£¨9·Ö£©½«0.6 molÍͶÈë100mLÒ»¶¨Å¨¶ÈµÄHNO3£¨×ãÁ¿£©ÖУ¬ÓÉÓÚHNO3µÄŨ¶È²»Í¬£¬¿ÉÄÜ·¢ÉúÁ½¸ö·´Ó¦£¨¼ÙÉèÿ¸ö·´Ó¦ÖвúÉúµÄÆøÌåÖ»ÓÐÒ»ÖÖ£©¡£
£¨1£©ÒÑÖªÁ½¸ö·´Ó¦ÖШD¸ö·´Ó¦µÄÀë×Ó·½³Ìʽ£¨·´Ó¦¢Ú£©£¬Ð´³öÁí¨D¸ö·´Ó¦£¨·´Ó¦¢Ù£©µÄÀë×Ó·½³Ìʽ£º
¢Ù____________________________________________________________________£»
¢Ú3Cu+8H++2NO3¡¥ ===== 3Cu2++2NO¡ü+4H2O¡£
£¨2£©¼ÙÉèÖ»·¢Éú·´Ó¦¢Ù£¬²úÉúµÄÆøÌåÔÚ±ê×¼×´¿öϵÄÌå»ýΪ_____L£»¼ÙÉèÖ»·¢Éú·´Ó¦¢Ú£¬²úÉúµÄÆøÌåÔÚ±ê×¼×´¿öϵÄÌå»ýΪ________L¡£
£¨3£©Êµ¼Ê²úÉúµÄÆøÌåÔÚ±ê×¼×´¿öϵÄÌå»ýΪ22.4 L£¬Ôò²Î¼Ó·´Ó¦¢ÙµÄÍÓë²Î¼Ó·´Ó¦¢ÚµÄ͵ÄÎïÖʵÄÁ¿Ö®±ÈΪ___________¡£
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£ºÔĶÁÀí½â
¡°ÇâÄÜ¡±½«ÊÇδÀ´×îÀíÏëµÄÐÂÄÜÔ´¡£
¢ñ.ʵÑé²âµÃ£¬1gÇâÆøȼÉÕÉú³ÉҺ̬ˮʱ·Å³ö142.9kJÈÈÁ¿£¬ÔòÇâÆøȼÉÕµÄÈÈ»¯Ñ§·½³ÌʽΪ_______¡££¨ÌîÐòºÅ£©
A.2H2£¨g£©+O2£¨g£© 2H2O£¨l£© ¡÷H= £142.9kJ¡¤mol¡ª1
B.H2£¨g£©+1/2 O2£¨g£© H2O£¨l£© ¡÷H= £285.8kJ¡¤mol¡ª1
C.2H2+O22H2O£¨l£© ¡÷H= £571.6kJ¡¤mol¡ª1
D.H2£¨g£©+1/2O2£¨g£© H2O£¨g£© ¡÷H= £285.8kJ¡¤mol¡ª1
¢ò.ij»¯Ñ§¼Ò¸ù¾Ý¡°Ô×Ó¾¼Ã¡±µÄ˼Ï룬Éè¼ÆÁËÈçÏÂÖƱ¸H2µÄ·´Ó¦²½Öè
¢ÙCaBr2+H2OCaO+2HBr ¢Ú2HBr+HgHgBr2+H2
¢ÛHgBr2+___________________ ¢Ü2HgO2Hg+O2¡ü
ÇëÄã¸ù¾Ý¡°Ô×Ó¾¼Ã¡±µÄ˼ÏëÍê³ÉÉÏÊö²½Öè¢ÛµÄ»¯Ñ§·½³Ìʽ£º____________¡£
¸ù¾Ý¡°ÂÌÉ«»¯Ñ§¡±µÄ˼ÏëÆÀ¹À¸Ã·½·¨ÖÆH2µÄÖ÷Ҫȱµã£º______________¡£
¢ó.ÀûÓúËÄÜ°ÑË®·Ö½âÖÆÇâÆø£¬ÊÇÄ¿Ç°ÕýÔÚÑо¿µÄ¿ÎÌâ¡£ÏÂͼÊÇÆäÖеÄÒ»ÖÖÁ÷³Ì£¬ÆäÖÐÓÃÁ˹ýÁ¿µÄµâ¡££¨Ìáʾ£º·´Ó¦¢ÚµÄ²úÎïÊÇO2¡¢SO2ºÍH2O£©
Íê³ÉÏÂÁз´Ó¦µÄ»¯Ñ§·½³Ìʽ£º
·´Ó¦¢Ù__________________________£»·´Ó¦¢Ú__________________________¡£
´Ë·¨ÖÆÈ¡ÇâÆøµÄ×î´óÓŵãÊÇ_______________________________________________¡£
¢ô.ÇâÆøͨ³£ÓÃÉú²úˮúÆøµÄ·½·¨ÖƵá£ÆäÖÐCO£¨g£©+ H2O£¨g£© CO2£¨g£©+ H2£¨g£©; ¡÷H<0¡£
ÔÚ850¡æʱ£¬K=1¡£
£¨1£©ÈôÉý¸ßζȵ½950¡æʱ£¬´ïµ½Æ½ºâʱK______1£¨Ìî¡°´óÓÚ¡±¡¢¡°Ð¡ÓÚ¡±»ò¡°µÈÓÚ¡±£©
£¨2£©850¡æʱ£¬ÈôÏòÒ»ÈÝ»ý¿É±äµÄÃܱÕÈÝÆ÷ÖÐͬʱ³äÈë 1.0 mol CO¡¢3.0molH2O¡¢1.0mol CO2 ºÍ x mol H2£¬Ôò£º
¢Ùµ±x=5.0ʱ£¬ÉÏÊöƽºâÏò___________£¨ÌîÕý·´Ó¦»òÄæ·´Ó¦£©·½Ïò½øÐС£
¢ÚÈôҪʹÉÏÊö·´Ó¦¿ªÊ¼Ê±ÏòÕý·´Ó¦·½Ïò½øÐУ¬ÔòxÓ¦Âú×ãµÄÌõ¼þÊÇ__________¡£
£¨3£©ÔÚ850¡æʱ£¬ÈôÉèx£½5.0molºÍx£½6.0mol£¬ÆäËüÎïÖʵÄͶÁϲ»±ä£¬µ±ÉÏÊö·´Ó¦´ïµ½Æ½ºâºó£¬²âµÃH2µÄÌå»ý·ÖÊý·Ö±ðΪa£¥¡¢b£¥£¬Ôòa_______ b£¨Ìî¡°´óÓÚ¡±¡¢¡°Ð¡ÓÚ¡±»ò¡°µÈÓÚ¡±£©¡£
¢õ.ÇâÆø»¹ÔÑõ»¯ÍËùµÃµÄºìÉ«¹ÌÌå¿ÉÄÜÊÇÍÓëÑõ»¯ÑÇ͵ĻìºÏÎÒÑÖªCu2OÔÚËáÐÔÈÜÒºÖпɷ¢Éú×ÔÉíÑõ»¯»¹Ô·´Ó¦£¬Éú³ÉCu2+ºÍµ¥ÖÊÍ¡£
£¨1£©ÏÖÓÐ8¿ËÑõ»¯Í±»ÇâÆø»¹Ôºó£¬µÃµ½ºìÉ«¹ÌÌå6.8¿Ë£¬ÆäÖꬵ¥ÖÊÍÓëÑõ»¯ÑÇ͵ÄÎïÖʵÄÁ¿Ö®±ÈÊÇ £»
£¨2£©Èô½«6.8¿ËÉÏÊö»ìºÏÎïÓë×ãÁ¿µÄÏ¡ÁòËá³ä·Ö·´Ó¦ºó¹ýÂË£¬¿ÉµÃµ½¹ÌÌå g£»
£¨3£©Èô½«6.8¿ËÉÏÊö»ìºÏÎïÓëÒ»¶¨Á¿µÄŨÏõËá³ä·Ö·´Ó¦£¬
¢ÙÉú³É±ê×¼×´¿öÏÂ1.568ÉýµÄÆøÌ壨²»¿¼ÂÇNO2µÄÈܽ⣬Ҳ²»¿¼ÂÇNO2ÓëN2O4µÄת»¯£©£¬Ôò¸ÃÆøÌåµÄ³É·ÖÊÇ £¬ÆäÎïÖʵÄÁ¿Ö®±ÈÊÇ £»
¢Ú°ÑµÃµ½µÄÈÜҺСÐÄÕô·¢Å¨Ëõ£¬°ÑÎö³öµÄ¾§Ìå¹ýÂË£¬µÃ¾§Ìå23.68g¡£¾·ÖÎö£¬ÔÈÜÒºÖеÄCu2+ÓÐ20%²ÐÁôÔÚĸҺÖС£ÇóËùµÃ¾§ÌåµÄ»¯Ñ§Ê½
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2012½ìÕã½Ê¡¶«ÑôÖÐѧ¸ßÈýÏÂѧÆڽ׶μì²â»¯Ñ§ÊÔ¾í ÌâÐÍ£ºÌî¿ÕÌâ
¡°ÇâÄÜ¡±½«ÊÇδÀ´×îÀíÏëµÄÐÂÄÜÔ´¡£
¢ñ.ʵÑé²âµÃ£¬1gÇâÆøȼÉÕÉú³ÉҺ̬ˮʱ·Å³ö142.9kJÈÈÁ¿£¬ÔòÇâÆøȼÉÕµÄÈÈ»¯Ñ§·½³ÌʽΪ_______¡££¨ÌîÐòºÅ£©
A£®2H2£¨g£©+O2£¨g£© 2H2O£¨l£©¡÷H= £142.9kJ¡¤mol¡ª1 |
B£®H2£¨g£©+1/2 O2£¨g£© H2O£¨l£©¡÷H= £285.8kJ¡¤mol¡ª1 |
C£®2H2+O22H2O£¨l£©¡÷H= £571.6kJ¡¤mol¡ª1 |
D£®H2£¨g£©+1/2 O2£¨g£© H2O£¨g£© ¡÷H= £285.8kJ¡¤mol¡ª1 |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2013-2014ѧÄê½Î÷Ê¡°ËУ¸ßÈýÏÂѧÆÚÁª¿¼Àí×Û»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÌî¿ÕÌâ
Ä¿Ç°£¬¡°µÍ̼¾¼Ã¡±±¸ÊܹØ×¢£¬CO2µÄ²úÉú¼°ÓÐЧ¿ª·¢ÀûÓóÉΪ¿Æѧ¼ÒÑо¿µÄÖØÒª¿ÎÌâ¡£ÊÔÔËÓÃËùѧ֪ʶ£¬½â¾öÏÂÁÐÎÊÌ⣺
£¨1£©ÒÑ֪ij·´Ó¦µÄƽºâ±í´ïʽΪ£ºËüËù¶ÔÓ¦µÄ»¯Ñ§·´Ó¦Îª£º__? ___
£¨2£©¡ª¶¨Ìõ¼þÏ£¬½«C(s)ºÍH2O(g)·Ö±ð¼ÓÈë¼×¡¢ÒÒÁ½¸öÃܱÕÈÝÆ÷ÖУ¬·¢Éú£¨1£©Öз´Ó¦£ºÆäÏà¹ØÊý¾ÝÈçϱíËùʾ£º
ÈÝÆ÷ | ÈÝ»ý/L | ζÈ/¡æ | ÆðʼÁ¿/mol | ƽºâÁ¿/mol[ | ´ïµ½Æ½ºâËùÐèʱ¼ä/min | |
C(s) | H2O(g) | H2(g) | ||||
¼× | 2 | T1 | 2 | 4 | 3.2 | 8 |
ÒÒ | 1 | T2 | 1 | 2 | 1,2 | 3 |
¢ÙT10Cʱ£¬¸Ã·´Ó¦µÄƽºâ³£ÊýK=_______
¢ÚÒÒÈÝÆ÷ÖУ¬µ±·´Ó¦½øÐе½1.5minʱ£¬H2O(g)µÄÎïÖʵÄÁ¿Å¨¶È_______ (ÌîÑ¡Ïî×Öĸ£©¡£
A£®=0.8 mol¡¤L-1??? B£®=1.4 mol¡¤L-1??? C£®<1.4 mol¡¤L-1??? D£®>1.4 mol¡¤L-1
¢Û±ûÈÝÆ÷µÄÈÝ»ýΪ1L£¬T1¡æʱ£¬°´ÏÂÁÐÅä±È³äÈëC(s)¡¢H2O(g)¡¢CO2(g)ºÍH2(g)£¬ ´ïµ½Æ½?? ºâʱ¸÷ÆøÌåµÄÌå»ý·ÖÊýÓë¼×ÈÝÆ÷ÍêÈ«ÏàͬµÄÊÇ_______(ÌîÑ¡Ïî×Öĸ£©¡£
A.0.6 mol¡¢1.0 mol¡¢0.5 mol¡¢1.0 mol??
B£® 0.6 mol¡¢2.0 mol¡¢0 mol¡¢0 mol
C.1.0 mol¡¢2.0 mol¡¢1.0 mol¡¢2.0 mol??????
D£® 0.25 mol¡¢0.5 mol¡¢0.75 mol¡¢1.5 mol
£¨3£©ÔÚÒ»¶¨Ìõ¼þÏ£¬¿Æѧ¼ÒÀûÓôÓÑ̵ÀÆøÖзÖÀë³öCO2ÓëÌ«ÑôÄܵç³Øµç½âË®²úÉúµÄH2ºÏ³É¼×´¼£¬ÒÑÖªCH3OH¡¢H2µÄȼÉÕÈÈ·Ö±ðΪ£º¡÷H£½£725.5kJ/mol¡¢¡÷H£½£285.8kJ/mol£¬Ð´³ö¹¤ÒµÉÏÒÔCO2¡¢H2ºÏ³ÉCH3OHµÄÈÈ»¯Ñ§·½³Ìʽ£º?????????????????????????????? ¡£
£¨4£©½«È¼Ãº·ÏÆøÖеÄCO2ת»¯Îª¼×Ãѵķ´Ó¦ÔÀíΪ£º
2CO2(g£©+ 6H2(g£©CH3OCH3(g£©+ 3H2O(g)
ÒÑÖªÒ»¶¨Ñ¹Ç¿Ï£¬¸Ã·´Ó¦ÔÚ²»Í¬Î¶ȡ¢²»Í¬Í¶ÁϱÈʱ£¬CO2µÄת»¯ÂʼûÏÂ±í£º
ͶÁϱÈ[n(H2£©/ n(CO2)] | 500 K | 600 K | 700 K | 800 K |
1.5 | 45% | 33% | 20% | 12% |
2.0 | 60% | 43% | 28% | 15% |
3.0 | 83% | 62% | 37% | 22% |
¢Ù¸Ã·´Ó¦µÄìʱä¡÷H?? 0£¬ìرä¡÷S___0£¨Ì¡¢£¼»ò£½£©¡£
¢ÚÓü×ÃÑ×÷ΪȼÁϵç³ØÔÁÏ£¬ÔÚ¼îÐÔ½éÖÊÖиõç³Ø¸º¼«µÄµç¼«·´Ó¦Ê½?????????????? ¡£ÈôÒÔ1.12 L¡¤min-1£¨±ê×¼×´¿ö£©µÄËÙÂÊÏò¸Ãµç³ØÖÐͨÈë¼×ÃÑ£¨·ÐµãΪ-24.9 ¡æ£©£¬Óøõç³Øµç½â500 mL 2 mol¡¤L-1 CuSO4ÈÜÒº£¬Í¨µç0.50 minºó£¬ÀíÂÛÉÏ¿ÉÎö³ö½ðÊôÍ???????? g¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2011-2012ѧÄêÕã½Ê¡¸ßÈýÏÂѧÆڽ׶μì²â»¯Ñ§ÊÔ¾í ÌâÐÍ£ºÌî¿ÕÌâ
¡°ÇâÄÜ¡±½«ÊÇδÀ´×îÀíÏëµÄÐÂÄÜÔ´¡£
¢ñ.ʵÑé²âµÃ£¬1gÇâÆøȼÉÕÉú³ÉҺ̬ˮʱ·Å³ö142.9kJÈÈÁ¿£¬ÔòÇâÆøȼÉÕµÄÈÈ»¯Ñ§·½³ÌʽΪ_______¡££¨ÌîÐòºÅ£©
A.2H2£¨g£©+O2£¨g£© 2H2O£¨l£© ¡÷H= £142.9kJ¡¤mol¡ª1
B.H2£¨g£©+1/2 O2£¨g£© H2O£¨l£© ¡÷H= £285.8kJ¡¤mol¡ª1
C.2H2+O22H2O£¨l£© ¡÷H= £571.6kJ¡¤mol¡ª1
D.H2£¨g£©+1/2 O2£¨g£© H2O£¨g£© ¡÷H= £285.8kJ¡¤mol¡ª1
¢ò.ij»¯Ñ§¼Ò¸ù¾Ý¡°Ô×Ó¾¼Ã¡±µÄ˼Ï룬Éè¼ÆÁËÈçÏÂÖƱ¸H2µÄ·´Ó¦²½Öè
¢ÙCaBr2+H2OCaO+2HBr ¢Ú2HBr+HgHgBr2+H2
¢ÛHgBr2+___________________ ¢Ü2HgO2Hg+O2¡ü
ÇëÄã¸ù¾Ý¡°Ô×Ó¾¼Ã¡±µÄ˼ÏëÍê³ÉÉÏÊö²½Öè¢ÛµÄ»¯Ñ§·½³Ìʽ£º____________¡£
¸ù¾Ý¡°ÂÌÉ«»¯Ñ§¡±µÄ˼ÏëÆÀ¹À¸Ã·½·¨ÖÆH2µÄÖ÷Ҫȱµã£º______________¡£
¢ó.ÀûÓúËÄÜ°ÑË®·Ö½âÖÆÇâÆø£¬ÊÇÄ¿Ç°ÕýÔÚÑо¿µÄ¿ÎÌâ¡£ÏÂͼÊÇÆäÖеÄÒ»ÖÖÁ÷³Ì£¬ÆäÖÐÓÃÁ˹ýÁ¿µÄµâ¡££¨Ìáʾ£º·´Ó¦¢ÚµÄ²úÎïÊÇO2¡¢SO2ºÍH2O£©
Íê³ÉÏÂÁз´Ó¦µÄ»¯Ñ§·½³Ìʽ£º
·´Ó¦¢Ù__________________________£»·´Ó¦¢Ú__________________________¡£
´Ë·¨ÖÆÈ¡ÇâÆøµÄ×î´óÓŵãÊÇ_______________________________________________¡£
¢ô.ÇâÆøͨ³£ÓÃÉú²úˮúÆøµÄ·½·¨ÖƵá£ÆäÖÐCO£¨g£©+ H2O£¨g£© CO2£¨g£©+ H2£¨g£©; ¡÷H<0¡£
ÔÚ850¡æʱ£¬K=1¡£
£¨1£©ÈôÉý¸ßζȵ½950¡æʱ£¬´ïµ½Æ½ºâʱK______1£¨Ìî¡°´óÓÚ¡±¡¢¡°Ð¡ÓÚ¡±»ò¡°µÈÓÚ¡±£©
£¨2£©850¡æʱ£¬ÈôÏòÒ»ÈÝ»ý¿É±äµÄÃܱÕÈÝÆ÷ÖÐͬʱ³äÈë 1.0 mol CO¡¢3.0molH2O¡¢1.0mol CO2 ºÍ x mol H2£¬Ôò£º
¢Ùµ±x=5.0ʱ£¬ÉÏÊöƽºâÏò___________£¨ÌîÕý·´Ó¦»òÄæ·´Ó¦£©·½Ïò½øÐС£
¢ÚÈôҪʹÉÏÊö·´Ó¦¿ªÊ¼Ê±ÏòÕý·´Ó¦·½Ïò½øÐУ¬ÔòxÓ¦Âú×ãµÄÌõ¼þÊÇ__________¡£
£¨3£©ÔÚ850¡æʱ£¬ÈôÉèx£½5.0 molºÍx£½6.0mol£¬ÆäËüÎïÖʵÄͶÁϲ»±ä£¬µ±ÉÏÊö·´Ó¦´ïµ½Æ½ºâºó£¬²âµÃH2µÄÌå»ý·ÖÊý·Ö±ðΪa£¥¡¢b£¥£¬Ôòa _______ b£¨Ìî¡°´óÓÚ¡±¡¢¡°Ð¡ÓÚ¡±»ò¡°µÈÓÚ¡±£©¡£
¢õ.ÇâÆø»¹ÔÑõ»¯ÍËùµÃµÄºìÉ«¹ÌÌå¿ÉÄÜÊÇÍÓëÑõ»¯ÑÇ͵ĻìºÏÎÒÑÖªCu2OÔÚËáÐÔÈÜÒºÖпɷ¢Éú×ÔÉíÑõ»¯»¹Ô·´Ó¦£¬Éú³ÉCu2+ºÍµ¥ÖÊÍ¡£
£¨1£©ÏÖÓÐ8¿ËÑõ»¯Í±»ÇâÆø»¹Ôºó£¬µÃµ½ºìÉ«¹ÌÌå6.8¿Ë£¬ÆäÖꬵ¥ÖÊÍÓëÑõ»¯ÑÇ͵ÄÎïÖʵÄÁ¿Ö®±ÈÊÇ £»
£¨2£©Èô½«6.8¿ËÉÏÊö»ìºÏÎïÓë×ãÁ¿µÄÏ¡ÁòËá³ä·Ö·´Ó¦ºó¹ýÂË£¬¿ÉµÃµ½¹ÌÌå g£»
£¨3£©Èô½«6.8¿ËÉÏÊö»ìºÏÎïÓëÒ»¶¨Á¿µÄŨÏõËá³ä·Ö·´Ó¦£¬
¢ÙÉú³É±ê×¼×´¿öÏÂ1.568ÉýµÄÆøÌ壨²»¿¼ÂÇNO2µÄÈܽ⣬Ҳ²»¿¼ÂÇNO2ÓëN2O4µÄת»¯£©£¬Ôò¸ÃÆøÌåµÄ³É·ÖÊÇ £¬ÆäÎïÖʵÄÁ¿Ö®±ÈÊÇ £»
¢Ú°ÑµÃµ½µÄÈÜҺСÐÄÕô·¢Å¨Ëõ£¬°ÑÎö³öµÄ¾§Ìå¹ýÂË£¬µÃ¾§Ìå23.68g¡£¾·ÖÎö£¬ÔÈÜÒºÖеÄCu2+ÓÐ20%²ÐÁôÔÚĸҺÖС£ÇóËùµÃ¾§ÌåµÄ»¯Ñ§Ê½
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£ºÕã½Ê¡Ä£ÄâÌâ ÌâÐÍ£ºÌî¿ÕÌâ
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