¡¾ÌâÄ¿¡¿ÄøÓëVA×åÔªËØÐγɵĻ¯ºÏÎïÊÇÖØÒªµÄ°ëµ¼Ìå²ÄÁÏ£¬Ó¦ÓÃ×î¹ã·ºµÄÊÇÉ黯ïØ(GaAs)£¬»Ø´ðÏÂÁÐÎÊÌ⣺
(1)»ù̬GaÔ×ӵĺËÍâµç×ÓÅŲ¼Ê½Îª_____£¬»ù̬AsÔ×ÓºËÍâÓÐ_________¸öδ³É¶Ôµç×Ó¡£
(2)ïØʧȥµç×ÓµÄÖ𼶵çÀëÄÜ(µ¥Î»£ºkJ¡¤mol-1)µÄÊýÖµÒÀ´ÎΪ577¡¢1984.5¡¢2961.8¡¢6192ÓÉ´Ë¿ÉÍÆÖªïصÄÖ÷Òª»¯ºÏ¼ÛΪ____ºÍ+3¡£ÉéµÄµç¸ºÐÔ±ÈÄø____(Ìî¡°´ó¡±»ò¡°Ð¡¡±)¡£
(3)±È½ÏÏÂÁÐïصı»¯ÎïµÄÈÛµãºÍ·Ðµã£¬·ÖÎöÆä±ä»¯¹æÂɼ°ÔÒò£º________________________¡£
ïصı»¯Îï | GaCl3 | GaBr3 | GaI3 |
ÈÛµã/¡æ | 77.75 | 122.3 | 211.5 |
·Ðµã/¡æ | 201.2 | 279 | 346 |
GaF3µÄÈ۵㳬¹ý1000 ¡æ£¬¿ÉÄܵÄÔÒò___________________________________________¡£
(4)¶þË®ºÏ²ÝËáïصĽṹÈçͼËùʾ£¬ÆäÖÐïØÔ×ÓµÄÅäλÊýΪ______£¬²ÝËá¸ùÖÐ̼Ô×ÓµÄÔÓ»¯·½Ê½Îª______________¡£
(5)É黯ïØÈÛµãΪ1238¡æ£¬Á¢·½¾§°û½á¹¹ÈçͼËùʾ£¬¾§°û²ÎÊýΪa=565 pm¡£¸Ã¾§ÌåµÄÀàÐÍΪ_________£¬¾§ÌåµÄÃܶÈΪ___________(ÉèNAΪ°¢·ü¼ÓµÂÂÞ³£ÊýµÄÊýÖµ,ÁгöËãʽ¼´¿É)g¡¤cm-3¡£
¡¾´ð°¸¡¿[Ar]3d104s24p1»ò1s22s22p63s23p63d104s24p1 3 +1 ´ó GaCl3¡¢GaBr3¡¢GaI3µÄÈ۷еãÒÀ´ÎÉí¸ß£¬ËüÃǾùΪ·Ö×Ó¾§Ì壬½á¹¹ÏàËÆ£¬Ïà¶Ô·Ö×ÓÖÊÁ¿ÒÀ´ÎÔö´ó£¬·Ö×Ó¼ä×÷ÓÃÁ¦ÒÀ´ÎÔöÇ¿ GaF3ΪÀë×Ó¾§Ìå 4 sp2 Ô×Ó¾§Ìå g/cm3
¡¾½âÎö¡¿
(1)GaµÄÔ×ÓÐòÊýΪ31£¬Æä»ù̬Ô×ӵĵç×ÓʽÅŲ¼Ê½Îª£º[Ar]3d104s24p1»ò1s22s22p63s23p63d104s24p1£¬AsµÄÔ×ÓÐòÊýΪ33£¬ÔòAsµÄ»ù̬Ô×ӵĵç×ÓÅŲ¼Ê½Îª£º[Ar]3d104s24p3£¬ËùÒÔ»ù̬AsÔ×ÓºËÍâÓÐ3¸öδ³É¶Ôµç×Ó£¬
¹Ê´ð°¸Îª£º[Ar]3d104s24p1»ò1s22s22p63s23p63d104s24p1£¬3£»
(2)µçÀëÄÜÊÇÆø̬Ô×Óʧȥµç×ÓËùÐèÒªµÄÄÜÁ¿£¬ÓÉïصÄÇ°Ëļ¶µçÀëÄÜ¿ÉÖª£¬ÆäÖ÷Òª»¯ºÏ¼ÛΪ+1£¬+3£¬ÓÉÓÚAsµÄ×îÍâ²ãµç×ÓÅŲ¼Îª4s24p3£¬ÊÇ°ëÂúÎȶ¨×´Ì¬£¬¶øGaµÄ×îÍâ²ãµç×ÓÅŲ¼Îª4s24p1ÌرðÊÇ4p1Ò×ʧµç×Ó£¬ËùÒÔAsµÄµç¸ºÐÔ±ÈGa´ó£¬
¹Ê´ð°¸Îª£º+1£¬´ó£»
(3)±íÖÐÊý¾Ý¿ÉÖª£¬ïصı»¯ÎïµÄÈÛµãºÍ·Ðµã¶¼²»¸ß£¬ÇÒ°´ÕÕÂÈ¡¢äå¡¢µâÒÀ´ÎÉý¸ß£¬ÓÉÓÚËüÃÇ×é³ÉÏàͬ£¬½á¹¹ÏàËÆ£¬¶¼ÊÇ·Ö×Ó¾§Ì壬ËùÒÔËæ×ÅÏà¶Ô·Ö×ÓÖÊÁ¿µÄÔö´ó£¬·Ö×Ó¼ä×÷ÓÃÁ¦Ôö´ó£¬¹ÊÈ۷еãÉý¸ß£¬GaF3µÄÈ۵㳬¹ý1000¡æ£¬ÊÇÓÉÓÚFµÄµç¸ºÐԺܴó£¬ÐγɵÄGaF3ÊÇÀë×Ó¾§Ì壬
¹Ê´ð°¸Îª£ºGaCl3¡¢GaBr3¡¢GaI3µÄÈ۷еãÒÀ´ÎÉí¸ß£¬ËüÃǾùΪ·Ö×Ó¾§Ì壬½á¹¹ÏàËÆ£¬Ïà¶Ô·Ö×ÓÖÊÁ¿ÒÀ´ÎÔö´ó£¬·Ö×Ó¼ä×÷ÓÃÁ¦ÒÀ´ÎÔöÇ¿£»GaF3ΪÀë×Ó¾§Ì壻
(4)ÓɶþË®ºÏ²ÝËáïصĽṹͼ¿ÉÖª£¬ïØÔ×ÓµÄÅäλÊýΪ4£¬²ÝËá¸ùÖÐ̼Ô×ÓÓëôÈ»ùÖеÄ̼Ô×ÓµÄÔÓ»¯·½Ê½Ïàͬ£¬ÐγɵĶ¼ÊÇƽÃæ½á¹¹£¬ËùÒÔÓ¦¸ÃÊÇsp2ÔÓ»¯£»
¹Ê´ð°¸Îª£º4£¬sp2£»
(5)ÓÉÓڸþ§ÌåµÄÈÛµã¸ß£¬ÇÒÉéºÍïض¼²»ÊÇ»îÆÃÔªËØ£¬ËùÒԸþ§ÌåÊÇÔ×Ó¾§Ì壬Æ仯ѧʽΪGa4As4£¬¸Ã¾§ÌåµÄÖÊÁ¿m=g£¬Ìå»ýΪV=£¨565¡Á10-10£©3cm3£¬ÔòÆäÃܶÈΪ£ºg/cm3£¬
¹Ê´ð°¸Îª£ºÔ×Ó¾§Ì壬g/cm3¡£
Ä꼶 | ¸ßÖÐ¿Î³Ì | Ä꼶 | ³õÖÐ¿Î³Ì |
¸ßÒ» | ¸ßÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ | ³õÒ» | ³õÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ |
¸ß¶þ | ¸ß¶þÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õ¶þ | ³õ¶þÃâ·Ñ¿Î³ÌÍƼö£¡ |
¸ßÈý | ¸ßÈýÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õÈý | ³õÈýÃâ·Ñ¿Î³ÌÍƼö£¡ |
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÓлúÎïHÊÇÒ»ÖÖÖØÒªµÄ¸ß·Ö×Ó»¯ºÏÎÆäºÏ³É·ÏßÈçÏ£º
ÒÑÖª£º
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©AµÄÃû³ÆÊÇ_______________£¬CÖк¬Ñõ¹ÙÄÜÍÅÃû³ÆÊÇ______________¡£
£¨2£©Ð´³ö·´Ó¦ÀàÐÍ£ºA¡úB______________£¬C¡úD__________________¡£
£¨3£©B¡úCµÄ·´Ó¦ÊÔ¼ÁºÍ·´Ó¦Ìõ¼þÊÇ______________________¡£
£¨4£©D+E¡úFµÄ·´Ó¦·½³ÌʽÊÇ_________________¡£
£¨5£©GµÄ·Ö×ÓʽÊÇ____________________¡£
£¨6£©Âú×ãÏÂÁÐÌõ¼þµÄFµÄͬ·ÖÒì¹¹Ìå¹²ÓÐ__________ÖÖ(²»¿¼ÂÇÁ¢ÌåÒì¹¹)¡£
a.±½»·ÉÏÓÐÁ½¸öÈ¡´ú»ù£¬ÎÞÆäËû»·×´½á¹¹£»b.º¬Ì¼Ì¼Èý¼ü£¬ÎÞ-C¡ÔCOH½á¹¹¡£
£¨7£©¶à»·»¯ºÏÎïÊÇÓлúÑо¿µÄÖØÒª·½Ïò£¬ÇëÉè¼ÆÓÉ¡¢.CH3CHO¡¢-CHOºÏ³É¶à»·»¯ºÏÎïµÄ·Ïß(ÎÞ»úÊÔ¼ÁÈÎÑ¡)______________
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿·ÏÄø´ß»¯¼ÁÖÐÖ÷ÒªÓÐNi£¬»¹ÓÐAl¡¢Fe¡¢SiO2¼°ÆäËü²»ÈÜÓÚËá¡¢¼îµÄÔÓÖÊ¡£ÏÖÓ÷ÏÄø´ß»¯¼ÁÖƱ¸NiSO4¡¤7H2O¾§Ì壬ÆäÁ÷³ÌÈçÏÂͼ£º
ÒÑÖª£ºKsp[Fe(OH)3]=8.0¡Á10£38£¬Ka sp[Fe(OH)2]=8.0¡Á10£16£¬K[Al(OH)3]=3.2¡Á10£34£¬Ksp[Ni(OH)2]=2.0¡Á10£15£¬1g2=0.3£¬
»Ø´ðÏÂÁÐÎÊÌâ
£¨1£©¡°¼î½þ¡±µÄÄ¿µÄÊÇ______¡£
£¨2£©¡°Ëá½þʱ·¢ÉúµÄÀë×Ó·½³ÌʽΪFe+2H+=Fe2++H2¡ü¡¢_______¡£
£¨3£©¡°¾»»¯³ýÔÓÐèÒªÏȼÓÈëH2O2ÈÜÒº£¬·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ______£¬È»ºóµ÷½ÚPH=_______ʱ£¬Ê¹ÔÓÖÊÀë×ÓÇ¡ºÃÍêÈ«³Áµí(×¢£ºÀë×ÓŨ¶ÈСÓÚ»òµÈÓÚ1¡Á10£5mol/Lʱ³ÁµíÍêÈ«)
£¨4£©¡°²Ù×÷AΪ¡±______¡£
£¨5£©Ni2+ÔÚÇ¿¼îÐÔÈÜÒºÖл¹¿É±»NaClOÑõ»¯ÎªNiOOH£¬¸Ã·´Ó¦Àë×Ó·½³ÌʽΪ______¡£
£¨6£© NiOOHÒ²¿É×÷ΪԵç³ØµÄµç¼«²ÄÁÏ£¬ÈôÔÚ¼îÐÔÌõ¼þÏÂÐγÉȼÁϵç³Ø£¬¸º¼«Í¨ÈëN2H4ÆøÌ壬Ôò¸º¼«µÄµç¼«·´Ó¦Ê½Îª______¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿´Ó¹ÅÖÁ½ñ£¬Ìú¼°Æ仯ºÏÎïÔÚÈËÀàÉú²úÉú»îÖеÄ×÷Ó÷¢ÉúÁ˾޴ó¸Ä±ä¡£
(1)¹Å´úÖйúËÄ´ó·¢Ã÷Ö®Ò»µÄÖ¸ÄÏÕëÊÇÓÉÌìÈ»´ÅʯÖƳɵÄ,ÆäÖ÷Òª³É·ÖÊÇ____(Ìî×ÖĸÐòºÅ)¡£
a£®Fe b£®FeO c£®Fe3O4 d£®Fe2O3
(2)ÁòËáÔüµÄÖ÷Òª»¯Ñ§³É·ÖΪ£ºSiO2Ô¼45%£¬Fe2O3Ô¼40%£¬Al2O3Ô¼10%£¬MgOÔ¼5%¡£Óø÷ÏÔüÖÆÈ¡Ò©Óø¨ÁϺìÑõ»¯ÌúµÄ¹¤ÒÕÁ÷³ÌÈçÏÂ(²¿·Ö²Ù×÷ºÍÌõ¼þÂÔ)£º
¢ÙÔÚ²½Öè¢ñÖвúÉúµÄÓж¾ÆøÌå¿ÉÄÜÓÐ_____________________¡£
¢ÚÔÚ²½Öè¢ó²Ù×÷ÖУ¬Òª³ýÈ¥µÄÀë×Ó֮һΪAl3+¡£Èô³£ÎÂʱKsp[Al(OH)3]=1.0¡Á10-32£¬´ËʱÀíÂÛÉϽ«Al3+ ³ÁµíÍêÈ«£¬ÔòÈÜÒºµÄpHΪ_________¡£(c(Al3+)¡Ü1.0¡Á10-5mol/L ÊÓΪAl3+³ÁµíÍêÈ«)
¢Û²½Öè¢ôÖУ¬Éú³ÉFeCO3µÄÀë×Ó·½³ÌʽÊÇ____________________________¡£
(3)ÂÈ»¯ÌúÈÜÒº³ÆΪ»¯Ñ§ÊÔ¼ÁÖеġ°¶àÃæÊÖ¡±¡£ÏòÂÈ»¯ÍºÍÂÈ»¯ÌúµÄ»ìºÏÈÜÒºÖмÓÈëÑõ»¯Í·ÛÄ©»á²úÉú³Áµí£¬Ð´³ö¸Ã³ÁµíµÄ»¯Ñ§Ê½______________¡£ÇëÓÃƽºâÒƶ¯µÄÔÀí£¬½áºÏ±ØÒªµÄÀë×Ó·½³Ìʽ£¬¶Ô´ËÏÖÏó×÷³ö½âÊÍ£º__________________________________________________________¡£
(4)¢Ù¹ÅÀ϶øÉñÆæµÄÀ¶É«È¾ÁÏÆÕ³ʿÀ¶µÄºÏ³É·½·¨ÈçÏ£º
¸´·Ö½â·´Ó¦¢òµÄÀë×Ó·½³ÌʽÊÇ____________________________________________¡£
¢ÚÈç½ñ»ùÓÚÆÕ³ʿÀ¶ºÏ³ÉÔÀí¿É¼ì²âʳƷÖÐÊÇ·ñº¬CN-£¬·½°¸ÈçÏ£º
ÈôÊÔÖ½½»À¶ÔòÖ¤Ã÷ʳƷÖк¬ÓÐCN-£¬Çë½âÊͼì²âʱÊÔÖ½ÖÐFeSO4µÄ×÷Óãº____________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿µâ»¯¼Ø¿ÉÓÃÓÚÓлúºÏ³É¡¢ÖÆÒ©µÈ£¬ËüÊÇÒ»ÖÖ°×É«Á¢·½½á¾§»ò·ÛÄ©£¬Ò×ÈÜÓÚË®¡£ÊµÑéÊÒÖƱ¸KI µÄʵÑé×°ÖÃÈçÏÂ
ʵÑé²½ÖèÈçÏ£º
¢ÙÔÚÉÏͼËùʾµÄCÖмÓÈë 127g ÑÐϸµÄµ¥ÖÊI2 ºÍ 195g 30%µÄ KOH ÈÜÒº£¬¾çÁÒ½Á°è¡£
¢ÚI2ÍêÈ«·´Ó¦ºó£¬´ò¿ªµ¯»É¼ÐÏòÆäÖÐͨÈë×ãÁ¿µÄ H2S¡£
¢Û½«×°Öà C ÖÐËùµÃÈÜÒºÓÃÏ¡ H2SO4Ëữºó£¬ÖÃÓÚˮԡÉϼÓÈÈ 10min¡£
¢ÜÔÚ×°Öà C µÄÈÜÒºÖмÓÈë BaCO3 £¬³ä·Ö½Á°èºó£¬¹ýÂË¡¢Ï´µÓ¡£
¢Ý½«ÂËÒºÓÃÇâµâËáËữ£¬Õô·¢Å¨ËõÖÁ±íÃæ³öÏֽᾧĤ£¬ ¡¢ ¡¢Ï´µÓ¡¢¸ÉÔï¡£
¢ÞµÃ²úÆ· 145g¡£
»Ø´ðÏÂÁÐÎÊÌ⣻
(1)²½Öè¢Ù½«µâÑÐϸµÄÄ¿µÄÊÇ_____________________________________¡£
(2)×°ÖÃA Öз¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ_______________________________ £»×°Öà B ÖеÄÊÔ¼ÁÊÇ____________________£¬C×°ÖõÄÃû³ÆÊÇ_____________¡£
(3)×°Öà C ÖÐI2Óë KOH ·´Ó¦²úÎïÖ®Ò»ÊÇ KIO3 £¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ_______________¡£
(4)²½Öè¢ÜµÄÄ¿µÄÊÇ___________________________________ ¡£
(5)²¹³äÍêÕû²½Öè¢Ý __________________¡¢_______________¡£
(6)±¾´ÎʵÑé²úÂÊΪ__________________ (±£ÁôËÄλÓÐЧÊý×Ö)¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÀûÓÃĤ¼¼ÊõÔÀíºÍµç»¯Ñ§ÔÀíÖƱ¸ÉÙÁ¿ÁòËáºÍÂÌÉ«Ïõ»¯¼ÁN2O5£¬×°ÖÃÈçͼËùʾ£¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£¨ £©
A. µç¼«b·´Ó¦Ê½ÊÇO2+4e-+2H2O=4OH-
B. µç½âºóÒÒ×°ÖÃdµç¼«¸½½üÈÜÒºµÄpH²»±ä
C. cµç¼«Éϵĵ缫·´Ó¦Ê½ÎªN2O4-2e-+H2O=N2O5+2H+
D. ¼×ÖÐÿÏûºÄ1mol SO2£¬ÒÒ×°ÖÃÖÐÓÐ1mol H+ͨ¹ý¸ôĤ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿£¨11·Ö£©ÒÑÖªX¡¢YºÍZÈýÖÖÔªËصÄÔ×ÓÐòÊýÖ®ºÍµÈÓÚ42¡£XÔªËØÔ×ÓµÄ4p¹ìµÀÉÏÓÐ3¸öδ³É¶Ôµç×Ó£¬YÔªËØÔ×ÓµÄ×îÍâ²ã2p¹ìµÀÉÏÓÐ2¸öδ³É¶Ôµç×Ó¡£X¸úY¿ÉÐγɻ¯ºÏÎïX2Y3£¬ZÔªËØ¿ÉÒÔÐγɸºÒ»¼ÛÀë×Ó¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺
(1)XÔªËØÔ×Ó»ù̬ʱµÄµç×ÓÅŲ¼Ê½Îª________£¬¸ÃÔªËصķûºÅÊÇ________¡£
(2)YÔªËØÔ×ӵļ۲ãµç×ӵĵç×ÓÅŲ¼Í¼Îª________£¬¸ÃÔªËصÄÃû³ÆÊÇ________¡£
(3)XÓëZ¿ÉÐγɻ¯ºÏÎïXZ3£¬¸Ã»¯ºÏÎïµÄ¿Õ¼ä¹¹ÐÍΪ________¡£
(4)ÒÑÖª»¯ºÏÎïX2Y3ÔÚÏ¡ÁòËáÈÜÒºÖпɱ»½ðÊôп»¹ÔΪXZ3£¬²úÎﻹÓÐZnSO4ºÍH2O£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ___________________________________________________________¡£
(5)±È½ÏXµÄÇ⻯ÎïÓëͬ×åµÚ¶þ¡¢ÈýÖÜÆÚÔªËØËùÐγɵÄÇ⻯ÎïÎȶ¨ÐÔ¡¢·Ðµã¸ßµÍ²¢ËµÃ÷ÀíÓÉ
_________________________________________________________________________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿¼ÓÈÈN2O5£¬ÒÀ´Î·¢Éú·Ö½â·´Ó¦£º¢ÙN2O5(g)N2O3(g)£«O2(g)¡¢¢ÚN2O3(g)N2O(g)£«O2(g)¡£ÔÚÌå»ýΪ2 LµÄºãÈÝÃܱÕÈÝÆ÷ÖгäÈë8 mol N2O5£¬¼ÓÈȵ½T ¡æʱ·´Ó¦´ïµ½Æ½ºâ״̬£¬´ËʱO2ºÍN2O3µÄÎïÖʵÄÁ¿·Ö±ðΪ9 mol¡¢3.4 mol£¬ÔòT ¡æʱ·´Ó¦¢ÙµÄƽºâ³£ÊýΪ
A. 10.7 B. 8.5 C. 9.6 D. 10.2
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÏÂÁÐÓйشӺ£´øÖÐÌáÈ¡µâµÄʵÑéÔÀíºÍ×°ÖÃÄܴﵽʵÑéÄ¿µÄµÄÊÇ
A. ÓÃ×°Öü××ÆÉÕË麣´ø
B. ÓÃ×°ÖÃÒÒ¹ýÂ˺£´ø»ÒµÄ½þÅÝÒº
C. ÓÃ×°ÖñûÖƱ¸ÓÃÓÚÑõ»¯½þÅÝÒºÖÐIµÄCl2
D. ÓÃ×°Öö¡ÎüÊÕÑõ»¯½þÅÝÒºÖÐIºóµÄCl2βÆø
²é¿´´ð°¸ºÍ½âÎö>>
°Ù¶ÈÖÂÐÅ - 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