ÌâÄ¿ÄÚÈÝ

 ÊµÑéÊÒÖÆ±¸äåÒÒÍ飨C2H5Br£©µÄ×°ÖúͲ½ÖèÈçͼ£º£¨ÒÑÖªäåÒÒÍéµÄ·Ðµã38£®4¡æ£©

¢Ù¼ì²é×°ÖÃµÄÆøÃÜÐÔ£¬Ïò×°ÖÃͼËùʾµÄUÐιܺʹó

ÉÕ±­ÖмÓÈë±ùË®£»

¢ÚÔÚÔ²µ×ÉÕÆ¿ÖмÓÈë10mL95%ÒÒ´¼¡¢28mL80%Ũ

ÁòËᣬȻºó¼ÓÈëÑÐϸµÄ13gä廝įºÍ¼¸Á£Ëé´ÉƬ£»

¢ÛСÐļÓÈÈ£¬Ê¹Æä³ä·Ö·´Ó¦¡£

»Ø´ðÏÂÁÐÎÊÌ⣺

   £¨1£©¸ÃʵÑéÖÆÈ¡äåÒÒÍéµÄ»¯Ñ§·½³ÌʽΪ£º

 

____________________________¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡£

   £¨2£©·´Ó¦Ê±Èôζȹý¸ß£¬¿É¿´µ½Óкì×ØÉ«ÆøÌå²úÉú£¬¸ÃÆøÌå·Ö×ÓʽΪ_______  ____£¬Í¬Ê±»¹Éú³ÉÁíÒ»ÖÖµÄÎÞÉ«ÆøÌå¡£

£¨3£©UÐ͹ÜÄڿɹ۲쵽µÄÏÖÏóÊÇ___________________________¡£

  £¨4£©·´Ó¦½áÊøºó£¬UÐιÜÖдÖÖÆµÄC2H5Br³Êר»ÆÉ«¡£ÎªÁ˳ýÈ¥´Ö²úÆ·ÖеÄÔÓÖÊ£¬¿ÉÑ¡ÔñÏÂÁÐÊÔ¼ÁÖеÄ_______________£¨Ìî×Öĸ£©

A£®NaOHÈÜÒº        B£®H2O      

C£®Na2SO3ÈÜÒº        D£®CCl4

ËùÐèµÄÖ÷Òª²£Á§ÒÇÆ÷ÊÇ______________£¨ÌîÒÇÆ÷Ãû³Æ£©¡£Òª½øÒ»²½ÖƵô¿¾»µÄC2H5Br£¬¿ÉÓÃˮϴ£¬È»ºó¼ÓÈëÎÞË®CaCl2£¬ÔÙ½øÐÐ_______________£¨Ìî²Ù×÷Ãû³Æ£©¡£

   £¨5£©ÏÂÁм¸ÏîʵÑé²½Ö裬¿ÉÓÃÓÚ¼ìÑéäåÒÒÍéÖÐäåÔªËØ£¬ÆäÕýÈ·µÄ²Ù×÷˳ÐòÊÇ£ºÈ¡ÉÙÁ¿äåÒÒÍ飬Ȼºó__________________£¨Ìî´úºÅ£©¡£

¢Ù¼ÓÈÈ£»¢Ú¼ÓÈëAgNO3ÈÜÒº£»¢Û¼ÓÈëÏ¡HNO3Ëữ£»¢Ü¼ÓÈëNaOHÈÜÒº£»¢ÝÀäÈ´

£¨6£©äåÒÒÍ飨C2H5Br£©¶àÒ»¸ö̼µÄͬϵÎïÔÚÇâÑõ»¯ÄƵĴ¼ÈÜÒºÖÐÄÜ·¢Éú·´Ó¦£¬Ð´³ö¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡£

 

 £¨16·Ö£¬Ã¿¿Õ2·Ö£©

£¨1£©C2H5OH+NaBr+H2SO4NaHSO4+C2H5Br+H2O

£¨2£©Br2                       £¨3£©ÓÐÓÍ×´ÒºÌåÉú³É¡£          

£¨4£©c£»  ·ÖҺ©¶·£»  ÕôÁó¡£        £¨5£©¢Ü¢Ù¢Ý¢Û¢Ú

 


£¨6£©CH3CH2CH2Br£«NaOH            CH3£­CH£½CH2¡ü£«NaBr£«H2O  

 

 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
¡¢ÊµÑéÊÒÖÆ±¸äåÒÒÍ飨C2H5Br£©µÄ×°ÖúͲ½ÖèÈçͼ£º£¨ÒÑÖªäåÒÒÍéµÄ·Ðµã38.4¡æ£©
¢Ù¼ì²é×°ÖÃµÄÆøÃÜÐÔ£¬Ïò×°ÖÃͼËùʾµÄUÐιܺʹóÉÕ±­ÖмÓÈë±ùË®£»
¢ÚÔÚÔ²µ×ÉÕÆ¿ÖмÓÈë10mL95%ÒÒ´¼¡¢28mL80%ŨÁòËᣬȻºó¼ÓÈëÑÐϸµÄ13gä廝įºÍ¼¸Á£Ëé´ÉƬ£»
¢ÛСÐļÓÈÈ£¬Ê¹Æä³ä·Ö·´Ó¦£®
»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©¸ÃʵÑéÖÆÈ¡äåÒÒÍéµÄ»¯Ñ§·½³ÌʽΪ£º
NaBr+H2SO4+C2H5OH
¡÷
NaHSO4+C2H5Br+H2O
NaBr+H2SO4+C2H5OH
¡÷
NaHSO4+C2H5Br+H2O

£¨2£©·´Ó¦Ê±Èôζȹý¸ß£¬¿É¿´µ½Óкì×ØÉ«ÆøÌå²úÉú£¬¸ÃÆøÌå·Ö×ÓʽΪ
Br2
Br2
£¬Í¬Ê±»¹Éú³ÉÁíÒ»ÖÖÎÞÉ«ÆøÌ壮
£¨3£©UÐ͹ÜÄڿɹ۲쵽µÄÏÖÏóÊÇ
ÓÐÓÍ×´ÒºÌåÉú³É
ÓÐÓÍ×´ÒºÌåÉú³É
£®
£¨4£©·´Ó¦½áÊøºó£¬UÐιÜÖдÖÖÆµÄC2H5Br³Êר»ÆÉ«£®ÎªÁ˳ýÈ¥´Ö²úÆ·ÖеÄÔÓÖÊ£¬¿ÉÑ¡ÔñÏÂÁÐÊÔ¼ÁÖеÄ
C
C
£¨Ìî×Öĸ£©
A£®NaOHÈÜÒº        B£®H2O         C£®Na2SO3ÈÜÒº        D£®CCl4
ËùÐèµÄÖ÷Òª²£Á§ÒÇÆ÷ÊÇ
·ÖҺ©¶·
·ÖҺ©¶·
£¨ÌîÒÇÆ÷Ãû³Æ£©£®Òª½øÒ»²½ÖƵô¿¾»µÄC2H5Br£¬¿ÉÓÃˮϴ£¬È»ºó¼ÓÈëÎÞË®CaCl2£¬ÔÙ½øÐÐ
ÕôÁó
ÕôÁó
£¨Ìî²Ù×÷Ãû³Æ£©£®
£¨5£©ÏÂÁм¸ÏîʵÑé²½Ö裬¿ÉÓÃÓÚ¼ìÑéäåÒÒÍéÖÐäåÔªËØ£¬ÆäÕýÈ·µÄ²Ù×÷˳ÐòÊÇ£ºÈ¡ÉÙÁ¿äåÒÒÍ飬Ȼºó
¢Ü¢Ù¢Ý¢Û¢Ú
¢Ü¢Ù¢Ý¢Û¢Ú
£¨Ìî´úºÅ£©£®
¢Ù¼ÓÈÈ£»¢Ú¼ÓÈëAgNO3ÈÜÒº£»¢Û¼ÓÈëÏ¡HNO3Ëữ£»¢Ü¼ÓÈëNaOHÈÜÒº£»¢ÝÀäÈ´
£¨6£©äåÒÒÍ飨C2H5Br£©¶àÒ»¸ö̼µÄͬϵÎïËùÓпÉÄܵÄÎïÖÊÔÚÇâÑõ»¯ÄƵĴ¼ÈÜÒºÖÐÄÜ·¢Éú·´Ó¦£¬Ð´³ö¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ
CH3CH2CH2Br+NaOH
´¼
¡÷
CH3CH=CH2+NaBr+H2O
CH3CH2CH2Br+NaOH
´¼
¡÷
CH3CH=CH2+NaBr+H2O
£¬
CH3CHBrCH3+NaOH
´¼
¡÷
CH3CH=CH2+NaBr+H2O
CH3CHBrCH3+NaOH
´¼
¡÷
CH3CH=CH2+NaBr+H2O
£®
ÒÑÖª£ºCH3CH2OH+NaBr+H2SO4£¨Å¨£©
¡÷
CH3CH2Br+NaHSO4+H2O£®
ʵÑéÊÒÖÆ±¸äåÒÒÍ飨·ÐµãΪ38.4¡æ£©µÄ×°ÖúͲ½ÖèÈçÏ£º
¢Ù°´ÈçͼËùʾÁ¬½ÓÒÇÆ÷£¬¼ì²é×°ÖÃµÄÆøÃÜÐÔ£¬È»ºóÏòUÐιܺʹóÉÕ±­Àï¼ÓÈë±ùË®£»
¢ÚÔÚÔ²µ×ÉÕÆ¿ÖмÓÈë10mL95%ÒÒ´¼¡¢28mLŨÁòËᣬȻºó¼ÓÈëÑÐϸµÄ13gä廝įºÍ¼¸Á£Ëé´ÉƬ£»
¢ÛС»ð¼ÓÈÈ£¬Ê¹Æä³ä·Ö·´Ó¦£®
ÊԻشðÏÂÁÐÎÊÌ⣺
£¨1£©·´Ó¦Ê±Èôζȹý¸ß¿É¿´µ½Óкì×ØÉ«ÆøÌå²úÉú£¬¸ÃÆøÌåµÄ»¯Ñ§Ê½Îª
Br2
Br2
£®
£¨2£©·´Ó¦½áÊøºó£¬UÐιÜÖдÖÖÆµÄäåÒÒÍé³Êר»ÆÉ«£®½«UÐιÜÖеĻìºÏÎïµ¹Èë·ÖҺ©¶·ÖУ¬¾²Ö㬴ýÒºÌå·Ö²ãºó£¬·ÖÒº£¬È¡
ϲã
ϲã
£¨Ìî¡°Éϲ㡱»ò¡°Ï²㡱£©ÒºÌ壮ΪÁ˳ýÈ¥ÆäÖеÄÔÓÖÊ£¬¿ÉÑ¡ÔñÏÂÁÐÊÔ¼ÁÖеÄ
AC
AC
£¨ÌîÐòºÅ£©£®
A£®Na2SO3ÈÜÒº         B£®H2O         C£®Na2CO3ÈÜÒº      D£®CCl4
£¨3£©Òª½øÒ»²½ÖƵô¿¾»µÄC2H5Br£¬¿ÉÔÙÓÃˮϴ£¬È»ºó¼ÓÈëÎÞË®CaCl2¸ÉÔÔÙ½øÐÐ
ÕôÁó
ÕôÁó
£¨Ìî²Ù×÷Ãû³Æ£©£®
£¨4£©ÏÂÁм¸ÏîʵÑé²½Ö裬¿ÉÓÃÓÚ¼ìÑéäåÒÒÍéÖеÄäåÔªËØ£¬ÆäÕýÈ·µÄ²Ù×÷˳ÐòÊÇ£ºÈ¡ÉÙÁ¿äåÒÒÍ飬Ȼºó
¢Ü¢Ù¢Ý¢Û¢Ú
¢Ü¢Ù¢Ý¢Û¢Ú
£¨ÌîÐòºÅ£©£®
¢Ù¼ÓÈÈ   ¢Ú¼ÓÈëAgNO3ÈÜÒº   ¢Û¼ÓÈëÏ¡HNO3Ëữ   ¢Ü¼ÓÈëNaOHÈÜÒº   ¢ÝÀäÈ´
Çëд³öÔڴ˹ý³ÌÖУ¬ÓÐäåÒÒÍé²Î¼ÓµÄÖ÷Òª·´Ó¦µÄ»¯Ñ§·½³Ìʽ
CH3CH2Br+NaOH
¡÷
CH3CH2OH+NaBr
CH3CH2Br+NaOH
¡÷
CH3CH2OH+NaBr
£®
äåÒÒÍéµÄ·ÐµãÊÇ38.4¡æ£¬ÃܶÈÊÇ1.46g/cm3£®ÈçͼΪʵÑéÊÒÖÆ±¸äåÒÒÍéµÄ×°ÖÃʾÒâͼ£¨¼Ð³ÖÒÇÆ÷ÒÑÂÔÈ¥£©£¬ÆäÖÐGÖÐÊ¢ÕôÁóË®£®ÊµÑéʱѡÓõÄÒ©Æ·ÓУºä廝į¡¢95%ÒÒ´¼¡¢Å¨H2SO4£®ÖƱ¸¹ý³ÌÖб߷´Ó¦±ßÕôÁó£¬Õô³öµÄäåÒÒÍéÓÃË®ÏÂÊÕ¼¯·¨»ñµÃ£®
¿ÉÄÜ·¢ÉúµÄ¸±·´Ó¦£ºH2SO4£¨Å¨£©+2HBr¡úBr2+SO2+2H2O
£¨1£©Îª·ÀÖ¹¸±·´Ó¦µÄ·¢Éú£¬ÏòÔ²µ×ÉÕÆ¿ÄÚ¼ÓÈëҩƷʱ£¬»¹ÐèÊÊÁ¿µÄ
Ë®
Ë®
£®
£¨2£©·´Ó¦²ÉÓÃˮԡ¼ÓÈȵÄÄ¿µÄÊÇ£º¢Ù·´Ó¦ÈÝÆ÷ÊÜÈȾùÔÈ£¬¢Ú
¿ØÖÆ·´Ó¦Î¶ÈÔÚ100¡æÒÔÄÚ
¿ØÖÆ·´Ó¦Î¶ÈÔÚ100¡æÒÔÄÚ
£®Î¶ȼÆÏÔʾµÄζÈ×îºÃ¿ØÖÆÔÚ
38.4
38.4
¡æ×óÓÒ£»
£¨3£©×°ÖÃBµÄ×÷ÓÃÊÇ£º¢ÙʹäåÒÒÍéÕôÁó³ö£¬¢Ú
ÀäÄý»ØÁ÷ÒÒ´¼ºÍË®
ÀäÄý»ØÁ÷ÒÒ´¼ºÍË®
£»
£¨4£©²ÉÈ¡±ß·´Ó¦±ßÕôÁóµÄ²Ù×÷·½·¨¿ÉÒÔÌá¸ßÒÒ´¼µÄת»¯ÂÊ£¬Ö÷ÒªÔ­ÒòÊÇ
ÕôÁóÄܼ°Ê±·ÖÀë³öÉú³ÉÎ´Ù½øÆ½ºâÏòÓÒÒÆ¶¯
ÕôÁóÄܼ°Ê±·ÖÀë³öÉú³ÉÎ´Ù½øÆ½ºâÏòÓÒÒÆ¶¯
£»
£¨5£©äåÒÒÍé¿ÉÓÃË®ÏÂÊÕ¼¯·¨µÄÒÀ¾ÝÊÇ
²»ÈÜÓÚË®
²»ÈÜÓÚË®
¡¢
ÃܶȱÈË®´ó
ÃܶȱÈË®´ó
£¬½ÓÒº¹Ü¿ÚÇ¡ºÃ½þÈëµ½ÒºÃæµÄÄ¿µÄÊÇ
·ÀÖ¹µ¹Îü
·ÀÖ¹µ¹Îü
£»
£¨6£©´Ö²úÆ·ÓÃˮϴµÓºóÓлú²ãÈԳʺìרɫ£¬Óû³ýÈ¥¸ÃÔÓÖÊ£¬¼ÓÈëµÄ×îºÃÊÔ¼ÁΪ
B
B
£¨Ìî±àºÅ£©
A£®µâ»¯¼ØÈÜÒº     B£®ÑÇÁòËáÇâÄÆÈÜÒº    C£®ÇâÑõ»¯ÄÆÈÜÒº£®
´¼ÓëÇâ±Ëá·´Ó¦ÊÇÖÆ±¸Â±´úÌþµÄÖØÒª·½·¨£®ÊµÑéÊÒÖÆ±¸äåÒÒÍéºÍ1-äå¶¡ÍéµÄ·´Ó¦ÈçÏ£º
¢ÙNaBr+H2SO4=HBr+NaHSO4£»¢ÚR-OH+HBr?R-Br+H2O
¿ÉÄÜ´æÔڵĸ±·´Ó¦ÓУº¼ÓÈȹý³ÌÖз´Ó¦»ìºÏÎï»á³ÊÏÖ»ÆÉ«»òºìרɫ£»´¼ÔÚŨÁòËáµÄ´æÔÚÏÂÍÑË®Éú³ÉÏ©ºÍÃѵȣ®ÓйØÊý¾ÝÁбíÈçÏ£»
ÒÒ´¼ äåÒÒÍé Õý¶¡´¼ 1-äå¶¡Íé
ÃܶÈ/g?cm-3 0.7893 1.4604 0.8098 1.2758
·Ðµã/¡æ 78.5 38.4 117.2 101.6
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©äå´úÌþµÄË®ÈÜÐÔ
СÓÚ
СÓÚ
£¨Ìî¡°´óÓÚ¡±¡¢¡°µÈÓÚ¡±»ò¡°Ð¡ÓÚ¡±£©ÏàÓ¦µÄ´¼£¬½«1-äå¶¡Íé´Ö²úÆ·ÖÃÓÚ·ÖҺ©¶·ÖмÓË®£¬Õñµ´ºó¾²Ö㬲úÎïÔÚ
ϲã
ϲã
£¨Ìî¡°Éϲ㡱¡¢¡°Ï²㡱»ò¡°²»·Ö²ã¡±£©
£¨2£©ÖƱ¸²Ù×÷ÖУ¬¼ÓÈëµÄŨÁòËá±ØÐë½øÐÐÊʵ±µÄÏ¡ÊÍ£¬ÆäÄ¿µÄÊÇ
ab
ab
£¨Ìî×Öĸ£©
a£®¼õÉÙ¸±²úÎïÏ©ºÍÃѵÄÉú³Éb£®¼õÉÙBr2µÄÉú³É  c£®Ë®ÊÇ·´Ó¦µÄ´ß»¯¼Á
£¨3£©ÔÚÖÆ±¸äåÒÒÍéʱ£¬²ÉÓñ߷´Ó¦±ßÕô³ö²úÎïµÄ·½·¨£¬ÓÐÀûÓÚäåÒÒÍéµÄÉú³É£»µ«ÔÚÖÆ±¸1-äå¶¡Íéʱȴ²»Äܱ߷´Ó¦±ßÕô³ö²úÎÆäÔ­ÒòÊÇ
1-äå¶¡ÍéºÍÕý¶¡´¼µÄ·ÐµãÏà²î²»´ó
1-äå¶¡ÍéºÍÕý¶¡´¼µÄ·ÐµãÏà²î²»´ó

£¨4£©µÃµ½µÄäåÒÒÍéÖк¬ÓÐÉÙÁ¿ÒÒ´¼£¬ÎªÁËÖÆµÃ´¿¾»µÄäåÒÒÍ飬¿ÉÓÃÕôÁóˮϴµÓ£¬·ÖÒººó£¬ÔÙ¼ÓÈëÎÞË®CaCl2½øÐеÄʵÑé²Ù×÷ÊÇ
b
b
£¨Ìî×Öĸ£©
a£®·ÖÒº    b£®ÕôÁó     c£®ÝÍÈ¡     d£®¹ýÂË
£¨5£©ÎªÁ˼ìÑéäåÒÒÍéÖк¬ÓÐäåÔªËØ£¬Í¨³£²ÉÓõķ½·¨ÊÇÈ¡ÉÙÁ¿äåÒÒÍ飬Ȼºó
¢Ü¢Ù¢Û¢Ú
¢Ü¢Ù¢Û¢Ú
£¨°´ÊµÑéµÄ²Ù×÷˳ÐòÑ¡ÌîÏÂÁÐÐòºÅ£©
¢Ù¼ÓÈÈ    ¢Ú¼ÓÈëAgNO3 ¢Û¼ÓÈëÏ¡HNO3  ¢Ü¼ÓÈëNaOHÈÜÒº£®
´¼ÓëÇâ±Ëá·´Ó¦ÊÇÖÆ±¸Â±´úÌþµÄÖØÒª·½·¨£®ÊµÑéÊÒÖÆ±¸äåÒÒÍéºÍ1-äå¶¡ÍéµÄ·´Ó¦ÈçÏ£º
NaBr+H2SO4HBr+NaHSO4                 ¢Ù
R-OH+HBr?R-Br+H2O                     ¢Ú
¿ÉÄÜ´æÔڵĸ±·´Ó¦ÓУº´¼ÔÚŨÁòËáµÄ´æÔÚÏÂÍÑË®Éú³ÉÏ©ºÍÃÑ£¬Br-±»Å¨ÁòËáÑõ»¯ÎªBr2µÈ£®ÓйØÊý¾ÝÁбíÈçÏ£»
ÒÒ´¼ äåÒÒÍé Õý¶¡´¼ 1-äå¶¡Íé
ÃܶÈ/g?cm-3 0.7893 1.4604 0.8098 1.2758
·Ðµã/¡æ 78.5 38.4 117.2 101.6
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÔÚäåÒÒÍéºÍ1-äå¶¡ÍéµÄÖÆ±¸ÊµÑéÖУ¬ÏÂÁÐÒÇÆ÷×î²»¿ÉÄÜÓõ½µÄÊÇ
 
£®£¨Ìî×Öĸ£©
a£®Ô²µ×ÉÕÆ¿    b£®Á¿Í²    c£®×¶ÐÎÆ¿    d£®²¼ÊÏ©¶·
£¨2£©äå´úÌþµÄË®ÈÜÐÔ
 
£¨Ìî¡°´óÓÚ¡±¡¢¡°µÈÓÚ¡±»ò¡°Ð¡ÓÚ¡±£©ÏàÓ¦µÄ´¼£»ÆäÔ­ÒòÊÇ
 

 
£®
£¨3£©½«1-äå¶¡Íé´Ö²úÆ·ÖÃÓÚ·ÖҺ©¶·ÖмÓË®£¬Õñµ´ºó¾²Ö㬲úÎïÔÚ
 
£¨Ìî¡°Éϲ㡱¡¢¡°Ï²㡱»ò¡°²»·Ö²ã¡±£©£®
£¨4£©ÖƱ¸²Ù×÷ÖУ¬¼ÓÈëµÄŨÁòËá±ØÐë½øÐÐÏ¡ÊÍ£¬ÆäÄ¿µÄÊÇ
 
£®£¨Ìî×Öĸ£©
a£®¼õÉÙ¸±²úÎïÏ©ºÍÃѵÄÉú³É         b£®¼õÉÙBr2µÄÉú³É
c£®¼õÉÙHBrµÄ»Ó·¢               d£®Ë®ÊÇ·´Ó¦µÄ´ß»¯¼Á
£¨5£©Óû³ýÈ¥äå´úÍéÖеÄÉÙÁ¿ÔÓÖÊBr2£¬ÏÂÁÐÎïÖÊÖÐ×îÊʺϵÄÊÇ
 
£®£¨Ìî×Öĸ£©
a£®NaI    b£®NaOH    c£®NaHSO3    d£®KCl
£¨6£©ÔÚÖÆ±¸äåÒÒÍéʱ£¬²ÉÓñ߷´Ó¦±ßÕô³ö²úÎïµÄ·½·¨£¬ÆäÓÐÀûÓÚ
 
£»µ«ÔÚÖÆ±¸1-äå¶¡Íéʱȴ²»Äܱ߷´Ó¦±ßÕô³ö²úÎÆäÔ­ÒòÊÇ
 
£®

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

¾«Ó¢¼Ò½ÌÍø