ÌâÄ¿ÄÚÈÝ
Ϊ¶¨ÐÔ̽¾¿ÒÒ´¼µÄÐÔÖÊ£¬Ä³»¯Ñ§¿ÎÍâС×éͨ¹ý²éÔÄ×ÊÁϲ¢Éè¼ÆÁËʵÑé·½°¸½øÐÐ̽¾¿£®
·½°¸¢ñ£ºÔÚÊ¢ÓÐÉÙÁ¿ÎÞË®ÒÒ´¼µÄÊÔ¹ÜÖУ¬¼ÓÈëÒ»Á£³ýȥúÓ͵ĽðÊôÄÆ£¬ÔÚÊԹܿÚѸËÙÈûÉÏÅäÓмâ×ìµ¼¹ÜµÄµ¥¿×Èû£¬µãȼ·Å³öµÄÆøÌ壬²¢°ÑÒ»¸ÉÔïµÄСÉÕ±ÕÖÔÚ»ðÑæÉÏ£¬ÔÚÉÕ±±ÚÉϳöÏÖÒºµÎºó£¬Ñ¸ËÙµ¹×ªÉÕ±£¬ÏòÉÕ±ÖмÓÈëÉÙÁ¿µÄ³ÎÇåʯ»ÒË®£¬¹Û²ìÓÐÎÞ»ì×DzúÉú£®
£¨1£©Çëд³öÒÒ´¼ÓëÄÆ·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º
£¨2£©ÒÔÉÏʵÑéÉè¼ÆÒòȱÉÙ±ØÒªµÄ²½Öè¶ø´æÔÚ°²È«Òþ»¼£¬ÇëÄãÖ¸³öËùȱÉÙ±ØÒªµÄ²½ÖèÊÇ
£¨3£©ÈôÏòÉÕ±ÖмÓÈëÉÙÁ¿³ÎÇåʯ»ÒË®ºó·¢ÏÖÓлì×Ç£¬ÔòȼÉÕ²úÉúCO2µÄÎïÖÊ×î¿ÉÄÜÊÇ
·½°¸¢ò£º£¨1£©È¡Ò»¸ùÍË¿£¬°ÑÆäÖÐÒ»¶ËÈƳÉÂÝÐý×´£¨Ôö´ó½Ó´¥Ãæ»ý£©£®µãȼһյ¾Æ¾«µÆ£¬°ÑÈƳÉÂÝÐý×´Ò»¶ËµÄÍË¿ÒƵ½¾Æ¾«µÆÍâÑæÉÏ×ÆÉÕ£¨ÈçÓÒͼ£©£¬¹Û²ìµ½µÄʵÑéÏÖÏó£º
£¨2£©°ÑÂÝÐý×´ÍË¿Íù¾Æ¾«µÆÄÚÑæÒƶ¯£¬¹Û²ìµ½µÄʵÑéÏÖÏó£º
·½°¸¢ñ£ºÔÚÊ¢ÓÐÉÙÁ¿ÎÞË®ÒÒ´¼µÄÊÔ¹ÜÖУ¬¼ÓÈëÒ»Á£³ýȥúÓ͵ĽðÊôÄÆ£¬ÔÚÊԹܿÚѸËÙÈûÉÏÅäÓмâ×ìµ¼¹ÜµÄµ¥¿×Èû£¬µãȼ·Å³öµÄÆøÌ壬²¢°ÑÒ»¸ÉÔïµÄСÉÕ±ÕÖÔÚ»ðÑæÉÏ£¬ÔÚÉÕ±±ÚÉϳöÏÖÒºµÎºó£¬Ñ¸ËÙµ¹×ªÉÕ±£¬ÏòÉÕ±ÖмÓÈëÉÙÁ¿µÄ³ÎÇåʯ»ÒË®£¬¹Û²ìÓÐÎÞ»ì×DzúÉú£®
£¨1£©Çëд³öÒÒ´¼ÓëÄÆ·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º
2Na+2CH3CH2OH=2CH3CH2ONa+H2¡ü
2Na+2CH3CH2OH=2CH3CH2ONa+H2¡ü
£®£¨2£©ÒÔÉÏʵÑéÉè¼ÆÒòȱÉÙ±ØÒªµÄ²½Öè¶ø´æÔÚ°²È«Òþ»¼£¬ÇëÄãÖ¸³öËùȱÉÙ±ØÒªµÄ²½ÖèÊÇ
µãȼ·Å³öµÄÆøÌå֮ǰûÓмìÑé´¿¶È
µãȼ·Å³öµÄÆøÌå֮ǰûÓмìÑé´¿¶È
£®£¨3£©ÈôÏòÉÕ±ÖмÓÈëÉÙÁ¿³ÎÇåʯ»ÒË®ºó·¢ÏÖÓлì×Ç£¬ÔòȼÉÕ²úÉúCO2µÄÎïÖÊ×î¿ÉÄÜÊÇ
ÒÒ´¼£¨ÕôÆø£©
ÒÒ´¼£¨ÕôÆø£©
£¨Ð´Ãû³Æ£©£®·½°¸¢ò£º£¨1£©È¡Ò»¸ùÍË¿£¬°ÑÆäÖÐÒ»¶ËÈƳÉÂÝÐý×´£¨Ôö´ó½Ó´¥Ãæ»ý£©£®µãȼһյ¾Æ¾«µÆ£¬°ÑÈƳÉÂÝÐý×´Ò»¶ËµÄÍË¿ÒƵ½¾Æ¾«µÆÍâÑæÉÏ×ÆÉÕ£¨ÈçÓÒͼ£©£¬¹Û²ìµ½µÄʵÑéÏÖÏó£º
ÍË¿ÓÉ×ϺìÉ«±äºÚ
ÍË¿ÓÉ×ϺìÉ«±äºÚ
£®£¨2£©°ÑÂÝÐý×´ÍË¿Íù¾Æ¾«µÆÄÚÑæÒƶ¯£¬¹Û²ìµ½µÄʵÑéÏÖÏó£º
ÍË¿ÓɺÚÉ«±ä³É×ϺìÉ«
ÍË¿ÓɺÚÉ«±ä³É×ϺìÉ«
£¬Óû¯Ñ§·½³Ìʽ±íʾ¸ÃÏÖÏó²úÉúµÄÔÀí£ºCH3CH2OH+CuO
CH3CHO+Cu+H2O
¡÷ |
CH3CH2OH+CuO
CH3CHO+Cu+H2O
£®¡÷ |
·ÖÎö£º·½°¸¢ñ£¨1£©ÄÆÄÜÖû»ôÇ»ùÉϵÄÇ⣮
£¨2£©µãȼ¿ÉȼÐÔÆøÌå֮ǰ±ØÐë¼ìÑé´¿¶È·ñÔò·¢Éú±¬Õ¨£®
£¨3£©ÒÒ´¼Ò×»Ó·¢¿ÉÄÜȼÉÕÉú³É¶þÑõ»¯Ì¼£®
·½°¸¢ò£¨1£©ÍË¿ÓëÑõÆø·´Ó¦Éú³ÉÑõ»¯Í£®
£¨2£©Ñõ»¯Í±»ÒÒ´¼»¹Ô³É͵¥ÖÊ£¬Í¬Ê±ÒÒ´¼±»Ñõ»¯ÎªÒÒÈ©£®
£¨2£©µãȼ¿ÉȼÐÔÆøÌå֮ǰ±ØÐë¼ìÑé´¿¶È·ñÔò·¢Éú±¬Õ¨£®
£¨3£©ÒÒ´¼Ò×»Ó·¢¿ÉÄÜȼÉÕÉú³É¶þÑõ»¯Ì¼£®
·½°¸¢ò£¨1£©ÍË¿ÓëÑõÆø·´Ó¦Éú³ÉÑõ»¯Í£®
£¨2£©Ñõ»¯Í±»ÒÒ´¼»¹Ô³É͵¥ÖÊ£¬Í¬Ê±ÒÒ´¼±»Ñõ»¯ÎªÒÒÈ©£®
½â´ð£º½â£º·½°¸¢ñ£¨1£©ÄÆÄÜÖû»ôÇ»ùÉϵÄÇ⣬2Na+2CH3CH2OH=2CH3CH2ONa+H2¡ü£»
£¨2£©È¼¿ÉȼÐÔÆøÌå֮ǰ±ØÐë¼ìÑé´¿¶È·ñÔò·¢Éú±¬Õ¨£¬¹Êµãȼ·Å³öµÄÆøÌå֮ǰûÓмìÑé´¿¶È£»
£¨3£©¿ÉȼÐÔÆøÌå³ýÁËÇâÆøÁíÍâÒ»ÖÖ¿ÉÄܾÍÊǻӷ¢³öµÄÒÒ´¼È¼ÉÕÉú³É¶þÑõ»¯Ì¼£¬¹ÊȼÉÕ²úÉúCO2µÄÎïÖÊ×î¿ÉÄÜÊÇÒÒ´¼£¨ÕôÆø£©£»
·½°¸¢ò£¨1£©ÍË¿ÓëÑõÆø·´Ó¦Éú³ÉºÚÉ«µÄÑõ»¯Í£¬¹Ê¹Û²ìµ½µÄʵÑéÏÖÏóΪÍË¿ÓÉ×ϺìÉ«±äºÚ£»
£¨2£©ºÚÉ«µÄÑõ»¯Í±»ÒÒ´¼»¹Ô³ÉͺìÉ«µÄ͵¥ÖÊ£¬Í¬Ê±ÒÒ´¼±»Ñõ»¯ÎªÒÒÈ©£¬¹ÊʵÑéÏÖÏóΪÍË¿ÓɺÚÉ«±ä³É×ϺìÉ«£¬»¯Ñ§·½³ÌʽΪ£ºCH3CH2OH+CuO
CH3CHO+Cu+H2O
¹Ê´ð°¸Îª£º·½°¸¢ñ£¨1£©2Na+2CH3CH2OH=2CH3CH2ONa+H2¡ü£»
£¨2£©µãȼ·Å³öµÄÆøÌå֮ǰûÓмìÑé´¿¶È£»
£¨3£©ÒÒ´¼£¨ÕôÆø£©£»
·½°¸¢ò£¨1£©ÍË¿ÓÉ×ϺìÉ«±äºÚ£»
£¨2£©ÍË¿ÓɺÚÉ«±ä³É×ϺìÉ«£¬CH3CH2OH+CuO
CH3CHO+Cu+H2O
£¨2£©È¼¿ÉȼÐÔÆøÌå֮ǰ±ØÐë¼ìÑé´¿¶È·ñÔò·¢Éú±¬Õ¨£¬¹Êµãȼ·Å³öµÄÆøÌå֮ǰûÓмìÑé´¿¶È£»
£¨3£©¿ÉȼÐÔÆøÌå³ýÁËÇâÆøÁíÍâÒ»ÖÖ¿ÉÄܾÍÊǻӷ¢³öµÄÒÒ´¼È¼ÉÕÉú³É¶þÑõ»¯Ì¼£¬¹ÊȼÉÕ²úÉúCO2µÄÎïÖÊ×î¿ÉÄÜÊÇÒÒ´¼£¨ÕôÆø£©£»
·½°¸¢ò£¨1£©ÍË¿ÓëÑõÆø·´Ó¦Éú³ÉºÚÉ«µÄÑõ»¯Í£¬¹Ê¹Û²ìµ½µÄʵÑéÏÖÏóΪÍË¿ÓÉ×ϺìÉ«±äºÚ£»
£¨2£©ºÚÉ«µÄÑõ»¯Í±»ÒÒ´¼»¹Ô³ÉͺìÉ«µÄ͵¥ÖÊ£¬Í¬Ê±ÒÒ´¼±»Ñõ»¯ÎªÒÒÈ©£¬¹ÊʵÑéÏÖÏóΪÍË¿ÓɺÚÉ«±ä³É×ϺìÉ«£¬»¯Ñ§·½³ÌʽΪ£ºCH3CH2OH+CuO
¡÷ |
¹Ê´ð°¸Îª£º·½°¸¢ñ£¨1£©2Na+2CH3CH2OH=2CH3CH2ONa+H2¡ü£»
£¨2£©µãȼ·Å³öµÄÆøÌå֮ǰûÓмìÑé´¿¶È£»
£¨3£©ÒÒ´¼£¨ÕôÆø£©£»
·½°¸¢ò£¨1£©ÍË¿ÓÉ×ϺìÉ«±äºÚ£»
£¨2£©ÍË¿ÓɺÚÉ«±ä³É×ϺìÉ«£¬CH3CH2OH+CuO
¡÷ |
µãÆÀ£º±¾Ì⿼²éÒÒ´¼µÄÎïÀíÐÔÖÊ¡¢»¯Ñ§ÐÔÖÊ£¬ÄѶȲ»´ó£¬×¢ÒâÒÒ´¼µÄÑõ»¯£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿