ÌâÄ¿ÄÚÈÝ

I¼×¡¢ÒÒ¡¢±ûÈýλͬѧ·Ö±ðÓÃÈçÏÂÈýÌ×ʵÑé×°Öü°»¯Ñ§Ò©Æ·£¨ÆäÖмîʯ»ÒΪ¹ÌÌåÇâÑõ»¯ÄƺÍÉúʯ»ÒµÄ»ìºÏÎÖÆÈ¡°±Æø¡£ÇëÄã²ÎÓë̽¾¿£¬²¢»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©ÈýλͬѧÖÆÈ¡°±ÆøµÄ»¯Ñ§·½³ÌʽΪ£º_____________________________________________¡£
£¨2£©ÈýλͬѧÓÃÉÏÊö×°ÖÃÖÆÈ¡°±Æøʱ,ÆäÖÐÒÒͬѧûÓÐÊÕ¼¯µ½°±£¨Èç¹ûËûÃǵÄʵÑé²Ù×÷¶¼ÕýÈ·£©£¬ÄãÈÏΪÊÕ¼¯²»µ½°±ÆøµÄÖ÷ÒªÔ­ÒòÊÇ_____________________________________(Óû¯Ñ§·½³Ìʽ±íʾ)¡£
£¨3£©¼ìÑé°±ÆøÊÇ·ñÊÕ¼¯ÂúµÄ·½·¨ÊÇ£¨¼òÊö²Ù×÷·½·¨¡¢ÏÖÏóºÍ½áÂÛ£©_______________________
IIijУ»¯Ñ§Ð¡×éѧÉúÉè¼ÆÈçͼװÖÃ(ͼÖÐÌú¼ÐµÈ¼Ð³Ö×°ÖÃÒÑÂÔÈ¥)½øÐа±Æø´ß»¯Ñõ»¯µÄʵÑé¡£

£¨4£©ÓÃ×°ÖÃAÖÆÈ¡´¿¾»¡¢¸ÉÔïµÄ°±Æø£¬´óÊÔ¹ÜÄÚÊÇ̼ËáÑΣ»¼îʯ»ÒµÄ×÷ÓÃÊÇ__________________________________¡£
£¨5£©½«²úÉúµÄ°±ÆøÓë¹ýÁ¿µÄÑõÆøͨµ½×°ÖÃB(´ß»¯¼ÁΪ²¬Ê¯ÃÞ)ÖУ¬Óþƾ«ÅçµÆ¼ÓÈÈ£º°±´ß»¯Ñõ»¯µÄ»¯Ñ§·½³ÌʽÊÇ_______________£»ÊÔ¹ÜÄÚÆøÌå±äΪºì×ØÉ«£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ_________________¡£
1£©2NH4Cl + Ca(OH)2 ==CaCl2 + 2NH3¡ü+ 2H2
(2) 2NH3 + H2SO4 =(NH4)2SO4   
£¨3£©½«ÊªµÄºìɫʯÈïÊÔÖ½·ÅÔڹܿڣ»±äÀ¶ÔòÒÑÂú 
£¨4£©ÎüÊÕCO2ºÍË®ÕôÆø
£¨5£©4NH3 + 5O2 ="==4NO" + 6H2O     2NO + O2 =2NO2

ÊÔÌâ·ÖÎö£º£¨1£©Èýλͬѧ¶¼ÊÇÀûÓÃÂÈ»¯ï§ÓëÇâÑõ»¯¸Æ·´Ó¦ÖƱ¸°±Æø£¬Æ仯ѧ·½³ÌʽΪ
2NH4Cl + Ca(OH)2 ==CaCl2 + 2NH3¡ü+ 2H2O£»£¨2£©°±ÆøÄܹ»ÓëŨÁòËá·´Ó¦£¬ËùÒÔÒÒͬѧÊÕ²»µ½°±Æø£¬¼´·¢ÉúÁËÈçÏ·´Ó¦£º2NH3 + H2SO4 =(NH4)2SO4  £»£¨3£©¼ìÑé°±ÆøÊÇ·ñÊÕ¼¯ÂúµÄ·½·¨ÊÇ£º½«ÊªµÄºìɫʯÈïÊÔÖ½·ÅÔڹܿڣ»±äÀ¶ÔòÒÑÂú£»£¨4£©ÓÉ̼ËáÑÎÓëÂÈ»¯ï§·´Ó¦Éú³ÉÁ˶þÑõ»¯Ì¼ºÍË®£¬ËùÒÔ¼îʯ»ÒµÄ×÷ÓÃÊÇÎüÊÕCO2ºÍË®ÕôÆø£»£¨5£©°±´ß»¯Ñõ»¯µÄ»¯Ñ§·½³ÌʽÊÇ£º4NH3 + 5O2 ="==4NO" + 6H2O£¬NO¿ÉÒÔ±»ÑõÆøÑõ»¯Îªºì×ØÉ«µÄ¶þÑõ»¯µª£¬Æ仯ѧ·½³ÌʽΪ2NO + O2 =2NO2¡£
µãÆÀ£º±¾Ì⿼²éÁËʵÑéÊÒÖƱ¸°±ÆøµÄ»ù´¡ÖªÊ¶£¬ÊǸ߿¼¿¼²éµÄÖص㣬±¾Ìâ²»ÄÑ¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø