¡¾ÌâÄ¿¡¿£¨1£©ÒÑÖª£º¢Ù CH3OH(g)£«H2O(g)===CO2(g)£«3H2(g)¡¡¦¤H£½£«49.0 kJ/mol

¢Ú CH3OH(g)£«3/2O2(g)===CO2(g)£«2H2O(g)¡¡¦¤H£½£­192.9 kJ/mol

ÓÉÉÏÊö·½³Ìʽ¿ÉÖª£ºCH3OHµÄȼÉÕÈÈ________(Ìî¡°´óÓÚ¡±¡°µÈÓÚ¡±»ò¡°Ð¡ÓÚ¡±)192.9 kJ/mol¡£

ÒÑ֪ˮµÄÆø»¯ÈÈΪ44 kJ/mol¡£Ôò±íʾÇâÆøȼÉÕÈȵÄÈÈ»¯Ñ§·½³ÌʽΪ__________¡£

£¨2£©ÒÔCO2ÓëNH3ΪԭÁϿɺϳɻ¯·ÊÄòËØ[»¯Ñ§Ê½ÎªCO(NH2)2]¡£

ÒÑÖª£º¢Ù 2NH3(g)£«CO2(g)===NH2CO2NH4(s)¡¡¦¤H£½£­159.5 kJ/mol

¢ÚNH2CO2NH4(s)===CO(NH2)2(s)£«H2O(g)¡¡¦¤H£½£«116.5 kJ/mol

¢ÛH2O(l)===H2O(g)¡¡¦¤H£½£«44.0 kJ/mol

д³öCO2ÓëNH3ºÏ³ÉÄòËغÍҺ̬ˮµÄÈÈ»¯Ñ§·½³ÌʽÊÇ______¡£

£¨3£©ÒÑÖª£º¢Ù Fe(s)£«1/2O2(g)===FeO(s)¡¡¦¤H1£½£­272.0 kJ/mol

¢Ú 2Al(s)£«3/2O2(g)===Al2O3(s)¡¡¦¤H2£½£­1675.7 kJ/mol

AlºÍFeO·¢ÉúÂÁÈÈ·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽÊÇ______¡£Ä³Í¬Ñ§ÈÏΪ£¬ÂÁÈÈ·´Ó¦¿ÉÓÃÓÚ¹¤ÒµÁ¶Ìú£¬ÄãµÄÅжÏÊÇ______(Ìî¡°ÄÜ¡±»ò¡°²»ÄÜ¡±)£¬ÄãµÄÀíÓÉÊÇ_____¡£

¡¾´ð°¸¡¿´óÓÚ H2(g)£«1/2O2(g)===H2O(l)¡¡¦¤H£½-124.6 kJ/mol 2NH3(g)£«CO2(g)===CO(NH2)2(s)£«H2O(l)¡¡¦¤H£½£­87.0 kJ/mol 3FeO(s)£«2Al(s)===Al2O3(s)£«3Fe(s)¡¡¦¤H£½£­859.7 kJ/mol ²»ÄÜ ¸Ã·´Ó¦µÄÒý·¢£¬ÐèÏûºÄ´óÁ¿ÄÜÁ¿£¬³É±¾½Ï¸ß

¡¾½âÎö¡¿

ÈÈ»¯Ñ§·´Ó¦ÊͷŵÄÄÜÁ¿Óë·´Ó¦µÄ¹ý³ÌÎ޹أ¬ÓëÎïÖʵÄʼ̬¡¢ÖÕ̬Óйأ¬ÀûÓøÇ˹¶¨ÂɽøÐмÆË㣻

£¨1£©CH3OHµÄȼÉÕÈÈΪ1molҺ̬¼×´¼ÍêȫȼÉÕÉú³ÉÎȶ¨µÄÑõ»¯ÎïÊͷŵÄÈÈÁ¿£¬¶Ô±È·´Ó¦¢Ú£¬ÔòȼÉÕÈÈ´óÓÚ192.9kJ/mol£»ÒÑ֪ˮµÄÆø»¯ÈÈΪ44 kJ/mol£¬Ôò¢ÛH2O(g)= H2O(l) ¦¤H£½£­44 kJ/mol£¬¸ù¾Ý¸Ç˹¶¨ÂÉ£¬+¢Û¿ÉµÃH2(g)£«1/2O2(g)=H2O(l) ¦¤H£½-124.6 kJ/mol£»

£¨2£©¸ù¾Ý¸Ç˹¶¨ÂÉ£¬¢Ù+¢Ú-¢Û¼´¿ÉµÃµ½2NH3(g)£«CO2(g)=CO(NH2)2(s)£«H2O(l) ¦¤H£½£«116.5£­159.5-44.0=-87.0 kJ/mol£»

£¨3£©¸ù¾Ý¸Ç˹¶¨ÂÉ£¬¢Ú-¢Ù¡Á3¿ÉµÃ2Al(s)£«3FeO(s) =Al2O3(s)+3Fe(s) ¦¤H£½£­859.7 kJ/mol£»ÂÁÈÈ·´Ó¦²»ÄÜÓÃÓÚ¹¤ÒµÁ¶Ìú£¬Òò¸Ã·´Ó¦µÄÒý·¢ÐèÒª½Ï¶àµÄÄÜÔ´£¬³É±¾½Ï¸ß¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿×¶ÐÎÆ¿ÄÚÊ¢ÓÐÆøÌåX£¬µÎ¹ÜÄÚÊ¢ÓÐÒºÌåY¡£Èô¼·Ñ¹½ºÍ·µÎ¹Ü£¬Ê¹ÒºÌåYµÎÈëÆ¿ÖУ¬Õñµ´£¬¹ýÒ»»á¿É¼ûСÆøÇòa¹ÄÆð¡£ÆøÌåXºÍÒºÌåY²»¿ÉÄÜÊÇ£¨ £©

A.XÊÇNH3£¬YÊÇË®

B.XÊÇSO2£¬YÊÇNaOHŨÈÜÒº

C.XÊÇCO2£¬YÊÇÏ¡ÁòËá

D.XÊÇHCl£¬YÊÇNaNO3Ï¡ÈÜÒº

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿2017ÄêÎÒ¹úÊ×ËÒʹÓÃÁËîѺϽð²ÄÁϵĹú²úº½Ä¸ÏÂË®¡£îÑ(Ti)³£ÎÂÏÂÓëËá¡¢¼î¾ù²»·´Ó¦£¬µ«¸ßÎÂÏÂÄܱ»¿ÕÆøÑõ»¯¡£ÓÉîÑÌú¿ó(Ö÷Òª³É·ÖÊÇFeOºÍTiO2)ÌáÈ¡½ðÊôîѵÄÖ÷Òª¹¤ÒÕÁ÷³ÌÈçͼ£¬ÏÂÁÐ˵·¨´íÎóµÄÊÇ

A. ²½ÖèIÖÐ̼×÷»¹Ô­¼Á

B. ²½ÖèIIÖÐδ·¢ÉúÑõ»¯»¹Ô­·´Ó¦

C. ²½ÖèIIIÐèÔÚë²Æø»·¾³ÖнøÐУ¬·ÀÖ¹½ðÊô±»¿ÕÆøÑõ»¯

D. ¿ÉÓÃÏ¡ÁòËá³ýÈ¥½ðÊôîÑÖеÄÉÙÁ¿Ã¾

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÈçͼÊÇζȺÍѹǿ¶Ô·´Ó¦X£«Y2ZÓ°ÏìµÄʾÒâͼ¡£Í¼Öкá×ø±ê±íʾζȣ¬×Ý×ø±ê±íʾƽºâ»ìºÏÆøÌåÖÐZµÄÌå»ý·ÖÊý¡£ÏÂÁÐÐðÊöÕýÈ·µÄÊÇ£¨ £©

A.ÉÏÊö¿ÉÄæ·´Ó¦µÄÕý·´Ó¦Îª·ÅÈÈ·´Ó¦B.X¡¢Y¡¢Z¾ùΪÆø̬

C.XºÍYÖÐ×î¶àÖ»ÓÐÒ»ÖÖΪÆø̬£¬ZΪÆø̬D.ÉÏÊö·´Ó¦µÄÄæ·´Ó¦µÄ¦¤H£¾0

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿FeCl3ÊÇÖÐѧʵÑéÊÒ³£ÓõÄÊÔ¼Á£¬¿ÉÒÔÓÃÀ´ÖƱ¸ÇâÑõ»¯Ìú½ºÌå¡£

£¨1£©ÏÂÁÐÖƱ¸ÇâÑõ»¯Ìú½ºÌåµÄ²Ù×÷·½·¨ÕýÈ·µÄÊÇ____________£¨Ìî×Öĸ£©£»

A£®Ïò±¥ºÍÂÈ»¯ÌúÈÜÒºÖеμÓÉÙÁ¿µÄÇâÑõ»¯ÄÆÏ¡ÈÜÒº

B£®¼ÓÈÈÖó·ÐÂÈ»¯Ìú±¥ºÍÈÜÒº

C£®ÔÚ°±Ë®ÖеμÓÂÈ»¯ÌúŨÈÜÒº

D£®ÔÚ·ÐË®Öеμӱ¥ºÍÂÈ»¯ÌúÈÜÒº£¬Öó·ÐÖÁÒºÌå³ÊºìºÖÉ«

£¨2£©Ð´³öÖƱ¸ÇâÑõ»¯Ìú½ºÌåµÄÀë×Ó·½³Ìʽ_______________£»

£¨3£©ÏÂÁÐÓ뽺ÌåÐÔÖÊÎ޹صÄÊÇ_____________£¨Ìî×Öĸ£©£»

A£®ºÓÁ÷È뺣¿Ú´¦ÐγÉɳÖÞ

B£®Ê¹ÓÃ΢²¨ÊÖÊõµ¶½øÐÐÍâ¿ÆÊÖÊõ£¬¿Éʹ¿ªµ¶´¦µÄѪҺѸËÙÄý¹Ì¶ø¼õÉÙʧѪ

C£®Éö¹¦ÄÜË¥½ßµÈ¼²²¡ÒýÆðµÄѪҺÖж¾£¬¿ÉÀûÓÃѪҺ͸Îö½øÐÐÖÎÁÆ

D£®ÔÚ±¥ºÍÂÈ»¯ÌúÈÜÒºÖеμÓNaOHÈÜÒº£¬²úÉúºìºÖÉ«³Áµí

E£®Ò±½ð³§³£Óøßѹµç³ýÈ¥Ñ̳¾

£¨4£©´ÓÈÜÒºÖзÖÀëÌá´¿Fe(OH)3½ºÌåµÄ·½·¨½Ð_____________£»

£¨5£©ÏòÇâÑõ»¯Ìú½ºÌåÖÐÖðµÎ¼ÓÈëÏ¡ÁòËáÖÁ¹ýÁ¿µÄÏÖÏóÊÇ_____________________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿NAΪ°¢·ü¼ÓµÂÂÞ³£ÊýµÄÖµ£¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ( )

A. ±ê×¼×´¿öÏ£¬22.4 LË®º¬ÓеÄË®·Ö×ÓÊýΪNA

B. ³£Î³£Ñ¹Ï£¬22 g CO2º¬ÓеÄCO2·Ö×ÓÊýΪ0.5NA

C. ±ê×¼×´¿öÏ£¬32 g O2ºÍCO2µÄ»ìºÏÆøÌ庬ÓеÄÑõÔ­×ÓÊýΪ2NA

D. 40 g NaOHÈܽâÔÚ1 LË®ÖУ¬µÃµ½ÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈΪ1 mol/L

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÏÂÁÐÊǶÔijÈÜÒº½øÐÐÀë×Ó¼ì²âµÄ·½·¨ºÍ½áÂÛ£¬ÆäÖÐÕýÈ·µÄÊÇ(¡¡¡¡)

A. Óò¬Ë¿ÕºÈ¡ÉÙÁ¿Ä³ÈÜÒº½øÐÐÑæÉ«·´Ó¦£¬»ðÑæ³Ê»ÆÉ«£¬¸ÃÈÜÒºÒ»¶¨ÊÇÄÆÑÎÈÜÒº

B. ¼ÓÈë×ãÁ¿µÄCaCl2ÈÜÒº£¬²úÉú°×É«³Áµí£¬ÔòÈÜÒºÖÐÒ»¶¨º¬ÓдóÁ¿µÄCO

C. ¼ÓÈëNaOHÈÜÒººó¼ÓÈȲúÉúÓд̼¤ÐÔÆøζµÄÆøÌ壬ÔòÈÜÒºÖÐÒ»¶¨º¬ÓдóÁ¿µÄNH

D. ÏȼÓÈëÊÊÁ¿µÄÑÎËáËữ£¬ÔÙ¼ÓÈëAgNO3ÈÜÒº£¬²úÉú°×É«³Áµí£¬ÔòÈÜÒºÖÐÒ»¶¨º¬ÓдóÁ¿µÄCl£­

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÒÑÖªÓÐÒÔÏÂÎïÖÊÏ໥ת»¯£º

ÊԻشð£º

(1)д³öBµÄ»¯Ñ§Ê½________£¬DµÄ»¯Ñ§Ê½________¡£

(2)д³öÓÉEת±ä³ÉFµÄ»¯Ñ§·½³Ìʽ________________________¡£

(3)д³öÓÃKSCN¼ø±ðGÈÜÒºµÄÀë×Ó·½³Ìʽ________________£»ÏòGÈÜÒº¼ÓÈëAµÄÓйØÀë×Ó·´Ó¦·½³Ìʽ________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿º¬ÓÐa mol FeBr2µÄÈÜÒºÖУ¬Í¨Èëx mol Cl2¡£ÏÂÁи÷ÏîΪͨCl2¹ý³ÌÖУ¬ÈÜÒºÄÚ·¢Éú·´Ó¦µÄÀë×Ó·½³Ìʽ£¬ÆäÖв»ÕýÈ·µÄÊÇ£¨ £©

A.x£½0.4a£¬2Fe2++Cl2£½2Fe3++2Cl-

B.x£½0.6a£¬2Br£­+Cl2£½Br2+2Cl£­

C.x=a£¬2Fe2£«+2Br£­+2Cl2£½Br2+2Fe3£«+4Cl£­

D.x=1.5a£¬2Fe2£«+4Br£­+3Cl2£½2Br2+2Fe3£«+6Cl£­

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸