Ñо¿CO2µÄÀûÓöԴٽøµÍ̼Éç»áµÄ¹¹½¨¾ßÓÐÖØÒªµÄÒâÒå¡£
(1)½«CO2Ó뽹̿×÷ÓÃÉú³ÉCO£¬CO¿ÉÓÃÓÚÁ¶ÌúµÈ¡£
ÒÑÖª£ºFe2O3(s)£«3C(s)=2Fe(s)£«3CO(g)£»¦¤H1£½£«489.0 kJ¡¤mol£­1
C(s)£«CO2(g)=2CO(g)£»¦¤H2£½£«172.5 kJ¡¤mol£­1¡£
ÔòCO»¹Ô­Fe2O3µÄÈÈ»¯Ñ§·½³ÌʽΪ________________________________
(2)ijʵÑ齫CO2ºÍH2³äÈëÒ»¶¨Ìå»ýµÄÃܱÕÈÝÆ÷ÖУ¬ÔÚÁ½ÖÖ²»Í¬Ìõ¼þÏ·¢Éú·´Ó¦£ºCO2(g)£«3H2(g)CH3OH(g)£«H2O(g)¡¡¦¤H£½£­49.0 kJ¡¤mol£­1£¬²âµÃCH3OHµÄÎïÖʵÄÁ¿Ëæʱ¼äµÄ±ä»¯ÈçͼËùʾ£¬Çë»Ø´ðÏÂÁÐÎÊÌ⣺

¢Ù¸Ã·´Ó¦µÄƽºâ³£ÊýµÄ±í´ïʽΪK£½________¡£
¢ÚÇúÏߢñ¡¢¢ò¶ÔÓ¦µÄƽºâ³£Êý´óС¹ØϵΪK¢ñ________K¢ò(Ìî¡°´óÓÚ¡±¡¢¡°µÈÓÚ¡±»ò¡°Ð¡ÓÚ¡±)¡£
¢ÛÔÚÏÂͼa¡¢b¡¢cÈýµãÖУ¬H2µÄת»¯ÂÊÓɸߵ½µÍµÄ˳ÐòÊÇ________(Ìî×Öĸ)¡£

(3)ÔÚÆäËûÌõ¼þ²»±äµÄÇé¿öÏ£¬½«ÈÝÆ÷Ìå»ýѹËõµ½Ô­À´µÄ1/2£¬ÓëԭƽºâÏà±È£¬ÏÂÁÐÓйØ˵·¨ÕýÈ·µÄÊÇ________(ÌîÐòºÅ)¡£
a£®ÇâÆøµÄŨ¶È¼õС
b£®Õý·´Ó¦ËÙÂʼӿ죬Äæ·´Ó¦ËÙÂÊÒ²¼Ó¿ì
c£®¼×´¼µÄÎïÖʵÄÁ¿Ôö¼Ó
d£®ÖØÐÂƽºâʱn(H2)/n(CH3OH)Ôö´ó

(1)Fe2O3(s)£«3CO(g)=2Fe(s)£«3CO2(g)£»¦¤H£½£­28.5 kJ¡¤mol£­1
(2)¢Ù¡¡¢Ú´óÓÚ¡¡¢Ûa¡¢b¡¢c
(3)bc

½âÎö

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

×÷Ϊ¹ú¼ÒÕýÔÚʵʩµÄ¡°Î÷Æø¶«Ê䡱¹¤³ÌÖÕµãÕ¾¡ª¡ªÉϺ£½«Ö𲽸ıäÒÔú¡¢Ê¯ÓÍΪÖ÷µÄÄÜÔ´½á¹¹£¬Õâ¶Ô½â¾ö³ÇÊл·¾³ÎÛȾÒâÒåÖØ´ó¡£
(1)Ä¿Ç°ÉϺ£´ó²¿·Ö³ÇÊоÓÃñËùʹÓõÄȼÁÏÖ÷ÒªÊǹܵÀúÆø£¬ÆÖ¶«ÐÂÇø¾ÓÃñ
¿ªÊ¼Ê¹Óö«º£ÌìÈ»Æø×÷ΪÃñÓÃȼÁÏ£¬¹ÜµÀúÆøµÄÖ÷Òª³É·ÖÊÇCO¡¢H2ºÍÉÙÁ¿
ÌþÀ࣬ÌìÈ»ÆøµÄÖ÷Òª³É·ÖÊÇCH4£¬ËüÃÇȼÉյĻ¯Ñ§·½³ÌʽΪ£º
2CO£«O2 2CO2¡¡2H2£«O22H2O¡¡CH4£«2O2CO2£«2H2O
¸ù¾ÝÒÔÉÏ»¯Ñ§·½³ÌʽÅжϣºÈ¼ÉÕÏàͬÌå»ýµÄ¹ÜµÀúÆøºÍÌìÈ»Æø£¬ÏûºÄ¿ÕÆøÌå»ý½Ï´óµÄÊÇ¡¡¡¡¡¡¡¡£¬Òò´ËȼÉչܵÀúÆøµÄÔî¾ßÈçÐè¸ÄÉÕÌìÈ»Æø£¬Ôî¾ßµÄ¸Ä½ø·½·¨ÊÇ¡¡¡¡¡¡½ø·ç¿Ú(Ìî¡°Ôö´ó¡±»ò¡°¼õС¡±)£¬Èç²»×÷¸Ä½ø¿ÉÄܲúÉúµÄ²»Á¼½á¹ûÊÇ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡£
(2)¹ÜµÀúÆøÖк¬ÓеÄÌþÀ࣬³ý¼×ÍéÍ⣬»¹ÓÐÉÙÁ¿ÒÒÍé¡¢±ûÍé¡¢¶¡ÍéµÈ£¬ËüÃǵÄijЩÐÔÖÊÈçÏ£º

 
ÒÒÍé
±ûÍé
¶¡Íé
ÈÛµã/¡æ
£­183.3
£­189.7
£­138.4
·Ðµã/¡æ
£­88.6
£­42.1
£­0.5
 
ÊÔ¸ù¾ÝÒÔÉÏij¸ö¹Ø¼üÊý¾Ý½âÊͶ¬ÌìÑϺ®µÄ¼¾½ÚÓÐʱ¹ÜµÀúÆø»ðÑæºÜС£¬²¢ÇҳʶÏÐø״̬µÄÔ­ÒòÊÇ___________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ÔÚÒ»¸öСÉÕ±­Àï¼ÓÈëÔ¼20 gÒÑÑÐÄ¥³É·ÛÄ©µÄÇâÑõ»¯±µ¾§Ìå[Ba£¨OH£©2¡¤8H2O]£¬½«Ð¡ÉÕ±­·ÅÔÚÊÂÏÈÒѵÎÓÐ3¡«4µÎË®µÄ²£Á§Æ¬ÉÏ£¬È»ºóÏòÉÕ±­ÄÚ¼ÓÈëÔ¼10 gÂÈ»¯ï§¾§Ì壬²¢Á¢¼´Óò£Á§°ôѸËÙ½Á°è¡£ÊԻشðÏÂÁÐÎÊÌ⣺
£¨1£©Ð´³ö·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º______________¡£
£¨2£©ÊµÑéÒªÁ¢¼´Óò£Á§°ôѸËÙ½Á°èµÄÔ­ÒòÊÇ_________________¡£
£¨3£©Èç¹ûʵÑéÖÐûÓп´µ½¡°½á±ù¡±ÏÖÏ󣬿ÉÄܵÄÔ­ÒòÊÇ__________________________£¨´ð³öÈý¸ö»òÈý¸öÒÔÉÏÔ­Òò£©¡£
£¨4£©Èç¹ûûÓп´µ½¡°½á±ù¡±ÏÖÏó£¬ÎÒÃÇ»¹¿ÉÒÔ²ÉÈ¡ÄÄЩ·½Ê½À´ËµÃ÷¸Ã·´Ó¦ÎüÈÈ£¿__________________________________________________________________________________________£¨´ð³öÁ½ÖÖ·½°¸£©¡£
£¨5£©¡°½á±ù¡±ÏÖÏó˵Ã÷·´Ó¦ÊÇÒ»¸ö________£¨Ìî¡°·Å³ö¡±»ò¡°ÎüÊÕ¡±£©ÄÜÁ¿µÄ·´Ó¦£¬¼´¶Ï¿ª¾É»¯Ñ§¼ü________£¨Ìî¡°·Å³ö¡±»ò¡°ÎüÊÕ¡±£©ÄÜÁ¿____________£¨Ìî¡°>¡±»ò¡°<¡±£©ÐγÉл¯Ñ§¼ü________£¨Ìî¡°ÎüÊÕ¡±»ò¡°·Å³ö¡±£©µÄÄÜÁ¿¡£
£¨6£©¸Ã·´Ó¦ÔÚ³£ÎÂϾͿÉÒÔ½øÐУ¬ËµÃ÷______________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ͨ³£ÈËÃǰѲð¿ª1 molij»¯Ñ§¼üËùÎüÊÕµÄÄÜÁ¿¿´³É¸Ã»¯Ñ§¼üµÄ¼üÄÜ¡£¼üÄܵĴóС¿ÉÒÔºâÁ¿»¯Ñ§¼üµÄÇ¿Èõ,Ò²¿ÉÒÔ¹ÀË㻯ѧ·´Ó¦µÄ·´Ó¦ÈÈ(¦¤H),»¯Ñ§·´Ó¦µÄ¦¤HµÈÓÚ·´Ó¦ÖжÏÁѾɻ¯Ñ§¼üµÄ¼üÄÜÖ®ºÍÓë·´Ó¦ÖÐÐγÉл¯Ñ§¼üµÄ¼üÄÜÖ®ºÍµÄ²î¡£

»¯Ñ§¼ü
Si¡ªO
Si¡ªCl
H¡ªH
H¡ªCl
Si¡ªSi
Si¡ªC
¼üÄÜ/kJ¡¤mol-1
460
360
436
431
176
347
 
Çë»Ø´ðÏÂÁÐÎÊÌâ:
(1)±È½ÏÏÂÁÐÁ½×éÎïÖʵÄÈÛµã¸ßµÍ(Ìî¡°>¡±»ò¡°<¡±)¡£
SiC¡¡¡¡¡¡¡¡Si;SiCl4¡¡¡¡¡¡¡¡SiO2¡£ 
(2)ÈçͼÁ¢·½ÌåÖÐÐĵġ°¡±±íʾ¹è¾§ÌåÖеÄÒ»¸öÔ­×Ó,ÇëÔÚÁ¢·½ÌåµÄ¶¥µãÓá°¡±±íʾ³öÓëÖ®½ôÁڵĹèÔ­×Ó¡£

(3)¹¤ÒµÉÏÓøߴ¿¹è¿Éͨ¹ýÏÂÁз´Ó¦ÖÆÈ¡:SiCl4(g)+2H2(g)Si(s)+4HCl(g),¸Ã·´Ó¦µÄ·´Ó¦ÈȦ¤H=¡¡¡¡¡¡kJ/mol¡£ 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

»ð¼ýÍƽøÆ÷ÖÐÊ¢ÓÐÇ¿»¹Ô­¼ÁҺ̬ëÂ(N2H4)ºÍÇ¿Ñõ»¯¼ÁË«ÑõË®¡£µ±ËüÃÇ»ìºÏ·´Ó¦Ê±£¬¼´²úÉú´óÁ¿µªÆøºÍË®ÕôÆø£¬²¢·Å³ö´óÁ¿ÈÈ¡£ÒÑÖª£º0£®4 molҺ̬ëÂÓë×ãÁ¿µÄË«ÑõË®·´Ó¦£¬Éú³ÉµªÆøºÍË®ÕôÆø£¬·Å³ö256£®652 kJµÄÈÈÁ¿¡£
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪ_______________________________________________¡£
£¨2£©ÓÖÒÑÖªH2O(l)=H2O(g)¡¡¦¤H£½£«44 kJ¡¤mol£­1£¬Ôò16 gҺ̬ëÂÓëË«ÑõË®·´Ó¦Éú³ÉҺ̬ˮʱ·Å³öµÄÈÈÁ¿ÊÇ________ kJ¡£
£¨3£©´Ë·´Ó¦ÓÃÓÚ»ð¼ýÍƽø£¬³ýÊÍ·Å´óÁ¿ÈȺͿìËÙ²úÉú´óÁ¿ÆøÌåÍ⻹ÓÐÒ»¸öºÜ´óµÄÓŵãÊÇ________________________________________________________________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ÒÑÖªµ¥ÖÊÁòÔÚͨ³£Ìõ¼þÏÂÒÔS8(б·½Áò)µÄÐÎʽ´æÔÚ£¬¶øÔÚÕôÆø״̬ʱ£¬º¬ÓÐS2¡¢S4¡¢S6¼°S8µÈ¶àÖÖͬËØÒìÐÎÌ壬ÆäÖÐS4¡¢S6ºÍS8¾ßÓÐÏàËƵĽṹÌص㣬Æä½á¹¹ÈçÏÂͼËùʾ£º

ÔÚÒ»¶¨Ìõ¼þÏ£¬S8(s)ºÍO2(g)·¢Éú·´Ó¦ÒÀ´Îת»¯ÎªSO2(g)ºÍSO3(g)¡£·´Ó¦¹ý³ÌºÍÄÜÁ¿¹Øϵ¿ÉÓÃÏÂͼ¼òµ¥±íʾ(ͼÖеĦ¤H±íʾÉú³É1 mol²úÎïµÄÊý¾Ý)¡£

£¨1£©Ð´³ö±íʾS8ȼÉÕÈȵÄÈÈ»¯Ñ§·½³Ìʽ___________________________________¡£
£¨2£©Ð´³öSO3·Ö½âÉú³ÉSO2ºÍO2µÄÈÈ»¯Ñ§·½³Ìʽ_______________________________________________________________¡£
£¨3£©»¯Ñ§ÉϹ涨£¬²ð¿ª»òÐγÉ1 mol»¯Ñ§¼üÎüÊÕ»ò·Å³öµÄÄÜÁ¿³ÆΪ¸Ã»¯Ñ§¼üµÄ¼üÄÜ£¬µ¥Î»kJ¡¤mol¡£ÈôÒÑÖªÁòÑõ¼üµÄ¼üÄÜΪd kJ¡¤mol£­1£¬ÑõÑõ¼üµÄ¼üÄÜΪe kJ¡¤mol£­1£¬ÔòS8·Ö×ÓÖÐÁòÁò¼üµÄ¼üÄÜΪ____________________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

±½ÒÒÏ©ÊÇÏÖ´úʯÓÍ»¯¹¤²úÆ·ÖÐ×îÖØÒªµÄµ¥ÌåÖ®Ò»¡£ÔÚ¹¤ÒµÉÏ£¬±½ÒÒÏ©¿ÉÓÉÒÒ±½ºÍCO2
´ß»¯ÍÑÇâÖƵá£×Ü·´Ó¦Ô­ÀíÈçÏ£º
   ¡÷H
»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÒÒ±½ÔÚCO2Æø·ÕÖеķ´Ó¦¿É·ÖÁ½²½½øÐУº
¡÷H1=£«117.6kJ¡¤mol£­1
H2 (g)£«CO2 (g)CO (g)£«H2O (g) ¡÷H2=£«41.2kJ¡¤mol£­1
ÓÉÒÒ±½ÖÆÈ¡±½ÒÒÏ©·´Ó¦µÄ             ¡£
£¨2£©ÔÚζÈΪT1ʱ£¬¸Ã·´Ó¦µÄƽºâ³£ÊýK=0.5mol/L¡£ÔÚ2LµÄÃܱÕÈÝÆ÷ÖмÓÈëÒÒ±½ÓëCO2£¬·´Ó¦µ½Ä³Ê±¿Ì²âµÃ»ìºÏÎïÖи÷×é·ÖµÄÎïÖʵÄÁ¿¾ùΪ1.0mol¡£
¢Ù¸Ãʱ¿Ì»¯Ñ§·´Ó¦          £¨Ìî¡°ÊÇ¡±»ò¡°²»ÊÇ¡±£©´¦ÓÚƽºâ״̬£»
¢ÚÏÂÁÐÐðÊöÄÜ˵Ã÷ÒÒ±½ÓëCO2ÔÚ¸ÃÌõ¼þÏ·´Ó¦ÒѴﵽƽºâ״̬µÄÊÇ          £¨ÌîÕýÈ·´ð°¸±àºÅ£©£»
a£®Õý¡¢Äæ·´Ó¦ËÙÂʵıÈÖµºã¶¨     b£®c(CO2)=c(CO)
c£®»ìºÏÆøÌåµÄÃܶȲ»±ä                    d£®CO2µÄÌå»ý·ÖÊý±£³Ö²»±ä
¢ÛÈô½«·´Ó¦¸ÄΪºãѹ¾øÈÈÌõ¼þϽøÐУ¬´ïµ½Æ½ºâʱ£¬ÔòÒÒ±½µÄÎïÖʵÄÁ¿Å¨¶È     £¨ÌîÕýÈ·´ð°¸±àºÅ£©

A£®´óÓÚ0.5mol/LB£®Ð¡ÓÚ0.5mol/L
C£®µÈÓÚ0.5mol/LD£®²»È·¶¨
£¨3£©ÔÚζÈΪT2ʱµÄºãÈÝÆ÷ÖУ¬ÒÒ±½¡¢CO2µÄÆðʼŨ¶È·Ö±ðΪ2.0mol/LºÍ3.0mol/L£¬É跴Ӧƽºâºó×ÜѹǿΪP¡¢ÆðʼѹǿΪ£¬Ôò·´Ó¦´ïµ½Æ½ºâʱ±½ÒÒÏ©µÄŨ¶ÈΪ      £¬           £¨¾ùÓú¬¡¢PµÄ±í´ïʽ±íʾ£©¡£
£¨4£©Ð´³öÓɱ½ÒÒÏ©ÔÚÒ»¶¨Ìõ¼þϺϳɾ۱½ÒÒÏ©µÄ»¯Ñ§·½³Ìʽ                       ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ÏÂͼÊÇú»¯¹¤²úÒµÁ´µÄÒ»²¿·Ö,ÊÔÔËÓÃËùѧ֪ʶ,½â¾öÏÂÁÐÎÊÌâ:

¢ñ.ÒÑÖª¸Ã²úÒµÁ´ÖÐij·´Ó¦µÄƽºâ³£Êý±í´ïʽΪ:K=,д³öËüËù¶ÔÓ¦·´Ó¦µÄ»¯Ñ§·½³Ìʽ:                                  
                                                     ¡¡¡£ 
¢ò.¶þ¼×ÃÑ(CH3OCH3)ÔÚδÀ´¿ÉÄÜÌæ´ú²ñÓͺÍÒº»¯Æø×÷Ϊ½à¾»ÒºÌåȼÁÏʹÓ᣹¤ÒµÉÏÒÔCOºÍH2ΪԭÁÏÉú²úCH3OCH3¡£¹¤ÒµÖƱ¸¶þ¼×ÃÑÔÚ´ß»¯·´Ó¦ÊÒÖÐ(ѹÁ¦2.0¡«10.0 MPa,ζÈ230¡«280 ¡æ)½øÐÐÏÂÁз´Ó¦:
¢ÙCO(g)+2H2(g)CH3OH(g)
¦¤H1="-90.7" kJ¡¤mol-1
¢Ú2CH3OH(g)CH3OCH3(g)+H2O(g)
¦¤H2="-23.5" kJ¡¤mol-1
¢ÛCO(g)+H2O(g)CO2(g)+H2(g)
¦¤H3="-41.2" kJ¡¤mol-1
(1)д³ö´ß»¯·´Ó¦ÊÒÖÐÈý¸ö·´Ó¦µÄ×Ü·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ:                          ¡£
(2)ÔÚijζÈÏÂ,2 LÃܱÕÈÝÆ÷Öз¢Éú·´Ó¦¢Ù,ÆðʼʱCO¡¢H2µÄÎïÖʵÄÁ¿·Ö±ðΪ2 molºÍ6 mol,3 minºó´ïµ½Æ½ºâ,²âµÃCOµÄת»¯ÂÊΪ60%,Ôò3 minÄÚCOµÄƽ¾ù·´Ó¦ËÙÂÊΪ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡£ÈôͬÑùÌõ¼þÏÂÆðʼʱCOÎïÖʵÄÁ¿Îª4 mol,´ïµ½Æ½ºâºóCH3OHΪ2.4 mol,ÔòÆðʼʱH2Ϊ¡¡¡¡¡¡¡¡mol¡£
(3)ÏÂÁÐÓйط´Ó¦¢ÛµÄ˵·¨ÕýÈ·µÄÊÇ¡¡¡¡¡¡¡¡¡£
A.ÔÚÌå»ý¿É±äµÄÃܱÕÈÝÆ÷ÖÐ,ÔÚ·´Ó¦¢Û´ïµ½Æ½ºâºó,Èô¼Óѹ,Ôòƽºâ²»Òƶ¯¡¢»ìºÏÆøÌåƽ¾ùÏà¶Ô·Ö×ÓÖÊÁ¿²»±ä¡¢»ìºÏÆøÌåÃܶȲ»±ä
B.Èô830 ¡æʱ·´Ó¦¢ÛµÄK=1.0,ÔòÔÚ´ß»¯·´Ó¦ÊÒÖз´Ó¦¢ÛµÄK£¾1.0
C.ijζÈÏÂ,ÈôÏòÒÑ´ïƽºâµÄ·´Ó¦¢ÛÖмÓÈëµÈÎïÖʵÄÁ¿µÄCOºÍH2O(g),ÔòƽºâÓÒÒÆ¡¢Æ½ºâ³£Êý±ä´ó
(4)ΪÁËÑ°ÕÒºÏÊʵķ´Ó¦Î¶È,Ñо¿Õß½øÐÐÁËһϵÁÐʵÑé,ÿ´ÎʵÑé±£³ÖÔ­ÁÏÆø×é³É¡¢Ñ¹Ç¿¡¢·´Ó¦Ê±¼äµÈÒòËز»±ä,ʵÑé½á¹ûÈçͼ,

ÔòCOת»¯ÂÊËæζȱ仯µÄ¹æÂÉÊÇ                                        ¡¡¡£
ÆäÔ­ÒòÊÇ                                                   ¡¡¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

°±ÊÇ×îÖØÒªµÄ»¯¹¤²úÆ·Ö®Ò»¡£
(1)ºÏ³É°±ÓõÄÇâÆø¿ÉÒÔ¼×ÍéΪԭÁÏÖƵá£Óйػ¯Ñ§·´Ó¦µÄÄÜÁ¿±ä»¯ÈçÏÂͼËùʾ¡£

·´Ó¦¢Ù¢Ú¢ÛΪ________·´Ó¦(Ìî¡°ÎüÈÈ¡±»ò¡°·ÅÈÈ¡±)¡£
CH4(g)ÓëH2O(g)·´Ó¦Éú³ÉCO(g)ºÍH2(g)µÄÈÈ»¯Ñ§·½³ÌʽΪ___________¡£
(2)Óð±ÆøÖÆÈ¡ÄòËØ[CO(NH2)2]µÄ·´Ó¦Îª£º2NH3(g)£«CO2(g)CO(NH2)2(l)£«H2O(g)¡£
¢ÙijζÈÏ£¬ÏòÈÝ»ýΪ10 LµÄÃܱÕÈÝÆ÷ÖÐͨÈë2 mol NH3ºÍ1 mol CO2£¬·´Ó¦´ïµ½Æ½ºâʱCO2µÄת»¯ÂÊΪ50%¡£¸Ã·´Ó¦µÄ»¯Ñ§Æ½ºâ³£Êý±í´ïʽΪK£½______¡£¸ÃζÈÏÂƽºâ³£ÊýKµÄ¼ÆËã½á¹ûΪ_____¡£
¢ÚΪ½øÒ»²½Ìá¸ßCO2µÄƽºâת»¯ÂÊ£¬ÏÂÁдëÊ©ÖÐÄܴﵽĿµÄµÄÊÇ________¡£

A£®Ìá¸ßNH3µÄŨ¶È B£®Ôö´óѹǿ
C£®¼°Ê±×ªÒÆÉú³ÉµÄÄòËØ D£®Ê¹Óøü¸ßЧµÄ´ß»¯¼Á

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸