ÌâÄ¿ÄÚÈÝ

ÒÑÖª£º¼×¡¢ÒÒ¡¢±ûΪ³£¼ûµ¥ÖÊ£¬¼×³£ÎÂÏÂΪ¹ÌÌ壬ÒÒΪһ»ÆÂÌÉ«ÆøÌ壬±ûΪÎÞÉ«ÆøÌ壮A¡¢B¡¢C¡¢D¾ùΪ»¯ºÏÎDΪһºìºÖÉ«³Áµí£¬ËüÃÇÖ®¼äÓÐÏÂͼת»¯¹Øϵ£®Çë»Ø´ðÒÔÏÂÎÊÌ⣮

£¨1£©Ð´³öÏÂÁÐÎïÖʵĻ¯Ñ§Ê½£»ÒÒ
Cl2
Cl2
£»B
FeCl2
FeCl2
£®
£¨2£©Ð´³ö¢Û·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º
4Fe£¨OH£©2+2H2O+O2=4Fe£¨OH£©3
4Fe£¨OH£©2+2H2O+O2=4Fe£¨OH£©3
£®
£¨3£©Ð´³ö¢Ú·´Ó¦µÄÀë×Ó·½³Ìʽ£º
Fe+2Fe3+=3Fe2+
Fe+2Fe3+=3Fe2+
£®
·ÖÎö£ºÒÀ¾ÝÒÒΪһ»ÆÂÌÉ«ÆøÌåËùÒÔΪÂÈÆø£¬DΪһºìºÖÉ«³Áµí˵Ã÷DΪÇâÑõ»¯Ìú£¬½áºÏת»¯Í¼DÊÇAºÍÇâÑõ»¯ÄÆ·´Ó¦Éú³É£¬ËùÒÔAÒ»¶¨ÊǺ¬ÌúµÄ»¯ºÏÎÍƶϼ×Ϊ½ðÊôÌú£¬AΪÂÈ»¯Ìú£»AºÍ¼×·´Ó¦Éú³ÉB£¬ÊÇÂÈ»¯ÌúºÍÌú·´Ó¦Éú³ÉÁËBΪÂÈ»¯ÑÇÌú£¬BºÍÇâÑõ»¯ÄÆ·´Ó¦Éú³ÉCΪÇâÑõ»¯ÑÇÌú³Áµí£¬CºÍ±ûÆøÌå·´Ó¦Éú³ÉDÇâÑõ»¯Ìú£»ËùÒÔ±ûΪÑõÆø£»×ÛºÏת»¯¹Øϵ·ÖÎöÍƶϸöÎïÖʵÄ×é³ÉºÍ·¢ÉúµÄ»¯Ñ§·´Ó¦£»
½â´ð£º½â£º±¾ÌâµÄ½âÌâ¹Ø¼üÊÇÒÒΪһ»ÆÂÌÉ«ÆøÌåÖ±½ÓÅжÏΪÂÈÆø£»DΪһºìºÖÉ«³ÁµíÅжÏΪÇâÑõ»¯Ìú£»½áºÏ¼×¡¢ÒÒ¡¢±ûΪ³£¼ûµ¥ÖÊ£¬¼×³£ÎÂÏÂΪ¹ÌÌ壬±ûΪÎÞÉ«ÆøÌ壮A¡¢B¡¢C¡¢D¾ùΪ»¯ºÏÎͨ¹ýת»¯¹ØϵͼÖðÒ»·ÖÎöÅжϸöÎïÖʵÄ×é³ÉºÍ·´Ó¦£»ÓÉDÊÇÇâÑõ»¯Ìú¿ÉÖª£¬A+NaOH¡úFe£¨OH£©3£¬ÍƶÏAÊǺ¬ÌúµÄÂÈ»¯Îȷ¶¨AΪFeCl3£¬¼×ΪFe£»×ª»¯Í¼ÖÐ A+¼×¡úB£¬ÊÇ·¢ÉúÁËFe+2Fe3+=3Fe2+£¬B+NaOH¡úC£¬·¢ÉúµÄÊÇFe2++2OH-=Fe£¨OH£©2¡ý£»C+±û¡úD£¬·¢ÉúµÄ·´Ó¦ÊÇ£º4Fe£¨OH£©2+2H2O+O2=4Fe£¨OH£©3£»
×ÛÉÏËùÊö£º×ª»¯¹ØϵÖеĸ÷ÎïÖÊ·Ö±ðΪ£¬
¼×£ºFe£»ÒÒ£ºCl2£»±û£ºO2£»A£ºFeCl3£»B£ºFeCl2£»C£ºFe£¨OH£©2£»D£ºFe£¨OH£©3£»
ÒÀ¾ÝÉÏÊö·ÖÎöºÍÅжϻشðÏÂÁÐÎÊÌ⣺
£¨1£©ÒҵĻ¯Ñ§Ê½ÎªCl2£»BµÄ»¯Ñ§Ê½FeCl2£»
¹Ê´ð°¸Îª£ºCl2    FeCl2     
£¨2£©µÚ¢Û·´Ó¦µÄ»¯Ñ§·½³ÌʽÊDZ»ÑõÆøÑõ»¯ÎªÇâÑõ»¯ÌúµÄ·´Ó¦£¬Ñõ»¯ÑÇÌú»¯Ñ§·½³ÌʽΪ4Fe£¨OH£©2+2H2O+O2=4Fe£¨OH£©3
¹Ê´ð°¸Îª£º4Fe£¨OH£©2+2H2O+O2=4Fe£¨OH£©3
£¨3£©µÚ¢Ú·´Ó¦µÄÀë×Ó·½³ÌʽÊÇÂÈ»¯ÌúºÍÌú·¢ÉúµÄÑõ»¯»¹Ô­·´Ó¦£¬ÈÜÒºÖз´Ó¦µÄÀë×Ó·½³ÌʽΪ£ºFe+2Fe3+=3Fe2+£»
¹Ê´ð°¸Îª£ºFe+2Fe3+=3Fe2+
µãÆÀ£º±¾Ì⿼²éÁËÂÈÆø¡¢Ìú¼°ÆäÌúµÄ»¯ºÏÎïµÄ »¯Ñ§ÐÔÖʵÄÓ¦ÓúÍÏÖÏóÅжϣ¬¹Ø¼üÊÇÎïÖʵÄÌØÊâÑÕÉ«ºÍÌØÕ÷·´Ó¦£¬ÕÒµ½ÍƶÏÎïÖʵÄÌâÑÛÊǽâ¾öÎÊÌâµÄÍ»ÆÆ¿Ú£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
A¡¢B¡¢C¡¢D¡¢E ÎåÖÖ¶ÌÖÜÆÚÖ÷×åÔªËØ·ÖÕ¼Èý¸öÖÜÆÚ£¬A¡¢B¡¢CΪͬһÖÜÆÚÒÀ´ÎÏàÁÚµÄ3ÖÖÔªËØ£¬AºÍCµÄÔ­×ÓÐòÊýÖ®±ÈΪ3£º4£¬EÔ­×ӵĵç×Ó²ãÊýµÈÓÚ×îÍâ²ãµç×ÓÊý£¬DµÄÔ­×ÓÐòÊýСÓÚE£®ÇëÓû¯Ñ§ÓÃÓï»Ø´ðÏà¹ØÎÊÌ⣺
£¨1£©AÔªËØÔÚÖÜÆÚ±íÖеÄλÖÃ
µÚ¶þÖÜÆÚ¢ôA×å
µÚ¶þÖÜÆÚ¢ôA×å
£®
£¨2£©±È½ÏCºÍE¼òµ¥Àë×Ӱ뾶´óС£º
O2-£¾Al3+
O2-£¾Al3+
£®
£¨3£©ÔªËØEµÄÒ»ÖÖ³£¼ûµÄ¿ÉÈÜÐÔÑÎÈÜÒº³Ê¼îÐÔ£¬ÆäÔ­ÒòÊÇ£¨ÓÃÀë×Ó·½³Ìʽ±íʾ£©£º
A1O2-+2H2O?Al£¨OH£©3+OH-
A1O2-+2H2O?Al£¨OH£©3+OH-
£»
X Y Z
¼× ÒÒ ±û
£¨4£©X¡¢Y¡¢Z¡¢¼×¡¢ÒÒ¡¢±ûÊÇÓÉA¡¢B¡¢C·Ö±ðÓëDÐγɵĻ¯ºÏÎÁùÖÖ»¯ºÏÎï¿ÉÒÔÅųÉÏÂ±í£¬ÆäÖÐͬһºáÐеķÖ×ÓÖеç×ÓÊýÏàͬ£¬Í¬Ò»×ÝÐеÄÎïÖÊËùº¬ÔªËØÖÖÀàÏàͬ£¬ÆäÖÐX¡¢Y¡¢¼×³£Î³£Ñ¹ÏÂΪÆøÌ壬Z¡¢ÒÒ¡¢±û³£Î³£Ñ¹ÏÂΪҺÌ壮
¢ÙÈô¼×µÄµç×ÓÊýÓëCH3OHÏàͬ£¬Ôò¼×µÄ·Ö×ÓʽΪ
C2H6
C2H6
£»±ûµÄµç×ÓʽΪ
£®
¢ÚÒҵķÖ×ÓʽΪB2D4£¬ÒҺͱû³£×÷»ð¼ýÍƽøÆ÷µÄȼÁÏ£¬·´Ó¦ºóµÄ²úÎïÎÞÎÛȾ£®ÒÑÖª8gҺ̬ÒÒÓë×ãÁ¿ÒºÌ¬±ûÍêÈ«·´Ó¦£¬²úÎï¾ùΪÆøÌåʱ£¬·Å³öÈÈÁ¿Îª160.35kJ£¬ÊÔд³ö¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ£º
N2H4£¨1£©+2H2O2£¨1£©=N2£¨g£©+4H2O£¨g£©¡÷H=-641.40kJ/mol
N2H4£¨1£©+2H2O2£¨1£©=N2£¨g£©+4H2O£¨g£©¡÷H=-641.40kJ/mol
£®
ÒÑÖªA¡¢B¡¢C¡¢D¡¢EÊǶÌÖÜÆÚÖÐÔ­×ÓÐòÊýÒÀ´ÎÔö´óµÄ5ÖÖÖ÷×åÔªËØ£¬ÆäÖÐÔªËØAµÄµ¥ÖÊÔÚ³£ÎÂϳÊÆø̬£¬ÔªËØBµÄÔ­×Ó×îÍâ²ãµç×ÓÊýÊÇÆäµç×Ó²ãÊýµÄ2±¶£¬ÔªËØCÔÚͬÖÜÆÚµÄÖ÷×åÔªËØÖÐÔ­×Ӱ뾶×î´ó£¬ÔªËØDµÄºÏ½ðÊÇÈÕ³£Éú»îÖг£ÓõĽðÊô²ÄÁÏ£®³£Î³£Ñ¹Ï£¬Eµ¥ÖÊÊǵ­»ÆÉ«¹ÌÌ壬³£ÔÚ»ðɽ¿Ú¸½½ü³Á»ý£®
£¨1£©B¡¢CÁ½ÖÖÔªËصÄÔªËØ·ûºÅ£ºB£º
 
£»  C£º
 
£®DÔÚÔªËØÖÜÆÚ±íÖеÄλÖÃΪ
 
£®
£¨2£©A2EµÄȼÉÕÈÈ¡÷H=-a kJ?mol-1£¬Ð´³öA2EȼÉÕ·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ£º
 
£®
£¨3£©CAµÄµç×ÓʽΪ
 
£»CAÓëË®·´Ó¦·Å³öÆøÌåµÄ»¯Ñ§·½³ÌʽΪ
 
£®
£¨4£©¼×¡¢ÒÒ¡¢±û·Ö±ðÊÇB¡¢D¡¢EÈýÖÖÔªËØ×î¸ß¼Ûº¬ÑõËáµÄÄÆÑΣ¬¼×¡¢ÒÒ¶¼ÄÜÓë±û·¢Éú·´Ó¦£¬ÇÒ±ûÓÃÁ¿²»Í¬£¬·´Ó¦µÄ²úÎﲻͬ£®»Ø´ðÎÊÌ⣺
¢ÙÏò±ûÈÜÒºÖлºÂýµÎ¼Ó¹ýÁ¿µÄÒÒÈÜÒº¹ý³ÌÖз¢Éú·´Ó¦µÄÀë×Ó·½³Ìʽ
 
£¬
 
£®
¢ÚÏò¼×ÈÜÒºÖлºÂýµÎ¼ÓµÈÎïÖʵÄÁ¿µÄ±ûÈÜÒººó£¬ËùµÃÈÜÒºÖÐÀë×ÓŨ¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòΪ
 
£®
£¨5£©ÓÃDµ¥ÖÊ×÷Ñô¼«£¬Ê¯Ä«×÷Òõ¼«£¬NaHCO3ÈÜÒº×÷µç½âÒº½øÐеç½â£¬Éú³ÉÄÑÈÜÎïR£¬RÊÜÈÈ·Ö½âÉú³É»¯ºÏÎïQ£®Ð´³öÑô¼«Éú³ÉRµÄµç¼«·´Ó¦Ê½£º
 
£»ÓÉRÉú³ÉQµÄ»¯Ñ§·½³Ìʽ£º
 
£®

¶ÌÖÜÆÚÔªËØR¡¢Q¡¢M¡¢TÔÚÔªËØÖÜÆÚ±íÖеÄÏà¶ÔλÖÃÈçÏÂ±í£¬ÒÑÖªRÔ­×Ó×îÍâ²ãµç×ÓÊýÓë´ÎÍâ²ãµç×ÓÊýÖ®±ÈΪ2£º1¡£

£¨1£©È˵ĺ¹ÒºÖÐ̨ÓÐTµÄ¼òµ¥Àë×Ó£¬ÆäÀë×ӽṹʾÒâͼΪ_________________£»ÔªËØMÔÚÔªËØÖÜÆÚ±íÖеÄλÖÃÊÇ___________________¡£

£¨2£©RµÄ×î¸ß¼ÛÑõ»¯ÎïËùº¬µÄ»¯Ñ§¼üÀàÐÍÊÇ__________¼ü(Ñ¡Ìî¡°Àë×Ó¡±»ò¡°¹²¼Û¡±)¡£

£¨3£©¼ÓÈÈʱ£¬QµÄ×î¸ß¼ÛÑõ»¯Îï¶ÔӦˮ»¯ÎïµÄŨÈÜÒºÓëµ¥ÖÊR·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ__________(ÓþßÌåµÄ»¯Ñ§Ê½±íʾ)¡£

£¨4£©ÔÚÒ»¶¨Ìõ¼þϼס¢ÒÒ¡¢±ûÓÐÈçÏÂת»¯£º¼×ÒÒ±û£¬ÈôÆäÖм×Êǵ¥ÖÊ£¬ÒÒ¡¢±ûΪ»¯ºÏÎxÊǾßÓÐÑõ»¯ÐÔµÄÎÞÉ«ÆøÌåµ¥ÖÊ£¬Ôò¼×µÄ»¯Ñ§×é³É²»¿ÉÄÜÊÇ________(Ñ¡ÌîÐòºÅ£¬ÏÂͬ)¡£

¢ÙR???? ¢ÚQ2?? ¢ÛM???? ¢ÜT2

£¨5£©¹¤ÒµÉÏ£¬³£ÀûÓá£ROÓëMO2·´Ó¦Éú³É¹Ì̬Mµ¥ÖʺÍRO2£¬´Ó¶øÏû³ýÕâÁ½ÖÖÆøÌå¶Ô´óÆøµÄÎÛȾ¡£

ÒÑÖª£º2RO(g)+O2(g)£½2RO2(g)? ¡÷H£½-akJ¡¤mol-l

M(s)+O2(g)£½MO2(g)????? ¡÷H£½-bkJ¡¤mol-l

Ôò·´Ó¦2RO(g)+MO2(g)£½2RO2(g)+M(s)?? ¡÷H£½___________¡£

£¨6£©ÔªËØTµÄº¬ÑõËáHTO¾ßÓÐƯ°×ÐÔ¡£Íù20mL 0.0lmol¡¤L-lµÄHTOÈÜÒºÖÐÖðµÎ¼ÓÈëÒ»¶¨Å¨¶ÈµÄÉÕ¼îÈÜÒº£¬²âµÃ»ìºÏÈÜÒºµÄζȱ仯ÈçÏÂͼËùʾ¡£¾Ý´ËÅжϣº¸ÃÉÕ¼îÈÜÒºµÄŨ¶ÈΪ______________mol¡¤L-l£»ÏÂÁÐÓйØ˵·¨ÕýÈ·µÄÊÇ_______________¡£

¢ÙHTOµÄµçÀë³£Êý£ºbµã£¾aµã

¢ÚÓÉË®µçÀë³öµÄc(OH¡ª)£ºbµã£¼cµã

¢Û´Óaµãµ½bµã£¬»ìºÏÈÜÒºÖпÉÄÜ´æÔÚ£ºc(TO¡ª)= c(Na+)

¢Ü´Óbµãµ½cµã£¬»ìºÏÈÜÒºÖÐÒ»Ö±´æÔÚ£ºc(Na£«)£¾c(TO£­)£¾c(OH£­)£¾c(H£«)

 

(16·Ö) ϱíΪԪËØÖÜÆÚ±íµÄÒ»²¿·Ö£¬Çë²ÎÕÕÔªËآ٣­¢àÔÚ±íÖеÄλÖã¬Óû¯Ñ§ÓÃÓï

»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©¢Ü¡¢¢Ý¡¢¢ÞµÄÔ­×Ӱ뾶ÓÉ´óµ½Ð¡µÄ˳ÐòΪ_______________________£¬¢Ú¡¢¢ß¡¢¢àµÄ×î¸ß¼Ûº¬ÑõËáµÄËáÐÔÓÉÇ¿µ½ÈõµÄ˳ÐòÊÇ______________________¡£

£¨2£©ÓɱíÖÐÁ½ÖÖÔªËصÄÔ­×Ó°´1£º1×é³ÉµÄ³£¼ûҺ̬»¯ºÏÎïµÄÏ¡ÒºÒ×±»´ß»¯·Ö½â£¬¿ÉʹÓõĴ߻¯¼ÁΪ£¨ÌîÐòºÅ£©_________________¡£

    a£®MnO2¡¡¡¡¡¡¡¡¡¡¡¡ b£®FeCl 3¡¡¡¡¡¡¡¡¡¡ c£®Na2SO3¡¡¡¡¡¡¡¡¡¡ d£®KMnO4

ÈôÒÑÖª1 ¿Ë¸ÃҺ̬»¯ºÏÎï·Ö½â³É¢ÛµÄµ¥ÖʺÍÒ»ÖÖ³£¼ûÒºÌåʱ£¬¿É·Å³ö2.9kJÄÜÁ¿£¬Ð´³ö¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ£º                                                     

£¨3£©¹¤ÒµÉϳ£²ÉÓõç½âAºÍ±ù¾§Ê¯(Na3AlF6)»ìºÏÎïµÄ·½·¨Ò±Á¶ÖƱ¸¢ÞµÄµ¥ÖÊ£¬Çë´ÓA¾§ÌåµÄÀàÐͺÍ΢Á£¼äµÄÏ໥×÷ÓÃÁ¦½Ç¶È½âÊͼӱù¾§Ê¯(Na3AlF6)µÄÔ­Òò                  ¡£                     ²¢Ð´³öµç½âʱµÄµç¼«·´Ó¦Ê½£º                                                   ¡£

£¨4£©¼×¡¢ÒÒ¡¢±ûÊÇÉÏÊö²¿·ÖÔªËØ×é³ÉµÄË«Ô­×Ó·Ö×Ó»ò¸º¶þ¼ÛË«Ô­×ÓÒõÀë×Ó£¬ÇҼס¢ÒÒ¡¢±ûµÄµç×Ó×ÜÊýÏàµÈ¡£¼×ÊÇÒ»ÖÖ¼«Ç¿µÄÑõ»¯ÐÔµ¥ÖÊ¡£±ûÓë¢ÝµÄÑôÀë×Ó¿ÉÐγÉÒ»ÖÖµ­»ÆÉ«¹ÌÌåB£¬¸Ã¹ÌÌå¸úË®·´Ó¦¿ÉµÃµ½¢ÛµÄµ¥ÖÊ¡£ÔòBµç×Óʽ           £¬ÒҵĽṹʽ      £¬¼×µÄ×é³ÉÔªËصÄÔ­×ӽṹʾÒâͼ             £¬ÄÜÖ¤Ã÷¼×µÄ×é³ÉÔªËطǽðÊôÐÔºÜÇ¿µÄÊÂʵ                                            ¡£(ÈξÙÒ»Àý¼´¿É)

 

ÒÑÖªA¡¢BΪ³£¼ûµÄ½ðÊôµ¥ÖÊ£¬C¡¢DΪ³£¼ûµÄ·Ç½ðÊôµ¥ÖÊ£¬¼×¡¢ÒÒ¡¢±ûΪÈýÖÖ³£¼ûµÄ»¯ºÏÎËüÃÇÖ®¼äµÄÏ໥ת»¯¹ØϵÈçÏÂͼËùʾ£¨²¿·Ö²úÎï¼°·´Ó¦Ìõ¼þûÓÐÁгö£©£º

£¨1£©Èô¼×ÊÇÖÐѧ»¯Ñ§Öг£¼ûµÄºì×ØÉ«·Ûĩ״¹ÌÌ壬Ôòµ¥ÖÊAÓëNaOHÈÜÒº·´Ó¦µÄÀë×Ó·½³ÌʽΪ                       £»BÔÚ³±ÊªµÄ¿ÕÆøÖÐÒ×·¢Éúµç»¯Ñ§¸¯Ê´£¬Ð´³öÆä·¢Éúµç»¯Ñ§¸¯Ê´Ê±Ô­µç³Ø¸º¼«µÄµç¼«·´Ó¦Ê½£º                               ¡£

£¨2£©ÈôÒÒÔÚË®ÈÜÒº³ÊÈõËáÐÔ£¬±ûÊÇÒ»ÖÖ´óÆøÎÛȾÎÓд̼¤ÐÔÆøζ¡£ÊÔд³öÒÒÔÚË®ÈÜÒºÖз¢ÉúµçÀëʱµÄµçÀë·½³Ìʽ                                              £»ÒÑÖª16g¹ÌÌåµ¥ÖÊDÍêȫȼÉÕת»¯³ÉÒÒʱ£¬·Å³ö148.4kJµÄÈÈÁ¿£¬Ôò¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪ                                          ¡£

£¨3£©ÈôÒÒÔÚË®ÈÜÒº³ÊÈõ¼îÐÔ£¬²¢¿ÉÓÃ×÷Å©Òµ»¯·Ê£¬D³£¿öÏÂΪÆøÌ壬Ôò½øÐÐÏÂÁÐÑо¿£º        ÏÖ½«0.40 mol CºÍ0.20 mol D³äÈë10LµÄÃܱÕÈÝÆ÷ÖУ¬ÔÚÒ»¶¨Ìõ¼þÏÂʹÆä·¢Éú·´Ó¦£¬ÓйØC¡¢D¡¢ÒÒÈýÕßµÄÎïÖʵÄÁ¿µÄ±ä»¯Óëʱ¼äµÄ¹ØϵÈçÓÒͼËùʾ£º

 ¢ÙÈôt1 = 10min£¬Ôò0ÖÁt1ʱ¼äÄÚCÎïÖʵÄƽ¾ù·´Ó¦ËÙÂÊΪ          £»¸Ã·´Ó¦ÔÚt2ʱ´ïµ½Æ½ºâ£¬Æ仯ѧ·´Ó¦·½³ÌʽΪ              £¬´ËζÈϵĸ÷´Ó¦µÄƽºâ³£ÊýΪ          ¡£

¢Ú¸ù¾ÝͼÖÐÇúÏ߱仯Çé¿öÅжϣ¬t1ʱ¿Ì¸Ä±äµÄ·´Ó¦Ìõ¼þ¿ÉÄÜÊÇ    £¨ÌîÏÂÁи÷ÏîÐòºÅ£©

a£®¼ÓÈëÁË´ß»¯¼Á    

b£®½µµÍÁË·´Ó¦µÄζȠ   

c£®ÏòÈÝÆ÷ÖÐÓÖ³äÈëÁËÆøÌåD

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

¾«Ó¢¼Ò½ÌÍø