ÌâÄ¿ÄÚÈÝ

£¨08ÎߺþÈýÄ££©»¯Ñ§ÓÃÓïÊÇѧϰ»¯Ñ§µÄÖØÒª¹¤¾ß£¬ÏÂÃæÓÃÀ´±íʾÎïÖʱ仯µÄ»¯Ñ§ÓÃÓïÖУ¬ÕýÈ·µÄÊÇ £¨£©

       A£®ÇâÆøȼÉÕµÄÈÈ»¯Ñ§·½³Ìʽ£º 2H2£¨g£©+ O2£¨g£©=2H2O£¨l£©£» ¡÷H= +570kJ?mol£­1

       B£®¼×Íé¼îÐÔȼÁϵç³ØµÄ¸º¼«·´Ó¦£º CH4+10OH£­£­8e£­= CO32£­+7H2O

       C£®±¥ºÍFeCl3ÈÜÒºµÎÈë·ÐË®Öеķ´Ó¦£º FeCl3+3H2O = Fe£¨OH£©3¡ý+3HCl¡ü

       D£®´ÎÂÈËá¸ÆÈÜÒºÖÐͨÈë¹ýÁ¿µÄSO2£º ClO£­+ H2O + SO2 =HSO3£­+ HClO

´ð°¸£ºB
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

2(08Õã½­Ê¡¿ª»¯ÖÐѧģÄâ)ʵÑéÊÒÖиù¾Ý2SO2£«O22SO3£»¦¤H=-393.2 kJ?mol-1Éè¼ÆÈçÏÂͼËùʾʵÑé×°ÖÃÀ´ÖƱ¸SO3¹ÌÌå¡£Çë»Ø´ðÏÂÁÐÎÊÌâ¡£  

 
 

 


£¨1£©ÊµÑéÇ°£¬±ØÐë½øÐеIJÙ×÷ÊÇ£¨Ìî²Ù×÷Ãû³Æ£¬²»±Øд¾ßÌå¹ý³Ì£©¡¡¡¡¡¡¡¡¡¡¡¡

£¨2£©ÔÚA×°ÖÃÖмÓÈëNa2SO3¹ÌÌåµÄͬʱ£¬»¹Ðè¼Ó¼¸µÎË®£¬È»ºóÔٵμÓŨÁòËá¡£¼Ó¼¸µÎË®µÄ×÷ÓÃÊÇ                           ¡¡                            

£¨3£©Ð¡ÊÔ¹ÜCµÄ×÷ÓÃÊÇ                                                   

£¨4£©¹ã¿ÚÆ¿DÄÚÊ¢µÄÊÔ¼ÁÊÇ                ¡£×°ÖÃDµÄÈý¸ö×÷ÓÃÊÇ       ¢Ù¡¡¡¡ ¡¡¡¡

            ¢Ú                   ¢Û                      

£¨5£©ÊµÑéÖе±Cr2O3±íÃæºìÈÈʱ£¬Ó¦½«¾Æ¾«µÆÒÆ¿ªÒ»»á¶ùÔÙ¼ÓÈÈ£¬ÒÔ·Àζȹý¸ß£¬ÕâÑù×öµÄÔ­ÒòÊÇ                          ¡¡                                   ¡¡

£¨6£©×°ÖÃFÖÐUÐ͹ÜÄÚÊÕ¼¯µ½µÄÎïÖʵÄÑÕÉ«¡¢×´Ì¬ÊÇ                         

£¨7£©×°ÖÃGµÄ×÷ÓÃÊÇ                                                      

£¨8£©´ÓG×°Öõ¼³öµÄβÆø´¦Àí·½·¨ÊÇ                                        

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

¾«Ó¢¼Ò½ÌÍø