Спутник технологический. KJ 1B. [Редактировать]

KJ 1 (Kongjian 1) это малый космический аппарат который был создан для целей орбитальной отработки технологий . Более подробные детали об аппарате не раскрывались.

Дополнительная классификация

#Наименования
1Тип оператора(владельца) - государственный
2Страна оператор(владелец) - Китай
3Страна производитель - Китай
4Тип орбиты - НОО

Найдено 1000 документов по запросу «KJ 1B». [Перейти к поиску]


Дата загрузки: 2016-12-24
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0.05/5
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:, #,% #)T0, &O$ 0/0401#, 10&UQ...#K,(%$.)%.$#, U*@10(60/*6 1*3:, )&%&$#K, '4* K0%, 1#,B&$8*$&'#1*0, .90%1&6#1#4*%*90()&+,*1B&$8#3**, &,@#%$#%#Q, )&8O#1*+, (&%&'&+, ('K@*A D((40/&'#12, &(&U011&(%*, *Q, /0K..., g[6 MmcglmpD• l,(%#%50,$ #((8&%$012, 7%#O2, *, 1#O$ #'401*K, *1B&$ 8#%*@#3** , (&3*#451&67)&1&8* 90( )&-&,*, *(%&$*90()&-& &U$#@&'#1*K:,O$#4*@*$&'#1, &O2...



Дата загрузки: 2017-03-18
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0.07/5
... ΔE ~ ΔH, effect of PΔV usually small ΔE = -2045 kJ ΔH = - 2043 kJ PΔV = +2 kJ C3H8(g) + 5 O2(g) –> 3 CO2(g) + 4 H2O... C3H8(g) + 5 O2(g) –> 3 CO2(g) + 4 H2O(l) ΔH = - 2219 kJ (End Test 2) Effect of quantity (coefficients) is significant ΔH = - 4438 kJ 2 C3H8(g) + 10 O2(g) –> 6 CO2(g) + 8 H2O...(g) –> 2.25 CO2(g) + 3 H2O(l ) ΔH = - 2219 kJ] ΔH = - 1664 kJ Ch 8-CH1100-Fall 2005 Start... Eqns 2 NH3(g) ΔH˚ = + 92.38 kJ ΔH˚ = – 92.38 kJ N2(g) + 3 H2(g) Enthalpy ΔH magnitude... through). NH3(g) ==> 1/2 N2(g) + 3/2 H2(g) ΔH˚ = + 46.2 kJ If you want the reverse.... 4 x [ 1/2N2(g) + 3/2 H2(g) ==> 1 NH3(g) – 46.2 kJ ] ΔH˚ = -184.8 kJ Ch 8-CH1100-Fall 2005 12... ΔHsub Ex: If ΔHfus = 10 kJ/mol. How much heat to... 0.7 mol? Sol’n: (0.7 mol) x (10 kJ/mol) = 7 kJ ΔHfusion solid ΔHsub = ΔHfusion + ΔHvap Ex: ΔH sub = 83 kJ/mol ΔHvap = 53 kJ/mol. What is ΔH to... of 100 g/mol? Ans:15 kJ Ch 8-CH1100-Fall 2005 13... ˚C) q = (350 g) (4.184 j/g˚C)(–23 ˚C) q = – 33681.2 J = – 33.7 kJ Ch 8-CH1100-Fall 2005 16... q p = ΔE + PΔV = ΔH (for liqs ΔV ~ 0) q rxn = ΔH = – 2790.2 J ~ – 2.79 kJ Ch 8-CH1100-Fall 2005 18... shown: 1 N2(g)+ 3 H2(g) => 2 NH3(g) ΔH˚ = - 92 kJ Use above rxn to find... NH3(g) : N2(g)+ 3H2(g) => 2 NH3(g) ΔH˚= - 92 kJ) ΔH˚ = - 92 kJ 2 NH3(g) => 1 N2(g)+ 3 H2(g) ΔH˚= +92...(g)+ 3 H2(g) ΔH˚= +92 kJ] 5 NH3(g) => 2.5 N2(g)+ 7.5 H2(g) ΔH˚ = - 92 kJ) ΔH˚ = -184 kJ (reverse) (multiply to get correct coef.) (5/2 x 92 kJ) = ΔH˚= +230 kJ Ch 8-CH1100-Fall 2005 21... C3H8(g) + 5 O2(g) => 3 CO2(g) + 4 H2O(g) ΔH˚ = -2219 kJ b) Draw PE diagram of that... tell you conversion factor of kJ ΔH rxn for the # mol in... you know: - 2219 kJ/ 1 mol C3H8 . Go to kJ for mols you... mol C3H8 x (- 2219 kJ/ 1 mol C3H8) = 777 kJ Ch 8-CH1100-Fall 2005...)+ 2 NH3(aq) + 10 H2O(l) ΔH˚ = +80.3 kJ Ch 8-CH1100-Fall 2005 8.7 23...: 1 C3H8(g) + 5 O2(g) => 3 CO2(g) + 4 H2O(g) ΔH˚ = - 2219 kJ Solution: Convert g to moles: (15... x (– 2219 kJ/1mol C3H8) => ΔH˚= - 776.7 kJ 1b) 0.35 C3H8(g)+ 1.75 O2(g) ΔH˚= - 776.7 kJ 1.05... rxn. 3 H2(g) + N2(g) ==> 2 NH3(g) ΔH˚ = -92.2 kJ Intermediates H2 + N2H4 Reactants 3H2 + N2 Step 1 ΔH˚ = +95.4 kJ Step 2 ΔH˚ = -187.6 kJ Final ΔH ˚ = -92.2 kJ Product 2 NH3 Hess...(g) --> CO2(g) ΔH = - 393.5 kJ • CO2(g) --> C(s) + O2(g) ΔH = + 393.5 kJ • 2 CO2(g) --> 2 C(s) + 2 O2(g) ΔH = +787 kJ • This MUST...(l) ΔH˚ = ? CH4(g) + O2(g) ==> CH2O(g) + H2O(g) ΔH˚ = - 284 kJ CH2O(g) + O2 (g) ==> CO2(g) + H2O(g) ΔH˚ = - 518 kJ H2O(l) ==> H2O(g) ΔH˚ = + 44 kJ Ch 8-CH1100-Fall 2005... (g) ==> CO2(g) + H2O(g) ΔH˚ = - 518 kJ 2 [ H2O(g) ==> H2O(l) ΔH˚ = + 44 kJ ] ΔH˚ = - 88 kJ CH4(g) + 2 O2(g) ==> CO2(g) + 2 H2O(l) ΔH˚ = - 890 kJ Ch 8-CH1100... data to find ΔH˚rxn (in kJ) for: C2H4(g) + 3 O2(g) ==> 2 CO2(g) + 2 H2O(g) Substance ΔHf˚ (kJ/mol) H2O(l) H2O(g) CO2(g) - 285... ΔG = ΔH - TΔS For a given reaction ΔH is +45 kJ and ΔS is +199 J/K. At what...? (226.1 K) A reaction has ΔG of - 55 kJ and ΔS of +145J/K at 25... is non-spontaneous? Answers: a) - 11.8 kJ b) Y c) Y d) Y Ch 8-CH1100-Fall 2005 e) None...



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0.04/5
... EH EN F FS G H HM HZ J KJ KW LB MB MS PF...). Each HF frequency is given a KJ designator, and these run from... EH EN F FS G H HM HZ J KJ KW LB MB MS PF...). Each HF frequency is given a KJ designator, and these run from..., 542, 824, 818 KA KG KJ KR KW - unknown 408, 390... contact with Colombo ACC (SEA-1B), USB, 1538 UTC, Pos. Rep... EH EN F FS G H HM HZ J KJ KW LB MB MS PF...). Each HF frequency is given a KJ designator, and these run from... EH EN F FS G H HM HZ J KJ KW LB MB MS PF...). Each HF frequency is given a KJ designator, and these run from... EH EN F FS G H HM HZ J KJ KW LB MB MS PF...). Each HF frequency is given a KJ designator, and these run from... EH EN F FS G H HM HZ J KJ KW LB MB MS PF...). Each HF frequency is given a KJ designator, and these run from... EH EN F FS G H HM HZ J KJ KW LB MB MS PF...). Each HF frequency is given a KJ designator, and these run from... with Kuala Lumpur ACC (SEA-1B), USB, 2249 UTC, Pos. Rep....0 05708.0 no WUN-v09 (SEA-1B), USB, 2308 UTC, Pos. Rep... contact with Colombo ACC (SEA-1B), 1900 UTC, Pos. Rep. and... contact with Colombo ACC (SEA-1B), 1900 UTC, Pos. Rep. and... EH EN F FS G H HM HZ J KJ KW LB MB MS PF...). Each HF frequency is given a KJ designator, and these run from... with Kuala Lumpur ACC(SEA-1B), USB, 2308 UTC, Pos. Rep... with Kuala Lumpur ACC(SEA-1B), USB, 2249 UTC, Pos. Rep... EH EN F FS G H HM HZ J KJ KW LB MB MS PF...). Each HF frequency is given a KJ designator, and these run from... with Kuala Lumpur ATC (SEA-1B), USB, 2141 UTC, Pos. Rep... with Kuala Lumpur ACC (SEA-1B), USB, 2127 UTC, Pos. Rep...) Hawk 75: 0021Z Hawk 75 [B-1B] call Andrews PP Dyess AFB...



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0.57/5
... inner amine nitrogen atoms, Figureshow 1b, tetradentate binding mode mode through... atoms,atoms, FigureFigure 1c, or 1b, tridentate binding mode mode with... 1, 2 and 3. 1 Hg–N(2) Hg–N(2A) Hg–N(1B) Hg–N(1C) 2.075(5) 2.075(5) 2.690(6) 2.690(6) N(2)–Hg–N(2A) N(2)–Hg–N(1B) N(2A)–Hg–N(1B) N(2)–Hg–N(1C) N(2A)–Hg–N(1C) N(1B)–Hg–N(1C) 180.0(2) 94.3(2) 85.7(2) ... was found to be 7.75 KJ/mol for 1’ at 0.0013 mmol..., 32.41, and 30.50 KJ/mol at near zero CO2... of liquefaction of CO2 (17 kJ/mol) but lower than the...-5 at zero coverage (ca. 50 kJ/mol) [23–25], indicating relatively... steadily above 4.35 KJ/mol for 1’ and 8.46 KJ/mol for 4’, respectively...



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0.13/5
... of' 6.3.10. (cf. Arabic bint 1B ) followed by a matronym. t ( It is... - 24 red; KMAH 1a KMAH 1b KMAH 1c KMAH Id KMAH... . DIll 37 1O(E 31 ~a=+&II-E)KJ DIB 36 DIe 57 H ... {a} ...... v+6·1I... hll'ln--- KMEH 1e a 1. 0 KMAH 1b {n} - - OG {n} - - - - - - T to} 0 x {n} - - - - - - {~} 1 KMHH 43c OGo1xa8xAOnX... Ah'1aMh- - {h} I 0 {a} - - h X 0 DII X a {DII} - )} {M} {h} - {h} KMAH 1b hi h KMEH 3a O} - - a f J.. 1 - {q} {h} {l} x - It I h {x} {~} X I 0 n - - - {h} - - DII... KHHH 36a r} {m} in} e8 KHHH 1b 0 i ~ 1'1 I in} 0 n {r} to} {e} 1'1 r ~ 0 f {I} r in} X e 1'1 S {I} n - Yi {OJ... KMHH 26a KMEH 4d KMEH 1b KMHH 12 KMEH 1c KMEH... Sb 20 KMHH 19c KMAH 1b a 1. 0 m n - - - {n} - - A} is} n - - 1. KMHH lSd KMHH 33d... IXI} Xn ~ KMHH 47b e! KMAH 1b rh e h hoe {rh} {X} {A} KMHH 24c...} {e} Y} e KMBH 7 KMHH 30b mHve KMEH 1b e Xh {m} X KMHH 30e m} - - n A {m} {Y} Y {A} {m} KMGH 10... ! I n {R} H I {H} m ~ I ! - n h X1 ~ ! KMHH 19a n - - - {h} - - IXI KMAH 1b mA I ! m 0 m m KMGH 10 X t to} + X KMHH... KMHH 27c KMHH Sb KMAH 1b in} A ~ ~ to} +} {h} w f to} 0 X in} - - - - - w {I} m I w 0 to... I KMHH 19b Y in} n0 KMEH 1b 0 '" {e} in} 0 l' A w to} KMHH 35c g - 0 M I 0 KMHH... 6 KMHH 33a n - - eASe!ln- KMAH 1b KMHH 45 8} II (I'l) I A KMAH 1e... n h Script 2 mtI a X0 mtI KMHH 1b {1} r W 0 1 {I} ~ m - m - {A} ~ {n} I ml't-n m1 mto} A W A I {w} m~ W {I} mH Xn... 0 mtI ~ r w 0 KMGH 6 KMHH 33b ~ I w} 1 {e} KMHH 1b I} {X} 1 w X! rl ~ Y I I {N} 0 KMHH 35d {R} o} 1 !h KMHH 40b... {n} KMHH 27a M' {M} 1 {a} {g} } h M) g I X a g M mh {I} - n M g - - {I} gOA I KMHH 1b KMHH 23d og Me8Ah{t}Mhh8g... KMAH 1c KMEH 1b mA KMHH 35a KMEH 1b - - I) r'I {It} rh No...~110A w n S {I} {r'I} No (~) } • {n} KMAH 1c - - 0 n KMEH 1b - a Y {m} {e} x l r KMHH1a q} {J} {I} w a {n} I KMGH 8 9 - 0 ~ KMGH 3f 1)KMAH...} X e 1'1 S {I} n - M , 0 I h {X} S A ~ A {S} c ~ , {M} Y {a} {X} J h d h e - I A a A f I {n} {I} M {~} {~} KMHH 34d m A X f {oJ + + , f n X0 m ~ ~ I s KMHH 1b 0 ~ f ~ R f J}nnh I Aa~ '1 oA ~ III KMHH... KHHH 33e KMAH 1a KHHH 1b Y{I} 17 ~ {h} d n {X} it} n n {J} J 8 J x r in} J x0 ~ r {h} I rn... KMHH 4 h {T} {}} {+} {a} {h} It ~ KMHH 11a KMEH 1b l - n It 0 ~ - It o {h} X e 1'1 1'1 o '1 0 n ~ 1 e {1} X 1 {e} D ~ KMHH 5 0 Y} I It h - 1 {~} till... S KMHH 11b KMHH 10 KMAH 1b KMGH 2 w---wa80 X KMHH 27b KMHH...> KMEH 0 KMHH 17 \ ~ KMEH KMEH 1b 6b KMEH I i ~ I 14c 8 I 't " KMEH \'7"" KMDH... ;f e"71 M I 0~ a SITE Dl 9 76 I~ J.0, Ff ~6 ~+ A 't. kJ ~-,. 80 79 I n rt I F\" cr .L ~::~ ~ 0 25...



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... die Entnahmen sein: • im Zeitpunkt t = 0: J – ∑g ⋅ x Kj j=1 1 4243 Kj0 Einzahlungsüberschüsse... interpretiert werden. in den Zeitpunkten t = 1, 2, …, T: J – ∑g ⋅ x Kj j=1 14243 Kjt Einzahlungsüberschüsse... Kapazitätsobergrenze (für j =1, 2, …. J) nicht verletzen: x Kj ≤ x max . Kj Ist eine Kapitalanlage- oder Kapitalaufnahmem... 33 , : : EN Ba → K ! EN Ba K , : (1) : : • t=0 : J – ∑g ⋅ x Kj j=1 1 4243 Kj0 + w K 0 ⋅ EN Ba ≤ K 14243...- und Finanzierungsobjekten w K 0 ⋅ EN Ba K t=0 t=0 • t = 1,2,...T : J – ∑g bK0 ⋅ x Kj j=1 1 4243 Kjt + w K t ⋅ EN Ba ≤ K 14243... Ba K bKt , (2) bKt = 0 : xKj x Kj ≤ x max Kj j (j =1, 2, …. J): 34 Bereich Wirtschaftswissenschaften (3) Nichtnegativität: Die... Entnahmestrom sollen nicht negativ werden: x Kj ≥ 0 EN K ≥ 0. Das Ergebnis dieses Ansatzes... noch nicht bekannten Preises P: J – ∑ g Kj0 ⋅ x Kj + P + w K 0 ⋅ EN Be ≤ bK 0 . K j=1 • in den...ücksichtigung der Unternehmenszahlungen gUKt: J – ∑ g Kjt ⋅ x Kj + w Kt ⋅ EN Be ≤ b Kt + g UKt... auszuhandelnde Preis P anfallen. Diskussionspapiere (3) 35 : : x Kj ≥ 0 EN K ≥ 0 max EN Ba K max... = (0; gUK1; gUK2; …; gUKT)30 : : P→ ! ( (1) ) : : : • t=0 P: J – ∑ g Kj0 ⋅ x Kj + P + w K 0 ⋅ EN Be ≤ bK 0 K j=1 • gUKt: t = 1, 2, …, T J – ∑ g Kjt ⋅ x Kj + w Kt ⋅ EN Be ≤ b Kt + g UKt K j=1 ... Kapazitätsobergrenze (für j =1, 2, …. J) nicht verletzen: x Kj ≤ x max . Kj Ist eine Kapitalanlage- oder Kapitalaufnahmem...äufer (negativer Kaufpreis) ausgeschlossen:31 x Kj ≥ 0 P ≥ 0. Zur Veranschaulichung wird nun die... EN Be ≥ EN Ba K K (3) : (j =1, 2, …. J): x Kj ≤ x max Kj (4) : , ( ):31 x Kj ≥ 0 P≥0 ( ) (t = 4) ( KU t = 0 t=0 ( ( 31 ) t = 0 (IF) ) (AK... J (1a) – ∑ g Kj0 ⋅ x Kj + P + w K0 ⋅ EN Be ≤ b K0 K (für t = 0) j=1 J (1b) – ∑ g Kjt ⋅ x Kj + w Kt ⋅ EN Be... ≥ EN Ba K K (3) Kapazitätsrestriktionen x Kj ≤ x max Kj (4) Nichtnegativität (4a) x Kj ≥ 0 (4b) EN Be...! (1) J (1a) – ∑ g Kj0 ⋅ x Kj + P + w K0 ⋅ EN Be ≤ b K0 K ( t = 0) j=1 J (1b) – ∑ g Kjt ⋅ x Kj + w Kt ⋅ EN Be ≤ b Kt + g UKt ( t = 1, …, T) K j=1 (2) max EN Be ≥ EN Ba K K (3) x Kj ≤ x max Kj (j = 1, …, J) (4a) x Kj ≥ 0 (j = 1, …, J) (4b) EN Be ≥0 K (4c...ätsrestriktionen 64748 J ∑x ⋅ u j → min! 14243 max Kj j=1 bewertete Kapazität Restriktion der Zahlungen... EN K + ∑ x max ⋅ u j → min! ∑ bKt ⋅ d t + ∑ g UKt ⋅ d t − δ1⋅4 Kj 243 t =0 t =1 j=1 1424 3 14243 valued withdrawal... ⋅ d t + ∑ g UKt ⋅ d t + ∑ x max ⋅ u j – EN Ba ⋅ ∑ w Kt ⋅ d t . Kj K In der optimalen Lösung des...ß diese einen nicht-negativen Kapitalwert C Kj ≥ 0 im Zeitpunkt t = 0 haben. Da C Be C Be Kj einen heutigen Geldbetrag verkörpert, folgt aus der Lenkpreistheorie Kj · d0 = uj Be und – wegen... die Identität von uj und C Kj . 38 Vgl. ROLLBERG, Unternehmensplanung (2001... ⋅ d t + ∑ g UKt ⋅ d t + ∑ x max ⋅ u j – EN Ba ⋅ ∑ w Kt ⋅ d t Kj K P = Pmax > 0 (1a) d0 = 1 Pmax ρ d0... Kt t=0 j (1) : T − ∑ g Kjt ⋅ d t + u j = 0 ⇔ t =0 T u j = ∑ g Kjt ⋅ d t t =0 C Be Kj ≥0 t=0 C Be Kj  · d0 = uj, 38 ROLLBERG, Unternehmensplanung...), 182-185 C Be Kj d0 = 1, 178 uj C Be Kj , HERING, Investitionstheorie (2003... ⋅ d t + ∑ g UKt ⋅ d t + ∑ x max ⋅ u j − EN Ba ⋅ ∑ w Kt ⋅ d t Kj K oder wegen dt d0 T =: ρBe... d 0 = 1 sowie C Be = ∑ g Kjt ⋅ ρBe Kt Kj Kt t =0 T Pmax = ∑ b Kt ⋅ ρ + ∑ g UKt ⋅ ρ + Be Kt t =0 t =0 T Be Kt t =1 ∑x C j >0 max Kj ⋅ C − EN Be Kj Ba max K T ⋅ ∑ w Kt ⋅ ρBe... bewertenden Unternehmens Be Kt max Kj Kapitalwert des sonstigen Bewertungsprogramms Be Kj programms Diese Formel besagt, daß ... Pmax = ∑ b Kt ⋅ d t + ∑ g UKt ⋅ d t + ∑ x max ⋅ ∑ w Kt ⋅ d t Kj ⋅ u j − EN K albo poprzez dt =:ρBe... ⋅ ρ + ∑ g UKt ⋅ ρ + t=0 Be Kt T Be CBe Kj = ∑ g Kjt ⋅ ρKt d 0 = 1 oraz Be Kt t =1 ∑x C j >0 max Kj t=0 ⋅ C − EN Be Kj T Ba max K ⋅ ∑ w Kt ⋅ ρBe... . ∑ g UKt ⋅ ρBeKt + ∑ bKt ⋅ ρBeKt + ∑ Kj Kj K Kt Be t =1 t =0 t =0 C Kj >0 14243 144442 24443 44443... − ∑ w Kt ⋅ EN Ba ⋅ ρBe . Kj K Kt t =1 t =0 C Kj >0 t =1 1442443 14444244443 144424443 n Pmax Zukunftserfolgswert... − ∑ w Kt ⋅ EN Ba ⋅ ρBe . Kj K Kt t =1 t =0 C Kj >0 t =1 1442443 14444244443 144424443 n Pmax 61...



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...{-) rnnjorf;:'J; !;;in Ol"n!)aroo, eJ ;;1b~,~coso cs rOCufrOnll=l, y ncllli::rllTll...



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....29 ϩ ͸ N ⌬hƒk ⌬G ƒЊ(298.15 K) ϭ 53.88 ϩ k (kJ molϪ1) (3-3.1) k (kJ molϪ1) (3-3.2) k ƒ k C pЊ(T ) ϭ ͭ͸ ͭ͸ ͭ͸ k k k ͮ ͭ͸ ͮ ͮ NkCpAk Ϫ 37.93....15 K) ϭ 53.88 ϩ k ƒk ϭ Ϫ25.73 kJ molϪ1 kϭ1 C Њp(700) ϭ ͭ͸ ͭ͸ ͭ͸ ͮ ͭ͸ NkCpAk Ϫ 37.93... Base Property ⌬G Њƒ(298.15 K) kJ molϪ1 ⌬H Њƒ(298.15 K) kJ molϪ1 C Њp(100 K) J mol....5981ͬ ␪ ) ϩ W ͸ M (C ) Ϫ 10.7983ͬ ␪ (3-4.1) (kJ molϪ1) (3-4.2) j pA2j j Nk(CpB1k k ϩ (kJ molϪ) j Nk(CpA1k....15 K) ϭ 10.835 ϩ Ϫ1 k kϭ1 ͸ N (gƒk) ϭ Ϫ23.597 kJ mol 5 ⌬G ƒЊ(298.15 K ) ϭ Ϫ14.83.... % Err. (2nd-Order) ⌬G ƒЊ(298.15 K), kJ molϪ1 Experimental Calculated (First Order... tested with the method Property ⌬G Њƒ kJ molϪ1 ⌬H Њƒ kJ molϪ1 C Њƒ(100 K) J molϪ1 KϪ1 C Њp(298... tested with the method Property ⌬G Њƒ kJ molϪ1 ⌬H Њƒ kJ molϪ1 C Њƒ(100 K) J molϪ1 KϪ1 C Њp(298... C Њp 500K C Њp 600K C Њp 800K C Њp 1000K C Њp 1500K kJ molϪ1 J molϪ1KϪ1 J molϪ1K... C Њp 500K C Њp 600K C Њp 800K C Њp 1000K C Њp 1500K kJ molϪ1 J molϪ1KϪ1 J molϪ1K... C Њp 500K C Њp 600K C Њp 800K C Њp 1000K C Њp 1500K kJ molϪ1 J molϪ1KϪ1 J molϪ1K... C Њp 500K C Њp 600K C Њp 800K C Њp 1000K C Њp 1500K kJ molϪ1 J molϪ1KϪ1 J molϪ1K... C Њp 500K C Њp 600K C Њp 800K C Њp 1000K C Њp 1500K kJ molϪ1 J molϪ1KϪ1 J molϪ1K... C Њp 500K C Њp 600K C Њp 800K C Њp 1000K C Њp 1500K kJ molϪ1 J molϪ1KϪ1 J molϪ1K... C Њp 500K C Њp 600K C Њp 800K C Њp 1000K C Њp 1500K kJ molϪ1 J molϪ1KϪ1 J molϪ1K... Joback % Error C/G % Error Benson C Њp(700K) kJ molϪ1 % Error Joback % Error C/G % Error....22 kJ mol 7 ⌬H Њƒ(298.15 K) ϭ k Ϫ1 ƒk kϭ1 ͸ N (S Њ) ϭ 0.397 kJ mol 7 S Њ(298.15 K) ϭ k Ϫ1 k KϪ1 kϭ1 ͸ N (S Њ) ϭ 0.801 kJ mol...) Ϫ S el Њ (298.15K)] ϭ Ϫ24.92 kJ molϪ1 ͸ N (C Њ ) ϭ 303.6 J mol 7 C Њp(800 K) ϭ k pk...) standard enthalpy of reaction at T, kJ molϪ1; Eqs. (3-1.2) and (3-1.4) standard (lower... at T and 1.01326 bar (1 atm.), kJ molϪ1 KϪ1; Eq. (3-1.8), Sec. 3-5 entropy contribution... of T and V Zϵ PV ϭ ƒP (T, P ) RT (4-2.1a) ϭ ƒV (T, V ) (4-2.1b) where V is the molar volume... Zϭ1ϩB ϭ1ϩ ͩ ͪ ͩ ͪ P P ϩ (C Ϫ B2) RT RT B C ϩ ϩ⅐⅐⅐ V V2 2 ϩ⅐⅐⅐ (4-5.1a) (4-5.1b) where the coefficients B, C, . . . are.... Equation (4-6.1) in the form Eq. (4-2.1b) is Zϭ V (⌰ / RT )V(V Ϫ ␩) Ϫ V Ϫ b (V Ϫ b)(V 2 ϩ ␦V ϩ ␧) (4-6.2) When it is... P Ϫ P0 ϭ ƒ (T, ␳, ␳0) (4-12.1a) ␳ Ϫ ␳0 ϭ ƒ (T, P, P0) (4-12.1b) where the reference density, ␳0, is..., J molϪ1 KϪ1 enthalpy change of boiling, kJ molϪ1 reaction equilibrium constant molecular... Zϭ1ϩB ϭ1ϩ ͩ ͪ ͩ ͪ P P ϩ (C Ϫ B2) RT RT 2 ϩ ... B C ϩ ϩ ... V V2 (5-4.1a) (5-4.1b) where the coefficients B, C, . . . are... ϭ 0.282 and ZRA2 ϭ 0.247. Equation (5-8.1b) then gives ZRAm ϭ 0.7 ϫ 0.282 ϩ 0.3 ϫ 0.247..., Eq. (5-8.3d ) weight fraction, Eq. (5-2.1b) mole fraction, Eq. (5.2-1a) compressibility.... (4-5.1). Zϭ1ϩB ϭ1ϩ ͩ ͪ ͩ ͪ P P ϩ (C Ϫ B2) RT RT 2 ϩ... B C ϩ ϩ... V V2 (4-5.1a) (4-5.1b) This is the only case... 2 RT B 2 ϩ... ϩ... P (C Ϫ B ) P Ϫ RT 2 RT ϪB P RT 2 2 2 Ϫ... 2 ϩ... Z (T, V ), Eq. (4-5.1b) ͩͪ 1ϪZ U ig Ϫ U RT 2 P P Ϫ (C Ϫ B2) RT RT... (5-4.2) Virial Equation ͫ͸ ln ␾i n (5-4.1a) & (5-4.2a) 2 jϭ1 (5-4.1b) & (5-4.2a,b) 2 V ͸yB n j jϭ1 ͬ yj Bij(T ) Ϫ B(T, {y}) ij... ij j j ͸ xk⌳ki N xj⌳ij ϩ 1 Ϫ k ͸ x⌳ ͸x␶ G ␶ Ϫ ͸G x N j kj j NRTL ͸␶G x g ϭ͸x RT ͸G x ͸␶G x xG ln ␥ ϭ ϩ͸ ͸G x ͸G x N N E ji... ͩ͸ ␪ ␶ ͪ N ln ␥i ϭ ln i ij kj k k ⌽i xi ϩ ΂ Ϫ j ji ⌽i k kj N kj k k i j ji j i j N j j j N j ij j N k kj k where ⌽i ϭ ri xi ͸rx N k k k and ␪i ϭ qi xi ͸qx N k k k i ΃ kj... group in molecule j ͸x␯ X ϭ ͸x ͸␯ j kj j (8-10.56) k j j kj k where ␯kj is the number of... 1 Ϫ ln (8-10.59) ͸xl ͩ͸ ͪ ͸ ͸ ͬ ␪j ␶ij ␪j ␶j i Ϫ j ␪k␶kj j k z li ϭ (ri Ϫ qi) Ϫ (ri Ϫ 1) z ϭ 10... 1 Ϫ ln ␪j␶ji Ϫ jϭ1 ͸ 3 jϭ1 ΂͸ ΃΅ ␪j␶ij (8-16.16) 3 ␪k␶kj kϭ1 where qi is the UNIQUAC... factor Greek ␣ ␣ij ␤ ␥i ⌫k ␦ ␦ ␧ ␰ ␩ ␪ ⍜i ␬ ␭ ␭ij ⌳ij ␯ (i) k ␯kj ␲* ␳ ␴2 ␶ ␶ij ␾i ⌽i ␹ ⌿mn ␺12 ␻ ⍀ parameter in...-Nies: Dechema Chemistry Series, 1: Part 1b, 108 (1974). Gmehling, J., and U. Onken... viscosity equation is: ͸ K ͩ1 ϩ 2 ͸ H n ␩m ϭ iϪ1 i iϭ1 ij ͸ ͸ n Kj ϩ jϭ1 n ͪ Hij Hik Kj Kk jϭ1 i kϭ1 i (9-5.1) where ␩m is the.... Data Ser., Vol. I, Parts 1 and 1b, DECHEMA, Frankfurt, 1977. Gordon, A. R.: J. Chem... (methyl amine) CAS # DelHf0, kJ / mol DelGf0, kJ / mol 7440-37-1 7726...-19-7 431-89-0 DelHf0, kJ / mol DelGf0, kJ / mol Ϫ264.40 Ϫ83....31 13838-16-9 DelHb, kJ / mol DelHm, kJ / mol 14.70 38...-2-pentene 2-methyl-2-butene DelHf0, kJ / mol DelGf0, kJ / mol 554-12-1 141....08 106.90 DelHb, kJ / mol DelHm, kJ / mol 32.15 35...) CAS # DelHf0, kJ / mol DelGf0, kJ / mol DelHb, kJ / mol DelHm, kJ / mol 1.88... 2,2,3,4-tetramethylpentane 2,2,4,4-tetramethylpentane 2,3,3,4-tetramethylpentane CAS # DelHf0, kJ / mol 540-84-1 Ϫ224.01....31 Ϫ116.59 DelHb, kJ / mol DelHm, kJ / mol 30.79 32... CAS # DelHf0, kJ / mol DelGf0, kJ / mol DelHb, kJ / mol DelHm, kJ / mol 138...-17-8 7440-63-3 DelHf0, kJ / mol DelGf0, kJ / mol 0.00 0.00 Ϫ241... 0.00 0.00 0.00 DelHb, kJ / mol DelHm, kJ / mol 0.90 40.66... 0.0729 hv1k hƒ1k gƒ1k w1k kJ molϪ1 kJ molϪ1 0.296 0.147 Ϫ0.071... in App. A. hƒ1k gƒ1k w1k kJ molϪ1 kJ molϪ1 0.263 0.500 X X X X 0.503...Ϫ1 0.297 0.292 Ϫ0.399 Ϫ0.720 kJ molϪ1 kJ molϪ1 1.252 1.041 Ϫ2.792 Ϫ2.092... w2j X 0.03654 0.21106 gƒ2j kJ molϪ1 kJ molϪ1 Ϫ9.874 Ϫ3.887 Ϫ24.125...



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... from a transient formaldehyde complex. Ph3P 1b H H H O + H2 PPh3 H Ru Ph3P CO... Scheme 1: Alcohol dehydrogenation catalysed by 1b. The decarbonylation side-reaction affords... mechanism for dehydrogenation catalysed by 1b, comparing methanol, ethanol and 2-propanol... H Dehydrogenation Pathway B H Ru Ph3P Ph3P 1b Initiation H PPh3 H H H O Ph3P - H2 - HCHO... transition states; (c) free energy profile (kJ/mol), with calculations carried at... driving force for 1b + MeOH  1b + H2CO + H2, G = + 50.5 kJ/mol),[6] formation of..., cf. the reaction: 1b + MeOH  22 + 3 H2, G = -84.5 kJ/mol (4) Table 1: Reaction... (G) for pathways E-G (all data in kJ/mol).a E EBSSE Esolvb EG G Overall... + H2 -14.4 0.0 11.9 -81.9 -84.5 1b + MeOH  22 + 3 H2 Pathway E (methanol) 69.9 -4.2 -0.2 -48.5 20.2 1b  2 + H2 -17.6 14.2 16.4 16... Pathway F (methanol) -28.0 -5.9 -87.7 94.6 1b + MeOH  13H + H2 + PPh3 216... small activation barriers (below 22.4 kJ/mol), a feature which is consistent... transition states; (c) free energy profile (kJ/mol), with calculations carried at... the overall barrier of G = 51.7 kJ/mol relative to 31 (Table 3, 18.8 kJ/mol + 32.9 kJ/mol) is easily overcome... transition states; (c) free energy profile (kJ/mol), with calculations carried at... H Ru H Ph3P 13H' H Me H O H - MeCHO 1b + PPh3 14H' Decarbonylation Pathway F - H2... transition states; (c) free energy profile (kJ/mol), with calculations carried at... F with ethanol (all data in kJ/mol).a E EBSSE Esolvb EG G Overall... + H2 -71.8 0.0 10.2 -76.8 -138.4 1b + EtOH  22 + 2 H2 + CH4 Pathway F (ethanol) 1b + EtOH  13H’ + H2 + PPh3 203... Bb Cb Cneut. Db F G Methanol 1b + MeO-  TS3-4 + H2 9 + MeOH  TS3Hb-7 + HCHO 1b + MeO-  TS13-14 + H2 + PPh3... + MeO1b + MeOH  TS9perp-19 + 2 H2 1b + MeOH  TS13H-23perp + H2 + PPh3...-  TS13-28 + H2 + PPh3 Ethanol f 1b + EtO-  TS’3-4 + H2 2 + EtOH  TS...’13-28 + H2 + PPh3 2-propanol h 1b + iPrO-  TS”3-4 + H2 2 + iPrOH  TS”3Hb-7 1b + iPrO-  TS”13-14 + H2... this case, as G = 116.8 kJ/mol for: 1b + EtOH  TS’23-24... barriers are lowered by 6-28 kJ/mol for dehydrogenation and by up to 9 kJ/mol for decarbonylation. This result... decarbonylation path (G), by ca. 30 kJ/mol). Given typical uncertainties of... XTOF (TDTS) Methanol 95 100 A 1b TS3-4 5 4 2 100 B 7 TS3Hb-7 98 9 100 92 C 1b TS13-14 TS14-15 a 8 100... overall free energy barrier (148.9 kJ/mol) than previously reported (134... (0.999) -H transfer 1b + MeO-  TS3-4 + H2 1b + EtO-  TS’3-4 + H2 1b + iPrO-  TS...  TS3Hb-7 + HCHO    flipping of HCHO 1b + MeO-  TS13-14 + H2 + PPh3 2  1 HCHO and rotation of H2 1b + MeOH  TS13H-14H + H2 + PPh3...-17 +HCHO+ MeO2 1 H-agostic interaction 1b + MeOH  TS9perp-19 + 2 H2 7  TS9perp-19 H2 dissociation 1b + MeOH  TS13H-23perp + H2 + PPh3... 1.88 (4.21) 1.012 (1.008) C-H +0.03 1b + MeO  TS13-28 + H2 + PPh3... intermediates, bottom: free energy profile (kJ/mol, B97-D2/ECP2 level... 20-perp, but only by +2.1 kJ/mol. [14] e.g.: a) B. B. Wayland, B. A. Woods and... to afford 1b. This process is very demanding thermodynamically (ΔG = +135 kJ/mol... 1b, EG = -95.8 kJ/mol at P = 1354 atm, whereas it is -120.9 kJ... the ECP3 level, in kJ/mol.(a) Global 1b (b) Methanol 2 (b) TS9perp-19 TS13H... B, D and E. Methanol E E’BSSEa Esolv EG G b B 1b + MeOH  TS3Hb-7 + H2 91.3 26...-7 21.8 28.9 21.3 62.3 134.3 D 1b + MeOH  TS16Hb-17 + PPh3 185.4 -18.5 -2.0 -63.4 101.5 E 1b + MeOH  TS9perp-19 + 2 H2 152....4 10.8 13.3 15.7 123.1 Ethanol c B 1b + EtOH  TS’3Hb-7 + H2 84...-7 + MeCHO 105.9 10.5 1.4 -22.5 95.3 D 1b + EtOH  TS’16Hb-17 + PPh3 178.6 -18.6 -2.7 -56.2 101.1 2-propanol d B 1b + iPrOH  TS”3Hb-7 + H2 74...-7 + Me2CO 51.3 13.8 -0.1 -38.6 26.4 D 1b + iPrOH  TS”16Hb-17 + PPh3... difference of intramolecular BSSE in 1b and the transition state (see... of Pathway F. E E’BSSEa Esolv EG G 1b + MeOH  TS13H-23perp + H2 + PPh3 251.2 -18.7 -7.8 -97.6 127.1 1b + MeOH  TS23-24 + 2 H2 + PPh3 270.2 -21.5 -7.9 -120.7 120.1 1b + MeOH  TS24-25 + 2 H2 + PPh3 276.8 -21.1 -7.7 -130.0 118.0 1b + EtOH  TS’13H-23 + H2 + PPh3 236.8 -18.0 -5.3 -95.6 117.8 1b + EtOH  TS’23-24 + 2 H2... 268.0 -17.5 -13.5 -120.2 116.8 1b + iPrOH  TS”13H-23perp + H2 + PPh3 245.9 -16.5 -3.8 93.3 132.3 1b + iPrOH  TS”23-24 + 2 H2... difference of intramolecular BSSE in 1b and the transition state (see... EH EG 13.5 -5.4 12.7 -6.2 13.5 A 1b + MeO-  TS3-4 + H2 -6.2 B 9 + MeOH  TS3Hb... -14.4 -11.0 -10.1 -7.1 -14.4 -11.0 C 1b + MeO-  TS13-14 + H2 + PPh3....5 -24.2 -84.4 -34.2 -91.5 Cneut 1b + MeOH  TS13H-14H + H2 + PPh3....8 15.7 -13.4 22.4 -22.8 15.7 F 1b + MeOH  TS13H-23perp + H2 + PPh3....5 -97.6 -22.7 -90.7 -33.5 -97.5 G 1b + MeO-  TS13-28 + H2 + PPh3....90 0.00 0.00 12.10 (1b) (2) (3) (4) (5) (1b) (TS1b-2) (no TS) (TS3-4) (no... 0.00 5.10 5.10 0.00 0.00 (1b) (2) (TS1b-2) (no TS) - 22 - Bühl... 2013 PPh3 H Ru Ph3P Ph3P 1b Dehydrogenation Pathway B H H Initiation PPh3 H H H Ru...



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...). HAI-1 has a splice variant, HAI-1B that contains extra residues in... Res 148: 295-311. Christie KJ, Turnley AM. 2012. Regulation of... hepatocyte growth factor activator inhibitor-1B as a potential physiological inhibitor of... hepatocyte growth factor activator inhibitor-1B (HAI-1B), a new splice variant of... hepatocyte growth factor activator inhibitor-1B (HAI-1B) and HAI-2. FEBS Lett... Pathol 36: 626-633. Lee KJ, Dietrich P, Jessell TM. 2000a. Genetic.... Development 134: 3203-3211. Yoon KJ, Koo BK, Im SK, Jeong...