(11·ÖÏÂÁпòͼËùʾµÄÎïÖÊת»¯¹ØÏµÖУ¬¼×ÊÇÈÕ³£Éú»îÖг£¼ûµÄ½ðÊô£¬ÒÒ¡¢±û¡¢¶¡Êdz£¼ûµÄÆøÌåµ¥ÖÊ¡£ÆøÌåBÓëÆøÌåCÏàÓö²úÉú´óÁ¿µÄ°×ÑÌÉú³ÉE£¬AÊÇÇ¿¼î£¬DÊǺ£Ë®ÖÐŨ¶È×î¸ßµÄÑÎ(²¿·Ö·´Ó¦ÎïºÍÉú³ÉÎï¼°Ë®ÒÑÂÔÈ¥)¡£

Çë»Ø´ðÏÂÁÐÎÊÌ⣺
(1)д³öÒÒÓë±ûÉú³ÉBµÄ»¯Ñ§·½³Ìʽ£º________________________________________¡£
(2)д³ö¼×ºÍAÈÜÒº·´Ó¦µÄÀë×Ó·½³Ìʽ£º_______________________________________¡£
(3)¼ìÑéEÖеÄÑôÀë×Óʱ£¬È¡ÉÙÁ¿EÓÚÊÔ¹ÜÖУ¬________________ÔòÖ¤Ã÷EÖÐÓиÃÑôÀë×Ó¡£
(4)д³öʵÑéÊÒÖÆÈ¡BµÄ»¯Ñ§·½³Ìʽ£º______________________________________¡£
(5)B³£ÓÃ×÷¹¤ÒµÖÆÄ³ËáµÄÔ­ÁÏ£¬Ð´³öÓÉBÖÆ¸ÃËá¹ý³ÌÖеĻ¯Ñ§·½³Ìʽ£º
________________________________________________________¡£

£¨11·Ö£©¡¡(1)N2£«3H22NH3        £¨2·Ö£©
(2)2Al£«2OH£­£«2H2O===2AlO2-£«3H2¡ü£¨2·Ö£©
(3)¼ÓÈëNaOHÈÜÒº£¬¼ÓÈÈ£¬ÓÃʪÈóµÄʯÈïÊÔÖ½¿¿½üÊԹܿڣ¬ÊÔÖ½±äÀ¶£¨2·Ö£©
(4)2NH4Cl£«Ca(OH)3CaCl2£«2NH3¡ü£«2H2O£¨2·Ö£©
(5)4NH3£«5O24NO£«6H2O£¬£¨Ã¿¸ö·½³Ìʽ1·Ö£©2NO£«O2===2NO2£¬3NO2£«H2O===2HNO3£«NO

½âÎöÊÔÌâ·ÖÎö£ºÆøÌåBÓëÆøÌåCÏàÓö²úÉú´óÁ¿µÄ°×ÑÌÉú³ÉE£¬Õâ˵Ã÷EÓ¦¸ÃÊÇÂÈ»¯ï§£¬BºÍCÓ¦¸ÃÊÇÂÈ»¯ÇâºÍ°±Æø¡£±ûºÍ¶¡µãȼÉú³ÉC£¬ËùÒÔCÊÇÂÈ»¯Ç⣬±ûÊÇÇâÆø¡¢¶¡ÊÇÂÈÆø£¬ÔòÒÒÊǵªÆø£¬BÊǰ±Æø¡£AÊÇÇ¿¼î£¬DÊǺ£Ë®ÖÐŨ¶È×î¸ßµÄÑΣ¬ÔòDÊÇÂÈ»¯ÄÆ£¬ËùÒÔA¾ÍÊÇÇâÑõ»¯ÄÆ¡£¼×ºÍÇâÑõ»¯ÄÆÈÜÒº·´Ó¦Éú³ÉÇâÆø£¬ËùÒÔ¼×Êǵ¥ÖÊAl¡£
¿¼µã£º¿¼²éÎÞ»ú¿òͼÌâµÄÓйØÅжÏ
µãÆÀ£º»¯Ñ§ÍƶÏÌâÊÇÒ»Àà×ÛºÏÐÔ½ÏÇ¿µÄÊÔÌ⣬ÈçÔªËØ¼°»¯ºÏÎïÐÔÖʺÍÉç»áÉú»î£¬»·¾³±£»¤£¬»¯Ñ§¼ÆËãµÈ֪ʶ£¬»¹¿ÉÒýÈëѧ¿Æ¼ä×ۺϡ£Ëü²»½ö¿É¿¼²ìѧÉú¶Ô»¯Ñ§ÖªÊ¶µÄÀí½â³Ì¶È£¬¸üÖØÒªµÄÊÇÅàÑøÑ§ÉúµÄ×ۺϷÖÎöÄÜÁ¦ºÍ˼ά·½·¨¡£½â¿òͼÌâµÄ·½·¨£º×î¹Ø¼üµÄÊÇѰÕÒ¡°Í»ÆÆ¿Ú¡±£¬¡°Í»ÆÆ¿Ú¡±¾ÍÊÇ×¥¡°ÌØ¡±×Ö£¬ÀýÈçÌØÊâÑÕÉ«¡¢ÌØÊâ״̬¡¢ÌØÊâÆøÎ¶¡¢ÌØÊâ·´Ó¦¡¢ÌØÊâÏÖÏó¡¢ÌØÊâÖÆ·¨¡¢ÌØÊâÓÃ;µÈ¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2012-2013ѧÄ긣½¨Ê¡¶«É½ÏصڶþÖÐѧ¸ß¶þµÚÒ»´ÎÔ¿¼»¯Ñ§ÊÔ¾í£¨´ø½âÎö£© ÌâÐÍ£ºÌî¿ÕÌâ

(11·Ö)ÒÀ¾ÝÑõ»¯»¹Ô­·´Ó¦£º2Ag+ (aq) + Cu(s) ="=" Cu2+ (aq) + 2Ag (s)Éè¼ÆµÄÔ­µç³ØÈçͼËùʾ¡£

Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©µç¼«XµÄ²ÄÁÏÊÇ             £»µç½âÖÊÈÜÒºYÊÇ                                   £»
£¨2£©Òøµç¼«Îªµç³ØµÄ             ¼«£»Òøµç¼«ÉÏ·¢ÉúµÄµç¼«·´Ó¦Ê½                      ¡£
£¨3£©Íâµç·Öеĵç×ÓÊÇ´Ó           µç¼«Á÷Ïò           µç¼«£¨Ìîµç¼«µÄ²ÄÁÏ£©¡£
£¨4£©ÀûÓ÷´Ó¦Zn £« 2 FeCl3 £½ ZnCl2 £« 2 FeCl2£¬Éè¼ÆÒ»¸öÔ­µç³Ø£¬ÔÚÏÂÃæ¿òͼÄÚ»­³öÓйصÄʵÑé×°ÖÃͼ£º

Õý¼«µÄµç¼«·´Ó¦Ê½Îª                                  ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2012-2013ѧÄêÁÉÄþ³¯ÑôÁø³Ç¸ß¼¶ÖÐѧ¸ßÈýÉÏѧÆÚµÚÈý´ÎÔ¿¼»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÌî¿ÕÌâ

(11·ÖÏÂÁпòͼËùʾµÄÎïÖÊת»¯¹ØÏµÖУ¬¼×ÊÇÈÕ³£Éú»îÖг£¼ûµÄ½ðÊô£¬ÒÒ¡¢±û¡¢¶¡Êdz£¼ûµÄÆøÌåµ¥ÖÊ¡£ÆøÌåBÓëÆøÌåCÏàÓö²úÉú´óÁ¿µÄ°×ÑÌÉú³ÉE£¬AÊÇÇ¿¼î£¬DÊǺ£Ë®ÖÐŨ¶È×î¸ßµÄÑÎ(²¿·Ö·´Ó¦ÎïºÍÉú³ÉÎï¼°Ë®ÒÑÂÔÈ¥)¡£

Çë»Ø´ðÏÂÁÐÎÊÌ⣺

(1)д³öÒÒÓë±ûÉú³ÉBµÄ»¯Ñ§·½³Ìʽ£º________________________________________¡£

(2)д³ö¼×ºÍAÈÜÒº·´Ó¦µÄÀë×Ó·½³Ìʽ£º_______________________________________¡£

(3)¼ìÑéEÖеÄÑôÀë×Óʱ£¬È¡ÉÙÁ¿EÓÚÊÔ¹ÜÖУ¬________________ÔòÖ¤Ã÷EÖÐÓиÃÑôÀë×Ó¡£

(4)д³öʵÑéÊÒÖÆÈ¡BµÄ»¯Ñ§·½³Ìʽ£º______________________________________¡£

(5)B³£ÓÃ×÷¹¤ÒµÖÆÄ³ËáµÄÔ­ÁÏ£¬Ð´³öÓÉBÖÆ¸ÃËá¹ý³ÌÖеĻ¯Ñ§·½³Ìʽ£º

________________________________________________________¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2014½ì¸£½¨Ê¡¸ß¶þµÚÒ»´ÎÔ¿¼»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÌî¿ÕÌâ

(11·Ö)ÒÀ¾ÝÑõ»¯»¹Ô­·´Ó¦£º2Ag+ (aq) + Cu(s) == Cu2+ (aq) + 2Ag (s)Éè¼ÆµÄÔ­µç³ØÈçͼËùʾ¡£

Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©µç¼«XµÄ²ÄÁÏÊÇ               £»µç½âÖÊÈÜÒºYÊÇ                                     £»

£¨2£©Òøµç¼«Îªµç³ØµÄ                ¼«£»Òøµç¼«ÉÏ·¢ÉúµÄµç¼«·´Ó¦Ê½                        ¡£

£¨3£©Íâµç·Öеĵç×ÓÊÇ´Ó             µç¼«Á÷Ïò             µç¼«£¨Ìîµç¼«µÄ²ÄÁÏ£©¡£

£¨4£©ÀûÓ÷´Ó¦Zn £« 2 FeCl3 £½ ZnCl2 £« 2 FeCl2£¬Éè¼ÆÒ»¸öÔ­µç³Ø£¬ÔÚÏÂÃæ¿òͼÄÚ»­³öÓйصÄʵÑé×°ÖÃͼ£º

Õý¼«µÄµç¼«·´Ó¦Ê½Îª                                   ¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸