³£ÎÂÏ£¬½«Ä³Ò»Ôª¼îBOHºÍHClÈÜÒºµÈÌå»ý»ìºÏ£¬Á½ÖÖÈÜÒºµÄŨ¶ÈºÍ»ìºÏºóËùµÃÈÜÒºµÄpHÈçÏÂ±í£º
ʵÑé±àºÅ | HClµÄÎïÖʵÄÁ¿Å¨¶È £¨mol¡¤L-1£© | BOHµÄÎïÖʵÄÁ¿Å¨¶È £¨mol¡¤L-1£© | »ìºÏÈÜÒºµÄpH |
¢Ù | 0.1 | 0.1 | pH=5 |
¢Ú | c | 0.2 | pH=7 |
¢Û | 0.1 | 0.2 | pH>7 |
£¨12·Ö£©£¨Ã¿¿Õ2·Ö£©
£¨1£© Èõ¼î£» 1¡Á10¡ª5
£¨2£©<£» =
£¨3£© >£» =
½âÎöÊÔÌâ·ÖÎö£º£¨1£©µÈÎïÖʵÄÁ¿µÄÒ»ÔªËá¡¢¼î»ìºÏʱÈÜÒº³ÊËáÐÔ£¬Ôò¼îÊÇÈõ¼î£¬ÑÎÈÜÒºÖÐÇâÀë×Ó¾ÍÊÇÈõ¸ùÀë×ÓË®½âµÃµç×ӵģ¬ËùÒÔÈÜÒºÖÐË®µçÀë³öµÄc£¨OH££©=c£¨H£«£©=1¡Á10-5mol¡¤L£1£¬¹Ê´ð°¸Îª£ºÈõ¼î£¬1¡Á10-5£»
£¨2£©BOHΪÈõ¼î£¬µÈÌå»ýµÈŨ¶È»ìºÏÈÜÒºµÄpHСÓÚ7£¬ÔòΪ±£Ö¤pH=7£¬Ó¦Ê¹ËáŨ¶ÈСÓÚ0.2mol¡¤L£1£¬ÈÜÒº³ÊÖÐÐÔ£¬ÈÜÒºÖÐc£¨OH££©=c£¨H£«£©£¬ÈÜÒº³ÊµçÖÐÐÔ£¬´æÔÚµçºÉÊغ㣬ËùÒÔc£¨B£«£©=c£¨Cl££©£¬¹Ê´ð°¸Îª£º£¼£¬=£»
£¨3£©ÓÉ¢Û×éʵÑé½á¹û¿ÉÖª£¬»ìºÏºóΪBOHÓëBClµÄ»ìºÏÒº£¬pH£¾7£¬¼îµÄµçÀë´óÓÚÑεÄË®½â£¬ÈÜÒºÖдæÔÚµçºÉÊغãc£¨H£«£©+c£¨B£« £©=c£¨OH££©+c£¨Cl£ £©£¬¸ù¾ÝÎïÁÏÊغãµÃc£¨B£«£©+c£¨BOH£©=2c£¨Cl£ £©£¬½«Á½¸öµÈʽÕûÀíµÃc£¨B£«£©-2c £¨OH££©=c£¨BOH£©-2c£¨H£«£©£¬¹Ê´ð°¸Îª£º£¾£¬=£®
¿¼µã£ºËá¼î»ìºÏʱµÄ¶¨ÐÔÅжϼ°ÓйØpHµÄ¼ÆËã
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£ºÌî¿ÕÌâ
ÒÑÖª£ºÔÚÊÒÎÂʱH2OH++OH-¡¡KW=10-14¡¡ CH3COOH H++ CH3COO£¡¡Ka=1.8¡Á10-5
£¨1£©È¡ÊÊÁ¿´×ËáÈÜÒº£¬¼ÓÈëÉÙÁ¿´×ËáÄƹÌÌ壬´ËʱÈÜÒºÖÐC£¨H+£©ÓëC£¨CH3COOH£©µÄ±ÈÖµ¡¡ ¡¡ £¨Ìî¡°Ôö´ó¡±»ò¡°¼õС¡±»ò¡°²»±ä¡±£©
£¨2£©´×ËáÄÆË®½âµÄÀë×Ó·½³ÌʽΪ¡¡¡¡¡¡¡¡ ¡¡¡¡¡¡¡¡ ¡£µ±Éý¸ßζÈʱ£¬C(OH¡ª)½«¡¡¡¡¡¡¡¡£¨Ìî¡°Ôö´ó¡±¡°¼õС¡±¡°²»±ä¡±£©£»
£¨3£©0£®5mol¡¤L-1´×ËáÄÆÈÜÒºpHΪm£¬ÆäË®½âµÄ³Ì¶È£¨ÒÑË®½âµÄ´×ËáÄÆÓëÔÓд×ËáÄƵıÈÖµ£©Îªa£»1mol¡¤L-1´×ËáÄÆÈÜÒºpHΪn£¬Ë®½âµÄ³Ì¶ÈΪb£¬ÔòmÓënµÄ¹ØϵΪ¡¡¡¡¡¡¡¡ £¬aÓëbµÄ¹ØϵΪ¡¡¡¡¡¡¡¡¡¡£¨Ìî¡°´óÓÚ¡±¡°Ð¡ÓÚ¡±¡°µÈÓÚ¡±£©£»
£¨4£©½«µÈÌå»ýµÈŨ¶ÈµÄ´×ËáºÍÇâÑõ»¯ÄÆÈÜÒº»ìºÏºó£¬ËùµÃÈÜÒºÖÐÀë×ÓŨ¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòÊÇ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡ ¡¡¡¡¡¡ ¡£
£¨5£©Èô´×ËáºÍÇâÑõ»¯ÄÆÈÜÒº»ìºÏºópH<7£¬Ôòc£¨Na+£©_______________ c£¨CH3COO££©£¨Ìî¡°´óÓÚ¡±¡¢¡°Ð¡ÓÚ¡±»ò¡°µÈÓÚ¡±£©£¬
£¨6£©ÊÒÎÂʱ£¬ÈôÓÉpH£½3µÄHAÈÜÒºV1mLÓëpH£½11µÄNaOHÈÜÒºV2 mL»ìºÏ£¬ÔòÏÂÁÐ˵·¨²»ÕýÈ·µÄÊÇ____________¡£
A£®Èô·´Ó¦ºóÈÜÒº³ÊÖÐÐÔ£¬Ôòc£¨H+£©+c£¨OH££©£½2¡Á10£7mol¡¤L£1 |
B£®ÈôV1=V2£¬·´Ó¦ºóÈÜÒºpHÒ»¶¨µÈÓÚ7 |
C£®Èô·´Ó¦ºóÈÜÒº³ÊËáÐÔ£¬ÔòV1Ò»¶¨´óÓÚV2 |
D£®Èô·´Ó¦ºóÈÜÒº³Ê¼îÐÔ£¬ÔòV1Ò»¶¨Ð¡ÓÚV2 |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
»Ø´ðÏÂÁÐÓйØNa2SÈÜÒºµÄÎÊÌâ¡£
£¨1£©Na2SÈÜҺʢװÔÚ´øÏð½ºÈûµÄÊÔ¼ÁÆ¿ÖУ¬×ÐϸÎÅÓгô¼¦µ°Æø棬ÇëÓÃÀë×Ó·½³Ìʽ½âÊÍ£º
£¬ ¡£
£¨2£©½«Na2SÈÜÒº¼ÓÈëAlCl3ÈÜÒºÖУ¬Óа×É«³ÁµíºÍ³ô¼¦µ°ÆøζµÄÆøÌåÉú³É£¬·¢ÉúµÄÀë×Ó·´Ó¦Îª£º
¡£
£¨3£©½«Na2SÈÜÒº¼ÓÈëAgClµÄ×ÇÒºÖУ¬Éú³ÉµÄºÚÉ«³ÁµíÊÇ £¨Ð´»¯Ñ§Ê½£©¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
ÄÑÈÜÐÔÔÓ±ʯ(K2SO4¡¤MgSO4¡¤2CaSO4¡¤2H2O)ÊôÓÚ¡°´ô¿ó¡±£¬ÔÚË®ÖдæÔÚÈçÏÂÈܽâƽºâ£º
K2SO4¡¤MgSO4¡¤2CaSO4¡¤2H2O(s)2Ca2£«£«2K£«£«Mg2£«£«4SO42-£«2H2O¡£ÎªÄܳä·ÖÀûÓüØ×ÊÔ´£¬Óñ¥ºÍCa(OH)2ÈÜÒºÈܽþÔÓ±ʯÖƱ¸ÁòËá¼Ø£¬¹¤ÒÕÁ÷³ÌÈçÏ£º
£¨1£©ÂËÔüÖ÷Òª³É·ÖÓÐ________ºÍCaSO4ÒÔ¼°Î´ÈÜÔÓ±ʯ¡£
£¨2£©Óû¯Ñ§Æ½ºâÒƶ¯ÔÀí½âÊÍCa(OH)2ÈÜÒºÄÜÈܽâÔÓ±ʯ½þ³öK£«µÄÔÒò£º
¡£
£¨3£©¡°³ýÔÓ¡±»·½ÚÖУ¬ÏȼÓÈë ÈÜÒº£¬¾½Á°èµÈ²Ù×÷ºó£¬¹ýÂË£¬ÔÙ¼ÓÈë ÈÜÒºµ÷ÂËÒºpHÖÁÖÐÐÔ¡£
£¨4£©²»Í¬Î¶ÈÏ£¬K£«µÄ½þ³öŨ¶ÈÓëÈܽþʱ¼äµÄ¹Øϵ¼ûÓÒͼ¡£ÓÉͼ¿ÉµÃ£¬Ëæ×ÅζÈÉý¸ß£¬
¢Ù £¬
¢Ú £¬
¢ÛÈܽþ³öµÄK£«µÄƽºâŨ¶ÈÔö´ó¡£
£¨5£©ÓÐÈËÒÔ¿ÉÈÜÐÔ̼ËáÑÎΪÈܽþ¼Á£¬ÔòÈܽþ¹ý³ÌÖлᷢÉú£ºCaSO4(s)£«CO32-CaCO3(s)£«SO42-¡£ÒÑÖª298 Kʱ£¬Ksp(CaCO3)£½2.80¡Á10£9£¬Ksp(CaSO4)£½4.90¡Á10£5£¬¼ÆËã´ËζÈϸ÷´Ó¦µÄƽºâ³£Êý£¬K£½ ¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
³£ÎÂÏ£¬Èç¹ûÈ¡0£®1 mol/L HAÈÜÒºÓë0£®1 mol/L NaOHÈÜÒºµÈÌå»ý»ìºÏ£¨ºöÂÔ»ìºÏºó
ÈÜÒºÌå»ýµÄ±ä»¯£©£¬²âµÃ»ìºÏÈÜÒºµÄpH=8,ÊԻشðÒÔÏÂÎÊÌ⣺
(1)»ìºÏÈÜÒºµÄpH=8µÄÔÒò£º £¨ÓÃÀë×Ó·½³Ìʽ±íʾ£©¡£
(2)»ìºÏÈÜÒºÖÐÓÉË®µçÀë³öµÄc(H+) £¨Ìî¡°>¡±¡°<¡±»ò¡°=¡±£©0£®1 mol/L NaOH
ÈÜÒºÖÐÓÉË®µçÀë³öµÄc(H+)¡£
(3)Çó³ö»ìºÏÒºÖÐÏÂÁÐËãʽµÄ¾«È·¼ÆËã½á¹û£¨Ìî¾ßÌåÊý×Ö£©£º
c(Na+£©£c(A-£©£½ mol/L,
c(OH-£©£c(HA) £½ mol/L¡£
(4)ÒÑÖªNH4AÈÜҺΪÖÐÐÔ£¬ÓÖÖªHAÈÜÒº¼Óµ½Na2CO3ÈÜÒºÖÐÓÐÆøÌå·Å³ö£¬ÊÔÍƶϣ¨NH4)2CO3
ÈÜÒºµÄpH £¨Ìî¡°´óÓÚ¡±¡°Ð¡ÓÚ¡±»ò¡°µÈÓÚ¡±)7£»½«Í¬Î¶ÈϵÈŨ¶ÈµÄËÄ
ÖÖÑÎÈÜÒº£º
A£®NH4HCO3 B£®NH4A C£®(NH4)2SO4 D£®NH4Cl
°´pHÓÉ´óµ½Ð¡µÄ˳ÐòÅÅÁÐÊÇ £¨ÌîÐòºÅ£©¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
ÎÒ¹úij´óÐ͵ç½âÍÉú²úÆóÒµ£¬ÆäÒ±Á¶¹¤ÒÕÖÐÍ¡¢Áò»ØÊÕÂÊ´ïµ½97£¥¡¢87£¥¡£ÏÂͼ±íʾÆäÒ±Á¶¼Ó¹¤µÄÁ÷³Ì£º
Ò±Á¶ÖеÄÖ÷Òª·´Ó¦£ºCu2S + O2 =" 2Cu" + SO2
£¨1£©ÑÌÆøÖеÄÖ÷Òª·ÏÆøÊÇ________________£¬´ÓÌá¸ß×ÊÔ´ÀûÓÃÂʺͼõÅÅ¿¼ÂÇ£¬Æä×ÛºÏÀûÓ÷½Ê½ÊÇÖÆ___________¡£
£¨2£©µç½â·¨Á¶Íʱ£¬Ñô¼«ÊÇ____________£¨Ìî¡°´¿Í°å¡±»ò¡°´ÖͰ塱£©£»´ÖÍÖк¬ÓеĽð¡¢ÒøÒÔµ¥ÖʵÄÐÎʽÔÚµç½â²Û_______________£¨Ìî¡°Ñô¼«¡±»ò¡°Òõ¼«¡±µÄ²Ûµ×£¬Òõ¼«µÄµç¼«·´Ó¦Ê½ÊÇ_________________________________________¡£
£¨3£©ÔÚ¾«Á¶ÍµÄ¹ý³ÌÖУ¬µç½âÖÊÈÜÒºÖÐc(Fe2+)¡¢c(Zn2+)»áÖð½¥Ôö´ó¶øÓ°Ïì½øÒ»²½µç½â¡£
¼¸ÖÖÎïÖʵÄÈܶȻý³£Êý£¨KSP£©£º
ÎïÖÊ | Fe(OH)2 | Fe(OH)3 | Zn(OH)2 | Cu(OH)2 |
KSP | 8.0¡Á10£16 | 4.0¡Á10£38 | 3.0¡Á10£17 | 2.2¡Á10£20 |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
ij¹¤Òµ·ÏË®Öк¬ÓÐCN-ºÍCr2OµÈÀë×Ó£¬Ðè¾ÎÛË®´¦Àí´ï±êºó²ÅÄÜÅÅ·Å£¬ÎÛË®´¦Àí³§ÄâÓÃÏÂÁÐÁ÷³Ì½øÐд¦Àí£º
»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©²½Öè¢ÚÖУ¬CN-±»ClO-Ñõ»¯ÎªCNO-µÄÀë×Ó·½³ÌʽΪ________________¡£
£¨2£©²½Öè¢ÛµÄ·´Ó¦ÎªS2O32££«Cr2O72££«H£«SO42££«Cr3£«£«H2O£¨Î´Åäƽ£©£¬ÔòÿÏûºÄ0.4mol Cr2O72£×ªÒÆ__________mol e-¡£
£¨3£©º¬Cr3+·ÏË®¿ÉÒÔ¼ÓÈëÊìʯ»Ò½øÒ»²½´¦Àí£¬Ä¿µÄÊÇ____________________¡£
£¨4£©ÔÚ25¡æÏ£¬½«amol/LµÄNaCNÈÜÒºÓë0.01mol/LµÄÑÎËáµÈÌå»ý»ìºÏ£¬·´Ó¦ºó²âµÃÈÜÒºpH£½7£¬Ôòa________0.01£¨Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°=¡±£©£»Óú¬aµÄ´úÊýʽ±íʾHCNµÄµçÀë³£ÊýKa£½_________________¡£
£¨5£©È¡¹¤Òµ·ÏˮˮÑùÓÚÊÔ¹ÜÖУ¬¼ÓÈëNaOHÈÜÒº¹Û²ìµ½ÓÐÀ¶É«³ÁµíÉú³É£¬¼ÌÐø¼ÓÖÁ²»ÔÙ²úÉúÀ¶É«³ÁµíΪֹ£¬ÔÙÏòÈÜÒºÖмÓÈë×ãÁ¿Na2SÈÜÒº£¬À¶É«³Áµíת»¯³ÉºÚÉ«³Áµí¡£¸Ã¹ý³ÌÖз´Ó¦µÄÀë×Ó·½³ÌÊÇ________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÎÊ´ðÌâ
£¨12·Ö£©D¡¢A¡¢B¡¢CΪËÄÖÖÔ×ÓÐòÊýÒÀ´ÎÔö´óµÄ¶ÌÖÜÆÚÔªËØ£¬A¡¢B¡¢CͬÖÜÆÚ£¬AµÄÔ×Ӱ뾶ÊÇͬÖÜÆÚÖÐ×î´óµÄ£»B¡¢DͬÖ÷×å¡£ÒÑÖªDÔªËصÄÒ»ÖÖµ¥ÖÊÊÇÈÕ³£Éú»îÖÐÒûË®»ú³£ÓõÄÏû¶¾¼Á£¬CÔªËصĵ¥ÖÊ¿ÉÒÔ´ÓA¡¢BÁ½ÔªËØ×é³ÉµÄ»¯ºÏÎïµÄË®ÈÜÒºÖÐÖû»³öBÔªËصĵ¥ÖÊ¡£
£¨1£©CÔªËØÔÚÖÜÆÚ±íÖеÄλÖà ÖÜÆÚ ×å¡£
£¨2£©A¡¢BÔªËØÐγɵij£¼û»¯ºÏÎïË®ÈÜÒºÏÔ ÐÔ£¬ÔÒòÊÇ£¨ÓÃÀë×Ó·½³Ìʽ±íʾ£© ÓÃʯī×öµç¼«µç½â¸Ã»¯ºÏÎïµÄË®ÈÜÒº£¬ÔòÒõ¼«·´Ó¦Ê½Îª £¬
£¨3£©A¡¢DÔªËØ¿ÉÒÔÐγɻ¯ºÏÎïA2D2£¬Ð´³öA2D2ÓëCO2·´Ó¦µÄ»¯Ñ§·½³Ìʽ £¨ÓÃÔªËØ·ûºÅ±íʾ£©¡£¸Ã·´Ó¦Öл¹Ô¼ÁÊÇ ¡£
£¨4£©BÔªËصĵ¥ÖÊÔÚ²»Í¬µÄÌõ¼þÏ¿ÉÒÔÓëO2·¢ÉúһϵÁз´Ó¦£º¢Ù B(s)+O2(g)=BO2(g)£»¡÷H=£296.8kJ/mol¢Ú2BO2(g)+O2(g) 2BO3(g)£»¡÷H=£196.6kJ/mol
Ôò1 mol BO3(g)ÈôÍêÈ«·Ö½â³ÉB£¨s£©£¬·´Ó¦¹ý³ÌÖеÄÈÈЧӦΪ ¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÊµÑéÌâ
£¨14·Ö£©
ÇâäåËáÔÚ¹¤ÒµºÍÒ½Ò©ÁìÓòÖÐÓÐ׏㷺µÄÓÃ;£¬ÏÂͼÊÇijÐËȤС×éÄ£Ä⹤³§ÖƱ¸ÇâäåËá´ÖÆ·²¢¾«ÖƵÄÁ÷³ÌÈçÏ£º
£¨1£©»ìºÏ¢ÙʹÓñùË®µÄÄ¿µÄÊÇ £»
£¨2£©²Ù×÷IIºÍIIIµÄÃû³ÆÊÇ , £»
£¨3£©»ìºÏ¢ÚÖз¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ £»
£¨4£©¹¤ÒµÉÏÓÃÇâäåËáºÍ´óÀíʯÖƵÃä廯¸ÆÖк¬ÓÐÉÙÁ¿Al3+¡¢Fe3+ÔÓÖÊ£¬¼ÓÈëÊÊÁ¿µÄÊÔ¼Á £¨Ìѧʽ£©ºó¿ØÖÆÈÜÒºµÄPHԼΪ8.0¼´¿É³ýÈ¥ÔÓÖÊ£¬¿ØÖÆÈÜÒºµÄPHԼΪ8.0µÄÄ¿µÄÊÇ_______________________________________________________£»
£¨5£©t¡æʱ£¬½«HBrͨÈëAgNO3ÈÜÒºÖÐÉú³ÉµÄAgBrÔÚË®ÖеijÁµíÈܽâƽºâÇúÏßÈçͼËùʾ£¬ÓÖÖªt¡æʱAgClµÄKsp=4¡Ál0-10£¬ÏÂÁÐ˵·¨²»ÕýÈ·µÄÊÇ £¨ £©
A£®Ïòº¬ÓÐCl-ºÍBr- µÄ»ìºÏÒºÖеμÓÏõËáÒøÈÜÒº£¬Ò»¶¨ÏȲúÉúAgBrµÄ³Áµí |
B£®ÔÚAgBr±¥ºÍÈÜÒºÖмÓÈëNaBr¹ÌÌ壬¿ÉʹÈÜÒºÓÉcµãµ½bµã |
C£®Í¼ÖÐaµã¶ÔÓ¦µÄÊÇAgBrµÄ²»±¥ºÍÈÜÒº |
D£®ÔÚt¡æʱ£¬AgCl(s)+Br-(aq)AgBr(s)+Cl-(aq)ƽºâ³£Êý¦ª¡Ö816 |
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