ÌâÄ¿ÄÚÈÝ

ÓÃÖкÍÈȵIJⶨ·½·¨£¬¼´±£Î¡¢¸ôÈȵÄÌõ¼þÏ£¬ÏòÊ¢ÓÐ20 mL¡¡2.08 mol/LµÄNaOHÈÜÒºµÄÊÔ¹ÜÖзÖÎå´Î¼ÓÈë5 mLδ֪Ũ¶ÈµÄH2SO4(±ß¼Ó±ßÕñµ´£¬Ã¿´Î¼Ó1 mL£¬(²»¿¼ÂÇH2SO4ÈÜÓÚË®µÄ·ÅÈÈЧӦ)ºó£¬²âµÃÈÜÒºµÄζȷֱðÊÇ1.4¡æ¡¢2.5¡æ¡¢4.2¡æ¡¢5.2¡æ¡¢5.18¡æ£¬Ôò¸ÃÁòËáµÄŨ¶ÈÔ¼ÊÇ

[¡¡¡¡]

A£®

20.8 mol/L

B£®

6.9 mol/L

C£®

5.2 mol/L

D£®

4.16 mol/L

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
£¨1£©ÏÂÁÐÓйØÊµÑé²Ù×÷»òÅжϲ»ÕýÈ·µÄÊÇ_____________£¨ÌîÐòºÅ£¬¶àÑ¡¿Û·Ö£©¡£

A.ÓÃ10 mLÁ¿Í²×¼È·Á¿È¡Ï¡ÁòËáÈÜÒº8.0 mL

B.ÓøÉÔïµÄpHÊÔÖ½²â¶¨ÂÈË®µÄpH

C.ÓüîʽµÎ¶¨¹ÜÁ¿È¡KMnO4ÈÜÒº19.60 mL

D.ʹÓÃÈÝÁ¿Æ¿ÅäÖÆÈÜҺʱ£¬¸©ÊÓÒºÃæ¶¨ÈݺóËùµÃÈÜÒºµÄŨ¶ÈÆ«´ó

E.½«±¥ºÍFeCl3ÈÜÒºµÎÈëÕôÁóË®Öм´µÃFe(OH)3½ºÌå

F.Ô²µ×ÉÕÆ¿¡¢×¶ÐÎÆ¿¡¢Õô·¢Ãó¼ÓÈÈʱ¶¼Ó¦µæÔÚʯÃÞÍøÉÏ

£¨2£©ÓÃ50 mL 0.5 mol¡¤L-1µÄÑÎËáÓë50 mL 0.55 mol¡¤L-1µÄNaOHÈÜÒº½øÐÐÖкÍÈȵIJⶨʵÑ飬Çë»Ø´ðÏÂÁÐÎÊÌ⣺

¢ÙÔÚʵÑéÖÐʹÓõÄÒÇÆ÷³ý´óÉÕ±­¡¢Ð¡ÉÕ±­¡¢»·Ðβ£Á§½Á°è°ô¡¢ÅÝÄ­ËÜÁÏ»òÖ½Ìõ¡¢ÅÝÄ­ËÜÁϰå»òÓ²Ö½°å£¨ÖÐÐÄÓÐÁ½¸öС¿×£©Í⣬»¹±ØÐëÒªÓõ½µÄÒÇÆ÷ÓÐ___________________________¡£

¢ÚÒªÏëÌá¸ßÖкÍÈȲⶨµÄ׼ȷÐÔ£¬¿É²ÉÓõĴëÊ©³ýÒÔÏÂÁгöµÄÈýÖÖÒÔÍ⣬ÇëÄãÔÙд³öÈýÖÖ£º

a.Èç¹û°´½Ì²ÄÖеķ½·¨×ö£¬Ò»¶¨ÒªÊ¹Ð¡ÉÕ±­±­¿ÚÓë´óÉÕ±­±­¿ÚÏàÆ½¡£

b.ÑÎËáºÍNaOHÈÜҺŨ¶ÈµÄÅäÖÆÒª×¼È·£¬ÇÒNaOHÈÜÒºµÄŨ¶ÈÐëÉÔ´óÓÚÑÎËáµÄŨ¶È¡£

c.ʵÑé²Ù×÷ʱ¶¯×÷Òª¿ì£¬×¢Òâ²»Òª½«ÈÜÒºÈ÷µ½ÍâÃæ¡£

d.__________________________________________________________¡£

e.__________________________________________________________¡£

f.__________________________________________________________¡£

¢ñ¡¢ÏÂÁÐʵÑé²Ù×÷»ò¶ÔʵÑéÊÂʵµÄÃèÊöÕýÈ·µÄÊÇ____________________

¢Ù ÓÃÁ¿Í²Á¿È¡Ï¡ÁòËáÈÜÒº8.0mL£»

¢ÚÖкÍÈȵIJⶨʵÑéÖУ¬¿ÉÓýðÊôË¿£¨°ô£©´úÌæ»·ÐνÁ°è²£Á§°ô£»

¢ÛÓÃÈȵÄŨÑÎËáÏ´µÓ¸½×ÅÓÐMnO2µÄÊԹܣ»

¢ÜÔÚÁòËáÍ­¾§Ìå½á¾§Ë®º¬Á¿µÄ²â¶¨ÖУ¬Èô¼ÓÈȺóµÄÎÞË®ÁòËáÍ­·ÛÄ©±íÃæ·¢ºÚ£¬ÔòËù²â½á¾§Ë®º¬Á¿¿ÉÄÜ»áÆ«¸ß £»

¢ÝFe(OH)3½ºÌåÓëFeCl3ÈÜÒº¿ÉÓùýÂ˵ķ½·¨·ÖÀ룻

¢ÞÓüîʽµÎ¶¨¹ÜÁ¿È¡KMnO4ÈÜÒº20.50mL £»

¢ß½«Ë®ÑØÉÕ±­ÄÚ±Ú»º»º×¢ÈëŨÁòËáÖУ¬²»¶ÏÓò£Á§°ô½Á°èÒÔÏ¡ÊÍŨÁòË᣻

¢àÓÃʪÈóµÄpHÊÔÖ½²âÁ¿Ä³ÈÜÒºpHʱ£¬²âÁ¿ÖµÒ»¶¨±ÈÕæÊµÖµÐ¡£»

¢áпºÍÒ»¶¨Á¿Ï¡ÁòËá·´Ó¦£¬Îª¼Ó¿ìËÙÂʶø²»Ó°ÏìH2µÄÁ¿¿ÉÏòÈÜÒºÖмÓÊÊÁ¿Cu(NO3)2¾§Ìå¡£

¢ò¡¢2013Äê³õ£¬Îíö²ÌìÆø¶à´ÎËÁŰÌì½ò¡¢±±¾©µÈµØÇø¡£ÆäÖУ¬È¼ÃººÍÆû³µÎ²ÆøÊÇÔì³É¿ÕÆøÎÛȾµÄÔ­ÒòÖ®Ò»¡£

£¨1£©Æû³µÎ²Æø¾»»¯µÄÖ÷ÒªÔ­ÀíΪ£º2NO(g) + 2CO(g)2CO2(g)+ N2(g)¡£¡÷H£¼0

¢Ù¸Ã·´Ó¦Æ½ºâ³£Êý±í´ïʽ                                       

¢ÚÈô¸Ã·´Ó¦ÔÚ¾øÈÈ¡¢ºãÈݵÄÃܱÕÌåϵÖнøÐУ¬ÏÂÁÐʾÒâͼÕýÈ·ÇÒÄÜ˵Ã÷·´Ó¦ÔÚ½øÐе½t1ʱ¿Ì´ïµ½Æ½ºâ״̬µÄÊÇ             £¨Ìî´úºÅ£©¡£

£¨2£©Ö±½ÓÅÅ·ÅúȼÉÕ²úÉúµÄÑÌÆø»áÒýÆðÑÏÖØµÄ»·¾³ÎÊÌâ¡£

úȼÉÕ²úÉúµÄÑÌÆøº¬µªµÄÑõ»¯ÎÓÃCH4´ß»¯»¹Ô­NOX¿ÉÒÔÏû³ýµªÑõ»¯ÎïµÄÎÛȾ¡£

ÒÑÖª£º¢ÙCH4(g)+2NO2(g)£½N2(g)£«CO2(g)+2H2O(g)¡¡ ¡÷H£½£­867 kJ/mol

¢Ú2NO2(g) N2O4(g)   ¡÷H£½£­56.9 kJ/mol

¢ÛH2O(g) £½ H2O(l)   ¦¤H £½ £­44.0 kJ£¯mol

д³öCH4´ß»¯»¹Ô­N2O4(g)Éú³ÉN2ºÍH2O(l)µÄÈÈ»¯Ñ§·½³Ìʽ£º                   ¡£

 

Çë°´ÒªÇóÌî¿Õ£º

£¨1£©ÏÂÁÐʵÑé²Ù×÷»ò¶ÔʵÑéÊÂʵµÄÐðÊöÖдæÔÚ´íÎóµÄÊÇ____________£¨ÌîÐòºÅ£©¡£

A.ÓÃ50 mLËáʽµÎ¶¨¹Ü׼ȷÁ¿È¡25.00 mL KMnO4ËáÐÔÈÜÒº£¬·ÅÈë×¶ÐÎÆ¿ÖдýÓÃ

B.ÔڲⶨÁòËáÍ­¾§ÌåµÄ½á¾§Ë®º¬Á¿Ê±£¬½«×ÆÉÕÁòËáÍ­¾§ÌåµÄÛáÛö·ÅÔÚ¿ÕÆøÖÐÀäÈ´£¬È»ºó³ÆÁ¿

C.ÖкÍÈȵIJⶨËùÐèµÄ²£Á§ÒÇÆ÷ÓÐÉÕ±­¡¢Î¶ȼơ¢Á¿Í²

D.ʵÑéÊÒÖнøÐеÄÏÂÁÐʵÑé¾ùÐèˮԡ¼ÓÈÈ£º

¢Ù±½µÄÏõ»¯·´Ó¦  ¢ÚÆÏÌÑÌǵÄÒø¾µ·´Ó¦  ¢ÛÒÒËáÒÒõ¥µÄÖÆ±¸  ¢Üµí·Û¡¢ÏËÎ¬ËØµÄË®½â

E.ÅäÖÆ1 mol¡¤L-1µÄNaOHÈÜҺʱ£¬ÏÂÁвÙ×÷¾ù»áÔì³ÉËùÅäŨ¶ÈÆ«µÍ£º¢ÙÈܽâºóÉÕ±­Î´¶à´ÎÏ´µÓ  ¢Ú¶¨ÈÝʱÑöÊӿ̶ÈÏß  ¢ÛÈÝÁ¿Æ¿ÖÐÔ­ÓÐÉÙÐíÕôÁóË®

F.ʵÑéÊÒÅäÖÆÂÈ»¯ÑÇÌúÈÜҺʱ£¬½«ÂÈ»¯ÑÇÌúÏÈÈܽâÔÚÑÎËáÖУ¬È»ºóÓÃÕôÁóˮϡÊͲ¢¼ÓÈëÉÙÁ¿Ìú·Û

£¨2£©Ä³Í¬Ñ§Éè¼ÆÁËÈçͼËùʾ²â¶¨ºìÁ×ÔÚÂÈÆøÖÐȼÉÕ²úÎïµÄ·Ö×ÓʽµÄ×°Ö㬸Ã×°ÖÃÖÐBÊǵ×Ãæ»ýΪ100 cm2µÄԲͲ״²£Á§ÈÝÆ÷£¨ÃÜ·â¸ÇÉÏ×°Óз§ÃÅ£©£¬ÉÏÃæ±êÓÐÀåÃ×µ¥Î»µÄ¿Ì¶È£¬ÆäËû¼Ð³Ö×°ÖÃÒÑÂÔÈ¥¡££¨²»¿¼ÂÇ¿ÉÄæ·´Ó¦£©

²Ù×÷²½Ö裺

¢Ù¼ì²é×°ÖÃµÄÆøÃÜÐÔ

¢Ú½«0.5 gºìÁׯ½ÆÌÔÚµçÈȰåÉÏ£¬¸Ç½ôÈÝÆ÷

¢Û´ò¿ª·§ÃÅ1¡¢·§ÃÅ2£¬´ÓAµÄÆ¿¿Ú¼ÓÈëÒºÌåC£¬Ê¹BÖÐÒºÃæÖÁ¿Ì¶È15.0 cm

¢ÜÏòBÖгäÈëÂÈÆø£¬´ýÂÈÆø³äÂúºó¹Ø±Õ·§ÃÅ1¡¢·§ÃÅ2£¬Í¨µç¼ÓÈȺìÁס­¡­

Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨i£©¼ì²é×°ÖÃÆøÃÜÐÔ¾ßÌå·½·¨ÊÇ____________________¡£

£¨ii£©ÊµÑé²½Öè¢Û¼ÓÈëµÄÒºÌåCÊÇ____________£¨ÌîÃû³Æ£©¡£

£¨iii£©·´Ó¦½áÊø²¢³ä·ÖÀäÈ´ºó£¬¶ÁÈ¡BÖÐÒºÃæ¿Ì¶È¶ÁÊýʱ£¬Ó¦×¢Òâ________________________£»ÈôÒºÃæÔڿ̶È5.6 cm´¦£¬´ËʱʵÑéÌõ¼þ½üËÆ¿´×÷±ê×¼×´¿ö£¬ÔòÉú³ÉÎïPClxÖÐxµÄÖµÊÇ____________£¨¼ÆËã½á¹û±£Áô1λСÊý£©¡£

  ÏÂÁÐʵÑé²Ù×÷»ò¶ÔʵÑéÊÂʵµÄÃèÊöÕýÈ·µÄÊÇ____________________

¢Ù ÓÃÁ¿Í²Á¿È¡Ï¡ÁòËáÈÜÒº8.0mL£»

¢ÚÖкÍÈȵIJⶨʵÑéÖУ¬¿ÉÓýðÊôË¿£¨°ô£©´úÌæ»·ÐνÁ°è²£Á§°ô£»

¢ÛÓÃÈȵÄŨÑÎËáÏ´µÓ¸½×ÅÓÐMnO2µÄÊԹܣ»

¢ÜÔÚÁòËáÍ­¾§Ìå½á¾§Ë®º¬Á¿µÄ²â¶¨ÖУ¬Èô¼ÓÈȺóµÄÎÞË®ÁòËáÍ­·ÛÄ©±íÃæ·¢ºÚ£¬ÔòËù²â½á¾§Ë®º¬Á¿¿ÉÄÜ»áÆ«¸ß £»

¢ÝFe(OH)3½ºÌåÓëFeCl3ÈÜÒº¿ÉÓùýÂ˵ķ½·¨·ÖÀ룻

¢ÞÓüîʽµÎ¶¨¹ÜÁ¿µÃKMnO4ÈÜÒº20.50mL £»

¢ß½«Ë®ÑØÉÕ±­ÄÚ±Ú»º»º×¢ÈëŨÁòËáÖУ¬²»¶ÏÓò£Á§°ô½Á°èÒÔÏ¡ÊÍŨÁòË᣻

¢àÓÃʪÈóµÄpHÊÔÖ½²âÁ¿Ä³ÈÜÒºpHʱ£¬²âÁ¿ÖµÒ»¶¨±ÈÕæÊµÖµÐ¡£»

¢áпºÍÒ»¶¨Á¿Ï¡ÁòËá·´Ó¦£¬Îª¼Ó¿ìËÙÂʶø²»Ó°ÏìH2µÄÁ¿¿ÉÏòÈÜÒºÖмÓÊÊÁ¿Cu(NO3)2¾§Ìå¡£

¢ò¡¢£¨5·Ö£©2013Äê³õ£¬Îíö²ÌìÆø¶à´ÎËÁŰÌì½ò¡¢±±¾©µÈµØÇø¡£ÆäÖУ¬È¼ÃººÍÆû³µÎ²ÆøÊÇÔì³É¿ÕÆøÎÛȾµÄÔ­ÒòÖ®Ò»¡£

£¨1£©Æû³µÎ²Æø¾»»¯µÄÖ÷ÒªÔ­ÀíΪ£º2NO(g) + 2CO(g)2CO2(g)+ N2(g)¡£¡÷H£¼0

¢Ù¸Ã·´Ó¦Æ½ºâ³£Êý±í´ïʽ                                        

¢ÚÈô¸Ã·´Ó¦ÔÚ¾øÈÈ¡¢ºãÈݵÄÃܱÕÌåϵÖнøÐУ¬ÏÂÁÐʾÒâͼÕýÈ·ÇÒÄÜ˵Ã÷·´Ó¦ÔÚ½øÐе½

t1ʱ¿Ì´ïµ½Æ½ºâ״̬µÄÊÇ             £¨Ìî´úºÅ£©¡£

£¨2£©Ö±½ÓÅÅ·ÅúȼÉÕ²úÉúµÄÑÌÆø»áÒýÆðÑÏÖØµÄ»·¾³ÎÊÌâ¡£

úȼÉÕ²úÉúµÄÑÌÆøº¬µªµÄÑõ»¯ÎÓÃCH4´ß»¯»¹Ô­NOX¿ÉÒÔÏû³ýµªÑõ»¯ÎïµÄÎÛȾ¡£

ÒÑÖª£ºCH4(g)+2NO2(g)£½N2(g)£«CO2(g)+2H2O(g)¡¡ ¡÷H£½£­867 kJ/mol

2NO2(g) N2O4(g)   ¡÷H£½£­56.9 kJ/mol

 H2O(g) £½ H2O(l)   ¦¤H £½ £­44.0 kJ£¯mol

д³öCH4´ß»¯»¹Ô­N2O4(g)Éú³ÉN2ºÍH2O(l)µÄÈÈ»¯Ñ§·½³Ìʽ£º

                                                                     ¡£

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

¾«Ó¢¼Ò½ÌÍø