ÌâÄ¿ÄÚÈÝ

Q¡¢W¡¢X¡¢Y¡¢ZÎåÖÖ¶ÌÖÜÆÚÔªËØ£¬Ô­×ÓÐòÊýÒÀ´ÎÔö´ó¡£QÔÚÔªËØÖÜÆÚ±íÖÐÔ­×Ӱ뾶×îС£¬WÔªËØ×î¸ßÕý¼ÛÓë×îµÍ¸º¼Û´úÊýºÍΪ0£»YÓëQͬÖ÷×壻X¡¢Z·Ö±ðÊǵؿÇÖк¬Á¿×î¸ßµÄ·Ç½ðÊôÔªËغͽðÊôÔªËØ¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©X¡¢Z¼òµ¥Àë×Ӱ뾶½Ï´óÊÇ               £¨ÓÃÀë×Ó·ûºÅ±íʾ£©¡£
£¨2£©ÓÉÕâÎåÖÖÔªËØÖеÄÈô¸ÉÖÖ×é³ÉµÄ»¯ºÏÎï¼×¡¢ÒÒ¡¢±û¡¢¶¡ÔÚË®ÈÜÒºÖÐÓÐÈçÏÂת»¯¹Øϵ£º£¬ÆäÖбûÊÇÈÜÓÚË®ÏÔËáÐÔµÄÆøÌ壬¶¡ÊÇÇ¿¼î¡£
¢ÙÈôÒÒ³£×÷Ϊ±ºÖƸâµãµÄ·¢½Í·Û£¬ÔòÒÒº¬ÓеĻ¯Ñ§¼üÀàÐÍÓР                    £»ÒÒÈÜÒºÖÐÑôÀë×ÓŨ¶ÈÖ®ºÍ       ÒõÀë×ÓŨ¶ÈÖ®ºÍ£¨Ìî¡°£¾¡±¡¢¡°£½¡±»ò¡°£¼¡±£©¡£
¢ÚÈôÒÒÊÇÄÑÈÜÎ¼×ÈÜÒºÓë¹ýÁ¿µÄ±ûÉú³ÉÒÒµÄÀë×Ó·½³ÌʽΪ£º                                ¡£

£¨1£©O2£­£¨2·Ö£©
£¨2£©¢ÙÀë×Ó¼ü¡¢¹²¼Û¼ü£¨2·Ö£© £¾£¨2·Ö£©
¢Ú[Al(OH)4]£­+CO2£½Al(OH)3¡ý+ HCO3£­£¨3·Ö£©

½âÎöÊÔÌâ·ÖÎö£ºQÔÚÔªËØÖÜÆÚ±íÖÐÔ­×Ӱ뾶×îС£¬QΪHÔªËØ£¬X¡¢Z·Ö±ðÊǵؿÇÖк¬Á¿×î¸ßµÄ·Ç½ðÊôÔªËغͽðÊôÔªËØ£¬XΪOÔªËØ¡¢ZΪAlÔªËØ£¬WÔªËØ×î¸ßÕý¼ÛÓë×îµÍ¸º¼Û´úÊýºÍΪ0£¬WΪ¢ôA×åÔªËØ£¬¸ù¾ÝÔ­×ÓÐòÊýÒÀ´ÎÔö´ó£¬ËùÒÔWΪCÔªËØ£¬YÓëQͬÖ÷×壬ÔòYΪNaÔªËØ¡£
£¨1£©µç×ÓÅŲ¼ÏàͬµÄÀë×Ó£¬Ô­×ÓÐòÊýÔ½´ó£¬Àë×Ӱ뾶ԽС£¬ËùÒÔO2?°ë¾¶´óÓÚAl3+°ë¾¶¡£
£¨2£©±ûÊÇÈÜÓÚË®ÏÔËáÐÔµÄÆøÌ壬¶¡ÊÇÇ¿¼î£¬Ôò±ûΪCO2,¡¢¶¡ÎªNaOH¡£
¢ÙÈôÒÒ³£×÷Ϊ±ºÖƸâµãµÄ·¢½Í·Û£¬ÔòÒÒΪNaHCO3£¬º¬ÓеĻ¯Ñ§¼üÓУºÀë×Ó¼ü¡¢¹²¼Û¼ü£»ÑôÀë×ÓΪH+¡¢Na+£¬ÒõÀë×ÓΪHCO3?¡¢OH?¡¢CO32?£¬¸ù¾ÝµçºÉÊغã¿ÉµÃ£ºc£¨H+£©+c£¨Na+£©=c£¨HCO3?£©+c£¨OH?£©+2c£¨CO32?£©£¬ËùÒÔÑôÀë×ÓŨ¶ÈÖ®ºÍ>ÒõÀë×ÓŨ¶ÈÖ®ºÍ¡£
¢ÚÈôÒÒÊÇÄÑÈÜÎÔòÒÒΪAl(OH)3£¬¼×ÈÜÒºÓë¹ýÁ¿µÄ±ûÉú³ÉÒҵķ´Ó¦ÎªNa[Al(OH)4]ÓëCO2·´Ó¦Éú³ÉAl(OH)3³ÁµíºÍNaHCO3£¬Àë×Ó·½³ÌʽΪ£º[Al(OH)4]£­+CO2£½Al(OH)3¡ý+ HCO3£­¡£
¿¼µã£º±¾Ì⿼²éÔªËØÓëÎïÖʵÄÍƶϡ¢Àë×Ӱ뾶±È½Ï¡¢»¯Ñ§¼üµÄÅжϡ¢Àë×ÓŨ¶È±È½Ï¡¢Àë×Ó·½³ÌʽµÄÊéд¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

ÏÂÃæÊÇÔªËØÖÜÆÚ±íµÄÒ»²¿·Ö£¬²ÎÕÕÔªËآ٣­¢àÔÚ±íÖеÄλÖã¬ÇëÓû¯Ñ§ÓÃÓï»Ø´ðÏÂÁÐÎÊÌ⣺

×å
ÖÜÆÚ
IA
 
0
1
¢Ù
¢òA
¢óA
¢ôA
¢õA
¢öA
¢÷A
 
2
 
 
 
¢Ú
¢Û
¢Ü
 
 
3
¢Ý
 
¢Þ
¢ß
 
 
¢à
 
 
£¨1£©¢Ü¡¢¢Ý¡¢¢ÞµÄÔ­×Ӱ뾶ÓÉ´óµ½Ð¡µÄ˳ÐòΪ(ÔªËØ·ûºÅ)________________________¡£
£¨2£©¢Ú¡¢¢Û¡¢¢ßµÄ×î¸ß¼Ûº¬ÑõËáµÄËáÐÔÓÉÇ¿µ½ÈõµÄ˳ÐòÊÇ(Ìѧʽ)________     ¡£
£¨3£©¢Ù¡¢¢Ü¡¢¢Ý¡¢¢àÖеÄijЩԪËØ¿ÉÐγɼȺ¬Àë×Ó¼üÓÖº¬¼«ÐÔ¹²¼Û¼üµÄ»¯ºÏÎд³öÆäÖÐÒ»ÖÖ»¯ºÏÎïµÄ»¯Ñ§Ê½£º_______________¡£
£¨4£©Óɢں͢Ü×é³ÉµÄ»¯ºÏÎïÓë¢ÝµÄͬÖÜÆÚÏàÁÚÖ÷×åÔªËصĵ¥ÖÊ·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ:_______¡£
£¨5£©¢Þµ¥ÖÊÓë¢ÝµÄ×î¸ß¼ÛÑõ»¯ÎïµÄË®»¯Îï·´Ó¦µÄÀë×Ó·½³ÌʽΪ                    ¡£
£¨6£©ÈôÓâ٢Ú×é³É×î¼òµ¥µÄÓлúÎï×÷ΪȼÁϵç³ØµÄÔ­ÁÏ£¬Çëд³öÔÚ¼îÐÔ½éÖÊÖÐȼÁϵç³Ø¸º¼«µÄµç¼«·´Ó¦Ê½:                                                         ¡£
£¨7£©È¼Ãº·ÏÆøÖеĺ¬ÓеªÑõ»¯ÎNOx£©¡¢¶þÑõ»¯Ì¼µÈÆøÌ壬³£ÓÃÏÂÁз½·¨¶Ôȼú·ÏÆø½øÐÐÍÑÏõ´¦Àíʱ£¬³£ÀûÓü×Íé´ß»¯»¹Ô­µªÑõ»¯Îï¡£
È磺CH4(g)£«4NO2(g)=4NO(g)£«CO2(g)£«2H2O(g) £¬ ¡÷H=£­574 kJ¡¤mol£­1
CH4(g)£«4NO(g)=2N2(g)£«CO2(g)£«2H2O(g) £¬ ¡÷H=£­1160 kJ¡¤mol£­1
ÔòCH4(g)½«NO2(g)»¹Ô­ÎªN2(g)µÈµÄÈÈ»¯Ñ§·½³ÌʽΪ                             ¡£

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø