ÌâÄ¿ÄÚÈÝ

Íß˹±¬Õ¨ÒѳÉΪµ¼ÖÂÎÒ¹úú¿óÌØ´ó¶ñÐÔʹʵġ°Í·ºÅɱÊÖ¡±£®
£¨1£©Íß˹´æÔÚÓÚú²ã¼°ÖÜΧÑÒ²ãÖУ¬ÊǾ®ÏÂÓк¦ÆøÌåµÄ×ܳƣ®Íß˹ÊôÓÚ
»ìºÏÎï
»ìºÏÎï
£¨Ñ¡Ìî¡°´¿¾»Î»ò¡°»ìºÏÎ£©£¬Íß˹µÄÖ÷Òª³É·ÖÊǼ×Í飬¼×ÍéµÄ»¯Ñ§Ê½Îª
CH4
CH4
£¬ÆäȼÉյĻ¯Ñ§·½³ÌʽΪ
CH4+2O2
 µãȼ 
.
 
CO2+2H2O
CH4+2O2
 µãȼ 
.
 
CO2+2H2O
£»
£¨2£©µ±·¢ÉúÍß˹±¬Õ¨Ê±£¬ÏÂÁÐ×ԾȴëÊ©²»µ±µÄÊÇ
A
A
£»
A£®Õ¾ÔÚÔ­µØ²»¶¯            B£®ÈôÑÛÇ°ÓÐË®£¬Ó¦¸©ÎÔ»ò²àÎÔÓÚË®ÖУ¬²¢ÓÃʪë½íÎæס¿Ú±Ç
C£®±³¶Ô±¬Õ¨µØµãѸËÙÎÔµ¹     D£®Ñ¡ÔñºÏÊÊͨµÀѸËÙÌÓÉú
£¨3£©Ãº¿óÍß˹±¬Õ¨ÓÐÈý¸ö±ØÐëÌõ¼þ£ºÍß˹Ũ¶È´ïµ½±¬Õ¨Ï޶ȡ¢
ÓöÃ÷»ð
ÓöÃ÷»ð
ºÍ×ã¹»µÄ
ÑõÆø
ÑõÆø
£¬¶ÔÈκÎÒ»¸öÌõ¼þµÄÓÐЧ¿ØÖƶ¼¿É±ÜÃâÍß˹±¬Õ¨£®¾Ý´Ë£¬ÎÒ¹ú°²È«¼à¹Ü²¿ÃÅÌá³öÁË¡°Ïȳéºó²É¡¢¼à²â¼à¿Ø¡¢ÒԷ綨²ú¡±ÈýÏî´ëÊ©£®Çë¶ÔÆäÖеġ°Ïȳéºó²É¡°´ëÊ©½øÐнâÊÍ£º
½µµÍÍß˹µÄŨ¶È
½µµÍÍß˹µÄŨ¶È
£®
·ÖÎö£º£¨1£©¸ù¾ÝÍß˹µÄÖ÷Òª³É·ÖÊǼ×ÍéÒÔ¼°¼×ÍéȼÉÕÉú³É¶þÑõ»¯Ì¼ºÍË®½øÐнâ´ð£»
£¨2£©¸ù¾Ý³£¼ûµÄ×ԾȴëÊ©½øÐнâ´ð£»
£¨3£©¸ù¾Ý·¢Éú±¬Õ¨µÄÌõ¼þ£ºÔÚÓÐÏ޵ĿռäÄÚ£¬¿ÉȼÆøÌå»ò·Û³¾Óë¿ÕÆø»ìºÏ£¬´ïµ½±¬Õ¨¼«ÏÞ½â´ð±¾Ì⣮
½â´ð£º½â£º£¨1£©Íß˹µÄÖ÷Òª³É·ÖÊǼ×Í飬»¹º¬ÓÐÆøÌåÔÓÖÊ£¬ËùÒÔÊôÓÚ»ìºÏÎ»¯Ñ§Ê½ÊÇCH4£»
¼×ÍéȼÉÕÉú³É¶þÑõ»¯Ì¼ºÍË®£¬»¯Ñ§·½³ÌʽΪCH4+2O2
 µãȼ 
.
 
CO2+2H2O£»
£¨2£©A¡¢Õ¾ÔÚÔ­µØ²»¶¯Ò×ÊÜÉ˺¦£¬Òª²ÉÈ¡Ö÷¶¯´ëÊ©£¬ËùÒÔ´íÎó£»
B¡¢ÈôÑÛÇ°ÓÐË®£¬Ó¦¸©ÎÔ»ò²àÎÔÓÚË®ÖУ¬²¢ÓÃʪë½íÎæס¿Ú±Ç£¬¿É·ÀÖ¹±¬Õ¨ÒýÆðÒ·þȼÉÕ¼°ÎüÈëÓж¾ÆøÌ壬ËùÒÔÕýÈ·£»
C¡¢±³¶Ô±¬Õ¨µØµãѸËÙÎÔµ¹£¬¿É½µµÍ±¬Õ¨Ê±¶ÔÉíÌåµÄÉ˺¦£¬ËùÒÔÕýÈ·£»
D¡¢Ñ¡ÔñºÏÊÊͨµÀѸËÙÌÓÉú£¬ÊÇ×ԾȵݲȫÓÐЧ·½·¨£¬ËùÒÔÕýÈ·£®
£¨3£©¸ù¾ÝȼÉպͷ¢Éú±¬Õ¨µÄÌõ¼þ£ºÔÚÓÐÏ޵ĿռäÄÚ£¬¿ÉȼÆøÌå»ò·Û³¾Óë¿ÕÆø»ìºÏ£¬´ïµ½±¬Õ¨¼«ÏÞ£¬Óöµ½Ã÷»ð£¬ËùÒÔú¿óÍß˹±¬Õ¨Èý¸ö±ØÐëÌõ¼þÍß˹Ũ¶È´ïµ½±¬Õ¨Ï޶ȣ¬Óöµ½Ã÷»ð£¬×ã¹»µÄÑõÆø£»ÎÒ¹ú°²È«¼à¹Ü²¿ÃÅÌá³öÁË¡°Ïȳéºó²É¡±µÄ´ëÊ©Ö÷ҪĿµÄÊÇ£º½µµÍÍß˹µÄŨ¶È£®
¹Ê´ð°¸Îª£º£¨1£©»ìºÏÎCH4£»CH4+2O2
 µãȼ 
.
 
CO2+2H2O£»
£¨2£©A£»
£¨3£©Óöµ½Ã÷»ð£»ÑõÆø£»½µµÍÍß˹µÄŨ¶È£®
µãÆÀ£º½â´ð±¾ÌâʱҪ³ä·ÖÀí½â¸÷ÖÖÎïÖʵÄ×é³É³É·Ö¼°ÆäÐÔÖÊ£¬Ö»ÓÐÕâÑù²ÅÄܶÔÎÊÌâ×ö³öÕýÈ·µÄÅжϣ®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
ÎÒ¹úú̿µÄ°²È«Éú²úÐÎʽ·Ç³£ÑϾþ£¬Ãº¿óÍß˹±¬Õ¨Ê¼þºÍú¿ó͸ˮʼþ½ÓÁ¬²»¶Ï£®2005Äê11Ô¡¢12ÔºÚÁú½­¡¢ºÓ±±µÈµØÓÖ·¢Éúú·Û±¬Õ¨¡¢Íß˹±¬Õ¨µÈÌØ´óʹʣ®Íß˹ÒѳÉΪµ¼ÖÂÎÒ¹úú¿óÌØ´ó¶ñÐÔʹʵġ°Í·ºÅɱÊÖ¡±£®
£¨1£©Íß˹´æÔÚÓÚú²ã¼°ÖÜΧÑÒ²ãÖУ¬ÊǾ®ÏÂÓк¦ÆøÌåµÄ×ܳƣ®Íß˹µÄÖ÷Òª³É·ÖÊǼ×Í飬Íß˹±¬Õ¨»á²úÉú¸ßΡ¢¸ßѹºÍ³å»÷¶É£¬²¢·Å³öÓж¾ÆøÌ壮²úÉúÓж¾ÆøÌåµÄ»¯Ñ§·½³Ìʽ¿ÉÄÜÊÇ
2CH4+3O2
 µãȼ 
.
 
2CO+4H2O
2CH4+3O2
 µãȼ 
.
 
2CO+4H2O
£®
£¨2£©µ±·¢ÉúÍß˹±¬Õ¨Ê±£¬ÏÂÁÐ×ԾȴëÊ©²»µ±µÄÊÇ
A
A
£®
A£®Õ¾ÔÚÔ­µØ²»¶¯   B£®±³¶Ô±¬Õ¨µØµãѸËÙÎÔµ¹
C£®ÈôÑÛÇ°ÓÐË®£¬Ó¦¸©ÎÔ»ò²àÎÔÓÚË®ÖУ¬²¢ÓÃʪë½íÎæס¿Ú±ÇD£®Ñ¡ÔñºÏÊÊͨµÀѸËÙÌÓÉú
£¨3£©Íß˹±¬Õ¨ºó£¬ÎªÁË·ÀÖ¹¶þ´Î±¬Õ¨£¬Ó¦²ÉÈ¡µÄ´ëÊ©ÊÇ
Á¢¼´ÇжÏͨÍùʹʵصãµÄÒ»ÇеçÔ´
Á¢¼´ÇжÏͨÍùʹʵصãµÄÒ»ÇеçÔ´
¡¢
»Ö¸´Í¨·ç
»Ö¸´Í¨·ç
£®£¨´ðÁ½Ìõ£©
£¨4£©Ãº¿óÍß˹±¬Õ¨ÓÐÈý¸ö±ØÐëÌõ¼þ--Íß˹Ũ¶È´ïµ½±¬Õ¨Ï޶ȡ¢
»ðÔ´
»ðÔ´
ºÍ×ã¹»µÄ
ÑõÆø
ÑõÆø
£¬¶ÔÈκÎÒ»¸öÌõ¼þµÄÓÐЧ¿ØÖƶ¼¿É±ÜÃâÍß˹±¬Õ¨£®¾Ý´Ë£¬ÎÒ¹ú°²È«¼à¹Ü²¿ÃÅÌá³öÁË¡°Ïȳéºó²É¡¢¼à²â¼à¿Ø¡¢ÒԷ綨²ú¡±ÈýÏî´ëÊ©£®Çë¶ÔÆäÖеÄÈÎÒâÒ»Ïî´ëÊ©½øÐнâÊÍ£º
¡°¼à²â¼à¿Ø¡±ÊÇÖ¸½¨Á¢ºÍÍêÉƼà²â¼à¿ØÊֶΣ¬×öµ½·À»¼ÓÚδȻ
¡°¼à²â¼à¿Ø¡±ÊÇÖ¸½¨Á¢ºÍÍêÉƼà²â¼à¿ØÊֶΣ¬×öµ½·À»¼ÓÚδȻ
£®
£¨5£©ÇëÄã·ÖÎöÒ»ÏÂÔì³ÉÎÒ¹úú¿ó°²È«Éú²úÐÎʽÑϾþµÄÔ­Òò£º
Ϊıȡ±©Àû£¬¿óÖ÷ÊÓ°²È«Îª¶ùÏ·£¬Î¥¹æÉú²ú
Ϊıȡ±©Àû£¬¿óÖ÷ÊÓ°²È«Îª¶ùÏ·£¬Î¥¹æÉú²ú
£®
£¨6£©£©¸ù¾ÝËùѧµÄ»¯Ñ§ÖªÊ¶ÖªµÀ£º»¯Ñ§ÎïÖÊ¿ÉÒÔÔ츣ÓÚÈËÀ࣬µ«ÓÐʱҲ»á¸øÈËÀà´øÀ´Î£º¦£®ÇëÄã¾Ù³öÉú»îÖеÄÒ»ÖÖÎïÖÊ˵Ã÷Õâ¸öÎÊÌâ
ËÜÁÏ´ü£¬ÓÐÀûµÄ·½ÃæÊÇ·½±ãÈËÃÇÊ¢×°ÎïÆ·£¬²»ÀûµÄÊÇ·ÏÆúËÜÁÏ´ü»áÔì³É»·¾³ÎÛȾ
ËÜÁÏ´ü£¬ÓÐÀûµÄ·½ÃæÊÇ·½±ãÈËÃÇÊ¢×°ÎïÆ·£¬²»ÀûµÄÊÇ·ÏÆúËÜÁÏ´ü»áÔì³É»·¾³ÎÛȾ
£®

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

¾«Ó¢¼Ò½ÌÍø