¹¤ÒµºÏ³É°±ÓëÖƱ¸ÏõËáÒ»°ã¿ÉÁ¬ÐøÉú²ú£¬Á÷³ÌÈçÏ£º
£¨1£©¹¤ÒµÉú²úʱ£¬ÖÆÈ¡ÇâÆøµÄÒ»¸ö·´Ó¦Îª£ºCO+H2O(g)CO2+H2
¢Ùt¡æʱ£¬Íù1LÃܱÕÈÝÆ÷ÖгäÈë0.2mol COºÍ0.3molË®ÕôÆø¡£·´Ó¦½¨Á¢Æ½ºâºó£¬ÌåϵÖÐc(H2)=0.12mol¡¤L£1¡£¸ÃζÈÏ´˷´Ó¦µÄƽºâ³£ÊýK=_____£¨Ìî¼ÆËã½á¹û£©¡£
¢Ú±£³ÖζȲ»±ä£¬ÏòÉÏÊöƽºâÌåϵÖÐÔÙ¼ÓÈë0.1molCO£¬µ±·´Ó¦ÖØн¨Á¢Æ½ºâʱ£¬Ë®ÕôÆøµÄת»¯ÂʦÁ(H2O)=________¡£
£¨2£©ºÏ³ÉËþÖз¢Éú·´Ó¦N2(g)+3H2(g)2NH3(g)£»¡÷H<0¡£Ï±íΪ²»Í¬Î¶Èϸ÷´Ó¦µÄƽºâ³£Êý¡£ÓÉ´Ë¿ÉÍÆÖª£¬±íÖÐ
T/K
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â ¹¤ÒµºÏ³É°±ÓëÖƱ¸ÏõËáÒ»°ã¿ÉÁ¬ÐøÉú²ú£¬Á÷³ÌÈçÏ£º £¨1£©¹¤ÒµÉú²úʱ£¬ÖÆÈ¡ÇâÆøµÄÒ»¸ö·´Ó¦Îª£ºCO+H2O£¨g£©?CO2+H2 ¢Ùt¡æʱ£¬Íù1LÃܱÕÈÝÆ÷ÖгäÈë0.2mol COºÍ0.3molË®ÕôÆø£®·´Ó¦½¨Á¢Æ½ºâºó£¬ÌåϵÖÐc£¨H2£©=0.12mol?L-1£®¸ÃζÈÏ´˷´Ó¦µÄƽºâ³£ÊýK= 1 1 £¨Ìî¼ÆËã½á¹û£©£®¢Ú±£³ÖζȲ»±ä£¬ÏòÉÏÊöƽºâÌåϵÖÐÔÙ¼ÓÈë0.1molCO£¬µ±·´Ó¦ÖØн¨Á¢Æ½ºâʱ£¬Ë®ÕôÆøµÄת»¯ÂʦÁ£¨H2O£©= 50% 50% £®£¨2£©ºÏ³ÉËþÖз¢Éú·´Ó¦N2£¨g£©+3H2£¨g£©?2NH3£¨g£©£»¡÷H£¼0£®Ï±íΪ²»Í¬Î¶Èϸ÷´Ó¦µÄƽºâ³£Êý£®ÓÉ´Ë¿ÉÍÆÖª£¬±íÖÐ T1 £¼ £¼ 573K£¨Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°=¡±£©£®£¨3£©NH3ºÍO2ÔÚ²¬Ïµ´ß»¯¼Á×÷ÓÃÏ´Ó145¡æ¾Í¿ªÊ¼·´Ó¦£º4NH3£¨g£©+5O2£¨g£©?4NO£¨g£©+6H2O£¨g£©¡÷H=-905kJ?mol-1£¬²»Í¬Î¶ÈÏÂNO²úÂÊÈçÓÒͼËùʾ£®Î¶ȸßÓÚ900¡æʱ£¬NO²úÂÊϽµµÄÔÒò ζȸßÓÚ900¡æʱ£¬Æ½ºâÏò×óÒƶ¯ ζȸßÓÚ900¡æʱ£¬Æ½ºâÏò×óÒƶ¯ £®£¨4£©·ÏË®ÖеÄN¡¢PÔªËØÊÇÔì³ÉË®Ì帻ӪÑø»¯µÄ¹Ø¼üÒòËØ£¬Å©Ò©³§ÅŷŵķÏË®Öг£º¬Óн϶àµÄNH4+ºÍPO43-£¬Ò»°ã¿ÉÒÔͨ¹ýÁ½ÖÖ·½·¨½«Æä³ýÈ¥£® ¢Ù·½·¨Ò»£º½«Ca£¨OH£©2»òCaO Ͷ¼Óµ½´ý´¦ÀíµÄ·ÏË®ÖУ¬Éú³ÉÁ×Ëá¸Æ£¬´Ó¶ø½øÐлØÊÕ£®µ±´¦ÀíºóµÄ·ÏË®ÖÐc£¨Ca2+£©=2¡Á10-7 mol/Lʱ£¬ÈÜÒºÖÐc£¨PO43-£©= 5¡Á10-7 5¡Á10-7 mol/L£®£¨ÒÑÖªKsp[Ca3£¨PO4£©2]=2¡Á10-33£© ¢Ú·½·¨¶þ£ºÔÚ·ÏË®ÖмÓÈëþ¿ó¹¤Òµ·ÏË®£¬¾Í¿ÉÒÔÉú³É¸ßƷλµÄÁ׿óʯ-Äñ·àʯ£¬·´Ó¦µÄ·½³ÌʽΪMg2++NH4++PO43-=MgNH4PO4¡ý£®¸Ã·½·¨ÖÐÐèÒª¿ØÖÆÎÛË®µÄpHΪ7.5¡«10£¬ÈôpH¸ßÓÚ10.7£¬Äñ·àʯµÄ²úÁ¿»á´ó´ó½µµÍ£®ÆäÔÒò¿ÉÄÜΪ µ±pH¸ßÓÚ10.7ʱ£¬ÈÜÒºÖеÄMg2+¡¢NH4+»áÓëOH-·´Ó¦£¬Æ½ºâÏòÄæ·´Ó¦·½ÏòÒƶ¯ µ±pH¸ßÓÚ10.7ʱ£¬ÈÜÒºÖеÄMg2+¡¢NH4+»áÓëOH-·´Ó¦£¬Æ½ºâÏòÄæ·´Ó¦·½ÏòÒƶ¯ £®Óë·½·¨Ò»Ïà±È£¬·½·¨¶þµÄÓŵãΪÄÜͬʱ³ýÈ¥·ÏË®Öеĵª£¬³ä·ÖÀûÓÃÁËþ¿ó¹¤Òµ·ÏË® ÄÜͬʱ³ýÈ¥·ÏË®Öеĵª£¬³ä·ÖÀûÓÃÁËþ¿ó¹¤Òµ·ÏË® £®²é¿´´ð°¸ºÍ½âÎö>> ¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â £¨2011?ʯ¾°É½Çøһģ£©¹¤ÒµºÏ³É°±ÓëÖƱ¸ÏõËáÒ»°ã¿ÉÁ¬ÐøÉú²ú£¬Á÷³ÌÈçÏ£º £¨1£©¹¤ÒµÉú²úʱ£¬ÖÆÈ¡ÇâÆøµÄÒ»¸ö·´Ó¦Îª£ºCO+H2O£¨g£©?CO2+H2£®t¡æʱ£¬Íù1LÃܱÕÈÝÆ÷ÖгäÈë0.2mol COºÍ0.3molË®ÕôÆø£®·´Ó¦½¨Á¢Æ½ºâºó£¬ÌåϵÖÐc£¨H2£©=0.12mol?L-1£®¸ÃζÈÏ´˷´Ó¦µÄƽºâ³£ÊýK= 1 1 £¨Ìî¼ÆËã½á¹û£©£®£¨2£©ºÏ³ÉÅàÖз¢Éú·´Ó¦N2£¨g£©+3H2£¨g£©?2NH3£¨g£©¡÷H£¼0£®Ï±íΪ²»Í¬Î¶Èϸ÷´Ó¦µÄƽºâ³£Êý£®ÓÉ´Ë¿ÉÍÆÖª£¬±íÖÐT1 £¼ £¼ 300¡æ£¨Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°=¡±£©£®
ζȸßÓÚ900¡æʱ£¬Æ½ºâÏò×óÒƶ¯ ζȸßÓÚ900¡æʱ£¬Æ½ºâÏò×óÒƶ¯ £®£¨4£©ÔÚÉÏÊöÁ÷³ÌͼÖУ¬Ñõ»¯Â¯Öз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ 4NH3+5O2
4NH3+5O2 £®
£¨5£©ÏõË᳧µÄβÆøº¬ÓеªµÄÑõ»¯ÎÈç¹û²»¾´¦ÀíÖ±½ÓÅŷŽ«ÎÛȾ¿ÕÆø£®Ä¿Ç°¿Æѧ¼Ò̽Ë÷ÀûÓÃȼÁÏÆøÌåÖеļ×ÍéµÈ½«µªµÄÑõ»¯ÎﻹÔΪµªÆøºÍË®£¬·´Ó¦»úÀíΪ£º CH4£¨g£©+4NO2£¨g£©¨T4NO£¨g£©+CO2£¨g£©+2H2O£¨g£©¡÷H=-574kJ?mol-1 CH4£¨g£©+4NO£¨g£©¨T2N2£¨g£©+CO2£¨g£©+2H2O£¨g£©¡÷H=-1160kJ?mol-1 Ôò¼×ÍéÖ±½Ó½«N02»¹ÔΪN2µÄÈÈ»¯Ñ§·½³ÌʽΪ£º CH4£¨g£©+2NO2£¨g£©¨TN2£¨g£©+CO2£¨g£©+2H2O£¨g£©¡÷H=-867kJ?mol-1 CH4£¨g£©+2NO2£¨g£©¨TN2£¨g£©+CO2£¨g£©+2H2O£¨g£©¡÷H=-867kJ?mol-1 £®£¨6£©°±ÆøÔÚ´¿ÑõÖÐȼÉÕ£¬Éú³ÉÒ»ÖÖµ¥ÖʺÍË®£¬ÊÔд³ö¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ 4NH3+5O2
4NH3+5O2 £¬¿Æѧ¼ÒÀûÓôËÔÀí£¬Éè¼Æ³É°±ÆøÒ»ÑõÆøȼÁϵç³Ø£¬ÔòͨÈë°±ÆøµÄµç¼«ÊÇ
¸º¼« ¸º¼« £¨Ìî¡°Õý¼«¡±»ò¡°¸º¼«¡±£©£»¼îÐÔÌõ¼þÏ£¬¸Ãµç¼«·¢Éú·´Ó¦µÄµç¼«·´Ó¦Ê½Îª2NH3-6e-+6OH-¡úN2+6H2O 2NH3-6e-+6OH-¡úN2+6H2O £®²é¿´´ð°¸ºÍ½âÎö>> ¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â £¨2011?Õò½Ò»Ä££©¹¤ÒµºÏ³É°±ÓëÖƱ¸ÏõËáÒ»°ã¿ÉÁ¬ÐøÉú²ú£¬Á÷³ÌÈçÏ£º Íê³ÉÏÂÁÐÎÊÌ⣺ ¢ñ£®ºÏ³É°± £¨1£©Ð´³ö×°ÖâÙÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º N2+3H2
N2+3H2 £®
£¨2£©ÒÑÖªÔÚÒ»¶¨µÄζÈϽøÈë×°Öâٵĵª¡¢Çâ»ìºÏÆøÌåÓë´ÓºÏ³ÉËþ³öÀ´µÄ»ìºÏÆøÌåѹǿ֮±ÈΪ5£º4£¬ÔòµªµÄת»¯ÂÊΪ 40% 40% £®¢ò£®°±µÄ½Ó´¥Ñõ»¯ÔÀí £¨3£©ÔÚ900¡æ×°ÖâÚÖз´Ó¦ÓУº? 4NH3£¨g£©+5O2£¨g£©?4NO£¨g£©+6H2O£¨g£©£»¡÷H=-905.5kJ?mol-1 K1=1¡Á1053 £¨900¡æ£© 4NH3£¨g£©+4O2£¨g£©?4N2O£¨g£©+6H2O£¨g£©£»¡÷H=-1103kJ?mol-1 K2=1¡Á1061 £¨900¡æ£© 4NH3£¨g£©+3O2£¨g£©?2N2£¨g£©+6H2O£¨g£©£»¡÷H=-1267kJ?mol-1 K3=1¡Á1067 £¨900¡æ£© ³ýÁËÉÏÁз´Ó¦Í⣬°±ºÍÒ»Ñõ»¯µªÏ໥×÷Ó㺠4NH3£¨g£©+6NO£¨g£©?5N2£¨g£©+6H2O£¨g£©£»¡÷H=-1804kJ?mol-1£¬»¹¿ÉÄÜ·¢Éú°±¡¢Ò»Ñõ»¯µªµÄ·Ö½â£® Íê³ÉÈÈ»¯Ñ§·½³Ìʽ£º2NO£¨g£©?N2£¨g£©+O2£¨g£©£»¡÷H= -180.75 kJ?mol-1 -180.75 kJ?mol-1 £®£¨4£©²¬-îîºÏ½ð´ß»¯¼ÁµÄ´ß»¯»úÀíΪÀë½âºÍ½áºÏÁ½¹ý³Ì£¬ÈçͼËùʾ£º ÓÉÓÚ²¬¶ÔNOºÍË®·Ö×ÓµÄÎü¸½Á¦½ÏС£¬ÓÐÀûÓÚµªÓëÑõÔ×Ó½áºÏ£¬Ê¹µÃNOºÍË®·Ö×ÓÔÚ²¬±íÃæÍѸ½£¬½øÈëÆøÏàÖУ®ÈôûÓÐʹÓò¬-îîºÏ½ð´ß»¯¼Á£¬°±Ñõ»¯½á¹û½«Ö÷ÒªÉú³É µªÆøºÍË®ÕôÆø µªÆøºÍË®ÕôÆø £®ËµÃ÷´ß»¯¼Á¶Ô·´Ó¦ÓÐÑ¡ÔñÐÔ Ñ¡ÔñÐÔ £®£¨5£©Î¶ȶÔÒ»Ñõ»¯µª²úÂʵÄÓ°Ïì µ±Î¶ȴóÓÚ900¡æʱ£¬NOµÄ²úÂÊϽµµÄÔÒò A¡¢B A¡¢B £¨Ñ¡ÌîÐòºÅ£©A£®´Ù½øÁËÒ»Ñõ»¯µªµÄ·Ö½â B£®´Ù½øÁË°±µÄ·Ö½â C£®Ê¹°±ºÍÒ»Ñõ»¯µªµÄ·´Ó¦Æ½ºâÒƶ¯£¬Éú³É¸ü¶àN2 £¨6£©ÏõËṤҵµÄβÆø³£ÓÃNa2CO3ÈÜÒº´¦Àí£¬Î²ÆøµÄNO¡¢NO2¿ÉÈ«²¿±»ÎüÊÕ£¬Ð´³öÓÃNa2CO3ÈÜÒºÎüÊյķ´Ó¦·½³Ìʽ NO+NO2+Na2CO3¨T2NaNO2+CO2¡¢2NO2+Na2CO3¨TNaNO2+NaNO3+CO2 NO+NO2+Na2CO3¨T2NaNO2+CO2¡¢2NO2+Na2CO3¨TNaNO2+NaNO3+CO2 £®²é¿´´ð°¸ºÍ½âÎö>> ¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º £¨2009?ËÞǨģÄ⣩¹¤ÒµºÏ³É°±ÓëÖƱ¸ÏõËáÒ»°ã¿ÉÁ¬ÐøÉú²ú£¬Á÷³ÌÈçͼ1£º ÔÚ800¡æ¡¢²¬´ß»¯¼Á´æÔÚÌõ¼þÏ£¬°±ÓëÑõÆø·´Ó¦µÄÖ÷Òª²úÎïÊÇNO ºÍH2O£®ÔÚʵ¼ÊÉú²úÖа±µÄÑõ»¯ÂÊÓë»ìºÏÆøÖÐÑõ°±±È£¨ÑõÆøÓë°±ÆøÎïÖʵÄÁ¿±È£¬ÒÔ¦Ã{n£¨O2£©/n£¨NH3£©}±íʾ£©µÄ¹ØϵÈçͼ2Ëùʾ£® Çë»Ø´ðÏÂÁи÷Ì⣺ £¨1£©Èô°±Ñõ»¯ÂÊ´ïµ½100%£¬ÀíÂÛÉϦÃ{n£¨O2£©/n £¨NH3£©}Ϊ1.25£¬¶øʵ¼ÊÉú²úÒª½«¦Ãֵά³ÖÔÚ1.7¡«2.2Ö®¼ä£¬ÆäÔÒòÊÇ Ò»·½ÃæÌá¸ßÑõÆøÁ¿£¬Ôö´ó°±µÄÑõ»¯ÂÊ£»Áí·½Ãæ¦ÃÖµÔڸ÷¶Î§£¬°±µÄÑõ»¯ÂÊÒѸߴï95¡«99%£¬ÔÙÌá¸ß£¬°±µÄÑõ»¯ÂÊÉÏÉý¿Õ¼äÒÑÓÐÏÞ£¬·´¶ø»áÔö¼ÓÄܺģ¬Ìá¸ßÉú²ú³É±¾ Ò»·½ÃæÌá¸ßÑõÆøÁ¿£¬Ôö´ó°±µÄÑõ»¯ÂÊ£»Áí·½Ãæ¦ÃÖµÔڸ÷¶Î§£¬°±µÄÑõ»¯ÂÊÒѸߴï95¡«99%£¬ÔÙÌá¸ß£¬°±µÄÑõ»¯ÂÊÉÏÉý¿Õ¼äÒÑÓÐÏÞ£¬·´¶ø»áÔö¼ÓÄܺģ¬Ìá¸ßÉú²ú³É±¾ £®£¨2£©Èôʹ°±µÄÑõ»¯ÂÊ´ïµ½95%£¬Ó¦¿ØÖÆ°±ÔÚ°±¡¢¿ÕÆø»ìºÏÆøÌåÖеÄÌå»ý·ÖÊýԼΪ 10.5% 10.5% £¨ÉèÑõÆøÕ¼¿ÕÆøµÄÌå»ý·ÖÊýΪ20%£©£®½«¦Ã=1.75µÄ°±¡¢¿ÕÆø»ìºÏÆøÌåͨÈë800¡æ¡¢Ê¢Óв¬´ß»¯¼ÁµÄÑõ»¯Â¯£¬³ä·Ö·´Ó¦ºóµ¼Èëµ½ÎüÊÕËþµÄÆøÌåµÄÖ÷Òª³É·ÖÊÇ N2¡¢NO2¡¢H2O N2¡¢NO2¡¢H2O £®£¨3£©ÏÖÒÔa mol NH3ºÍ×ãÁ¿¿ÕÆøΪÔÁÏ£¨²»¿¼ÂÇN2µÄ·´Ó¦£©×î´ó³Ì¶ÈÖÆÈ¡NH4NO3£¬¾¹ýһϵÁÐת»¯·´Ó¦ºó£¬Ïò·´Ó¦ºóµÄ»ìºÏÎïÖмÓÈëb gË®£¬µÃµ½ÃܶÈΪ¦Ñ g?mL-1µÄÈÜÒº£¬¼ÆËã¸ÃÈÜÒºÖÐNH4NO3ÎïÖʵÄÁ¿Å¨¶È¿ÉÄܵÄ×î´óÖµ
²é¿´´ð°¸ºÍ½âÎö>> ¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º ¹¤ÒµºÏ³É°±ÓëÖƱ¸ÏõËáÒ»°ã¿ÉÁ¬ÐøÉú²ú£¬Á÷³ÌÈçͼ£º £¨1£©¹¤ÒµÉú²úʱ£¬ÖÆÈ¡ÇâÆøµÄÒ»¸ö·´Ó¦Îª£ºCO£¨g£©+H2O£¨g£©?CO2£¨g£©+H2£¨g£©£®t¡æʱ£¬Íù10LÃܱÕÈÝÆ÷ÖгäÈë2mol COºÍ3molË®ÕôÆø£®·´Ó¦½¨Á¢Æ½ºâºó£¬ÌåϵÖÐc£¨H2£©=0.12mol?L-1£®Ôò¸ÃζÈÏ´˷´Ó¦µÄƽºâ³£ÊýK= £¨2£©ºÏ³ÉËþÖз¢Éú·´Ó¦N2£¨g£©+3H2£¨g£©?2NH3£¨g£©¡÷H£¼0£®Ï±íΪ²»Í¬Î¶Èϸ÷´Ó¦µÄƽºâ³£Êý£®ÓÉ´Ë¿ÉÍÆÖª£¬±íÖÐT1
£¨4£©Óð±ÆøÑõ»¯¿ÉÒÔÉú²úÏõËᣬµ«Î²ÆøÖеÄNOx»áÎÛȾ¿ÕÆø£®Ä¿Ç°¿Æѧ¼Ò̽Ë÷ÀûÓÃȼÁÏÆøÌåÖеļ×ÍéµÈ½«µªµÄÑõ»¯ÎﻹÔΪµªÆøºÍË®£¬·´Ó¦»úÀíΪ£º CH4£¨g£©+4NO2£¨g£©¨T4NO£¨g£©+CO2£¨g£©+2H2O£¨g£©¡÷H=-574kJ?mol-1 CH4£¨g£©+4NO£¨g£©¨T2N2£¨g£©+CO2£¨g£©+2H2O£¨g£©¡÷H=-1160kJ?mol-1 Ôò¼×ÍéÖ±½Ó½«NO2»¹ÔΪN2µÄÈÈ»¯Ñ§·½³ÌʽΪ £¨5£©Ä³Ñо¿Ð¡×éÔÚʵÑéÊÒÒÔ¡°Ag-ZSM-5¡±Îª´ß»¯¼Á£¬²âµÃ½«NOת»¯ÎªN2µÄת»¯ÂÊËæζȱ仯Çé¿öÈçÏÂͼ£®¾Ýͼ·ÖÎö£¬Èô²»Ê¹ÓÃCO£¬Î¶ȳ¬¹ý775¡æ£¬·¢ÏÖNOµÄת»¯ÂʽµµÍ£¬Æä¿ÉÄܵÄÔÒòΪ
²é¿´´ð°¸ºÍ½âÎö>> ͬ²½Á·Ï°²á´ð°¸ °Ù¶ÈÖÂÐÅ - Á·Ï°²áÁбí - ÊÔÌâÁÐ±í 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°æȨÉùÃ÷£º±¾Õ¾ËùÓÐÎÄÕ£¬Í¼Æ¬À´Ô´ÓÚÍøÂ磬Öø×÷Ȩ¼°°æȨ¹éÔ×÷ÕßËùÓУ¬×ªÔØÎÞÒâÇÖ·¸°æȨ£¬ÈçÓÐÇÖȨ£¬Çë×÷ÕßËÙÀ´º¯¸æÖª£¬ÎÒÃǽ«¾¡¿ì´¦Àí£¬ÁªÏµqq£º3310059649¡£ ICP±¸°¸ÐòºÅ: »¦ICP±¸07509807ºÅ-10 ¶õ¹«Íø°²±¸42018502000812ºÅ |