ÌâÄ¿ÄÚÈÝ

¸ù¾ÝÓйØÐÅÏ¢£º2SO2£¨g£©+O2£¨g£©   2SO3£¨g£©+196.7KJ£»Ä³ÖÖÏõËá¸Æ¾§Ìå¿É±íʾΪCa£¨NO3£©2¡¤8H2O£»25¡æÊ±NaHCO3µÄÈܽâ¶ÈΪ9.6g/100gË®£»¿ÕÆøÖÐÑõÆøµÄÌå»ý·ÖÊýΪ21%£¬ÅжÏÏÂÁÐÓйØÊý¾ÝÕýÈ·µÄÊÇ


  1. A.
    ÊÒÎÂÏ£¬ÅäÖÆ2.5mol/LµÄNaHCO3ÈÜÒº
  2. B.
    ±¬Õ¨×îÇ¿ÁҵĿӵÀÆøÖм×ÍéµÄÌå»ý·ÖÊýΪ10.5%
  3. C.
    ÓÃ1molSO2ÆøÌåÓë×ãÁ¿ÑõÆø·´Ó¦·Å³öµÄÈÈÁ¿Îª98.35KJ
  4. D.
    ÓëCaO×÷ÓÃÖ±½ÓÖÆCa£¨NO3£©2¡¤8H2OËùÓÃÏõËáµÄÖÊÁ¿·ÖÊýΪ50%
D
±¾Ìâ×ۺϿ¼²é³£¹æ¼ÆËã¡£25¡æÊ±NaHCO3µÄÈܽâ¶ÈΪ9.6g/100gË®£¬ËùÒÔ±¥ºÍNaHCO3ÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈԼΪ1mol/L£¬ËùÒÔÎÞ·¨ÅäÖÆ2.5mol/LµÄNaHCO3ÈÜÒº£¬AÑ¡Ïî´íÎó£»±¬Õ¨×îÇ¿ÁÒµÄÇé¿öÊǼ×ÍéÓëÑõÆøÍêÈ«·´Ó¦£¬Éè¼×ÍéÌå»ýx£¬Ôò²Î¼Ó·´Ó¦ÑõÆøµÄÌå»ýΪ2x£¬Ôò¼×ÍéµÄÌå»ý·ÖÊýΪ£ºx/(+x)¡Á100%=9.5%£¬BÑ¡Ïî´íÎó£»SO2ÓëÑõÆøµÄ·´Ó¦ÊÇ¿ÉÄæ·´Ó¦£¬ËùÒÔ1molSO2ÆøÌå²»¿ÉÄÜÍêȫת»¯£¬Ôò·Å³öµÄÈÈÁ¿Ð¡ÓÚ98.35KJ£¬CÑ¡Ïî´íÎó¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø