ÌâÄ¿ÄÚÈÝ


ÓлúÎï±ûÊÇÒ»ÖÖÏãÁÏ£¬ÆäºÏ³É·ÏßÈçͼK29­4Ëùʾ£¬Ï©ÌþAµÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª56£¬ÆäºË´Å¹²ÕñÇâÆ×ÏÔʾֻÓÐÁ½×é·å£»·¼Ïã×廯ºÏÎïD¿ÉÒÔ·¢ÉúÒø¾µ·´Ó¦£¬ÔÚ´ß»¯¼Á´æÔÚµÄÌõ¼þÏ£¬1 mol DÓë2 mol H2·´Ó¦¿ÉÒÔÉú³ÉÒÒ£»±ûÖк¬ÓÐÁ½¸ö¡ªCH3¡£

ͼK29­4

ÒÑÖª£ºR¡ªCH===CH2R¡ªCH2CH2OH

(1)AµÄ½á¹¹¼òʽΪ________________¡£

(2)DÖÐËùº¬¹ÙÄÜÍŵÄÃû³ÆÊÇ________________£¬D²»ÄÜ·¢ÉúµÄ·´Ó¦ÀàÐÍÓÐ________(ÌîÐòºÅ)¡£

a£®È¡´ú·´Ó¦  b£®¼Ó¾Û·´Ó¦

c£®ÏûÈ¥·´Ó¦  d£®Ñõ»¯·´Ó¦

(3)¼×ÓëÒÒ·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ________________________________________________________________________

________________________________________________________________________¡£

(4)д³öÓëÒÒ»¥ÎªÍ¬·ÖÒì¹¹ÌåÇÒÂú×ãÏÂÁÐÌõ¼þµÄÓлúÎïµÄ½á¹¹¼òʽ£º________________________________________________________________________¡£

a£®ÓöFeCl3ÈÜÒºÏÔ×ÏÉ«

b£®±½»·ÉϵÄÒ»äåÈ¡´úÎïÖ»ÓÐ2ÖÖ


(1)(CH3)2C===CH2

(2)̼̼˫¼ü¡¢È©»ù¡¡c

(3)(CH3)2CHCOOH£«¡ªCH2CH2CH2OH (CH3)2CHCOOCH2CH2CH2¡ª£«H2O

(4)HO¡¢HO

[½âÎö] (1)¸ù¾ÝÏ©ÌþAµÄÏà¶Ô·Ö×ÓÖÊÁ¿¿ÉÖª£¬AµÄ·Ö×ÓʽΪC4H8£¬µ±·Ö×ÓÖк¬ÓÐÁ½ÀàHÔ­×Óʱ£¬¸ÃÏ©ÌþµÄ½á¹¹¼òʽΪCH3CH===CHCH3»òCH2===C(CH3)2£¬½áºÏÌâ¸ÉÖеÄÒÑÖªÌõ¼þ¿ÉÖªAµÄ½á¹¹¼òʽΪºóÕß¡£½áºÏAµÄ½á¹¹¼òʽ£¬ÁªÏµÌâÄ¿Öеġ°ÒÑÖª¡±£¬¹ÊBµÄ½á¹¹¼òʽΪ(CH3)2CHCH2OH£»CΪ´¼BÑõ»¯ºóµÃµ½µÄÈ©£¬Æä½á¹¹¼òʽΪ(CH3)2CHCHO£»¼×ΪCÖеÄÈ©»ù±»Ñõ»¯ºóµÃµ½µÄ²úÎ¹Ê¼×µÄ½á¹¹¼òʽΪ(CH3)2CHCOOH£»¸ù¾Ý¼×ÓëÒÒ·´Ó¦Éú³É±ûµÄÌõ¼þ¿ÉÖª£¬¸Ã·´Ó¦Îªõ¥»¯·´Ó¦£¬Òò±ûÖÐÖ»º¬ÓÐÁ½¸ö¼×»ù£¬ÔòÒªÇóÒÒ·Ö×ÓÖÐÎÞ¼×»ù£¬Ôò±ûµÄ½á¹¹¼òʽΪ(CH3)2CHCOOCH2CH2CH2¡ª£¬ÒҵĽṹ¼òʽΪ¡ªCH2CH2CH2OH£»ÒòDÄÜ·¢ÉúÒø¾µ·´Ó¦£¬¹Ê·Ö×ÓÖк¬ÓÐÈ©»ù£¬¶ø 1 mol DÄÜÓë2 mol H2·¢Éú¼Ó³É·´Ó¦Éú³ÉÒÒ£¬¹ÊDµÄ½á¹¹¼òʽΪCHCHCHO¡£

(2)ÒòΪDµÄ½á¹¹¼òʽΪ¡ªCH===CHCHO£¬¹ÊÆäËùº¬µÄ¹ÙÄÜÍŵÄÃû³ÆÎªÌ¼Ì¼Ë«¼üºÍÈ©»ù£»·Ö×ÓÖб½»·ÉϵÄHÔ­×Ó¿ÉÒÔ·¢ÉúÈ¡´ú·´Ó¦£»·Ö×ÓÖк¬ÓÐ̼̼˫¼ü£¬ÄÜ·¢Éú¼Ó¾Û·´Ó¦£»·Ö×ÓÖÐÎÞôÇ»ù»ò±ԭ×Ó£¬²»ÄÜ·¢ÉúÏûÈ¥·´Ó¦£»·Ö×ÓÖеÄ̼̼˫¼üºÍÈ©»ù¶¼ÄÜ·¢ÉúÑõ»¯·´Ó¦¡£

(3)¼×·Ö×ÓÖк¬ÓÐôÈ»ù£¬ÒÒ·Ö×ÓÖк¬Óд¼ôÇ»ù£¬¹Ê¶þÕß·¢Éúõ¥»¯·´Ó¦¡£

(4)ÓöFeCl3ÈÜÒºÏÔ×ÏÉ«£¬ÒªÇó·Ö×ÓÖк¬ÓзÓôÇ»ù£»±½»·ÉϵÄÒ»äåÈ¡´úÎïÖ»ÓÐ2ÖÖʱ£¬ÒªÇó¸Ã·Ö×ÓÖк¬ÓÐ2¸ö´¦ÓÚ¶ÔλµÄÈ¡´ú»ù£¬¶ø±û»ùÓÐÁ½ÖÖ£º¡ªCH2CH2CH3ºÍ(CH3)2CH¡ª¡£¹Ê¸Ãͬ·ÖÒì¹¹ÌåµÄ½á¹¹¼òʽΪHO¡¢HO¡£


Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

    ½«º£Ë®µ­»¯ÓëŨº£Ë®×ÊÔ´»¯½áºÏÆðÀ´ÊÇ×ÛºÏÀûÓú£Ë®µÄÖØÒªÍ¾¾¶Ö®Ò»¡£Ò»°ãÊÇÏȽ«º£Ë®µ­»¯»ñµÃµ­Ë®£¬ÔÙ´ÓÊ£ÓàµÄŨº£Ë®ÖÐͨ¹ýһϵÁй¤ÒÕÁ÷³ÌÌáÈ¡ÆäËû²úÆ·¡£

»Ø´ðÏÂÁÐÎÊÌ⣺

  (1)ÏÂÁиĽøºÍÓÅ»¯º£Ë®×ÛºÏÀûÓù¤ÒÕµÄÉèÏëºÍ×ö·¨¿ÉÐеÄÊÇ________(ÌîÐòºÅ)¡£

  ¢ÙÓûìÄý·¨»ñÈ¡µ­Ë®

¢ÚÌá¸ß²¿·Ö²úÆ·µÄÖÊÁ¿

¢ÛÓÅ»¯ÌáÈ¡²úÆ·µÄÆ·ÖÖ

¢Ü¸Ä½ø¼Ø¡¢ä塢þµÈµÄÌáÈ¡¹¤ÒÕ

(2)²ÉÓá°¿ÕÆø´µ³ö·¨¡±´ÓŨº£Ë®´µ³öBr2£¬²¢Óô¿¼îÎüÊÕ¡£¼îÎüÊÕäåµÄÖ÷Òª·´Ó¦ÊÇBr2£«Na2CO3£«H2O¡úNaBr£«NaBrO3£«NaHCO3£¬ÎüÊÕ1 mol Br2ʱ£¬×ªÒƵĵç×ÓÊýΪ________mol¡£

    (3)º£Ë®ÌáþµÄÒ»¶Î¹¤ÒÕÁ÷³ÌÈçÏÂͼ£º

Ũº£Ë®µÄÖ÷Òª³É·ÖÈçÏ£º

Àë×Ó

Na£«

Mg2£«

Cl£­

SO

Ũ¶È/(g¡¤L£­1)

63.7

28.8

144.6

46.4

¸Ã¹¤ÒÕ¹ý³ÌÖУ¬ÍÑÁò½×¶ÎÖ÷Òª·´Ó¦µÄÀë×Ó·½³ÌʽΪ______________________________£¬²úÆ·2µÄ»¯Ñ§Ê½Îª__________£¬1 LŨº£Ë®×î¶à¿ÉµÃµ½²úÆ·2µÄÖÊÁ¿Îª________g¡£

(4)²ÉÓÃʯīÑô¼«¡¢²»Ðâ¸ÖÒõ¼«µç½âÈÛÈÚµÄÂÈ»¯Ã¾£¬·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ________________________£»µç½âʱ£¬ÈôÓÐÉÙÁ¿Ë®´æÔÚ»áÔì³É²úƷþµÄÏûºÄ£¬Ð´³öÓйط´Ó¦µÄ»¯Ñ§·½³Ìʽ£º____________________________________________________¡£

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø