(I) ÏÂͼ¼×ºÍÒÒÊÇijѧУÍõÀÏÊ¦ÖÆ±¸NH3 ²¢½øÐÐÐÔÖÊʵÑéʱµÄ¸Ä½ø×°Öᣰ´Í¼¼×°ÑÒÇÆ÷°²×°ºÃ£¬³ÆÈ¡2g ¹ÌÌåÂÈ»¯ï§×°ÈëÊԹܵײ¿£¬ÔÙ¿ìËÙ³ÆÈ¡2g ÇâÑõ»¯ÄƸ²¸ÇÔÚÂÈ»¯ï§ÉÏ·½£»Á¢¼´ÓôøÓеιܵÄÈû×ÓÈû½ô£¨µÎ¹ÜÔ¤ÏÈÎüÈëÔ¼2mL Ũ°±Ë®£©£»ÉÕ±ÄÚÊ¢µÎÓзÓ̪ÊÔÒºµÄË®£º°ÑŨ°±Ë®µÎÈëÊÔ¹ÜÀ¿ÉÁ¢¼´¹Û²ìµ½ÊÔ¹ÜÄÚ·¢Éú¾çÁÒ·´Ó¦£¬ÓдóÁ¿ÆøÅÝ.
![]()
½«ÉÏÊöÖÐÊÕÂúNH3 µÄÔ²µ×ÉÕÆ¿È¡Ï£¬¸Ä×°³ÉͼÒÒËùʾµÄ×°Ö㬽ºÍ·µÎ¹ÜÄÚÊÂÏÈÔ¤ÎüÈë2mLH2O £¬´ËÊ±Ð¡ÆøÇòϵÔÚ²£Á§°ôÉϳÊ×ÔÈ»ËɳÛ״̬£»½«µÎ¹ÜÄÚµÄË®ÂýÂýµÎÈëÉÕÆ¿ÖУ¬ÇáÇá»Î¶¯ÉÕÆ¿£¬Í¨¹ý¹Û²ìʵÑéÏÖÏó±ã¿ÉÒÔÑéÖ¤NH3 µÄij¸öÐÔÖÊ¡£°´ÒªÇ󻨴ðÏÂÁÐÎÊÌ⣺
(1)ÖÐѧ»¯Ñ§½Ì²ÄÖÐÊÇÓÃÓëÖÆO2ÏàͬµÄÖÆ±¸×°ÖÃÀ´¼ÓÈÈÖÆÈ¡NH3µÄ£¬¸Ã»¯Ñ§·½³ÌʽΪ:__________________________________________________________
(2)ÏÂÃæÊÇijͬѧ¹ØÓÚÍõÀÏʦÄÜÓÃͼ¼×ÖÆÈ¡NH3µÄÔÒò·ÖÎö£¬ÓеÀÀíµÄÊÇ__________¡£
¢ÙÔÚNH3 ¡¤ H2O ÖÐÓÐÆ½ºâ
ʹƽºâÏò×óÒÆ¶¯
¢ÚÔÚNH3¡¤ H2O ÖÐÓÐÆ½ºâ
ʹƽºâÏò×óÒÆ¶¯
¢ÛNa0H ÈÜÓÚˮʱ·ÅÈÈ£¬Ê¹ÌåϵµÄζÈÉý¸ß£¬NH3 µÄÈܽâ¶È¼õС
¢ÜNH4ClÓëNaOH ÔÚ´ËÇé¿öÏ¿ɷ´Ó¦Éú³ÉNH3£¬![]()
¢ÝNH4Cl »á·Ö½âÊͷųöNH3
(3)ͼ¼×ÖеÄNH4Cl ÓëNaOH ¹ÌÌå»ìºÏÎïÄÜ·ñÓÃCaO ¹ÌÌå´úÌæ (ÌÄÜ¡±Ó롱²»ÄÜ¡±)
(4)ÈçºÎÅжÏͼ¼×ÖÐÉÕÆ¿¼ºÊÕÂúNH3 ?_____________________________________________
(5)ͼÒÒÖнºÍ·µÎ¹ÜÖеÄË®¼·ÈëÉÕÆ¿ºó£¬¹Û²ìµ½µÄÏÖÏóÊÇ Ëü˵Ã÷ÁËNH3
![]()
(¢ò)ÈçÉÏͼËùʾ£ºÔÚB²ÛÖÐ×°ÓÐ500 mLË®£¬ÈÝ»ýΪa mLµÄÊÔ¹ÜA³äÂúÁËNO2ºÍNOµÄ»ìºÏÆøÌ壨±ê×¼×´¿ö£©£¬½«ÊÔ¹ÜAµ¹²åÈëB²ÛµÄË®ÖС£³ä·Ö·´Ó¦ºó£¬ÊÔ¹ÜAÖÐÓàÏÂÆøÌåµÄÌå»ýΪ0.5a mL£¬ÔòÔ»ìºÏÆøÌåÖÐNO2ºÍNOµÄÎïÖʵÄÁ¿Ö®±ÈΪ
ͨ¹ýµ¼Æø¹ÜCÍùÓàÏÂ0.5a mLÆøÌåµÄÊÔ¹ÜAÖгÖÐøÍ¨ÈëÑõÆø£¬AÖпÉÄܹ۲쵽µÄÏÖÏóÊÇ£º
___________________________________________________
Óйط´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º___________________________________________
µ±ÊÔ¹ÜAÖгäÂúÆøÌåʱֹͣͨÈëÑõÆø£¬È»ºó½«ÊÔ¹ÜÈ¡³öË®²Û£¬Ôò¹²Í¨ÈëÑõÆøµÄÌå»ýΪ ________mL£¬Ë®²ÛBÖÐÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈΪ mol¡¤L-1(ÉèÈÜÒºµÄÌå»ýÈÔΪ500 mL)
£¨20·Ö£©
£¨¢ñ£©£¨1£©
£¨2·Ö£©
£¨2£©¢Ù¢Ú¢Û¢Ü £¨2·Ö£© £¨3£©ÄÜ £¨1·Ö£© £¨4£© ÉÕ±ÄÚµÄÈÜÒº±äºì£¨1·Ö£©
£¨5£©ÆøÇòÖð½¥±ä´ó£¨1·Ö£©£»¼«Ò×ÈÜÓÚË®£¨1·Ö£©
£¨¢ò£©£¨1£©3£º1 £¨2·Ö£©
£¨2£©ÎÞÉ«ÆøÌå±äΪºì×ØÉ«ÆøÌ壨1·Ö£©£¬ÊÔ¹ÜÖÐÒºÃæ²»¶ÏÉÏÉýÖÁÈ«³äÂú£¨1·Ö£©£¬¼ÌÐøÍ¨ÈëÑõÆøºó£¬ÊÔ¹ÜÖÐÒºÃæÏ½µ£¨1·Ö£©£¬×îºó³äÂúÎÞÉ«ÆøÌ壨1·Ö£©¡£[2NO£«O2£½2NO2£¨1·Ö£© 3NO2£«H2O£½2HNO3£«NO £¨1·Ö£©]»ò 4NO£«3O2£«2H2O£½4HNO3 £¨2·Ö£©£¨3£©1.375a £¨2·Ö£© a/11200£¨2·Ö£©
| Äê¼¶ | ¸ßÖÐ¿Î³Ì | Äê¼¶ | ³õÖÐ¿Î³Ì |
| ¸ßÒ» | ¸ßÒ»Ãâ·Ñ¿Î³ÌÍÆ¼ö£¡ | ³õÒ» | ³õÒ»Ãâ·Ñ¿Î³ÌÍÆ¼ö£¡ |
| ¸ß¶þ | ¸ß¶þÃâ·Ñ¿Î³ÌÍÆ¼ö£¡ | ³õ¶þ | ³õ¶þÃâ·Ñ¿Î³ÌÍÆ¼ö£¡ |
| ¸ßÈý | ¸ßÈýÃâ·Ñ¿Î³ÌÍÆ¼ö£¡ | ³õÈý | ³õÈýÃâ·Ñ¿Î³ÌÍÆ¼ö£¡ |
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
£¨Ã¿¿Õ2·Ö£¬¹²22·Ö£©
I.ʵÑé²âµÃ0.01mol/LµÄKMnO4µÄÁòËáÈÜÒººÍ0.1mol/LµÄH2C2O4ÈÜÒºµÈÌå»ý»ìºÏºó£¬·´Ó¦ËÙÂʦÔ[mol/(L ¡¤ s)]Ó뷴Ӧʱ¼ät£¨s£©µÄ¹ØÏµÈçͼËùʾ¡£»Ø´ðÈçÏÂÎÊÌ⣺
£¨1£©¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º
£¨2£©0¡út2ʱ¼ä¶ÎÄÚ·´Ó¦ËÙÂÊÔö´óµÄÔÒòÊÇ£º £¬
£¨3£©t2¡útʱ¼ä¶ÎÄÚ·´Ó¦ËÙÂʼõСµÄÔÒòÊÇ£º £¬
£¨4£©Í¼ÖÐÒõÓ°²¿·Ö¡°Ãæ»ý¡±±íʾt1¡út3ʱ¼äÀï ¡£
£Á£®Mn2+ÎïÖʵÄÁ¿Å¨¶ÈµÄÔö´ó B£®Mn2+ÎïÖʵÄÁ¿µÄÔö¼Ó
C£®SO42-ÎïÖʵÄÁ¿Å¨¶È D£®MnO4-ÎïÖʵÄÁ¿Å¨¶ÈµÄ¼õС
II. Ϊ±È½ÏFe3+ºÍCu2+¶ÔH2O2·Ö½âµÄ´ß»¯Ð§¹û£¬Ä³»¯Ñ§Ñо¿Ð¡×éµÄͬѧ·Ö±ðÉè¼ÆÁËÈçÏÂͼ¼×¡¢ÒÒËùʾµÄʵÑé¡£Çë»Ø´ðÏà¹ØÎÊÌ⣺
£¨1£©¶¨ÐÔ·ÖÎö£ºÈçÉÏͼ¼×¿É¹Û²ì £¬¶¨ÐԱȽϵóö½áÂÛ¡£ÓÐͬѧÌá³ö½«FeCl3¸ÄΪFe2(SO4)3¸üΪºÏÀí£¬ÆäÀíÓÉÊÇ ¡£
£¨2£©¶¨Á¿·ÖÎö£ºÈçÉÏͼÒÒËùʾ£¬ÊµÑéʱ¾ùÒÔÉú³É40mLÆøÌåΪ׼£¬ÆäËû¿ÉÄÜÓ°ÏìʵÑéµÄÒòËØ¾ùÒѺöÂÔ¡£Í¼ÖÐÒÇÆ÷AµÄÃû³ÆÎª £¬ÊµÑéÖÐÐèÒª²âÁ¿µÄÊý¾ÝÊÇ
£¨3£©¼ÓÈë0.01mol MnO2·ÛÄ©ÓÚ60mL H2O2ÈÜÒºÖУ¬ÔÚ±ê×¼×´¿öÏ·ųöÆøÌåÌå»ýºÍʱ¼äµÄ¹ØÏµÈçͼËùʾ¡£Éè·Å³öÆøÌåµÄ×ÜÌå»ýΪV mL¡£
¢Ù·Å³öV/3 mLÆøÌåʱËùÐèʱ¼äΪ min¡£
¢Ú ¸ÃH2O2ÈÜÒºµÄŨ¶ÈΪ
¢ÛA¡¢B¡¢C¡¢D¸÷µã·´Ó¦ËÙÂÊ¿ìÂýµÄ˳ÐòΪ > > >
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£ººÓ±±Ê¡¼½ÖÝÖÐѧ10-11ѧÄê¸ßÒ»ÏÂѧÆÚÆÚÄ©¿¼ÊÔ£¨»¯Ñ§Àí£©B¾í ÌâÐÍ£ºÊµÑéÌâ
£¨Ã¿¿Õ2·Ö£¬¹²22·Ö£©
I.ʵÑé²âµÃ0.01mol/LµÄKMnO4µÄÁòËáÈÜÒººÍ0.1mol/LµÄH2C2O4ÈÜÒºµÈÌå»ý»ìºÏºó£¬·´Ó¦ËÙÂʦÔ[mol/(L ¡¤ s)]Ó뷴Ӧʱ¼ät£¨s£©µÄ¹ØÏµÈçͼËùʾ¡£»Ø´ðÈçÏÂÎÊÌ⣺![]()
£¨1£©¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º
£¨2£©0¡út2ʱ¼ä¶ÎÄÚ·´Ó¦ËÙÂÊÔö´óµÄÔÒòÊÇ£º £¬
£¨3£©t2¡útʱ¼ä¶ÎÄÚ·´Ó¦ËÙÂʼõСµÄÔÒòÊÇ£º £¬
£¨4£©Í¼ÖÐÒõÓ°²¿·Ö¡°Ãæ»ý¡±±íʾt1¡út3ʱ¼äÀï ¡£
£Á£®Mn2+ÎïÖʵÄÁ¿Å¨¶ÈµÄÔö´ó B£®Mn2+ÎïÖʵÄÁ¿µÄÔö¼Ó
C£®SO42-ÎïÖʵÄÁ¿Å¨¶È D£®MnO4-ÎïÖʵÄÁ¿Å¨¶ÈµÄ¼õС
II.Ϊ±È½ÏFe3+ºÍCu2+¶ÔH2O2·Ö½âµÄ´ß»¯Ð§¹û£¬Ä³»¯Ñ§Ñо¿Ð¡×éµÄͬѧ·Ö±ðÉè¼ÆÁËÈçÏÂͼ¼×¡¢ÒÒËùʾµÄʵÑé¡£Çë»Ø´ðÏà¹ØÎÊÌ⣺![]()
£¨1£©¶¨ÐÔ·ÖÎö£ºÈçÉÏͼ¼×¿É¹Û²ì £¬¶¨ÐԱȽϵóö½áÂÛ¡£ÓÐͬѧÌá³ö½«FeCl3¸ÄΪFe2(SO4)3¸üΪºÏÀí£¬ÆäÀíÓÉÊÇ ¡£
£¨2£©¶¨Á¿·ÖÎö£ºÈçÉÏͼÒÒËùʾ£¬ÊµÑéʱ¾ùÒÔÉú³É40mLÆøÌåΪ׼£¬ÆäËû¿ÉÄÜÓ°ÏìʵÑéµÄÒòËØ¾ùÒѺöÂÔ¡£Í¼ÖÐÒÇÆ÷AµÄÃû³ÆÎª £¬ÊµÑéÖÐÐèÒª²âÁ¿µÄÊý¾ÝÊÇ
£¨3£©¼ÓÈë0.01mol MnO2·ÛÄ©ÓÚ60mL H2O2ÈÜÒºÖУ¬ÔÚ±ê×¼×´¿öÏ·ųöÆøÌåÌå»ýºÍʱ¼äµÄ¹ØÏµÈçͼËùʾ¡£Éè·Å³öÆøÌåµÄ×ÜÌå»ýΪV mL¡£![]()
¢Ù·Å³öV/3 mLÆøÌåʱËùÐèʱ¼äΪ min¡£
¢Ú¸ÃH2O2ÈÜÒºµÄŨ¶ÈΪ
¢ÛA¡¢B¡¢C¡¢D¸÷µã·´Ó¦ËÙÂÊ¿ìÂýµÄ˳ÐòΪ > > >
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2012-2013ѧÄê¹ã¶«Ê¡¶«Ý¸ÊиßÈýÉÏѧÆÚµ÷ÑвâÊÔÀí×Û»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£º¼ÆËãÌâ
I£®¸ß¯Á¶ÌúÊÇÒ±Á¶ÌúµÄÖ÷Òª·½·¨£¬·¢ÉúµÄÖ÷Òª·´Ó¦Îª£º
Fe2O3(s) + 3CO(g)
2Fe(s)+3CO2(g)
¡÷H
£¨1£©ÒÑÖª£º¢ÙFe2O3(s) + 3C(ʯī)=2Fe(s) + 3CO(g) ¡÷H1
¢ÚC(ʯ+ CO2(g) = 2CO(g) ¡÷H2
Ôò¡÷H___________________(Óú¬¡÷H1 ¡¢¡÷H2µÄ´úÊýʽ±íʾ)¡£
£¨2£©¸ß¯Á¶Ìú·´Ó¦µÄƽºâ³£Êý±í´ïʽK=____________________________¡£
£¨3£©ÔÚijζÈʱ£¬¸Ã·´Ó¦µÄƽºâ³£ÊýK=64£¬ÔÚ2LºãÈÝÃܱÕÈÝÆ÷¼×ºÍÒÒÖУ¬·Ö±ð°´Ï±íËùʾ¼ÓÈëÎïÖÊ£¬·´Ó¦¾¹ýÒ»¶Îʱ¼äºó´ïµ½Æ½ºâ¡£
|
|
Fe2O3 |
CO |
Fe |
CO2 |
|
¼×/mol |
1.0 |
1.0 |
1.0 |
1.0 |
|
ÒÒ/mol |
1.0 |
1.5 |
1.0 |
1.0 |
¢Ù¼×ÈÝÆ÷ÖÐCOµÄƽºâת»¯ÂÊΪ_______________________¡£
¢ÚÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ____________________(Ìî±àºÅ£©¡£
A£®ÈôÈÝÆ÷ѹǿºã¶¨£¬·´Ó¦´ïµ½Æ½ºâ״̬
B£®ÈôÈÝÆ÷ÄÚÆøÌåÃܶȺ㶨£¬·´Ó¦´ïµ½Æ½ºâ״̬
C£®¼×ÈÝÆ÷ÖÐCOµÄƽºâת»¯ÂÊ´óÓÚÒÒµÄ
D£®Ôö¼ÓFe2O3¾ÍÄÜÌá¸ßCOµÄת»¯ÂÊ
II£®ÄÉÃ×MgO¿ÉÓÃÄòËØÓëÂÈ»¯Ã¾ºÏ³É¡£Ä³Ð¡×éÑо¿¸Ã·´Ó¦ÔÚζÈΪ378¡«398KʱµÄ·´Ó¦Ê±¼ä¡¢·´Ó¦ÎïÅä±ÈµÈÒòËØ¶ÔÆä²úÂʵÄÓ°Ïì¡£ÇëÍê³ÉÒÔÏÂʵÑéÉè¼Æ±í£º
|
񅧏 |
ζÈ/K |
·´Ó¦Ê±¼ä/h |
·´Ó¦ÎïÎïÖʵÄÁ¿Åä±È |
ʵÑéÄ¿µÄ |
|
¢Ù |
378 |
4 |
3¡Ã1 |
ʵÑé¢ÚºÍ¢Ü̽¾¿________ ______________________ ʵÑé¢ÚºÍ__________̽¾¿ ·´Ó¦Ê±¼ä¶Ô²úÂʵÄÓ°Ïì¡£ |
|
¢Ú |
378 |
4 |
4¡Ã1 |
|
|
¢Û |
378 |
3 |
_______ |
|
|
¢Ü |
398 |
4 |
4¡Ã1 |
ÏÂͼΪζȶÔÄÉÃ×MgO²úÂÊ£¨ÇúÏßa£©ºÍÁ£¾¶(ÇúÏßb)µÄÓ°Ï죬Çë¹éÄɳöζȶÔÄÉÃ×MgOÖÆ±¸µÄÓ°Ïì¹æÂÉ£¨Ð´³öÒ»Ìõ£©£º
___________________________________________¡£
![]()
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£ººÓ±±Ê¡10-11ѧÄê¸ßÒ»ÏÂѧÆÚÆÚÄ©¿¼ÊÔ£¨»¯Ñ§Àí£©B¾í ÌâÐÍ£ºÊµÑéÌâ
£¨Ã¿¿Õ2·Ö£¬¹²22·Ö£©
I.ʵÑé²âµÃ0.01mol/LµÄKMnO4µÄÁòËáÈÜÒººÍ0.1mol/LµÄH2C2O4ÈÜÒºµÈÌå»ý»ìºÏºó£¬·´Ó¦ËÙÂʦÔ[mol/(L ¡¤ s)]Ó뷴Ӧʱ¼ät£¨s£©µÄ¹ØÏµÈçͼËùʾ¡£»Ø´ðÈçÏÂÎÊÌ⣺
![]()
£¨1£©¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º
£¨2£©0¡út2ʱ¼ä¶ÎÄÚ·´Ó¦ËÙÂÊÔö´óµÄÔÒòÊÇ£º £¬
£¨3£©t2¡útʱ¼ä¶ÎÄÚ·´Ó¦ËÙÂʼõСµÄÔÒòÊÇ£º £¬
£¨4£©Í¼ÖÐÒõÓ°²¿·Ö¡°Ãæ»ý¡±±íʾt1¡út3ʱ¼äÀï ¡£
£Á£®Mn2+ÎïÖʵÄÁ¿Å¨¶ÈµÄÔö´ó B£®Mn2+ÎïÖʵÄÁ¿µÄÔö¼Ó
C£®SO42-ÎïÖʵÄÁ¿Å¨¶È D£®MnO4-ÎïÖʵÄÁ¿Å¨¶ÈµÄ¼õС
II. Ϊ±È½ÏFe3+ºÍCu2+¶ÔH2O2·Ö½âµÄ´ß»¯Ð§¹û£¬Ä³»¯Ñ§Ñо¿Ð¡×éµÄͬѧ·Ö±ðÉè¼ÆÁËÈçÏÂͼ¼×¡¢ÒÒËùʾµÄʵÑé¡£Çë»Ø´ðÏà¹ØÎÊÌ⣺
![]()
£¨1£©¶¨ÐÔ·ÖÎö£ºÈçÉÏͼ¼×¿É¹Û²ì £¬¶¨ÐԱȽϵóö½áÂÛ¡£ÓÐͬѧÌá³ö½«FeCl3¸ÄΪFe2(SO4)3¸üΪºÏÀí£¬ÆäÀíÓÉÊÇ ¡£
£¨2£©¶¨Á¿·ÖÎö£ºÈçÉÏͼÒÒËùʾ£¬ÊµÑéʱ¾ùÒÔÉú³É40mLÆøÌåΪ׼£¬ÆäËû¿ÉÄÜÓ°ÏìʵÑéµÄÒòËØ¾ùÒѺöÂÔ¡£Í¼ÖÐÒÇÆ÷AµÄÃû³ÆÎª £¬ÊµÑéÖÐÐèÒª²âÁ¿µÄÊý¾ÝÊÇ
£¨3£©¼ÓÈë0.01mol MnO2·ÛÄ©ÓÚ60mL H2O2ÈÜÒºÖУ¬ÔÚ±ê×¼×´¿öÏ·ųöÆøÌåÌå»ýºÍʱ¼äµÄ¹ØÏµÈçͼËùʾ¡£Éè·Å³öÆøÌåµÄ×ÜÌå»ýΪV mL¡£
![]()
¢Ù·Å³öV/3 mLÆøÌåʱËùÐèʱ¼äΪ min¡£
¢Ú ¸ÃH2O2ÈÜÒºµÄŨ¶ÈΪ
¢ÛA¡¢B¡¢C¡¢D¸÷µã·´Ó¦ËÙÂÊ¿ìÂýµÄ˳ÐòΪ > > >
²é¿´´ð°¸ºÍ½âÎö>>
¹ú¼ÊѧУÓÅÑ¡ - Á·Ï°²áÁбí - ÊÔÌâÁбí
ºþ±±Ê¡»¥ÁªÍøÎ¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨Æ½Ì¨ | ÍøÉÏÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | µçÐÅթƾٱ¨×¨Çø | ÉæÀúÊ·ÐéÎÞÖ÷ÒåÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | ÉæÆóÇÖȨ¾Ù±¨×¨Çø
Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com