¡¾ÌâÄ¿¡¿NH4Al(SO4)2ÊÇʳƷ¼Ó¹¤ÖÐ×îΪ¿ì½ÝµÄʳƷÌí¼Ó¼Á£¬ÓÃÓÚ±º¿¾Ê³Æ·ÖУ»NH4HSO4ÔÚ·ÖÎöÊÔ¼Á¡¢Ò½Ò©¡¢µç×Ó¹¤ÒµÖÐÓÃ;¹ã·º¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©NH4Al(SO4)2¿É×÷¾»Ë®¼Á£¬ÆäÀíÓÉÊÇ_____________________________(ÓñØÒªµÄ»¯Ñ§ÓÃÓïºÍÏà¹ØÎÄ×Ö˵Ã÷)¡£
£¨2£©ÏàͬÌõ¼þÏ£¬0.1 mol¡¤L£1NH4Al(SO4)2ÖÐc()________0.1 mol¡¤L£1NH4HSO4ÖÐc(
)¡£(Ìî¡°µÈÓÚ¡±¡¢¡°´óÓÚ¡±»ò¡°Ð¡ÓÚ¡±)
£¨3£©ÓõÈʽ±íʾ0.1 mol¡¤L£1NH4Al(SO4)2ÈÜÒºÖи÷Àë×ÓŨ¶È¹Øϵ____________¡£
£¨4£©ÈçͼÊÇ0.1 mol¡¤L£1µç½âÖÊÈÜÒºµÄpHËæζȱ仯µÄͼÏñ¡£
ÆäÖзûºÏ0.1 mol¡¤L£1NH4Al(SO4)2µÄpHËæζȱ仯µÄÇúÏßÊÇ________(Ìîд×Öĸ)£¬µ¼ÖÂpHËæζȱ仯µÄÔÒòÊÇ_____________________________£»
£¨5£©ÊÒÎÂʱ£¬Ïò100 mL 0.1 mol¡¤L£1HClÈÜÒºÖеμÓ0.1 mol¡¤L£1°±Ë®£¬µÃµ½ÈÜÒºpHÓ백ˮÌå»ýµÄ¹ØϵÇúÏßÈçͼËùʾ£º
¢ÙÊÔ·ÖÎöͼÖÐa¡¢b¡¢c¡¢dËĸöµã£¬Ë®µÄµçÀë³Ì¶È×î´óµÄÊÇ____________£»
¢ÚÔÚbµã£¬ÈÜÒºÖи÷Àë×ÓŨ¶ÈÓÉ´óµ½Ð¡µÄÅÅÁÐ˳ÐòÊÇ__________________;
¢Ûд³öaµã»ìºÏÈÜÒºÖÐÏÂÁÐËãʽµÄ¾«È·½á¹û£¨²»ÄܽüËƼÆË㣩£º
c£¨Cl-£©- c£¨NH4+£©=____________£¬c£¨H+£©- c£¨NH3¡¤H2O£©=____________¡£
¡¾´ð°¸¡¿A13+Ë®½âÉú³ÉµÄA1(OH)3½ºÌå¾ßÓÐÎü¸½ÐÔ£¬¼´A13++3H2O= A1(OH)3(½ºÌå)+3H+£¬A1(OH)3½ºÌåÎü¸½Ðü¸¡¿ÅÁ£Ê¹Æä³Á½µ´Ó¶ø¾»»¯Ë® СÓÚ c(NH4+)+3c(Al3+)+c(H+)=2c(SO42-)+c(OH-) 1 NHAl(SO4)2Ë®½â£¬ÈÜÒº³ÊËáÐÔ£¬Éý¸ßζÈÆäË®½â³Ì¶ÈÔö´ó£¬pH¼õС a c(Cl-)= c(NH4+)£¾c(OH-)=c(H+) 10-6-10-8£» 10-8
¡¾½âÎö¡¿
(1)Al3+Ë®½âÉú³ÉµÄAl(OH)3½ºÌ壬¾ßÓÐÎü¸½ÐÔ£»
(2)NH4Al(SO4)2ÖÐAl3+Ë®½â³ÊËáÐÔÒÖÖÆNH4+Ë®½â£¬HSO4-µçÀë³öH+ͬÑùÒÖÖÆNH4+Ë®½â£»
(3)ÈÜÒºÖдæÔÚµçºÉÊغã·ÖÎö£»
(4)NH4Al(SO4)2Ë®½â£¬ÈÜÒº³ÊËáÐÔ£¬Éý¸ßζÈÆäË®½â³Ì¶ÈÔö´ó£»
(5)¢Ùa¡¢b¡¢c¡¢dËĸöµã£¬¸ù¾Ý·´Ó¦Á¿µÄ¹Øϵ£¬aµãÑÎËáºÍ°±Ë®Ç¡ºÃÍêÈ«·´Ó¦£¬ÈÜÒºÖÐÖ»ÓÐNH4Cl£»b¡¢c¡¢dÈýµãÈÜÒº¾ùº¬ÓÐNH3H2O£¬¾Ý´Ë·ÖÎöÅжϣ»¢ÚbµãÈÜÒºÖк¬ÓÐNH4ClºÍNH3H2O£¬³ÊÖÐÐÔ£¬c(OH-)=c(H+)£¬½áºÏµçºÉÊغã·ÖÎö½â´ð£»¢ÛaµãΪÂÈ»¯ï§ÈÜÒº£¬¸ù¾ÝµçºÉÊغãºÍÖÊ×ÓÊغã¼ÆËã¡£
(1)Al3+Ë®½âÉú³ÉµÄAl(OH)3½ºÌ壬ÇâÑõ»¯ÂÁ½ºÌå±íÃæ»ý´ó£¬¾ßÓÐÎü¸½ÐÔ£¬Äܹ»Îü¸½Ë®ÖÐÐü¸¡ÔÓÖÊ´ïµ½¾»Ë®µÄÄ¿µÄ£¬Ë®½âµÄÀë×Ó·½³ÌʽΪ£ºAl3++3H2O=Al(OH)3½ºÌå+3H+£¬¹Ê´ð°¸Îª£ºAl3+Ë®½âÉú³ÉµÄAl(OH)3½ºÌ壬¾ßÓÐÎü¸½ÐÔ£¬¼´Al3++3H2O=Al(OH)3½ºÌå+3H+£¬Al(OH)3Îü¸½Ðü¸¡¿ÅÁ£Ê¹Æä³Á½µ´Ó¶ø¾»»¯Ë®£»
(2)NH4Al(SO4)2ÓëNH4HSO4ÖеÄNH4+¾ù·¢ÉúË®½â£¬µ«ÊÇNH4Al(SO4)2ÖÐAl3+Ë®½â³ÊËáÐÔ»áÒÖÖÆNH4+Ë®½â£¬HSO4-µçÀë³öH+ͬÑùÒÖÖÆNH4+Ë®½â£¬ÒòΪHSO4-µçÀëÉú³ÉµÄH+Ũ¶È±ÈAl3+Ë®½âÉú³ÉµÄH+Ũ¶È´ó£¬ËùÒÔNH4HSO4ÖÐNH4+Ë®½â³Ì¶È±ÈNH4Al(SO4)2ÖеÄС£¬¹Ê´ð°¸Îª£ºÐ¡ÓÚ£»
(3)¸ù¾ÝµçºÉÊغ㣬ÓÐc(NH4+)+3c(Al3+)+c(H+)=2c(SO42-)+c(OH-)£¬¹Ê´ð°¸Îª£ºc(NH4+)+3c(Al3+)+c(H+)=2c(SO42-)+c(OH-)£»
(4)NH4Al(SO4)2Ë®½â£¬ÈÜÒº³ÊËáÐÔ£¬pH£¼7£¬Éý¸ßζȣ¬Ë®½â³Ì¶ÈÔö´ó£¬pH¼õС£¬·ûºÏµÄÇúÏßΪ¢ñ£¬¹Ê´ð°¸Îª£º¢ñ£»NH4Al(SO4)2Ë®½â£¬ÈÜÒº³ÊËáÐÔ£¬Éý¸ßζȣ¬ÆäË®½â³Ì¶ÈÔö´ó£¬pH¼õС£»
(5)¢Ùa¡¢b¡¢c¡¢dËĸöµã£¬¸ù¾Ý·´Ó¦Á¿µÄ¹Øϵ£¬aµãÑÎËáºÍ°±Ë®Ç¡ºÃÍêÈ«·´Ó¦£¬ÈÜÒºÖÐÖ»ÓÐNH4Cl£»b¡¢c¡¢dÈýµãÈÜÒº¾ùº¬ÓÐNH3H2O£¬NH4ClË®½â´Ù½øË®µÄµçÀ룬¶øNH3H2OÒÖÖÆË®µÄµçÀ룬Òò´ËË®µÄµçÀë³Ì¶È×î´óµÄÊÇaµã£¬¹Ê´ð°¸Îª£ºa£»
¢ÚbµãÈÜÒºÖк¬ÓÐNH4ClºÍNH3H2O£¬³ÊÖÐÐÔ£¬c(OH-)=c(H+)£¬¸ù¾ÝµçºÉÊغ㣬c(NH4+)= c(Cl-)£¬Ôòbµãʱc(Cl-)= c(NH4+)£¾c(OH-)=c(H+)£¬¹Ê´ð°¸Îª£ºc(Cl-)= c(NH4+)£¾c(OH-)=c(H+)£»
¢ÛaµãΪÂÈ»¯ï§ÈÜÒº£¬¸ù¾ÝµçºÉÊغ㣬c(Cl-)- c(NH4+)= c(H+)-c(OH-)=10-6-=10-6-10-8£¬¸ù¾ÝÖÊ×ÓÊغãc(H+)=c(NH3¡¤H2O)+ c(OH-)£¬Òò´Ëc(H+)-c(NH3¡¤H2O)= c(OH-)=10-8£¬¹Ê´ð°¸Îª£º10-6-10-8£»10-8¡£
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¿¡¿Èçͼ¼×ÊÇÒ»ÖÖÀûÓÃ΢ÉúÎォ·ÏË®ÖеÄÄòËØ(K2NCONH2)µÄ»¯Ñ§ÄÜÖ±½Óת»¯ÎªµçÄÜ£¬²¢Éú³É¶Ô»·¾³ÎÞº¦ÎïÖʵÄ×°ÁD£»Í¬Ê±ÀûÓôË×°ÖõĵçÄÜÔÚÌúÉ϶ÆÍ£¬ÏÂÁÐ˵·¨ÖÐÕýÈ·µÄÊÇ
A. H+͸¹ýÖÊ×Ó½»»»Ä¤ÓÉÓÒÏò×óÒƶ¯
B. ͵缫ӦÓëXÏàÁ¬½Ó
C. Mµç¼«·´Ó¦Ê½£ºH2NCONH2+H2O-6e-== CO2¡ü+N2¡ü+6H+
D. µ±Nµç¼«ÏûºÄ0.25 molÆøÌåʱ£¬Ìúµç¼«ÔöÖØ16 g
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿·úËû°·GÊÇÒ»ÖÖ¿ÉÓÃÓÚÖÎÁÆÖ×ÁöµÄÒ©ÎʵÑéÊÒÓÉ·¼ÏãÌþAÖƱ¸GµÄºÏ³É·ÏßÈçͼ£º
»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©AµÄ½á¹¹¼òʽΪ__¡£CµÄ»¯Ñ§Ãû³ÆÊÇ__¡£
£¨2£©¢ÛµÄ·´Ó¦ÊÔ¼ÁºÍ·´Ó¦Ìõ¼þ·Ö±ðÊÇ__£¬¸Ã·´Ó¦µÄÀàÐÍÊÇ__¡£
£¨3£©¢ÝµÄ·´Ó¦·½³ÌʽΪ__¡£ßÁà¤ÊÇÒ»ÖÖÓлú¼î£¬Æä×÷ÓÃÊÇ__¡£
£¨4£©GµÄ·Ö×ÓʽΪ__¡£
£¨5£©HÊÇGµÄͬ·ÖÒì¹¹Ì壬Æä±½»·ÉϵÄÈ¡´ú»ùÓëGµÄÏàͬµ«Î»Öò»Í¬£¬ÔòH¿ÉÄܵĽṹÓÐ__ÖÖ¡£
£¨6£©4¼×Ñõ»ùÒÒõ£±½°·()ÊÇÖØÒªµÄ¾«Ï¸»¯¹¤ÖмäÌ壬д³öÓɱ½¼×ÃÑ(
)ÖƱ¸4¼×Ñõ»ùÒÒõ£±½°·µÄºÏ³É·Ïß__(ÆäËûÊÔ¼ÁÈÎÑ¡)¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿½«Å¨¶ÈΪ0.1mol¡¤L-1HFÈÜÒº¼ÓË®²»¶ÏÏ¡ÊÍ£¬ÏÂÁи÷Á¿Ê¼ÖÕ±£³ÖÔö´óµÄÊÇ
A.c£¨H+£©B.Ka£¨HF£©C.c(F-£©/c(H+£©D.c(F-£©/c(HF£©
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÏÂÁÐËĸöͼÏñËù·´Ó³µÄÄÚÈÝÓëÏàÓ¦·´Ó¦·ûºÏµÄÊÇ(a¡¢b¡¢c¡¢d¾ù´óÓÚ0)£¨ £©
A. 4NH3(g)£«5O2(g)4NO(g)£«6H2O(g)¦¤H£½-akJ¡¤mol-1
B. N2(g)£«3H2(g)2NH3(g)¦¤H£½-bkJ¡¤mol-1
C. 2SO3(g)2SO2(g)£«O2(g)¦¤H£½+ckJ¡¤mol-1
D. 2N2O5(g)4NO2(g)£«O2(g)¦¤H£½+dkJ¡¤mol-1
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÏÂÁÐÓйØ˵·¨ÖÐÕýÈ·µÄÊÇ( )
A. ÒÑÖª25 ¡æʱNH4CNÈÜÒºÏÔ¼îÐÔ,Ôò25 ¡æʱµÄµçÀë³£ÊýK(NH3¡¤H2O)>K(HCN)
B. ÊÒÎÂʱ, ÏàͬÌå»ý¡¢ÏàͬpHµÄÑÎËáºÍ´×ËáÈÜÒºÖÐÓÉË®µçÀë³öµÄc(H+):ÑÎËáСÓÚ´×ËáÈÜÒº
C. ÒÑÖªKsp(AgCl)=1.56¡Á10-10,Ksp(Ag2CrO4)=9.0¡Á10-12,Ïòº¬ÓÐCl-¡¢CrO42-ÇÒŨ¶È¾ùΪ0.010 mol¡¤L-1ÈÜÒºÖÐÖðµÎ¼ÓÈë0.010 mol¡¤L-1µÄAgNO3ÈÜҺʱ, CrO42-ÏȲúÉú³Áµí
D. ³£ÎÂÏÂpH=7µÄCH3COOHºÍNaOH»ìºÏÈÜÒºÖÐ,c(Na+)>c(CH3COO-)
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿BaCl2¡¤xH2OÊÇÓÃ;¹ã·ºµÄ»ù´¡»¯¹¤²úÆ·¡£ÎÒ¹úÄ¿Ç°Ö÷ÒªÊÇÓÃÑÎËáºÍÁò»¯±µ(º¬ÉÙÁ¿Mg2£«¡¢Fe3£«µÈ)·´Ó¦Éú²úBaCl2¡¤xH2O£¬¹¤ÒÕÁ÷³ÌÈçͼËùʾ¡£Çë»Ø´ð£º
ÒÑÖª£ºÊÒÎÂʱKsp[Mg(OH)2]=1.8¡Á10£11£¬Ksp[Fe(OH)3]=4.0¡Á10£38
£¨1£©·´Ó¦IÖÐÉú³ÉµÄH2SÓÃ×ãÁ¿°±Ë®ÎüÊÕ£¬Ò»¶¨Ìõ¼þÏÂÏòËùµÃÈÜÒºÖÐͨÈË¿ÕÆø£¬Óֿɵõ½µ¥ÖÊÁò²¢Ê¹ÎüÊÕÒºÔÙÉú£¬ÔÙÉú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ_________¡£
£¨2£©ËùµÃÂÈ»¯±µ´ÖÒºÖк¬Áò»¯ºÏÎï(H2S¡¢HS£µÈ)Ó°Ïì²úÆ·ÖÊÁ¿£¬¿É¹ÄÈËÔ¤ÈȺóµÄ¿ÕÆø´µ³ö£¬Ô¤ÈÈ¿ÕÆøµÄÄ¿µÄÊÇ_________£»³ÁµíAµÄÖ÷Òª³É·ÖÊÇ_________¡£
£¨3£©ÈÈ¿ÕÆø´µ³ýʱ»áÓв¿·ÖHS£×ª±äΪS2O32££¬Ê¹²úÆ·ÈÔ²»ÄÜ´ïµ½´¿¶ÈÒªÇ󣬿ÉÔÙÐÐËữÍÑÁò£¬ËữÍÑÁòʱµÄÀë×Ó·½³ÌʽΪ_________¡£
£¨4£©ÊÒÎÂʱ£¬ÎªÊ¹Mg2£«¡¢Fe3£«ÍêÈ«³Áµí(µ±ÈÜÒºÖÐÀë×ÓŨ¶ÈСÓÚ1¡Ál0£5molʱÈÏΪ¸ÃÀë×ÓÍêÈ«³Áµí)£¬Ó¦½«ÈÜÒºµÄpHµ÷ÖÁ_________(Ö»ÁÐËãʽ)ÒÔÉÏ¡£
£¨5£©ÊµÑéÊҲⶨ²úÆ·ÖÐxµÄ²½ÖèÈçÏ£º
¢Ù׼ȷ³ÆÈ¡12.23 g BaCl2¡¤xH2OÑùÆ·£¬¼ÓÈël00 mLÏ¡ÑÎËá¼ÓÈÈÈܽ⣻
¢Ú±ß½Á°è£¬±ßÖðµÎ¼ÓÈë0.lmolH2SO4ÈÜÒº£¬´ýBaSO4ÍêÈ«³Áµí£¬¹ýÂË£¬³Á½µºóÏ´µÓ³Áµí2¡ª3´Î£»
¢Û½«³Áµí³ä·Ö¸ÉÔïºó³ÆÁ¿£¬ÖÊÁ¿Îª11.65 g£¬¼ìÑéBaSO4³ÁµíÊÇ·ñÏ´µÓ¸É¾»µÄ·½·¨ÊÇ_______£»¾¼ÆËãxµÄÊýֵΪ_________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÓÐA¡¢B¡¢C¡¢DËÄÖÖ³£¼ûµÄ½ðÊôµ¥ÖÊ£¬AÔÚ¿ÕÆøÖÐȼÉÕÉú³Éµ»ÆÉ«¹ÌÌ壻BΪºìÉ«¹ÌÌ壬Ðâʴʱ±äΪÂÌÉ«£»CÔÚ¿ÕÆøÖмÓÈÈÈÛ»¯µ«²»µÎÂ䣻DÔÚÑõÆøÖÐȼÉÕ£¬»ðÐÇËÄÉä¡£
¸ù¾ÝÒÔÉÏÐÅÏ¢»Ø´ðÏÂÁÐÎÊÌ⣺
(1)д³ö¶ÔÓ¦»¯Ñ§Ê½£ºA______£¬B______£¬C______£¬D______¡£
(2)AÔÚÊÒÎÂÏÂÓëÑõÆø·´Ó¦Ê±Éú³É__________£¨Ð´»¯Ñ§Ê½£¬ÏÂͬ£©£¬DÔÚ¿ÕÆøÖÐÐâÊ´Éú³É__________¡£
(3)д³öÏÂÁл¯Ñ§·½³Ìʽ£º
¢ÙAÔÚ¿ÕÆøÖÐȼÉÕ______________________£»
¢ÚBÓëÑõÆøÔÚ¼ÓÈÈÌõ¼þÏ·´Ó¦______________________£»
¢ÛCÔÚÑõÆøÖÐȼÉÕ________________________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿Ô×ÓÐòÊýÒÀ´ÎÔö´óµÄËÄÖÖÔªËØA¡¢B¡¢C¡¢D·Ö±ð´¦ÓÚµÚ1¡«4ÖÜÆÚ£¬ÆäÖÐAÔ×ÓºËÊÇÒ»¸öÖÊ×Ó£»BÔ×ÓºËÍâµç×ÓÓÐ6ÖÖ²»Í¬µÄÔ˶¯×´Ì¬£¬BÓëC¿ÉÐγÉÕýËÄÃæÌåÐηÖ×Ó£¬DÔ×ÓÍâΧµç×ÓÅŲ¼Îª3d104s1¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÕâËÄÖÖÔªËØÖе縺ÐÔ×î´óµÄÊÇ____(ÌîÔªËØ·ûºÅ£¬ÏÂͬ)£¬µÚÒ»µçÀëÄÜ×îСµÄÊÇ____£»
£¨2£©CËùÔÚµÄÖ÷×åÔªËØÆø̬Ç⻯ÎïÖУ¬·Ðµã×îµÍµÄÊÇ____(Ìѧʽ)£»
£¨3£©BÔªËØ¿ÉÐγɶàÖÖµ¥ÖÊ£¬ÆäÖС±Ö»ÓÐÒ»²ãÔ×Óºñ¡±µÄÎïÖÊ£¬±»¹«ÈÏΪĿǰÊÀ½çÉÏÒÑÖªµÄ×¡¢×î¼áÓ²¡¢´«µ¼µç×ÓËÙ¶È×î¿ìµÄÐÂÐͲÄÁÏ£¬¸Ã²ÄÁϾ§Ìå½á¹¹ÈçͼËùʾ£¬ÆäÔ×ÓµÄÔÓ»¯ÀàÐÍΪ______£»
£¨4£©DµÄ´×ËáÑξ§Ìå¾Ö²¿½á¹¹Èçͼ£¬¸Ã¾§ÌåÖк¬ÓеĻ¯Ñ§¼üÊÇ____(ÌîÑ¡ÏîÐòºÅ)£»
¢Ù¼«ÐÔ¼ü¡¡ ¢Ú·Ç¼«ÐÔ¼ü ¢ÛÅäλ¼ü¡¡¡¡ ¢Ü½ðÊô¼ü
£¨5£©Ä³Ñ§ÉúËù×öµÄÓйØDÔªËصÄʵÑéÁ÷³ÌÈçÏÂͼ£º
Dµ¥ÖÊ×ØÉ«µÄÑÌ
ÂÌÉ«ÈÜÒº
À¶É«³Áµí
À¶É«ÈÜÒº
ºÚÉ«³Áµí
ÇëÊéдµÚ¢Ý²½·´Ó¦µÄÀë×Ó·½³Ìʽ£º_____________¡£
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