¡¾ÌâÄ¿¡¿ÊµÑéÊÒÒªÅäÖÆ100 mL 2 mol/L NaClÈÜÒº£¬Çë»Ø´ðÏÂÁÐÎÊÌ⣺
(1)ÅäÖƹý³ÌÖÐÐèҪʹÓõÄÖ÷Òª²£Á§ÒÇÆ÷°üÀ¨ÉÕ±¡¢²£Á§°ô¡¢½ºÍ·µÎ¹Ü¡¢Á¿Í²ºÍ___________¡£
(2)ÓÃÍÐÅÌÌìƽ³ÆÈ¡ÂÈ»¯ÄƹÌÌ壬ÆäÖÊÁ¿Îª__________ g¡£
(3)ÏÂÁÐÖ÷Òª²Ù×÷²½ÖèµÄÕýȷ˳ÐòÊÇ___________(ÌîÐòºÅ)¡£
¢Ù³ÆÈ¡Ò»¶¨ÖÊÁ¿µÄÂÈ»¯ÄÆ£¬·ÅÈëÉÕ±ÖУ¬ÓÃÊÊÁ¿ÕôÁóË®Èܽ⣻
¢Ú¼ÓË®ÖÁÒºÃæÀëÈÝÁ¿Æ¿¾±¿Ì¶ÈÏßÏÂ1~2ÀåÃ×ʱ£¬¸ÄÓýºÍ·µÎ¹ÜµÎ¼ÓÕôÁóË®ÖÁ°¼ÒºÃæÓë¿Ì¶ÈÏßÏàÇУ»
¢Û½«ÈÜҺתÒƵ½ÈÝÁ¿Æ¿ÖУ»
¢Ü¸ÇºÃÆ¿Èû£¬·´¸´ÉÏϵߵ¹£¬Ò¡ÔÈ£»
¢ÝÓÃÉÙÁ¿ÕôÁóˮϴµÓÉÕ±ÄڱںͲ£Á§°ô2~3´Î£¬Ï´µÓҺתÒƵ½ÈÝÁ¿Æ¿ÖС£
(4)Èç¹ûʵÑé¹ý³ÌÖÐȱÉÙ²½Öè¢Ý£¬»áÔì³ÉËùÅäÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶È_______(Ìî¡°Æ«¸ß¡±»ò¡°Æ«µÍ¡±»ò¡°ÎÞÓ°Ï족£¬)
¡¾´ð°¸¡¿100mLÈÝÁ¿Æ¿ 11.7 ¢Ù¢Û¢Ý¢Ú¢Ü Æ«µÍ
¡¾½âÎö¡¿
±¾ÌâÖ÷Òª¿¼²éÅäÖÆÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈµÄÈÜҺʵÑé¡£
£¨1£©ÒÀ¾ÝÓùÌÌåÅäÖÆÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈÈÜÒºµÄÒ»°ã²½ÖèÑ¡ÔñÐèÒªµÄÒÇÆ÷£»
£¨2£©ÒÀ¾Ým=cVM¼ÆËãÐèÒªÂÈ»¯ÄƵÄÖÊÁ¿£»
£¨3£©ÒÀ¾ÝÓùÌÌåÅäÖÆÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈÈÜÒºµÄÒ»°ã²½ÖèÅÅÐò£»
£¨4£©·ÖÎö²Ù×÷¶ÔÈÜÖʵÄÎïÖʵÄÁ¿ºÍÈÜÒºÌå»ýµÄÓ°Ï죬ÒÀ¾Ýc=½øÐÐÎó²î·ÖÎö¡£
£¨1£©ÓùÌÌåÅäÖÆÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈÈÜÒºµÄÒ»°ã²½Ö裺¼ÆËã¡¢³ÆÁ¿¡¢Èܽ⡢ÀäÈ´¡¢ÒÆÒº¡¢Ï´µÓ¡¢¶¨ÈÝ¡¢Ò¡Ôȵȣ¬Óõ½µÄÒÇÆ÷£ºÍÐÅÌÌìƽ¡¢Ò©³×¡¢ÉÕ±¡¢²£Á§°ô¡¢ÈÝÁ¿Æ¿¡¢½ºÍ·µÎ¹Ü£¬ÅäÖÆ100mL2mol/LNaClÈÜÒº£¬Ó¦Ñ¡Ôñ100mLÈÝÁ¿Æ¿£¬ËùÒÔ»¹È±ÉÙµÄÒÇÆ÷£º100 mLÈÝÁ¿Æ¿£»
£¨2£©ÅäÖÆ100mL2mol/LNaClÈÜÒº£¬ÐèÒªÈÜÖʵÄÖÊÁ¿Îª£º0.1L¡Á2mol/L¡Á58.5g/mol=11.7g£»
£¨3£©ÓùÌÌåÅäÖÆÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈÈÜÒºµÄÒ»°ã²½Ö裺¼ÆËã¡¢³ÆÁ¿¡¢Èܽ⡢ÀäÈ´¡¢ÒÆÒº¡¢Ï´µÓ¡¢¶¨ÈÝ¡¢Ò¡ÔÈ¡¢×°Æ¿¡¢Ìù±êÇ©£¬ËùÒÔÕýÈ·µÄ²Ù×÷²½ÖèΪ£º¢Ù¢Û¢Ý¢Ú¢Ü£»
£¨4£©Èç¹ûʵÑé¹ý³ÌÖÐȱÉÙ²½Öè¢Ý£¬»áÔì³É²¿·ÖÈÜÖÊËðʧ£¬ÈÜÖʵÄÎïÖʵÄÁ¿Æ«Ð¡£¬ËùÅäÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈÆ«µÍ¡£
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¿¡¿ÖÎÀíSO2¡¢CO¡¢NOxÎÛȾÊÇ»¯Ñ§¹¤×÷ÕßÑо¿µÄÖØÒª¿ÎÌâ¡£
¢ñ.ÁòË᳧´óÁ¿Åŷź¬SO2µÄβÆø»á¶Ô»·¾³Ôì³ÉÑÏÖØΣº¦¡£
£¨1£©¹¤ÒµÉÏ¿ÉÀûÓ÷ϼîÒº£¨Ö÷Òª³É·ÖΪNa2CO3£©´¦ÀíÁòË᳧βÆøÖеÄSO2£¬µÃµ½Na2SO3ÈÜÒº£¬¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽΪ__________________________________¡£
¢ò.Á¤Çà»ìÄýÍÁ¿É×÷Ϊ·´Ó¦£»2CO(g)+O2(g)2CO2(g)µÄ´ß»¯¼Á¡£Í¼¼×±íʾÔÚÏàͬµÄºãÈÝÃܱÕÈÝÆ÷¡¢ÏàͬÆðʼŨ¶È¡¢Ïàͬ·´Ó¦Ê±¼ä¶ÎÏÂ,ʹÓÃͬÖÊÁ¿µÄ²»Í¬Á¤Çà»ìÄýÍÁ£¨¦ÁÐÍ¡¢¦ÂÐÍ£©´ß»¯Ê±£¬COµÄת»¯ÂÊÓëζȵĹØϵ¡£
£¨2£©a¡¢b¡¢c¡¢d ËĵãÖУ¬´ïµ½Æ½ºâ״̬µÄÊÇ__________________________________¡£
£¨3£©ÒÑÖªcµãʱÈÝÆ÷ÖÐO2Ũ¶ÈΪ0.02 mol/L£¬Ôò50¡æʱ£¬ÔÚ¦ÁÐÍÁ¤Çà»ìÄýÍÁÖÐCOת»¯·´Ó¦µÄƽºâ³£ÊýK=____________£¨Óú¬xµÄ´úÊýʽ±íʾ£©¡£
£¨4£©ÏÂÁйØÓÚͼ¼×µÄ˵·¨ÕýÈ·µÄÊÇ_____________¡£
A.COת»¯·´Ó¦µÄƽºâ³£ÊýK(a)
B.ÔÚ¾ùδ´ïµ½Æ½ºâ״̬ʱ£¬Í¬ÎÂϦÂÐÍÁ¤Çà»ìÄýÍÁÖÐCOת»¯ËÙÂʱȦÁÐÍÒª´ó
C.bµãʱCOÓëO2·Ö×ÓÖ®¼ä·¢ÉúÓÐЧÅöײµÄ¼¸ÂÊÔÚÕû¸öʵÑé¹ý³ÌÖÐ×î¸ß
D.eµãת»¯ÂʳöÏÖÍ»±äµÄÔÒò¿ÉÄÜÊÇζÈÉý¸ßºó´ß»¯¼Áʧȥ»îÐÔ
¢ó.ijº¬îÜ´ß»¯¼Á¿ÉÒÔ´ß»¯Ïû³ý²ñÓͳµÎ²ÆøÖеÄ̼ÑÌ(C)ºÍNOx¡£²»Í¬Î¶ÈÏ£¬½«Ä£ÄâβÆø£¨³É·ÖÈçϱíËùʾ£©ÒÔÏàͬµÄÁ÷ËÙͨ¹ý¸Ã´ß»¯¼Á£¬²âµÃËùÓвúÎCO2¡¢N2¡¢N2O£©ÓëNOµÄÏà¹ØÊý¾Ý½á¹ûÈçͼÒÒËùʾ¡£
Ä£ÄâβÆø | ÆøÌå(10mol) | ̼ÑÌ | ||
NO | O2 | He | ||
ÎïÖʵÄÁ¿(mol) | 0.025 | 0.5 | 9.475 | n |
£¨5£©375¡æʱ£¬²âµÃÅųöµÄÆøÌåÖк¬0.45 molO2ºÍ0.0525 mol CO2£¬ÔòYµÄ»¯Ñ§Ê½Îª______________¡£
£¨6£©ÊµÑé¹ý³ÌÖвÉÓÃNOÄ£ÄâNOx£¬¶ø²»²ÉÓÃNO2µÄÔÒòÊÇ____________________________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿Ä³Ï¡ÁòËáºÍÏ¡ÏõËáµÄ»ìºÏÈÜÒº200 mL£¬Æ½¾ù·Ö³ÉÁ½·Ý¡£ÏòÆäÖÐÒ»·ÝÖÐÖð½¥¼ÓÈëÍ·Û£¬×î¶àÄÜÈÜ19.2 g(ÒÑÖªÏõËáÖ»±»»¹ÔΪNOÆøÌå)¡£ÏòÁíÒ»·ÝÖÐÖð½¥¼ÓÈëÌú·Û£¬²úÉúÆøÌåµÄÁ¿ËæÌú·ÛÖÊÁ¿Ôö¼ÓµÄ±ä»¯ÈçÏÂͼËùʾ¡£ÏÂÁзÖÎö»ò½á¹û²»ÕýÈ·µÄÊÇ
A. ͼÖУ¬AB¶ÎµÄ·´Ó¦ÎªFe£«2Fe3£«===3Fe2£«£¬ÈÜÒºÖÐÈÜÖÊ×îÖÕΪÁòËáÑÇÌú
B. ÔÈÜÒºÖÐÁòËáŨ¶ÈΪ2.5 mol¡¤L£1
C. Ô»ìºÏÈÜÒºÖÐÏõËá¸ùÀë×ÓµÄÎïÖʵÄÁ¿Îª0.2 mol
D. ͼÖУ¬OA¶Î²úÉúµÄÆøÌåÊÇÒ»Ñõ»¯µª£¬BC¶Î²úÉúµÄÆøÌåÊÇÇâÆø
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿Ä³Í¬Ñ§ÔÚʵÑéÖв»Ð¡ÐĽ«¼¸µÎÐÂÖÆÂÈË®µÎÈëÊ¢ÓÐä廯ÑÇÌúÈÜÒºµÄÊÔ¼ÁÆ¿ÖУ¬·¢ÏÖÈÜÒº±ä³ÉÁË»ÆÉ«£¬¸ù¾ÝÒÑÓл¯Ñ§ÖªÊ¶£¬ÇëÄã²ÎÓëÉÏÊöÏÖÏóÐγÉÔÒòµÄ·ÖÎöÓë̽¾¿£º
£¨1£©Ìá³öÎÊÌâ²ÂÏ룺
²ÂÏëÒ»£ºÈÜÒº³Ê»ÆÉ«ÊÇÒò·¢ÉúÀë×Ó·´Ó¦¢Ù_________________________________________(ÌîÀë×Ó·½³Ìʽ)ËùÖ¡£
²ÂÏë¶þ£ºÈÜÒº³Ê»ÆÉ«ÊÇÒò·¢ÉúÀë×Ó·´Ó¦¢Ú______________________________________(ÌîÀë×Ó·½³Ìʽ)ËùÖ¡£
£¨2£©Éè¼ÆʵÑé²¢ÑéÖ¤
ΪÑéÖ¤¢ÙÓë¢ÚÖÐÊÇÄĸöÔÒòµ¼ÖÂÁËÈÜÒº±ä»ÆÉ«£¬Éè¼Æ²¢½øÐÐÁËÒÔÏÂʵÑé¡£Çë¸ù¾ÝÒÔÏÂËù¸øÊÔ¼Á£¬½øÐкÏÀíÑ¡Óã¬Íê³ÉʵÑé·½°¸1ºÍ·½°¸2£º__________________¡¢______________¡¢__________________¡¢______________¡£
£¨3£©ÊµÑé½áÂÛ£ºÒÔÉÏʵÑé²»½öÑéÖ¤ÁËÈÜÒº±ä»ÆµÄÕæʵÔÒò£¬Í¬Ê±Ö¤Ã÷ÁËFe2£«µÄ»¹ÔÐÔ±ÈBr£______________(Ìî¡°Ç¿¡±»ò¡°Èõ¡±)¡£
£¨4£©ÊµÑ鷴˼
¢ñ.¸ù¾ÝÉÏÊöʵÑéÍƲ⣬ÈôÔÚä廯ÑÇÌúÈÜÒºÖеÎÈë×ãÁ¿ÂÈË®£¬ÔÙ¼ÓÈëCCl4²¢³ä·ÖÕñµ´ºó¾²Ö¹£¬²úÉúµÄÏÖÏóÊÇ______________________________________¡£
¢ò.ÔÚ100 mL FeBr2ÈÜÒºÖÐͨÈë2.24 L Cl2(±ê×¼×´¿ö)£¬ÈÜÒºÖÐÓÐ1/2µÄBr£±»Ñõ»¯³Éµ¥ÖÊBr2£¬ÔòÔFeBr2ÈÜÒºÖÐFeBr2µÄÎïÖʵÄÁ¿Å¨¶ÈΪ__________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿2017Äê5Ôº£µ×ÌìÈ»ÆøË®ºÏÎï(Ë׳ơ°¿Éȼ±ù¡±)ÊԲɳɹ¦£¬ÕâÊÇÎÒ¹úÄÜÔ´¿ª·¢µÄÒ»´ÎÀúÊ·ÐÔÍ»ÆÆ¡£Ò»¶¨Ìõ¼þÏ£¬CH4ºÍCO2¶¼ÄÜÓëH2OÐγÉÈçÏÂͼËùʾµÄÁý×´½á¹¹(±íÃæµÄСÇòÊÇË®·Ö×Ó£¬ÄÚ²¿µÄ´óÇòÊÇCH4·Ö×Ó»òCO2·Ö×Ó£»¡°¿Éȼ±ù¡±ÊÇCH4ÓëH2OÐγɵÄË®ºÏÎï)£¬ÆäÏà¹Ø²ÎÊý¼ûÏÂ±í¡£
(1)CH4ºÍCO2Ëùº¬µÄÈýÖÖÔªËص縺ÐÔ´Ó´óµ½Ð¡µÄ˳ÐòΪ_______£»Ì¼Ô×ÓµÄ×î¸ßÄܼ¶µÄ·ûºÅÊÇ_______£¬Æäµç×ÓÔÆÐÎ×´ÊÇ_______¡£
(2)CO2·Ö×ÓÖÐ̼Ô×ÓµÄÔÓ»¯¹ìµÀÀàÐÍΪ_______£¬·Ö×ÓµÄÁ¢Ìå¹¹ÐÍΪ_____¡£CO2ÓëSO2ÏàͬÌõ¼þÏÂÔÚË®ÖеÄÈܽâ¶È½Ï´óµÄÊÇSO2£¬ÀíÓÉÊÇ________¡£
(3)Ϊ¿ª²Éº£µ×µÄ¡°¿Éȼ±ù¡±£¬ÓпÆѧ¼ÒÌá³öÓÃCO2Öû»CH4µÄÉèÏë¡£ÒÑÖªÉÏͼÖÐÁý×´½á¹¹µÄ¿Õǻֱ¾¶Îª0.586nm£¬¸ù¾ÝÉÏÊöͼ±í£¬´ÓÎïÖʽṹ¼°ÐÔÖʵĽǶȷÖÎö£¬¸ÃÉèÏëµÄÒÀ¾ÝÊÇ_______¡£
(4)¡°¿Éȼ±ù¡±ÖзÖ×Ó¼ä´æÔÚµÄ×÷ÓÃÁ¦ÊÇÇâ¼üºÍ_________£¬ÉÏͼÖÐ×îСµÄ»·ÖÐÁ¬½ÓµÄÔ×Ó×ÜÊýÊÇ_______¡£
(5)Ë®ÔÚ²»Í¬µÄζȺÍѹÁ¦Ìõ¼þÏ¿ÉÐγÉ11ÖÖ²»Í¬½á¹¹µÄ¾§Ì壬ÃܶȴӱÈË®ÇáµÄ0.92g/cm3µ½Ô¼ÎªË®µÄ1.5 ±¶¡£±ùÊÇÈËÃÇÆù½ñÒÑÖªµÄÓÉÒ»ÖÖ¼òµ¥·Ö×Ӷѻý³ö½á¹¹»¨Ñù×î¶àµÄ»¯ºÏÎï¡£ÆäÖбù- ¢÷µÄ¾§Ìå½á¹¹ÎªÒ»¸öÈçͼËùʾµÄÁ¢·½¾§°û£¬Ã¿¸öË®·Ö×ÓÓëÖÜΧ4¸öË®·Ö×ÓÒÔÇâ¼ü½áºÏ¡£ÉèO-H¡O¾àÀëΪapm£¬°¢·ü¼ÓµÂÂÞ³£ÊýµÄֵΪNA£¬Ôò¸Ã±ù- ¢÷¾§ÌåµÄÃܶÈΪ____ g/cm3(Áгö¼ÆËãʽ¼´¿É)¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿NAΪ°¢·ü¼ÓµÂÂÞ³£Êý¡£ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ
A. ÊÊÁ¿Na2O2ÈÜÓÚË®³ä·Ö·´Ó¦£¬Éú³É0.1mol O2£¬Ôòµç×ÓתÒÆÊýĿΪ0.4NA
B. 1mol°±»ù(-NH2)Öк¬Óеç×ÓµÄÊýĿΪ9NA
C. 42gÓлúÎïC3H6Öк¬ÓÐË«¼üÊýĿΪNA
D. 1mol/LµÄNaClOÈÜÒºÖк¬ÓÐClO-µÄÊýĿСÓÚNA
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿¶ÌÖÜÆÚÖ÷×åÔªËØX¡¢Y¡¢Z¡¢R¡¢TµÄÔ×Ӱ뾶ÓëÔ×ÓÐòÊý¹ØϵÈçÏÂͼËùʾ¡£RÔ×Ó×îÍâ²ãµç×ÓÊýÊǵç×Ó²ãÊýµÄ2±¶£¬YÓëZÄÜÐγÉZ2Y¡¢Z2Y2ÐÍÀë×Ó»¯ºÏÎZÓëTÐγɵĻ¯ºÏÎïZ2TÄÜÆÆ»µË®µÄµçÀëƽºâ¡£ÏÂÁÐÍƶÏÕýÈ·µÄÊÇ
A. Ô×Ӱ뾶ºÍÀë×Ӱ뾶¾ùÂú×㣺Y<Z
B. Ç⻯ÎïµÄ·Ðµã²»Ò»¶¨ÊÇ£ºY>R
C. ×î¸ß¼ÛÑõ»¯Îï¶ÔӦˮ»¯ÎïµÄËáÐÔ£ºT<R
D. ÓÉX¡¢R¡¢Y¡¢ZËÄÖÖÔªËØ×é³ÉµÄ»¯ºÏÎïË®ÈÜÒºÒ»¶¨ÏÔ¼îÐÔ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿°ÑCa(OH)2¹ÌÌå·ÅÈëÒ»¶¨Á¿µÄÕôÁóË®ÖУ¬Ò»¶¨Î¶ÈÏ´ﵽƽºâ£ºCa(OH)2£¨s£©Ca2+£¨aq£©+2OH££¨aq£©£®µ±ÏòÐü×ÇÒºÖмÓÉÙÁ¿Éúʯ»Òºó£¬Èôζȱ£³Ö²»±ä£¬ÏÂÁÐÅжÏÕýÈ·µÄÊÇ £¨ £©
A. ÈÜÒºÖÐCa2+ÊýÄ¿Ôö¶à B. ÈÜÒºÖÐc(Ca2+)Ôö´ó
C. ÈÜÒºpHÖµ²»±ä D. ÈÜÒºpHÖµÔö´ó
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÏÂͼµÄ¸÷·½¿ò±íʾÓйصÄÒ»ÖÖ·´Ó¦Îï»òÉú³ÉÎï(ijЩÎïÖÊÒѾÂÔÈ¥)£¬ÆäÖг£ÎÂÏÂA¡¢C¡¢D¡¢EΪÎÞÉ«ÆøÌ壬CÄÜʹʪÈóµÄºìɫʯÈïÊÔÖ½±äÀ¶£¬B³£ÎÂÏÂΪÎÞÉ«ÒºÌ壬FeÓöŨµÄGÈÜÒº¶Û»¯¡£
(1)д³öÏÂÁи÷ÎïÖʵĻ¯Ñ§Ê½£º
B£º____________£»F£º____________£»G£º___________¡£
(2)д³öÏÂÁб仯µÄ·´Ó¦·½³Ìʽ£º
A¡úD£º________________________________________£»
G¡úE£º________________________________________¡£
(3)ʵÑéÊÒÀ³£ÓüÓÈÈ_____________________µÄ»ìºÏÎïµÄ·½·¨ÖÆÈ¡ÆøÌåC£¬³£²ÉÓÃ____________·¨À´ÊÕ¼¯¡£
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