¹¤Òµ²ÉÈ¡°±Ñõ»¯·¨À´ÖÆÈ¡ÏõËᣬÖ÷Òª·´Ó¦Îª£º

4NH3£«5O24NO£«6H2O£»4NO£«3O2£«2H2O4HNO3

Çë»Ø´ðÏÂÁÐÎÊÌ⣺

(1)

Éè¿ÕÆøÖÐÑõÆøµÄÌå»ý·ÖÊýΪ0.20£¬µªÆøÌå»ý·ÖÊýΪ0.80£¬ÎªÊ¹NH3Ç¡ºÃÍêÈ«Ñõ»¯ÎªÒ»Ñõ»¯µª£¬°±Ò»¿ÕÆø»ìºÏÎïÖа±µÄÌå»ý·ÖÊýΪ________(±£Áô2λСÊý)£®

(2)

Èô½«20.0molµÄNH3ºÍÒ»¶¨Á¿´¿¾»µÄÑõÆø³ä·Ö·´Ó¦ºó£¬ÔÙת»¯ÎªHNO3£®

¢ÙÔÚÓÒͼÖл­³öÉú³ÉHNO3µÄÎïÖʵÄÁ¿n(HNO3)ºÍ·´Ó¦ÏûºÄµÄÑõÆøµÄÎïÖʵÄÁ¿n(O2)¹ØϵµÄÀíÂÛÇúÏߣ®

¢Úд³öµ±25.0mol¡Ün(C2)¡Ü40.0molʱ£¬n(HNO3)ºÍn(O2)µÄ¹Øϵʽ£º________£®

´ð°¸£º1£®0.14;
½âÎö£º

(1)

[½âÌâ˼·]

Ïȹ̶¨NH3µÄÌå»ý£¬ÔÙ±íʾ³öËùÐèO2µÄÌå»ý£¬ÔÙÓÿÕÆøÖÐO2µÄÌå»ý·ÖÊýת»¯³ÉËùÐè¿ÕÆøµÄÌå»ý£¬×îºóÇó³ö°±—¿ÕÆ
Ìáʾ£º

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÏõËáÊÇÒ»ÖÖÖØÒªµÄ»¯¹¤Ô­ÁÏ£¬¹¤ÒµÉÏͨ³£²ÉÓð±Ñõ»¯·¨ÖÆÈ¡¡£Ä³Ð£»¯Ñ§ÐËȤС×éµÄͬѧÒÔÂÈ»¯ï§ºÍÇâÑõ»¯¸ÆΪÖ÷ÒªÔ­Áϲ¢Éè¼ÆÁËÏÂÁÐ×°ÖÃÀ´ÖÆÏõËᣨÈýÑõ»¯¶þ¸õΪ´ß»¯¼Á£¬¼ÓÈȼ°¼Ð³Ö×°ÖÃδ»­³ö£©£º

Íê³ÉÏÂÁÐÎÊÌ⣺

£¨1£©ÊµÑéʱ£¬A¡¢CÁ½×°ÖþùÐè¼ÓÈÈ£¬Ó¦ÏȼÓÈÈ__________×°Öã¬Ô­ÒòÊÇ_______________£»

£¨2£©D×°ÖÃÖÐÓ¦Ìî³äµÄÎïÖÊÊÇ__________£¬¸ÃÎïÖʵÄÖ÷Òª×÷ÓÃÊÇ______________________£»

£¨3£©E×°ÖõÄ×÷ÓÃÊÇ__________£¬F¡¢G×°ÖÃÖеÄÎïÖÊ·Ö±ðÊÇ_____________£¬__________£»

£¨4£©Èô±£ÁôÉÏͼÖкÚÉ«´ÖÏß¿òÄÚµÄ×°Öõ«È¥µôͨ¿ÕÆøµÄµ¼¹ÜB£¬½«C×°ÖÃÖеÄË«¿×ÏðƤÈû»»³Éµ¥¿×ÏðƤÈû£¬ÇëÄãÓÃͼʾµÄ·½·¨Éè¼ÆÒ»¸ö×î¼òµ¥µÄʵÑé·½°¸Í¬ÑùÍê³ÉÏõËáµÄÖÆÈ¡£¨ÔÚÏÂÃæµÄ·½¿òÖл­³ö×°ÖÃͼ²¢×¢Ã÷ËùÓÃÒ©Æ·µÄÃû³Æ£©¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÏõËáÊÇÒ»ÖÖÖØÒªµÄ»¯¹¤Ô­ÁÏ£¬¹¤ÒµÉÏͨ³£²ÉÓð±Ñõ»¯·¨ÖÆÈ¡¡£Ä³Ð£»¯Ñ§ÐËȤС×éµÄͬѧÒÔÂÈ»¯ï§ºÍÇâÑõ»¯¸ÆΪÖ÷ÒªÔ­Áϲ¢Éè¼ÆÁËÏÂÁÐ×°ÖÃÀ´ÖÆÏõËá(ÈýÑõ»¯¶þ¸õΪ´ß»¯¼Á£¬¼ÓÈȼ°¼Ð³Ö×°ÖÃδ»­³ö)£º

Íê³ÉÏÂÁÐÎÊÌ⣺

(1)ʵÑéʱ£¬A¡¢CÁ½×°ÖþùÐè¼ÓÈÈ£¬Ó¦ÏȼÓÈÈ_________×°Öã¬Ô­ÒòÊÇ______________________¡£

(2)D×°ÖÃÖÐÓ¦Ìî³äµÄÎïÖÊÊÇ_________£¬¸ÃÎïÖʵÄÖ÷Òª×÷ÓÃÊÇ__________________________¡£

(3)E×°ÖõÄ×÷ÓÃÊÇ__________________£¬F¡¢G×°ÖÃÖеÄÎïÖÊ·Ö±ðÊÇ____________________¡£

(4)Èô±£ÁôÉÏͼÖкÚÉ«´ÖÏß¿òÄÚµÄ×°Ö㬵«È¥µôͨ¿ÕÆøµÄµ¼¹ÜB£¬½«C×°ÖÃÖеÄË«¿×ÏðƤÈû»»³Éµ¥¿×ÏðƤÈû£¬ÇëÄãÓÃͼʾµÄ·½·¨Éè¼ÆÒ»¸ö×î¼òµ¥µÄʵÑé·½°¸Í¬ÑùÍê³ÉÏõËáµÄÖÆÈ¡(ÔÚÏÂÃæµÄ·½¿òÖл­³ö×°ÖÃͼ²¢×¢Ã÷ËùÓÃÒ©Æ·µÄÃû³Æ)¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â

(¢ñ)µÂ¹úÈ˹þ²®ÔÚ1905Äê·¢Ã÷µÄºÏ³É°±·´Ó¦Ô­ÀíΪ:N2(g)+3H2(g) 2NH3(g);ÒÑÖª298 Kʱ,¦¤H=-92.4 kJ¡¤mol-1,¦¤S=-198.2 J¡¤mol-1¡¤K-1.ÊԻشðÏÂÁÐÎÊÌâ:

(1)Çë¸ù¾ÝÕý·´Ó¦µÄìʱäºÍìرä¼ÆËã·ÖÎö298 KϺϳɰ±·´Ó¦ÄÜ×Ô·¢½øÐÐ(ÁгöËãʽ¼´¿É)___________________.ÆäŨ¶ÈìØ(Qc) __________________»¯Ñ§Æ½ºâ³£Êý(Kc)(Ìî¡°´óÓÚ¡±¡°µÈÓÚ¡±»ò¡°Ð¡ÓÚ¡±£©Ê±£¬·´Ó¦ÏòÓÒ½øÐÐ.

(2)ÔÚʵ¼Ê¹¤ÒµºÏ³É°±Éú²úÖвÉÈ¡µÄ´ëÊ©ÊÇ__________________ (ÌîÐòºÅ).

A.²ÉÓýϵÍѹǿ

B.²ÉÓÃ800 K×óÓҵĸßÎÂ

C.ÓÃÌú´¥Ã½×÷´ß»¯¼Á

D.½«Éú³ÉµÄ°±Òº»¯²¢¼°Ê±´ÓÌåϵÖзÖÀë³öÀ´,N2ºÍH2Ñ­»·µ½ºÏ³ÉËþÖв¢²¹³äN2ºÍH2

(3)ÈçÏÂͼËùʾÊÇʵÑéÊÒÄ£Ä⹤ҵ·¨ºÏ³É°±µÄ¼òÒ××°ÖÃ,¼òÊö¼ìÑéÓа±ÆøÉú³ÉµÄ·½·¨

___________________________________________________________________________

___________________________________________________________________________.

 (¢ò)ÏõËáÊÇÒ»ÖÖÖØÒªµÄ»¯¹¤Ô­ÁÏ,¹¤ÒµÉÏͨ³£²ÉÓð±Ñõ»¯·¨ÖÆÈ¡.ijУ»¯Ñ§ÐËȤС×éµÄͬѧÉè¼ÆÁËÈçÏÂͼËùʾװÖÃÀûÓÃÖÐѧʵÑéÊÒ³£¼ûÊÔ¼ÁÖÆÈ¡NH3,²¢ÒÔ¿ÕÆø,NH3ΪԭÁÏÄ£Ä⹤ҵÖÆHNO3(ÈýÑõ»¯¶þ¸õΪ´ß»¯¼Á,¼ÓÈȼ°¼Ð³Ö×°ÖÃδ»­³ö):

»Ø´ðÏÂÁÐÎÊÌâ:

(1)ʵÑéʱ,A,CÁ½×°ÖþùÐè¼ÓÈÈ,Ó¦ÏȼÓÈÈ______________×°ÖÃ,Ô­ÒòÊÇ___________________;

(2)D×°ÖÃÖÐÓ¦Ìî³äµÄÎïÖÊÊÇ______________,¸ÃÎïÖʵÄÖ÷Òª×÷ÓÃÊÇ______________;

(3)E×°ÖõÄ×÷ÓÃÊÇ______________,F,G×°ÖÃÖеÄÎïÖÊ·Ö±ðÊÇ______________¡¢______________;

(4)Èô±£ÁôÉÏͼÖкÚÉ«´ÖÏß¿òÄÚµÄ×°Öõ«È¥µôͨ¿ÕÆøµÄµ¼¹ÜB,½«C×°ÖÃÖеÄË«¿×ÏðƤÈû»»³Éµ¥¿×ÏðƤÈû,ÇëÄãÓÃͼʾµÄ·½·¨Éè¼ÆÒ»¸ö×î¼òµ¥µÄʵÑé·½°¸Í¬ÑùÍê³ÉÏõËáµÄÖÆÈ¡(ÔÚÈçÏÂͼËùʾµÄ·½¿òÖл­³ö×°ÖÃͼ²¢×¢Ã÷ËùÓÃÒ©Æ·µÄÃû³Æ).

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2012Ä곿ưæ¸ßÖл¯Ñ§Ñ¡ÐÞ6 4.2 ʵÑéÊÒÖƱ¸»¯¹¤Ô­ÁÏÁ·Ï°¾í£¨½âÎö°æ£© ÌâÐÍ£ºÌî¿ÕÌâ

ÏõËáÊÇÒ»ÖÖÖØÒªµÄ»¯¹¤Ô­ÁÏ£¬¹¤ÒµÉÏͨ³£²ÉÓð±Ñõ»¯·¨ÖÆÈ¡¡£Ä³Ð£»¯Ñ§ÐËȤС×éµÄͬѧÒÔÂÈ»¯ï§ºÍÇâÑõ»¯¸ÆΪÖ÷ÒªÔ­Áϲ¢Éè¼ÆÁËÏÂÁÐ×°ÖÃÀ´ÖÆÏõËá(ÈýÑõ»¯¶þ¸õΪ´ß»¯¼Á£¬¼ÓÈȼ°¼Ð³Ö×°ÖÃδ»­³ö)£º

»Ø´ðÏÂÁÐÎÊÌ⣺

(1)ʵÑéʱ£¬A¡¢CÁ½×°ÖþùÐè¼ÓÈÈ£¬Ó¦ÏȼÓÈÈ________×°Öã¬Ô­ÒòÊÇ

________________________________________________________________________¡£

(2)D×°ÖÃÖÐÓ¦Ìî³äµÄÎïÖÊÊÇ________£¬¸ÃÎïÖʵÄÖ÷Òª×÷ÓÃÊÇ____________________¡£

(3)E×°ÖõÄ×÷ÓÃÊÇ__________________£¬F¡¢G×°ÖÃÖеÄÎïÖÊ·Ö±ðÊÇ________¡¢________¡£

(4)Èô±£ÁôÉÏͼÖкÚÉ«´ÖÏß¿òÄÚµÄ×°Öõ«È¥µôͨ¿ÕÆøµÄµ¼¹ÜB£¬½«C×°ÖÃÖеÄË«¿×ÏðƤÈû»»³Éµ¥¿×ÏðƤÈû£¬ÇëÄãÓÃͼʾµÄ·½·¨Éè¼ÆÒ»¸ö×î¼òµ¥µÄʵÑé·½°¸Í¬ÑùÍê³ÉÏõËáµÄÖÆÈ¡(ÔÚÏÂÃæµÄ·½¿òÖл­³ö×°ÖÃͼ²¢×¢Ã÷ËùÓÃÒ©Æ·µÄÃû³Æ)¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸