ÌâÄ¿ÄÚÈÝ
¡¾ÌâÄ¿¡¿X¡¢Y¡¢Z¡¢W¡¢RÊÇÔªËØÖÜÆÚ±íÇ°ËÄÖÜÆÚÖеij£¼ûÔªËØ£¬ÆäÏà¹ØÐÅÏ¢ÈçÏÂ±í£º
ÔªËØ | Ïà¹ØÐÅÏ¢ |
X | ×é³Éµ°°×ÖʵĻù´¡ÔªËØ,Æä×î¸ßÕý»¯ºÏ¼ÛÓë×îµÍ¸º»¯ºÏ¼ÛµÄ´úÊýºÍΪ2 |
Y | µØ¿ÇÖк¬Á¿×î¸ßµÄÔªËØ |
Z | ´æÔÚÖÊÁ¿ÊýΪ23£¬ÖÐ×ÓÊýΪ12µÄºËËØ |
W | Éú»îÖдóÁ¿Ê¹ÓÃÆäºÏ½ðÖÆÆ·£¬¹¤ÒµÉÏ¿ÉÓõç½âÈÛÈÚÑõ»¯ÎïµÄ·½·¨ÖƱ¸Æäµ¥ÖÊ |
R | ÓжàÖÖ»¯ºÏ¼Û£¬Æä°×É«ÇâÑõ»¯ÎïÔÚ¿ÕÆøÖлáѸËÙ±ä³É»ÒÂÌÉ«£¬×îºó±ä³ÉºìºÖÉ« |
(1)RÔÚÔªËØÖÜÆÚ±íÖеÄλÖÃΪ______£»X¡¢Z¡¢WÈýÖÖÔªËصÄÔ×Ӱ뾶´Ó´óµ½Ð¡µÄ˳ÐòÊÇ________(ÓÃÔªËØ·ûºÅ±íʾ)¡£
(2)XÓëÇâÁ½ÔªËØ°´Ô×ÓÊýÄ¿±È1¡Ã3ºÍ2¡Ã4¹¹³É·Ö×ÓAºÍB£¬AµÄ´ß»¯Ñõ»¯·½³ÌʽΪ_______£»BµÄ½á¹¹Ê½Îª_______£»»¯ºÏÎïZ2YÖдæÔڵĻ¯Ñ§¼üÀàÐÍΪ________¡£
(3)WÈÜÓÚZµÄ×î¸ß¼ÛÑõ»¯ÎïµÄË®»¯ÎïµÄÀë×Ó·½³Ìʽ_______________¡£
(4)ÓÃRCl3ÈÜÒº¸¯Ê´ÍÏß·°åµÄÀë×Ó·½³ÌʽΪ______¡£¼ìÑéÈÜÒºÖÐR3+³£ÓõÄÊÔ¼ÁÊÇ________£¬¿ÉÒԹ۲쵽µÄÏÖÏóÊÇ________________¡£
¡¾´ð°¸¡¿ µÚËÄÖÜÆÚµÚ¢ø×å Na£¾Al£¾N 4NH3+5O2=4NO+6H2O Àë×Ó¼ü 2Al+2OH-+2H2O=2AlO2-+3H2¡ü Cu+2Fe3+=Cu2++2Fe2+ KSCN ±äΪѪºìÉ«ÈÜÒº
¡¾½âÎö¡¿XÊÇ×é³Éµ°°×ÖʵĻù´¡ÔªËØ£¬Æä×î¸ßÕý»¯ºÏ¼ÛÓë×îµÍ¸º»¯ºÏ¼ÛµÄ´úÊýºÍΪ2£¬ÔòXΪNÔªËØ£»YÊǵؿÇÖк¬Á¿×î¸ßµÄÔªËØ£¬ÔòYÊÇOÔªËØ£»Z´æÔÚÖÊÁ¿ÊýΪ23£¬ÖÐ×ÓÊýΪ12µÄºËËØ£¬ZµÄÖÊ×ÓÊýÊÇ11£¬ÎªNaÔªËØ£»WÊÇÉú»îÖдóÁ¿Ê¹ÓÃÆäºÏ½ðÖÆÆ·£¬¹¤ÒµÉÏ¿ÉÓõç½âÈÛÈÚÑõ»¯ÎïµÄ·½·¨ÖƱ¸Æäµ¥ÖÊ£¬ÔòWÊÇAlÔªËØ£»RÓжàÖÖ»¯ºÏ¼Û£¬Æä°×É«ÇâÑõ»¯ÎïÔÚ¿ÕÆøÖлáѸËÙ±ä³É»ÒÂÌÉ«£¬×îºó±ä³ÉºìºÖÉ«£¬ÔòRÊÇFeÔªËØ£»
(1)FeµÄºËµçºÉÊýΪ26£¬ÔÚÔªËØÖÜÆÚ±íÖеÄλÖÃΪµÚËÄÖÜÆÚµÚ¢ø×壻µç×Ó²ãÊýÔ½¶àÔ×Ӱ뾶Խ´ó£¬Í¬Ò»ÖÜÆÚÖУ¬Ô×Ӱ뾶Ëæ×ÅÔ×ÓÐòÊýµÄÔö´ó¶ø¼õС£¬ËùÒÔN¡¢Na¡¢AlËÄÖÖÔªËصÄÔ×Ӱ뾶´Ó´óµ½Ð¡µÄ˳ÐòÊÇNa£¾Al£¾N£»
(2)NÓëÇâÁ½ÔªËØ°´Ô×ÓÊýÄ¿±È1£º3ºÍ2£º4¹¹³É·Ö×ÓAºÍB£¬AºÍB·Ö±ðÊÇNH3¡¢N2H4£¬AÔÚÑõÆøÖд߻¯Ñõ»¯µÄ»¯Ñ§·½³ÌʽΪ4NH3+5O2=4NO+6H2O£¬BµÄ½á¹¹Ê½Îª £¬»¯ºÏÎïNa2OÖдæÔڵĻ¯Ñ§¼üÀàÐÍΪÀë×Ó¼ü£»
(3)AlÈÜÓÚNaOHÈÜÒºÉú³ÉNaAlO2ºÍÇâÆø£¬·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ2Al+2OH-+2H2O=2AlO2-+3H2¡ü£»
(4)FeCl3ÈÜÒº¸¯Ê´ÍÏß·°åµÄÀë×Ó·½³ÌʽΪ2Fe3++Cu=Cu2++2Fe2+£¬¼ìÑéÈÜÒºÖÐFe3+³£ÓõÄÊÔ¼ÁÊÇKSCNÈÜÒº£¬¿ÉÒԹ۲쵽µÄÏÖÏóÊÇÈÜÒº³ÊºìÉ«¡£
