¡¾ÌâÄ¿¡¿¢ñ.£¨1£©25¡æ 101 kPaʱ£¬ÇâÆøºÍÑõÆø·´Ó¦Éú³É1molË®ÕôÆø·ÅÈÈ241.8kJ£¬¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪ____¡£

£¨2£©Èô1gË®ÕôÆøת»¯ÎªÒºÌ¬Ë®·ÅÈÈ2.444 kJ£¬Ôò·´Ó¦2H2(g)£«O2(g)=2H2O(l)µÄ¦¤H£½___£¬ÓÉ´Ë¿ÉÖªÇâÆøµÄȼÉÕÈÈΪ____¡£(½á¹û±£ÁôСÊýµãºóһλ)

¢ò.ÒÑÖªH£«(aq)£«OH£­(aq)=H2O(l) ¦¤H£½£­57.3kJ/mol£¬»Ø´ðÏÂÁÐÓйØÖкͷ´Ó¦µÄÎÊÌ⣺

£¨1£©ÓÃ0.1molBa(OH)2Åä³ÉÏ¡ÈÜÒºÓë×ãÁ¿Ï¡ÏõËá·´Ó¦£¬Äܷųö___kJµÄÄÜÁ¿¡£

£¨2£©ÈçͼËùʾװÖÃÖУ¬ÒÇÆ÷AµÄÃû³ÆÊÇ___£¬×÷ÓÃÊÇ__£»ËéÅÝÄ­ËÜÁϵÄ×÷ÓÃÊÇ___¡£

£¨3£©Í¨¹ýʵÑé²â¶¨µÄÖкÍÈȵĦ¤H³£³£´óÓÚ£­57.3kJ/mol£¬ÆäÔ­Òò¿ÉÄÜÊÇ___¡£

£¨4£©ÓÃÏàͬŨ¶ÈºÍÌå»ýµÄ°±Ë®(NH3¡¤H2O)´úÌæNaOHÈÜÒº½øÐÐÉÏÊöʵÑ飬²âµÃµÄÖкÍÈȵÄÊýÖµ____(Ìî¡°Æ«´ó¡±¡¢¡°Æ«Ð¡¡±»ò¡°ÎÞÓ°Ï족)¡£

¡¾´ð°¸¡¿H2(g)£«O2(g)=H2O(g) ¦¤H£½£­241.8 kJ/mol £­571.6kJ/mol 285.8kJ/mol 11.46 »·Ðβ£Á§½Á°è°ô ½Á°è£¬Ê¹ÈÜÒº³ä·Ö»ìºÏ ±£Î¡¢¸ôÈÈ¡¢¼õÉÙʵÑé¹ý³ÌÖеÄÈÈÁ¿Ëðʧ ʵÑéÖв»¿É±ÜÃâÓÐÉÙÁ¿ÈÈÁ¿Ëðʧ ƫС

¡¾½âÎö¡¿

¢ñ.(1)¸ù¾ÝÇâÆøºÍÑõÆø·´Ó¦Éú³É1molË®ÕôÆø·ÅÈÈ241.8kJ£¬Êéд·´Ó¦µÄÈÈ»¯Ñ§·´Ó¦·½³Ìʽ£»

(2)¸ù¾Ý1gË®ÕôÆøת»¯³ÉҺ̬ˮ·ÅÈÈ2.444kJ£¬¼ÆËã1molË®ÕôÆøת»¯ÎªÒºÌ¬Ë®·Å³öµÄÈÈÁ¿£¬ÔÙ¸ù¾Ý¸Ç˹¶¨ÂɼÆËãÆäìʱ䣻

¢ò. (1)ÓÉH+(aq)+OH-(aq)¨TH2O(l)¡÷H=-57.3kJmol-1¿ÉÖªÉú³É1molH2O·Å³öÈÈÁ¿Îª57.3kJ£¬È»ºó¸ù¾ÝË®µÄÎïÖʵÄÁ¿ÓëÈÈÁ¿³ÉÕý±ÈÇó³öÈÈÁ¿£»

(2)¸ù¾ÝͼʾÅжÏÒÇÆ÷Ãû³Æ£»ÖкÍÈȲⶨʱӦ¾¡Á¿¼õÉÙÈÈÁ¿É¢Ê§£»

(3)ʵ¼ÊʵÑé¹ý³ÌÖÐÒ»¶¨ÓÐÈÈÁ¿É¢Ê§£»

(4)NH3H2OΪÈõµç½âÖÊ£¬µçÀëÐèÒªÎüÈÈ£¬¾Ý´Ë·ÖÎöÅжϡ£

¢ñ.(1)ÓÉÇâÆøºÍÑõÆø·´Ó¦Éú³É1molË®ÕôÆø·ÅÈÈ241.8kJ£¬Éú³É2molË®ÕôÆø·Å³öÈÈÁ¿Îª483.6kJ£¬¸Ã·´Ó¦ÈÈ»¯Ñ§·´Ó¦·½³ÌʽΪH2(g)£«O2(g)=H2O(g) ¦¤H£½£­241.8 kJ/mol»ò2H2(g)+O2(g)=2H2O(g)¡÷H=-483.6kJ/mol£¬¹Ê´ð°¸Îª£ºH2(g)£«O2(g)=H2O(g) ¦¤H£½£­241.8 kJ/mol»ò2H2(g)+O2(g)=2H2O(g)¡÷H=-483.6kJ/mol£»

(2)Èô1gË®ÕôÆøת»¯³ÉҺ̬ˮ·ÅÈÈ2.444kJ£¬Ôò1molË®ÕôÆøת»¯ÎªÒºÌ¬Ë®·Å³öÈÈÁ¿==43.992kJ/mol£¬¸ù¾Ý¸Ç˹¶¨ÂÉÖªÆäìʱä=-483.6kJ/mol+(-43.992kJ/mol)¡Á2=-571.58kJ/mol¡Ö-571.6kJ/mol£¬ÔòÇâÆøµÄȼÉÕÈÈ==285.8kJ/mol£¬¹Ê´ð°¸Îª£º-571.6kJ/mol£»285.8kJ/mol£»

¢ò.(1)ÓÉH+(aq)+OH-(aq)¨TH2O(l)¡÷H=-57.3kJmol-1¿ÉÖªÉú³É1molH2O·Å³öÈÈÁ¿Îª57.3kJ£¬0.1mol Ba(OH)2Åä³ÉÏ¡ÈÜÒºÓë×ãÁ¿Ï¡ÏõËá·´Ó¦¿ÉµÃ0.2molH2O£¬ËùÒԷųöµÄÈÈÁ¿Îª57.3kJ¡Á0.2=11.46kJ£¬¹Ê´ð°¸Îª£º11.46£»

(2)¸ù¾Ýͼʾ£¬ÒÇÆ÷AÊÇ»·Ðβ£Á§½Á°è°ô£¬¿ÉÆðµ½½Á°è£¬Ê¹ÈÜÒº³ä·Ö»ìºÏµÄÄ¿µÄ£»ÖкÍÈȲⶨʵÑé³É°ÜµÄ¹Ø¼üÊDZ£Î¹¤×÷£¬´óСÉÕ±­Ö®¼äÌîÂúËéÅÝÄ­ËÜÁÏ¿ÉÒÔÆðµ½±£Î¡¢¸ôÈÈ£¬¼õÉÙÈÈÁ¿É¢Ê§µÄ×÷Ó㬹ʴð°¸Îª£º»·Ðβ£Á§½Á°è°ô£»½Á°è£¬Ê¹ÈÜÒº³ä·Ö»ìºÏ£»±£Î¡¢¸ôÈÈ£¬¼õÉÙʵÑé¹ý³ÌÖеÄÈÈÁ¿É¢Ê§£»

(3) ʵ¼ÊʵÑé¹ý³ÌÖÐÒ»¶¨Óв¿·ÖÈÈÁ¿É¢Ê§£¬Òò´ËÇóµÃµÄÖкÍÈÈ¡÷H´óÓÚ-57.3kJmol-1£¬¹Ê´ð°¸Îª£ºÊµÑé¹ý³ÌÖв»¿É±ÜÃâÓÐÈÈÁ¿É¢Ê§£»

(4)NH3H2OΪÈõ¼î£¬µçÀë¹ý³ÌΪÎüÈȹý³Ì£¬ÓÃÏàͬŨ¶ÈºÍÌå»ýµÄ°±Ë®(NH3H2O)´úÌæNaOHÈÜÒº½øÐÐÉÏÊöʵÑ飬²âµÃµÄÖкÍÈȵÄÊýֵƫС£¬¹Ê´ð°¸Îª£ºÆ«Ð¡¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÒÑÖªAΪµ­»ÆÉ«¹ÌÌ壬RÊǵؿÇÖк¬Á¿×î¶àµÄ½ðÊôÔªËصĵ¥ÖÊ£¬TΪÉú»îÖÐʹÓÃ×î¹ã·ºµÄ½ðÊôµ¥ÖÊ£¬DÊǾßÓдÅÐԵĺÚÉ«¾§Ìå £¬C¡¢FÊÇÎÞÉ«ÎÞζµÄÆøÌ壬HÊÇ°×É«³Áµí£¬WÈÜÒºÖеμÓKSCNÈÜÒº³öÏÖºìÉ«¡£

(1)ÎïÖÊDµÄ»¯Ñ§Ê½Îª___________£¬HµÄÃû³Æ___________¡£

(2)ÔÚ³±Êª¿ÕÆøÖбä³ÉMµÄʵÑéÏÖÏóÊÇ______________£¬»¯Ñ§·½³ÌʽΪ___________¡£

(3)ÓëWÈÜÒºÒ²ÄÜ·¢Éú·´Ó¦£¬Æä·´Ó¦µÄÀàÐÍΪ______(ÌîÐòºÅ)¡£

»¯ºÏ·´Ó¦ Öû»·´Ó¦ ¸´·Ö½â·´Ó¦ Ñõ»¯»¹Ô­·´Ó¦

(4)ºÍRÔÚÈÜÒºÖз´Ó¦Éú³ÉFµÄÀë×Ó·½³ÌʽΪ____________________¡£

(5)½«Í¶Èëµ½EÈÜÒºÖУ¬¿ÉÒԹ۲쵽µÄÏÖÏóÊÇ£º___________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿Ïû³ýβÆøÖеÄNOÊÇ»·¾³¿ÆѧÑо¿µÄÈȵã¿ÎÌâ¡£

I£®NOÑõ»¯»úÀí

ÒÑÖª£º2NO(g)+O2(g) 2NO2(g)¡÷H=-110kJ¡¤mol-1

25¡æʱ£¬½«NOºÍO2°´ÎïÖʵÄÁ¿Ö®±ÈΪ2:1³äÈëºãÈÝ·´Ó¦ÈÝÆ÷ÖУ¬Óòâѹ·¨Ñо¿Æä·´Ó¦µÄ½øÐÐÇé¿ö¡£ÌåϵµÄ×ÜѹǿpËæʱ¼ätµÄ±ä»¯ÈçϱíËùʾ(ºöÂÔNO2ÓëN2O4µÄת»¯)

t/min

0

80

160

p/kPa

75.0

63.0

55.0

55.0

£¨1£©0~80min£¬v(O2)=_____kPa/min£»Ëæ×Å·´Ó¦½øÐУ¬·´Ó¦ËÙÂÊÖð½¥¼õСµÄÔ­ÒòÊÇ______________¡£

ÓÃƽºâ·Öѹ´úÌæƽºâŨ¶ÈËùµÃµ½µÄƽºâ³£ÊýÓÃK(p)±íʾ£¬25¡æʱ£¬K(p)µÄֵΪ_______(±£Áô3λÓÐЧÊý×Ö)¡£

£¨2£©²éÔÄ×ÊÁÏ£¬¶ÔÓÚ×Ü·´Ó¦2NOg)+O2(g) 2NO2(g)ÓÐÈçÏÂÁ½²½Àú³Ì

µÚÒ»²½2NO(g) N2O2(g) ¿ìËÙ·´Ó¦

µÚ¶þ²½N2O2(g)+O2(g) 2NO2(g) Âý·´Ó¦

×Ü·´Ó¦ËÙÂÊÖ÷ÒªÓɵÚ_____²½¾ö¶¨£»ÈôÀûÓ÷Ö×Ó²¶»ñÆ÷Êʵ±¼õÉÙ·´Ó¦ÈÝÆ÷ÖеÄN2O2£¬×Ü·´Ó¦µÄƽºâ³£ÊýK(p)½«___(Ìî¡°Ôö´ó¡±¡¢¡°¼õС¡±»ò¡°²»±ä¡±)£»ÈôÌá¸ß·´Ó¦Î¶ÈÖÁ35¡æ£¬ÔòÌåϵѹǿP(35¡æ)______P(25¡æ)(Ìî¡°´óÓÚ¡±¡¢¡°µÈÓÚ¡±»ò¡°Ð¡ÓÚ¡±)¡£

II. ijζÈÏÂÒ»ÃܱÕÈÝÆ÷ÖгäÈëÒ»¶¨Á¿µÄNO2£¬²âµÃNO2Ũ¶ÈËæʱ¼ä±ä»¯µÄÇúÏßÈçÉÏͼËùʾ¡£

£¨1£©·´Ó¦Ìåϵ´ïƽºâºóѹǿΪP1£¬ÈôÉý¸ßζȣ¬Ôٴδïƽºâºó£¬»ìºÏÆøÌåµÄƽ¾ùÏà¶Ô·Ö×ÓÖÊÁ¿______Ìî¡°Ôö´ó¡±¡¢¡°¼õС¡±»ò¡°²»±ä¡±£»

£¨2£©ÈôÔÚºãκãÈÝÌõ¼þÏ£¬ÏòƽºâÌåϵÖгäÈëÒ»¶¨Á¿O2£¬Ôٴδïƽºâºó£¬²âµÃѹǿΪP2£¬c(O2)=0.09mol/L£¬ÔòP2£ºP1=______

£¨3£©¸ÃζÈÏ·´Ó¦2NO(g)+O2(g) 2NO2(g)µÄ»¯Ñ§Æ½ºâ³£ÊýKΪ______¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿³£ÎÂÏ£¬ÔÚpH=5µÄCH3COOHÈÜÒºÖдæÔÚÈçϵçÀëƽºâ£ºCH3COOHCH3COO-+H+£¬¶ÔÓÚ¸Ãƽºâ£¬ÏÂÁÐÐðÊöÕýÈ·µÄÊÇ

A£®¼ÓÈëˮʱ£¬Æ½ºâÏòÓÒÒƶ¯£¬CH3COOHµçÀë³£ÊýÔö´ó

B£®¼ÓÈëÉÙÁ¿CH3COONa¹ÌÌ壬ƽºâÏòÓÒÒƶ¯

C£®¼ÓÈëÉÙÁ¿NaOH¹ÌÌ壬ƽºâÏòÓÒÒƶ¯£¬c(H+)¼õС

D£®¼ÓÈëÉÙÁ¿pH=5µÄÁòËᣬÈÜÒºÖÐc(H+)Ôö´ó

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿£¨Ò»£©ÊµÑéÊÒÓÃͼ¼××°ÖÃÖƱ¸SO3²¢²â¶¨SO2´ß»¯Ñõ»¯ÎªSO3µÄת»¯ÂÊ¡£

ÒÑÖª£ºSO3ÈÛµãΪ16.8¡æ£¬·ÐµãΪ44.8¡æ£¬¼ÙÉèÆøÌå½øÈë×°ÖÿÉÒÔ±»ÍêÈ«ÎüÊÕ£¬²»¿¼ÂÇ¿ÕÆøµÄÓ°Ïì¡£

£¨1£©AÖÐʹÓÃŨÁòËáµÄÖÊÁ¿·ÖÊýΪ70%µÄÔ­ÒòÊÇ___£¬BÖÐŨH2SO4µÄ×÷ÓÃÊÇ___¡£

£¨2£©µ±ÊµÑéֹͣͨÈëSO2£¬Ï¨Ãð¾Æ¾«µÆºó£¬»¹ÐèÒª¼ÌÐøͨһ¶Îʱ¼äµÄÑõÆø£¬ÆäÄ¿µÄÊÇ___¡£

£¨3£©ÊµÑé½áÊøºó²âµÃ×°ÖÃDÔö¼ÓÁËag£¬×°ÖÃEÖеijÁµíÏ´µÓºæ¸ÉºóÆäÖÊÁ¿Îªbg¡£ÔòEÖеijÁµíµÄ»¯Ñ§³É·ÖÊÇ___£¨Ìîд»¯Ñ§Ê½£©£¬±¾ÊµÑéÖÐSO2ת»¯ÂÊΪ___£¨ÓôúÊýʽ±íʾ£¬²»Óû¯¼ò£©¡£

£¨¶þ£©SO3ÈÜÓÚŨÁòËáºó¿ÉµÃµ½·¢ÑÌÁòËᣬ¹¤ÒµÉϰѸÉÔïµÄÂÈ»¯ÇâÆøÌåͨÈëµ½·¢ÑÌÁòËáÖпÉÒԵõ½HSO3Cl¡£HSO3ClÊÇÒ»ÖÖÎÞÉ«ÒºÌ壬·ÐµãΪ152¡æ£¬ÓÐÇ¿¸¯Ê´ÐÔ£¬Óöʪ¿ÕÆø²úÉúÇ¿Áҵİ×Îí¡£ÏÖÓÃͼÒÒËùʾµÄ×°ÖÃÖÆÈ¡HSO3Cl£¨¼Ð³Ö¼°¼ÓÈÈ×°ÖÃÂÔÈ¥£©¡£

£¨1£©HSO3ClÓöʪ¿ÕÆø²úÉúÇ¿Áҵİ×Îí£¬Çë½áºÏÓû¯Ñ§·½³Ìʽ½âÊÍÆäÔ­Òò___¡£

£¨2£©·ÖҺ©¶·Ï·½½ÓµÄëϸ¹Ü£¬Æä×÷ÓÃÊÇ___£»Èô²»ÓÃëϸ¹Ü¶øÖ±½ÓÓ÷ÖҺ©¶·×¢ÈëŨÑÎËᣬ¿ÉÄÜ·¢ÉúµÄÏÖÏóÊÇ___¡£

£¨3£©×°ÖÃFµÄ×÷ÓÃÊÇ___¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÎªÌ½¾¿Ä³ÄÑÈÜÐÔÑÎX(½öº¬ÈýÖÖ³£¼ûÔªËØ)µÄ×é³É£¬Éè¼Æ²¢Íê³ÉÒÔÏÂʵÑé(Á÷³ÌÖв¿·ÖÎïÖÊÒÑÂÔÈ¥)£º

ÒÑÖª£ºÆøÌåAºÍÆøÌåBËùº¬ÔªËØÏàͬ£¬¶¼ÊÇÎÞÉ«ÎÞζÆøÌ壬¹ÌÌåCΪ´¿¾»ÎïÇÒ¾ßÓдÅÐÔ£¬µ¥ÖÊDÊÇÄ¿Ç°½¨ÖþÐÐÒµÓ¦ÓÃ×î¹ã·ºµÄ½ðÊô¡£¸ù¾ÝÉÏÊöÐÅÏ¢£¬»Ø´ðÏÂÁÐÎÊÌ⣺

(1)ÑÎXµÄ»¯Ñ§Ê½Îª________________¡£

(2)ÎÞË®Ìõ¼þÏ£¬ÉÙÁ¿NaH¾ÍÄÜÓë¹ÌÌåC·´Ó¦²¢·Å³ö´óÁ¿µÄÈÈ£¬Ð´³ö¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ___________________¡£

(3)½«²úÉúµÄÆøÌåAÈ«²¿±»100 mL 0.35 mol¡¤L£­1ÇâÑõ»¯ÄÆÈÜÒº³ä·ÖÎüÊÕ£¬·´Ó¦µÄ×ÜÀë×Ó·½³ÌʽΪ____________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÏÂÁз´Ó¦ÊôÓÚ·ÇÑõ»¯»¹Ô­·´Ó¦µÄÊÇ£¨¡¡¡¡£©

A. Fe2O3+3CO2Fe+3CO2 B. NH4NO3N2O¡ü+2H2O

C. 2NaHCO3Na2CO3+CO2¡ü+H2O D. CuO+CO¨TCu+CO2

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÏÂͼÊÇÁòËáÊÔ¼ÁÆ¿±êÇ©ÉϵÄÄÚÈÝ£º

(1)ij»¯Ñ§ÐËȤС×é½øÐÐÁòËáÐÔÖʵÄʵÑé̽¾¿Ê±£¬ÐèÒª490mL 4.6mol¡¤L-1µÄÏ¡ÁòËᣬÏÖÒªÅäÖƸÃŨ¶ÈµÄÈÜÒºËùÐèµÄ²£Á§ÒÇÆ÷³ýÁ¿Í²¡¢ÉÕ±­¡¢²£Á§°ô¡¢½ºÍ·µÎ¹ÜÍ⣬»¹ÐèÒª__________(ÌîÒÇÆ÷Ãû³Æ)£»ÐèÒªÁ¿È¡98%ŨÁòËá____________mL½øÐÐÅäÖÆ£»

(2)ÅäÖÆÈÜҺʱÓÐÈçϲÙ×÷£ºa.Ï¡ÊÍÈܽâb.Ò¡ÔÈc.Ï´µÓd.ÀäÈ´e.Á¿È¡f.½«ÈÜÒºÒÆÖÁÈÝÁ¿Æ¿g.¶¨ÈÝ£¬ÊµÑé²Ù×÷˳ÐòÕýÈ·µÄÊÇ£¨___________£©¡£

A. e¡úa¡úf¡úd¡úc¡úf¡úg¡úb B. e¡úa¡úd¡úf¡úc¡úf¡úg¡úb

C. e¡úa¡úf¡úd¡úc¡úf¡úb¡úg D. e¡úa¡úd¡úf¡úc¡úf¡úb¡úg

(3)ÏÂÁÐΪÅäÖƹý³ÌÖв¿·Ö²Ù×÷µÄʾÒâͼ£¬ÆäÖÐÓдíÎóµÄÊÇ____(ÌîÐòºÅ)£»

(4)ÔÚÅäÖÆ4.6mol¡¤L-1Ï¡ÁòËáµÄ¹ý³ÌÖУ¬ÏÂÁÐÇé¿ö»áÒýÆðÅäÖÆËùµÃµÄÁòËáÈÜÒºÎïÖʵÄÁ¿Å¨¶ÈÆ«¸ßµÄÊÇ___£»

A.δ¾­ÀäÈ´³ÃÈȽ«ÈÜҺעÈëÈÝÁ¿Æ¿ÖÐ B.ÈÝÁ¿Æ¿Ï´µÓºó£¬Î´¸ÉÔï´¦Àí

C.¶¨ÈÝʱÑöÊÓ¹Û²ìÒºÃæ D.δϴµÓÉÕ±­ºÍ²£Á§°ô

(5)ΪÖкÍ100mL 2.3 mol¡¤L-1KOHÈÜÒººóÏÔÖÐÐÔ£¬ÐèÒª¼ÓÈë________mL 4.6mol¡¤L-1Ï¡ÁòËá¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿25¡æ¡¢101kPaÏ£¬Ì¼¡¢ÇâÆø¡¢¼×ÍéºÍÆÏÌÑÌǵÄȼÉÕÈÈÒÀ´ÎÊÇ393.5kJ/mol¡¢285.8kJ/mol¡¢890.3kJ/mol¡¢2800kJ/mol£¬ÔòÏÂÁÐÈÈ»¯Ñ§·½³ÌʽÕýÈ·µÄÊÇ

A.C£¨s£©+O2£¨g£©£½CO£¨g£© ¦¤H=-393.5kJ/mol

B.2H2£¨g£©+O2£¨g£©£½2H2O£¨l£© ¦¤H =+571.6kJ/mol

C.CH4£¨g£©+2O2£¨g£©£½ CO2£¨g£©+2H2O£¨g£© ¦¤H=-890.3kJ/mol

D.C6H12O6£¨s£©+6O2£¨g£©£½6CO2£¨g£©+6H2O£¨l£© ¦¤H=-2800kJ/mol

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸