»¯Ñ§Ð¡×é¸ù¾Ý°±Æø»¹Ô­Ñõ»¯Í­µÄ·´Ó¦£¬Éè¼ÆÊµÑé²â¶¨CuÔªËØµÄÏà¶ÔÔ­×ÓÖÊÁ¿¡£ÒÑÖª£º¢Ù 2NH4Cl+Ca(OH)2=CaCl2+2NH3¡ü+2H2O   ¢Ú °±Æø£¨NH3£©ÊǼîÐÔÆøÌå

Çë½áºÏÏÂͼ»Ø´ðÎÊÌâ¡£

£¨1£©½«´¿¾»¸ÉÔïµÄ°±ÆøÍ¨ÈëBÖУ¬¹Û²ìµ½²£Á§¹ÜÄÚºÚÉ«¹ÌÌå±äΪÁÁºìÉ«£¬¹Ü¿ÚÓÐÒºµÎ£¬Í¬Ê±Éú³É¿ÕÆøÖк¬Á¿×î¶àµÄÆøÌ壬д³öBÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ          ¡£

£¨2£©²â¶¨CuÔªËØÏà¶ÔÔ­×ÓÖÊÁ¿µÄʵÑé¹ý³ÌΪ£ºÏȳÆÁ¿CuOµÄÖÊÁ¿£¬ÍêÈ«·´Ó¦ºó²â¶¨Éú³ÉË®µÄÖÊÁ¿£¬Óɴ˼ÆËã³öCuÔªËØµÄÏà¶ÔÔ­×ÓÖÊÁ¿¡£

¢ñ.С×éͬѧÀûÓÃÉÏͼ½øÐÐʵÑ飬ÏÂÁÐ×°ÖÃÁ¬½ÓºÏÀíµÄÊÇ£¨ÌîÐòºÅ£¬×°ÖÿÉÖØ¸´Ê¹Óã©      ¡£

¢Ù ACBDC           ¢Ú ADBCD        ¢Û ADBDC            ¢Ü ABDC

¢ò.ÔÚ±¾ÊµÑéÖУ¬Ê¹²â¶¨½á¹ûÆ«´óµÄÔ­Òò¿ÉÄÜÊÇ_______________ (ÌîÐòºÅ)£»

¢Ù CuOδÍêÈ«Æð·´Ó¦               ¢Ú CuO²»¸ÉÔï

¢Û CuOÖлìÓв»·´Ó¦µÄÔÓÖÊ         ¢Ü NH4C1ÓëCa(OH)2»ìºÏÎï²»¸ÉÔï   

¢ó.ÔÚ±¾ÊµÑéÖУ¬»¹Í¨¹ý²â¶¨___________________________µÄÖÊÁ¿´ïµ½ÊµÑéÄ¿µÄ¡£ 

¡¾½âÎö¡¿£¨1£©¸ù¾Ý°±Æø»¹Ô­Ñõ»¯Í­·´Ó¦µÄÏÖÏó£¬ÅжϷ´Ó¦µÄÉú³ÉÎд³ö·´Ó¦µÄ»¯Ñ§·½³Ìʽ£¬ÆäÖÐ¿ÕÆøÖк¬Á¿×î¶àµÄÆøÌåΪµªÆø£»

£¨2£©¢Ù¸ù¾ÝʵÑéÉèÏ룬ÔÚËùÌṩµÄ×°ÖÃÖÐÑ¡Ôñ¼òµ¥¡¢ºÏÀíµÄ×°ÖÃ×é×°ÄÜʵÏÖʵÑéÄ¿µÄʵÑé×°Öã»

¢Ú·ÖÎöËùÌṩÒòËØ¶ÔʵÑé½á¹ûµÄÓ°Ï죬ÅÐ¶ÏÆäÖлᵼÖ²ⶨ½á¹ûÆ«´óµÄÒòËØ£»

¢Û¸ù¾Ý¶ÔʵÑéµÄÀí½â£¬Îª´ïµ½²â¶¨Ä¿µÄ£¬Åжϻ¹¿ÉÒÔ²ÉÈ¡µÄ²âÁ¿Öµ£®

 

¡¾´ð°¸¡¿

£¨1£©3CuO + 2NH3 ¼ÓÈÈ 3Cu + N2 + 3H2O

£¨2£©¢ñ. ¢Û    ¢ò.¢Ù¢Û £¨´Ë¿Õ2·Ö£¬Â©Ñ¡1·Ö£¬ÓдíÎóÑ¡Ïî²»¸ø·Ö£©

     ¢ó.CuOºÍCu

 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â

СÃ÷ÉÏÍøÔĶÁʱµÃÖª¸ù¾Ý°±Æø»¹Ô­Ñõ»¯Í­µÄ·´Ó¦£¬¿ÉÉè¼Æ²â¶¨Í­ÔªËØÏà¶ÔÔ­×ÓÖÊÁ¿£¨Ar£©µÄʵÑ飮СÃ÷ºÜ¸ÐÐËȤ£¬ºÍ»¯Ñ§ÐËȤС×éµÄͬѧÉè¼ÆÁËÈçÏÂ̽¾¿»î¶¯£¬Çë½áºÏͼ»Ø´ðÎÊÌ⣮
¾«Ó¢¼Ò½ÌÍø
£¨1£©ÊµÑéÊÒÀûÓÃÂÈ»¯ï§ºÍÇâÑõ»¯¸Æ¹ÌÌåÖ®¼äµÄ¸´·Ö½â·´Ó¦ÖÆÈ¡°±Æø£®Ð¡Ã÷½«AÓëEÁ¬½Ó£¬¹Û²ìµ½ÎÞÉ«·Ó̪ÊÔÒº±äºì£¬ËµÃ÷°±ÆøÏÔ
 
£¨Ìî¡°ËáÐÔ¡±¡¢¡°ÖÐÐÔ¡±¡¢»ò¡°¼îÐÔ¡±£©£®AÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ
 
£®
£¨2£©Ð¡¸Õ½«´¿¾»¸ÉÔïµÄ°±ÆøÍ¨ÈëBÖУ¬¹Û²ìµ½²£Á§¹ÜÄÚºÚÉ«¹ÌÌå±äΪÁÁºìÉ«£¬¹Ü¿ÚÓÐÒºµÎ£¬Í¬Ê±Éú³É¿ÕÆøÖк¬Á¿×î¶àµÄÆøÌ壬BÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ
 
£®
£¨3£©Ð¡×éͬѧ²â¶¨Í­ÔªËØÏà¶ÔÔ­×ÓÖÊÁ¿£¨Ar£©µÄʵÑéÉèÏëÊÇ£ºÏȳÆÁ¿·´Ó¦ÎïCuOµÄÖÊÁ¿M£¨CuO£©£¬·´Ó¦ÍêÈ«ºó²â¶¨Éú³ÉÎïË®µÄÖÊÁ¿M£¨H2O£©£¬Óɴ˼ÆËã³öCuµÄÏà¶ÔÔ­×ÓÖÊÁ¿£®
¢ÙС×éÿλͬѧ´ÓËùÌṩµÄÒÇÆ÷ÖÐÑ¡Ôñ²¢×é×°ÁËÒ»Ì×ÓÃÓÚ±¾ÊµÑéµÄ×°Öã¬ÆäÖмòµ¥¡¢ºÏÀíµÄÊÇ£¨ÌîÐòºÅ£¬×°ÖÿÉÖØ¸´Ê¹Óã©
 
£®
A£®acdbbeecd       B£®aeebbcdee        C£®aeebbeecd             D£®abbeecd
¢ÚÔÚ±¾ÊµÑéÖУ¬Ê¹²â¶¨½á¹ûAr£¨Cu£©Æ«´óµÄÊÇ
 
 £¨ÌîÐòºÅ£©£»
A£®CuOδÍêÈ«Æð·´Ó¦               B£®CuO²»¸ÉÔï       C£®ÇâÑõ»¯ÄƹÌÌå²»¸ÉÔï
D£®CuOÖлìÓв»·´Ó¦µÄÔÓÖÊ         E£®NH4ClÓëCa£¨OH£©2»ìºÏÎï²»¸ÉÔï
¢ÛÔÚ±¾ÊµÑéÖУ¬»¹Í¨¹ý²â¶¨
 
ºÍ
 
µÄÖÊÁ¿´ïµ½Ä¿µÄ£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨2012?³¯ÑôÇø¶þÄ££©»¯Ñ§Ð¡×é¸ù¾Ý°±Æø»¹Ô­Ñõ»¯Í­µÄ·´Ó¦£¬Éè¼ÆÊµÑé²â¶¨CuÔªËØµÄÏà¶ÔÔ­×ÓÖÊÁ¿£®
ÒÑÖª£º¢Ù2NH4Cl+Ca£¨OH£©2=CaCl2+2NH3¡ü+2H2O   ¢Ú°±Æø£¨NH3£©ÊǼîÐÔÆøÌå
Çë½áºÏÏÂͼ»Ø´ðÎÊÌ⣮

£¨1£©½«´¿¾»¸ÉÔïµÄ°±ÆøÍ¨ÈëBÖУ¬¹Û²ìµ½²£Á§¹ÜÄÚºÚÉ«¹ÌÌå±äΪÁÁºìÉ«£¬¹Ü¿ÚÓÐÒºµÎ£¬Í¬Ê±Éú³É¿ÕÆøÖк¬Á¿×î¶àµÄÆøÌ壬д³öBÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ
3CuO+2NH3
  ¡÷  
.
 
3Cu+N2+3H2O
3CuO+2NH3
  ¡÷  
.
 
3Cu+N2+3H2O
£®
£¨2£©²â¶¨CuÔªËØÏà¶ÔÔ­×ÓÖÊÁ¿µÄʵÑé¹ý³ÌΪ£ºÏȳÆÁ¿CuOµÄÖÊÁ¿£¬ÍêÈ«·´Ó¦ºó²â¶¨Éú³ÉË®µÄÖÊÁ¿£¬Óɴ˼ÆËã³öCuÔªËØµÄÏà¶ÔÔ­×ÓÖÊÁ¿£®
¢ñ£®Ð¡×éͬѧÀûÓÃÉÏͼ½øÐÐʵÑ飬ÏÂÁÐ×°ÖÃÁ¬½ÓºÏÀíµÄÊÇ£¨ÌîÐòºÅ£¬×°ÖÿÉÖØ¸´Ê¹Óã©
¢Û
¢Û
£®
¢ÙACBDC           ¢ÚADBCD        ¢ÛADBDC            ¢ÜABDC
¢ò£®ÔÚ±¾ÊµÑéÖУ¬Ê¹²â¶¨½á¹ûÆ«´óµÄÔ­Òò¿ÉÄÜÊÇ
¢Ù¢Û
¢Ù¢Û
 £¨ÌîÐòºÅ£©£»
¢ÙCuOδÍêÈ«Æð·´Ó¦               ¢ÚCuO²»¸ÉÔï
¢ÛCuOÖлìÓв»·´Ó¦µÄÔÓÖÊ         ¢ÜNH4C1ÓëCa£¨OH£©2»ìºÏÎï²»¸ÉÔï
¢ó£®ÔÚ±¾ÊµÑéÖУ¬»¹Í¨¹ý²â¶¨
CuOºÍCu
CuOºÍCu
µÄÖÊÁ¿´ïµ½ÊµÑéÄ¿µÄ£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º2012½ì±±¾©Êг¯ÑôÇøÖп¼¶þÄ£»¯Ñ§ÊÔ¾í£¨´ø½âÎö£© ÌâÐÍ£ºÌ½¾¿Ìâ

»¯Ñ§Ð¡×é¸ù¾Ý°±Æø»¹Ô­Ñõ»¯Í­µÄ·´Ó¦£¬Éè¼ÆÊµÑé²â¶¨CuÔªËØµÄÏà¶ÔÔ­×ÓÖÊÁ¿¡£ÒÑÖª£º¢Ù 2NH4Cl+Ca(OH)2=CaCl2+2NH3¡ü+2H2O  ¢Ú °±Æø£¨NH3£©ÊǼîÐÔÆøÌå
Çë½áºÏÏÂͼ»Ø´ðÎÊÌâ¡£

£¨1£©½«´¿¾»¸ÉÔïµÄ°±ÆøÍ¨ÈëBÖУ¬¹Û²ìµ½²£Á§¹ÜÄÚºÚÉ«¹ÌÌå±äΪÁÁºìÉ«£¬¹Ü¿ÚÓÐÒºµÎ£¬Í¬Ê±Éú³É¿ÕÆøÖк¬Á¿×î¶àµÄÆøÌ壬д³öBÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ         ¡£
£¨2£©²â¶¨CuÔªËØÏà¶ÔÔ­×ÓÖÊÁ¿µÄʵÑé¹ý³ÌΪ£ºÏȳÆÁ¿CuOµÄÖÊÁ¿£¬ÍêÈ«·´Ó¦ºó²â¶¨Éú³ÉË®µÄÖÊÁ¿£¬Óɴ˼ÆËã³öCuÔªËØµÄÏà¶ÔÔ­×ÓÖÊÁ¿¡£
¢ñ.С×éͬѧÀûÓÃÉÏͼ½øÐÐʵÑ飬ÏÂÁÐ×°ÖÃÁ¬½ÓºÏÀíµÄÊÇ£¨ÌîÐòºÅ£¬×°ÖÿÉÖØ¸´Ê¹Óã©     ¡£
¢Ù ACBDC          ¢Ú ADBCD       ¢Û ADBDC           ¢Ü ABDC
¢ò.ÔÚ±¾ÊµÑéÖУ¬Ê¹²â¶¨½á¹ûÆ«´óµÄÔ­Òò¿ÉÄÜÊÇ_______________ (ÌîÐòºÅ)£»
¢Ù CuOδÍêÈ«Æð·´Ó¦              ¢Ú CuO²»¸ÉÔï
¢Û CuOÖлìÓв»·´Ó¦µÄÔÓÖÊ        ¢Ü NH4C1ÓëCa(OH)2»ìºÏÎï²»¸ÉÔï   
¢ó.ÔÚ±¾ÊµÑéÖУ¬»¹Í¨¹ý²â¶¨___________________________µÄÖÊÁ¿´ïµ½ÊµÑéÄ¿µÄ¡£ 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º2012Äê±±¾©Êг¯ÑôÇøÖп¼»¯Ñ§¶þÄ£ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£º½â´ðÌâ

»¯Ñ§Ð¡×é¸ù¾Ý°±Æø»¹Ô­Ñõ»¯Í­µÄ·´Ó¦£¬Éè¼ÆÊµÑé²â¶¨CuÔªËØµÄÏà¶ÔÔ­×ÓÖÊÁ¿£®
ÒÑÖª£º¢Ù2NH4Cl+Ca£¨OH£©2=CaCl2+2NH3¡ü+2H2O   ¢Ú°±Æø£¨NH3£©ÊǼîÐÔÆøÌå
Çë½áºÏÏÂͼ»Ø´ðÎÊÌ⣮

£¨1£©½«´¿¾»¸ÉÔïµÄ°±ÆøÍ¨ÈëBÖУ¬¹Û²ìµ½²£Á§¹ÜÄÚºÚÉ«¹ÌÌå±äΪÁÁºìÉ«£¬¹Ü¿ÚÓÐÒºµÎ£¬Í¬Ê±Éú³É¿ÕÆøÖк¬Á¿×î¶àµÄÆøÌ壬д³öBÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ______ 3Cu+N2+3H2O

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸