ÌâÄ¿ÄÚÈÝ

½«Á½¸ö²¬µç¼«²åÈËKOHÈÜÒºÖУ¬ÏòÁ½¼«·Ö±ðͨÈËCH4ºÍO2£¬¹¹³É¼×ÍéȼÁϵç³Ø¡£ÒÑÖª¡£Í¨ÈËCH4µÄÒ»¼«£¬Æäµç¼«·´Ó¦Ê½ÊÇ£ºCH4 + 10OH-  - 8e-£½CO32-£«7 H2O£»Í¨Èë£Ï2µÄÁíÒ»¼«£¬Æäµç¼«·´Ó¦Ê½ÊÇ£º2O2 + 4H2O £«8e- = 8OH-¡£ÏÂÁÐÐðÊö²»ÕýÈ·µÄÊÇ(  )¡£

A£®Í¨ÈËCH4µÄµç¼«Îª¸º¼«

B£®Õý¼«·¢ÉúÑõ»¯·´Ó¦

C£®È¼Áϵç³Ø¹¤×÷ʱ£¬ÈÜÒºÖеÄOH-Ïò¸º¼«Òƶ¯ 

D£®¸Ãµç³ØʹÓÃÒ»¶Îʱ¼äºóÓ¦²¹³äKOH

 

¡¾´ð°¸¡¿

B

¡¾½âÎö¡¿

ÊÔÌâ·ÖÎö£ºÔ­µç³ØÖиº¼«Ê§È¥µç×Ó£¬·¢ÉúÑõ»¯·´Ó¦¡£Õý¼«µÃµ½µç×Ó£¬·¢Éú»¹Ô­·´Ó¦£¬ËùÒÔ¼×ÍéÔÚ¸º¼«Í¨È룬ÑõÆøÔÚÕý¼«Í¨È룬Òò´ËAÕýÈ·£¬B²»ÕýÈ·£»¸ù¾Ýµç¼«·´Ó¦Ê½¿ÉÖª£¬Ñ¡ÏîCºÍD¶¼ÊÇÕýÈ·µÄ£¬´ð°¸Ñ¡B¡£

¿¼µã£º¿¼²éÔ­µç³ØµÄÓйØÅжϡ£

µãÆÀ£ºÔÚÔ­µç³ØÖнϻîÆõĽðÊô×÷¸º¼«£¬Ê§È¥µç×Ó£¬·¢ÉúÑõ»¯·´Ó¦¡£µç×Ó¾­µ¼Ïß´«µÝµ½Õý¼«ÉÏ£¬ËùÒÔÈÜÒºÖеÄÑôÀë×ÓÏòÕý¼«Òƶ¯£¬ÒõÀë×ÓÏò¸º¼«Òƶ¯¡£Õý¼«µÃµ½µç×Ó£¬·¢Éú»¹Ô­·´Ó¦£¬¾Ý´Ë¿ÉÒÔ½øÐÐÓйصÄÅжϺͼÆËã¡£

 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
£¨2010?˳ÒåÇøһģ£©½üÄêÀ´£¬Ì¼ºÍ̼µÄ»¯ºÏÎïÔÚÉú²úÉú»îʵ¼ÊÖÐÓ¦Óù㷺£®
£¨1£©¼×ÍéȼÉշųö´óÁ¿µÄÈÈ£¬¿É×÷ΪÄÜÔ´ÓÃÓÚÈËÀàµÄÉú²úºÍÉú»î£®
ÒÑÖª ¢Ù2CH4£¨g£©+3O2£¨g£©=2CO £¨g£©+4H2O£¨l£©¡÷H1=-1214.6KJ/mol
¢Ú2CO £¨g£©+O2£¨g£©=2CO2£¨g£©¡÷H2=-566kJ/mol
Ôò·´Ó¦CH4£¨g£©+2O2£¨g£©=CO2£¨g£©+2H2O £¨l£© µÄ¡÷H=
-890.3KJ/mol
-890.3KJ/mol
£®
£¨2£©½«Á½¸öʯīµç¼«²åÈËKOHÈÜÒºÖУ¬ÏòÁ½¼«·Ö±ðͨÈëCH4ºÍO2£¬¹¹³É¼×ÍéȼÁϵç³Ø£®Í¨ÈëCH4µÄÒ»¼«£¬Æäµç¼«·´Ó¦Ê½ÊÇ£ºCH4+10OH--8e-=CO32-+7H2O£»Í¨ÈëO2µÄÒ»¼«£¬Æäµç¼«·´Ó¦Ê½ÊÇ
O2+4e_+2H2O=4OH-
O2+4e_+2H2O=4OH-
£®
£¨3£©ÈôÓÃʯī×öµç¼«µç½â500ml±¥ºÍʳÑÎË®£¬Ð´³öµç½â·´Ó¦µÄÀë×Ó·½³ÌʽΪ£º
2Cl-+2H2O
 Í¨µç 
.
 
H2¡ü+Cl2¡ü+2OH-
2Cl-+2H2O
 Í¨µç 
.
 
H2¡ü+Cl2¡ü+2OH-
£»µç½âÒ»¶Îʱ¼äºóÁ½¼«¹²ÊÕ¼¯µ½±ê×¼×´¿öϵÄÆøÌå1.12L£¨²»¿¼ÂÇÆøÌåµÄÈܽ⣩£®Í£Ö¹Í¨µç£¬¼ÙÉ跴ӦǰºóÈÜÒºÌå»ý²»±ä£¬ÔòËùµÃÈÜÒºµÄpH=
13
13
£®
£¨4£©½«²»Í¬Á¿µÄCO £¨g£© ºÍH2O £¨g£© ·Ö±ðͨÈëµ½Ìå»ýΪ2LµÄºãÈÝÃܱÕÈÝÆ÷ÖУ¬½øÐз´Ó¦
CO £¨g£©+H2O £¨g£©?CO2£¨g£©+H2£¨g£©£¬µÃµ½ÈçÏÂÈý×éÊý¾Ý£º
ʵÑé×é ζÈ/¡æ ÆðʼÁ¿/mol ƽºâÁ¿/mol ´ïµ½Æ½ºâËù
Ðèʱ¼ä/min
H2O CO CO2 CO
1 650 2 4 1.6 2.4 5
2 900 1 2 0.4 1.6 3
3 900 a b c d t
¢ÙʵÑé1ÖÐÒÔ¦Ô£¨H2£© ±íʾµÄ·´Ó¦ËÙÂÊΪ
0.16mol?£¨L?min£©-1
0.16mol?£¨L?min£©-1
£®
¢ÚʵÑé2ÖеÄƽºâ³£ÊýÊÇ
0.17
0.17
 £¨¼ÆËã½á¹û±£ÁôÁ½Î»Ð¡Êý£©£®
¢Û¸Ã·´Ó¦µÄÕý·´Ó¦Îª
·Å
·Å
£¨Ìî¡°Îü¡±»ò¡°·Å¡±£©ÈÈ·´Ó¦£®
¢ÜÈôʵÑé3Òª´ïµ½ÓëʵÑé2ÏàͬµÄƽºâ״̬£¨¼´¸÷ÎïÖʵÄÖÊÁ¿·ÖÊý·Ö±ðÏàµÈ£©£¬Ôòa¡¢bÓ¦Âú×ãµÄ¹ØϵÊÇ
b=2a£¾l£¨»òa£¾0.5£¬b=2a£»»òb£¾1£¬b=2a£©
b=2a£¾l£¨»òa£¾0.5£¬b=2a£»»òb£¾1£¬b=2a£©
 £¨Óú¬a¡¢bµÄÊýѧʽ±íʾ£©£®
ÓÐЧµØÀûÓÃÏÖÓÐÐÂÄÜÔ´ºÍ¿ª·¢ÐÂÄÜÔ´ÒÑÊܵ½¸÷¹úµÄÖØÊÓ£®
£¨1£©¿ÉÓøĽøÆûÓÍ×é³ÉµÄ°ì·¨À´¸ÄÉÆÆûÓ͵ÄȼÉÕÐÔÄÜ£®ÀýÈ磬ÔÚÆûÓÍÖмÓÈëÒÒ´¼À´Éú²ú¡°ÎÞǦÆûÓÍ¡±£®ÒÒ´¼µÄ·Ö×ÓʽΪC2H6O£¬ÊÔ¸ù¾ÝC¡¢H¡¢O³É¼üµÄÌص㣬д³öC2H6OËùÓпÉÄܵĽṹʽ»ò½á¹¹¼òʽ
CH3CH2OH¡¢CH3OCH3
CH3CH2OH¡¢CH3OCH3
£®
£¨2£©ÌìÈ»ÆøµÄÖ÷ÒªµÄ³É·ÖÊǼ×Í飬ÆäȼÉÕ²úÎïÎÞ¶¾¡¢ÈÈÖµ¸ß¡¢¹ÜµÀÊäËÍ·½±ã£¬½«³ÉΪÎÒ¹úÎ÷²¿¿ª·¢µÄÖصãÖ®Ò»£®ÄÜ˵Ã÷¼×ÍéÊÇÕýËÄÃæÌå¶ø·ÇÕý·½ÐÎƽÃæ½á¹¹µÄÀíÓÉÊÇ
¢Ú
¢Ú
£®£¨Ìîд±àºÅ£¬¶àÑ¡µ¹¿Û·Ö£©
¢ÙÆäÒ»ÂÈÈ¡´úÎï²»´æÔÚͬ·ÖÒì¹¹Ìå  ¡¡¡¡¡¡¡¡¡¡¡¡¢ÚÆä¶þÂÈÈ¡´úÎï²»´æÔÚͬ·ÖÒì¹¹Ìå
¢ÛÆäÈýÂÈÈ¡´úÎï²»´æÔÚͬ·ÖÒì¹¹Ìå              ¢ÜÆäËÄÂÈÈ¡´úÎï²»´æÔÚͬ·ÖÒì¹¹Ìå
£¨3£©½«Á½¸öʯīµç¼«²åÈËKOHÈÜÒºÖУ¬ÏòÁ½¼«·Ö±ðͨÈëCH4ºÍO2£¬¹¹³É¼×ÍéȼÁϵç³Ø£®Í¨ÈëCH4µÄÒ»¼«µç¼«·´Ó¦Ê½ÊÇ£ºCH4+10OH--8e-=CO32-+7H2O£»Í¨ÈëO2µÄÒ»¼«£¬Æäµç¼«·´Ó¦Ê½ÊÇ
2O2+4H2O+8e-=8OH-£¨»òO2+2H2O+4e-=4OH-£©
2O2+4H2O+8e-=8OH-£¨»òO2+2H2O+4e-=4OH-£©
£®ÒÑÖª4g¼×ÍéÍêȫȼÉÕÉú³ÉCO2ÆøÌåºÍҺ̬ˮʱ·Å³ö222.5kJµÄÈÈÁ¿£¬Ôò¼×ÍéȼÉÕµÄÈÈ»¯Ñ§·½³ÌʽΪ
CH4£¨g£©+2O2£¨g£©=CO2£¨g£©+2H2O£¨l£©¡÷H=-890kJ/mol
CH4£¨g£©+2O2£¨g£©=CO2£¨g£©+2H2O£¨l£©¡÷H=-890kJ/mol
£»
£¨4£©ÇâÄÜÊÇÈËÀàδÀ´µÄÀíÏëÄÜÔ´£®1980ÄêÎÒ¹úÊ×´ÎÖƳÉÒ»Á¾È¼ÇâÆû³µ£¬³ËÔ±12ÈË£¬ÒÔ50km/hÐÐÊ»ÁË40km£®ÎªÁËÓÐЧ·¢Õ¹ÃñÓÃÇâÄÜÔ´£¬Ê×ÏȱØÐëÖƵÃÁ®¼ÛµÄÇâÆø£®ÏÂÁмȿɹ©¿ª·¢ÓÖÏûºÄ½ÏµÍ¾­¼ÃµÄÖÆÇâ·½·¨ÊÇ
¢Û
¢Û
£¨Ìîд±àºÅ£¬¶àÑ¡µ¹¿Û·Ö£©
¢Ùµç½âË®     ¡¡¡¡¡¡¢ÚпºÍÏ¡ÁòËá·´Ó¦     ¡¡¡¡¢Û¹â½âº£Ë®
Æä´Î£¬ÖƵô¿ÇâÆøºó£¬»¹ÐèÒª½â¾öµÄÎÊÌâÊÇ
Öü´æ»òÔËÊä
Öü´æ»òÔËÊä
£®£¨Ð´³öÆäÖеÄÒ»¸ö£©

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

¾«Ó¢¼Ò½ÌÍø