ÌâÄ¿ÄÚÈÝ

¢ñ£®´×ËáÊÇÎÒÃǺÜÊìϤµÄÒ»ÖÖÓлúÎ
£¨1£©Éú»îÖпÉʹÓô×Ëá³ýůƿÖеÄË®¹¸£¬ÓÉ´Ë˵Ã÷´×Ëá¾ßÓеÄÐÔÖÊÊÇ
 
£®
A£®»Ó·¢ÐÔ     B£®ËáÐÔ±È̼ËáÇ¿ C£®ÄѵçÀë     D£®Ò×ÈÜÓÚË®
£¨2£©Ä³Í¬Ñ§ÎªÁ˽â´×ËáÈÜÒºµÄµ¼µçÐÔ½øÐÐÁËÏàӦʵÑ飬ÈçͼΪ±ù´×ËᣨÎÞË®´×Ëá¾§Ì壩ÔÚÏ¡Ê͹ý³ÌÖÐÈÜÒºµÄµ¼µçÐԱ仯¹ØÏµÍ¼?£®
Çë»Ø´ð£º
¢ÙÏ¡Ê͹ý³ÌÖУ¬a¡¢b¡¢cÈý´¦ÈÜÒºÖУ¬ÓÉË®µçÀë³öµÄc£¨H+£©ÓÉСµ½´ó˳ÐòÊÇ
 
£®£¨Ìîa¡¢b¡¢c£©
¢Ú´Óbµãµ½cµã£¬ÈÜÒºÖÐ
c(CH3COO-)
c(CH3COOH)
µÄ±ÈÖµ
 
£¨Ìî¡°Ôö´ó¡±¡°¼õС¡±»ò¡°²»±ä¡±£©£®
¢ò£®ÊÒÎÂÏ£¬½«µÈÌå»ýCH3COOHÈÜÒººÍNaOHÈÜÒº»ìºÏ£®
£¨3£©Èô»ìºÏÈÜÒºÖÐc£¨H+£©=c£¨OH-£©£¬»ìºÏÈÜÒºÖÐÀë×ÓŨ¶ÈµÄ´óС˳ÐòΪ
 
£¬»ìºÏǰc£¨CH3COOH£©
 
c£¨NaOH£©£¨Óá°´óÓÚ¡±»ò¡°Ð¡ÓÚ¡±»ò¡°µÈÓÚ¡±Ìî¿Õ£©£®
£¨4£©Èô»ìºÏÈÜÒºµÄÈÜÖÊΪCH3COONaºÍNaOH£¬ÔòÏÂÁйØÏµ²»ÕýÈ·µÄÊÇ
 
£®
¢Ùc£¨Na+£©£¾c£¨CH3COO-£©£¾c£¨OH-£©£¾c£¨H+£©
¢Úc£¨Na+£©£¾c£¨OH-£©£¾c£¨CH3COO-£©£¾c£¨H+£©
¢Ûc£¨CH3COO-£©£¾c£¨Na+£©£¾c£¨OH-£©£¾c£¨H+£©
¢Üc£¨Na+£©£¾c£¨CH3COO-£©£¾c£¨H+£©£¾c£¨OH-£©
¿¼µã£ºËá¼î»ìºÏʱµÄ¶¨ÐÔÅжϼ°ÓйØphµÄ¼ÆËã,Èõµç½âÖÊÔÚË®ÈÜÒºÖеĵçÀëÆ½ºâ
רÌ⣺µçÀëÆ½ºâÓëÈÜÒºµÄpHרÌâ
·ÖÎö£º£¨1£©´×ËáÄܺÍ̼Ëá¸Æ·´Ó¦Éú³É´×Ëá¸ÆºÍ¶þÑõ»¯Ì¼¡¢Ë®£¬¸ù¾ÝÇ¿ËáÖÆÈ¡ÈõËá·ÖÎö£»?
£¨2£©¢ÙËáÒÖÖÆË®µçÀ룬ËáÖÐÇâÀë×ÓŨ¶ÈÔ½´ó£¬ÆäÒÖÖÆË®µçÀë³Ì¶ÈÔ½´ó£»    
¢Ú¼ÓˮϡÊÍ´Ù½ø´×ËáµçÀ룬Ôòn£¨CH3COO-£©Ôö´ó¡¢n£¨CH3COOH£©¼õС£»
£¨3£©Èô»ìºÏÈÜÒºÖÐc£¨H+£©=c£¨OH-£©£¬¸ù¾ÝµçºÉÊØºãÅжÏc£¨Na+£©¡¢c£¨CH3COO-£©Ïà¶Ô´óС£¬Ëá¸ùÀë×ÓË®½â³Ì¶È½ÏС£»´×ËáÄÆÈÜÒº³Ê¼îÐÔ£¬ÒªÊ¹»ìºÏÈÜÒº³ÊÖÐÐÔ£¬Ôò´×ËáÎïÖʵÄÁ¿Ó¦¸Ã´óЩ£»
£¨4£©Èô»ìºÏÈÜÒºµÄÈÜÖÊΪCH3COONaºÍNaOH£¬NaOHµçÀëºÍ´×Ëá¸ùÀë×ÓË®½â¶øÊ¹ÈÜÒº³Ê¼îÐÔ£¬ÔÙ½áºÏÎïÁÏÊØºãÅжϣ®
½â´ð£º ½â£º£¨1£©´×ËáÄܺÍ̼Ëá¸Æ·´Ó¦Éú³É´×Ëá¸ÆºÍ¶þÑõ»¯Ì¼¡¢Ë®£¬Ç¿ËáÄܺÍÈõËáµÄÑη¢Éú¸´·Ö½â·´Ó¦£¬ËùÒԸ÷´Ó¦ÌåÏÖ´×ËáËáÐÔ´óÓÚ̼Ëᣬ¹ÊÑ¡B£»
£¨2£©¢ÙÈÜÒºµÄµ¼µçÄÜÁ¦ÓëÇâÀë×ÓŨ¶È³ÉÕý±È£¬¸ù¾ÝÈÜÒºµ¼µçÄÜÁ¦Öª£¬ÇâÀë×ÓŨ¶È´óС˳ÐòÊÇb£¾a£¾c£¬ËáÒÖÖÆË®µçÀ룬ËáÖÐÇâÀë×ÓŨ¶ÈÔ½´ó£¬ÆäÒÖÖÆË®µçÀë³Ì¶ÈÔ½´ó£¬ËùÒÔÓÉË®µçÀë³öµÄc£¨H+£©ÓÉСµ½´ó˳ÐòÊÇb¡¢a¡¢c£¬¹Ê´ð°¸Îª£ºb¡¢a¡¢c£»
¢Ú¼ÓˮϡÊÍ´Ù½ø´×ËáµçÀ룬Ôòn£¨CH3COO-£©Ôö´ó¡¢n£¨CH3COOH£©¼õС£¬ËùÒÔ´Óbµãµ½cµã£¬ÈÜÒºÖÐ
c(CH3COO-)
c(CH3COOH)
µÄ±ÈÖµÔö´ó£¬¹Ê´ð°¸Îª£ºÔö´ó£»
£¨3£©Èô»ìºÏÈÜÒºÖÐc£¨H+£©=c£¨OH-£©£¬¸ù¾ÝµçºÉÊØºãµÃc£¨Na+£©=c£¨CH3COO-£©£¬Ëá¸ùÀë×ÓË®½â³Ì¶ÈºÜС£¬ËùÒÔÈÜÒºÖÐÀë×ÓŨ¶È´óС˳ÐòÊÇc£¨Na+£©=c£¨CH3COO-£©£¾c£¨H+£©=c£¨OH-£©£¬´×ËáÄÆÈÜÒº³Ê¼îÐÔ£¬ÒªÊ¹»ìºÏÈÜÒº³ÊÖÐÐÔ£¬Ôò´×ËáµÄÎïÖʵÄÁ¿ÉÔ΢´óÓÚÇâÑõ»¯ÄÆ£¬ËùÒÔ»ìºÏǰc£¨CH3COOH£©´óÓÚc£¨NaOH£©£¬
¹Ê´ð°¸Îª£ºc£¨Na+£©=c£¨CH3COO-£©£¾c£¨H+£©=c£¨OH-£©£»´óÓÚ£»
£¨4£©Èô»ìºÏÈÜÒºµÄÈÜÖÊΪCH3COONaºÍNaOH£¬
¢Ù´×Ëá¸ùÀë×ÓË®½â¡¢NaOHµçÀë¶øÊ¹ÈÜÒº³Ê¼îÐÔ£¬Ôòc£¨OH-£©£¾c£¨H+£©£¬¸ù¾ÝµçºÉÊØºãµÃc£¨Na+£©£¾c£¨CH3COO-£©£¬µ«ÈÜÒºÖÐÇâÑõ»¯ÄÆÁ¿½ÏÉÙʱ£¬ÔòÈÜÒºÖÐÀë×ÓŨ¶È´óС˳ÐòÊÇc£¨Na+£©£¾c£¨CH3COO-£©£¾c£¨OH-£©£¾c£¨H+£© £¬¹ÊÕýÈ·£»
¢Úµ±ÈÜÒºÖÐNaOHµÄÎïÖʵÄÁ¿½Ï´óʱ£¬¸ù¾ÝÎïÁÏÊØºãµÃc£¨Na+£©£¾c£¨OH-£©£¾c£¨CH3COO-£©£¾c£¨H+£©£¬¹ÊÕýÈ·£»
¢Û¸ù¾ÝÎïÁÏÊØºãµÃc£¨CH3COO-£©£¼c£¨Na+£©£¬¹Ê´íÎó£»
¢ÜNaOHµçÀëºÍ´×Ëá¸ùÀë×ÓË®½â¶øÊ¹ÈÜÒº³Ê¼îÐÔ£¬Ôòc£¨H+£©£¼c£¨OH-£©£¬¹Ê´íÎó£»
¹ÊÑ¡¢Û¢Ü£®
µãÆÀ£º±¾Ì⿼²éÁËÀë×ÓŨ¶È´óС±È½Ï¡¢Èõµç½âÖʵĵçÀëµÈ֪ʶµã£¬¸ù¾ÝÈõµç½âÖʵĵçÀëÌØµã½áºÏÈÜÒºÖÐÈÜÖʵÄÐÔÖÊ·ÖÎö½â´ð£¬×¢ÒâÈÜÒºµ¼µçÄÜÁ¦ÓëÀë×ÓŨ¶ÈÓйأ¬Óëµç½âÖÊÇ¿ÈõÎ޹أ¬ÎªÒ×´íµã£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
¶ÌÖÜÆÚÔªËØA¡¢B¡¢C¡¢D£¬ÆäÓйØÐÔÖÊ»ò½á¹¹ÐÅÏ¢Èç±íËùʾ£®
ÔªËØ A B C D
ÓйØÐÔÖÊ»ò½á¹ûÐÅÏ¢ ÄÜÐγÉ+7¼ÛµÄ»¯ºÏÎï »ù̬ԭ×ÓºËÍâsÄܼ¶Éϵĵç×ÓÊýÊÇpÄܼ¶Éϵç×ÓÊýµÄ2±¶ ¸ÃÔªËØÐγɵĵ¥ÖÊͨ³£ÓÐÁ½ÖÖÍ¬Î»ËØÒìÐÎÌ壬ÆäÖÐÒ»ÖÖµ¥ÖÊ·Ö²¼Ôڸ߿գ¬Æð×Å×èÖ¹×ÏÍâ·øÉäµÄ×÷Óà ͨ³£Çé¿öÏÂÄÜÐγɶÌÖÜÆÚÖÐ×îÎȶ¨µÄ˫ԭ×Ó·Ö×Ó
£¨1£©BλÓÚÔªËØÖÜÆÚ±íÖеÚ
 
ÖÜÆÚµÚ
 
×壬AÔªËØµÄ¼Ûµç×ÓÅŲ¼Ê½
 
£¬ÔªËØBºÍÔªËØCÏà±È£¬µç¸ºÐÔ½ÏСµÄÊÇ
 
 £¨Ð´ÔªËØÃû³Æ£©£®»¯ºÏÎïBC2¾§ÌåÓëBµÄµ¥Öʾ§ÌåÊô
 
 £¨ÌîͬһÀàÐÍ»ò²»Í¬ÀàÐÍ£©£¬ÀíÓÉ
 
£®
£¨2£©CºÍDµÄÆøÌ¬Ç⻯ÎïÖУ¬½ÏÎȶ¨µÄÊÇ
 
£¨Ð´»¯Ñ§Ê½£©£¬BµÄÔ­×Ó£®CµÄÔ­×ÓºÍÇâÔ­×ÓÐγÉÒ»ÖÖº¬ÓÐÇâ¼üÇÒÏà¶Ô·Ö×ÓÖÊÁ¿×îСµÄÓлúÎïÃû³Æ
 
£®
£¨3£©CÓëD¿ÉÒÔ×é³É¶àÖÖÐÎʽµÄ»¯ºÏÎÆäÖÐD2C5Äܹ»Ë®·´Ó¦Éú³ÉÎïÖÊX£®Ð´³öÎïÖÊXµÄÏ¡ÈÜÒºÓë¹ýÁ¿µÄFe·´Ó¦µÄÏÖÏó
 
£¬Àë×Ó·½³Ìʽ£º
 
£®
£¨4£©AµÄ×î¸ß¼ÛÑõ»¯ÎïΪÎÞɫҺÌ壬9.15g¸ÃÎïÖÊÓë×ãÁ¿µÄË®»ìºÏ£¬µÃµ½Ò»ÖÖÏ¡ÈÜÒº£¬²¢·Å³öQKJÈÈÁ¿£¬¸Ã·´Ó¦µÄÈÈ»¯Ñ§·´Ó¦·½³ÌʽΪ
 
£®
2014Äê10Ô³õ£¬Îíö²ÌìÆø¶à´ÎËÁŰºÓ±±¡¢Ìì½ò¡¢±±¾©µÈµØÇø£®ÆäÖУ¬È¼ÃººÍÆû³µÎ²ÆøÊÇÔì³É¿ÕÆøÎÛȾµÄÔ­ÒòÖ®Ò»£®
£¨1£©Æû³µÎ²Æø¾»»¯µÄÖ÷ÒªÔ­ÀíΪ£º2NO£¨g£©+2CO£¨g£©
´ß»¯¼Á
2CO2£¨g£©+N2£¨g£©£®¡÷H£¼0
¢Ù¸Ã·´Ó¦µÄƽºâ³£Êý±í´ïʽ
 

¢ÚÈô¸Ã·´Ó¦ÔÚ¾øÈÈ¡¢ºãÈݵÄÃܱÕÌåϵÖнøÐУ¬ÏÂÁÐʾÒâͼÕýÈ·ÇÒÄÜ˵Ã÷·´Ó¦ÔÚ½øÐе½t1ʱ¿Ì´ïµ½Æ½ºâ״̬µÄÊÇ
 
 £¨Ìî´úºÅ£©£®
£¨2£©Ö±½ÓÅÅ·ÅúȼÉÕ²úÉúµÄÑÌÆø»áÒýÆðÑÏÖØµÄ»·¾³ÎÊÌ⣮úȼÉÕ²úÉúµÄÑÌÆøº¬µªµÄÑõ»¯ÎÓÃCH4´ß»¯»¹Ô­
NOx¿ÉÒÔÏû³ýµªÑõ»¯ÎïµÄÎÛȾ£®
ÒÑÖª£ºCH4£¨g£©+2NO2£¨g£©=N2£¨g£©+CO2£¨g£©+2H2O£¨g£©¡÷H=-867kJ/mol
2NO2£¨g£©?N2O4£¨g£©¡÷H=-56.9kJ/mol
H2O£¨g£©=H2O£¨l£©¡÷H=-44.0kJ/mol
д³öCH4´ß»¯»¹Ô­N2O4£¨g£©Éú³ÉN2ºÍH2O£¨l£©µÄÈÈ»¯Ñ§·½³Ìʽ£º
 
£®
£¨3£©ÔÚÒ»¶¨Ìõ¼þÏ£¬Ò²¿ÉÒÔÓÃNH3´¦ÀíNOx£®ÒÑÖªNOÓëNH3·¢Éú·´Ó¦Éú³ÉN2ºÍH2O£¬ÏÖÓÐNOºÍNH3µÄ»ìºÏÎï1mol£¬³ä·Ö·´Ó¦ºóµÃµ½µÄ»¹Ô­²úÎï±ÈÑõ»¯²úÎï¶à1.4g£¬ÔòÔ­·´Ó¦»ìºÏÎïÖÐNOµÄÎïÖʵÄÁ¿¿ÉÄÜÊÇ
 
£®
£¨4£©ÒÔ¼×ÍéΪԭÁÏÖÆÈ¡ÇâÆøÊǹ¤ÒµÉϳ£ÓõÄÖÆÇâ·½·¨£®Ôò2molCH4Óë×ãÁ¿H2O£¨g£©·´Ó¦×î¶à¿ÉÉú³É
 
mol H2£¬Ð´³ö¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ
 
£®
£¨5£©ÉÏÊö·½·¨ÖƵõÄH2¿ÉÒÔºÍCOÔÚÒ»¶¨Ìõ¼þϺϳɼ״¼ºÍ¶þ¼×ÃÑ£¨CH3OCH3£©¼°Ðí¶àÌþÀàÎïÖÊ£®µ±Á½ÕßÒÔÎïÖʵÄÁ¿1£º1´ß»¯·´Ó¦£¬ÆäÔ­×ÓÀûÓÃÂÊ´ï100%£¬ºÏ³ÉµÄÎïÖÊ¿ÉÄÜÊÇ
 
£®
a£®ÆûÓÍ        b£®¼×´¼            c£®¼×È©            d£®ÒÒËᣮ

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

¾«Ó¢¼Ò½ÌÍø