ÌâÄ¿ÄÚÈÝ
£¨8·Ö£©ÏÂͼÊÇijͬѧÉè¼ÆµÄľ̿ºÍŨÁòËá·´Ó¦²¢¼ìÑéËùµÃÈ«²¿²úÎïµÄʵÑé×°Öá£ÒÑÖªËáÐÔ¸ßÃÌËá¼ØÈÜÒº¾ßÓÐÇ¿Ñõ»¯ÐÔ£¬¿É½«SÔªËØ´Ó+4¼ÛÑõ»¯Îª+6¼Û¡£Çë»Ø´ð£º
![]()
£¨1£©Å¨ÁòËáÓëľ̿ÔÚ¼ÓÈÈÌõ¼þÏ·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌΪ £»
Èô·´Ó¦Öй²×ªÒÆÁË0.4molµç×Ó£¬Ôò²Î¼Ó·´Ó¦µÄÁòËáµÄÎïÖʵÄÁ¿Îª mol¡£
£¨2£©Í¼ÖÐ4¸öÏ´ÆøÆ¿×°ÓеÄÊÔ¼ÁÇë´ÓÏÂÁÐÒ©Æ·ÖÐÑ¡Ôñ£º³ÎÇåʯ»ÒË®£»ËáÐÔ¸ßÃÌËá¼ØÈÜÒº£»Æ·ºìÈÜÒº£»£¨¿ÉÖØ¸´Ñ¡Óã©ÊÔÖ¸³ö¸÷Ï´ÆøÆ¿ÄÚÊÔ¼ÁÃû³ÆºÍ×÷Óãº[À´Ô´:ZXXK]
¢Ù £¬ £»¢Ú £¬ £»
¢Û £¬ £» ¢Ü £¬ ¡£
£¨8·Ö£©£¨1£©£ºC +2H2SO4£¨Å¨£©== CO2¡ü+2SO2¡ü+2H2O£¬ 0.2 £¨¸÷2·Ö£©
£¨2£©£¨Ã¿¿Õ1·Ö È«¶Ô¸ø·Ö£©
¢Ù Æ·ºìÈÜÒº ¼ìÑ鯸ÌåÖк¬SO2; ¢Ú ËáÐÔ¸ßÃÌËá¼ØÈÜÒº£»³ýÈ¥SO2ÆøÌå
¢Û Æ·ºìÈÜÒº ¼ìÑ鯸ÌåÖк¬SO2ÊÇ·ñ³ý¾¡£»¢Ü ³ÎÇåʯ»ÒË® ¼ìÑéCO2ÆøÌåµÄ´æÔÚ
¡¾½âÎö¡¿