ÌâÄ¿ÄÚÈÝ

ÏÂÁÐʵÑé²Ù×÷ÓëʵÑéÄ¿µÄ»ò½áÂÛÒ»ÖµÄÊÇ£¨¡¡¡¡£©
Ñ¡Ïî ʵÑé²Ù×÷ ʵÑéÄ¿µÄ»ò½áÂÛ
A ÏòijÈÜÒºÖÐÏȵÎÈë¼ÓÏ¡ÏõËáËữ£¬ÔٵμÓBaCl2ÈÜÒº£¬Óа×É«³ÁµíÉú³É ¸ÃÈÜÒºÖÐÒ»¶¨º¬ÓÐAg+
B ½«»ìºÏÆøÌåͨ¹ý±¥ºÍNa2CO3ÈÜÒº ³ýÈ¥CO2ÖлìÓÐHCl
C ÅäÖÆSnCl2ÈÜҺʱ£¬ÏȽ«SnCl2ÈÜÓÚÊÊÁ¿Ï¡ÑÎËᣬÔÙÓÃÕôÁóˮϡÊÍ£¬±£´æʱÔÚÊÔ¼ÁÆ¿ÖмÓÈëÉÙÁ¿µÄÎýÁ£ ÒÖÖÆSn2+Ë®½â£¬²¢·ÀÖ¹Sn2+±»Ñõ»¯ÎªSn4+
D ½«±½µÎÈëäåË®ÖУ¬Õñµ´£¬¾²Öã¬äåË®²ãÍÊÉ« äåºÍ±½·¢Éú¼Ó³É·´Ó¦
·ÖÎö£ºA£®¼ÓÏõËáËữ£¬ÈÜÒºÖдæÔÚÁòËá¸ùÀë×Ó»òÑÇÁòËá¸ùÀë×Ó¾ù²úÉúÏàͬµÄÏÖÏó£»
B£®¶þÑõ»¯Ì¼¡¢HCl¾ùÓë̼ËáÄÆÈÜÒº·´Ó¦£»
C£®¼ÓËáÒÖÖƽðÊôÀë×ÓµÄË®½â£¬¼ÓÉÙÁ¿µÄÎýÁ££¬¾ßÓл¹Ô­ÐÔ£¬·ÀÖ¹Sn2+±»Ñõ»¯£»
D£®äå²»Ò×ÈÜÓÚË®£¬Ò×ÈÜÓÚ±½£®
½â´ð£º½â£ºA£®¼ÓÏõËáËữ£¬ÈÜÒºÖдæÔÚÁòËá¸ùÀë×Ó»òÑÇÁòËá¸ùÀë×Ó£¬ÔٵμÓBaCl2ÈÜÒº£¬¾ùÓа×É«³ÁµíÉú³É£¬²»ÄÜÈ·¶¨º¬ÓÐAg+£¬Ó¦ÏȼÓÑÎËáËữ£¬¹ÊA´íÎó£»
B£®¶þÑõ»¯Ì¼¡¢HCl¾ùÓë̼ËáÄÆÈÜÒº·´Ó¦£¬Ôòѡ̼ËáÇâÄÆÈÜÒºÀ´³ýÈ¥CO2ÖлìÓÐHCl£¬¹ÊB´íÎó£»
C£®Òò½ðÊôÀë×ÓË®½â£¬ÇÒÒ×±»Ñõ»¯£¬Ôò¼ÓËáÒÖÖƽðÊôÀë×ÓµÄË®½â£¬¼ÓÉÙÁ¿µÄÎýÁ££¬¾ßÓл¹Ô­ÐÔ£¬·ÀÖ¹Sn2+±»Ñõ»¯£¬¹ÊCÕýÈ·£»
D£®äå²»Ò×ÈÜÓÚË®£¬Ò×ÈÜÓÚ±½£¬Ôò½«±½µÎÈëäåË®ÖУ¬Õñµ´£¬¾²Öã¬äåË®²ãÍÊÉ«£¬·¢ÉúÝÍÈ¡£¬ÎªÎïÀí±ä»¯£¬¹ÊD´íÎó£»
¹ÊÑ¡C£®
µãÆÀ£º±¾Ì⿼²éʵÑé·½°¸µÄÆÀ¼Û£¬Éæ¼°µ½Àë×ӵļìÑé¡¢ÝÍÈ¡¡¢ÎïÖʵijýÔÓ¡¢ÑÎÀàË®½â¡¢ÊÔ¼ÁµÄ´æ·ÅµÈ֪ʶµã£¬ÊôÓÚ³£¼ûµÄÀäÆ´ÊÔÌ⣬¿¼²éµã½Ï¶à£¬Ñ§ÉúӦעÒâ˼άµÄ¼°Ê±×ª»»À´½â´ð£¬AΪ½â´ðµÄÒ×´íµã£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
ijͬѧÀûÓÃÏÂÁÐ×°ÖÃʵÏÖÍ­ÓëŨÏõËᡢϡÏõËá·´Ó¦£¬¹ý³ÌÈçͼËùʾ£º
I£® È¡Ò»¶ÎÍ­Ë¿£¬ÓÃÏ¡ÁòËá³ýȥͭÐâ[Ö÷Òª³É·ÖÊÇCu2£¨OH£©2CO3]£®
¢ò£®½«Ï´µÓºóµÄÍ­Ë¿×ö³ÉÔÑÊý½Ï¶àµÄÂÝÐý×´£®
¢ó£®°´ÈçͼËùʾװÖÃÁ¬½ÓÒÇÆ÷¡¢¼ì²éÆøÃÜÐÔ¡¢×°È뻯ѧÊÔ¼Á£®
£¨1£©¹ý³ÌI·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ
Cu2£¨OH£©2CO3+4H+¨T2Cu2++CO2¡ü+3H2O
Cu2£¨OH£©2CO3+4H+¨T2Cu2++CO2¡ü+3H2O
£®
£¨2£©Ð´³ö¹ý³Ì¢óÖмì²éÆøÃÜÐԵķ½·¨
´ò¿ªbºÍa£¬¹Ø±Õ·ÖҺ©¶·µÄ»îÈû£¬½«×ó±ßµ¼¹Ü²åÈëʢˮµÄÉÕ±­ÖУ¬ÇáÇáÀ­¶¯×¢ÉäÆ÷»îÈû£¬Èôµ¼¹ÜÖÐÒºÃæÉÏÉýÔò˵Ã÷ÆøÃÜÐÔºÃ
´ò¿ªbºÍa£¬¹Ø±Õ·ÖҺ©¶·µÄ»îÈû£¬½«×ó±ßµ¼¹Ü²åÈëʢˮµÄÉÕ±­ÖУ¬ÇáÇáÀ­¶¯×¢ÉäÆ÷»îÈû£¬Èôµ¼¹ÜÖÐÒºÃæÉÏÉýÔò˵Ã÷ÆøÃÜÐÔºÃ
£®
£¨3£©¹ý³Ì¢óµÄºóÐø²Ù×÷ÈçÏ£º
¢Ù´ò¿ªÖ¹Ë®¼ÐaºÍb£¬ÇáÍÆ×¢ÉäÆ÷£¬Ê¹Å¨ÏõËáÓëÍ­Ë¿½Ó´¥£¬¹Û²ìµ½µÄÏÖÏóÊÇ
²úÉúºì×ØÉ«ÆøÌå
²úÉúºì×ØÉ«ÆøÌå
£¬Ò»¶Îʱ¼äºóʹ·´Ó¦Í£Ö¹µÄ²Ù×÷ÊÇ
ÇáÇὫעÉäÆ÷»îÈûÏòÓÒÀ­Ê¹Í­Ë¿ºÍÈÜÒº·Ö¿ª
ÇáÇὫעÉäÆ÷»îÈûÏòÓÒÀ­Ê¹Í­Ë¿ºÍÈÜÒº·Ö¿ª
£¬¹Ø±Õa£¬È¡ÏÂ×¢ÉäÆ÷£®
¢Ú´ò¿ªbºÍ·ÖҺ©¶·»îÈû£¬µ±²£Á§¹Ü³äÂúÏ¡ÏõËáºó£¬¹Ø±ÕbºÍ·ÖҺ©¶·»îÈû£¬´ò¿ªa£¬¹Û²ìµ½ÓÐÆøÅݲúÉú£®Ï¡ÏõËá³äÂú²£Á§¹ÜµÄʵÑéÄ¿ÊÇ
½«²£Á§¹ÜÖеÄNO2ºÍ¿ÕÆøÅųö
½«²£Á§¹ÜÖеÄNO2ºÍ¿ÕÆøÅųö
£¬¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ
3Cu+8H++2NO3-=3Cu2++2NO¡ü+4H2O
3Cu+8H++2NO3-=3Cu2++2NO¡ü+4H2O
£®
£¨4£©ÁíÈ¡3֧ʢÂúNO2ÆøÌåµÄСÊԹֱܷðµ¹ÖÃÔÚÊ¢Óг£ÎÂË®¡¢ÈÈË®ºÍ±ùË®µÄ3Ö»ÉÕ±­ÖУ¬·¢ÏÖÒºÃæÉÏÉýµÄ¸ß¶ÈÃ÷ÏÔ²»Ò»Ö£®½á¹ûÈç±íËùʾ£¨ºöÂÔζȶÔÆøÌåÌå»ýµÄÓ°Ï죩£º
ʵÑé±àºÅ Ë®ÎÂ/¡æ ÒºÃæÉÏÉý¸ß¶È
1 25 ³¬¹ýÊԹܵÄ
2
3
2 50 ²»×ãÊԹܵÄ
2
3
3 0 ÒºÃæÉÏÉý³¬¹ýʵÑé1
¢Ù¸ù¾ÝÉϱíµÃ³öµÄ½áÂÛÊÇζÈÔ½
µÍ
µÍ
£¨Ìî¡°¸ß¡±»ò¡°µÍ¡±£©£¬½øÈëÊÔ¹ÜÖеÄÈÜÒºÔ½¶à£®
¢Ú²éÔÄ×ÊÁÏ£º
a£®NO2ÓëË®·´Ó¦µÄʵ¼Ê¹ý³ÌΪ£º2NO2+H2O=HNO3+HNO2   3HNO2=HNO3+2NO¡ü+H2O£»
b£®HNO2²»Îȶ¨£®
Ôò²úÉúÉÏÊöÏÖÏóµÄÔ­ÒòÊÇ
ζȵͣ¬HNO2·Ö½âÁ¿¼õÉÙ£¬·Ö½â²úÉúµÄNOÆøÌåÁ¿¼õÉÙ£¬ËùÒÔ½øÈëÊԹܵÄÈÜÒº¶à
ζȵͣ¬HNO2·Ö½âÁ¿¼õÉÙ£¬·Ö½â²úÉúµÄNOÆøÌåÁ¿¼õÉÙ£¬ËùÒÔ½øÈëÊԹܵÄÈÜÒº¶à
£®
¾«Ó¢¼Ò½ÌÍøË×»°Ëµ£¬¡°³Â¾ÆÀÏ´×ÌرðÏ㡱£¬ÆäÔ­ÒòÊǾÆÔÚ´¢´æ¹ý³ÌÖÐÉú³ÉÁËÓÐÏãζµÄÒÒËáÒÒõ¥£¬ÔÚʵÑéÊÒÀïÎÒÃÇÒ²¿ÉÒÔÓÃÈçͼËùʾµÄ×°ÖÃÀ´Ä£Äâ¸Ã¹ý³Ì£®Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©±¥ºÍ̼ËáÄÆÈÜÒºµÄÖ÷Òª×÷ÓÃÊÇ
 
£®
£¨2£©×°ÖÃÖÐͨÕôÆøµÄµ¼¹ÜÖ»Äܲ嵽±¥ºÍ̼ËáÄÆÈÜÒºµÄÒºÃæ´¦£¬²»ÄܲåÈëÈÜÒºÖУ¬Ä¿
 
£¬³¤µ¼¹ÜµÄ×÷ÓÃÊÇ
 
£®
£¨3£©ÈôÒª°ÑÖƵõÄÒÒËáÒÒõ¥·ÖÀë³öÀ´£¬Ó¦²ÉÓõÄʵÑé²Ù×÷ÊÇ
 
£®
£¨4£©½øÐиÃʵÑéʱ£¬×îºÃÏòÊԹܼ×ÖмÓÈ뼸¿éËé´ÉƬ£¬ÆäÄ¿µÄÊÇ
 
£®
£¨5£©ÊµÑéÊÒ¿ÉÓÃÒÒ´¼À´ÖÆÈ¡ÒÒÏ©£¬½«Éú³ÉµÄÒÒϩͨÈëäåµÄËÄÂÈ»¯Ì¼ÈÜÒº£¬·´Ó¦ºóÉú³ÉÎïµÄ½á¹¹¼òʽÊÇ
 
£®
£¨6£©Éú³ÉÒÒËáÒÒõ¥µÄ·´Ó¦ÊÇ¿ÉÄæ·´Ó¦£¬·´Ó¦Îï²»ÄÜÍêȫת»¯ÎªÉú³ÉÎ·´Ó¦Ò»¶Îʱ¼äºó£¬¾Í´ïµ½Á˸÷´Ó¦µÄÏ޶ȣ¬¼´´ïµ½»¯Ñ§Æ½ºâ״̬£®ÏÂÁÐÃèÊöÄÜ˵Ã÷¸Ã·´Ó¦ÒÑ´ïµ½»¯Ñ§Æ½ºâ״̬µÄÊÇ£¨ÌîÐòºÅ£©
 
£®
¢Ùµ¥Î»Ê±¼äÀÉú³É1molÒÒËáÒÒõ¥£¬Í¬Ê±Éú³É1molË®
¢Úµ¥Î»Ê±¼äÀÉú³É1molÒÒËáÒÒõ¥£¬Í¬Ê±Éú³É1molÒÒËá
¢Ûµ¥Î»Ê±¼äÀÏûºÄ1molÒÒ´¼£¬Í¬Ê±ÏûºÄ1molÒÒËá
¢ÜÕý·´Ó¦µÄËÙÂÊÓëÄæ·´Ó¦µÄËÙÂÊÏàµÈ£®

(16·Ö)ijͬѧÀûÓÃÏÂÁÐ×°ÖÃʵÏÖÍ­ÓëŨÏõËᡢϡÏõËá·´Ó¦£¬¹ý³ÌÈçÏ£º
I£®È¡Ò»¶ÎÍ­Ë¿£¬ÓÃÏ¡ÁòËá³ýȥͭÐâ[Ö÷Òª³É·ÖÊÇCu2(OH)2CO3]¡£
¢ò£®½«Ï´µÓºóµÄÍ­Ë¿×ö³ÉÔÑÊý½Ï¶àµÄÂÝÐý×´¡£
III£®°´ÈçͼËùʾװÖÃÁ¬½ÓÒÇÆ÷¡¢¼ì²éÆøÃÜÐÔ¡¢×°È뻯ѧÊÔ¼Á¡£

(1)¹ý³ÌI·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ                      ¡£
(2)д³ö¹ý³ÌIIIÖмì²éÆøÃÜÐԵķ½·¨                     ¡£
¢Ç¹ý³ÌIIIµÄºóÐø²Ù×÷ÈçÏ£º
¢Ù´ò¿ªÖ¹Ë®¼ÐaºÍb£¬ÇáÍÆ×¢ÉäÆ÷£¬Ê¹Å¨ÏõËáÓëÍ­Ë¿½Ó´¥£¬¹Û²ìµ½µÄÏÖÏóÊÇ                    £¬Ò»¶Îʱ¼äºóʹ·´Ó¦Í£Ö¹µÄ²Ù×÷ÊÇ                              £¬¹Ø±Õa£¬È¡ÏÂ×¢ÉäÆ÷¡£
¢Ú´ò¿ªbºÍ·ÖҺ©¶·»îÈû£¬µ±²£Á§¹Ü³äÂúÏ¡ÏõËáºó£¬¹Ø±ÕbºÍ·ÖҺ©¶·»îÈû£¬´ò¿ªa£¬¹Û²ìµ½ÓÐÆøÅݲúÉú¡£Ï¡ÏõËá³äÂú²£Á§¹ÜµÄʵÑéÄ¿ÊÇ                           £¬¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ                         ¡£
(4)ÁíÈ¡3֧ʢÂúNO2ÆøÌåµÄСÊԹֱܷðµ¹ÖÃÔÚÊ¢Óг£ÎÂË®¡¢ÈÈË®ºÍ±ùË®µÄ3Ö»ÉÕ±­ÖУ¬·¢ÏÖÒºÃæÉÏÉýµÄ¸ß¶ÈÃ÷ÏÔ²»Ò»Ö¡£½á¹ûÈçϱíËùʾ(ºöÂÔζȶÔÆøÌåÌå»ýµÄÓ°Ïì)£º

¢Ù¸ù¾ÝÉϱíµÃ³öµÄ½áÂÛÊÇζÈÔ½             (Ìî¡°¸ß¡±»ò¡°µÍ¡±)£¬½øÈëÊÔ¹ÜÖеÄÈÜÒºÔ½¶à¡£
¢Ú²éÔÄ×ÊÁÏ£º
a£®NO2ÓëË®·´Ó¦µÄʵ¼Ê¹ý³ÌΪ£º2NO2+H2O=HNO3+HNO2  3HNO2 =HNO3+2NO­+H2O£»
b£®HNO2²»Îȶ¨¡£
Ôò²úÉúÉÏÊöÏÖÏóµÄÔ­ÒòÊÇ                                                            ¡£

(16·Ö)ijͬѧÀûÓÃÏÂÁÐ×°ÖÃʵÏÖÍ­ÓëŨÏõËᡢϡÏõËá·´Ó¦£¬¹ý³ÌÈçÏ£º

I£® È¡Ò»¶ÎÍ­Ë¿£¬ÓÃÏ¡ÁòËá³ýȥͭÐâ[Ö÷Òª³É·ÖÊÇCu2(OH)2CO3]¡£

¢ò£®½«Ï´µÓºóµÄÍ­Ë¿×ö³ÉÔÑÊý½Ï¶àµÄÂÝÐý×´¡£

III£®°´ÈçͼËùʾװÖÃÁ¬½ÓÒÇÆ÷¡¢¼ì²éÆøÃÜÐÔ¡¢×°È뻯ѧÊÔ¼Á¡£

(1)¹ý³ÌI·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ                       ¡£

(2)д³ö¹ý³ÌIIIÖмì²éÆøÃÜÐԵķ½·¨                      ¡£

¢Ç¹ý³ÌIIIµÄºóÐø²Ù×÷ÈçÏ£º

¢Ù´ò¿ªÖ¹Ë®¼ÐaºÍb£¬ÇáÍÆ×¢ÉäÆ÷£¬Ê¹Å¨ÏõËáÓëÍ­Ë¿½Ó´¥£¬¹Û²ìµ½µÄÏÖÏóÊÇ                     £¬Ò»¶Îʱ¼äºóʹ·´Ó¦Í£Ö¹µÄ²Ù×÷ÊÇ                               £¬¹Ø±Õa£¬È¡ÏÂ×¢ÉäÆ÷¡£

¢Ú´ò¿ªbºÍ·ÖҺ©¶·»îÈû£¬µ±²£Á§¹Ü³äÂúÏ¡ÏõËáºó£¬¹Ø±ÕbºÍ·ÖҺ©¶·»îÈû£¬´ò¿ªa£¬¹Û²ìµ½ÓÐÆøÅݲúÉú¡£Ï¡ÏõËá³äÂú²£Á§¹ÜµÄʵÑéÄ¿ÊÇ                            £¬¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ                          ¡£

(4)ÁíÈ¡3֧ʢÂúNO2ÆøÌåµÄСÊԹֱܷðµ¹ÖÃÔÚÊ¢Óг£ÎÂË®¡¢ÈÈË®ºÍ±ùË®µÄ3Ö»ÉÕ±­ÖУ¬·¢ÏÖÒºÃæÉÏÉýµÄ¸ß¶ÈÃ÷ÏÔ²»Ò»Ö¡£½á¹ûÈçϱíËùʾ(ºöÂÔζȶÔÆøÌåÌå»ýµÄÓ°Ïì)£º

¢Ù¸ù¾ÝÉϱíµÃ³öµÄ½áÂÛÊÇζÈÔ½              (Ìî¡°¸ß¡±»ò¡°µÍ¡±)£¬½øÈëÊÔ¹ÜÖеÄÈÜÒºÔ½¶à¡£

¢Ú²éÔÄ×ÊÁÏ£º

a£®NO2ÓëË®·´Ó¦µÄʵ¼Ê¹ý³ÌΪ£º2NO2+H2O=HNO3+HNO2   3HNO2 =HNO3+2NO­+H2O£»

b£®HNO2²»Îȶ¨¡£

Ôò²úÉúÉÏÊöÏÖÏóµÄÔ­ÒòÊÇ                                                             ¡£

 

ijͬѧÀûÓÃÏÂÁÐ×°ÖÃʵÏÖÍ­ÓëŨÏõËᡢϡÏõËá·´Ó¦£¬¹ý³ÌÈçÏ£º

I£® È¡Ò»¶ÎÍ­Ë¿£¬ÓÃÏ¡ÁòËá³ýȥͭÐâ[Ö÷Òª³É·ÖÊÇCu2(OH)2CO3]¡£

¢ò£®½«Ï´µÓºóµÄÍ­Ë¿×ö³ÉÔÑÊý½Ï¶àµÄÂÝÐý×´¡£

III£®°´ÈçͼËùʾװÖÃÁ¬½ÓÒÇÆ÷¡¢¼ì²éÆøÃÜÐÔ¡¢×°È뻯ѧÊÔ¼Á¡£

(1)¹ý³ÌI·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ                       ¡£

(2)д³ö¹ý³ÌIIIÖмì²éÆøÃÜÐԵķ½·¨                      ¡£

¢Ç¹ý³ÌIIIµÄºóÐø²Ù×÷ÈçÏ£º

¢Ù´ò¿ªÖ¹Ë®¼ÐaºÍb£¬ÇáÍÆ×¢ÉäÆ÷£¬Ê¹Å¨ÏõËáÓëÍ­Ë¿½Ó´¥£¬¹Û²ìµ½µÄÏÖÏóÊÇ          £¬Ò»¶Îʱ¼äºóʹ·´Ó¦Í£Ö¹µÄ²Ù×÷ÊÇ                               £¬¹Ø±Õa£¬È¡ÏÂ×¢ÉäÆ÷¡£

¢Ú´ò¿ªbºÍ·ÖҺ©¶·»îÈû£¬µ±²£Á§¹Ü³äÂúÏ¡ÏõËáºó£¬¹Ø±ÕbºÍ·ÖҺ©¶·»îÈû£¬´ò¿ªa£¬¹Û²ìµ½ÓÐÆøÅݲúÉú¡£Ï¡ÏõËá³äÂú²£Á§¹ÜµÄʵÑéÄ¿ÊÇ               £¬¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ                     ¡£

(4)ÁíÈ¡3֧ʢÂúNO2ÆøÌåµÄСÊԹֱܷðµ¹ÖÃÔÚÊ¢Óг£ÎÂË®¡¢ÈÈË®ºÍ±ùË®µÄ3Ö»ÉÕ±­ÖУ¬·¢ÏÖÒºÃæÉÏÉýµÄ¸ß¶ÈÃ÷ÏÔ²»Ò»Ö¡£½á¹ûÈçϱíËùʾ(ºöÂÔζȶÔÆøÌåÌå»ýµÄÓ°Ïì)£ºks5u

¢Ù¸ù¾ÝÉϱíµÃ³öµÄ½áÂÛÊÇζÈÔ½              (Ìî¡°¸ß¡±»ò¡°µÍ¡±)£¬½øÈëÊÔ¹ÜÖеÄÈÜÒºÔ½¶à¡£

¢Ú²éÔÄ×ÊÁÏ£ºa£®NO2ÓëË®·´Ó¦µÄʵ¼Ê¹ý³ÌΪ£º2NO2+H2O=HNO3+HNO2   

3HNO2 =HNO3+2NO­+H2O£»   b£®HNO2²»Îȶ¨¡£

Ôò²úÉúÉÏÊöÏÖÏóµÄÔ­ÒòÊÇ                                              ¡£

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

¾«Ó¢¼Ò½ÌÍø