ÌâÄ¿ÄÚÈÝ

ʵÑéÊÒÒÔ¹¤ÒµÌ¼Ëá¸Æ£¨º¬ÉÙÁ¿Na+¡¢Al3+¡¢Fe3+µÈÔÓÖÊ£©ÎªÔ­ÁÏÖÆÈ¡CaCl2¡¤H2OºÍCaO2µÄÖ÷ÒªÁ÷³ÌÈçÏ£º

£¨1£©¼ÓÈëÊÔ¼ÁX£¬µ÷½ÚÈÜÒºpHΪ¼îÐÔ»·¾³£¬ÒÔ³ýÈ¥ÈÜÒºÖÐAl3+ºÍFe3+£¬ÂËÔüµÄÖ÷Òª³É·ÖÊÇ___________¡£ÊÔ¼ÁX¿ÉÒÔÑ¡ÔñÏÂÁеÄ________£¨Ìî±àºÅ£©¡£

A£®CaO B£®CaCO3 C£®NH3¡¤H2D£®Ba(OH)2 
£¨2£©²Ù×÷IIÖнøÐÐÕô·¢Å¨Ëõʱ£¬³ýÈý½Ç¼Ü¡¢¾Æ¾«µÆÍ⣬»¹ÐèÒªµÄÒÇÆ÷ÓÐ__________¡£
£¨3£©ÓÉCaCl2ÖÆÈ¡CaO2µÄ·´Ó¦ÖУ¬Î¶Ȳ»ÒËÌ«¸ßµÄÔ­ÒòÊÇ_______________¡£
£¨4£©ÓÃÏÂÁÐ×°Öòⶨ¹¤ÒµÌ¼Ëá¸ÆµÄÖÊÁ¿·ÖÊý

¢Ù¼ìÑé×°ÖÃBÆøÃÜÐÔÁ¼ºÃµÄʵÑé¹ý³ÌÊÇ__________________________¡£
¢Ú°´A¡ªB¡ªC¡ªD˳ÐòÁ¬½Ó£¬È»ºó´ÓA×°ÖÃͨÈë¿ÕÆø£¬Ä¿µÄÊÇ_______________¡£
¢Û×°ÖÃDµÄ×÷ÓÃΪ______________________¡£
¢ÜʵÑéʱ£¬×¼È·³ÆÈ¡10.00g¹¤ÒµÌ¼Ëá¸Æ3·Ý£¬½øÐÐ3´Î²â¶¨£¬²âµÃBaCO3³ÁµíµÄƽ¾ùÖÊÁ¿Îª17.73g£¬ÔòÑùÆ·ÖÐCaCO3µÄÖÊÁ¿·ÖÊýΪ__________________¡£

£¨³ýÈ¥×¢Ã÷Í⣬ÿ¿Õ2·Ö£¬¹²14·Ö£©£¨1£©Fe(OH)3ºÍAl(OH)3   AC £¨2£©Õô·¢Ãó¡¢²£Á§°ô£¨ÛáÛöǯ£©
£¨3£©·Àֹζȹý¸ßH2O2·Ö½â
£¨4£©¢Ù¹Ø±Õ·ÖҺ©¶·µÄ»îÈû£¬ÓÃֹˮ¼Ð¼Ðס×ó±ß£¨»òÓұ߻ò²»Óã©ÏðƤ¹Ü´¦£¬½«Óұߣ¨»ò×ó±ß»òÁ½±ß£©µ¼Æø¹Ü²åÈëË®ÖУ¬Î¢ÈÈB×°Ö㬷¢ÏÖµ¼Æø¹ÜÓÐÆøÅݲúÉú£¬ÀäÈ´ºó£¬µ¼Æø¹ÜÓÐÒ»¶ÎÒºÖùÐγɣ¬×¢Ã÷ÆøÃÜÐÔÁ¼ºÃ£¨ºÏÀí¼´¿É£©   ¢ÚÅž»×°ÖÃÖжþÑõ»¯Ì¼£¨1·Ö£©
¢Û·ÀÖ¹¿ÕÆøÖÐCO2ÓëBa(OH)2ÈÜÒº·´Ó¦£¨1·Ö£©  ¢Ü90%

½âÎöÊÔÌâ·ÖÎö£º£¨1£©ÔÚ¼îÐÔÌõ¼þÏ£¬Al3+¡¢Fe3+·Ö±ðÉú³ÉÇâÑõ»¯ÂÁºÍÇâÑõ»¯Ìú³Áµí£¬ËùÒÔÂËÔüµÄÖ÷Òª³É·ÖÊÇFe(OH)3ºÍAl(OH)3£»ÓÉÓÚʵÑéÊÒÒÔ¹¤ÒµÌ¼Ëá¸ÆΪԭÁÏÖÆÈ¡CaCl2¡¤H2OºÍCaO2µÄ£¬ËùÒÔÊÔ¼ÁX²»ÄÜÑ¡ÓÃ̼Ëá¸Æ¡£ÓÖÒòΪ²»ÄÜÒýÈëеÄÔÓÖÊ£¬ËùÒÔ²»ÄÜÑ¡ÔñÇâÑõ»¯±µ£¬¿ÉÒÔÑ¡ÔñÑõ»¯¸Æ¡£ÓÉÓÚºóÐøʵÑéÖмÓÈë̼Ëá泥¬ËùÒÔX»¹¿ÉÒÔÊÇ°±Ë®£¬¼´´ð°¸Ñ¡AC¡£
£¨2£©ÓÉÓÚÂÈ»¯¸ÆÒ×ÈÜÓÚË®£¬ËùÒÔÒªµÃµ½ÂÈ»¯¸Æ¾§Ì壬Ӧ¸ÃÊÇÕô·¢Å¨ËõÀäÈ´½á¾§£¬ËùÒÔ²Ù×÷IIÖнøÐÐÕô·¢Å¨Ëõʱ£¬³ýÈý½Ç¼Ü¡¢¾Æ¾«µÆÍ⣬»¹ÐèÒªµÄÒÇÆ÷ÓÐÕô·¢Ãó¡¢²£Á§°ô£¨ÛáÛöǯ£©¡£
£¨3£©ÓÉCaCl2ÖÆÈ¡CaO2µÄ·´Ó¦ÖУ¬ÐèҪ˫ÑõË®²Î¼Ó¡£ÓÉÓÚË«ÑõË®Ò׷ֽ⣬ËùÒÔζȲ»ÒËÌ«¸ßµÄÔ­ÒòÊÇ£©·Àֹζȹý¸ßH2O2·Ö½â¡£
£¨4£©¢ÙÓÉÓÚB×°ÖÃÖк¬ÓзÖҺ©¶·£¬ËùÒÔÒª¼ìÑéÆäÆøÃÜÐÔ£¬ÕýÈ·µÄ²Ù×÷Ó¦¸ÃÊǹرշÖҺ©¶·µÄ»îÈû£¬ÓÃֹˮ¼Ð¼Ðס×ó±ß£¨»òÓұ߻ò²»Óã©ÏðƤ¹Ü´¦£¬½«Óұߣ¨»ò×ó±ß»òÁ½±ß£©µ¼Æø¹Ü²åÈëË®ÖУ¬Î¢ÈÈB×°Ö㬷¢ÏÖµ¼Æø¹ÜÓÐÆøÅݲúÉú£¬ÀäÈ´ºó£¬µ¼Æø¹ÜÓÐÒ»¶ÎÒºÖùÐγɣ¬×¢Ã÷ÆøÃÜÐÔÁ¼ºÃ¡£
¢ÚʵÑéÔ­ÀíÊÇÀûÓÃÑÎËáºÍ̼Ëá¸Æ·´Ó¦Éú³ÉCO2£¬CO2±»ÇâÑõ»¯±µÎüÊÕÉú³É̼Ëá±µ³Áµí£¬È»ºóͨ¹ý³ÆÁ¿C×°Öõļ´¿É¡£ÓÉÓÚÔÚ·´Ó¦ÖÐ×°ÖÃÖлá²ÐÁôCO2£¬ËùÒÔͨÈë¿ÕÆøµÄÄ¿µÄÊÇÅž»×°ÖÃÖжþÑõ»¯Ì¼¡£
¢ÛÓÉÓÚ¿ÕÆøÖк¬ÓÐCO2£¬Òò´ËD×°ÖÃÖмîʯ»ÒµÄ×÷ÓÃÊÇ·ÀÖ¹¿ÕÆøÖÐCO2ÓëBa(OH)2ÈÜÒº·´Ó¦¡£
¢Ü¸ù¾Ý̼ԭ×ÓÊغã¿ÉÖª£¬CaCO3µÄÖÊÁ¿ÊÇ¡Á100g/mol£½9.00g£¬ËùÒÔÑùÆ·ÖÐCaCO3µÄÖÊÁ¿·ÖÊýΪ¡Á100%£½90%¡£
¿¼µã£º¿¼²éÎïÖʵijýÔÓ¡¢ÊÔ¼ÁºÍÒÇÆ÷µÄÑ¡Ôñ¡¢·´Ó¦Ìõ¼þ¿ÉÖª¡¢×°ÖÃÆøÃÜÐÔ¼ìÑé¡¢ÎïÖʺ¬Á¿µÄ²â¶¨µÈ

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

ÂÈÆøÊÇÒ»ÖÖÖØÒªµÄ»¯¹¤Ô­ÁÏ¡£Ä³Ñ§Ï°Ð¡×éÔÚʵÑéÊÒÖÐÀûÓÃÏÂͼËùʾװÖÃÖÆÈ¡ÂÈÆø²¢Ì½¾¿ÆäÐÔÖÊ¡£

£¨1£©ÊµÑéÊÒÓöþÑõ»¯Ã̺ÍŨÑÎËá¼ÓÈÈÖÆÈ¡ÂÈÆø£¬·¢Éú×°ÖÃÖгýÔ²µ×ÉÕÆ¿ºÍµ¼¹ÜÍ⻹ÐèÓõ½µÄ²£Á§ÒÇÆ÷ÓÐ
                             £»
£¨2£©×°ÖÃAÖÐÊ¢ÓеÄÊÔ¼ÁÊÇ       £¬×÷ÓÃÊÇ                          ¡£
£¨3£©ÈôDÖÐÆ·ºìÈÜÒºÍÊÉ«£¬ÔòB×°ÖÃÖз¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ                     
£¨4£©Ö¤Ã÷FeBr2ÓëCl2·¢ÉúÁËÑõ»¯»¹Ô­·´Ó¦µÄʵÑé·½·¨ÊÇ             £¨Ìî²Ù×÷·½·¨£©¡£
ijÑо¿ÐÔѧϰС×éÓøÕÎüÊÕ¹ýÉÙÁ¿SO2µÄNaOHÈÜÒºÎüÊÕ´¦ÀíÉÏÊöʵÑéºóµÄβÆø¡£¾­·ÖÎöÎüÊÕβÆøÒ»¶Îʱ¼äºó£¬ÎüÊÕÒº£¨³ÊÇ¿¼îÐÔ£©Öп϶¨´æÔÚCl£­¡¢OH£­¡¢CO32- ºÍSO42£­£¬¶ÔÓÚ¿ÉÄÜ´æÔÚµÄÆäËûÒõÀë×Ó£¬Ñо¿Ð¡×éÌá³öÒÔÏÂ3ÖÖ¼ÙÉè¡£¼ÙÉè1£ºÖ»´æÔÚSO32£­£»¼ÙÉè2£ºÖ»´æÔÚClO£­£»¼ÙÉè3£º¼È²»´æÔÚSO32£­£¬Ò²²»´æÔÚClO£­¡£
£¨5£©Ñ§Ï°Ð¡×éÅжÏͬʱ´æÔÚSO32£­ºÍClO£­ÊDz»¿ÉÄܵÄÀíÓÉÊÇ                ¡£
38.ÏÖÏÞÑ¡ÒÔÏÂÊÔ¼Á£¬Éè¼ÆʵÑé·½°¸£¬½øÐÐʵÑ飬Çëд³öʵÑé²½ÖèÒÔ¼°Ô¤ÆÚÏÖÏóºÍ½áÂÛ¡£
a£®3 mol/L H2SO4
b£®0.01 mol/L KMnO4
c£®1 mol/L BaCl2ÈÜÒº
d£®µí·Ûµâ»¯¼ØÈÜÒº
e£®·Ó̪ÊÔÒº
²½ÖèÒ»£»È¡ÉÙÁ¿ÎüÊÕÒºÓÚÊÔ¹ÜÖУ¬µÎ¼Ó3 mol/L H2SO4ÖÁÈÜÒº³ÊËáÐÔ£¬È»ºó½«ËùµÃÈÜÒº·Ö×°ÓÚA¡¢B
Á½ÊÔ¹ÜÖС£
²½Öè¶þ£ºÏòAÊÔ¹ÜÖеμÓÉÙÁ¿___________ (ÌîÐòºÅ)£¬ÈôÈÜÒº_________________£¨ÌîÏÖÏ󣩣¬Ôò¼ÙÉè
1³ÉÁ¢¡£
²½ÖèÈý£ºÏòBÊÔ¹ÜÖеμÓÉÙÁ¿___________£¨ÌîÐòºÅ£©£¬ÈôÈÜÒº_________________£¨ÌîÏÖÏ󣩣¬Ôò¼ÙÉè2
³ÉÁ¢¡£

µâÔªËØÓС°ÖÇÁ¦ÔªËØ¡±Ö®³Æ¡£Ñо¿ÐÔѧϰС×é×öÁËÈçÏÂʵÑé̽¾¿º£´øÖеâÔªËØ´æÔÚ²¢²â¶¨ÆäÖеâÔªËصĺ¬Á¿¡£

£¨1£©²Ù×÷IΪ×ÆÉÕ£¬Ôò×ÆÉÕʱÓÃ____Ê¢×°º£´ø£¬²Ù×÷IIΪ____________________________£»
£¨2£©Ë®½þʱͨ³£Òª½«Ðü×ÇÒºÖó·Ð2¡«3min£¬Ä¿µÄÊÇ______________________________£»
£¨3£©²Ù×÷III£¬ÊÇͬѧÃǶÔÈÜÒºAÖеâÔªËصĴæÔÚÐÎʽ½øÐеÄ̽¾¿ÊµÑé¡£
[ÍƲâ]£º¢ÙÒÔIO3¡¥ÐÎʽ´æÔÚ£» ¢ÚÒÔI¡¥ÐÎʽ´æÔÚ
[²éÔÄ×ÊÁÏ]£ºIO3¡¥¾ßÓнÏÇ¿µÄÑõ»¯ÐÔ£¬I2+2S2O32¡¥=2I¡¥+S4O62¡¥
½«ÉÏÊöÈÜҺϡÊÍÅäÖƳÉ200mLÈÜÒº£¬ÇëÍê³ÉÏÂÁÐʵÑé̽¾¿¡£ÏÞÑ¡ÊÔ¼Á£º3%H2O2ÈÜÒº¡¢KSCNÈÜÒº¡¢FeCl2ÈÜÒº¡¢Ï¡ÁòËá¡£

ÐòºÅ
ʵÑé·½°¸
ʵÑéÏÖÏó
½áÂÛ
¢Ù
È¡ÉÙÁ¿Ï¡ÊͺóµÄÈÜÒºA¼ÓÈëµí·ÛºóÔÙÓÃÁòËáËữ£¬·Ö×°ÓÚÊÔ¹ÜI¡¢II
ÎÞÏÖÏó
 
¢Ú
ÍùÊÔ¹ÜIÖмÓÈë______
ÎÞÏÖÏó
Ö¤Ã÷²»ÊÇÒÔIO3¡¥ÐÎʽ´æÔÚ
¢Û
ÍùÊÔ¹ÜIIÖмÓÈë_______
_______________
Ö¤Ã÷ÒÔÐÎʽ´æÔÚ
 
£¨4£©¶¨Á¿¼ìÑ麣´øÖеĵ⺬Á¿£º
¢ÙÈ¡20mLÏ¡ÊͺóÈÜÒºA·Ö±ðÓÚ׶ÐÎÆ¿£¬·Ö±ðÓÃËáʽµÎ¶¨¹ÜµÎ¼Ó0.01mol/LKMnO4ÈÜÒºÖÁÈÜÒº¸ÕÏÔdzºìÉ«£¬½«I¡¥Ñõ»¯ÎªI2²¢µÃµ½ÈÜÒºB£»
¢ÚÔÚÈÜÒºB¼ÓÈëÁ½µÎµí·ÛÈÜÒº£¬ÓÃ0.01mol/LNa2S2O3ÈÜÒº£¬µÎ¶¨ÖÁÖյ㣬ÖÕµãÏÖÏóΪ___________£¬¼Ç¼Êý¾Ý£¬Öظ´Éϲⶨ²½Öè¢Ù¡¢¢ÚÁ½´Î£¬Èý´Îƽ¾ùÏûºÄNa2S2O3ÈÜÒºÌå»ýΪVmL£¬¼ÆË㺣´øÖеâÔªËصİٷֺ¬Á¿_________________£®£¨¼ÙÉè²Ù×÷I¡¢II¹ý³ÌÖеⲻËðʧ£¬Ïà¶ÔÔ­×ÓÁ¿I£­127£©

±µÑÎÉú²úÖÐÅųö´óÁ¿µÄ±µÄà[Ö÷Òªº¬BaCO3¡¢BaSO3¡¢Ba(FeO2)2µÈ]£¬Ä³Ö÷ÒªÉú²úBaCO3µÄ»¯¹¤³§ÀûÓñµÄàÖÆÈ¡Ba(NO3)2¾§Ìå¼°ÆäËû¸±²úÎÆ䲿·Ö¹¤ÒÕÁ÷³ÌÈçÏ£º

ÒÑÖª£º¢Ù Fe(OH)3ºÍFe(OH)2ÍêÈ«³Áµíʱ£¬ÈÜÒºµÄpH·Ö±ðΪ3.2ºÍ9.7
¢Ú Ba(NO3)2ÔÚÈÈË®ÖÐÈܽâ¶È½Ï´ó£¬ÔÚÀäË®ÖÐÈܽâ¶È½ÏС
¢Û KSP(BaSO4)=1.1¡Á10£­10£¬KSP (BaCO3)=5.1¡Á10£­9
£¨1£©¸Ã³§Éú²úµÄBaCO3Òòº¬ÓÐÉÙÁ¿BaSO4¶ø²»´¿£¬Ìá´¿µÄ·½·¨ÊÇ£º½«²úÆ·¼ÓÈë×ãÁ¿µÄ±¥ºÍNa2CO3ÈÜÒºÖУ¬³ä·Ö½Á°è£¬¹ýÂË£¬Ï´µÓ¡£ÓÃÀë×Ó·½³Ìʽ˵Ã÷Ìá´¿Ô­Àí£º                                  ¡£
£¨2£©ÉÏÊöÁ÷³ÌËáÈÜʱ£¬Ba(FeO2)2ÓëHNO3·´Ó¦Éú³ÉÁ½ÖÖÏõËáÑΣ¬»¯Ñ§·½³ÌʽΪ£º
                                                                    ¡£
£¨3£©¸Ã³§½áºÏ±¾³§Êµ¼Ê£¬Ñ¡ÓõÄXΪ       £»

A£®BaCl2 B£®BaCO3 C£®Ba(NO3)2 D£®Ba(OH)2 
£¨4£©·ÏÔü2Ϊ                    ¡£
£¨5£©²Ù×÷IIIΪ                                 ¡£
£¨6£©¹ýÂË3ºóµÄĸҺӦѭ»·µ½ÈÝÆ÷         ÖС££¨Ìî¡°a¡±¡¢¡°b¡±»ò¡°c¡±£©
£¨7£©³ÆÈ¡w¿Ë¾§ÌåÈÜÓÚÕôÁóË®£¬¼ÓÈë×ãÁ¿µÄÁòËᣬ³ä·Ö·´Ó¦ºó£¬¹ýÂË¡¢Ï´µÓ¡¢¸ÉÔ³ÆÁ¿³ÁµíÖÊÁ¿Îªm¿Ë£¬Ôò¸ÃBa(NO3)2µÄ´¿¶ÈΪ             £¨Ïà¶Ô·Ö×ÓÖÊÁ¿£ºBa(NO3)2Ϊ261£¬BaSO4Ϊ233£©¡£

1£­ÒÒÑõ»ùÝÁÊÇÒ»ÖÖÎÞÉ«ÒºÌ壬ÃܶȱÈË®´ó£¬²»ÈÜÓÚË®£¬Ò×ÈÜÓÚÒÒ´¼£¬ÈÛµã5.5¡æ£¬·Ðµã267¡æ¡£1£­ÝÁ·Ó£¨ÐÔÖÊÓë±½·ÓÏàËÆ£©ÈÛµã96¡æ£¬·Ðµã278¡æ£¬Î¢ÈÜÓÚË®£¬Ò×ÈÜÓÚÒÒ´¼£¬ÒÒ´¼µÄ·ÐµãΪ78.5¡æ¡£1£­ÒÒÑõ»ùÝÁ³£ÓÃ×÷ÏãÁÏ£¬Ò²¿ÉºÏ³ÉÆäËûÏãÁÏ¡£ÊµÑéÊÒÖƱ¸1£­ÒÒÑõ»ùÝÁµÄ¹ý³ÌÈçÏ£º
£«C2H5OH£«H2O
1£­ÝÁ·Ó                      1£­ÒÒÑõ»ùÝÁ
£¨1£©½«72g1£­ÝÁ·ÓÈÜÓÚ100mLÎÞË®ÒÒ´¼ÖУ¬¼ÓÈë5mLŨÁòËá»ìºÏ¡£½«»ìºÏÒºÖÃÓÚÈçͼËùʾµÄÈÝÆ÷ÖмÓÈȳä·Ö·´Ó¦¡£ÊµÑéÖÐʹÓùýÁ¿ÒÒ´¼µÄÔ­ÒòÊÇ                            ¡£ÉÕÆ¿ÉÏÁ¬½Ó³¤Ö±²£Á§¹ÜµÄÖ÷Òª×÷ÓÃÊÇ                                     ¡£

£¨2£©·´Ó¦½áÊø£¬½«ÉÕÆ¿ÖеÄÒºÌåµ¹ÈëÀäË®ÖУ¬¾­´¦ÀíµÃµ½Óлú²ã¡£ÎªÌá´¿²úÎïÓÐÒÔÏÂËIJ½²Ù×÷£º¢ÙÕôÁ󣻢Úˮϴ²¢·ÖÒº£»¢ÛÓÃ10%µÄNaOHÈÜÒº¼îÏ´²¢·ÖÒº£»¢ÜÓÃÎÞË®ÂÈ»¯¸Æ¸ÉÔï²¢¹ýÂË¡£ÕýÈ·µÄ˳ÐòÊÇ       £¨ÌîÐòºÅ£©¡£
A£®¢Û¢Ú¢Ü¢Ù     B£®¢Ù¢Ú¢Û¢Ü    C£®¢Ú¢Ù¢Û¢Ü
£¨3£©ÊµÑé²âµÃ1£­ÒÒÑõ»ùÝÁµÄ²úÁ¿Ó뷴Ӧʱ¼ä¡¢Î¶ȵı仯ÈçͼËùʾ£¬Ê±¼äÑÓ³¤¡¢Î¶ÈÉý¸ß£¬1£­ÒÒÑõ»ùÝÁµÄ²úÁ¿Ï½µµÄÔ­Òò¿ÉÄÜÊÇ                               ¡¢                            ¡£

£¨4£©Ä³Í¬Ñ§ÍƲ⾭Ìá´¿µÄ²úÆ·¿ÉÄÜ»¹º¬ÓÐ1£­ÝÁ·Ó¡¢ÒÒ´¼¡¢ÁòËáºÍË®µÈÔÓÖÊ£¬Éè¼ÆÁËÈçÏ·½°¸½øÐмìÑ飬ÇëÔÚ´ðÌ⿨ÉÏÍê³É±íÖÐÄÚÈÝ¡£

ʵÑéÄ¿µÄ
ʵÑé²Ù×÷
Ô¤ÆÚÏÖÏóºÍ½áÂÛ
¢ÙÓýðÊôÄƼìÑé1£­ÒÒÑõ»ùÝÁÊÇ·ñ´¿¾»
È¡ÉÙÁ¿¾­Ìá´¿µÄ²úÆ·ÓÚÊÔ¹ÜAÖУ¬¼ÓÈë½ðÊôÄÆ
Èô               £¬Ôò²úÆ·´¿¾»£»
Èô               £¬Ôò²úÆ·²»´¿¡£
¢Ú¼ìÑé¾­Ìá´¿µÄ²úÆ·ÊÇ·ñº¬ÓÐ1£­ÝÁ·Ó
                      
                      
Èô             £¬Ôòº¬ÓÐ1£­ÝÁ·Ó£»
Èô              £¬Ôò²»º¬1£­ÝÁ·Ó¡£
 

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

¾«Ó¢¼Ò½ÌÍø