£¨15·Ö£©°´ÒªÇóÌî¿Õ¡£
£¨1£©¾­²â¶¨Ä³º¬ÑõÓлúÎïµÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª94¡£ÏÖÈ¡9.4g¸Ãº¬ÑõÓлúÎï³ä·ÖȼÉÕºóµÄ²úÎïͨ¹ýŨÁòËáºóÔöÖØ5.4g£¬²¢ÊÕ¼¯µ½±ê×¼×´¿öÏÂCO2ÆøÌå13.44L¡£Ôò£º
¢Ù¸Ãº¬ÑõÓлúÎïµÄ·Ö×ÓʽΪ                 £»
¢ÚÇëд³ö¸Ãº¬ÑõÓлúÎïÖйÙÄÜÍŵÄÃû³Æ                   ¡£
£¨2£©Íê³ÉÏÂÁб仯µÄ»¯Ñ§·½³Ìʽ£º
¢Ù±½ÓëŨÁòËᡢŨÏõËáµÄ»ìºÏËáÔÚ50-60¡æʱµÄ·´Ó¦                         ¡£
¢Ú±½·ÓÓëŨäåË®·´Ó¦                                     ¡£
¢Û¶þÑõ»¯Ì¼Í¨Èë±½·ÓÄÆÈÜÒº£¬²úÉú°×É«»ë×Ç                         ¡£
¢Ü¼ÒÓÃÒº»¯ÆøµÄÖ÷Òª³É·ÖÖ®Ò»ÊǶ¡Í飬µ±1g¶¡ÍéÍêȫȼÉÕ²¢Éú³É¶þÑõ»¯Ì¼ºÍҺ̬ˮʱ£¬·Å³öÈÈÁ¿Îª50kJ¡£¶¡ÍéȼÉÕµÄÈÈ»¯Ñ§·½³ÌʽΪ                   ¡£

£¨1£©C6H6O  ôÇ»ù»ò·ÓôÇ»ù£¨2£«1·Ö£©
£¨2£©C4H10£¨g£©£«6.5O2(g)¡ú4CO2(g)£«5H2O(1)µÄ¡÷H=£­2900kJ/mol

½âÎö

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â

ÖÐѧ»¯Ñ§³£¼ûµÄÎïÖÊA¡¢B¡¢C¡¢DÖ®¼ä´æÔÚÈçÏÂת»¯¹Øϵ£ºA+B¡úC+D+H2O£¨Ã»ÓÐÅäƽ£©£®Çë°´ÒªÇóÌî¿Õ£º
£¨1£©ÈôAΪ¶ÌÖÜÆÚÔªËØ×é³ÉµÄºÚÉ«¹ÌÌåµ¥ÖÊ£¬ÓëBµÄŨÈÜÒº¹²ÈÈʱ£¬²úÉúC¡¢DÁ½ÖÖÆøÌ壮C¡¢DÁ½ÖÖÆøÌå¾ùÄÜʹ³ÎÇåʯ»ÒË®±ä»ë×Ç£¬Ôò¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ£º
C+2H2SO4£¨Å¨£©
  ¡÷  
.
 
CO2¡ü+2SO2¡ü+2H2O
C+2H2SO4£¨Å¨£©
  ¡÷  
.
 
CO2¡ü+2SO2¡ü+2H2O
£¬¼ø±ðÕâÁ½ÖÖÆøÌå²»ÄÜÑ¡ÓõÄÊÔ¼ÁÊÇ
a
a

a£®BaCl2ÈÜÒº  b£®KMnO4ÈÜÒº  c£®Æ·ºìÈÜÒº   d£®ËữµÄBa£¨NO3£©2ÈÜÒº
Ïò500mL2mol/LµÄNaOHÈÜÒºÖÐͨÈë0.8molÎÞÉ«ÎÞζµÄCÆøÌ壬ǡºÃÍêÈ«·´Ó¦£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ
5NaOH+4CO2¨TNa2CO3+3NaHCO3+H2O
5NaOH+4CO2¨TNa2CO3+3NaHCO3+H2O
£®´ËʱÈÜÒºÖеÄÀë×Ó°´ÕÕŨ¶ÈÓÉ´óµ½Ð¡ÅÅÁеÄ˳ÐòÊÇ
c£¨Na+£©£¾c£¨HCO3-£©£¾c£¨CO3 2- £©£¾c£¨OH-£©£¾c£¨H+£©
c£¨Na+£©£¾c£¨HCO3-£©£¾c£¨CO3 2- £©£¾c£¨OH-£©£¾c£¨H+£©
£®
£¨2£©ÈôAΪºìÉ«½ðÊôµ¥ÖÊ£¬ÓëÊÊÁ¿BµÄÈÜÒºÔÚ³£ÎÂÏÂÇ¡ºÃÍêÈ«·´Ó¦£¬Éú³ÉµÄÎÞÉ«ÆøÌåCÓö¿ÕÆøѸËÙ±ä³É¾­×ØÉ«£®Èô±»»¹Ô­µÄBÎïÖʵÄÁ¿Îª2molʱ£¬²úÉúCÆøÌåµÄÌå»ýÊÇ
44.8
44.8
 L£¨±ê¿ö£©£®½«Éú³ÉµÄºì×ØÉ«ÆøÌåͨÈëÒ»¸öÉÕÆ¿ÀÈû½ôÆ¿Èûºó£¬½«ÉÕÆ¿½þÈë±ùË®ÖУ¬ÉÕÆ¿ÖÐÆøÌåµÄÑÕÉ«±ädz£¬ÇëÓû¯Ñ§·½³ÌʽºÍ±ØÒªµÄÎÄ×Ö½âÊÍÑÕÉ«±ä»¯µÄÔ­Òò
2NO2£¨ºì×ØÉ«£©?N2O4£¨ÎÞÉ«£©£»¡÷H£¼0£¬½µÎ£¬Æ½ºâÏòÕý·´Ó¦·½ÏòÒƶ¯£¬ºì×ØÉ«NO2Ũ¶È¼õС£¬»ìºÏÆøÌåµÄÑÕÉ«±ädz
2NO2£¨ºì×ØÉ«£©?N2O4£¨ÎÞÉ«£©£»¡÷H£¼0£¬½µÎ£¬Æ½ºâÏòÕý·´Ó¦·½ÏòÒƶ¯£¬ºì×ØÉ«NO2Ũ¶È¼õС£¬»ìºÏÆøÌåµÄÑÕÉ«±ädz
£®
£¨3£©ÈôAÔÚË®ÖеÄÈܽâ¶ÈËæζȵÄÉý¸ß¶ø½µµÍ£»BΪ¶ÌÖÜÆڷǽðÊôµ¥ÖÊ£»DÊÇƯ°×·ÛµÄ³É·ÖÖ®Ò»£®C·¢ÉúË®½â·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ
ClO-+H2O?HClO+OH-
ClO-+H2O?HClO+OH-

£¨4£©ÈôAΪÎåºË10µç×ÓµÄÑôÀë×ÓÓëµ¥ºË18µç×ÓµÄÒõÀë×Ó¹¹³ÉµÄÎÞÉ«¾§Ì壬ÊÜÈÈÒ׷ֽ⣬·Ö½âºóÉú³ÉÁ½ÖÖ¼«Ò×ÈÜÓÚË®µÄÆøÌ壮¼ìÑéAÖÐÒõÀë×ӵķ½·¨ÊÇ
È¡ÉÙÁ¿AÓÚÒ»½à¾»ÊÔ¹ÜÖУ¬¼ÓÕôÁóË®Èܽâºó£¬µÎ¼ÓÏõËáÒøÈÜÒº£¬²úÉú°×ɫûµí£¬ÔٵμÓÏ¡ÏõËᣬÕñµ´£¬³Áµí²»Èܽ⣬ÔòAµÄÒõÀë×ÓΪCl-
È¡ÉÙÁ¿AÓÚÒ»½à¾»ÊÔ¹ÜÖУ¬¼ÓÕôÁóË®Èܽâºó£¬µÎ¼ÓÏõËáÒøÈÜÒº£¬²úÉú°×ɫûµí£¬ÔٵμÓÏ¡ÏõËᣬÕñµ´£¬³Áµí²»Èܽ⣬ÔòAµÄÒõÀë×ÓΪCl-
£¨¼òÊö²Ù×÷¹ý³Ì¼°½áÂÛ£©£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨15·Ö£©°´ÒªÇóÌî¿Õ¡£

   £¨1£©¾­²â¶¨Ä³º¬ÑõÓлúÎïµÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª94¡£ÏÖÈ¡9.4g¸Ãº¬ÑõÓлúÎï³ä·ÖȼÉÕºóµÄ²úÎïͨ¹ýŨÁòËáºóÔöÖØ5.4g£¬²¢ÊÕ¼¯µ½±ê×¼×´¿öÏÂCO2ÆøÌå13.44L¡£Ôò£º

        ¢Ù¸Ãº¬ÑõÓлúÎïµÄ·Ö×ÓʽΪ                 £»

        ¢ÚÇëд³ö¸Ãº¬ÑõÓлúÎïÖйÙÄÜÍŵÄÃû³Æ                   ¡£

   £¨2£©Íê³ÉÏÂÁб仯µÄ»¯Ñ§·½³Ìʽ£º

        ¢Ù±½ÓëŨÁòËᡢŨÏõËáµÄ»ìºÏËáÔÚ50-60¡æʱµÄ·´Ó¦                         ¡£

        ¢Ú±½·ÓÓëŨäåË®·´Ó¦                                     ¡£

        ¢Û¶þÑõ»¯Ì¼Í¨Èë±½·ÓÄÆÈÜÒº£¬²úÉú°×É«»ë×Ç                         ¡£

        ¢Ü¼ÒÓÃÒº»¯ÆøµÄÖ÷Òª³É·ÖÖ®Ò»ÊǶ¡Í飬µ±1g¶¡ÍéÍêȫȼÉÕ²¢Éú³É¶þÑõ»¯Ì¼ºÍҺ̬ˮʱ£¬·Å³öÈÈÁ¿Îª50kJ¡£¶¡ÍéȼÉÕµÄÈÈ»¯Ñ§·½³ÌʽΪ                   ¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2010Ä긣½¨Ê¡¸ß¶þµÚ¶þѧÆÚÆÚÄ©¿¼ÊÔ»¯Ñ§ÊÔÌâ ÌâÐÍ£ºÌî¿ÕÌâ

£¨15·Ö£©°´ÒªÇóÌî¿Õ¡£

   £¨1£©¾­²â¶¨Ä³º¬ÑõÓлúÎïµÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª94¡£ÏÖÈ¡9.4g¸Ãº¬ÑõÓлúÎï³ä·ÖȼÉÕºóµÄ²úÎïͨ¹ýŨÁòËáºóÔöÖØ5.4g£¬²¢ÊÕ¼¯µ½±ê×¼×´¿öÏÂCO2ÆøÌå13.44L¡£Ôò£º

        ¢Ù¸Ãº¬ÑõÓлúÎïµÄ·Ö×ÓʽΪ                  £»

        ¢ÚÇëд³ö¸Ãº¬ÑõÓлúÎïÖйÙÄÜÍŵÄÃû³Æ                    ¡£

   £¨2£©Íê³ÉÏÂÁб仯µÄ»¯Ñ§·½³Ìʽ£º

        ¢Ù±½ÓëŨÁòËᡢŨÏõËáµÄ»ìºÏËáÔÚ50-60¡æʱµÄ·´Ó¦                          ¡£

        ¢Ú±½·ÓÓëŨäåË®·´Ó¦                                      ¡£

        ¢Û¶þÑõ»¯Ì¼Í¨Èë±½·ÓÄÆÈÜÒº£¬²úÉú°×É«»ë×Ç                          ¡£

        ¢Ü¼ÒÓÃÒº»¯ÆøµÄÖ÷Òª³É·ÖÖ®Ò»ÊǶ¡Í飬µ±1g¶¡ÍéÍêȫȼÉÕ²¢Éú³É¶þÑõ»¯Ì¼ºÍҺ̬ˮʱ£¬·Å³öÈÈÁ¿Îª50kJ¡£¶¡ÍéȼÉÕµÄÈÈ»¯Ñ§·½³ÌʽΪ                    ¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

°´ÒªÇóÌî¿Õ¡£

   £¨1£©¾­²â¶¨Ä³º¬ÑõÓлúÎïµÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª94¡£ÏÖÈ¡9.4g¸Ãº¬ÑõÓлúÎï³ä·ÖȼÉÕºóµÄ²úÎïͨ¹ýŨÁòËáºóÔöÖØ5.4g£¬²¢ÊÕ¼¯µ½±ê×¼×´¿öÏÂCO2ÆøÌå13.44L¡£Ôò£º

        ¢Ù¸Ãº¬ÑõÓлúÎïµÄ·Ö×ÓʽΪ                  £»

        ¢ÚÇëд³ö¸Ãº¬ÑõÓлúÎïÖйÙÄÜÍŵÄÃû³Æ                    ¡£

   £¨2£©Íê³ÉÏÂÁб仯µÄ»¯Ñ§·½³Ìʽ£º

        ¢Ù±½ÓëŨÁòËᡢŨÏõËáµÄ»ìºÏËáÔÚ50-60¡æʱµÄ·´Ó¦                          ¡£

        ¢Ú±½·ÓÓëŨäåË®·´Ó¦                                      ¡£

        ¢Û¶þÑõ»¯Ì¼Í¨Èë±½·ÓÄÆÈÜÒº£¬²úÉú°×É«»ë×Ç                          ¡£

        ¢Ü¼ÒÓÃÒº»¯ÆøµÄÖ÷Òª³É·ÖÖ®Ò»ÊǶ¡Í飬µ±1g¶¡ÍéÍêȫȼÉÕ²¢Éú³É¶þÑõ»¯Ì¼ºÍҺ̬ˮʱ£¬·Å³öÈÈÁ¿Îª50kJ¡£¶¡ÍéȼÉÕµÄÈÈ»¯Ñ§·½³ÌʽΪ                    ¡£

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸