¡¾ÌâÄ¿¡¿Ï±íÊÇÔªËØÖÜÆÚ±íµÄÒ»²¿·Ö£¬±íÖеÄÊý×Ö±íʾһÖÖ¶ÌÖÜÆÚÔªËØ£¬»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©»³ö¢ÛµÄÔ×ӽṹʾÒâͼ£º____________________¡£
£¨2£©¢Û¡¢¢Ü¡¢¢ß¡¢¢àËÄÖÖÔªËØËùÐγɵÄÆø̬Ç⻯ÎïÖÐ×îÎȶ¨µÄÊÇ__________£¨Ìѧʽ£©¡£
£¨3£©¢ÙÔÚÔªËØÖÜÆÚ±íÖеÄλÖÃÊÇ____________________________¡£
£¨4£©¢à¡¢¢áÔªËصÄ×î¸ß¼ÛÑõ»¯Îï¶ÔӦˮ»¯ÎïµÄËáÐÔ£º____>____£¨Óû¯Ñ§Ê½±íʾ£©¡£_________
£¨5£©ÓÉ¢Û¡¢¢ÝºÍ¢â×é³ÉµÄ»¯ºÏÎïËùº¬»¯Ñ§¼üÀàÐÍΪ__________£¬µç×ÓʽΪ____________¡£
£¨6£©¢Ü¡¢¢Ý¡¢¢ÞµÄÔ×Ӱ뾶ÓÉ´óµ½Ð¡µÄ˳ÐòÊÇ£º____>____>____£¨ÓÃÔªËØ·ûºÅ±íʾ£©¡£__________
£¨7£©¢Þµ¥ÖÊÓëÇâÑõ»¯ÄÆÈÜÒº·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ__________________________________¡£
£¨8£©ÒÑÖª»¯ºÏÎïA¡¢B¡¢C¡¢D¡¢E¡¢F¡¢GºÍµ¥Öʼס¢ÒÒËùº¬ÔªËؾùΪÉϱíÖÐÔªËØ×é³É¡£
¢ÙAµÄŨÈÜÒºÓë¼×ÄÜ·¢ÉúÈçÏÂͼËùʾµÄ·´Ó¦¡£
¢Ú¼×Êdz£¼ûµÄºÚÉ«¹ÌÌåµ¥ÖÊ£¬¿ÉΪÉú²úÉú»îÌṩÈÈÄÜ¡£
¢ÛÒÒÊdz£¼ûµÄÎÞÉ«ÆøÌåµ¥ÖÊ¡£
¢ÜBÊÇÎÞÉ«Óд̼¤ÐÔÆøζµÄÆøÌ壬ÊÇÖ÷ÒªµÄ´óÆøÎÛȾÎïÖ®Ò»¡£
¢Ý³£ÎÂÏ£¬CÊÇÒ»ÖÖÎÞÉ«ÒºÌå¡£
»Ø´ðÎÊÌ⣺
¢Ùд³öÏÂÁÐÎïÖʵĻ¯Ñ§Ê½£ºA_____£¬E_____£¬G_____¡£
¢Úд³öÏÂÁз´Ó¦µÄ»¯Ñ§·½³Ìʽ£º
C£«E¡úF£«ÒÒ________________________________¡£
B£«C£«ÒÒ¡úA________________________________¡£
¡¾´ð°¸¡¿ HF µÚ¶þÖÜÆÚIVA×å HClO4 H2SO4 Àë×Ó¼ü¡¢¹²¼Û¼ü Na Al F 2Al£«2OH££«2H2O£½2AlO2££«3H2¡ü H2SO4 Na2O2 Na2CO3 2Na2O2£«2H2O£½4NaOH£«O2¡ü 2SO2£«O2£«2H2O£½2H2SO4
¡¾½âÎö¡¿
¸ù¾ÝÔªËØÖÜÆÚ±íÖª£¬¢Ù¢Ú¢Û¢Ü¢Ý¢Þ¢ß¢à¢á¢âËù´ú±íµÄÔªËØ·Ö±ðΪC¡¢N¡¢O¡¢F¡¢Na¡¢Al¡¢P¡¢S¡¢Cl¡¢H¡£
£¨1£©¢ÛÊÇOÔªËØ£¬OÔ×ÓÖÊ×ÓÊý=ºËÍâµç×ÓÊý=8£¬K²ã2¸öµç×Ó¡¢L²ã6¸öµç×Ó£¬Ô×ӽṹʾÒâͼ£º£¬¹Ê´ð°¸Îª£º£»
£¨2£©¢Û¡¢¢Ü¡¢¢ß¡¢¢àËÄÖÖÔªËØ·Ö±ðÊÇO¡¢F¡¢P¡¢S£¬·Ç½ðÊôÐÔԽǿÆäÆø̬Ç⻯ÎïÔ½Îȶ¨£¬ÆäÖзǽðÊôÐÔ×îÇ¿µÄÔªËØÊÇF£¬Òò´ËHF×îÎȶ¨£¬¹Ê´ð°¸Îª£ºHF£»
£¨3£©¢ÙÊÇCÔªËØ£¬ÔÚÔªËØÖÜÆÚ±íÖеÄλÖÃÊǵڶþÖÜÆÚIVA×壬¹Ê´ð°¸Îª£ºµÚ¶þÖÜÆÚIVA×壻
£¨4£©¢à¡¢¢áÔªËØ·Ö±ðÊÇS¡¢Cl£¬·Ç½ðÊôÐÔCl£¾S£¬·Ç½ðÊôÐÔԽǿ£¬×î¸ß¼ÛÑõ»¯Îï¶ÔӦˮ»¯ÎïµÄËáÐÔԽǿ£¬Òò´ËËáÐÔ£ºHClO4£¾H2SO4£¬¹Ê´ð°¸Îª£ºHClO4£»H2SO4£»
£¨5£©¢Û¡¢¢ÝºÍ¢âÔªËØ·Ö±ðÊÇO¡¢Na¡¢H£¬ÓÉ¢Û¡¢¢ÝºÍ¢â×é³ÉµÄ»¯ºÏÎïÊÇNaOH£¬NaOHÊÇÀë×Ó»¯ºÏÎËùº¬»¯Ñ§¼üÀàÐÍΪÀë×Ó¼üºÍ¹²¼Û¼ü£¬µç×ÓʽΪ£º£¬¹Ê´ð°¸Îª£ºÀë×Ó¼ü¡¢¹²¼Û¼ü£»£»
£¨6£©¢Ü¡¢¢Ý¡¢¢Þ·Ö±ðÊÇF¡¢Na¡¢Al£¬¸ù¾ÝͬÖÜÆÚ×Ô×óµ½ÓÒÔ×Ӱ뾶Öð½¥¼õС£¬Ô×Ӱ뾶£ºNa£¾Al£¾Cl£¬Í¬Ö÷×å×ÔÉ϶øÏÂÔ×Ӱ뾶Öð½¥Ôö´ó£¬Ô×Ӱ뾶£ºCl£¾F£¬Òò´ËÔ×Ӱ뾶£ºNa£¾Al£¾F£¬¹Ê´ð°¸Îª£ºNa£»Al£»F£»
£¨7£©¢ÞÊÇAlÔªËØ£¬AlÓëÇâÑõ»¯ÄÆÈÜÒº·´Ó¦µÄÀë×Ó·½³ÌʽΪ2Al£«2OH££«2H2O£½2AlO2££«3H2¡ü£»
£¨8£©¢ÙÓÉ·´Ó¦¹Øϵͼ¿ÉÖª£º¼×Êdz£¼ûµÄºÚÉ«¹ÌÌåµ¥ÖÊ£¬¿ÉΪÉú²úÉú»îÌṩÈÈÄÜ£¬¿ÉÖª¼×ÊÇC£¬BÊÇÎÞÉ«Óд̼¤ÐÔÆøζµÄÆøÌ壬ÊÇÖ÷ÒªµÄ´óÆøÎÛȾÎïÖ®Ò»£¬ÎªSO2ÆøÌ壬Òò´ËAΪŨÁòËáÈÜÒº£¬³£ÎÂÏ£¬CÊÇÒ»ÖÖÎÞÉ«ÒºÌåÊÇË®£¬DΪCO2£¬¶þÑõ»¯ÁòºÍÑõÆøºÍË®·´Ó¦Éú³ÉÁòËᣬ¹ÊÒÒΪO2£¬Ë®¡¢¶þÑõ»¯Ì¼¾ùÓëE·´Ó¦Éú³ÉÑõÆø£¬ËµÃ÷EΪNa2O2£¬ÔòFΪNaOH£¬GΪNa2CO3£¬¹Ê´ð°¸Îª£ºH2SO4£»Na2O2£»Na2CO3£»
¢ÚË®ºÍ¹ýÑõ»¯ÄÆ·´Ó¦Éú³ÉÇâÑõ»¯ÄƺÍÑõÆø£¬»¯Ñ§·½³ÌʽΪ£º2Na2O2+2H2O=4NaOH+O2¡ü£¬¶þÑõ»¯ÁòºÍÑõÆøºÍË®·´Ó¦Éú³ÉÁòËᣬ»¯Ñ§·½³ÌʽΪ£º2SO2+O2+2H2O=2H2SO4£¬¹Ê´ð°¸Îª£º2Na2O2+2H2O=
4NaOH+O2¡ü£»2SO2+O2+2H2O=2H2SO4¡£
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¿¡¿ÓлúÒÒżÒö´æÔÚÓÚÆ¡¾ÆÖУ¬ÊǾÆÀàµ÷ÏãÖÐÒ»¸ö¼«ÆäÖØÒªµÄÆ·ÖÖ£¬Ä³Ñо¿ÐÔѧϰС×éΪȷ¶¨ÒÒżÒöµÄ½á¹¹£¬½øÐÐÈçÏÂ̽¾¿¡£
²½ÖèÒ»£º½«ÒÒżÒöÕôÆøͨ¹ýÈȵÄÑõ»¯Í(´ß»¯¼Á)Ñõ»¯³É¶þÑõ»¯Ì¼ºÍË®£¬ÔÙÓÃ×°ÓÐÎÞË®ÂÈ»¯¸ÆºÍ¹ÌÌåÇâÑõ»¯ÄƵÄÎüÊÕ¹ÜÍêÈ«ÎüÊÕ£¬Èçͼ1¡£2.64 gÒÒżÒöµÄÕôÆøÑõ»¯²úÉú5.28 g¶þÑõ»¯Ì¼ºÍ2.16 gË®¡£
²½Öè¶þ£ºÉýÎÂʹÒÒżÒöÆû»¯£¬²âÆäÃܶÈÊÇÏàͬÌõ¼þÏÂH2µÄ44±¶
²½ÖèÈý£ºÓú˴Ź²ÕñÒDzâ³öÒÒżÒöµÄºË´Å¹²ÕñÇâÆ×Èçͼ2£¬Í¼ÖÐ4¸ö·åµÄÃæ»ý±ÈΪ1¡Ã3¡Ã1¡Ã3¡£
²½ÖèËÄ£ºÀûÓúìÍâ¹âÆ×ÒDzâµÃÒÒżÒö·Ö×ӵĺìÍâ¹âÆ×Èçͼ3¡£
£¨1£©Í¼1×°ÖÃÖÐÁ½Ö§UÐ͹ܲ»ÄÜ»¥»»µÄÀíÓÉÊÇ__________________________£®
£¨2£©ÒÒżÒöµÄĦ¶ûÖÊÁ¿Îª____________¡£
£¨3£©ÒÒżÒöµÄ·Ö×ÓʽΪ____________________¡£
£¨4£©ÒÒżÒöµÄ½á¹¹¼òʽΪ ________________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÏÂÁйØÓÚ (ÂÝ[2£¬2]ÎìÍé)µÄ˵·¨ÕýÈ·µÄÊÇ
A. Ò»ÂÈ´úÎïµÄ½á¹¹Ö»ÓÐÒ»ÖÖ
B. ÓëÎìÏ©»¥ÎªÍ¬·ÖÒì¹¹Ìå
C. ËùÓÐ̼Ô×Ó¾ù´¦Í¬Ò»Æ½Ãæ
D. ÄÜʹËáÐÔ¸ßÃÌËá¼ØÈÜÒºÍÊÉ«
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿¶þ¼×ÃÑ(CH3OCH3)±»³ÆΪ21ÊÀ¼ÍµÄÐÂÐÍȼÁÏ£¬ÒÔCOºÍH2ΪÔÁÏÉú²ú¶þ¼×ÃÑÖ÷Òª·¢ÉúÒÔÏÂÈý¸ö·´Ó¦£º
£¨1£©¸Ã¹¤ÒÕµÄ×Ü·´Ó¦Îª3CO(g)£«3H2(g)CH3OCH3(g)£«CO2(g) ¦¤H£½_________£¬»¯Ñ§Æ½ºâ³£ÊýK£½______________(Óú¬K1¡¢K2¡¢K3µÄ´úÊýʽ±íʾ)¡£
»¯Ñ§·´Ó¦·½³Ìʽ | »¯Ñ§Æ½ºâ³£Êý | |
¢ÙCO(g)£«2H2(g)CH3OH(g) | ¦¤H1=-99 kJmol-1 | K1 |
¢Ú2CH3OH(g)CH3OCH3(g)£«H2O(g) | ¦¤H2£½£24 kJmol-1 | K2 |
¢ÛCO(g)£«H2O(g)CO2(g)£«H2(g) | ¦¤H3£½£41 kJmol-1 | K3 |
£¨2£©Ä³Î¶ÈÏ£¬½«8.0molH2ºÍ4.0molCO³äÈëÈÝ»ýΪ2LµÄÃܱÕÈÝÆ÷ÖУ¬·¢Éú·´Ó¦£º4H2(g)+2CO(g) CH3OCH3(g)+H2O(g)£¬10 ·ÖÖÓºó·´Ó¦´ïƽºâ£¬²âµÃ¶þ¼×ÃѵÄÌå»ý·ÖÊýΪ25%£¬ÔòÓÃH2±íʾµÄ·´Ó¦ËÙÂÊΪ_________£¬COµÄת»¯ÂÊΪ________¡£
£¨3£©ÏÂÁдëÊ©ÖУ¬ÄÜÌá¸ßCH3OCH3²úÂʵÄÓÐ________¡£
A£®·ÖÀë³ö¶þ¼×ÃÑ B£®½µµÍÎÂ¶È C£®¸ÄÓøßЧ´ß»¯¼ÁD£®Ôö´óѹǿ
£¨4£©¸Ã¹¤ÒÕÖз´Ó¦¢ÛµÄ·¢ÉúÌá¸ßÁËCH3OCH3µÄ²úÂÊ£¬ÔÒòÊÇ________________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÏÖÓÐËÄ×é·Öɢϵ£º¢ÙÆûÓͺÍË®ÐγɵĻìºÏÒº¢Úº¬ÓÐÄàɳµÄʳÑÎË®¢ÛÈÜÓеâ(I2)µÄÂÈ»¯¼ØÈÜÒº¢ÜÒÒ¶þ´¼ºÍ±ûÈý´¼»ìºÏÈÜÒº(ÒÒ¶þ´¼ºÍ±ûÈý´¼µÄ²¿·ÖÎïÀíÐÔÖʼû±í)¡£
ÎïÖÊ | ÈÛµã/¡æ | ·Ðµã/¡æ | ÃܶÈ/gcm-3 | ÈܽâÐÔ |
ÒÒ¶þ´¼ | 11.5 | 198 | 1.11 | Ò×ÈÜÓÚË®ºÍÒÒ´¼ |
±ûÈý´¼ | 17.9 | 290 | 1.26 | ÄܸúË®¡¢¾Æ¾«ÒÔÈÎÒâ±È»¥ÈÜ |
ÇëÓÃÈçͼËùʾµÄÒÇÆ÷·ÖÀëÒÔÉϸ÷»ìºÏÒº£¬ÒÇÆ÷ºÍ·½·¨²»ÄܶÔÓ¦µÄÊÇ£¨ £©
A. ¢Ù¡ªc¡ª·ÖÒºB. ¢Ú¡ªa¡ªÝÍÈ¡
C. ¢Û¡ªc¡ªÝÍÈ¡D. ¢Ü¡ªa¡ªÕôÁó
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿Ï±íÊÇÔªËØÖÜÆÚ±íµÄÒ»²¿·Ö£¬Çë»Ø´ðÓйØÎÊÌâ:
£¨1£©±íÖл¯Ñ§ÐÔÖÊ×î²»»îÆõÄÔªËØ£¬ÆäÔ×ӽṹʾÒâͼΪ______¡£
£¨2£©ÔªËØ¢ßÓë¢àµÄÔ×Ӱ뾶´óС¹ØϵÊÇ£º¢ß______¢à(Ìî¡°£¾¡±»ò¡°£¼¡±£©¡£
£¨3£©¢Ü¢ÝÁ½ÔªËØÏà±È½Ï£¬½ðÊôÐÔ½ÏÇ¿µÄÊÇ______ (ÌîÔªËØÃû³Æ£©¡£
£¨4£©ÔªËآٵÄ×î¸ß¼ÛÑõ»¯ÎïµÄË®»¯ÎïµÄ»¯Ñ§Ê½Îª________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÏÂÁÐ˵·¨ÖÐÕýÈ·µÄÊÇ£¨ £©
A. °Ñ100mL3mo/LH2SO4ÈÜÒºÓë100mL1mo/LBaCl2ÈÜÒº»ìºÏ£¬ËùµÃÈÜÒºc(SO42-)±äΪ1mol/L
B. °Ñ200mL3mol/LBaCl2ÈÜÒºÓë100mL3mol/LKCl ÈÜÒº»ìºÏºó£¬ËùµÃÈÜÒºc(Cl-)ÈÔΪ3mol/L
C. °Ñ100mL20%µÄNaOHÈÜÒºÓë100mLH2O»ìºÏºó£¬ËùµÃÈÜÒºÖÐNaOHµÄÖÊÁ¿·ÖÊýΪ10%
D. °Ñ100g20%µÄNaClÈÜÒºÓë100mLH2O»ìºÏºó£¬ËùµÃÈÜÒºÖÐNaClµÄÖÊÁ¿·ÖÊýΪ10%
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿¸ß·Ö×Ó»¯ºÏÎïGÊǿɱ»Ï¸¾ú×÷Ϊ̼ԴºÍÄÜÔ´ÀûÓõľۺÏÎÊôÓÚÒ»ÖÖÉúÎï¿É½µ½â¸ß·Ö×Ó²ÄÁÏ£¬ÔÚʳƷ¡¢Ò©Æ·°ü×°·½Ãæ¾ßÓжÀÌØÓÅÊÆ¡£ÒÑÖªA~G¾ùΪÓлú»¯ºÏÎÒÔÏÂΪ¸ß·Ö×Ó»¯ºÏÎïGµÄÒ»ÖֺϳÉ·Ïߣº
»Ø´ðÒÔÏÂÎÊÌ⣺
(1)ÓÉCaC2ÖƱ¸AµÄ»¯Ñ§·½³ÌʽΪ_________________________________¡£
(2)AÉú³ÉBµÄ·´Ó¦ÀàÐÍΪ______________________¡£
(3)CµÄ»¯Ñ§Ãû³ÆÊÇ___________£¬C¡úDËùÐèÊÔ¼ÁºÍÌõ¼þ·Ö±ðÊÇ___________¡¢___________¡£
(4)EµÄ½á¹¹¼òʽΪ______________________¡£
(5)·¼Ïã×廯ºÏÎïHÊÇDµÄͬ·ÖÒì¹¹Ì壬ÔòH¿ÉÄܵĽṹ¹²ÓÐ___________ÖÖ(²»°üÀ¨D)£¬Ð´³öºË´Å¹²ÕñÇâÆ×ÓÐÈý×é·åÇÒ·åÃæ»ýÖ®±ÈΪ1©U2©U6µÄHµÄͬ·ÖÒì¹¹ÌåµÄ½á¹¹¼òʽ___________(ÈÎдһÖÖ)¡£
(6)ÓÉFÉú³ÉGµÄ»¯Ñ§·½³ÌʽΪ_________________________________¡£
(7)²Î¿¼ÌâÖÐÐÅÏ¢£¬Éè¼ÆÓÉ1£¬2¶þÂÈÒÒÍéºÍ±ù´×ËáΪÔÁÏÖÆÈ¡µÄºÏ³ÉÏß·ͼ(ÎÞ»úÊÔ¼ÁÈÎÑ¡)__________________¡£
ÒÑÖª£º£OHÓë̼̼˫¼üÁ½¶ËµÄ̼Ô×ÓÖ±½ÓÏàÁ¬²»Îȶ¨£¬»á×Ô±ä³É¡ªCHO¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿Ò»¶¨Î¶Èʱ£¬ÏòÈÝ»ýΪ 2 L µÄÃܱÕÈÝÆ÷ÖгäÈëÒ»¶¨Á¿µÄ SO2(g)ºÍ O2(g)£¬·¢Éú·´Ó¦£º2SO2(g) + O2(g) 2SO3(g) ¡÷H = -196 kJ/mol¡£Ò»¶Îʱ¼äºó·´Ó¦´ïµ½Æ½ºâ״̬£¬·´Ó¦¹ý³ÌÖвⶨµÄ²¿·ÖÊý¾ÝÈç±íËùʾ¡£
·´Ó¦Ê±¼ä/min | n(SO2)/mol | n(O2)/mol |
0 | 2 | 1 |
5 | 1.2 | |
10 | 0.4 | |
15 | 0.8 |
ÏÂÁÐ˵·¨²»ÕýÈ·µÄÊÇ
A. Ç°5 minµÄƽ¾ù·´Ó¦ËÙÂÊΪ¦Ô(SO2) = 0.08 mol/(L¡¤min)
B. ±£³ÖζȲ»±ä£¬ÏòƽºâºóµÄÈÝÆ÷ÖÐÔÙ³äÈë 0.2 mol SO2(g)ºÍ0.2 mol SO3(g)ʱ£¬¦ÔÕý> ¦ÔÄæ
C. ÏàͬζÈÏ£¬ÆðʼʱÏòÈÝÆ÷ÖгäÈë 1.5 mol SO3(g)£¬´ïµ½Æ½ºâ״̬ʱ SO3 µÄת»¯ÂÊΪ40%
D. ±£³ÖÆäËûÌõ¼þ²»±ä£¬ÈôÆðʼʱÏòÈÝÆ÷ÖгäÈë 2 mol SO3(g)£¬´ïµ½Æ½ºâ״̬ʱÎüÊÕ 78.4 kJ µÄÈÈÁ¿
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