³ãÈȵįÌÅÄÚÓз´Ó¦£ºC(s)£«O2(g)CO2(g)¡¡¦¤H£½£392 kJ¡¤mol£1£¬Íù¯ÌÅÄÚͨÈëË®ÕôÆøʱ£¬ÓÐÈçÏ·´Ó¦£ºC(s)£«H2O(g)CO(g)£«H2(g)¡¡¦¤H£½£«131 kJ¡¤mol£1£¬CO(g)£«1/2O2(g)CO2(g)¦¤H£½£282 kJ¡¤mol£1£¬H2(g)£«1/2O2(g)H2O(g)¡¡¦¤H£½£241 kJ¡¤mol£1£¬ÓÉÒÔÉÏ·´Ó¦ÍƶÏÍù³ãÈȵįÌÅÄÚͨÈëË®ÕôÆøʱ | |
[¡¡¡¡] | |
A£® |
²»ÄܽÚʡȼÁÏ£¬µ«ÄÜʹ¯»ð˲¼ä¸üÍú |
B£® |
Ëä²»ÄÜʹ¯»ð˲¼ä¸üÍú£¬µ«¿ÉÒÔ½ÚʡȼÁÏ |
C£® |
¼ÈÄÜʹ¯»ð˲¼ä¸üÍú£¬ÓÖÄܽÚʡȼÁÏ |
D£® |
¼È²»ÄÜʹ¯»ð˲¼ä¸üÍú£¬ÓÖ²»ÄܽÚʡȼÁÏ |
±¾ÌâÓ¦´ÓÁ½¸ö·½Ã濼ÂÇ£¬Ò»ÊÇÄÜ·ñʹ¯»ð˲¼ä¸üÍú£¬ÓÉÓÚÍù¯ÌÅÄÚͨÈëË®ÕôÆøʱ£¬ÓÐÈçÏ·´Ó¦·¢Éú£ºC(s)£«H2O(g)CO(g)£«H2(g)£¬Éú³ÉµÄCOºÍH2¶¼ÊÇ¿ÉȼÐÔÆøÌ壬¹ÊÄÜʹ¯»ð˲¼ä¸üÍú£®¶þÊÇÄÜ·ñ½ÚʡȼÁÏ£¬¸ù¾Ý¸Ç˹¶¨ÂÉ£¬C(s)£«H2O(g)CO(g)£«H2(g)¡¡¦¤H£½£«131 kJ¡¤mol£1£¬CO(g)£«1/2O2(g)CO2(g)¦¤H£½£282 kJ¡¤mol£1£¬H2(g)£«1/2O2(g)H2O(g)¡¡¦¤H£½£241 kJ¡¤mol£1£¬Èý¸ö·½³Ì¼ÓÔÚÒ»Æð¼´µÃ×Ü·´Ó¦Ê½C(s)£«O2(g)CO2(g)¡¡¦¤H£½£392 kJ¡¤mol£1£¬¹ÊÓëÏàͬÁ¿µÄ̿ȼÉշųöµÄÈÈÁ¿Ïàͬ£¬Òò´Ë²»ÄܽÚÊ¡ÔÁÏ£® |
Ä꼶 | ¸ßÖÐ¿Î³Ì | Ä꼶 | ³õÖÐ¿Î³Ì |
¸ßÒ» | ¸ßÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ | ³õÒ» | ³õÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ |
¸ß¶þ | ¸ß¶þÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õ¶þ | ³õ¶þÃâ·Ñ¿Î³ÌÍƼö£¡ |
¸ßÈý | ¸ßÈýÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õÈý | ³õÈýÃâ·Ñ¿Î³ÌÍƼö£¡ |
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£ºÑ§Ï°Öܱ¨¡¡»¯Ñ§¡¡È˽̿αê¸ß¶þ°æ(Ñ¡ÐÞ4)¡¡2009£2010ѧÄê¡¡µÚ2ÆÚ¡¡×ܵÚ158ÆÚ È˽̿αê°æ(Ñ¡ÐÞ4) ÌâÐÍ£º013
³ãÈȵįÌÅÄÚÓз´Ó¦£º C(s)£«O2(g)CO2(g)¡¡¦¤H£½£392 kJ/molÍù¯ÌÅÄÚͨÈëË®ÕôÆøʱ£¬ÓÐÈçÏ·´Ó¦£º C(s)£«H2O(g)CO(g)£«H2(g)¡¡¦¤H£½£«131 kJ/mol CO(g)£«O2(g)CO2(g)¡¡¦¤H£½£282 kJ/mol H2(g)£«O2(g)H2O(g)¡¡¦¤H£½£241 kJ/mol ÓÉÒÔÉÏ·´Ó¦ÍƶÏÍù³ãÈȵįÌÅÄÚͨÈëË®ÕôÆøʱ | |
[¡¡¡¡] | |
A£® |
²»ÄÜʹ¯»ð˲¼ä¸üÍú£¬µ«¿ÉÒÔ½ÚʡȼÁÏ |
B£® |
¼ÈÄÜʹ¯»ð˲¼ä¸üÍú£¬ÓÖÄܽÚʡȼÁÏ |
C£® |
²»ÄܽÚʡȼÁÏ£¬µ«ÄÜʹ¯»ð˲¼ä¸üÍú |
D£® |
¼È²»ÄÜʹ¯»ð˲¼ä¸üÍú£¬ÓÖ²»ÄܽÚʡȼÁÏ |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
A.²»ÄܽÚʡȼÁÏ£¬µ«ÄÜʹ¯»ð˲¼ä¸üÍú?
B.Ëä²»ÄÜʹ¯»ð˲¼ä¸üÍú£¬µ«¿ÉÒÔ½ÚʡȼÁÏ?
C.¼ÈÄÜʹ¯»ð˲¼ä¸üÍú£¬ÓÖÄܽÚʡȼÁÏ?
D.¼È²»ÄÜʹ¯»ð˲¼ä¸üÍú£¬ÓÖ²»ÄܽÚʡȼÁÏ?
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
³ãÈȵįÌÅÄÚÓз´Ó¦£ºC(s)£«O2(g)£½CO2(g)£»¡÷H£½£392kJ/mol£¬Íù¯ÌÅÄÚͨÈëË®ÕôÆøʱ£¬ÓÐÈçÏ·´Ó¦£ºC(s)£«H2O(g)£½CO(g)£«H2(g)£»¡÷H£½£«131kJ/mol£¬CO(g)£«1/2O2(g)£½CO2(g)£»¡÷H£½£282kJ/mol£¬H2(g)£«1/2O2(g)£½H2O(g)£»¡÷H£½£241kJ/mol£¬ÓÉÒÔÉÏ·´Ó¦ÍƶÏÍù³ãÈȵįÌÅÄÚͨÈëË®ÕôÆøʱ £¨ £©
A.²»ÄܽÚʡȼÁÏ£¬µ«ÄÜʹ¯»ð˲¼ä¸üÍú B.Ëä²»ÄÜʹ¯»ð˲¼ä¸üÍú£¬µ«¿ÉÒÔ½ÚʡȼÁÏ
C.¼ÈÄÜʹ¯»ð˲¼ä¸üÍú£¬ÓÖÄܽÚʡȼÁÏ D.¼È²»ÄÜʹ¯»ð˲¼ä¸üÍú£¬ÓÖ²»ÄܽÚʡȼÁÏ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
³ãÈȵįÌÅÄÚÓз´Ó¦£ºC(s)£«O2(g)£½CO2(g)£»¡÷H£½£392kJ/mol£¬Íù¯ÌÅÄÚͨÈëË®ÕôÆøʱ£¬ÓÐÈçÏ·´Ó¦£ºC(s)£«H2O(g)£½CO(g)£«H2(g)£»¡÷H£½£«131kJ/mol£¬CO(g)£«1/2O2(g)£½CO2(g)£»¡÷H£½£282kJ/mol£¬H2(g)£«1/2O2(g)£½H2O(g)£»¡÷H£½£241kJ/mol£¬ÓÉÒÔÉÏ·´Ó¦ÍƶÏÍù³ãÈȵįÌÅÄÚͨÈëË®ÕôÆøʱ£¨ £©
A.²»ÄܽÚʡȼÁÏ£¬µ«ÄÜʹ¯»ð˲¼ä¸üÍú B.Ëä²»ÄÜʹ¯»ð˲¼ä¸üÍú£¬µ«¿ÉÒÔ½ÚʡȼÁÏ
C.¼ÈÄÜʹ¯»ð˲¼ä¸üÍú£¬ÓÖÄܽÚʡȼÁÏ D.¼È²»ÄÜʹ¯»ð˲¼ä¸üÍú£¬ÓÖ²»ÄܽÚʡȼÁÏ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2014½ì¸£½¨Ê¡ËĵØÁùУ¸ß¶þµÚÒ»´ÎÔ¿¼»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÑ¡ÔñÌâ
³ãÈȵįÌÅÄÚÓз´Ó¦£ºC(s)+O2£¨g£©=CO2£¨g£©,¡÷H=-392kJ¡¤mol¡ª1 Íù¯ÌÅÄÚͨÈëË®ÕôÆûʱ£¬ÓÐÈçÏ·´Ó¦£ºC(s)+H2O(g)=CO(g)+H2(g)£¬¡÷H£½+131 kJ¡¤mol¡ª1 CO(g) + 1/2O2£¨g£©= CO2(g)£¬¦¤H=-282 kJ¡¤mol¡ª1¡¡H2£¨g£©+1/2O2£¨g£©=H2O£¨g£©£¬¡÷H=-241kJ¡¤mol¡ª1£¬ÓÉÒÔÉÏ·´Ó¦ÍƶÏÍù³ãÈȵįÌÅÄÚͨÈëË®ÕôÆûʱ£¨¡¡¡¡¡¡£©
A¡¢¼ÈÄÜʹ¯»ð¸üÍú£¬ÓÖÄܽÚʡȼÁÏ
B¡¢Ëä²»ÄÜʹ¯»ð¸üÍú£¬µ«¿ÉÒÔ½ÚʡȼÁÏ
C¡¢²»ÄܽÚʡȼÁÏ£¬µ«ÄÜʹ¯»ð˲¼ä¸üÍú
D ¡¢¼È²»ÄÜʹ¯»ð¸üÍú£¬ÓÖ²»ÄܽÚʡȼÁÏ
²é¿´´ð°¸ºÍ½âÎö>>
°Ù¶ÈÖÂÐÅ - Á·Ï°²áÁбí - ÊÔÌâÁбí
ºþ±±Ê¡»¥ÁªÍøÎ¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨Æ½Ì¨ | ÍøÉÏÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | µçÐÅթƾٱ¨×¨Çø | ÉæÀúÊ·ÐéÎÞÖ÷ÒåÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | ÉæÆóÇÖȨ¾Ù±¨×¨Çø
Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com