¸ù¾Ý»¯Ñ§·´Ó¦A+B==C+DÖУ¬Ä³Ñ§Éú×÷ÁËÈçÏÂËÄÖÖÐðÊö£º¢ÙÈômgAÓëB³ä·Ö·´Ó¦ºóÉú³ÉngCºÍwgD£¬Ôò²Î¼Ó·´Ó¦µÄBµÄÖÊÁ¿Îª(m+n-w)g£»¢ÚÈômgAºÍngBÇ¡ºÃÍêÈ«·´Ó¦£¬Éú³ÉµÄCºÍDµÄ×ÜÖÊÁ¿ÊÇ(m+n)g£»¢ÛÈôÈ¡A¡¢B¸÷mg£¬Ôò·´Ó¦Éú³ÉCºÍDµÄÖÊÁ¿×ܺͲ»Ò»¶¨ÊÇ2mg£»¢Ü·´Ó¦ÎïA¡¢BµÄÖÊÁ¿±ÈÒ»¶¨µÈÓÚC¡¢DµÄÖÊÁ¿±È¡£ÆäÖÐÕýÈ·µÄÊÇ(    )
A. ¢Ù¢Ú¢Û         B. ¢Ú¢Û¢Ü          C. ¢Ù¢Ü         D. ¢Ú¢Û
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

£¨4·Ö£©£¨1£©ÏÂÁз´Ó¦ÖУ¬ÊôÓÚ·ÅÈÈ·´Ó¦µÄÊÇ         £¬ÊôÓÚÎüÈÈ·´Ó¦µÄÊÇ           ¡£
¢ÙȼÉÕ·´Ó¦¢ÚÕ¨Ò©±¬Õ¨¢ÛËá¼îÖкͷ´Ó¦¢Ü¶þÑõ»¯Ì¼Í¨¹ý³ãÈȵÄ̼
¢ÝʳÎïÒòÑõ»¯¶ø¸¯°Ü¢ÞBa(OH)2¡¤8H2OÓëNH4Cl·´Ó¦¢ßÌú·ÛÓëÏ¡ÑÎËá·´Ó¦
£¨2£©Ð´³ö£¨1£©ÖТ޵Ļ¯Ñ§·½³Ìʽ                                        ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

25¡æ¡¢101 kPaÏ£¬CO¡¢HCOOH¡¢CH4ºÍCH3OHµÄȼÉÕÈÈÒÀ´ÎÊÇ67.6 kJ/mol¡¢62.8kJ/mol¡¢890.3 kJ/mol¡¢726kJ/mol£¬ÔòÏÂÁÐÈÈ»¯Ñ§·½³Ìʽ²»ÕýÈ·µÄÊÇ(   )
A£®CO(g)£«O2(g)£½CO2(g)£»DH£½£­67.6 kJ/mol
B£®HCOOH(1)£«O2(g)£½CO2(g)£«H2O(1)£»DH£½£­62.8 kJ/mol
C£®CH4g)£«2O2(g)£½CO2(g)£«2H2O(g)£»DH£½£­890.3 kJ/mol
D£®2CH3OH(l)£«3O2(g)£½2CO2(g)£«4H2O(l)£»DH£½£­1452 kJ/mol

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£º¼ÆËãÌâ

ˮúÆøÊÇÖØҪȼÁϺͻ¯¹¤Ô­ÁÏ£¬¿ÉÓÃË®ÕôÆøͨ¹ý³ãÈȵÄÌ¿²ãÖƵãº
C (s) + H2O(g) CO (g) +H2 (g)   ¡÷H£½ +131.3 kJ?mol£­1£¨¡¤¡¤¡¤¡¤¡¤¡¤¢Ù
£¨1£©Êµ¼Ê¹¤ÒµÉú²úÖУ¬ÏòÌ¿²ã¼ä¸ô½»ÌæͨÈëË®ÕôÆøºÍ¿ÕÆø£¬ÆäÖÐͨÈë¿ÕÆøµÄÔ­ÒòÊÇÓÉÓڸ÷´Ó¦ÊÇÎüÈÈ£¬µ¼ÖÂÌ¿²ãζȽµµÍ£¬Ð뼰ʱͨÈ븻Ñõ¿ÕÆø´Ù½øÌ¿²ãµÄȼÉÕ·ÅÈÈ£º
C (s) + O2(g)= CO2 (g)£»¡÷H = £­393.5kJ¡¤mo1£­1    ¡¤¡¤¡¤¡¤¡¤¡¤¢Ú
Ϊ±£³ÖÉú²úµÄÁ¬ÐøÐÔ£¬Èô²»¿¼ÂÇÆäËüÈÈÁ¿µÄ²úÉúºÍËðºÄ£¬Ôòÿ¼ä¸ôӦͨÈëµÄË®ÕôÆøºÍ¿ÕÆøµÄÌå»ý±È£¨Í¬ÎÂͬѹ£©Ô¼Îª¶àÉÙ£¿£¨Éè¿ÕÆøÖÐÑõÆøµÄÌå»ýÕ¼1/5£©
£¨2£©Ò»¶¨Î¶ÈÏ£¬Èý¸öÈÝÆ÷Öоù½øÐÐ×ÅÉÏÊö·´Ó¦¢Ù£¬¸÷ÈÝÆ÷ÖÐÌ¿×ãÁ¿£¬ÆäËüÎïÖʵÄÎïÖʵÄÁ¿Å¨¶È¼°ÕýÄæ·´Ó¦ËÙÂʹØϵÈçϱíËùʾ¡£ÇëÌîд±íÖÐÏàÓ¦µÄ¿Õ¸ñ¡£

£¨3£©ÉúÎïÒÒ´¼¿ÉÓɵí·Û»òÏËάËصÈÉúÎïÖÊÔ­ÁÏ·¢½Í»ñµÃ¡£ÀûÓÃÒÒ´¼¿É½ø¶ø»ñµÃºÏ³ÉÆø£¨CO¡¢H2£©¡£ÓÃÒÒ´¼Éú²úºÏ³ÉÆøÓÐÈçÏÂÁ½Ìõ·Ïߣº
a¡¢Ë®ÕôÆø´ß»¯ÖØÕû£ºCH3CH2OH(g)£«H2O(g)¡ú4H2(g)£«2CO(g)
b¡¢²¿·Ö´ß»¯Ñõ»¯£ºCH3CH2OH(g)£«1/2O2(g)¡ú3H2(g)£«2CO(g)
ijÉúÎïÖÊÄÜÑо¿Ëù×¼±¸ÀûÓÃÒÒ´¼µÃµ½µÄºÏ³ÉÆøºÏ³ÉÒ»ÖÖÉúÎïÆûÓÍ¡£ÒÒ´¼¸÷·ÖÒ»°ë°´a¡¢bÁ½Ê½·´Ó¦¡£ºÏ³ÉÆøºÏ³ÉÉúÎïÆûÓ͵ķ´Ó¦Îª£º2mCO£«(2m£«n)H2¡ú2CmHn£«2mH2O¡£¼Ù¶¨ºÏ³ÉµÄÉúÎïÆûÓÍÖк¬ÓÐX¡¢YÁ½Öֳɷ֣¬ÇÒX¡¢Y¶¼ÊÇÓÐ8¸ö̼ԭ×ÓµÄÌþ£¬XÊDZ½µÄͬϵÎYÊÇÍéÌþ¡£
¢ÙXµÄ·Ö×ÓʽΪ               £¬YµÄ·Ö×ÓʽΪ                 ¡£
¢Ú50¶ÖÖÊÁ¿·ÖÊýΪ92%µÄÒÒ´¼¾­ÉÏÊöת»¯£¨¼Ù¶¨¸÷²½×ª»¯ÂʾùΪ100%£©£¬Ôò×îÖÕ¿É»ñµÃXµÄÖÊÁ¿Îª¶àÉÙ¶Ö£¿

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÏÂÁйØÓÚ»¯Ñ§·´Ó¦Ëµ·¨Öв»ÕýÈ·µÄÊÇ
A£®»¯Ñ§·´Ó¦¶¼°éËæ×ÅÈÈÄÜ¡¢¹âÄÜ¡¢µçÄܵı仯B£®»¯Ñ§·´Ó¦Ò»¶¨ÓÐÐÂÎïÖÊÉú³É
C£®»¯Ñ§·´Ó¦ÊǾɼüµÄ¶ÏÁÑ£¬Ð¼üµÄÉú³ÉµÄ¹ý³ÌD£®Öкͷ´Ó¦Îª·ÅÈÈ·´Ó¦

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÏÂÁÐ˵·¨ÖÐÕýÈ·µÄÊÇ£¨   £©
A£®ÓÉHÔ­×ÓÐγÉ1 mol H¡ªH¼üÒªÎüÊÕÈÈÁ¿
B£®ËùÓÐȼÉÕ·´Ó¦¶¼ÊÇ·ÅÈÈ·´Ó¦
C£®ÔÚÏ¡ÈÜÒºÖУ¬1 molËáÓë1 mol¼î·¢ÉúÖкͷ´Ó¦Éú³ÉˮʱËùÊͷŵÄÈÈÁ¿³ÆΪÖкÍÈÈ
D£®·²¾­¼ÓÈȶø·¢ÉúµÄ»¯Ñ§·´Ó¦¶¼ÊÇÎüÈÈ·´Ó¦

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£¨   £©
A£®ÓÉHÐγÉ1 mol H£­H¼üÒªÎüÊÕÈÈÁ¿
B£®ËùÓÐȼÉÕ·´Ó¦¶¼ÊÇ·ÅÈÈ·´Ó¦
C£®16g O3µ¥ÖÊÖк¬ÓеķÖ×Ó¸öÊýΪNA
D£®·²¾­¼ÓÈȶø·¢ÉúµÄ»¯Ñ§·´Ó¦¶¼ÊÇÎüÈÈ·´Ó¦

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÏÂÁл¯Ñ§·´Ó¦µÄ¡÷HµÄ¾ø¶ÔÖµ(·ÅÈÈÁ¿)×î´óµÄÊÇ
A£®CH3COOH(aq) + NaOH(aq) ="=" CH3COONa(aq) + H2O(l)£»¡÷H1
B£®NaOH(aq) + 1/2H2SO4(aq) ="=" 1/2Na2SO4(aq) + H2O(l)£»¡÷H2
C£®NaOH(aq) + HCl(aq) ="=" NaCl(aq) + H2O(l)£»¡÷H3
D£®NaOH(aq) + 1/2H2SO4(Ũ) ="=" 1/2Na2SO4(aq) + H2O(l)£»¡÷H4

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

Ä¿Ç°£¬ÎÒ¹úÒѾ­¾ß±¸ÁË´óÁ¦ÍƹãȼÁÏÒÒ´¼µÄ»ù±¾Ìõ¼þ.2001ÄêºÓÄÏÊ¡Ö£ÖÝ¡¢ÂåÑô¡¢ÄÏÑôµÈ³ÇÊÐÒÑÔÚÆûÓÍÖÐÌí¼ÓȼÁÏÒÒ´¼¡£ÏÂÁжÔÓÚÒÒ´¼×÷ΪȼÁϵÄÐðÊöÖУ¬´íÎóµÄÊÇw.
w.
A£®ÒÒ´¼´¦ÓÚ°ëÑõ»¯Ì¬£¨ºÄÑõÁ¿±ÈÆûÓÍÉÙ£©£¬²»Ðè¸ÄÔì·¢¶¯»ú±ã¿Éʱ¼äÁãÎÛȾÅÅ·Å
B£®Ê¹ÓÃȼÁÏÒÒ´¼¿ÉÒÔ×ۺϽâ¾ö¹ú¼ÒʯÓͶÌȱ¡¢Á¸Ê³¹ýÊ£¼°»·¾³¶ñ»¯Èý´óÈȵãÎÊÌâ
C£®ÒÒ´¼µÄÕû¸öÉú²úºÍÏû·Ñ¹ý³Ì¿ÉÐγÉÎÞÎÛȾºÍ½à¾»µÄ±Õ·ѭ»·¹ý³Ì£¬ÊôÓÚ¿ÉÔÙÉúÄÜÔ´
D£®Ê¹ÓÃȼÁÏÒÒ´¼¶ÔÒÖÖÆ¡°ÎÂÊÒЧӦ¡±ÒâÒåÖØ´ó£¬Òò´ËȼÁÏÒÒ´¼Ò²ÊôÓÚ¡°Çå½àȼÁÏ¡±

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸