ÌâÄ¿ÄÚÈÝ
ϱíÊÇÔªËØÖÜÆÚ±íµÄÒ»²¿·Ö£¬A¡¢B¡¢C¡¢D¡¢E¡¢XÊDZíÖиø³öÔªËØ×é³ÉµÄ³£¼ûµ¥ÖÊ»ò»¯ºÏÎï¡£ÒÑÖªA¡¢B¡¢C¡¢D¡¢E¡¢X´æÔÚÈçͼËùʾת»¯¹ØÏµ£¨²¿·ÖÉú³ÉÎïºÍ·´Ó¦Ìõ¼þÂÔÈ¥£©
(1)¢ÙÈôEΪÑõ»¯ÎÔòAÓëË®·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ___________________¡£
¢Úµ±XÊǼîÐÔÑÎÈÜÒº£¬C·Ö×ÓÖÐÓÐ22¸öµç×Óʱ£¬CµÄµç×ÓʽΪ___________¡£Ïò¿ÕÆøÖдóÁ¿ÅÅ·ÅC¿ÉÄܵ¼Öµĺó¹ûÊÇ_________________¡£
¢Ûµ±XΪ½ðÊôµ¥ÖÊʱ£¬XÓëBµÄÏ¡ÈÜÒº·´Ó¦Éú³ÉCµÄÀë×Ó·½³ÌʽΪ________________¡£
(2)ÈôEÎªÆøÌåµ¥ÖÊ£¬DΪ°×É«³Áµí£¬AµÄ»¯Ñ§Ê½¿ÉÄÜÊÇ__________£¬Bº¬ÓеĻ¯Ñ§¼üÀàÐÍΪ_________¡£
(3)ÈôBÎªÆøÌåµ¥ÖÊ£¬D¿ÉÓëË®ÕôÆøÔÚÒ»¶¨Ìõ¼þÏ·¢Éú¿ÉÄæ·´Ó¦£¬Éú³ÉCºÍÒ»ÖÖ¿ÉȼÐÔÆøÌåµ¥ÖÊ£¬Ð´³ö¸Ã¿ÉÄæ·´Ó¦µÄ»¯Ñ§·½³Ìʽ_________________¡£
¢Úµ±XÊǼîÐÔÑÎÈÜÒº£¬C·Ö×ÓÖÐÓÐ22¸öµç×Óʱ£¬CµÄµç×ÓʽΪ___________¡£Ïò¿ÕÆøÖдóÁ¿ÅÅ·ÅC¿ÉÄܵ¼Öµĺó¹ûÊÇ_________________¡£
¢Ûµ±XΪ½ðÊôµ¥ÖÊʱ£¬XÓëBµÄÏ¡ÈÜÒº·´Ó¦Éú³ÉCµÄÀë×Ó·½³ÌʽΪ________________¡£
(2)ÈôEÎªÆøÌåµ¥ÖÊ£¬DΪ°×É«³Áµí£¬AµÄ»¯Ñ§Ê½¿ÉÄÜÊÇ__________£¬Bº¬ÓеĻ¯Ñ§¼üÀàÐÍΪ_________¡£
(3)ÈôBÎªÆøÌåµ¥ÖÊ£¬D¿ÉÓëË®ÕôÆøÔÚÒ»¶¨Ìõ¼þÏ·¢Éú¿ÉÄæ·´Ó¦£¬Éú³ÉCºÍÒ»ÖÖ¿ÉȼÐÔÆøÌåµ¥ÖÊ£¬Ð´³ö¸Ã¿ÉÄæ·´Ó¦µÄ»¯Ñ§·½³Ìʽ_________________¡£
(1)¢Ù3NO2+H2O=2HNO3+NO£»¢Ú
£»Ê¹µØÇòζÈÉý¸ß£¨»òÎÂÊÒЧӦ£©£»¢ÛFe+4H++NO3-=Fe3+
+NO¡ü+2H2O
(2)Na¡¢Na2O2»òNaH£»Àë×Ó¼üºÍ¼«ÐÔ¹²¼Û¼ü
(3)CO+H2O(g)
CO2+H2
+NO¡ü+2H2O
(2)Na¡¢Na2O2»òNaH£»Àë×Ó¼üºÍ¼«ÐÔ¹²¼Û¼ü
(3)CO+H2O(g)
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿