ÌâÄ¿ÄÚÈÝ
ÏõËṤҵ²úÉúµÄβÆø(NO¡¢NO2µÈ)¶Ô»·¾³Ôì³ÉÎÛȾ£¬Í¬Ê±Æû³µÎ²Æø(º¬ÌþÀà¡¢CO¡¢SO2ÓëNOµÈ)ÊdzÇÊпÕÆøµÄÎÛȾԴ¡£(1)Æû³µÒÔÆûÓÍ×÷Ϊ·¢¶¯»úȼÁÏ£¬ÔÚ²»¶ÏÆô¶¯´ò»ðµÄ¹ý³ÌÖвúÉúÁËCO¡¢NOµÈÓж¾ÆøÌ壬Èô¹©Ñõ²»×㣬Ö÷Òª²úÉúµÄÎÛȾÎïÊÇ________£¬Èô¹©Ñõ³ä×㣬Ö÷Òª²úÉúµÄÓж¾ÆøÌåÊÇ______________¡£
(2)¼òÊöCO¡¢NO¶ÔÈËÌå²úÉúµÄÓ°Ï죺________________________________________________¡£
(3)̸һ̸Æû³µÎ²ÆøÎÛȾÎï¶Ô»·¾³Ôì³ÉµÄÓ°Ïì________________________________________¡£
(4)ÄãÄÜ·ñÓ¦ÓõªÔªËصĵ¥Öʺͻ¯ºÏÎïÖ®¼äµÄת»¯¹Øϵ£¬Éè¼Æ³ýÈ¥Æû³µÎ²ÆøÖеªµÄÑõ»¯Îï(NO¡¢NO2)µÄÀíÂÛ·½°¸?_________________________________________________________________
(5)¿ØÖƳÇÊпÕÆøÎÛȾµÄ·½·¨¿ÉÒÔÓÐ( )
A.¿ª·¢ÇâÄÜÔ´ B.ʹÓõ綯³µ C.Ö²Ê÷ÔìÁÖ D.´÷ÉϺôÎüÃæ¾ß
˼·Óë´ð°¸£º(1)µ±¹©Ñõ²»×ãʱ£¬Ìþ(CxHy)²»ÍêȫȼÉÕ£¬²úÉúCOÓж¾ÆøÌ壬µ±¹©Ñõ³ä×ãʱ£¬²úÉúNO¡¢NO2µÈÓж¾ÆøÌå¡£
(2)CO¡¢NO¼«Ò×ÓëѪºìµ°°×½áºÏ£¬ÄÑ·ÖÀ룬ʹѪҺʧȥ¹©ÑõÄÜÁ¦£¬Ê¹ÈËÌåÖж¾£¬Í¬Ê±NOÔÚÈËÌåÖк¬Á¿²»Õý³££¬¿Éµ¼ÖÂÐÄÄÔѪ¹Ü²¡¡¢¸ßѪѹ¡¢Öз硢ÌÇÄò²¡µÈ¼²²¡¡£
(3)Æû³µÎ²ÆøÅŷŹý¶à£¬¿ÉÐγɹ⻯ѧÑÌÎí¶øÎÛȾ»·¾³¡£ÒýÆðºôÎüÑÏÖØÕÏ°£¬µ¼ÖÂÐÄ·ÎË¥½ß¡£
(4)¸ù¾ÝβÆø´¦ÀíÔÔò£¬NOºÍNO2Ӧת»¯ÎªN2£¬±ØÐëÑ¡Ôñ»¹Ô¼Á£¬ÄǾÍÊǵªµÄ»¯ºÏÎï°±£¬·½³ÌʽÊÇ£º6NO+4NH35N2+6H2O£¬6NO2+8NH37N2+12H2O¡£
(5)¿ØÖƳÇÊÐÎÛȾԴ£¬±ÜÃâÆû³µÈ¼ÉÕÆûÓÍ£¬ËùÒÔ±ØÐ뿪·¢ÐµÄÄÜÔ´£¬È翪·¢ÇâÄÜÔ´£¬Ê¹Óõ綯³µµÈ£¬´Ó¸ùÔ´ÉÏÏû³ýÎÛȾ¡£
A¡¢x=1.5ʱ£¬Ö»Éú³ÉNaNO2 | B¡¢2£¾x£¾1.5ʱ£¬Éú³ÉNaNO2ºÍNaNO3 | C¡¢x£¼1.5ʱ£¬Ðè²¹³äO2 | D¡¢x=2ʱ£¬Ö»Éú³ÉNaNO3 |