(9·Ö)ÏÂÁÐÎïÖÊÖ®¼äÄܹ»·¢ÉúÈçͼËùʾµÄ»¯Ñ§·´Ó¦£¬ºÏ½ðÓÉÁ½ÖÖ½ðÊô×é³É£¬È¡CÈÜÒº½øÐÐÑæÉ«·´Ó¦Ôò»ðÑæ³Ê»ÆÉ«¡£ÔÚ·´Ó¦ÖвúÉúµÄË®¾ùδÔÚͼÖбê³ö¡£ÆäÖÐG¡úHµÄ»¯Ñ§·½³ÌʽΪ£º4Fe(OH)2+O2+2H2O=4Fe(OH)3

£¨1£©Ð´³öÏÂÁÐÎïÖʵĻ¯Ñ§Ê½£ºA£º        D£º            F£º          
£¨2£©±ê³öÏÂÁз´Ó¦µÄµç×ÓתÒÆÇé¿ö£º4Fe(OH)2  +   O2   +   2H2O  ===  4Fe(OH)3
£¨3£©Ð´³öÏÂÁз´Ó¦µÄÀë×Ó·½³Ìʽ£º
A¡úB£º                                                            ¡¢
K¡úD£º                                                            ¡£

£¨1£©A£ºNa  D£ºNaAlO2  F£ºFeCl2

£¨3£©A¡úB£º 2Na+2H2O=2Na++2OH-+H2¡ü
K¡úD£º Al(OH)3+OH-=AlO2-+2H2O

½âÎöÊÔÌâ·ÖÎö£º¸ù¾ÝÌâÒâ¡°G¡úHµÄ»¯Ñ§·½³ÌʽΪ£º4Fe(OH)2+O2+2H2O=4Fe(OH)3¡±¿ÉÍÆÖª£¬ºÏ½ðº¬Ìú¡£FΪÂÈ»¯ÑÇÌú£¬ÓÉ¡°CÈÜÒº½øÐÐÑæÉ«·´Ó¦Ôò»ðÑæ³Ê»ÆÉ«¡±¿ÉÍÆÖª£ºAΪÄÆ£¬ÓÉת»¯Í¼D®K®L®D®Kת»¯¹ý³Ì£¬¿ÉÍÆÖª:DΪƫÂÁËáÄÆ£¬KΪÇâÑõ»¯ÂÁ£¬LΪÂÈ»¯ÂÁ£¬
£¨1£©A£ºNa  D£ºNaAlO2  F£ºFeCl2
£¨2£©Fe»¯ºÏ¼ÛÓÉ+2ÉýÖÁ+3£¬OÓÉ0¼Û½µÖÁ-2¼Û,תÒƵĵç×ÓÊýΪ4e£¬

£¨3£©A¡úBΪÄÆÓëË®·´Ó¦£º 2Na+2H2O=2Na++2OH-+H2¡ü
K¡úDÇâÑõ»¯ÂÁÓëÇâÑõ»¯ÄÆ·´Ó¦£º Al(OH)3+OH-=AlO2-+2H2O
¿¼µã£ºÄÆ¡¢ÂÁ¡¢Ìú½ðÊôµÄ»¯Ñ§ÐÔÖÊ
µãÆÀ£ºÊìÁ·ÕÆÎÕÏà¹ØÎïÖʵĻ¯Ñ§ÐÔÖʺÍÌØÕ÷·´Ó¦ÊÇ×öºÃÍƶÏÌâµÄ¹Ø¼ü¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â

ÏÂͼΪijЩ³£¼ûÎïÖÊÖ®¼äµÄת»¯¹Øϵ£®ÒÑÖª£ºA¡¢B¡¢IÖк¬ÓÐÏàͬµÄÑôÀë×ÓÇÒ¶¼ÊÇXY2ÐÍ»¯ºÏÎÇÒIÊÇʵÑéÊÒ³£ÓõĸÉÔï¼Á£»CΪֱÏßÐÍ·Ö×Ó£»E¡¢FΪ·Ç½ðÊôÆøÌåµ¥ÖÊ£®
Çë°´ÒªÇóÌî¿Õ£º

£¨1£©¢ÙBµÄµç×ÓʽÊÇ
£¬¢ÚKµÄ½á¹¹Ê½ÊÇ
H-O-Cl
H-O-Cl
£»
£¨2£©DÓëG·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ
2NH4Cl+Ca£¨OH£©2
  ¡÷  
.
 
CaCl2+2NH3¡ü+2H2O
2NH4Cl+Ca£¨OH£©2
  ¡÷  
.
 
CaCl2+2NH3¡ü+2H2O
£»
£¨3£©ÒÑÖªCµÄȼÉÕÈÈÊÇ1300kJ/mol£¬±íʾCµÄȼÉÕÈȵÄÈÈ»¯Ñ§·½³ÌʽÊÇ
C2H2£¨g£©+
5
2
O2£¨g£©=2CO2£¨g£©+H2O£¨l£©¡÷H=-1300kJ/mol
C2H2£¨g£©+
5
2
O2£¨g£©=2CO2£¨g£©+H2O£¨l£©¡÷H=-1300kJ/mol
£»
£¨4£©½«GÈÜÓÚË®Åä³ÉÈÜÒº£¬¼òÊö¼ìÑé¸ÃÈÜÒºGÖÐËùº¬ÑôÀë×ӵIJÙ×÷·½·¨£º
È¡ÉÙÁ¿µÄGÈÜÒºÓÚÊÔ¹ÜÖУ¬¼ÓÈëÊÊÁ¿µÄÇâÑõ»¯ÄÆÈÜÒº£¬¼ÓÈÈÊԹܣ¬½«ÊªÈóµÄºìɫʯÈïÊÔÖ½·ÅÔÚÊԹܿڣ¬ÈôÊÔÖ½±ä³ÉÀ¶É«£¬Ôòº¬ÓÐNH4+£¬£¨ÆäËûºÏÀí´ð°¸¾ù¸ø·Ö£©
È¡ÉÙÁ¿µÄGÈÜÒºÓÚÊÔ¹ÜÖУ¬¼ÓÈëÊÊÁ¿µÄÇâÑõ»¯ÄÆÈÜÒº£¬¼ÓÈÈÊԹܣ¬½«ÊªÈóµÄºìɫʯÈïÊÔÖ½·ÅÔÚÊԹܿڣ¬ÈôÊÔÖ½±ä³ÉÀ¶É«£¬Ôòº¬ÓÐNH4+£¬£¨ÆäËûºÏÀí´ð°¸¾ù¸ø·Ö£©
£»
£¨5£©³£ÎÂÏÂ0.1mol/LµÄJÈÜÒºÖÐc£¨H+£©/c£¨OH-£©=1¡Á10-8£¬ÏÂÁÐÐðÊö´íÎóµÄÊÇ
BCE
BCE
£»
A£®¸ÃÈÜÒºµÄpH=11£»
B£®¸ÃÈÜÒºÖеÄÈÜÖʵçÀë³öµÄÑôÀë×ÓŨ¶È0.1mol/L
C£®¸ÃÈÜÒºÖÐË®µçÀë³öµÄc£¨H+£©Óëc£¨OH-£©³Ë»ýΪ1¡Á10-22
D£®pH=3µÄÑÎËáÈÜÒºV1 LÓë¸Ã0.1mol/LµÄJÈÜÒºV2 L»ìºÏ£¬Èô»ìºÏÈÜÒºpH=7£¬Ôò£ºV1£¾V2
E£®½«ÒÔÉÏÈÜÒº¼ÓˮϡÊÍ100±¶ºó£¬pHֵΪ9£»
£¨6£©µ¥ÖÊFÔÚ¹¤ÒµÉÏÓÐÖØÒªµÄÓÃ;ÊÇ
ÖÆƯ°×·Û£¬ÖÆÑÎËᣨÖÆƯ°×ÒºµÈ£©
ÖÆƯ°×·Û£¬ÖÆÑÎËᣨÖÆƯ°×ÒºµÈ£©
£®£¨ÖÁÉÙÁ½ÖÖ£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2012½ì¸ßÈýÉÏѧÆÚ»¯Ñ§Ò»ÂÖ¸´Ï°¡¶´Ó¿óÎïµ½»ù´¡²ÄÁÏ¡·×¨Ìâ×ۺϲâÊÔ£¨Ëս̰棩 ÌâÐÍ£ºÌî¿ÕÌâ

(9·Ö) A¡«I·Ö±ð±íʾÖÐѧ»¯Ñ§Öг£¼ûµÄÒ»ÖÖÎïÖÊ£¬ËüÃÇÖ®¼äÏ໥¹ØϵÈçÏÂͼËùʾ(²¿·Ö·´Ó¦Îï¡¢Éú³ÉÎïûÓÐÁгö)¡£ÒÑÖªHΪÖ÷×åÔªËصĹÌ̬Ñõ»¯ÎFÊǺìºÖÉ«ÄÑÈÜÓÚË®µÄ³Áµí£¬ÇÒA¡¢B¡¢C¡¢D¡¢E¡¢FÁùÖÖÎïÖÊÖоùº¬Í¬Ò»ÖÖÔªËØ¡£

ÇëÌîдÏÂÁпհףº
(1)A¡¢B¡¢C¡¢D¡¢E¡¢FÁùÖÖÎïÖÊÖÐËùº¬Í¬Ò»ÖÖÔªËØÔÚÔªËØÖÜÆÚ±íÖÐλÖÃ________¡£
(2)д³öC¡¢HÎïÖʵĻ¯Ñ§Ê½£ºC________£¬H________¡£
(3)д³ö·´Ó¦¢Ù¢ßµÄ»¯Ñ§·½³Ìʽ£º
·´Ó¦¢Ù£º______________________________________________________________¡£
·´Ó¦¢ß£º______________________________________________________________¡£
(4)·´Ó¦¢Þ¹ý³ÌÖеÄÏÖÏóÊÇ______________________________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2015½ì¹ã¶«Ê¡¸ßÒ»12ÔÂÔ¿¼»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÍƶÏÌâ

 (9·Ö)ÏÂÁÐÎïÖÊÖ®¼äÄܹ»·¢ÉúÈçͼËùʾµÄ»¯Ñ§·´Ó¦£¬ºÏ½ðÓÉÁ½ÖÖ½ðÊô×é³É£¬È¡CÈÜÒº½øÐÐÑæÉ«·´Ó¦Ôò»ðÑæ³Ê»ÆÉ«¡£ÔÚ·´Ó¦ÖвúÉúµÄË®¾ùδÔÚͼÖбê³ö¡£ÆäÖÐG¡úHµÄ»¯Ñ§·½³ÌʽΪ£º4Fe(OH)2+O2+2H2O=4Fe(OH)3

£¨1£©Ð´³öÏÂÁÐÎïÖʵĻ¯Ñ§Ê½£ºA£º        D£º            F£º          

£¨2£©±ê³öÏÂÁз´Ó¦µÄµç×ÓתÒÆÇé¿ö£º4Fe(OH)2  +   O2   +   2H2O  ===  4Fe(OH)3

£¨3£©Ð´³öÏÂÁз´Ó¦µÄÀë×Ó·½³Ìʽ£º

A¡úB£º                                                            ¡¢

K¡úD£º                                                            ¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º¹ãÎ÷Ê¡¹ðÁÖÖÐѧ2010-2011ѧÄê¸ßÈýÄ꼶8ÔÂÔ¿¼»¯Ñ§ ÌâÐÍ£ºµ¥Ñ¡Ìâ

 [s1] ÏÂÁÐÎïÖÊÖ®¼äµÄÏ໥¹Øϵ²»ÕýÈ·µÄÊÇ            £¨    £©

    A£®CH3¡ªCH2¡ªNO2ºÍH2N¡ªCH2¡ªCOOH»¥ÎªÍ¬·ÖÒì¹¹Ìå

    B£®O2ºÍO3»¥ÎªÍ¬ËØÒìÐÎÌå

    C£®H¡¢D¡¢T»¥ÎªÍ¬Î»ËØ

    D£®¸É±ùºÍ±ùΪͬһÖÖÎïÖÊ

 


 [s1]9£®

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸