ÏÂÁÐʵÑé²Ù×÷ÓëʵÑéÄ¿µÄ»ò½áÂÛÒ»ÖµÄÊÇ£¨¡¡¡¡£©
Ñ¡Ïî ʵÑé²Ù×÷ ʵÑéÄ¿µÄ»ò½áÂÛ
A ÏòÕáÌÇÈÜÒºÖмÓÈëÏ¡ÁòËᣬ¼ÓÈÈ£¬ÔÙ¼ÓÈëÉÙÁ¿Òø°±ÈÜÒº£¬¼ÓÈÈ£¬ÎÞÒø¾µÉú³É ˵Ã÷ÕáÌÇδ·¢ÉúË®½â
B ½«ÓÃÒÒËáºÍÒÒ´¼ÖÆÈ¡µÄÒÒËáÒÒõ¥´ÖÆ·¼ÓÈë±¥ºÍNa2CO3ÈÜÒºÖÐÏ´µÓ¡¢·ÖÒº µÃµ½½Ï´¿¾»µÄÒÒËáÒÒõ¥
C Ïòº¬Óб½·ÓµÄ±½ÖмÓÈëŨäåË®£¬Õñµ´£¬¹ýÂË ³ýÈ¥±½Öеı½·Ó
D Ïò»ìÓÐBr2µÄC2H5Br´ÖÆ·ÖмÓÈëNaHSO3±¥ºÍÈÜÒº£¬³ä·ÖÕñµ´ÔÙ·ÖÒº ³ýÈ¥ÆäÖеÄBr2
·ÖÎö£ºA£®ÕáÌÇÔÚËáÐÔÌõ¼þÏÂË®½âÉú³ÉÆÏÌÑÌÇ£¬ÆÏÌÑÌÇÓ¦ÔÚ¼îÐÔÌõ¼þÏ·¢ÉúÒø¾µ·´Ó¦£»
B£®ÒÒËáÒÒõ¥²»ÈÜÓÚ±¥ºÍ̼ËáÄÆÈÜÒº£¬±¥ºÍ̼ËáÄÆÈÜÒº¿ÉÎüÊÕÒÒËáºÍÒÒ´¼£»
C£®±½·ÓºÍäåË®·´Ó¦Éú³ÉÈýäå±½·Ó£¬Èýäå±½·ÓÈÜÓÚ±½ÖУ»
D£®Br2ºÍNaHSO3·¢ÉúÑõ»¯»¹Ô­·´Ó¦£®
½â´ð£º½â£ºA£®ÕáÌÇÔÚËáÐÔÌõ¼þÏÂË®½âÉú³ÉÆÏÌÑÌÇ£¬ÆÏÌÑÌÇÓ¦ÔÚ¼îÐÔÌõ¼þÏ·¢ÉúÒø¾µ·´Ó¦£¬ÌâÖÐË®½âºóûÓе÷½ÚÈÜÒºÖÁ¼îÐÔ£¬²»ÄÜ·¢ÉúÒø¾µ·´Ó¦£¬¹ÊA´íÎó£»
B£®ÒÒËáÒÒõ¥²»ÈÜÓÚ±¥ºÍ̼ËáÄÆÈÜÒº£¬±¥ºÍ̼ËáÄÆÈÜÒº¿ÉÎüÊÕÒÒËáºÍÒÒ´¼£¬¿ÉµÃµ½´¿¾»µÄÒÒËáÒÒõ¥£¬¹ÊBÕýÈ·£»
C£®±½·ÓºÍäåË®·´Ó¦Éú³ÉÈýäå±½·Ó£¬Èýäå±½·ÓÈÜÓÚ±½ÖУ¬²»Äܵõ½´¿¾»µÄ±½£¬Ó¦ÓÃÇâÑõ»¯ÄÆÈÜÒº³ýÔÓ£¬¹ÊC´íÎó£»
D£®Br2ºÍNaHSO3·¢ÉúÑõ»¯»¹Ô­·´Ó¦£¬¿ÉÓÃÓÚ³ýÈ¥ÆäÖеÄBr2£¬¹ÊDÕýÈ·£®
¹ÊÑ¡BD£®
µãÆÀ£º±¾Ì⿼²é½ÏΪ×ۺϣ¬Éæ¼°ÎïÖʵķÖÀë¡¢Ìá´¿¡¢¼ìÑéµÈÎÊÌ⣬²àÖØÓÚѧÉúµÄ·ÖÎöÄÜÁ¦¡¢ÊµÑéÄÜÁ¦ºÍÆÀ¼ÛÄÜÁ¦µÄ¿¼²é£¬Îª¸ß¿¼³£¼ûÌâÐͺ͸ßƵ¿¼µã£¬×¢Òâ³ýÔÓʱ²»ÄÜÒýÈëеÄÔÓÖÊ£¬°ÑÎÕÎïÖʵÄÐÔÖʵÄÒìͬΪ½â´ð¸ÃÌâµÄ¹Ø¼ü£¬ÄѶȲ»´ó£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â

ijͬѧÀûÓÃÏÂÁÐ×°ÖÃʵÏÖÍ­ÓëŨÏõËᡢϡÏõËá·´Ó¦£¬¹ý³ÌÈçͼËùʾ£º
I£® È¡Ò»¶ÎÍ­Ë¿£¬ÓÃÏ¡ÁòËá³ýȥͭÐâ[Ö÷Òª³É·ÖÊÇCu2£¨OH£©2CO3]£®
¢ò£®½«Ï´µÓºóµÄÍ­Ë¿×ö³ÉÔÑÊý½Ï¶àµÄÂÝÐý×´£®
¢ó£®°´ÈçͼËùʾװÖÃÁ¬½ÓÒÇÆ÷¡¢¼ì²éÆøÃÜÐÔ¡¢×°È뻯ѧÊÔ¼Á£®
£¨1£©¹ý³ÌI·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ
Cu2£¨OH£©2CO3+4H+¨T2Cu2++CO2¡ü+3H2O
Cu2£¨OH£©2CO3+4H+¨T2Cu2++CO2¡ü+3H2O
£®
£¨2£©Ð´³ö¹ý³Ì¢óÖмì²éÆøÃÜÐԵķ½·¨
´ò¿ªbºÍa£¬¹Ø±Õ·ÖҺ©¶·µÄ»îÈû£¬½«×ó±ßµ¼¹Ü²åÈëʢˮµÄÉÕ±­ÖУ¬ÇáÇáÀ­¶¯×¢ÉäÆ÷»îÈû£¬Èôµ¼¹ÜÖÐÒºÃæÉÏÉýÔò˵Ã÷ÆøÃÜÐÔºÃ
´ò¿ªbºÍa£¬¹Ø±Õ·ÖҺ©¶·µÄ»îÈû£¬½«×ó±ßµ¼¹Ü²åÈëʢˮµÄÉÕ±­ÖУ¬ÇáÇáÀ­¶¯×¢ÉäÆ÷»îÈû£¬Èôµ¼¹ÜÖÐÒºÃæÉÏÉýÔò˵Ã÷ÆøÃÜÐÔºÃ
£®
£¨3£©¹ý³Ì¢óµÄºóÐø²Ù×÷ÈçÏ£º
¢Ù´ò¿ªÖ¹Ë®¼ÐaºÍb£¬ÇáÍÆ×¢ÉäÆ÷£¬Ê¹Å¨ÏõËáÓëÍ­Ë¿½Ó´¥£¬¹Û²ìµ½µÄÏÖÏóÊÇ
²úÉúºì×ØÉ«ÆøÌå
²úÉúºì×ØÉ«ÆøÌå
£¬Ò»¶Îʱ¼äºóʹ·´Ó¦Í£Ö¹µÄ²Ù×÷ÊÇ
ÇáÇὫעÉäÆ÷»îÈûÏòÓÒÀ­Ê¹Í­Ë¿ºÍÈÜÒº·Ö¿ª
ÇáÇὫעÉäÆ÷»îÈûÏòÓÒÀ­Ê¹Í­Ë¿ºÍÈÜÒº·Ö¿ª
£¬¹Ø±Õa£¬È¡ÏÂ×¢ÉäÆ÷£®
¢Ú´ò¿ªbºÍ·ÖҺ©¶·»îÈû£¬µ±²£Á§¹Ü³äÂúÏ¡ÏõËáºó£¬¹Ø±ÕbºÍ·ÖҺ©¶·»îÈû£¬´ò¿ªa£¬¹Û²ìµ½ÓÐÆøÅݲúÉú£®Ï¡ÏõËá³äÂú²£Á§¹ÜµÄʵÑéÄ¿ÊÇ
½«²£Á§¹ÜÖеÄNO2ºÍ¿ÕÆøÅųö
½«²£Á§¹ÜÖеÄNO2ºÍ¿ÕÆøÅųö
£¬¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ
3Cu+8H++2NO3-=3Cu2++2NO¡ü+4H2O
3Cu+8H++2NO3-=3Cu2++2NO¡ü+4H2O
£®
£¨4£©ÁíÈ¡3֧ʢÂúNO2ÆøÌåµÄСÊԹֱܷðµ¹ÖÃÔÚÊ¢Óг£ÎÂË®¡¢ÈÈË®ºÍ±ùË®µÄ3Ö»ÉÕ±­ÖУ¬·¢ÏÖÒºÃæÉÏÉýµÄ¸ß¶ÈÃ÷ÏÔ²»Ò»Ö£®½á¹ûÈç±íËùʾ£¨ºöÂÔζȶÔÆøÌåÌå»ýµÄÓ°Ï죩£º
ʵÑé±àºÅ Ë®ÎÂ/¡æ ÒºÃæÉÏÉý¸ß¶È
1 25 ³¬¹ýÊԹܵÄ
2
3
2 50 ²»×ãÊԹܵÄ
2
3
3 0 ÒºÃæÉÏÉý³¬¹ýʵÑé1
¢Ù¸ù¾ÝÉϱíµÃ³öµÄ½áÂÛÊÇζÈÔ½
µÍ
µÍ
£¨Ìî¡°¸ß¡±»ò¡°µÍ¡±£©£¬½øÈëÊÔ¹ÜÖеÄÈÜÒºÔ½¶à£®
¢Ú²éÔÄ×ÊÁÏ£º
a£®NO2ÓëË®·´Ó¦µÄʵ¼Ê¹ý³ÌΪ£º2NO2+H2O=HNO3+HNO2   3HNO2=HNO3+2NO¡ü+H2O£»
b£®HNO2²»Îȶ¨£®
Ôò²úÉúÉÏÊöÏÖÏóµÄÔ­ÒòÊÇ
ζȵͣ¬HNO2·Ö½âÁ¿¼õÉÙ£¬·Ö½â²úÉúµÄNOÆøÌåÁ¿¼õÉÙ£¬ËùÒÔ½øÈëÊԹܵÄÈÜÒº¶à
ζȵͣ¬HNO2·Ö½âÁ¿¼õÉÙ£¬·Ö½â²úÉúµÄNOÆøÌåÁ¿¼õÉÙ£¬ËùÒÔ½øÈëÊԹܵÄÈÜÒº¶à
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¾«Ó¢¼Ò½ÌÍøË×»°Ëµ£¬¡°³Â¾ÆÀÏ´×ÌرðÏ㡱£¬ÆäÔ­ÒòÊǾÆÔÚ´¢´æ¹ý³ÌÖÐÉú³ÉÁËÓÐÏãζµÄÒÒËáÒÒõ¥£¬ÔÚʵÑéÊÒÀïÎÒÃÇÒ²¿ÉÒÔÓÃÈçͼËùʾµÄ×°ÖÃÀ´Ä£Äâ¸Ã¹ý³Ì£®Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©±¥ºÍ̼ËáÄÆÈÜÒºµÄÖ÷Òª×÷ÓÃÊÇ
 
£®
£¨2£©×°ÖÃÖÐͨÕôÆøµÄµ¼¹ÜÖ»Äܲ嵽±¥ºÍ̼ËáÄÆÈÜÒºµÄÒºÃæ´¦£¬²»ÄܲåÈëÈÜÒºÖУ¬Ä¿
 
£¬³¤µ¼¹ÜµÄ×÷ÓÃÊÇ
 
£®
£¨3£©ÈôÒª°ÑÖƵõÄÒÒËáÒÒõ¥·ÖÀë³öÀ´£¬Ó¦²ÉÓõÄʵÑé²Ù×÷ÊÇ
 
£®
£¨4£©½øÐиÃʵÑéʱ£¬×îºÃÏòÊԹܼ×ÖмÓÈ뼸¿éËé´ÉƬ£¬ÆäÄ¿µÄÊÇ
 
£®
£¨5£©ÊµÑéÊÒ¿ÉÓÃÒÒ´¼À´ÖÆÈ¡ÒÒÏ©£¬½«Éú³ÉµÄÒÒϩͨÈëäåµÄËÄÂÈ»¯Ì¼ÈÜÒº£¬·´Ó¦ºóÉú³ÉÎïµÄ½á¹¹¼òʽÊÇ
 
£®
£¨6£©Éú³ÉÒÒËáÒÒõ¥µÄ·´Ó¦ÊÇ¿ÉÄæ·´Ó¦£¬·´Ó¦Îï²»ÄÜÍêȫת»¯ÎªÉú³ÉÎ·´Ó¦Ò»¶Îʱ¼äºó£¬¾Í´ïµ½Á˸÷´Ó¦µÄÏ޶ȣ¬¼´´ïµ½»¯Ñ§Æ½ºâ״̬£®ÏÂÁÐÃèÊöÄÜ˵Ã÷¸Ã·´Ó¦ÒÑ´ïµ½»¯Ñ§Æ½ºâ״̬µÄÊÇ£¨ÌîÐòºÅ£©
 
£®
¢Ùµ¥Î»Ê±¼äÀÉú³É1molÒÒËáÒÒõ¥£¬Í¬Ê±Éú³É1molË®
¢Úµ¥Î»Ê±¼äÀÉú³É1molÒÒËáÒÒõ¥£¬Í¬Ê±Éú³É1molÒÒËá
¢Ûµ¥Î»Ê±¼äÀÏûºÄ1molÒÒ´¼£¬Í¬Ê±ÏûºÄ1molÒÒËá
¢ÜÕý·´Ó¦µÄËÙÂÊÓëÄæ·´Ó¦µÄËÙÂÊÏàµÈ£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2013½ìÕã½­Ê¡º¼ÖÝÊиßÈýÉÏѧÆÚÆÚÖÐÆßУÁª¿¼»¯Ñ§ÊÔ¾í£¨´ø½âÎö£© ÌâÐÍ£ºÊµÑéÌâ

(16·Ö)ijͬѧÀûÓÃÏÂÁÐ×°ÖÃʵÏÖÍ­ÓëŨÏõËᡢϡÏõËá·´Ó¦£¬¹ý³ÌÈçÏ£º
I£®È¡Ò»¶ÎÍ­Ë¿£¬ÓÃÏ¡ÁòËá³ýȥͭÐâ[Ö÷Òª³É·ÖÊÇCu2(OH)2CO3]¡£
¢ò£®½«Ï´µÓºóµÄÍ­Ë¿×ö³ÉÔÑÊý½Ï¶àµÄÂÝÐý×´¡£
III£®°´ÈçͼËùʾװÖÃÁ¬½ÓÒÇÆ÷¡¢¼ì²éÆøÃÜÐÔ¡¢×°È뻯ѧÊÔ¼Á¡£

(1)¹ý³ÌI·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ                      ¡£
(2)д³ö¹ý³ÌIIIÖмì²éÆøÃÜÐԵķ½·¨                     ¡£
¢Ç¹ý³ÌIIIµÄºóÐø²Ù×÷ÈçÏ£º
¢Ù´ò¿ªÖ¹Ë®¼ÐaºÍb£¬ÇáÍÆ×¢ÉäÆ÷£¬Ê¹Å¨ÏõËáÓëÍ­Ë¿½Ó´¥£¬¹Û²ìµ½µÄÏÖÏóÊÇ                    £¬Ò»¶Îʱ¼äºóʹ·´Ó¦Í£Ö¹µÄ²Ù×÷ÊÇ                              £¬¹Ø±Õa£¬È¡ÏÂ×¢ÉäÆ÷¡£
¢Ú´ò¿ªbºÍ·ÖҺ©¶·»îÈû£¬µ±²£Á§¹Ü³äÂúÏ¡ÏõËáºó£¬¹Ø±ÕbºÍ·ÖҺ©¶·»îÈû£¬´ò¿ªa£¬¹Û²ìµ½ÓÐÆøÅݲúÉú¡£Ï¡ÏõËá³äÂú²£Á§¹ÜµÄʵÑéÄ¿ÊÇ                           £¬¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ                         ¡£
(4)ÁíÈ¡3֧ʢÂúNO2ÆøÌåµÄСÊԹֱܷðµ¹ÖÃÔÚÊ¢Óг£ÎÂË®¡¢ÈÈË®ºÍ±ùË®µÄ3Ö»ÉÕ±­ÖУ¬·¢ÏÖÒºÃæÉÏÉýµÄ¸ß¶ÈÃ÷ÏÔ²»Ò»Ö¡£½á¹ûÈçϱíËùʾ(ºöÂÔζȶÔÆøÌåÌå»ýµÄÓ°Ïì)£º

¢Ù¸ù¾ÝÉϱíµÃ³öµÄ½áÂÛÊÇζÈÔ½             (Ìî¡°¸ß¡±»ò¡°µÍ¡±)£¬½øÈëÊÔ¹ÜÖеÄÈÜÒºÔ½¶à¡£
¢Ú²éÔÄ×ÊÁÏ£º
a£®NO2ÓëË®·´Ó¦µÄʵ¼Ê¹ý³ÌΪ£º2NO2+H2O=HNO3+HNO2  3HNO2 =HNO3+2NO­+H2O£»
b£®HNO2²»Îȶ¨¡£
Ôò²úÉúÉÏÊöÏÖÏóµÄÔ­ÒòÊÇ                                                            ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2012-2013ѧÄêÕã½­Ê¡º¼ÖÝÊиßÈýÉÏѧÆÚÆÚÖÐÆßУÁª¿¼»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÊµÑéÌâ

(16·Ö)ijͬѧÀûÓÃÏÂÁÐ×°ÖÃʵÏÖÍ­ÓëŨÏõËᡢϡÏõËá·´Ó¦£¬¹ý³ÌÈçÏ£º

I£® È¡Ò»¶ÎÍ­Ë¿£¬ÓÃÏ¡ÁòËá³ýȥͭÐâ[Ö÷Òª³É·ÖÊÇCu2(OH)2CO3]¡£

¢ò£®½«Ï´µÓºóµÄÍ­Ë¿×ö³ÉÔÑÊý½Ï¶àµÄÂÝÐý×´¡£

III£®°´ÈçͼËùʾװÖÃÁ¬½ÓÒÇÆ÷¡¢¼ì²éÆøÃÜÐÔ¡¢×°È뻯ѧÊÔ¼Á¡£

(1)¹ý³ÌI·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ                       ¡£

(2)д³ö¹ý³ÌIIIÖмì²éÆøÃÜÐԵķ½·¨                      ¡£

¢Ç¹ý³ÌIIIµÄºóÐø²Ù×÷ÈçÏ£º

¢Ù´ò¿ªÖ¹Ë®¼ÐaºÍb£¬ÇáÍÆ×¢ÉäÆ÷£¬Ê¹Å¨ÏõËáÓëÍ­Ë¿½Ó´¥£¬¹Û²ìµ½µÄÏÖÏóÊÇ                     £¬Ò»¶Îʱ¼äºóʹ·´Ó¦Í£Ö¹µÄ²Ù×÷ÊÇ                               £¬¹Ø±Õa£¬È¡ÏÂ×¢ÉäÆ÷¡£

¢Ú´ò¿ªbºÍ·ÖҺ©¶·»îÈû£¬µ±²£Á§¹Ü³äÂúÏ¡ÏõËáºó£¬¹Ø±ÕbºÍ·ÖҺ©¶·»îÈû£¬´ò¿ªa£¬¹Û²ìµ½ÓÐÆøÅݲúÉú¡£Ï¡ÏõËá³äÂú²£Á§¹ÜµÄʵÑéÄ¿ÊÇ                            £¬¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ                          ¡£

(4)ÁíÈ¡3֧ʢÂúNO2ÆøÌåµÄСÊԹֱܷðµ¹ÖÃÔÚÊ¢Óг£ÎÂË®¡¢ÈÈË®ºÍ±ùË®µÄ3Ö»ÉÕ±­ÖУ¬·¢ÏÖÒºÃæÉÏÉýµÄ¸ß¶ÈÃ÷ÏÔ²»Ò»Ö¡£½á¹ûÈçϱíËùʾ(ºöÂÔζȶÔÆøÌåÌå»ýµÄÓ°Ïì)£º

¢Ù¸ù¾ÝÉϱíµÃ³öµÄ½áÂÛÊÇζÈÔ½              (Ìî¡°¸ß¡±»ò¡°µÍ¡±)£¬½øÈëÊÔ¹ÜÖеÄÈÜÒºÔ½¶à¡£

¢Ú²éÔÄ×ÊÁÏ£º

a£®NO2ÓëË®·´Ó¦µÄʵ¼Ê¹ý³ÌΪ£º2NO2+H2O=HNO3+HNO2   3HNO2 =HNO3+2NO­+H2O£»

b£®HNO2²»Îȶ¨¡£

Ôò²úÉúÉÏÊöÏÖÏóµÄÔ­ÒòÊÇ                                                             ¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ijͬѧÀûÓÃÏÂÁÐ×°ÖÃʵÏÖÍ­ÓëŨÏõËᡢϡÏõËá·´Ó¦£¬¹ý³ÌÈçÏ£º

I£® È¡Ò»¶ÎÍ­Ë¿£¬ÓÃÏ¡ÁòËá³ýȥͭÐâ[Ö÷Òª³É·ÖÊÇCu2(OH)2CO3]¡£

¢ò£®½«Ï´µÓºóµÄÍ­Ë¿×ö³ÉÔÑÊý½Ï¶àµÄÂÝÐý×´¡£

III£®°´ÈçͼËùʾװÖÃÁ¬½ÓÒÇÆ÷¡¢¼ì²éÆøÃÜÐÔ¡¢×°È뻯ѧÊÔ¼Á¡£

(1)¹ý³ÌI·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ                       ¡£

(2)д³ö¹ý³ÌIIIÖмì²éÆøÃÜÐԵķ½·¨                      ¡£

¢Ç¹ý³ÌIIIµÄºóÐø²Ù×÷ÈçÏ£º

¢Ù´ò¿ªÖ¹Ë®¼ÐaºÍb£¬ÇáÍÆ×¢ÉäÆ÷£¬Ê¹Å¨ÏõËáÓëÍ­Ë¿½Ó´¥£¬¹Û²ìµ½µÄÏÖÏóÊÇ          £¬Ò»¶Îʱ¼äºóʹ·´Ó¦Í£Ö¹µÄ²Ù×÷ÊÇ                               £¬¹Ø±Õa£¬È¡ÏÂ×¢ÉäÆ÷¡£

¢Ú´ò¿ªbºÍ·ÖҺ©¶·»îÈû£¬µ±²£Á§¹Ü³äÂúÏ¡ÏõËáºó£¬¹Ø±ÕbºÍ·ÖҺ©¶·»îÈû£¬´ò¿ªa£¬¹Û²ìµ½ÓÐÆøÅݲúÉú¡£Ï¡ÏõËá³äÂú²£Á§¹ÜµÄʵÑéÄ¿ÊÇ               £¬¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ                     ¡£

(4)ÁíÈ¡3֧ʢÂúNO2ÆøÌåµÄСÊԹֱܷðµ¹ÖÃÔÚÊ¢Óг£ÎÂË®¡¢ÈÈË®ºÍ±ùË®µÄ3Ö»ÉÕ±­ÖУ¬·¢ÏÖÒºÃæÉÏÉýµÄ¸ß¶ÈÃ÷ÏÔ²»Ò»Ö¡£½á¹ûÈçϱíËùʾ(ºöÂÔζȶÔÆøÌåÌå»ýµÄÓ°Ïì)£ºks5u

¢Ù¸ù¾ÝÉϱíµÃ³öµÄ½áÂÛÊÇζÈÔ½              (Ìî¡°¸ß¡±»ò¡°µÍ¡±)£¬½øÈëÊÔ¹ÜÖеÄÈÜÒºÔ½¶à¡£

¢Ú²éÔÄ×ÊÁÏ£ºa£®NO2ÓëË®·´Ó¦µÄʵ¼Ê¹ý³ÌΪ£º2NO2+H2O=HNO3+HNO2   

3HNO2 =HNO3+2NO­+H2O£»   b£®HNO2²»Îȶ¨¡£

Ôò²úÉúÉÏÊöÏÖÏóµÄÔ­ÒòÊÇ                                              ¡£

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸