ÌâÄ¿ÄÚÈÝ


ÒÒ»ùÏã²ÝÈ©£¨£©ÊÇʳƷÌí¼Ó¼ÁµÄÔöÏãÔ­ÁÏ¡£

£¨1£©Ð´³öÒÒ»ùÏã²ÝÈ©·Ö×ÓÖÐÁ½ÖÖº¬Ñõ¹ÙÄÜÍŵÄÃû³Æ                     ¡£

£¨2£©ÓëÒÒ»ùÏã²ÝÈ©»¥ÎªÍ¬·ÖÒì¹¹Ì壬ÄÜÓëNaHCO3ÈÜÒº·´Ó¦·Å³öÆøÌ壬ÇÒ±½»·ÉÏÖ»ÓÐÒ»¸ö²àÁ´£¨²»º¬R£­O£­R¡¯¼°R£­O£­COOH½á¹¹£©µÄÓР         ÖÖ¡£

£¨3£©AÊÇÉÏÊöͬ·ÖÒì¹¹ÌåÖеÄÒ»ÖÖ£¬¿É·¢ÉúÒÔϱ仯£º

ÒÑÖª£ºi. RCH2OHRCHO£»

ii. Óë±½»·Ö±½ÓÏàÁ¬µÄ̼ԭ×ÓÉÏÓÐÇâʱ£¬´Ë̼ԭ×Ӳſɱ»ËáÐÔKMnO4ÈÜÒºÑõ»¯ÎªôÈ»ù¡£

¢Ùд³öAµÄ½á¹¹¼òʽ£º              £¬·´Ó¦¢Ú·¢ÉúµÄÌõ¼þÊÇ                 ¡£

¢ÚÓÉA¿ÉÖ±½ÓÉú³ÉD£¬·´Ó¦¢ÙºÍ¢ÚµÄÄ¿µÄÊÇ                                   ¡£

¢Ûд³ö·´Ó¦·½³Ìʽ£ºA¡úB                                             £¬

CÓëNaOHË®ÈÜÒº¹²ÈÈ£º                                              ¡£

£¨4£©ÒÒ»ùÏã²ÝÈ©µÄÁíÒ»ÖÖͬ·ÖÒì¹¹ÌåE()ÊÇÒ»ÖÖÒ½Ò©ÖмäÌå¡£ÓÉÜîÏãÈ©()ºÏ³ÉE£¨ÆäËûÔ­ÁÏ×ÔÑ¡£©£¬Éæ¼°µÄ·´Ó¦ÀàÐÍÓУ¨°´·´Ó¦Ë³ÐòÌîд£©                         ¡£

 


£¨1£©È©»ù¡¢£¨·Ó£©ôÇ»ù£¨»òÃѼü£©£»

£¨2£©4ÖÖ£»

£¨3£©¢Ù  £¬NaOH´¼ÈÜÒº£¬¡÷

¢Ú¼õÉÙÈçõ¥»¯¡¢³ÉÃѵȸ±·´Ó¦µÄ·¢Éú

¢Û

£¨»ò£©£»

 

£¨4£©Ñõ»¯·´Ó¦¡¢õ¥»¯·´Ó¦£¨»òÈ¡´ú·´Ó¦£©


Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

Ñ̵ÀÆøÖеÄCO2¿ÉÓÃÓںϳɼ״¼ºÍ¼×ÃÑ¡£

(1)ÏÂͼËùʾΪÆäÖеÄÒ»ÖÖ·½·¨£º

¢ÙÓùâµç³Øµç½âˮʱ£¬Ñô¼«µÄµç¼«·´Ó¦Ê½Îª__________________________________¡£

¢Ú15¡«20%µÄÒÒ´¼°·(HOCH2CH2NH2)Ë®ÈÜÒº¾ßÓÐÈõ¼îÐÔ£¬ÔòHOCH2CH2NH3ClµÄË®ÈÜÒºÏÔ________(Ìî¡°Ëᡱ¡°¼î¡±»ò¡°ÖС±)ÐÔ¡£

¢ÛÒÑÖª£¬CH3OH(l)£«O2(g)===CO2(g)£«2H2O(l)  ¦¤H£½£­725.5 kJ¡¤mol£­1

  H2(g)£«O2(g)===H2O(l)¡¡¦¤H£½£­285.8 kJ¡¤mol£­1

Ôò¹¤ÒµÉÏÒÔCO2(g)¡¢H2(g)ΪԭÁϺϳÉCH3OH(l)µÄÈÈ»¯Ñ§·½³ÌʽΪ

________________________________________________________________________¡£

(2)½«CO2ת»¯Îª¼×Ãѵķ´Ó¦Ô­ÀíΪ2CO2(g)£«6H2(g)CH3OCH3(g)£«3H2O(l)

¢ÙÔÚºãΡ¢ºãÈݵÄÃܱÕÈÝÆ÷ÖУ¬ÏÂÁÐÃèÊöÄÜ˵Ã÷ÉÏÊö·´Ó¦ÒѴﻯѧƽºâ״̬µÄÊÇ________(ÌîÐòºÅ)¡£

a£®Éú³É1 mol CH3OCH3(g)µÄͬʱ£¬Éú³É3 mol H2O(l)

b£®ÌåϵÖлìºÏÆøÌåµÄÃܶȲ»ËæÊ±¼ä¶ø±ä»¯

c£®ÌåϵÖÐCO2ÓëH2µÄÌå»ý±ÈΪ1¡Ã3

d£®ÌåϵÖÐµÄÆøÌåµÄ×Üѹǿ²»ËæÊ±¼ä¶ø±ä»¯

¢ÚÒÑÖªÒ»¶¨Ñ¹Ç¿Ï£¬¸Ã·´Ó¦ÔÚ²»Í¬Î¶ȡ¢²»Í¬Í¶ÁϱÈʱ£¬CO2µÄת»¯ÂʼûÏÂ±í£º

ͶÁϱÈ

500 K

600 K

700 K

1.5

45%

33%

20%

x

a

b

c

ÉÏÊö·´Ó¦µÄ»¯Ñ§Æ½ºâ³£ÊýµÄ±í´ïʽΪ__________¡£¸Ã·´Ó¦µÄìʱ䦤H________0(Ìî¡°>¡±¡¢¡°<¡±»ò¡°£½¡±£¬ÏÂͬ)£¬Èôʹa>45%£¬Ôòx________1.5¡£

¢ÛÔÚÒ»¶¨Î¶ÈÏ£¬ÏòÌå»ýΪ0.5 LµÄÃܱÕÈÝÆ÷ÖмÓÈë2 mol CO2ºÍ6 mol H2,20 minºó´ïµ½Æ½ºâ£¬²âµÃƽºâʱCH3OCH3Ϊ0.5 mol£¬Ôò¸Ã·´Ó¦µÄ·´Ó¦ËÙÂÊv(CO2)£½____mol¡¤L£­1¡¤min£­1£¬H2µÄת»¯ÂʦÁ(H2)£½______£»¼ÓÈë´ß»¯¼Á£¬v(CO2)________(Ìî¡°Ôö´ó¡±¡¢¡°¼õС¡±»ò¡°²»±ä¡±£¬ÏÂͬ)£¬¦Á(H2)________¡£

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

¾«Ó¢¼Ò½ÌÍø