ÌâÄ¿ÄÚÈÝ

3£®ÔÚÈÝ»ýΪ2LµÄÃܱÕÈÝÆ÷ÖнøÐÐÈçÏ·´Ó¦£ºA£¨g£©+2B£¨g£©?3C£¨g£©+nD£¨g£©£¬¿ªÊ¼Ê±AΪ4mol£¬BΪ6mol£»5minĩʱ²âµÃCµÄÎïÖʵÄÁ¿Îª3mol£¬ÓÃD±íʾµÄ»¯Ñ§·´Ó¦ËÙÂÊv£¨D£©Îª0.2mol/£¨L•min£©£®
¼ÆË㣺£¨1£©5minÄ©AµÄÎïÖʵÄÁ¿Å¨¶ÈΪ1.5 mol/L£»
£¨2£©Ç°5minÄÚÓÃB±íʾµÄ»¯Ñ§·´Ó¦ËÙÂÊv£¨B£©Îª0.2 mol/£¨L•min£©£»
£¨3£©»¯Ñ§·½³ÌʽÖÐnֵΪ2£»
£¨4£©´Ë·´Ó¦ÔÚËÄÖÖ²»Í¬Çé¿öϵķ´Ó¦ËÙÂÊ·Ö±ðΪ£º
¢Ùv£¨A £©=5mol/£¨ L•min £©             
¢Úv£¨ B £©=6mol/£¨ L•min £©
¢Ûv£¨C£©=4mol/£¨ L•min £©              
¢Üv£¨D £©=8mol/£¨ L•min £©
Ôò¸Ã·´Ó¦µÄËÙÂÊÓÉ´óµ½Ð¡µÄ˳ÐòÊÇ¢Ù£¾¢Ü£¾¢Ú£¾¢Û£¨ÓñàºÅÌîд£©£®

·ÖÎö £¨1£©¸ù¾ÝCµÄÎïÖʵÄÁ¿¼ÆËã·´Ó¦µÄAµÄÎïÖʵÄÁ¿£¬´Ó¶øÖªµÀΪ·´Ó¦µÄAµÄÎïÖʵÄÁ¿£¬ÔÙÀûÓÃŨ¶È¹«Ê½¼ÆË㣻
£¨2£©¸ù¾ÝCµÄÎïÖʵÄÁ¿¼ÆËã·´Ó¦µÄBµÄÎïÖʵÄÁ¿£¬¸ù¾Ý·´Ó¦ËÙÂʹ«Ê½¼ÆË㣻
£¨3£©¸ù¾Ýͬһ·´Ó¦ÖС¢Í¬Ò»Ê±¼ä¶ÎÄÚ£¬¸÷ÎïÖʵķ´Ó¦ËÙÂÊÖ®±ÈµÈÓÚ¼ÆÁ¿ÊýÖ®±ÈÈ·¶¨nÖµ£»
£¨4£©°Ñ²»Í¬ÎïÖʵķ´Ó¦ËÙÂÊ»»Ëã³ÉͬһÎïÖʵķ´Ó¦ËÙÂʽøÐбȽϣ®

½â´ð ½â£º£¨1£©A£¨g£©+2B£¨g£©?3C£¨g£©+nD£¨g£©£¬
·´Ó¦¿ªÊ¼ 4mol 6mol 0 0
·´Ó¦ 1 mol 2 mol 3 mol
5minĩ 3 mol 4mol 3 mol
C£¨A£©=$\frac{n}{V}$=1.5 mol/L
¹Ê´ð°¸Îª£º1.5 mol/L£»
£¨2£©v£¨B£©=$\frac{¡÷n}{V¡÷t}$=$\frac{2}{2¡Á5}$=0.2 mol/£¨L•min£©
¹Ê´ð°¸Îª£º0.2 mol/£¨L•min£©£»
£¨3£©¸ù¾Ýͬһ·´Ó¦ÖС¢Í¬Ò»Ê±¼ä¶ÎÄÚ£¬¸÷ÎïÖʵķ´Ó¦ËÙÂÊÖ®±ÈµÈÓÚ¼ÆÁ¿ÊýÖ®±È£»
ËùÒÔv£¨B£©£ºv£¨D£©=0.2 mol/£¨L•min£©£º0.2mol/£¨L•min£©=2£ºn£¬n=2
¹Ê´ð°¸Îª£º2£»
£¨4£©°ÑËùÓÐËÙÂʶ¼»»Ëã³ÉAµÄ·´Ó¦ËÙÂÊ£»
¢Ùv£¨A £©=5mol/£¨ L•min £©
¢ÚÓÉv£¨ B £©=6 mol/£¨ L•min £©Öª£¬v£¨A £©=3mol/£¨ L•min £©
¢ÛÓÉv£¨C£©=4.5mol/£¨ L•min £©Öª£¬v£¨A £©=1.5 mol/£¨ L•min £©
¢ÜÓÉv£¨D £©=8mol/£¨ L•min £©Öª£¬v£¨A £©=4mol/£¨ L•min £©£¬Ôò¸Ã·´Ó¦µÄËÙÂÊÓÉ´óµ½Ð¡µÄ˳ÐòÊÇ£¬
¹Ê´ð°¸Îª£º¢Ù£¾¢Ü£¾¢Ú£¾¢Û£®

µãÆÀ ±¾Ì⿼²éÁË»¯Ñ§Æ½ºâ¼ÆËã·ÖÎö£¬·´Ó¦ËÙÂÊ¡¢Å¨¶ÈµÄ¼ÆËãÓ¦Óã¬ÕÆÎÕ»ù´¡Êǹؼü£¬ÌâÄ¿½Ï¼òµ¥£¬½â´ð£¨4£©µÄ·½·¨ÊÇ£º°ÑËùÓÐËÙÂʶ¼»»Ëã³ÉͬһÎïÖʵķ´Ó¦ËÙÂÊ£¬È»ºó±È½Ï´óС£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø