ORGANIC AND PHYSICAL CHEMISTRY OF POLYMERS phần 2 ppt

... the enthalpy of mixing is established as follows: H mix = z  n 2 1 ε 11 2( n 1 +n 2 X) + n 2 2 X 2 ε 22 2( n 1 +n 2 X) + n 1 n 2 Xε 12 (n 1 +n 2 X)  −  n 1 ε 11 2 + n 2 Xε 22 2  or H mix = ... expressions delivers u = (2/ N a )  1 2 −χ 12  V 2 2 M 2 2 V 0 1 F(Y) (4.49a) and Y = [2/ N a ]  1 2 −χ 12  V 2 2 M 2 2 V 0 1  3 4πs 2  3 /2...

Ngày tải lên: 14/08/2014, 10:20

... expression and since K 1 and α 1 are known for the standard polymer 1 and K 2 and α 2 for the polymer analyzed, one can easily deduce M 2 with the following expression: ln M 2 = 1 1 +α 2 ln K 1 K 2 + 1 ... variation of g  with the number of branches (f ) have been proposed. The Fixman–Stockmayer model gives g  = f 2/ 3 [2 −f +2 1 /2 (f −1)] −3 and for the Zimm...

Ngày tải lên: 14/08/2014, 10:20

... HOOC–CH 2 –CH(NH 2 )–COOH Glutamic acid HOOC–CH 2 –CH 2 –CH(NH 2 )–COOH Alanine H 2 N–CH(CH 3 )–COOH Arginine H 2 N–C(NH)–NH–(CH 2 ) 3 –CH(NH 2 )–COOH Asparagine H 2 N–CO–CH 2 –CH(NH 2 )–COOH Cysteine ... HS–CH 2 –CH(NH 2 )–COOH Glutamine H 2 N–CO–CH 2 –CH 2 –CH(NH 2 )–COOH Glycine H 2 N–CH 2 –COOH Histidine Imidazolyl–CH(NH 2 )–COOH Isoleucine CH 3 –...

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... (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)...

Ngày tải lên: 14/08/2014, 10:20

... δ CH 3 = 0.80 ppm, δ CH 2 = 1 .20 –1.35 ppm, δ CH =1.6ppm. 3 J [CH–CH 3 ] = 6.5 Hz, 3 J [CH–CH 2 ] = 8 Hz, 2 J [CH 2 ] =−10 Hz 3 J [CH–CH 2 ] = 8 Hz, 2 J [CH 2 ] =−10 Hz. ppm CH 3 CH 2 CH 1.5 1.0 0.5 Figure ... transitions and relaxations of polymers, their mechanical properties and their rheology. These thirteen chapters are rounded off by monographs (chapters XIV to...

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... expressed as C 2 = N 2 M 2 /V 1 = N 2 M 2 /N 1 V 0 1 ∼ f 2 M 2 /V 0 1 where m 2 represents the mass of the solute, V 1 and V 2 are the volumes occupied by the solvent and the solute, N 1 and N 2 are the ... equal to −f 2 . Equation (6.3) can now be written as V 0 1  =−RTf 2 (6.4) Themolefraction(f 2 ) and the mass concentration of the solute (C 2 ) can...

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... 28 1 n CH 2 CH 2 ~~~~CH 2 CH 2 ~~~~CH 2 ~~~~CH 2 ~~~~CH 2 ~~~~CH 2 CH 2 CH 3 CH CH 2 CH 2 CH 2 CH 2 CH 2 ~~~~CH 2 CH 3 ~~~~CH 2 CH etc. •• • Transfer to polyethylene occurs because of the very high reactivity of the ... needs of the reaction medium, one can use either hydrophilic, hydrophobic, or mixed systems of initiation; for example, S 2 O 8 2...

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... coordination of the double bond ANIONIC POLYMERIZATION 327 and mainly covalent active species. + n H 3 C H 3 C H 2 C H 3 C H 2 C C CH 3 CH 3 H 3 C CH 3 CH 2 CH 2 CH 3 CH 3 CH 3 CH 3 CH 3 CH 3 CH 2 CH 2 CH 2 C CO COOR 1 COOR 1 COOR 2 COOR 1 COOR 2 OR 2 OR 2 OR 2 OR 2 OR 2 , C CO propagation C n C , C CO CC O ... n H 3 C H 3 C H 2 C H 3 C H 2 C C CH 3 CH 3 H...

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... + CH 2 Bu CH 2 Bu −− Li + Li + − Li + − Li + n CH CH CH 2 CH 2 Li n CH CH CH 2 CH 2 Li " ;2& quot; 2n + " ;2& quot; H 2 CCH 2 2n + " ;2& quot; O (i) (ii) H CH 2 H 2 C CH 2 CH 2 OH CH 2 CH 2 OH + Whatever ... initiator and the addition of ethylene oxide after cyclization permits the difunctionalization of the macrocyclics formed: CH CH CH 2 ][ C...

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... −λ −1 ( 12. 35) In such a context the second term of expression ( 12. 24) is equal to 0 and the first term (λ x 2 +λ y 2 +λ z 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 to ε = ε 2 −ε 2 s 11,0 e 1 ∆e 1 ∆e 2 e 2 e e 2- 1 s 11,1 es t 1 t 2 t 1 t 2 Figure 12. 8. Ill...

Ngày tải lên: 14/08/2014, 10:20