ÔÚ20¡æʱ£¬½«12.1g¾§ÌåÈÜÓÚ44.3gË®ÖУ¬»òÔÚÏàͬζÈϽ«4.7gÎÞË®ÑÎÈÜÓÚ23.5gË®ÖУ¬¾ù¿ÉÖƵøÃÑεı¥ºÍÈÜÒº£®½«ÉÏÊöÁ½²¿·ÖÈÜÒº»ìºÏ£¬ÔÙ¼ÓˮϡÊÍΪ0.1mol/LµÄÈÜÒº£¬ÓöèÐԵ缫µç½â£¬µ±µç¼«ÉÏͨ¹ý0.15molµç×ÓµçÁ¿Ê±£¬ÈÜÒºÖеÄÇ¡ºÃÈ«²¿ÒÔµ¥ÖÊÎö³ö(ÉèÎÞÉú³É)£®Çó£º

(1)20¡æʱµÄÈܽâ¶È£®

(2)µÄĦ¶ûÖÊÁ¿¼°xµÄÖµ£®

(3)MµÄÏà¶ÔÔ­×ÓÖÊÁ¿£®

´ð°¸£º
½âÎö£º

´ð°¸£º(1)20g;(2)242g/mol£¬;(3)64

½âÎö£º(1)ÉèÈܽâ¶ÈΪS£¬Ôò(g)

(2)Éè12.1g¾§ÌåÖк¬Ë®a g£¬Ôòg£¬¼´º¬Ë®0.15mol£¬Ôò12.1g¾§ÌåΪ0.05mol£¬µÄĦ¶ûÖÊÁ¿Îª£®»ìºÏºóÈÜÒºÖк¬Îª£¬

ÓɵÃ

¡¡ 1¡¡¡¡

0.075¡¡0.15

(3)ÉèMµÄÏà¶ÔÔ­×ÓÖÊÁ¿Îªb£¬Ôò£¬¼´Ô­¾§ÌåΪ£®


Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÔÚ100¡æʱ£¬½«0.1molµÄËÄÑõ»¯¶þµªÆøÌå³äÈë1L³é¿ÕµÄÃܱÕÈÝÆ÷ÖУ¬¸ôÒ»¶¨Ê±¼ä¶Ô¸ÃÈÝÆ÷ÄÚµÄÎïÖʽøÐзÖÎö£¬µÃµ½ÈçÏÂÊý¾Ý£º
Ũ¶È     Ê±¼ä£¨S£© 0 20 40 60 80 100
C£¨N2O4£©/mol?L-1 0.1 c1 0.05 C3 a b
C£¨NO2£©/mol?L-1 0 0.06 C2 0.12 0.12 0.12
£¨1£©¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ
N2O4=2NO2
N2O4=2NO2
£¬´ïµ½Æ½ºâʱËÄÑõ»¯¶þµªµÄת»¯ÂÊΪ
60
60
%£¬±íÖÐC2
£¾
£¾
C3
=
=
a
=
=
b£¨Ñ¡Ìî¡°£¾¡±¡¢¡°£¼¡±¡¢¡°¨T¡±£©£®
£¨2£©20sʱËÄÑõ»¯¶þµªµÄŨ¶ÈC1=
0.07
0.07
mol?l-1£¬ÔÚ0s¡«20sÄÚËÄÑõ»¯¶þµªµÄƽ¾ù·´Ó¦ËÙÂÊΪ
0.0015
0.0015
mol?£¨L?s£©-1£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÔÚ80¡æʱ£¬½«0.4molµÄËÄÑõ»¯¶þµªÆøÌå³äÈë2LÒѳé¿ÕµÄ¹Ì¶¨ÈÝ»ýµÄÃܱÕÈÝÆ÷ÖУ¬¸ôÒ»¶Îʱ¼ä¶Ô¸ÃÈÝÆ÷ÄÚµÄÎïÖʽøÐзÖÎö£¬µÃµ½ÈçÏÂÊý¾Ý£º·´Ó¦½øÐÐÖÁ100sºó½«·´Ó¦»ìºÏÎïµÄζȽµµÍ£¬·¢ÏÖÆøÌåµÄÑÕÉ«±ädz£®
ʱ¼ä£¨s£©
C£¨mol/L£©

0

20

40

60

80

100
C£¨N2O4£© 0.20 a 0.10 c d e
C£¨NO2£© 0.00 0.12 b 0.22 0.22 0.22
£¨1£©¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ
N2O4?2NO2
N2O4?2NO2
£¬±íÖÐb
£¾
£¾
c£¨Ìî¡°£¼¡±¡¢¡°=¡±¡¢¡°£¾¡±£©£®
£¨2£©20sʱ£¬N2O4µÄŨ¶ÈΪ
0.14
0.14
mol/L£¬0¡«20sÄÚN2O4µÄƽ¾ù·´Ó¦ËÙÂÊΪ
0.003mol/L?s
0.003mol/L?s
£®
£¨3£©¸Ã·´Ó¦µÄƽºâ³£Êý±í´ïʽK=
c2(NO2)
c(N2O4)
c2(NO2)
c(N2O4)
£¬ÔÚ80¡æʱ¸Ã·´Ó¦µÄƽºâ³£ÊýKֵΪ
0.54
0.54
£¨±£Áô2λСÊý£©£®
£¨4£©ÔÚÆäËûÌõ¼þÏàͬʱ£¬¸Ã·´Ó¦µÄKÖµÔ½´ó£¬±íÃ÷½¨Á¢Æ½ºâʱ
ABD
ABD
£®A¡¢N2O4µÄת»¯ÂÊÔ½¸ß                 B¡¢NO2µÄ²úÁ¿Ô½´óC¡¢N2O4ÓëNO2µÄŨ¶ÈÖ®±ÈÔ½´ó          D¡¢Õý·´Ó¦½øÐеij̶ÈÔ½´ó£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

µªÊÇ´óÆøÖк¬Á¿·á¸»µÄÒ»ÖÖÔªËØ£¬µª¼°Æ仯ºÏÎïÔÚÉú²ú¡¢Éú»°ÖÐÓÐ×ÅÖØÒª×÷Ó㬼õÉÙµªÑõ»¯ÎïµÄÅÅ·ÅÊÇ»·¾³±£»¤µÄÖØÒªÄÚÈÝÖ®Ò»£®Çë»Ø´ðÏÂÁеª¼°Æ仯ºÏÎïµÄÏà¹ØÎÊÌ⣺
£¨1£©¾Ý±¨µÀ£¬Òâ´óÀû¿Æѧ¼Ò»ñµÃÁ˼«¾ßÑо¿¼ÛÖµµÄN4£¬Æä·Ö×ӽṹÓë°×Á×·Ö×ÓµÄÕýËÄÃæÌå½á¹¹ÏàËÆ£®ÒÑÖª¶ÏÁÑ1molN-N¼üÎüÊÕ167kJÈÈÁ¿£¬Éú³É1molN¡ÔN¼ü·Å³ö942kJÈÈÁ¿£¬Çëд³öN4ÆøÌåת±äΪN2·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ£º
N4£¨g£©=2N2£¨g£©¡÷H=-882KJ/mol
N4£¨g£©=2N2£¨g£©¡÷H=-882KJ/mol
£®
£¨2£©ÒÑÖªNH3ÔÚ´¿ÑõÆøÖÐȼÉÕ¿ÉÉú³ÉN2ºÍË®£®¾Ý±¨µÀ£¬NH3¿ÉÖ±½ÓÓÃ×÷³µÓüîÐÔȼÁϵç³Ø£¬Ð´³ö¸Ãµç³ØµÄ¸º¼«·´Ó¦Ê½£º
2NH3+6OH--6e-=N2+6H2O
2NH3+6OH--6e-=N2+6H2O
£®
£¨3£©ÔÚT1¡æʱ£¬½«5mol N2O5ÖÃÓÚ10L¹Ì¶¨ÈÝ»ýµÄÃܱÕÈÝÆ÷Öз¢ÉúÏÂÁз´Ó¦£º2N2O5£¨g£©?4NO2£¨g£©+O2£¨g£©£»¡÷H£¾0£®·´Ó¦ÖÁ5·ÖÖÓʱ¸÷ÎïÖʵÄŨ¶È²»ÔÙ·¢Éú±ä»¯£¬²âµÃNO2µÄÌå»ý·ÖÊýΪ50%£®
¢ÙÉÏÊöƽºâÌåϵÖÐO2µÄÌå»ý·ÖÊýΪ
12.5%
12.5%
£¬N2O5µÄÎïÖʵÄÁ¿Îª
3
3
mol£®
¢ÚÓÃO2±íʾ´Ó0¡«5minÄڸ÷´Ó¦µÄƽ¾ùËÙÂÊv£¨O2£©=
0.02mol/£¨L?min£©
0.02mol/£¨L?min£©
£®
¢Û¸ÃζÈϵÄƽºâ³£ÊýK=
0.1¡Á(0.4)4
(0.3)2
0.1¡Á(0.4)4
(0.3)2
 £¨´úÈëÊý×Ö±í´ïʽ£¬²»±Ø¼ÆË㻯¼ò£©
¢Ü½«ÉÏÊöƽºâÌåϵµÄζȽµÖÁT2¡æ£¬ÃܱÕÈÝÆ÷ÄÚ¼õСµÄÎïÀíÁ¿ÓÐ
AC
AC
£®
A£®Ñ¹Ç¿    B£®ÃܶȠ   C£®·´Ó¦ËÙÂÊD£®N2O5µÄŨ¶È£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÔÚ80¡æʱ£¬½«0.4molµÄËÄÑõ»¯¶þµªÆøÌå³äÈë2LÒѳé¿ÕµÄ¹Ì¶¨ÈÝ»ýµÄÃܱÕÈÝÆ÷ÖУ¬¸ôÒ»¶Îʱ¼ä¶Ô¸ÃÈÝÆ÷ÄÚµÄÎïÖʽøÐзÖÎö£¬µÃµ½ÈçÏÂÊý¾Ý£º
C£¨mol/L£©Ê±¼ä£¨s£© O 20 40 60 80 100
C£¨N2O4£© 0.20 a 0.10 c d e
C£¨NO2£© 0.00 0.12 b 0.22 0.22 0.22
·´Ó¦½øÐÐÖÁ100sºó½«·´Ó¦»ìºÏÎïµÄζȽµµÍ£¬·¢ÏÖÆøÌåµÄÑÕÉ«±ädz£®
£¨1£©±íÖÐb
£¾
£¾
c£¨Ìî¡°£¼¡±¡¢¡°=¡±¡¢¡°£¾¡±£©£®
£¨2£©20sʱ£¬N2O4µÄŨ¶ÈΪ
0.14
0.14
mol/L£¬0--20sÄÚN2O4µÄƽ¾ù·´Ó¦ËÙÂÊΪ
0.003mol/£¨L£®s£©
0.003mol/£¨L£®s£©
£®
£¨3£©ÔÚ80¡æʱ¸Ã·´Ó¦µÄƽºâ³£ÊýK=
c2(NO2)
c(N2O4)
c2(NO2)
c(N2O4)
£¨Ð´±í´ïʽ£©£®
£¨4£©ÔÚÆäËûÌõ¼þÏàͬʱ£¬¸Ã·´Ó¦µÄKÖµÔ½´ó£¬±íÃ÷½¨Á¢Æ½ºâʱ
ABD
ABD
£®
A¡¢N2O4µÄת»¯ÂÊÔ½¸ß                 B¡¢NO2µÄ²úÁ¿Ô½´ó
C¡¢N2O4ÓëNO2µÄŨ¶ÈÖ®±ÈÔ½´ó          D¡¢Õý·´Ó¦½øÐеij̶ÈÔ½´ó£®

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸