ÌâÄ¿ÄÚÈÝ


Éú»îÖеÄÐí¶àÏÖÏóÓ뻯ѧÓÐ×ÅÃÜÇеÄÁªÏµ¡£ÔÚS02¡¢NaOH¡¢·Û³¾¡¢CO2¡¢¼×´¼¡¢COÁùÖÖÎïÖÊÖУ¬Ñ¡ÔñÕýÈ·µÄÌîÈëÏÂÁпոñÄÚ£º

(1)ÒûÓüپÆÒýÆðÖж¾µ¼ÖÂÑÛ¾¦Ê§Ã÷£¬ÊÇÓÉÓÚ¼Ù¾ÆÖк¬ÓР         µÄÔ­Òò¡£

(2)ÓÃú¯ȡůʱ·¢ÉúÖж¾£¬Ô­ÒòÖ®Ò»ÊÇÓÉÓÚÊÒÄÚ           ÅŷŲ»³©¶øÒýÆðµÄ¡£

(3)ÈËÃÇÔÚ½øÈë¸ÉºÔÉǰ±ØÐë×öµÆ»ðʵÑ飬ÕâÊÇÒòΪÉî¾®ÖпÉÄÜ»ý´æÓдóÁ¿µÄ        £¬

(4)úÖÐÒ»°ã¶¼º¬ÓÐÁòÔªËØ£¬ÃºÈ¼ÉÕʱ»á²úÉúÎÛȾ¿ÕÆøµÄ          ÆøÌå¡£


¼×´¼¡¢Ò»Ñõ»¯Ì¼¡¢¶þÑõ»¯Ì¼¡¢¶þÑõ»¯Áò


Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

ÔĶÁ×ÊÁÏ£º

(1)°±Æø£¨NH3£©¾ßÓл¹Ô­ÐÔ£¬¿ÉÒÔ»¹Ô­Ñõ»¯Í­£¬Éú³ÉÍ­¡¢µªÆøºÍË®£»

(2)ʵÑéÊÒÖÆ°±Æø£º2NH4Cl + Ca(OH)2  CaCl2 + 2NH3¡ü + 2H2O

(3)°×É«µÄÎÞË®ÁòËáÍ­ÓöË®»á±ä³ÉÀ¶É«£¬¿ÉÒÔÓÃÓÚ¼ì²â΢Á¿Ë®µÄ´æÔÚ£»

(4)¼îʯ»Ò¿ÉÒÔÎüÊÕË®·Ö£¬¿ÉÒÔ¸ÉÔï°±Æø£»

ÀûÓð±Æø»¹Ô­Ñõ»¯Í­µÄ·´Ó¦£¬¿ÉÉè¼Æ²â¶¨Í­ÔªËØÏà¶ÔÔ­×ÓÖÊÁ¿½üËÆÖµµÄʵÑé¡£Ô­ÀíÊÇÏȳÆÁ¿·´Ó¦ÎïÑõ»¯Í­µÄÖÊÁ¿£¬·´Ó¦ÍêÈ«ºó²â¶¨Éú³ÉÎïË®µÄÖÊÁ¿£¬Óɴ˼ÆËãÍ­ÔªËØÏà¶ÔÔ­×ÓÖÊÁ¿¡£Îª´ËÉè¼ÆÁËÏÂÁÐʵÑé×°Öã¨Í¼Öв¿·Ö¼Ð³ÖÒÇÆ÷Ê¡ÂÔ£¬¼ÓÈëµÄNH4C1ÓëCa(OH)2µÄÁ¿×ãÒÔ²úÉúʹCuOÍêÈ«»¹Ô­µÄ°±Æø)£º

ͼ9

 

Çë»Ø´ðÏÂÁÐÎÊÌâ

£¨1£©Ð´³ö°±Æø»¹Ô­Ñõ»¯Í­µÄ»¯Ñ§·½³Ìʽ                                

£¨2£©B×°ÖõÄ×÷ÓÃÊÇ                            

D×°ÖõÄ×÷ÓÃÊÇ            __________

£¨3£©ÎªÁËÈ·ÈÏûÓÐË®½øÈëC×°ÖÃÖУ¬ÐèÔÚÕûÌ××°ÖÃÖÐÌí¼Ó×°ÖÃMÓÚ          £¨Ñ¡ÌîÐòºÅ£©¡£

      a. A£­B¼ä      b. B£­C¼ä      c.C£­D¼ä

£¨4£©ÔÚ±¾ÊµÑéÖУ¬Èô²âµÃË®µÄÖÊÁ¿Îªa g£¬·´Ó¦ÎïÑõ»¯Í­µÄÖÊÁ¿Îªb g£¬ÔòÍ­ÔªËØµÄÏà¶ÔÔ­×ÓÖÊÁ¿ÊÇ____¡£      

£¨5£©ÔÚ±¾ÊµÑéÖУ¬»¹¿Éͨ¹ý²â¶¨______ºÍ_____µÄÖÊÁ¿£¬´ïµ½ÊµÑéÄ¿µÄ¡£

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

¾«Ó¢¼Ò½ÌÍø