ÌâÄ¿ÄÚÈÝ

ÓÐX¡¢Y¡¢Z¡¢W¡¢MÎåÖÖ¶ÌÖÜÆÚÔªËØ£¬ÆäÖÐX¡¢Y¡¢Z¡¢WͬÖÜÆÚ£¬Z¡¢MͬÖ÷×壻X£«ÓëM2£­¾ßÓÐÏàͬµÄµç×Ó²ã½á¹¹£»Àë×Ӱ뾶£ºZ2£­>W£­£»YµÄÄÚ²ãµç×Ó×ÜÊýÊÇ×îÍâ²ãµç×ÓÊýµÄ5±¶¡£ÏÂÁÐ˵·¨Öв»ÕýÈ·µÄÊÇ(¡¡¡¡)

A£®W¡¢MµÄijÖÖµ¥ÖÊ¿É×÷Ϊˮ´¦ÀíÖеÄÏû¶¾¼Á

B£®µç½âYW2µÄÈÛÈÚÎï¿ÉÖƵÃYºÍWµÄµ¥ÖÊ

C£®ÏàͬÌõ¼þÏ£¬WµÄÇ⻯ÎïË®ÈÜÒºËáÐÔ±ÈZÈõ

D£®X¡¢MÁ½ÖÖÔªËØÐγɵÄÒ»ÖÖ»¯ºÏÎïÓëË®·´Ó¦¿ÉÉú³ÉMµÄijÖÖµ¥ÖÊ

 

¡¾½âÎö¡¿YµÄÄÚ²ãµç×Ó×ÜÊýÊÇ×îÍâ²ãµç×ÓÊýµÄ5±¶¿ÉÖªYΪþԪËØ£¬ÓÉÌâÄ¿Ëù¸øÐÅÏ¢¿ÉÖªÆäËû¼¸ÖÖÔªËØÔÚÖÜÆÚ±íÖеÄÏà¶ÔλÖÃΪ£¬¿ÉÖªXΪÄÆÔªËØ£¬ZΪÁòÔªËØ£¬WΪÂÈÔªËØ£¬MΪÑõÔªËØ¡£O3ºÍCl2¾ù¿É×÷Ϊˮ´¦ÀíÖеÄÏû¶¾¼Á£¬AÏîÕýÈ·£»µç½âÈÛÈÚµÄMgCl2¿ÉµÃMgºÍCl2£¬BÏîÕýÈ·£»HClÈÜÓÚË®ËáÐÔ±ÈH2SÇ¿£¬CÏî´íÎó£»Na2O2ÓëH2O·´Ó¦¿ÉÉú³ÉO2£¬DÏîÕýÈ·

 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

CH4¡¢H2¡¢C¶¼ÊÇÓÅÖʵÄÄÜÔ´ÎïÖÊ£¬ËüÃÇȼÉÕµÄÈÈ»¯Ñ§·½³ÌʽΪ£º

¢ÙCH4(g)£«2O2(g)=CO2(g)£«2H2O(l)¡¡¦¤H£½£­890.3 kJ¡¤mol£­1£¬

¢Ú2H2(g)£«O2(g)=2H2O(l)¡¡¦¤H£½£­571.6 kJ¡¤mol£­1£¬

¢ÛC(s)£«O2(g)=CO2(g)¡¡¦¤H£½£­393.5 kJ¡¤mol£­1¡£

(1)ÔÚÉÖдæÔÚÒ»ÖÖ¼×Íéϸ¾ú£¬ËüÃÇÒÀ¿¿Ã¸Ê¹¼×ÍéÓëO2×÷ÓòúÉúµÄÄÜÁ¿´æ»î£¬¼×Íéϸ¾úʹ1 mol¼×ÍéÉú³ÉCO2ÆøÌåÓëҺ̬ˮ£¬·Å³öµÄÄÜÁ¿________(Ìî¡°£¾¡±¡°£¼¡±»ò¡°£½¡±)890.3 kJ¡£

(2)¼×ÍéÓëCO2¿ÉÓÃÓںϳɺϳÉÆø(Ö÷Òª³É·ÖÊÇÒ»Ñõ»¯Ì¼ºÍÇâÆø)£ºCH4£«CO2=2CO£«2H2£¬1 g CH4ÍêÈ«·´Ó¦¿ÉÊÍ·Å15.46 kJµÄÈÈÁ¿£¬Ôò£º

¢ÙÏÂͼÄܱíʾ¸Ã·´Ó¦¹ý³ÌÖÐÄÜÁ¿±ä»¯µÄÊÇ________(Ìî×Öĸ)¡£

 

¢ÚÈô½«ÎïÖʵÄÁ¿¾ùΪ1 molµÄCH4ÓëCO2³äÈëijºãÈÝÃܱÕÈÝÆ÷ÖУ¬Ìåϵ·Å³öµÄÈÈÁ¿Ëæ×Åʱ¼äµÄ±ä»¯ÈçͼËùʾ£¬ÔòCH4µÄת»¯ÂÊΪ________¡£

 

(3)C(s)ÓëH2(g)²»·´Ó¦£¬ËùÒÔC(s)£«2H2(g)=CH4(g)µÄ·´Ó¦ÈÈÎÞ·¨Ö±½Ó²âÁ¿£¬µ«Í¨¹ýÉÏÊö·´Ó¦¿ÉÇó³ö£¬C(s)£«2H2(g)=CH4(g)µÄ·´Ó¦ÈȦ¤H£½________¡£

(4)Ä¿Ç°¶ÔÓÚÉÏÊöÈýÖÖÎïÖʵÄÑо¿ÊÇȼÁÏÑо¿µÄÖص㣬ÏÂÁйØÓÚÉÏÊöÈýÖÖÎïÖʵÄÑо¿·½ÏòÖпÉÐеÄÊÇ________(Ìî×Öĸ)¡£

A£®Ñ°ÕÒÓÅÖÊ´ß»¯¼Á£¬Ê¹CO2ÓëH2O·´Ó¦Éú³ÉCH4ÓëO2£¬²¢·Å³öÈÈÁ¿

B£®Ñ°ÕÒÓÅÖÊ´ß»¯¼Á£¬ÔÚ³£Î³£Ñ¹ÏÂʹCO2·Ö½âÉú³É̼ÓëO2

C£®Ñ°ÕÒÓÅÖÊ´ß»¯¼Á£¬ÀûÓÃÌ«ÑôÄÜʹ´óÆøÖеÄCO2Ó뺣µ×¿ª²ÉµÄCH4ºÏ³ÉºÏ³ÉÆø(CO¡¢H2)

D£®½«¹Ì̬̼ºÏ³ÉΪC60£¬ÒÔC60×÷ΪȼÁÏ

 

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø