Thông tin tài liệu
! ! " # " $ %&%'%(' ) " *+, !-, . / ! *)/%/ 01 20, !30 " ! % 4 " #$!. " 5 2 56 % )1 0 " 5 0 " 5- # " % )1 7 ! " " 1 - " 4 2 8 9 0 " 4 : 1 ;295 0 " :2< 5 ) " &49(< = 1 - 3 3 ; :2< 9 44 < × + × B 8 " #%1 5 > 50 " ? )% ! # " / ? ? ) 50 " ? )5 8 + → 5 5 7 ! " " :2@) 1 -%1 A24@ 5? -, !0 " 5 . " % )1 %/ 1 ) " 8;2@&@2@@(42@ 7 - % 35 B5 C5 3 5B % 5 3 A24@ < 3:2@3:28;@) 5 5 C5 8 ↓ 5 5 8 35 3:28;@) 5 8 3:28;@::38;2@ 7 ! " " 1 -)// - ! " # 1 2). 5 %0 " 29 5 ! " # " $ %&%'%(' 53 29 < 3:;>:2@+, ! ) " % 7 ! " " 1 -)// - ! " # 1 24). 5 %0 " @2 5 ) " 9 0 " 8 < & 8 < 0 " 4 : 4 : 0 " @ ( @ 0 " 9 4 D!), 53 @2 < 324)E5 3) 5>5 ⇒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⇒ 535 +, ! !%' = 1 -%. " %0 " %' " ? !/ -, 0 " " ? )J, ! 1 - " !. " ) " , " 0 " <: :K& ) 4 7 ! " " 1 - 1 :29)5 %0 " %' -, ) " 9 2 4 & 8 < 2 8 9 4 : 2 4 < ( @ 2 @ : %' 3& 3 <:: ::9: = :2) 8 5 C5 5 :2:2 6 :23 :29 = :28 ⇒ 38 ⇒ 8 9 9 ) #%5 > 50 " % ! 35 I 5 7 ! " " +). % %/ %. 1 5 0 " 5 % )1 @2L/ ? -, ! 5 1 4@%/ + ) " 8 92;). &24). 424<). &8289). 5 35 8 3 4@ :: = 4 4A29 824 + = → = :24@) 53 @2 :24@44 < − = :28) % 35 I 53:24@I:283:2@) + % 3:2@2438289). -, %) " & 8 4 4 9 ( @ < 5 38 % +, !% 8!/ 8 4 7 ! " " +). %%1 :2< 5 L/ , ? -, !, - / 0 " . " 1 0 . " % )1 . " #@:24+ ) " 8289). &24). 92;). (424<). L1 0, - 5 0 " 5 5 53@:24E5 3@:24I:2<38A29 5 3 8A29 44 = :2A) % 35 I 53 :2< :2A 44< −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→ = ) " 8 < 0 " 4 : 7 ! " " 1 - 1 2)// - ! " # 1 24). 5 %0 " @2 5 -, ) " 4 2 9 & 9 2 8 < 8 < 2 4 : ( 4 : 2 @ 4 1 -%') " " # )// - 1 & , !) " , " 0 " 94& -, %') " @ 4 2 8 9 & 8 < 2 4 : 4 : 2 @ ( @ : 2 9 6! )/ ? )%' 1 -) " &8( <2< :2 44 = 94 :24 9: !" #$ = = = 4 8@ :24 !" = = E 4 8@ 2@ = → = 7 ) " 4 0 " 8 9 :2 1 -%. " 4 0 " %' " # )/ / -1 2, !% )1 . " #;2 " 1 " / . 1 - -, %') " 4 2 8 9 & 8 9 2 4 : 4 < 2 @ : ( @ : 2 9 G, " #/ . %') " @K28@K&:K28:K @K2@K(4:K:K 4 4 !" !" = → = ; !" = E 4 :2 ; :2 9 − = = E ; 4 2@ :2 = →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→ = = = # 03 ? )5 3 ? ) 5' 2 ('+ → 1 : 2 '+ → ? )5 # " ? )5 % ' " ? ) 5 1 . " / " 1D ) 5 1# " ? ) 0 " '' 7 ! 1 -'21 1 :24)5 " " ' " !, " :2) 1 1 -1 21 7 ! " " 1 -1 . " ? ) 51 ) " :24)&:29):2<)(:28) 7 ! 1 -'1 :24)5 . " 1 :24 ) 5 ' , /:2) . " ? ) 1 1 ') " :2)+, !? ) 5 % !1 ) " :24:23:29) 8 & ' . )/ 5 34 OB5 3 ; 1 -50 " (, " 1 -1 ?- 0 " . " 1 ) / 1, ) 0 " , / 1 2, !% )1 . " #2< J, !. " 01 L5 8 PL 8 1 29H )1 8 5 - 1 - 5) " <28&A28:28(2:8 B5 : )* ' → 8 5 8 + - ) " 2< 5 5 8 ) → 5 54 ↓ 29 :2:@ 4 4 :< + = = × = = 5 3:2:@8:32@E 8 2< 2@ :28 = − = 1 - " :2)550 " :2)5 / 01 L5 8 PL 8 . " % )1 1 ) " :<&:2<9(29 :2)55C:2) :2)5C:2<) C7 - , 1Q " -, " " 225 K5@828% )1 H / - 12 " )QC4)6G6 Q) " 5&5 5 ( 4 5 & ' + , 00 ' $ & ' )56"5 7 1 -@1 1 01 D5 4 24: : . " ? '' 1 ) " :&@(; - ( ( + ''C1 @'' 782< 1 - " 81 1 01 D5 4 # 24: : C 1 - '' ? )# " 0 " % )1 ) " 2D )'' ) " :2)&:2):28)(:24) 7 1 -81 9'' 6'7J&6HJ 1 3 '' C 382<I2329 < ( 29 2 < "'" = = = ⇒ ∑ ∑ '' 3 2 :2 9 = & ' % . ( ),'- ( 0 " ? # % )1 %!/ " , " !? , % / N % )1 1 -!, / ( 0 " -. " ? ! 0/ " % )1 )C)& # !/ " N)C!)&01 N2!) " . )/ ,# " - 6. " ? ! % )1 C&' " 1 R) , - !/ " ? -, 6 " . 1 ? ) , - 0 " 1 ) ./ *01 *$21 Q' - 1 01 H ( ( ( 3 (4 3 4 + → + # O5HCO5H 6'-, ! )1 01 H ) ). " % )1 #8AI38< +, !/ / " % )1 1 0 " % )1 ). " / . 1 ? ) 1 2 0 " " N 6G61 ./ *01 *5"*'N' - 1 ' OI5 5 : 8 2) ' → OI55 6'-, ! )'' 1C)N ⇒ ∆ 3 4@IA39+, !/ / " ' 2 N C ' 2 C 6G6' ./ *01 *(*'Q' - 01 %/ " O55 N NL5CO55L N N 5 # O55L5CO55L 5 )C)C ∆ ↑ 3 ./ *01 *"&'"N' - N " - " O55O S L5CO55LO S 5 )C)C ∆ ↑ 38I= O S ./ *01 **(*'N' - 01 ) 55BOBL )C55BOBL 8 ) )C)C ∆ ↑ 3892@ :2@ 1 -N1 0 " 01 L 5 8 . " 1 +). 5 %0 " . " 1 <2A9 F +) " 42<4). &424<). 24). (24). TH/ % F . " N) " 3 − A G- 3 − L 5 8 C 3 )− 5 ↑ 5 6'-)C)) ⇒ ∆ 38B344 6'/ " " H )1 #<2A9I:2@3<2< CD )5 3 <2< :2 44 = C6/ . 5 +3:2243424<). : 1 -1 1 %/ / - ! " # 0 " 01 L%) 424, # 0 " +). %. %+ ) " 2). &24). 8289). (424<). 6'--)1 - C)):2@) . " % )1 # 9 9 24 42 ;< 4@ ) ) − = + ∆ = 8B3 +, !', " " ) )0 " :2@) !. " # 424I:3424CD ) 3 424:2@ :2 = C6/ . +3:224324). H !-, " " :2:@)'?' N1 01 1 1 / / @29H5=# % 2% !-,@24;@'?' . " / / 42H50 " 1 92@ +, !66'?') " 55 @ &55 8 55 8 (H/ % +. " H5 3 '?' C'?' " N 0 " 1 1 7# '?') " O55O S O55O S H5CO55H O S 5 ))C). " 38A2IO S ⇒ :2:8;@)::;@)C:2:8;@). " 392@I@24;@3 :2;@ C:2:8;@;<IO S 3:2;@CO S 3ACO S 3 @ B = '?' 3 @24;@ 49 :2:8;@ = C= O 44A349C= O 3: +, ! '?') " 55 @ 8 & ' )56& ' )56"5 : [...]... hơ ̣p X gồ m các olefin Nế u đố t cháy hoàn toàn Y thì thu đươ ̣c 0,66g CO 2 Vâ ̣y khi đố t cháy hoàn toàn X thì tổ ng khố i lươ ̣ng CO2 và H2O là: A 0,903g B 0,39g C 0,94g D 0,93g 14 Phương pháp nhóm nguyên tử trung binh: ̀ Nhóm ở đây có thể là số nhóm -OH, -NH2, NO2 Thí dụ1: Nitro hóa benzen thu đươ ̣c 14,1g hỗn hơ ̣p gồ m 2 chấ t nitro có khố i lươ ̣ng phân tử
Ngày đăng: 13/09/2013, 04:10
Xem thêm: phương pháp hóa học THPT, phương pháp hóa học THPT
