¡¾ÌâÄ¿¡¿Æû³µÎ²ÆøÖÐÅŷŵÄNOxºÍCOÎÛȾ»·¾³£¬ÔÚÆû³µÎ²ÆøϵͳÖÐ×°Öô߻¯×ª»¯Æ÷£¬¿ÉÓÐЧ½µµÍNOxºÍCOµÄÅÅ·Å¡£

ÒÑÖª£º¢Ù2CO(g)£«O2(g) 2CO2(g) ¦¤H£½566.0 kJ¡¤mol1

¢ÚN2(g)£«O2(g) 2NO(g) ¦¤H£½+180.5 k J¡¤mol1

¢Û2NO(g)£«O2(g) 2NO2(g) ¦¤H£½116.5 k J¡¤mol1

»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©COµÄȼÉÕÈÈΪ_________¡£Èô1 mol N2(g)¡¢1 mol O2(g) ·Ö×ÓÖл¯Ñ§¼ü¶ÏÁÑʱ·Ö±ðÐèÒªÎüÊÕ946 kJ¡¢498 kJµÄÄÜÁ¿£¬Ôò1 mol NO(g) ·Ö×ÓÖл¯Ñ§¼ü¶ÏÁÑʱÐèÎüÊÕµÄÄÜÁ¿Îª___________kJ¡£

£¨2£©CO½«NO2»¹Ô­Îªµ¥ÖʵÄÈÈ»¯Ñ§·½³ÌʽΪ_______¡£

£¨3£©ÎªÁËÄ£Äâ·´Ó¦2NO(g)£«2CO(g) N2(g)£«2CO2(g)ÔÚ´ß»¯×ª»¯Æ÷ÄڵŤ×÷Çé¿ö£¬¿ØÖÆÒ»¶¨Ìõ¼þ£¬È÷´Ó¦ÔÚºãÈÝÃܱÕÈÝÆ÷ÖнøÐУ¬Óô«¸ÐÆ÷²âµÃ²»Í¬Ê±¼äNOºÍCOµÄŨ¶ÈÈçÏÂ±í£º

ʱ¼ä/s

0

1

2

3

4

5

c(NO)/(10-4mol/L)

10.0

4.50

2.50

1.50

1.00

1.00

c(CO)/(10-3mol/L)

3.60

3.05

2.28

2.75

2.70

2.70

¢ÙÇ°2 sÄÚµÄƽ¾ù·´Ó¦ËÙÂÊv(N2)£½___________,´ËζÈÏ£¬¸Ã·´Ó¦µÄƽºâ³£ÊýK£½________¡£

¢ÚÄÜ˵Ã÷ÉÏÊö·´Ó¦´ïµ½Æ½ºâ״̬µÄÊÇ_________¡£

A£®2n(CO2)£½n(N2) B£®»ìºÏÆøÌåµÄƽ¾ùÏà¶Ô·Ö×ÓÖÊÁ¿²»±ä

C£®ÆøÌåÃܶȲ»±ä D£®ÈÝÆ÷ÄÚÆøÌåѹǿ²»±ä

¢Ûµ±NOÓëCOŨ¶ÈÏàµÈʱ£¬ÌåϵÖÐNOµÄƽºâת»¯ÂÊÓëζȡ¢Ñ¹Ç¿µÄ¹ØϵÈçÓÒͼËùʾ£¬ÔòNOµÄƽºâת»¯ÂÊËæζÈÉý¸ß¶ø¼õСµÄÔ­ÒòÊÇ___________ £¬Í¼ÖÐѹǿ£¨p1£¬p2¡¢p3£©µÄ´óС˳ÐòΪ_________ ¡£

¡¾´ð°¸¡¿ 283KJ/mol 631.75 2NO2(g)+4CO(g)=N2(g)+4CO(g) ¦¤H=-1196kJ/mol 1.875¡Á10-4mol/(L¡¤s) 5000£¨»ò5000L/mol£© BD ¸Ã·´Ó¦µÄÕý·´Ó¦·ÅÈÈ£¬Éý¸ßζȣ¬Æ½ºâÄæÏòÒƶ¯£¬NOµÄת»¯ÂʼõС£¨»òÕý·´Ó¦·ÅÈÈ£¬Î¶ÈÔ½¸ß£¬Ô½²»ÀûÓÚ·´Ó¦ÕýÏò½øÐУ¬NOµÄƽºâת»¯ÂÊԽС£© p1>p2>p3

¡¾½âÎö¡¿£¨1£©ÓÉ¢Ù2CO(g)£«O2(g) 2CO2(g) ¦¤H£½566.0 kJ¡¤mol1¿ÉÖª£¬ 2molCOÍêȫȼÉշųö566.0 kJµÄÈÈÁ¿£¬ËùÒÔ1molCOÍêȫȼÉշųö283kJµÄÈÈÁ¿£¬ËùÒÔCOµÄȼÉÕÈÈΪ283KJ/mol¡£Èô1 mol N2(g)¡¢1 mol O2(g) ·Ö×ÓÖл¯Ñ§¼ü¶ÏÁÑʱ·Ö±ðÐèÒªÎüÊÕ946 kJ¡¢498 kJµÄÄÜÁ¿£¬Éè1 mol NO(g) ·Ö×ÓÖл¯Ñ§¼ü¶ÏÁÑʱÐèÎüÊÕµÄÄÜÁ¿Îªx£¬ÓÉ¢ÚN2(g)£«O2(g) 2NO(g) ¦¤H£½946k +498k -2x=+180.5£¬½âÖ®µÃx=631.75£¬ËùÒÔ1 mol NO(g) ·Ö×ÓÖл¯Ñ§¼ü¶ÏÁÑʱÐèÎüÊÕµÄÄÜÁ¿Îª631.75kJ¡£

£¨2£©ÓÉ¢Ù2CO(g)£«O2(g) 2CO2(g) ¦¤H£½566.0 kJ¡¤mol1£¬¢ÚN2(g)£«O2(g) 2NO(g) ¦¤H£½=+180.5kJ/mol£¬¢Û2NO(g)£«O2(g) 2NO2(g) ¦¤H£½116.5 kJ¡¤mol1£¬¢Ù2-¢Ú-¢Û¿ÉµÃ£º2NO2(g)+4CO(g)=N2(g)+4CO(g) ¦¤H=(566.0 kJ¡¤mol1)-(+180.5kJ/mol)-( 116.5 k J¡¤mol1)=-1196kJ/mol £¬ËùÒÔCO½«NO2»¹Ô­Îªµ¥ÖʵÄÈÈ»¯Ñ§·½³ÌʽΪ2NO2(g)+4CO(g)=N2(g)+4CO(g) ¦¤H=-1196kJ/mol¡£

£¨3£©¢ÙÓɱíÖÐÊý¾Ý¿ÉÖª£¬Ç°2 sÄÚ£¬NOµÄ±ä»¯Á¿Îª(10.0-2.50)=7.50mol/L£¬ÓÉNÔ­×ÓÊغã¿ÉµÃN2µÄ±ä»¯Á¿Îª3.75mol/L£¬ËùÒÔÇ°2 sÄÚƽ¾ù·´Ó¦ËÙÂÊv(N2)£½1.875¡Á10-4mol/(L¡¤s),´ËζÈÏ£¬·´Ó¦ÔÚµÚ4s´ïƽºâ״̬ £¬¸÷×é·ÖµÄƽºâŨ¶È·Ö±ðΪc(NO)=1.00mol/L¡¢c(CO)=2.70mol/L¡¢c()==4.5mol/L¡¢c()==9/span>mol/L£¬ËùÒԸ÷´Ó¦µÄƽºâ³£ÊýK£½ ==5000¡£

¢ÚA£®ÔÚ·´Ó¦¹ý³ÌÖйØϵʽ2n(CO2)£½n(N2) ºã³ÉÁ¢£¬ËùÒÔ²»ÄÜ˵Ã÷ÊÇƽºâ״̬£» B£®ËäÈ»·´Ó¦¹ý³ÌÖÐÆøÌåµÄ×ÜÖÊÁ¿ºã²»±ä£¬µ«ÊÇÆøÌåµÄ×ÜÎïÖʵÄÁ¿±äС£¬ËùÒÔ»ìºÏÆøÌåµÄƽ¾ùĦ¶ûÖÊÁ¿ÔÚ·´Ó¦¹ý³ÌÖбä´ó£¬Ïà¶Ô·Ö×ÓÖÊÁ¿Ò²±ä´ó£¬Òò´Ëµ±ÆøÌåµÄ¬jÏà¶Ô·Ö×ÓÖÊÁ¿²»±äʱ£¬ÄÜ˵Ã÷ÊÇƽºâ״̬£»C£®·´Ó¦¹ý³ÌÖÐÆøÌåµÄ×ÜÌå»ýºÍ×ÜÖÊÁ¿¶¼²»±ä£¬ËùÒÔÆøÌåÃܶȲ»±ä £¬Òò´Ë¸Ã˵·¨Ò²²»ÄÜ˵Ã÷ÊÇƽºâ״̬£» D£®Õý·´Ó¦ÊÇÆøÌåÌå»ý¼õСµÄ·´Ó¦£¬ËùÒÔ·´Ó¦¹ý³ÌÖÐÆøÌåµÄѹǿÊDZäÁ¿£¬µ±ÈÝÆ÷ÄÚÆøÌåѹǿ²»±äʱ£¬ËµÃ÷·´Ó¦´ïƽºâ״̬¡£×ÛÉÏËùÊö£¬ÄÜ˵Ã÷ÉÏÊö·´Ó¦´ïµ½Æ½ºâ״̬µÄÊÇBD¡£

¢Ûµ±NOÓëCOŨ¶ÈÏàµÈʱ£¬ÌåϵÖÐNOµÄƽºâת»¯ÂÊÓëζȡ¢Ñ¹Ç¿µÄ¹ØϵÈçÓÒͼËùʾ£¬ÔòNOµÄƽºâת»¯ÂÊËæζÈÉý¸ß¶ø¼õСµÄÔ­ÒòÊǸ÷´Ó¦µÄÕý·´Ó¦·ÅÈÈ£¬Éý¸ßζȣ¬Æ½ºâÄæÏòÒƶ¯£¬NOµÄת»¯ÂʼõС£¨»òÕý·´Ó¦·ÅÈÈ£¬Î¶ÈÔ½¸ß£¬Ô½²»ÀûÓÚ·´Ó¦ÕýÏò½øÐУ¬NOµÄƽºâת»¯ÂÊԽС£©£»ÓÉ·´Ó¦·½³Ìʽ¿ÉÖª£¬ÔÚÏàͬζÈÏÂѹǿԽ´ó£¬Ô½ÓÐÀûÓÚ·´Ó¦ÕýÏò½øÐУ¬ÔòNOµÄת»¯ÂÊÔ½´ó£¬ËùÒÔͼÖÐѹǿ£¨p1£¬p2¡¢p3£©µÄ´óС˳ÐòΪp1>p2>p3 ¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÏÂÁÐÎïÖÊÖУ¬ÄÑÈÜÓÚCCl4µÄÊÇ £¨ £©

A.µâµ¥ÖÊB.Ë®C.±½D.ÆûÓÍ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿µ±¹âÊøͨ¹ýÏÂÁзÖɢϵʱ£¬ÄܲúÉú¶¡´ï¶ûÏÖÏóµÄÊÇ

A.Fe(OH)3½ºÌåB.ÂÈ»¯ÄÆÈÜÒºC.ÑÎËáD.ÁòËáÈÜÒº

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÏÂÁÐÐðÊöÕýÈ·µÄÊÇ(¡¡¡¡)

A£®½ºÌåÇø±ðÓÚÆäËû·ÖɢϵµÄ¸ù±¾Ô­ÒòÊǽºÌåÓж¡´ï¶ûÏÖÏó

B£®·ÖɢϵÖзÖÉ¢ÖÊÁ£×ÓÖ±¾¶ÓÉСµ½´óµÄÕýȷ˳ÐòÊÇ£ºÈÜÒº<½ºÌå<×ÇÒº

C£®¹âÊøͨ¹ý½ºÌåºÍ×ÇҺʱ¶¼¿ÉÒÔ¿´µ½Ò»Ìõ¹âÁÁµÄͨ·£¬¶øÈÜÒº²»ÄÜ

D£®½ºÌåµÄ·ÖÉ¢ÖÊ¿ÉÒÔͨ¹ý¹ýÂË´Ó·ÖÉ¢¼ÁÖзÖÀë³öÀ´

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿Âé»ÆËØÊÇÖÐÊàÉñ¾­Ð˷ܼÁ£¬ÆäºÏ³É·ÏßÈçͼËùʾ¡£NBSÊÇÒ»ÖÖÑ¡ÔñÐÔäå´úÊÔ¼Á¡£

ÒÑÖª£º

£¨1£©ÆäÖÐAΪÌþ£¬Ïà¶Ô·Ö×ÓÖÊÁ¿Îª92£¬AµÄ½á¹¹¼òʽÊÇ___________£»EÖк¬Ñõ¹ÙÄÜÍŵÄÃû³ÆΪ__¡£

£¨2£©·´Ó¦B¡úCµÄ·´Ó¦Ìõ¼þºÍÊÔ¼ÁÊÇ_____________________£¬

£¨3£©FµÄ½á¹¹¼òʽÊÇ____________________________¡£

£¨4£©Ð´³öC¡úDµÄ»¯Ñ§·½³Ìʽ_____________________________________________________¡£

£¨5£©»¯ºÏÎïFµÄ·¼Ïã×åͬ·ÖÒì¹¹ÌåÓжàÖÖ£¬MºÍNÊÇÆäÖеÄÁ½À࣬ËüÃǵĽṹºÍÐÔÖÊÈçÏ£º

¢ÙÒÑÖªMÓöFeCl3ÈÜÒº·¢ÉúÏÔÉ«·´Ó¦£¬ÄܺÍÒø°±ÈÜÒº·¢ÉúÒø¾µ·´Ó¦£¬±½»·ÉÏÖ»ÓÐÁ½¸ö¶Ôλȡ´ú»ù£¬ÔòMµÄ½á¹¹¼òʽ¿ÉÄÜΪ(дÁ½ÖÖ)____________¡¢______________________¡£

¢ÚÒÑÖªN·Ö×ÓÖк¬Óм׻ù£¬ÄÜ·¢ÉúË®½â·´Ó¦£¬±½»·ÉÏÖ»ÓÐÒ»¸öÈ¡´ú»ù£¬ÔòNµÄ½á¹¹ÓÐ_____ÖÖ¡£

£¨6£©²ÎÕÕÉÏÊöºÏ³É·Ïß¼°ÐÅÏ¢£¬Éè¼ÆÒ»ÌõÒÔÒÒ´¼ÎªÔ­ÁÏÖƱ¸CH3CHOHCOCH3µÄÁ÷³Ì:

CH3CH2OH ¡ú________________________________________

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿Ä³Ñо¿ÐÔѧϰС×éΪÁËÁ˽âHCl¡¢NaHSO4¡¢NaHCO3ÔÚË®ÈÜÒºÖеĵçÀëÇé¿ö£¬½øÐÐÏÂÁÐʵÑ飺

¢Ù·Ö±ð²â¶¨ÁË0.1 mol¡¤L£­1µÄHCl¡¢NaHSO4¡¢NaHCO3 ÈÜÒºÖÐH£«µÄÎïÖʵÄÁ¿Å¨¶È£¬HClÈÜÒº ÖÐc(H£«)£½0.1 mol¡¤L£­1£¬NaHSO4ÈÜÒºÖÐc(H£«)£½0.1 mol¡¤L£­1£¬¶øNaHCO3ÈÜÒºÖÐH£«µÄÎï ÖʵÄÁ¿Å¨¶ÈԶԶСÓÚ0.1 mol¡¤L£­1¡£

¢ÚÈ¡ÉÙÁ¿NaHSO4ÈÜÒºÓÚÊÔ¹ÜÖУ¬µÎÈëBaCl2ÈÜÒºÓв»ÈÜÓÚÏ¡ÏõËáµÄ°×É«³ÁµíÉú³É¡£

¢ÛÈ¡ÉÙÁ¿NaHCO3ÈÜÒºÓÚÊÔ¹ÜÖУ¬µÎÈ뼸µÎBaCl2ÈÜÒºÎÞÃ÷ÏÔÏÖÏó¡£

Çë¸ù¾ÝʵÑé½á¹û»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©·Ö±ðд³öHCl¡¢NaHSO4¡¢NaHCO3ÔÚË®ÈÜÒºÖеĵçÀë·½³Ìʽ£ºHCl£º__________________________________________________£»

NaHSO4£º__________________________________________________£»

NaHCO3£º__________________________________________________¡£

£¨2£©NaHSO4ÊôÓÚ¡°Ëᡱ¡°¼î¡±¡°ÑΡ±ÖеÄ________________________¡£

£¨3£©Ð´³ö¢ÚÖз´Ó¦µÄ»¯Ñ§·½³Ìʽ£º_________________________________________________¡£

£¨4£©Èô½«NaHSO4ÓëBa(OH)2ÔÚÈÜÒºÖа´ÕÕÎïÖʵÄÁ¿Ö®±È1¡Ã1»ìºÏ£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪ__________________________________________________________________________¡£

£¨5£©Èô½«NaHSO4ÖðµÎ¼ÓÈëBa(OH)2ÈÜÒºÖÐÖÁBa2£«Ç¡ºÃÍêÈ«³Áµí£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪ_________________________________________________________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÏÂÁÐÐðÊöÖÐÕýÈ·µÄÊÇ

A. ³ýÁã×åÔªËØÍ⣬¶ÌÖÜÆÚÔªËصÄ×î¸ß»¯ºÏ¼ÛÔÚÊýÖµÉ϶¼µÈÓÚ¸ÃÔªËØËùÊôµÄ×åÐòÊý

B. ³ý¶ÌÖÜÆÚÍ⣬ÆäËûÖÜÆÚ¾ùÓÐ18ÖÖÔªËØ

C. ¸±×åÔªËØÖÐûÓзǽðÊôÔªËØ

D. ¼î½ðÊôÔªËØÊÇÖ¸IA×åµÄËùÓÐÔªËØ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÊéдÈÈ»¯Ñ§·½³ÌʽҪעÃ÷ÎïÖʵľۼ¯×´Ì¬£¬Ô­ÒòÊÇ

A.¾ßÌå˵Ã÷·´Ó¦µÄÇé¿ö

B.ÎïÖʳÊÏÖµÄ״̬Óë·´Ó¦ìʱäÓйØ

C.˵Ã÷·´Ó¦Ìõ¼þ

D.ÎïÖʳÊÏÖµÄ״̬ÓëÉú³ÉʲôÎïÖÊÓйØ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿Îª³ýÈ¥´ÖÑÎÖеÄCaCl2¡¢MgCl2¡¢Na2SO4 ÒÔ¼°ÄàɳµÈÔÓÖÊ£¬Ä³Í¬Ñ§Éè¼ÆÁËÒ»ÖÖÖƱ¸¾«ÑεÄʵÑé·½°¸£¬²½ÖèÈçÏÂ(ÓÃÓÚ³ÁµíµÄÊÔ¼ÁÉÔ¹ýÁ¿)£º

£¨1£©ÅжÏBaCl2ÒѹýÁ¿µÄ·½·¨ÊÇ___________________________________¡£

£¨2£©µÚ¢Ü²½ÖУ¬Ïà¹ØµÄÀë×Ó·½³ÌʽÊÇ_______________________________¡£

£¨3£©Îª¼ìÑ龫Ñδ¿¶È£¬ÐèÅä230mL0.2mol/L NaCl(¾«ÑÎ)ÈÜÒº£¬ÔòÐèÓÃÍÐÅÌÌìƽ³ÆÈ¡¾«ÑιÌÌåµÄÖÊÁ¿Îª_______g¡£

£¨4£© ÅäÖÆNaCl(¾«ÑÎ)ÈÜҺʱÐèÓÃÈÝÁ¿Æ¿£¬ÆäÔÚʹÓÃÇ°±ØÐë______ ¡£

£¨5£© ÅäÖÆNaCl(¾«ÑÎ)ÈÜҺʱ£¬Èô³öÏÖÏÂÁвÙ×÷£¬Æä½á¹ûÆ«¸ßµÄÊÇ__________

A.³ÆÁ¿Ê±NaClÒѳ±½â B.ÌìƽµÄíÀÂëÒÑÐâÊ´

C.¶¨ÈÝÒ¡ÔȺó£¬ÒºÃæϽµÓÖ¼ÓË® D.¶¨ÈÝʱ¸©Êӿ̶ÈÏß

£¨6£©½«ÒÑÅäÖƺõÄŨ¶ÈΪC1 mol¡¤L£­1NaClÈÜÒºÓëµÈÖÊÁ¿µÄË®»ìºÏºó£¬´ËʱÈÜÒºµÄŨ¶ÈΪC2 mol¡¤L£­1 £¬ÔòC1 ÓëC2¶þÕߵĹØϵΪ______¡£

A. C1=2C2 B.C1<2C2 C.C1>2C2 D.2C1<C2

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸