£¨11·Ö£©Ñо¿È¼ÁϵÄȼÉպͶÔÎÛȾÆøÌå²úÎïµÄÎÞº¦»¯´¦Àí£¬¶ÔÓÚ·ÀÖ¹´óÆøÎÛȾÓÐÖØÒªÒâÒå¡£
£¨1£©½«Ãº×ª»¯ÎªÇå½àÆøÌåȼÁÏ£º
ÒÑÖª£ºH2(g)+1/2O2(g)=H2O(g) H= ?241£®8kJ/mol 
C(s)+1/2O2(g)=CO(g)  H= ?110£®5kJ/mol
д³ö½¹Ì¿ÓëË®ÕôÆø·´Ó¦ÖÆH2ºÍCOµÄÈÈ»¯Ñ§·½³Ìʽ                         ¡£
£¨2£©Ò»¶¨Ìõ¼þÏ£¬ÔÚÃܱÕÈÝÆ÷ÄÚ£¬SO2±»Ñõ»¯³ÉSO3µÄÈÈ»¯Ñ§·½³ÌʽΪ£º2SO2(g)+O2(g)  2SO3(g)£»
¡÷H=?a kJ/mo1£¬ÔÚÏàͬÌõ¼þÏÂÒªÏëµÃµ½2akJÈÈÁ¿£¬¼ÓÈë¸÷ÎïÖʵÄÎïÖʵÄÁ¿¿ÉÄÜÊÇ            
A£®4mo1 SO2ºÍ2mol O2¡¡¡¡¡¡¡¡¡¡ B£®4mol SO2¡¢2mo1 O2ºÍ2mol SO3
C£®4mol SO2ºÍ4mo1 O2¡¡¡¡¡¡     D£®6mo1 SO2ºÍ4mo1 O2
£¨3£©Æû³µÎ²ÆøÖÐNOxºÍCOµÄÉú³É¼°×ª»¯£º
¢ÙÒÑÖªÆø¸×ÖÐÉú³ÉNOµÄ·´Ó¦Îª£ºN2(g)+O2(g)  2NO(g) H£¾0
ÔÚÒ»¶¨Î¶ÈϵĶ¨ÈÝÃܱÕÈÝÆ÷ÖУ¬ÄÜ˵Ã÷´Ë·´Ó¦ÒÑ´ïƽºâµÄÊÇ          
A£®Ñ¹Ç¿²»±ä                   B£®»ìºÏÆøÌåƽ¾ùÏà¶Ô·Ö×ÓÖÊÁ¿²»±ä
C£®2vÕý(N2)£½vÄæ(NO)           D£® N2µÄÌå»ý·ÖÊý²»Ôٸıä
¢ÚÆû³µÈ¼ÓͲ»ÍêȫȼÉÕʱ²úÉúCO£¬ÓÐÈËÉèÏë°´ÏÂÁз´Ó¦³ýÈ¥CO£º2CO(g)=2C(s)+O2(g) H£¾0£¬
¼òÊö¸ÃÉèÏëÄÜ·ñʵÏÖµÄÒÀ¾Ý                                                             ¡£
£¨4£©È¼ÁÏCO¡¢H2ÔÚÒ»¶¨Ìõ¼þÏ¿ÉÒÔÏ໥ת»¯£ºCO(g)£«H2O(g)  CO2(g)£«H2(g)¡£ÔÚ420¡æʱ£¬Æ½ºâ³£ÊýK=9¡£Èô·´Ó¦¿ªÊ¼Ê±£¬CO¡¢H2OµÄŨ¶È¾ùΪ0£®1mol/L£¬ÔòCOÔÚ´Ë·´Ó¦Ìõ¼þϵÄת»¯ÂÊΪ              ¡£
£¨1£©C(s)+H2O(g)=CO(g)+H2(g) H=+131£®3kJ/mol  (3·Ö)(·½³Ìʽ¼°Ï±êÕýÈ·¿É¸ø1·Ö)  
£¨2£©D  (2·Ö)
£¨3£©¢ÙCD    (2·Ö)     
¢Ú¸Ã·´Ó¦ÊÇìÊÔö¡¢ìؼõµÄ·´Ó¦¡£¸ù¾ÝG=H¡ªTS, G£¾0,²»ÄÜʵÏÖ¡£  (2·Ö)
£¨4£©75%  (2·Ö)

ÊÔÌâ·ÖÎö£º£¨1£©ÒÑÖª£ºH2(g)+1/2O2(g)=H2O(g) H1= ?241£®8kJ/mol £»C(s)+1/2O2(g)="CO(g)" H2= ?110£®5kJ/mol¸ù¾Ý¸Ç˹¶¨ÂÉ£¬Ó÷½³Ìʽ2¼õÈ¥·½³Ìʽ1£¬¿ÉµÃ£ºÐ´³ö½¹Ì¿ÓëË®ÕôÆø·´Ó¦ÖÆH2ºÍCOµÄÈÈ»¯Ñ§·½³Ìʽ£¨1£©C(s)+H2O(g)=CO(g)+H2(g) H=+131£®3kJ/mol ¡££¨2£©ÓÉ·´Ó¦·½³Ìʽ¿ÉÖªÉú³É2molµÄSO3ʱ·ÅÈÈa kJ£¬ÏëµÃµ½2a kJÈÈÁ¿£¬ÔòÐèÉú³É4molSO3£»ÓÉÓÚ·´Ó¦ÊÇ¿ÉÄæ·´Ó¦£¬²»¿ÉÄÜÍêÈ«½øÐе½µ×£¬ËùÒÔÒªµÃµ½4molSO3£¬SO2ÓëO2ÎïÖʵÄÁ¿±ØÐëÒª¶àÓÚ4molºÍ 2mol£¬¹ÊD·ûºÏ£¬BÑ¡ÏîÖмÈÓÐÕý·´Ó¦ÓÖÓÐÄæ·´Ó¦£¬×îÖշųöµÄÈÈÁ¿Ð¡ÓÚ2akJ£¬¹ÊÑ¡D£®£¨3£©¢ÙA£®ÓÉÓÚ·´Ó¦Ç°ºóÌå»ý²»±ä£¬¹ÊѹǿʼÖÕ²»±ä²»ÄÜ×öΪƽºâÅжÏÒÀ¾Ý£¬´íÎó£» B£®ÓÉÓÚ·´Ó¦Ç°ºó×ÜÖÊÁ¿ºÍ×ÜÎïÖʵÄÁ¿¶¼²»¸Ä±ä£¬¹Ê»ìºÏÆøÌåƽ¾ùÏà¶Ô·Ö×ÓÖÊÁ¿²»±ä£¬¹Ê²»ÄÜ×öΪƽºâÅжÏÒÀ¾Ý£¬´íÎó£»Ñ¡CD¡£¢ÚÓÉ·´Ó¦2CO(g)=2C(s)+O2(g) H£¾0£¬¿ÉÖª¸Ã·´Ó¦ÊÇìÊÔö¡¢ìؼõµÄ·´Ó¦¡£¸ù¾ÝG=H¡ªTS, G£¾0,²»ÄÜʵÏÖ¡£
£¨4£©½â£ºÁîƽºâʱCOµÄŨ¶È±ä»¯Á¿Îªxmol/L£¬Ôò£º
CO£¨g£©+H2O£¨g£©?CO2£¨g£©+H2£¨g£©
¿ªÊ¼£¨mol/L£©£º0£®1        0£®1         0        0
±ä»¯£¨mol/L£©£ºc          c            c        c
ƽºâ£¨mol/L£©£º0£®1-c     0£® 1-c       c        c
¹Ê=9  ½âµÃc=0£®075£¬¹ÊÒ»Ñõ»¯Ì¼ÔÚ´ËÌõ¼þϵÄƽºâת»¯ÂÊ=75%£¬¹Ê´ð°¸Îª£º75%£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

ÒÀ¾ÝÊÂʵ£¬Ð´³öÏÂÁз´Ó¦µÄÈÈ»¯Ñ§·´Ó¦·½³Ìʽ¡£
£¨1£©ÔÚ25¡æ¡¢101kPaÏ£¬1g¼×´¼ÍêȫȼÉÕÉú³ÉCO2ºÍҺ̬ˮʱ·ÅÈÈ22£®68kJ¡£Ôò±íʾ¼×´¼È¼ÉÕÈȵÄÈÈ»¯Ñ§·½³ÌʽΪ                                 ¡£
£¨2£©ÈôÊÊÁ¿µÄN2ºÍO2ÍêÈ«·´Ó¦£¬Ã¿Éú³É23gNO2ÐèÒªÎüÊÕ16£®95kJÈÈÁ¿£¬Ôò±íʾ¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪ                                   ¡£
£¨3£©ÓÃNA±íʾ°¢·ü¼ÓµÂÂÞ³£Êý£¬ÔÚC2H2£¨Æø̬£©ÍêȫȼÉÕÉú³ÉCO2ºÍҺ̬ˮµÄ·´Ó¦ÖУ¬Ã¿ÓÐ5NA¸öµç×ÓתÒÆʱ£¬·Å³ö650kJµÄÈÈÁ¿£¬Ôò±íʾ¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪ   _________________________________________________¡£
£¨4£©ÒÑÖª²ð¿ª1molH¡ªH¼ü£¬1molN¡ªH¼ü£¬1molN¡ÔN¼ü·Ö±ðÐèÒªµÄÄÜÁ¿ÊÇ436kJ¡¢391kJ¡¢946kJ£¬ÔòN2ÓëH2·´Ó¦Éú³ÉNH3µÄÈÈ»¯Ñ§·½³ÌʽΪ                              ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

£¨1£©ÒÑÖª£º¢ÙFe(s)+1£¯2O2(g)=FeO(s)      ¡÷H1=-272.0KJ¡¤mol-1
¢Ú2Al(s)+3£¯2(g)=Al2O3(s)    ¡÷H2=-1675.7KJ¡¤mol-1
AlºÍFeO·¢ÉúÂÁÈÈ·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽÊÇ__                                 __¡£
ijͬѧÈÏΪ£¬ÂÁÈÈ·´Ó¦¿ÉÓÃÓÚ¹¤ÒµÁ¶Ìú£¬ÄãµÄÅжÏÊÇ_   (Ìî¡°ÄÜ¡±»ò¡°²»ÄÜ¡±)£¬ÄãµÄÀíÓÉ                                                                          

£¨2£©·´Ó¦ÎïÓëÉú³ÉÎï¾ùΪÆø̬µÄij¿ÉÄæ·´Ó¦ÔÚ²»Í¬Ìõ¼þϵķ´Ó¦Àú³Ì·Ö±ðΪA£®B£¬ÈçͼËùʾ¡£¢Ù¾ÝͼÅжϸ÷´Ó¦ÊÇ  (Ìî¡°Îü¡±»ò¡°·Å¡± )ÈÈ·´Ó¦£¬µ±·´Ó¦´ïµ½Æ½ºâºó£¬ÆäËûÌõ¼þ²»±ä£¬£¬Éý¸ßζȣ¬·´Ó¦ÎïµÄת»¯Âʽ«      (Ìî¡°Ôö´ó¡±¡¢ ¡°¼õС¡±»ò¡°²»±ä¡±)¡£
¢ÚÆäÖÐBÀú³Ì±íÃ÷´Ë·´Ó¦²ÉÓõÄÌõ¼þΪ    
(Ìî×Öĸ)¡£
A.Éý¸ßζȠ       B.Ôö´ó·´Ó¦ÎïµÄŨ¶È
C.½µµÍζȠ       D.ʹÓô߻¯¼Á

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

µª¼°Æ仯ºÏÎïÓëÈËÀà¸÷·½ÃæÓÐ×ÅÃÜÇеÄÁªÏµ¡£¢ñÏÖÓÐÒ»Ö§15mLµÄÊԹܣ¬³äÂúNOµ¹ÖÃÓÚË®²ÛÖУ¬ÏòÊÔ¹ÜÖлº»ºÍ¨ÈëÒ»¶¨Á¿ÑõÆø£¬µ±ÊÔ¹ÜÄÚÒºÃæÎȶ¨Ê±£¬Ê£ÓàÆøÌå3mL¡£ÔòͨÈëÑõÆøµÄÌå»ý¿ÉÄÜΪ                mL¡£
¢òÄ¿Ç°£¬Ïû³ýµªÑõ»¯ÎïÎÛȾÓжàÖÖ·½·¨¡£
£¨1£©ÓÃCH4´ß»¯»¹Ô­µªÑõ»¯Îï¿ÉÒÔÏû³ýµªÑõ»¯ÎïµÄÎÛȾ¡£ÒÑÖª£º
¢ÙCH4(g)+4NO2(g)= 4NO(g)+CO2(g)+2H2O(g) ¡÷H="-574kJ/mol"
¢ÚCH4(g)+4NO(g)= 2N2(g)+CO2(g)+2H2O(g) ¡÷H="-1160kJ/mol"
¢ÛH2O(g)= H2O(l)                     ¡÷H=-44kJ/mol
д³öCH4(g)ÓëNO2(g)·´Ó¦Éú³ÉN2(g) ¡¢CO2(g)ºÍH2O(l)µÄÈÈ»¯Ñ§·½³Ìʽ                           ¡£  
£¨2£©ÓûîÐÔÌ¿»¹Ô­·¨´¦ÀíµªÑõ»¯ÎÓйط´Ó¦Îª£º
ijÑо¿Ð¡×éÏòºãÈÝÃܱÕÈÝÆ÷¼ÓÈëÒ»¶¨Á¿µÄ»îÐÔÌ¿ºÍNO£¬ºãΣ¨T0C)Ìõ¼þÏ·´Ó¦£¬·´Ó¦½øÐе½²»Í¬Ê±¼ä²âµÃ¸÷ÎïÖʵÄŨ¶ÈÈçÏ£º

¢Ù²»ÄÜ×÷ΪÅжϷ´Ó¦´ïµ½»¯Ñ§Æ½ºâ״̬µÄÒÀ¾ÝÊÇ_______ £»£¨Ñ¡Ìî×Öĸ´úºÅ£©
A£®ÈÝÆ÷ÄÚCO2µÄŨ¶È±£³Ö²»±ä
B£®vÕý(N2)="2" vÕý(NO)
C£®ÈÝÆ÷ÄÚѹǿ±£³Ö²»±ä
D£®»ìºÏÆøÌåµÄÃܶȱ£³Ö²»±ä
E£®»ìºÏÆøÌåµÄƽ¾ùÏà¶Ô·Ö×ÓÖÊÁ¿±£³Ö²»±ä
¢ÚÇ°20·ÖÖÓ£¬Æ½¾ù·´Ó¦ËÙÂÊv(NO)=                      ¡£v(NO)=£¨0£®1- 0£®04£©/ 20 = 0£®003mol¡¤L-1¡¤ min-1
¢ÛÔÚT0Cʱ£¬¸Ã·´Ó¦µÄƽºâ³£ÊýΪ_______(±£ÁôÁ½Î»Ð¡Êý)£»
¢ÜÔÚ30 min£¬¸Ä±äijһÌõ¼þ·´Ó¦ÖØдﵽƽºâ£¬Ôò¸Ä±äµÄÌõ¼þÊÇ_______                      ¡£
£¨3£©¿Æѧ¼ÒÕýÔÚÑо¿ÀûÓô߻¯¼¼Êõ½«³¬ÒôËÙ·É»úβÆøÖеÄNOºÍCOת±ä³ÉCO2ºÍN2,Æ䷴ӦΪ£º

Ñо¿±íÃ÷£ºÔÚʹÓõÈÖÊÁ¿´ß»¯¼Áʱ£¬Ôö´ó´ß»¯¼ÁµÄ±È±íÃæ»ý¿ÉÌá¸ß»¯Ñ§·´Ó¦ËÙÂÊ¡£ÎªÁË·Ö±ðÑé֤ζȡ¢´ß»¯¼ÁµÄ±È±íÃæ»ý¶Ô»¯Ñ§·´Ó¦ËÙÂʵÄÓ°Ïì¹æÂÉ£¬Ä³Í¬Ñ§Éè¼ÆÁËÈý×éʵÑ飬²¿·ÖʵÑéÌõ¼þÒѾ­ÌîÔÚϱíÖС£
ʵÑé±àºÅ
T(0C)
NO³õʼŨ¶È
£¨mol/L£©
CO³õʼŨ¶È
£¨mol/L£©
´ß»¯¼ÁµÄ±È
±íÃæ»ý(m2/g)
¢ñ
280
1£®20¡Á10-3
5£®80¡Á10-3
82
¢ò
a
b
c
124
¢ó
350
d
e
124
 
ÉϱíÖУºa=_______,b=________,e=________ ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

µªÊǵØÇòÉϺ¬Á¿·á¸»µÄÒ»ÖÖÔªËØ£¬µª¼°Æ仯ºÏÎïÔÚ¹¤Å©ÒµÉú²ú¡¢Éú»îÖÐÓÐ×ÅÖØÒª×÷Óá£Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÓÒͼÊÇN2ºÍH2·´Ó¦Éú³É2 mol NH3¹ý³ÌÖÐÄÜÁ¿±ä
»¯Ê¾Òâͼ£¬Ð´³öÉú³ÉNH3µÄÈÈ»¯Ñ§·½³Ìʽ£º
_____________________________________________
___________________________¡£
£¨2£©ÓÉÆø̬»ù̬ԭ×ÓÐγÉ1 mol»¯Ñ§¼üÊͷŵÄ×îµÍÄÜÁ¿½Ð¼üÄÜ¡£´Ó»¯Ñ§¼üµÄ½Ç¶È·ÖÎö£¬»¯
ѧ·´Ó¦µÄ¹ý³Ì¾ÍÊÇ·´Ó¦ÎïµÄ»¯Ñ§¼üµÄÆÆ»µºÍÉú³ÉÎïµÄ»¯Ñ§¼üµÄÐγɹý³Ì¡£ÔÚ»¯Ñ§·´Ó¦¹ý
³ÌÖУ¬²ð¿ª»¯Ñ§¼üÐèÒªÏûºÄÄÜÁ¿£¬Ðγɻ¯Ñ§¼üÓÖ»áÊÍ·ÅÄÜÁ¿¡£
ÒÑÖª·´Ó¦N2(g)£«3H2(g)??2NH3(g)¡¡¦¤H£½a kJ¡¤mol£­1¡£
ÊÔ¸ù¾Ý±íÖÐËùÁмüÄÜÊý¾Ý¹ÀËãaµÄÊýÖµ£º________¡£
»¯Ñ§¼ü
H¡ªH
N¡ªH
N¡ÔN
¼üÄÜkJ¡¤mol£­1
436
391
945
 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÒÑÖª£º
£¨1£©Fe2O3(s) £«C(s)=CO2(g)£«2Fe(s) ¦¤H£½£«234£®1 kJ/mol
£¨2£©C(s)£«O2(g)=CO2(g) ¦¤H£½£­393£®5 kJ/mol
Ôò2Fe(s)£«O2(g)=Fe2O3(s) µÄ¦¤HÊÇ£¨  £©
A£®£­824£®4 kJ/molB£®£­627£®6 kJ/mol
C£®£­744£®7 kJ/molD£®£­169£®4 kJ/mol

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

¼×´¼ÊÇÒ»ÖÖÓÅÖÊȼÁÏ£¬¿ÉÖÆ×÷ȼÁϵç³Ø¡£
£¨1£©ÎªÌ½¾¿ÓÃCO2Éú²úȼÁϼ״¼µÄ·´Ó¦Ô­Àí£¬ÏÖ½øÐÐÈçÏÂʵÑ飺ÔÚÌå»ýΪ1LµÄÃܱÕÈÝÆ÷ÖУ¬³äÈë1molCO2ºÍ3molH2£¬Ò»¶¨Ìõ¼þÏ·¢Éú·´Ó¦£ºCO2(g)£«3H2(g)CH3OH(g)£«H2O(g)    ¡÷H£½£­49.0kJ/mol
²âµÃCO2ºÍCH3OH(g)µÄŨ¶ÈËæʱ¼ä±ä»¯ÈçÓÒͼ¡£Çë»Ø´ð£º

¢Ù´Ó·´Ó¦¿ªÊ¼µ½Æ½ºâ£¬ÇâÆøµÄ·´Ó¦ËÙÂÊ£ºv(H2)£½                        ¡£
¢ÚÄܹ»ËµÃ÷¸Ã·´Ó¦ÒѴﵽƽºâµÄÊÇ_________¡£
A£®ºãΡ¢ºãÈÝʱ£¬ÈÝÆ÷ÄÚµÄѹǿ²»Ôٱ仯
B£®ºãΡ¢ºãÈÝʱ£¬ÈÝÆ÷ÄÚ»ìºÏÆøÌåµÄÃܶȲ»Ôٱ仯
C£®Ò»¶¨Ìõ¼þÏ£¬CO¡¢H2ºÍCH3OHµÄŨ¶È±£³Ö²»±ä
D£®Ò»¶¨Ìõ¼þÏ£¬µ¥Î»Ê±¼äÄÚÏûºÄ3molH2µÄͬʱÉú³É1molCH3OH
¢ÛÏÂÁдëÊ©ÖÐÄÜʹƽºâ»ìºÏÎïÖÐn(CH3OH)/n(CO2)Ôö´óµÄÊÇ           ¡£
A£®¼ÓÈë´ß»¯¼Á                B£®³äÈëHe(g)£¬Ê¹ÌåϵѹǿÔö´ó
C£®½«H2O(g)´ÓÌåϵÖзÖÀë      D£®½µµÍζÈ
¢ÜÇó´ËζÈ(T1)ϸ÷´Ó¦µÄƽºâ³£ÊýK1£½                (¼ÆËã½á¹û±£ÁôÈýλÓÐЧÊý×Ö)¡£
¢ÝÁíÔÚζÈ(T2)Ìõ¼þϲâµÃƽºâ³£ÊýK2£¬ÒÑÖªT2£¾T1£¬ÔòK2     (Ìî¡°£¾¡±¡¢¡°£½¡±»ò¡°£¼¡±)K1¡£
£¨2£©ÒÔCH3OHΪȼÁÏ(ÒÔKOHÈÜÒº×÷µç½âÖÊÈÜÒº)¿ÉÖƳÉCH3OHȼÁϵç³Ø(µç³Ø×Ü·´Ó¦Ê½£º2CH3OH£«3O2£«4OH£­£½2CO32£­£«6H2O)£¬Ôò³äÈëCH3OHµÄµç¼«Îª  ¼«£¬³äÈëO2µÄµç¼«·´Ó¦Ê½              ¡£
£¨3£©ÒÑÖªÔÚ³£Î³£Ñ¹Ï£º
¢Ù2CH3OH(l)£«3O2(g)£½2CO2(g)£«4H2O(g)  ¡÷H1
¢Ú2CO(g)+O2(g)£½2CO2(g) ¡÷H2
Ôò1mol¼×´¼²»ÍêȫȼÉÕÉú³ÉÒ»Ñõ»¯Ì¼ºÍÆø̬ˮʱ·´Ó¦µÄ¡÷H£½                   ¡££¨Óú¬¡÷H1¡¢¡÷H2µÄʽ×Ó±íʾ£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

ÈËÃÇÒѾ­ÑÐÖƳöÒÔ±ûÍéΪȼÁϵÄÐÂÐÍȼÁϵç³Ø£¬µç½âÖÊΪÈÛÈÚ̼ËáÑΣ¬µç³Ø×Ü·´Ó¦·½³ÌʽΪ£ºC3H8+5O2=3CO2+4H2O¡£
£¨1£©ÒÑÖª£º2C3H8£¨g£©+7O2£¨g£©=6CO£¨g£©+8H2O£¨l£©
C£¨s£©+O2£¨g£©=CO2£¨g£©            
2C£¨s£©+O2£¨g£©=2CO£¨g£©           
Ôò·´Ó¦C3H8£¨g£©+5O2£¨g£©=3CO2£¨g£©+4H2O£¨1£©µÄ¡÷H___________________¡£. 
£¨2£©¸Ãµç³ØµÄÕý¼«Í¨ÈëO2ºÍCO2£¬¸º¼«Í¨Èë±ûÍ飬ÔòÕý¼«µÄµç¼«·´Ó¦Ê½Îª_________________£¬µç³Ø¹¤×÷ʱCO32¡ªÒÆÏò_____________¼«¡£
£¨3£©Óøõç³Øµç½â1L 1 mol¡¤L¡ª1µÄAgNO3ÈÜÒº£¬´Ëµç½â³Ø·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ______________________£»µ±¸Ãµç³ØÏûºÄ0.005molC3H8ʱ£¬ËùµÃÈÜÒºµÄpHΪ__________£¨ÈÜÒºÌå»ý±ä»¯ºöÂÔ²»¼Æ£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÒÑÖª2H2(g)+O2(g) ¡ú2H2O(l£©+571.6kJ£¬2H2(g)+O2(g) ¡ú2H2O(g)+483.6kJ¡£ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ
A£®2 molH2(g)ºÍ1molO2(g)µÄ×ÜÄÜÁ¿Ð¡ÓÚ2mol H2O (l)µÄÄÜÁ¿
B£®1 mol H2O (g)·Ö½â³ÉH2(g)ºÍO2 (g)£¬ÎüÊÕ241.8kJÄÜÁ¿
C£®1mol H2O (l)±ä³É1mo1 H2O (g)£¬ÎüÊÕ88 kJÄÜÁ¿
D£®µÈÖÊÁ¿µÄH2O (g)±ÈH2O(l£©Ëùº¬ÄÜÁ¿µÍ

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸