ÌâÄ¿ÄÚÈÝ

ʵÑ飺½«ÁòËáÑÇÌú¾§Ìå¼ÓÈëÈçͼ£¨1£©ÖÐËùʾµÄ¸ÉÔïÊÔ¹ÜAÀ¸ô¾ø¿ÕÆø¼ÓÇ¿Èȳä·Ö·´Ó¦ºó£¬ÊÔ¹ÜAÖеòÐÁô¹ÌÌå¼×£¬ÔÚÊÔ¹ÜBÄڵõ½ÎÞÉ«ÒºÌåÒÒ£®È¡³ö¼×·ÅÔÚͼ£¨2£©ÖеÄʯӢ¹ÜCµÄÖв¿£¬Á¬½ÓºÃÆäËû²¿·ÖµÄÒÇÆ÷ºó£¬¿ªÆôÐýÈûK£¬ÖðµÎµÎ¼ÓÒÒ´¼Ê¹C¹Ü×ó²¿Ê¯ÃÞÈÞ£¨Ò»ÖÖÏËά״¹èËáÑβÄÁÏ£¬ºÜÎȶ¨£¬²»È¼ÉÕ£©½þÈóÎü×㣬¹Ø±ÕK£¬È»ºó¼ÓÈÈ£¬¼¸·ÖÖÓ×óÓÒ£¨¼ÓÈȹý³ÌÖл¹¿ÉÒԽϿìµØ¿ªÆô¡¢¹Ø±ÕÐýÈûKÀ´²¹³äÒÒ´¼²¢Ê¹Ö®Í¨ÈëʯÃÞÈÞÀ£¬¿É¹Û²ìµ½¼×µÄÑÕÉ«Óɺì×ØÉ«Öð½¥±äΪºÚÉ«£®·´Ó¦Í£Ö¹ºó£¬È¡ÊÔ¹ÜEÖеÄÒºÌå0.5mLÖðµÎ¼ÓÈëµ½º¬ÐÂÅäÖƵÄÒø°±ÈÜÒºµÄÊÔ¹ÜFÖУ¬Õñµ´ºó£¬°ÑÊԹܷÅÔÚˮԡÖÐÎÂÈÈ£¬²»¾Ã»á¹Û¿´µ½ÊÔ¹ÜÄÚ±ÚÉϸ½×ÅÒ»²ã¹âÁÁÈç¾µµÄ½ðÊôÒø£®

»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Ð´³öÈçͼ£¨1£©ËùʾµÄ¸ÉÔïÊÔ¹ÜAÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º______ Fe2O3+SO2¡ü+SO3+14H2O
¡¾´ð°¸¡¿·ÖÎö£º£¨1£©½«ÁòËáÑÇÌú¾§Ì壨FeSO4?7H2O£©¼ÓÈëÈçͼ£¨1£©ÖÐËùʾµÄ¸ÉÔïÊÔ¹ÜAÀ¸ô¾ø¿ÕÆø¼ÓÇ¿Èȳä·Ö·´Ó¦ºó£¬ÊÔ¹ÜAÖеòÐÁô¹ÌÌå¼×£¬Í¼2×°ÖÃÖмÓÈÈ£¬¼×µÄÑÕÉ«Óɺì×ØÉ«Öð½¥±äΪºÚÉ«£¬Åжϼ×ΪFe2O3£»ËµÃ÷ÁòËáÑÇÌú¾§ÌåÊÜÈÈ·Ö½âÊÇ·¢ÉúÁËÑõ»¯»¹Ô­·´Ó¦£¬ÁòÔªËØ»¯ºÏ¼Û½µµÍ£¬ÌúÔªËØ»¯ºÏ¼ÛÉý¸ß£¬½áºÏÑõ»¯»¹Ô­·´Ó¦µç×ÓÊغã·ÖÎöÉú³É²úÎïÓжþÑõ»¯Áò£¬½áºÏ»¯Ñ§·½³ÌʽÅäƽд³ö£»
£¨2£©Í¼1×°Ö÷ֽâ²úÎïÖÐÉú³ÉÁ˶þÑõ»¯ÁòÎÛȾÆøÌ壬µ«È±ÉÙβÆø´¦Àí×°Öã»
£¨3£©ÒÀ¾Ý×°ÖÃÆøÃÜÐÔ¼ì²é·½·¨½øÐлشð£¬Ö÷ÒªÊÇÀûÓÃѹǿ¹ØϵÅжÏÆøÃÜÐԱ仯£»
£¨4£©ÒÀ¾ÝÒÒ´¼µÄ´ß»¯Ñõ»¯·´Ó¦Ô­ÀíÍƶÏΪÑõ»¯ÌúºÍÒÒ´¼·´Ó¦Éú³ÉÒÒÈ©£»
£¨5£©×°ÖÃDÊÇ°²È«×°Ö㬿ÉÒÔÆðµ½·Àµ¹ÎüµÄ×÷Óã»
£¨6£©FÊÔ¹ÜÖз´Ó¦ÎªÒÒÈ©±»Òø°±ÈÜÒºÑõ»¯µÄ»¯Ñ§·½³Ìʽ£®
½â´ð£º½â£º£¨1£©½«ÁòËáÑÇÌú¾§Ì壨FeSO4?7H2O£©¼ÓÈëÈçͼ£¨1£©ÖÐËùʾµÄ¸ÉÔïÊÔ¹ÜAÀ¸ô¾ø¿ÕÆø¼ÓÇ¿Èȳä·Ö·´Ó¦ºó£¬ÊÔ¹ÜAÖеòÐÁô¹ÌÌå¼×£¬Í¼2×°ÖÃÖмÓÈÈ£¬¼×µÄÑÕÉ«Óɺì×ØÉ«Öð½¥±äΪºÚÉ«£¬Åжϼ×ΪFe2O3£»ËµÃ÷ÁòËáÑÇÌú¾§ÌåÊÜÈÈ·Ö½âÊÇ·¢ÉúÁËÑõ»¯»¹Ô­·´Ó¦£¬ÁòÔªËØ»¯ºÏ¼Û½µµÍ£¬ÌúÔªËØ»¯ºÏ¼ÛÉý¸ß£¬½áºÏÑõ»¯»¹Ô­·´Ó¦µç×ÓÊغã·ÖÎöÉú³É²úÎïÓжþÑõ»¯Áò£¬ÒÀ¾ÝÔ­×ÓÊغãºÍµç×ÓÊغãÅäƽÊéд³öµÄ»¯Ñ§·½³ÌʽΪ£º2£¨FeSO4?7H2O£©Fe2O3+SO2¡ü+SO3+14H2O£¬
¹Ê´ð°¸Îª£º2£¨FeSO4?7H2O£©Fe2O3+SO2¡ü+SO3+14H2O£»
£¨2£©ÁòËáÑÇÌú·Ö½âÉú³ÉµÄ²úÎﺬÎÛȾÐÔÆøÌå¶þÑõ»¯Áò£¬²»ÄÜÅŷŵ½´óÆøÖУ¬ËùÒÔ×°ÖõÄȱµãΪ£ºSO2Ϊ´óÆøÎÛȾÎȱÉÙβÆø´¦Àí×°Öã¬
¹Ê´ð°¸Îª£ºSO2Ϊ´óÆøÎÛȾÎȱÉÙβÆø´¦Àí×°Öã»
£¨3£©×°ÖÃ2ÊÇÐèÒªÒÒ´¼Æø»¯ºÍÑõ»¯Ìú·´Ó¦£¬Éú³ÉÆøÌåÒÒÈ©£¬ÊÕ¼¯ÑéÖ¤£¬ÓÐÆøÌåµÄ×°ÖÃÖÐ ÐèÒª¼ì²é×°ÖÃÆøÃÜÐÔ£¬Ö÷ÒªÊÇÀûÓÃ×°ÖÃÖеÄѹǿ¹Øϵ·ÖÎö£¬¼ì²é·½·¨Îª£º¹Ø±Õ»îÈûK£¬½«EµÄµ¼Æø¹Ü²åÈëË®ÖУ¬Î¢ÈÈC¹Ü£¬EÖÐÓÐÆøÅݲúÉú£¬Í£Ö¹¼ÓÈȺó£¬E¹Ü³öÏÖÒ»¶ÎË®Öù£¬ËµÃ÷×°Öò»Â©Æø£¬
¹Ê´ð°¸Îª£º¹Ø±Õ»îÈûK£¬½«EµÄµ¼Æø¹Ü²åÈëË®ÖУ¬Î¢ÈÈC¹Ü£¬EÖÐÓÐÆøÅݲúÉú£¬Í£Ö¹¼ÓÈȺó£¬E¹Ü³öÏÖÒ»¶ÎË®Öù£¬ËµÃ÷×°Öò»Â©Æø£»
£¨4£©ÊÔ¹ÜCÖз´Ó¦ÊÇÒÒ´¼Æø»¯ºÍÑõ»¯ÌúµÄ·´Ó¦Éú³ÉÌú¡¢ÒÒÈ©ºÍË®£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º3CH3CH2OH+Fe2O3  2Fe+3CH3CHO+3H2O£¬¹Ê´ð°¸Îª£º3CH3CH2OH+Fe2O3 2Fe+3CH3CHO+3H2O£»
£¨5£©×°ÖÃDÊÇ°²È«×°Ö㬷ÀÖ¹ÒºÌåµ¹Îü£¬¹Ê´ð°¸Îª£º·ÀÖ¹µ¹Îü£»
£¨6£©ÊÔ¹ÜEÖеÄÒºÌå0.5mLÖðµÎ¼ÓÈëµ½º¬ÐÂÅäÖƵÄÒø°±ÈÜÒºµÄÊÔ¹ÜFÖУ¬Õñµ´ºó£¬°ÑÊԹܷÅÔÚˮԡÖÐÎÂÈÈ£¬²»¾Ã»á¹Û¿´µ½ÊÔ¹ÜÄÚ±ÚÉϸ½×ÅÒ»²ã¹âÁÁÈç¾µµÄ½ðÊôÒø£®Ö¤Ã÷Éú³ÉÁËÒÒÈ©£¬ÒÒÈ©ºÍÒø°±ÈÜÒº·´Ó¦³öÏÖÒø¾µ·´Ó¦ÏÖÏ󣬷´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£ºCH3CHO+2Ag£¨NH3£©2OH  CH3COONH4+2Ag¡ý+3NH3+H2O£¬
¹Ê´ð°¸Îª£ºCH3CHO+2Ag£¨NH3£©2OH CH3COONH4+2Ag¡ý+3NH3+H2O£®
µãÆÀ£º±¾Ì⿼²éÁËÎïÖÊת»¯¹ØϵµÄÓ¦Óã¬ÎïÖÊÐÔÖʵÄÓ¦Óã¬ÊµÑé×°Ö÷ÖÎöºÍʵÑé»ù±¾²Ù×÷µÄÓ¦Óã¬Ö÷ÒªÊÇÁòËáÑÇÌú·Ö½â²úÎïµÄÅжϣ¬²úÎïµÄÐÔÖÊÑéÖ¤£¬²úÎïµÄ´ß»¯×÷Óã¬ÒÒ´¼µÄ´ß»¯Ñõ»¯·´Ó¦£¬ÒÒÈ©µÄÑõ»¯·´Ó¦£¬Òø¾µ·´Ó¦µÈÐÔÖʵÄ×ÛºÏÓ¦Óã¬ÌâÄ¿ÄѶÈÖеȣ®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
ʵÑ飺½«ÁòËáÑÇÌú¾§Ì壨FeSO4?7H2O£©¼ÓÈëÈçͼ£¨1£©ÖÐËùʾµÄ¸ÉÔïÊÔ¹ÜAÀ¸ô¾ø¿ÕÆø¼ÓÇ¿Èȳä·Ö·´Ó¦ºó£¬ÊÔ¹ÜAÖеòÐÁô¹ÌÌå¼×£¬ÔÚÊÔ¹ÜBÄڵõ½ÎÞÉ«ÒºÌåÒÒ£®È¡³ö¼×·ÅÔÚͼ£¨2£©ÖеÄʯӢ¹ÜCµÄÖв¿£¬Á¬½ÓºÃÆäËû²¿·ÖµÄÒÇÆ÷ºó£¬¿ªÆôÐýÈûK£¬ÖðµÎµÎ¼ÓÒÒ´¼Ê¹C¹Ü×ó²¿Ê¯ÃÞÈÞ£¨Ò»ÖÖÏËά״¹èËáÑβÄÁÏ£¬ºÜÎȶ¨£¬²»È¼ÉÕ£©½þÈóÎü×㣬¹Ø±ÕK£¬È»ºó¼ÓÈÈ£¬¼¸·ÖÖÓ×óÓÒ£¨¼ÓÈȹý³ÌÖл¹¿ÉÒԽϿìµØ¿ªÆô¡¢¹Ø±ÕÐýÈûKÀ´²¹³äÒÒ´¼²¢Ê¹Ö®Í¨ÈëʯÃÞÈÞÀ£¬¿É¹Û²ìµ½¼×µÄÑÕÉ«Óɺì×ØÉ«Öð½¥±äΪºÚÉ«£®·´Ó¦Í£Ö¹ºó£¬È¡ÊÔ¹ÜEÖеÄÒºÌå0.5mLÖðµÎ¼ÓÈëµ½º¬ÐÂÅäÖƵÄÒø°±ÈÜÒºµÄÊÔ¹ÜFÖУ¬Õñµ´ºó£¬°ÑÊԹܷÅÔÚˮԡÖÐÎÂÈÈ£¬²»¾Ã»á¹Û¿´µ½ÊÔ¹ÜÄÚ±ÚÉϸ½×ÅÒ»²ã¹âÁÁÈç¾µµÄ½ðÊôÒø£®

»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Ð´³öÈçͼ£¨1£©ËùʾµÄ¸ÉÔïÊÔ¹ÜAÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º
2£¨FeSO4?7H2O£©
  ¡÷  
.
 
Fe2O3+SO2¡ü+SO3+14H2O
2£¨FeSO4?7H2O£©
  ¡÷  
.
 
Fe2O3+SO2¡ü+SO3+14H2O
£»
£¨2£©¸ù¾Ý·´Ó¦Ô­ÀíÇëÄãÆÀ¼Ûͼ£¨1£©ÊµÑé×°ÖõÄÖ÷ҪȱµãÊÇ£º
SO2Ϊ´óÆøÎÛȾÎȱÉÙβÆø´¦Àí×°ÖÃ
SO2Ϊ´óÆøÎÛȾÎȱÉÙβÆø´¦Àí×°ÖÃ
£»
£¨3£©¼òÊöÈçºÎ¼ìÑéͼ£¨2£©ËùʾװÖÃÆøÃÜÐÔ£º
¹Ø±Õ»îÈûK£¬½«EµÄµ¼Æø¹Ü²åÈëË®ÖУ¬Î¢ÈÈC¹Ü£¬EÖÐÓÐÆøÅݲúÉú£¬Í£Ö¹¼ÓÈȺó£¬E¹Ü³öÏÖÒ»¶ÎË®Öù£¬ËµÃ÷×°Öò»Â©Æø
¹Ø±Õ»îÈûK£¬½«EµÄµ¼Æø¹Ü²åÈëË®ÖУ¬Î¢ÈÈC¹Ü£¬EÖÐÓÐÆøÅݲúÉú£¬Í£Ö¹¼ÓÈȺó£¬E¹Ü³öÏÖÒ»¶ÎË®Öù£¬ËµÃ÷×°Öò»Â©Æø
£»
£¨4£©ÊÔ¹ÜCÖз´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º
3CH3CH2OH+Fe2O3
¡÷
2Fe+3CH3CHO+3H2O
3CH3CH2OH+Fe2O3
¡÷
2Fe+3CH3CHO+3H2O
£»
£¨5£©ÊÔ¹ÜDµÄ×÷ÓÃÊÇ£º
·ÀÖ¹µ¹Îü
·ÀÖ¹µ¹Îü
£»
£¨6£©Ð´³öFÊÔ¹ÜÖз´Ó¦µÄ»¯Ñ§·½³Ìʽ£º
CH3CHO+2Ag£¨NH3£©2OH
¡÷
CH3COONH4+2Ag¡ý+3NH3+H2O
CH3CHO+2Ag£¨NH3£©2OH
¡÷
CH3COONH4+2Ag¡ý+3NH3+H2O
£®

ʵÑ飺½«ÁòËáÑÇÌú¾§Ì壨FeSO4?7H2O£©¼ÓÈëÈçͼ£¨1£©ÖÐËùʾµÄ¸ÉÔïÊÔ¹ÜAÀ¸ô¾ø¿ÕÆø¼ÓÇ¿Èȳä·Ö·´Ó¦ºó£¬ÊÔ¹ÜAÖеòÐÁô¹ÌÌå¼×£¬ÔÚÊÔ¹ÜBÄڵõ½ÎÞÉ«ÒºÌåÒÒ£®È¡³ö¼×·ÅÔÚͼ£¨2£©ÖеÄʯӢ¹ÜCµÄÖв¿£¬Á¬½ÓºÃÆäËû²¿·ÖµÄÒÇÆ÷ºó£¬¿ªÆôÐýÈûK£¬ÖðµÎµÎ¼ÓÒÒ´¼Ê¹C¹Ü×ó²¿Ê¯ÃÞÈÞ£¨Ò»ÖÖÏËά״¹èËáÑβÄÁÏ£¬ºÜÎȶ¨£¬²»È¼ÉÕ£©½þÈóÎü×㣬¹Ø±ÕK£¬È»ºó¼ÓÈÈ£¬¼¸·ÖÖÓ×óÓÒ£¨¼ÓÈȹý³ÌÖл¹¿ÉÒԽϿìµØ¿ªÆô¡¢¹Ø±ÕÐýÈûKÀ´²¹³äÒÒ´¼²¢Ê¹Ö®Í¨ÈëʯÃÞÈÞÀ£¬¿É¹Û²ìµ½¼×µÄÑÕÉ«Óɺì×ØÉ«Öð½¥±äΪºÚÉ«£®·´Ó¦Í£Ö¹ºó£¬È¡ÊÔ¹ÜEÖеÄÒºÌå0.5mLÖðµÎ¼ÓÈëµ½º¬ÐÂÅäÖƵÄÒø°±ÈÜÒºµÄÊÔ¹ÜFÖУ¬Õñµ´ºó£¬°ÑÊԹܷÅÔÚˮԡÖÐÎÂÈÈ£¬²»¾Ã»á¹Û¿´µ½ÊÔ¹ÜÄÚ±ÚÉϸ½×ÅÒ»²ã¹âÁÁÈç¾µµÄ½ðÊôÒø£®

»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Ð´³öÈçͼ£¨1£©ËùʾµÄ¸ÉÔïÊÔ¹ÜAÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º______£»
£¨2£©¸ù¾Ý·´Ó¦Ô­ÀíÇëÄãÆÀ¼Ûͼ£¨1£©ÊµÑé×°ÖõÄÖ÷ҪȱµãÊÇ£º______£»
£¨3£©¼òÊöÈçºÎ¼ìÑéͼ£¨2£©ËùʾװÖÃÆøÃÜÐÔ£º______£»
£¨4£©ÊÔ¹ÜCÖз´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º______£»
£¨5£©ÊÔ¹ÜDµÄ×÷ÓÃÊÇ£º______£»
£¨6£©Ð´³öFÊÔ¹ÜÖз´Ó¦µÄ»¯Ñ§·½³Ìʽ£º______£®

ʵÑ飺½«ÁòËáÑÇÌú¾§Ì壨FeSO4¡¤7H2O£©¼ÓÈëÈçͼ1ËùʾµÄ¸ÉÔïÊÔ¹ÜAÀ¸ô¾ø¿ÕÆø¼ÓÇ¿Èȳä·Ö·´Ó¦ºó£¬ÊÔ¹ÜAÖеõ½²ÐÁô¹ÌÌå¼×£¬ÔÚÊÔ¹ÜBÄڵõ½ÎÞÉ«ÒºÌåÒÒ¡£È¡³ö¼×·ÅÔÚͼ2ÖеÄʯӢ¹ÜCµÄÖв¿£¬Á¬½ÓºÃÆäËû²¿·ÖµÄÒÇÆ÷ºó£¬¿ªÆôÐýÈûK£¬ÖðµÎµÎ¼ÓÒÒ´¼Ê¹C¹Ü×ó²¿Ê¯ÃÞÈÞ£¨Ò»ÖÖÏËά״¹èËáÑβÄÁÏ£¬ºÜÎȶ¨£¬²»È¼ÉÕ£©½þÈóÎü×㣬¹Ø±ÕK£¬È»ºó¼ÓÈÈ£¬¼¸·ÖÖÓ×óÓÒ£¨¼ÓÈȹý³ÌÖл¹¿ÉÒԽϿìµØ¿ªÆô¡¢¹Ø±ÕÐýÈûKÀ´²¹³äÒÒ´¼²¢Ê¹Ö®Í¨ÈëʯÃÞÈÞÀ£¬¿É¹Û²ìµ½¼×µÄÑÕÉ«Óɺì×ØÉ«Öð½¥±äΪºÚÉ«¡£·´Ó¦Í£Ö¹ºó£¬È¡ÊÔ¹ÜEÖеÄÒºÌå0.5 mLÖðµÎ¼ÓÈëµ½º¬ÐÂÅäÖƵÄÒø°±ÈÜÒºµÄÊÔ¹ÜFÖУ¬Õñµ´ºó£¬°ÑÊԹܷÅÔÚˮԡÖÐÎÂÈÈ£¬²»¾Ã»á¿´µ½ÊÔ¹ÜÄÚ±ÚÉϸ½×ÅÒ»²ã¹âÁÁÈç¾µµÄ½ðÊôÒø¡£

ͼ1                                                                    ͼ2

»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©Ð´³öÈçͼ1ËùʾµÄ¸ÉÔïÊÔ¹ÜAÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³Ìʽ___________________________£»

£¨2£©¸ù¾Ý·´Ó¦Ô­Àí£¬ÇëÄãÆÀ¼Ûͼ1ʵÑé×°ÖõÄÖ÷Ҫȱµã£º____________________________£»

£¨3£©ÊÔ¹ÜCÖз´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º_________________________________________£»

£¨4£©¼òÊöÈçºÎ¼ìÑéͼ2ËùʾװÖõÄÆøÃÜÐÔ£º_____________________________________£»

£¨5£©ÊÔ¹ÜD°²×°Ôڴ˵Ä×÷ÓÃÊÇ_____________________________________________£»

£¨6£©Ð´³öFÊÔ¹ÜÖз´Ó¦µÄ»¯Ñ§·½³Ìʽ£º_______________________________________¡£

ʵÑ飺½«ÁòËáÑÇÌú¾§Ìå(FeSO4?7H2O)¼ÓÈëÈçͼ1ÖÐËùʾµÄ¸ÉÔïÊÔ¹ÜAÀ¸ô¾ø¿ÕÆø¼ÓÇ¿

 

³ä·Ö·´Ó¦ºó£¬ÊÔ¹ÜAÖеòÐÁô¹ÌÌå¼×£¬ÔÚÊÔ¹ÜBÄڵõ½ÎÞÉ«ÒºÌåÒÒ¡£È¡³ö¼×·ÅÔÚͼ2ÖеÄʯӢ¹ÜCÖУ¬Á¬½ÓºÃÆäËû²¿·ÖµÄÒÇÆ÷ºó£¬¿ªÆôÐýÈûK£¬ÖðµÎµÎ¼ÓÒÒ´¼Ê¹C×ó²¿Ê¯ÃÞÈÞ(Ò»ÖÖÏËά״¹èËáÑβÄÁÏ£¬ºÜÎȶ¨£¬²»È¼ÉÕ)½þÈóÎü×㣬¹Ø±ÕK£¬È»ºó¼ÓÈÈ£¬¼¸·ÖÖÓ×óÓÒ(¼ÓÈȹý³ÌÖл¹¿ÉÒԽϿìµØ¿ªÆô¡¢¹Ø±ÕÐýÈûKÀ´²¹³äÒÒ´¼²¢Ê¹Ö®Í¨ÈëʯÃÞÈÞÀï)£¬¿É¹Û²ìµ½¼×µÄÑÕÉ«Óɺì×ØÉ«Öð½¥±äΪºÚÉ«¡£·´Ó¦Í£Ö¹ºó£¬È¡ÊÔ¹ÜEÖеÄÒºÌå0.5ml£¬ÖðµÎ¼ÓÈëµ½º¬ÐÂÅäÖƵÄÒø°±ÈÜÒºµÄÊÔ¹ÜFÖУ¬Õñµ´ºó£¬°ÑÊԹܷÅÔÚˮԡÖÐÎÂÈÈ£¬²»¾Ã»á¹Û¿´µ½ÊÔ¹ÜÄÚ±ÚÉϸ½×ÅÒ»²ã¹âÁÁÈç¾µµÄ½ðÊôÒø¡£

»Ø´ðÏÂÁÐÎÊÌ⣺

(1)д³öÈçͼ1ËùʾµÄ¸ÉÔïÊÔ¹ÜAÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º_____________________________¡£

(2)¸ù¾Ý·´Ó¦Ô­ÀíÇëÄãÆÀ¼Ûͼ1ʵÑé×°ÖõÄÖ÷Ҫȱµã_____________________________________¡£

(3)ÊÔ¹ÜCÖз´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º________________________________________¡£

(4)¼òÊöÈçºÎ¼ìÑéͼ2ËùʾװÖÃÆøÃÜÐÔ£º__________________________________¡£

(5)ÊÔ¹ÜD°²×°Ôڴ˵Ä×÷ÓÃÊÇ£º_______________________________________¡£

(6)д³öEÊÔ¹ÜÖз´Ó¦µÄ»¯Ñ§·½³Ìʽ£º__________________________________¡£

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

¾«Ó¢¼Ò½ÌÍø