ÌâÄ¿ÄÚÈÝ

7£®¡°Î¢¹Û-ºê¹Û-·ûºÅ¡±ÈýÖØ±íÕ÷ÊÇ»¯Ñ§¶ÀÌØµÄ±íʾÎïÖʼ°Æä±ä»¯µÄ·½·¨£¬Çë½áºÏͼʾÍê³ÉÏÂÁÐÎÊÌ⣺
£¨1£©Í¼1ÖУ¬¡°Cu¡±·ûºÅ±íʾ¶àÖÖÐÅÏ¢£¬Èç±íÊ¾Í­ÔªËØ¡¢½ðÊôÍ­µ¥ÖÊ£¬»¹Äܱíʾһ¸öÍ­Ô­×Ó£»Í­Äܳé³É˿˵Ã÷Í­¾ßÓÐÁ¼ºÃµÄÑÓÕ¹ÐÔ£»
£¨2£©´Ó΢Á£µÄ½Ç¶È˵Ã÷ͼ2·´Ó¦µÄʵÖÊÊÇÇâÀë×ÓÓë̼Ëá¸ùÀë×Ó½áºÏÉú³ÉË®ºÍ¶þÑõ»¯Ì¼£¨»òH+ºÍCO32-½áºÏÉú³ÉH2OºÍCO2£©£»
£¨3£©Èçͼ3Ëùʾ£¬ÔÚÒ»¶¨Ìõ¼þÏ£¬AÓëB·´Ó¦Éú³ÉCºÍD£®ËÄÖÖÎïÖÊÖÐÊôÓÚÑõ»¯ÎïµÄΪCD£¨ÌîÐòºÅ£©£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ4NH3+5O2$\frac{\underline{\;Ò»¶¨Ìõ¼þ\;}}{\;}$4NO+6H2O£®

·ÖÎö £¨1£©¸ù¾ÝÔªËØ·ûºÅµÄÒâÒå½øÐзÖÎö£»
£¨2£©¸ù¾Ýͼ·ÖÎö·´Ó¦µÄʵÖÊ£»
£¨3£©¸ù¾Ý΢¹ÛʾÒâͼµÄº¬ÒåºÍÑõ»¯ÎïµÄ¸ÅÄîÀ´·ÖÎö£®

½â´ð ½â£º£¨1£©¡°Cu¡±±íʾ¶àÖÖÐÅÏ¢£¬Èç±íÊ¾Í­ÔªËØ¡¢½ðÊôÍ­µ¥ÖÊ£¬»¹Äܱíʾһ¸öÍ­Ô­×Ó£»Í­¿ÉÒÔ³é³É˿˵Ã÷¾ßÓÐÁ¼ºÃµÄÑÓÕ¹ÐÔ£»
¹ÊÌһ¸öÍ­Ô­×Ó£»ÑÓÕ¹£» 
£¨2£©´Óͼ¿ÉÒÔ¿´³ö¸Ã·´Ó¦µÄʵÖÊÊÇÇâÀë×ÓºÍ̼Ëá¸ùÀë×Ó½áºÏÉú³ÉË®·Ö×ӺͶþÑõ»¯Ì¼·Ö×Ó£»
¹ÊÌÇâÀë×ÓÓë̼Ëá¸ùÀë×Ó½áºÏÉú³ÉË®ºÍ¶þÑõ»¯Ì¼£¨»òH+ºÍCO32-½áºÏÉú³ÉH2OºÍCO2£©£»
£¨3£©ÓÉ·´Ó¦µÄ½á¹¹Ê¾ÒâͼºÍÄ£ÐͱíʾµÄÔ­×ÓÖÖÀ࣬¿ÉÅжÏAΪNH3£¬BΪO2£¬CΪNO£¬DΪH2O£¬Ñõ»¯ÎïΪCD£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ4NH3+5O2$\frac{\underline{\;Ò»¶¨Ìõ¼þ\;}}{\;}$4NO+6H2O£¬
¹ÊÌC¡¢D£»4NH3+5O2$\frac{\underline{\;Ò»¶¨Ìõ¼þ\;}}{\;}$4NO+6H2O£®

µãÆÀ ±¾ÌâÖ÷Òª¿¼²éÁË»¯Ñ§·´Ó¦µÄ΢¹ÛÄ£Ðͱíʾ£¬Íê³É´ËÌ⣬¹Ø¼üÊǸù¾Ý·´Ó¦µÄʵÖʽáºÏÄ£Ð͵Ľṹ³ä·ÖÀí½âͼÖеÄÐÅÏ¢£¬Ö»ÓÐÕâÑù²ÅÄܶÔÎÊÌâ×ö³öÕýÈ·µÄÅжϣ®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø