(12·Ö) CH3CHBrOCOOC2H5£¨1-äåÒÒÑõ»ù̼ËáÒÒõ¥£©ÊÇÍ·æß¿¹¾úËØÍ·æßß»ÐÂõ¥µÄÖмäÌ壬ͷæßß»ÐÂõ¥2001ÄêÈ«ÇòÏúÊÛ¶î¸ß´ï4.7ÒÚÃÀÔª£¬Ä¿Ç°¹úÄÚÒ²Óжà¼ÒÆóÒµÉú²ú¡£¸ÃÖмäÌåÓÐËÄÌõºÏ³É·Ïߣº

ºÏ³É·ÏßÒ»£ºÓÉÒÒÑõ»ù¼×õ£ÂÈÓëäå½øÐÐ×ÔÓÉ»ù·´Ó¦µÃµ½1-äåÒÒÑõ»ù¼×õ£ÂÈ£¬ºóÕßÓë´¼½øÐÐÈ¡´ú·´Ó¦µÃµ½£»

ºÏ³É·Ï߶þ£º¶þÒÒ»ù̼Ëáõ¥Ö±½Óä廯µÃµ½£¬¸Ã·¨¸±²úÆ·Ì«¶à£»

ºÏ³É·ÏßÈý£º1-ÂÈÒÒÑõ»ù̼Ëáõ¥Óë¹ýÁ¿äåÑνøÐÐÈ¡´ú·´Ó¦£¬³£ÓÃäåÑÎÓÐä廯ﮡ¢ä廯ËÄÒÒ»ùﮡ¢ä廯ËĶ¡»ù﮵ȣ»

ºÏ³É·ÏßËÄ£ºÒÒÏ©»ù̼ËáÒÒõ¥Óëä廯Çâ·´Ó¦£¬¸Ã·¨²»Ê¹ÓÃÈܼÁ£¬¶øÇÒ²úÂʺÍÑ¡ÔñÐÔ¶¼±È½Ï²»´í¡£

ÒÑÖª£º £¨XΪ±ËØÔ­×Ó£©³ÆΪõ£Â±¼ü

»Ø´ðÏÂÁÐÎÊÌ⣺

(1)£®ºÏ³É·ÏßÒ»ÖÐÓë´¼½øÐÐÈ¡´ú·´Ó¦£¬¸Ã´¼µÄ½á¹¹¼òʽΪ­­­­­              

(2)£®ºÏ³É·Ï߶þ¸±²úÆ·Ì«¶àµÄÔ­ÒòΪ                                

(3)£®ºÏ³É·ÏßÈýµÄ·´Ó¦·½³ÌʽΪ£¨ÒÔä廯ï®ÎªÀý£¬²»ÐèҪдÌõ¼þ£©£º

                                                                 

(4)£®ºÏ³É·ÏßËĵķ´Ó¦·½³ÌʽΪ£¨²»ÐèҪдÌõ¼þ)

                                                                

(5)1-äåÒÒÑõ»ù̼ËáÒÒõ¥ÓжàÖÖͬ·ÖÒì¹¹Ì壬ÆäÖк¬ÓÐR1¨DOCOO¨DR2(R1 R2²»ÎªH)ͬ·ÖÒì¹¹ÌåÓР       ÖÖ£¨²»º¬¿Õ¼äÒì¹¹£©¡£

(1)C2H5OH£¨»òCH3CH2OH£©£¬·ÖÖµ2·Ö¡£

(2) ¶þÒÒ»ù̼Ëáõ¥ºÍäåÈ¡´ú£¬È¡´úλÖúÍÈ¡´úÇâÔ­×Ó¸öÊý¶¼²»È·¶¨£¬Òò´Ë²úÎï²»´¿£¬·ÖÖµ2·Ö¡£

(3) CH3CHClOCOOC2H5+LiBr¡úCH3CHBrOCOOC2H5+LiCl£¬·ÖÖµ3·Ö¡£

(4) CH2=CHOCOOC2H5+HBr¡úCH3CHBrOCOOC2H5£¬·ÖÖµ3·Ö¡£

(5) 10£¬·ÖÖµ2·Ö¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

н»Í¨·¨¹æ¶¨¾Æ¼ÝÒ»´Î¿Û12·Ö£®ÈçͼΪ½»¾¯¶Ô¼ÝʻԱÊÇ·ñÒû¾Æ½øÐмì²â£®ÆäÔ­ÀíÈçÏ£º2CrO3£¨ºìÉ«£©+3C2H5OH+3H2SO4=Cr2£¨SO4£©3£¨ÂÌÉ«£©+3CH3CHO+6H2O¸Ã·´Ó¦µÄÑõ»¯¼ÁÊÇ£¨¡¡¡¡£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨ÒÕ¸ßÍê³ÉÆո߲»×ö£¬12·Ö£©Ô­×ÓÐòÊýΪ12µÄÔªËØ£¬ÔÚÖÜÆÚ±íÖÐλÓÚ
Èý
Èý
ÖÜÆÚµÚ
¢òA
¢òA
×壻¸ÃÔªËØΪ
½ðÊô
½ðÊô
£¨½ðÊôÔªËØ»ò·Ç½ðÊôÔªËØ£©£¬ÔÚ»¯Ñ§·´Ó¦ÖÐÈÝÒ×
ʧ
ʧ
£¨Ê§»òµÃ£©µç×Ó£»Ô­×ÓÐòÊýΪ17µÄÔªËØ£¬ÔªËØ·ûºÅΪ
Cl
Cl
£¬¸ÃÔ­×ӵĵç×ÓʽΪ
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÐÂÐ޸ĵġ¶»ú¶¯³µ¼Ýʻ֤ÉêÁìºÍʹÓù涨¡·ÓÚ2010Äê4ÔÂ1ÈÕÆðʵʩ£¬Ð¹æ¹æ¶¨¾Æ¼ÝÒ»´Î¿Û12·Ö£¬³ö´ËÖØÈ­ÖÎÀí¾Æ¼ÝÊÇÒòΪ¾Æºó¼Ý³µÊÇÒý·¢½»Í¨Ê¹ʵÄÖØÒªÔ­Òò£®ÈçͼΪ½»¾¯¶Ô¼ÝʻԱÊÇ·ñÒû¾Æ½øÐмì²â£®ÆäÔ­ÀíÈçÏ£º
2CrO3£¨ºìÉ«£©+3C2H5OH+3H2SO4=Cr2£¨SO4£©3£¨ÂÌÉ«£©+3CH3CHO+6H2O
¸ù¾ÝÉÏÊöÐÅÏ¢»Ø´ðÎÊÌâ
£¨1£©¸Ã·´Ó¦Öб»¼ì²âµÄ³É·ÖΪ
CH3CH2OH
CH3CH2OH
£¬¸Ã·´Ó¦µÄÑõ»¯¼ÁÊÇ
CrO3
CrO3
£¬»¹Ô­²úÎïÊÇ
Cr2£¨SO4£©3
Cr2£¨SO4£©3
£¬´Ë·´Ó¦¿ÉÓÃÓÚ¼ì²â¼ÝʻԱÊÇ·ñÊǾƺó¼Ý³µµÄÔ­ÒòÊÇ
¼ì²â¼ÁÑÕÉ«ÓɺìÉ«±äΪÂÌÉ«£¬ÑÕÉ«±ä»¯Ã÷ÏÔ£¬Ò×ÓÚ¹Û²ì
¼ì²â¼ÁÑÕÉ«ÓɺìÉ«±äΪÂÌÉ«£¬ÑÕÉ«±ä»¯Ã÷ÏÔ£¬Ò×ÓÚ¹Û²ì

£¨2£©ÊµÑéÊÒÐèÒªÓÃÃܶÈΪ1.84g?cm-3¡¢ÖÊÁ¿·ÖÊýΪ98%µÄŨÁòËáÅäÖÆ1.0mol?L-1µÄÁòËá100mL£¬Á¿È¡Å¨ÁòËáʱÐèÒªÓõ½Á¿Í²×îºÃÑ¡ÓÃ
C
C

A£®50mL B£®20mL C£®10mL D£®100mL£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º½­Î÷Ê¡2011½ì¸ßÈýÉÏѧÆÚ¿ªÑ§Ä£Ä⿼ÊÔ»¯Ñ§ÊÔÌâ ÌâÐÍ£ºÌî¿ÕÌâ

£¨Ã¿¿Õ2·Ö£¬¹²12·Ö£©.

£¨1£©ÏÂÁÐ˵·¨²»ÕýÈ·µÄÊÇ                     ¡£

      A.ÓÃ100mLµÄÈÝÁ¿Æ¿×¼È·Á¿È¡100mLÒºÌå

      B.·ÖҺʱ£¬·ÖҺ©¶·ÖÐϲãÒºÌå´ÓÏ¿ÚÁ÷³ö£¬ÉϲãÒºÌå´ÓÉÏ¿Úµ¹³ö

      C.ÓÃÍÐÅÌÌìƽ³ÆÁ¿NaCl¹ÌÌåʱ£¬NaCl·ÅÔÚ³ÆÁ¿Ö½ÉÏ£¬³ÆÁ¿NaOH¹ÌÌåʱ£¬NaOH·ÅÔÚСÉÕ±­Àï

  £¨2£©»Ø´ðÏÂÁÐÎÊÌ⣨ÌîÐòºÅ£©£º

  ÏÂÁÐÒÇÆ÷ÖУº¢Ù©¶·¢ÚÈÝÁ¿Æ¿¢ÛÕôÁóÉÕÆ¿¢ÜÌìƽ¢Ý·ÖҺ©¶·¢ÞÁ¿Í²¢ßȼÉճס£

³£ÓÃÓÚÎïÖÊ·ÖÀëµÄÊÇ        £¬ÆäÖиù¾ÝÎïÖʷе㲻ͬÀ´·ÖÀëÎïÖʵÄÒÇÆ÷ÊÇ    ¡£

  £¨3£©ÓÃÊʵ±µÄÊÔ¼Á»ò·½·¨³ýÈ¥ÏÂÁÐÎïÖÊÖÐËù»ìÓеÄÉÙÁ¿ÔÓÖÊ£¬Ð´³öÓйصķ´Ó¦·½³Ìʽ¡£

     Ìú·ÛÖлìÓÐÉÙÁ¿ÂÁ·Û                        £»

FeCl3ÖлìÓÐÉÙÁ¿FeCl2                      £»

     ¹ÌÌå Na2CO3ÖлìÓÐÉÙÁ¿NaHCO3             ¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º½­ËÕÊ¡2010-2011ѧÄê¸ßÈý»¯Ñ§Ò»ÂÖ¹ý¹Ø²âÊÔ£¨2£© ÌâÐÍ£ºÌî¿ÕÌâ

(12·Ö)¹¤ÒµÉÏ´ÓÂÁÍÁ¿ó(º¬Ñõ»¯ÂÁ¡¢Ñõ»¯ÌúµÈ)ÖÆÈ¡ÂÁµÄÁ÷³ÌÈçͼËùʾ £º

¾­·ÖÎö£¬Éú²úÔ­ÁÏ(ÂÁÍÁ¿ó)ºÍÌáÈ¡Al2O3ºóµÄ²ÐÔü(³àÄà)µÄ²¿·Ö³É·Ö¼ûϱí(ÒÔÑõ»¯Îï±íʾ)ÇÒÒÑÖªÂÁÍÁ¿óÖеÄFe2O3È«²¿×ªÈë³àÄà¡£

(1)Éú²úÖÐÿÏûºÄ1tÂÁÍÁ¿ó½«²úÉú¶àÉÙ¶Ö³àÄࣿ

(2)ÊÔ¼ÆËã³öÿÁ¶1tÂÁ£¬ÀíÂÛÉÏÐèÒª¶àÉÙ¶ÖÂÁÍÁ¿ó£¿

 

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸