2, 26 9, 27 2, 27 9, 28 2, 29 2, 322 .Acetals...

ORGANIC AND PHYSICAL CHEMISTRY OF POLYMERS phần 1 docx

... δCH3= 0.80 ppm, δCH 2 = 1 .20 –1.35 ppm, δCH=1.6ppm.3J[CH–CH3]=6.5 Hz,3J[CH–CH 2 ]= 8 Hz, 2 J[CH 2 ]=−10 Hz3J[CH–CH 2 ]= 8 Hz, 2 J[CH 2 ]=−10 Hz.ppmCH3CH 2 CH1.5 1.0 0.5Figure ... transitions and relaxations of polymers, their mechanical properties and their rheology. These thirteen chapters are rounded off by monographs (chaptersXIV to XVI) of natural polymers and of some ... between dipoles, R and T being the gas constant and absolute Organic and Physical Chemistry of Polymers, by Yves Gnanou and Michel FontanilleCopyright  20 08 John Wiley & Sons, Inc.134 INTRODUCTIONTable...

ORGANIC AND PHYSICAL CHEMISTRY OF POLYMERS phần 3 ppsx

... expressed asC 2 = N 2 M 2 /V1= N 2 M 2 /N1V01∼ f 2 M 2 /V01where m 2 represents the mass of the solute, V1 and V 2 are the volumes occupiedby the solvent and the solute, N1 and N 2 are the ... equal to −f 2 . Equation (6.3)can now be written asV01 =−RTf 2 (6.4)Themolefraction(f 2 ) and the mass concentration of the solute (C 2 ) can be related:C 2 = m 2 /(V1+V 2 ) ∼ m 2 /V1which ... mass, one obtainsfor P(θ)P(θ) = 1 −q 2 s 2 3+···corresponding toP(θ) = 1 −16π 2 3λ 2 s 2 sin 2 θ 2 which gives for P−1(θ)P−1(θ) = 1 + q 2 s 2 3after observing that11−y∼=(1...

ORGANIC AND PHYSICAL CHEMISTRY OF POLYMERS phần 5 potx

... 28 1nCH 2 CH 2 ~~~~CH 2 CH 2 ~~~~CH 2 ~~~~CH 2 ~~~~CH 2 ~~~~CH 2 CH 2 CH3CHCH 2 CH 2 CH 2 CH 2 CH 2 ~~~~CH 2 CH3~~~~CH 2 CHetc.•••Transfer to polyethylene occurs because of the very high reactivity of the ... needs of the reaction medium, one can use eitherhydrophilic, hydrophobic, or mixed systems of initiation; for example,S 2 O8 2 +S 2 O3 2 −→ SO4 2 +SO4−•+S 2 O3−•Fe 2+ +H 2 O 2 −→ ... through the use of thiols(mercaptans):ktrC 12 H 25 S•~~CH 2 -CH=CH-CH 2 •+C 12 H 25 SH~~~CH 2 -CH=CH-CH3+This family of compounds, as well as sulﬁdes, disulﬁdes, and alkyl halides...

ORGANIC AND PHYSICAL CHEMISTRY OF POLYMERS phần 6 doc

... coordination of the double bondANIONIC POLYMERIZATION 327 and mainly covalent active species.+ nH3CH3CH 2 CH3CH 2 CCCH3CH3H3CCH3CH 2 CH 2 CH3CH3CH3CH3CH3CH3CH 2 CH 2 CH 2 CCOCOOR1COOR1COOR 2 COOR1COOR 2 OR 2 OR 2 OR 2 OR 2 OR 2 ,CCOpropagationCnC,CCOCCO ... nH3CH3CH 2 CH3CH 2 CCCH3CH3H3CCH3CH 2 CH 2 CH3CH3CH3CH3CH3CH3CH 2 CH 2 CH 2 CCOCOOR1COOR1COOR 2 COOR1COOR 2 OR 2 OR 2 OR 2 OR 2 OR 2 ,CCOpropagationCnC,CCOCCO SiMe3~~~or TMSDormant speciesCnCCC OSiMe3+NBu4+NBu4−In the case of ... strength of various heterocycles withvariable number of linksNumber of linksHeterocycleCH 2 O SNOO3 115 117 77.8 96.1 —4 109 — 79.0 — —5 25 .5 28 .0 4.1 — 30.56 0.4 — 9 .2 — 12. 17 25 .5 —...

ORGANIC AND PHYSICAL CHEMISTRY OF POLYMERS phần 7 docx

... +CH 2 BuCH 2 Bu−−Li+Li+−Li+−Li+nCH CH CH 2 CH 2 LinCH CH CH 2 CH 2 Li" ;2& quot;2n + " ;2& quot;H 2 CCH 2 2n + " ;2& quot;O(i)(ii) HCH 2 H 2 CCH 2 CH 2 OHCH 2 CH 2 OH+Whatever ... initiator and the addition of ethylene oxide after cyclization permitsthe difunctionalization of the macrocyclics formed:CH CH CH 2 ][ CH 2 CH CH CH 2 ][ CH 2 1 2 n butadiene1CH 2 CH 2 2 BuLi ... (10%)CO–O(CH 2 )5nBr–(CH 2 ) 12 –O–AlEt 2 Br–(CH 2 ) 12 –O–[CO–(CH 2 )5–O]n–HH 2 N–(CH 2 ) 12 –O–[CO–(CH 2 )5–O]n–HWhatever the nature of the end functional groups, these α-orα,ω-difunctionalized polymers are difﬁcult...

ORGANIC AND PHYSICAL CHEMISTRY OF POLYMERS phần 8 ppsx

... −λ−1( 12. 35)In such a context the second term of expression ( 12. 24) is equal to 0 and the ﬁrstterm (λx 2 +λy 2 +λz 2 −3) becomes(λ 2 +λ 2 2) = (λ −λ−1) 2 ( 12. 36)Thus the variation of the ... to ε 2 at t 2 and the difference ε 2 then corresponds toε = ε 2 −ε 2 s11,0e1∆e1∆e 2 e 2 ee 2- 1s11,1est1t 2 t1t 2 Figure 12. 8. Illustration of the Boltzmann superposition principle ... Variation of η against time at constant value of ˙γ.hR1R 2 RRR0wwR1R 2 wr0CL2RR0aC-PRR0g gR1R 2 gr0Figure 13.4. Rate proﬁle of the melt (ω) and proﬁle of the rate of shearing...

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