ÌâÄ¿ÄÚÈÝ

2001Äê6ÔÂ21ÈÕ£¬ºÓÄϵÄÖ£ÖÝ¡¢ÂåÑô¼°ÄÏÑôÊÐÂÊÏÈʹ²¿·ÖÆû³µ²ÉÓ÷â±ÕÔËÐз½Ê½£¬ÊÔÓÃеÄÆû³µÈ¼ÁÏ¡ª¡ª³µÓÃÒÒ´¼ÆûÓÍ¡£ÒÒ´¼£¬Ë×Ãû¾Æ¾«£¬ËüÊÇÒÔÓñÃס¢Ð¡Âó¡¢ÊíÀàµÈΪԭÁϾ­·¢½Í¡¢ÕôÁó¶øÖƳɵġ£ÒÒ´¼½øÒ»²½ÍÑË®£¬ÔÙ¼ÓÉÏÊÊÁ¿ÆûÓͺóÐγɱäÐÔȼÁÏÒÒ´¼¡£¶ø³µÓÃÒÒ´¼ÆûÓ;ÍÊǰѱäÐÔȼÁÏÒÒ´¼ºÍÆûÓÍ°´Ò»¶¨±ÈÀý»ìÅäÐγɵijµÓÃȼÁÏ¡£½áºÏÓйØ֪ʶ£¬Íê³ÉÒÔÏÂÎÊÌ⣺
(1)ÒÒ´¼µÄ½á¹¹¼òʽΪ_____________¡£ÆûÓÍÊÇÓÉʯÓÍ·ÖÁóËùµÃµÄµÍ·ÐµãÍéÌþ£¬Æä·Ö×ÓÖеÄ̼ԭ×ÓÊýÒ»°ãÔÚC5¡ªC11·¶Î§ÄÚ£¬ÈçÎìÍ飬Æä·Ö×ÓʽΪ__________________£¬½á¹¹¼òʽ¼°Æäͬ·ÖÒì¹¹Ìå·Ö±ðΪ_____________¡¢_____________¡¢_____________¡£
(2)ÒÒ´¼¿ÉÓɺ¬µí·Û¡²(C6H10O5)n¡³µÄÅ©²úÆ·£¬ÈçÓñÃס¢Ð¡Âó¡¢ÊíÀàµÈ¾­·¢½Í¡¢ÕôÁó¶øµÃ¡£Çëд³öÓɵí·ÛÖÆÒÒ´¼µÄ»¯Ñ§·½³Ìʽ£º
¢Ùµí·Û+Ë®ÆÏÌÑÌÇ(C6H12O6)
¢ÚÆÏÌÑÌÇÒÒ´¼
(3)µí·Û¿ÉÓÉÂÌÉ«Ö²Îï¾­¹âºÏ×÷ÓõÈһϵÁÐÉúÎﻯѧ·´Ó¦µÃµ½£¬¼´Ë®ºÍ¶þÑõ»¯Ì¼¾­¹âºÏ×÷ÓÃÉú³ÉÆÏÌÑÌÇ£¬ÓÉÆÏÌÑÌÇÔÙÉú³Éµí·Û¡£½øÐйâºÏ×÷Óõij¡ËùÊÇ_____________£¬·¢Éú¹âºÏ×÷ÓÃÉú³ÉÆÏÌÑÌǵĻ¯Ñ§·½³ÌʽÊÇ
__________________________________________________________¡£
(4)ÒÒ´¼³ä·ÖȼÉյIJúÎïΪ__________________ºÍ__________________¡£
(5)³µÓÃÒÒ´¼ÆûÓͳÆΪ»·±£È¼ÁÏ£¬ÆäÔ­ÒòÊÇ__________________________________________¡£
£¨1£©C2H5OH  C5H12  CH3¡ªCH2¡ªCH2¡ªCH2¡ªCH3 

£¨2£©¢Ù(C6H10O5)n+nH2OnC6H12O6
¢ÚC6H12O62C2H5OH+2CO2
£¨3£©Ò¶ÂÌÌå  6CO2+6H2OC6H12O6+6O2
£¨4£©CO2  H2O
£¨5£©ÄÜÓÐЧ½µµÍÆû³µÎ²Æø´øÀ´µÄÑÏÖØ´óÆøÎÛȾ£¬¸ÄÉÆ»·¾³ÖÊÁ¿
±¾ÌâÊÇ»¯Ñ§ÓëÉúÎïÏà½áºÏµÄÌâÄ¿£¬ÒÒ´¼¡¢ÆûÓÍȼÉÕ£¬Éú³ÉCO2ºÍË®£¬²»»á¶Ô»·¾³Ôì³ÉÎÛȾ¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
£¨12·Ö£©Ä³ÊµÑéС×éÏòäåË®ÖмÓÈë×ãÁ¿µÄÒÒÈ©£¬·¢ÏÖäåË®ÍÊÉ«¡£¸ÃʵÑéС×é¶Ô·´Ó¦²úÎï½øÐÐ̽¾¿¡£
£¨1£©Ìá³ö¼ÙÉè
¢ÙÒÒÈ©ÓëäåË®·¢Éú¼Ó³É·´Ó¦Éú³É
¢ÚÒÒÈ©ÓëäåË®·¢ÉúÈ¡´ú·´Ó¦Éú³ÉCH2BrCHO£»
¢ÛÒÒÈ©ÓëäåË®·¢ÉúÑõ»¯·´Ó¦Éú³É    A     ¡£
£¨2£©»ùÓÚ¼ÙÉè¢ÛÉè¼ÆʵÑé·½°¸Ö¤Ã÷ÄãµÄ¼ÙÉ裨²»ÒªÔÚ´ðÌ⿨ÉÏ×÷´ð£©
£¨3£©ÊµÑé¹ý³Ì
¸ù¾Ý£¨2£©µÄʵÑé·½°¸£¬½øÐÐʵÑé¡£ÇëÔÚ´ðÌ⿨ÉÏ°´Ï±í¸ñʽд³öʵÑé²Ù×÷²½Öè¡¢Ô¤ÆÚÏÖÏóÓë½áÂÛ¡££¨ÏÞѡʵÑéÒÇÆ÷ÓëÊÔ¼Á£ºÊԹܡ¢²£Á§°ô¡¢µÎ¹Ü¡¢¾Æ¾«µÆ¡¢Ìú¼Ų̈¡¢µ¼¹Ü¡¢½ºÈû¡¢ÒÒ´¼¡¢Å¨ÁòËá¡¢±¥ºÍ̼ËáÄÆÈÜÒº¡¢ÇâÑõ»¯ÄÆÈÜÒº¡££©
ʵÑé²Ù×÷
Ô¤ÆÚÏÖÏóÓë½áÂÛ
²½Öè1£ºÈ¡Ò»¶¨Á¿   B  ÓÚÊÔ¹ÜÖУ¬È»ºó±ßÕðµ´ÊԹܱßÂýÂý¼ÓÈëÊÊÁ¿   C   ºÍ    D    ¡£       
 
²½Öè2£º¹Ì¶¨ºÃ×°Öã¬Óþƾ«µÆ»ºÂý¼ÓÈÈ¡£½«²úÉúµÄÕôÆøͨµ½     E     ÈÜÒºµÄÒºÃæÉÏ¡£
             F              ,
         G          ¡£
£¨4£©·´Ë¼ÓëÆÀ¼Û£º¢ÙÓÐͬѧÈÏΪֱ½ÓÓÃpHÊÔÖ½¼ìÑé·´Ó¦ºóÈÜÒºµÄpH´óС¼´¿ÉÑéÖ¤·´Ó¦²úÎÄãÈÏΪÊÇ·ñºÏÀí£º   H   £¨ÌîºÏÀí»ò²»ºÏÀí£©¡£Ô­ÒòÊÇ       I       ¡£
¢ÚÒÒÈ©±»äåË®Ñõ»¯µÄ·´Ó¦·½³ÌʽÊÇ              K             ¡£
£¨16·Ö£©Ä³°àѧÉúÔÚÀÏʦָµ¼ÏÂ̽¾¿µªµÄ»¯ºÏÎïµÄijЩÐÔÖÊ¡£
£¨1£©Í¬Ñ§¼×ÔÚʵÑéÊÒÀûÓÃÏÂÁÐ×°Ö㨺óÃæÓÐͼ£©ÖÆÈ¡°±ÆøºÍÑõÆøµÄ»ìºÏÆøÌ壬²¢Íê³É°±µÄ´ß»¯Ñõ»¯¡£
AÖмÓÈëŨ°±Ë®£¬DÖмÓÈë¼îʯ»Ò£¬EÄÚ·ÅÖô߻¯¼Á(²¬Ê¯ÃÞ)£¬Çë»Ø´ð£º
¢ÙÒÇÆ÷BµÄÃû³Æ£º__________¡£BÄÚÖ»Ðè¼ÓÈëÒ»ÖÖ¹ÌÌåÊÔ¼Á£¬¸ÃÊÔ¼ÁµÄÃû³ÆΪ_________£¬BÖÐÄܲúÉú°±ÆøºÍÑõÆø»ìºÏÆøÌåµÄÔ­Òò(½áºÏ»¯Ñ§·½³Ìʽ»Ø´ð)_                 __¡£
¢Ú°´ÆøÁ÷·½ÏòÁ¬½Ó¸÷ÒÇÆ÷                                     £¨Ìî½Ó¿Ú×Öĸ£©

£¨2£©Í¬Ñ§ÒÒÄâÓü×ͬѧµÃµ½µÄ»ìºÏÆøÌåX(NO¼°¹ýÁ¿µÄNH3)£¬ÑéÖ¤NOÄܱ»°±Æø»¹Ô­²¢²âËãÆäת»¯ÂÊ£¨ºöÂÔ×°ÖÃÄÚ¿ÕÆøµÄÓ°Ï죩¡£×°ÖÃÈçÏ£º

¢Ù×°ÖÃCµÄ×÷ÓÿÉÄÜÊÇ ____________¡£
¢ÚÈô½øÈë×°ÖÃAµÄNO¹²268.8mL£¨ÒÑÕÛËãΪ±ê×¼×´¿ö£¬ÏÂͬ£©£¬°±Æø¹ýÁ¿£¬×îºóÊÕ¼¯µ½±ê×¼×´¿öÏÂ190.4 mL N2£¬ÔòNOµÄת»¯ÂÊΪ             ¡£
£¨3£©N2O3ÊÇÒ»ÖÖÐÂÐÍÏõ»¯¼Á¡£Ò»¶¨Î¶ÈÏ£¬ÔÚºãÈÝÃܱÕÈÝÆ÷ÖÐN2O3¿É·¢ÉúÏÂÁз´Ó¦£º2N2O3+O24NO2(g)£»¡÷H>0£¬Ï±íΪ·´Ó¦ÔÚijζÈϵIJ¿·ÖʵÑéÊý¾Ý:
t/s
0
500
1000
c(N2O3)/mol¡¤L£­1
5.00
3.52
2.48
 
¼ÆËãÔÚt=500sʱ£¬NO2µÄ·´Ó¦ËÙÂÊΪ                     ¡£

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

¾«Ó¢¼Ò½ÌÍø