ÌâÄ¿ÄÚÈÝ
¡¾ÌâÄ¿¡¿1906Ä꣬¹þ²®ÔÚ60¡æ¸ßΡ¢200Õ×ÅÁ¸ßѹµÄÌõ¼þÏ£¬ÓÃï°(Os)×÷´ß»¯¼Á£¬Ê״γɹ¦µÃµ½ÁË°±£¬µ«²úÂʽϵ͡£Ëæ×Å¿ÆѧµÄ½ø²½ÒÔ¼°¿Æѧ¼ÒÃǶԴ߻¯¼ÁµÄÑо¿¸Ä½ø£¬ÏÖÔÚ¹¤ÒµÉÏÆÕ±é²ÉÓÃÌú´¥Ã½£¨Ö÷Òª³É·ÖΪFe3O4£¬Öú´ß»¯¼Á£ºK2O¡¢Al2O3¡¢CaO¡¢MgO¡¢CoOµÈ£©×öºÏ³É°±µÄ´ß»¯¼Á£¬´ó´óÌá¸ßÁ˺ϳɰ±µÄ²úÂÊ¡£»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÒÑÖªÔªËØCo(îÜ£©µÄÔ×ÓºËÄÚÓÐ27¸öÖÊ×Ó£¬Ôò¸ÃÔªËØ»ù̬Ô×Ó¼Ûµç×ÓÅŲ¼Ê½Îª___¡£
£¨2£©ÔªËØFeµÄ»ù̬Ô×ÓºËÍâδ³É¶Ôµç×ÓÊýΪ___£»Fe2+ÓëFe3+Ïà±È½Ï£¬___¸üÎȶ¨¡£
£¨3£©C¡¢N¡¢O´¦ÓÚͬһÖÜÆÚ£¬ÆäÖеÚÒ»µçÀëÄÜ×î´óµÄ___£¬µç¸ºÐÔ×î´óµÄÊÇ___£¬ËüÃÇÓëÇâÐγɵļòµ¥Ç⻯Îï·Ðµã´Ó¸ßµ½µÍµÄ˳ÐòΪ___¡£
£¨4£©NH3·Ö×ÓÖÐNÔ×ÓµÄÔÓ»¯·½Ê½Îª___£»°±Æø¼«Ò×ÈÜÓÚË®£¨Èܽâ¶È1£º700)£¬³ýÁË°±ÆøÄÜÓëË®·¢Éú·´Ó¦Í⣬»¹ÓÐÁ½¸öÔÒò·Ö±ðÊÇ___¡¢___¡£
£¨5£©ÎÒ¹ú³¤Õ÷ϵÁÐÔËÔØ»ð¼ýʹÓõÄҺ̬ȼÁÏÖ÷ÒªÊÇÆ«¶þ¼×ë£۽ṹ¼òʽ£º(CH3)2NNH2£¬¿É¿´×÷ÊÇëÂ(NH2NH2)ÖÐͬһµªÔ×ÓÉϵÄÁ½¸öÇâÔ×Ó±»¼×»ùÈ¡´ú£ÝºÍËÄÑõ»¯¶þµª£¬È¼ÉÕʱ·¢Éú·´Ó¦£º(CH3)2NNH2+2N2O4=2CO2+4H2O+3N2¡£µ±¸Ã·´Ó¦ÏûºÄ1molN2O4ʱ½«ÐγÉ___mol¦Ð¼ü¡£
¡¾´ð°¸¡¿3d74s2 4 Fe3+ N O H2O£¾NH3£¾CH4 sp3 °±ÆøÓëË®¾ùΪ¼«ÐÔ·Ö×Ó»òÏàËÆÏàÈÜ °±Æø·Ö×ÓÓëË®·Ö×ÓÖ®¼ä¿ÉÒÔÐγÉÇâ¼ü 5
¡¾½âÎö¡¿
(1)ÒÑÖªÔªËØCo(îÜ)µÄÔ×ÓºËÄÚÓÐ27¸öÖÊ×Ó£¬ÖÊ×ÓÊýµÈÓÚÔ×ÓºËÍâµç×ÓÊý£¬Ôò¸ÃÔªËØ»ù̬Ô×ÓºËÍâµç×ÓÅŲ¼Ê½Îª£º[Ar]3d74s2£¬¼Ûµç×ÓÅŲ¼Ê½Îª3d74s2£»
(2)ÔªËØFeµÄ»ù̬Ô×ÓºËÍâÅŲ¼Ê½Îª1s22s22p63s23p63d64s2£¬3d¹ìµÀÉÏÓÐ6¸öµç×Ó£¬¸ù¾ÝºéÌعæÔò£¬µç×ÓÓÅÏÈÕ¼¾ÝÒ»¸ö¿Õ¹ìµÀ£¬ÔòºËÍâδ³É¶Ôµç×ÓÊýΪ4£»Fe2+¼Û²ãµç×ÓÅŲ¼Îª1s22s22p63s23p63d6£¬Fe3+¼Û²ãµç×ÓÅŲ¼Îª1s22s22p63s23p63d5£¬Fe3+µÄ3d¹ìµÀÉÏ´¦ÓÚ°ë³äÂúÎȶ¨×´Ì¬£¬Fe3+¸üÎȶ¨£»
(3)C¡¢N¡¢O´¦ÓÚͬһÖÜÆÚ£¬ËæºËµçºÉÊýÔö´ó£¬µÚÒ»µçÀëÄÜÖð½¥Ôö´ó£¬µ«NÔ×ÓºËÍâµç×Ó2p¹ìµÀÉϵç×Ó´¦ÓÚ°ë³äÂúÎȶ¨×´Ì¬£¬µÚÒ»µçÀë±ÈÏàÁÚÔªËØ´ó£¬ÆäÖеÚÒ»µçÀëÄÜ×î´óµÄΪNÔªËØ£»Í¬Ò»ÖÜÆÚ£¬ËæºËµçºÉÊýÔö´ó£¬µç¸ºÐÔÖð½¥Ôö´ó£¬µç¸ºÐÔ×î´óµÄÊÇO£»C¡¢N¡¢OÈýÖÖÔªËصÄ×î¼òµ¥Ç⻯Îï·Ö±ðΪ£ºCH4¡¢NH3¡¢H2O£¬ÓÉÓÚ°±Æø¡¢Ë®¾ùÄÜÐγÉÇâ¼ü£¬¹ÊÈýÕ߷еã´Ó¸ßµ½µÍµÄ˳ÐòΪ£ºH2O£¾NH3£¾CH4£»
(4)NH3·Ö×ÓÖÐNÔ×ӵļ۵ç×Ó¶ÔÊý=3+=4£¬ÔòNÔÓ»¯·½Ê½Îªsp3£»°±Æø¼«Ò×ÈÜÓÚË®(Èܽâ¶È1£º700)£¬³ýÁË°±ÆøÄÜÓëË®·¢Éú·´Ó¦Í⣬»¹ÓÐÁ½¸öÔÒò·Ö±ðÊÇ°±ÆøÓëË®¾ùΪ¼«ÐÔ·Ö×Ó»òÏàËÆÏàÈÜ¡¢°±Æø·Ö×ÓÓëË®·Ö×ÓÖ®¼ä¿ÉÒÔÐγÉÇâ¼ü£»
(5)Ò»¸öN2·Ö×ÓÓɵªµªÈý¼ü½áºÏ¶ø³É£¬Èý¼üÖк¬ÓÐÒ»¸ö¦Ò¼üºÍ2¸ö¦Ð¼ü£¬Ò»¸öCO2·Ö×ÓÖк¬ÓÐÁ½¸ö̼ÑõË«¼ü£¬Ã¿¸öË«¼üÖÐÒ»¸ö¦Ò¼üºÍ1¸ö¦Ð¼ü£¬¸ù¾Ý·´Ó¦(CH3)2NNH2+2N2O4=2CO2+4H2O+3N2£¬µ±¸Ã·´Ó¦ÏûºÄ1molN2O4ʱ½«Éú³É1mol CO2ºÍmol N2£¬ÔòÐγɦмüµÄÎïÖʵÄÁ¿=1mol¡Á2+mol¡Á2=5mol¡£