ÌâÄ¿ÄÚÈÝ

£¨10·Ö£©Ñ§Ï°»¯Ñ§Ó¦¸ÃÃ÷È·¡°´ÓÉú»îÖÐÀ´£¬µ½Éú»îÖÐÈ¥¡±µÀÀí¡£ÔÚÉú²úÉú»îÖУ¬ÎÒÃÇ»áÓöµ½¸÷ÖÖ¸÷ÑùµÄ»¯Ñ§ÎïÖʺͻ¯Ñ§·´Ó¦¡£
£¨1£©¹¤ÒµÉÏÉú²ú²£Á§¡¢Ë®ÄàºÍÁ¶Ìú¶¼ÒªÓõ½µÄÔ­ÁÏÊÇ£¨ÌîÃû³Æ£©    £¬¹¤ÒµÁ¶ÌúÖ÷ÒªÉ豸ÊÇ     
£¨2£©ÇëÄãд³ö¹¤ÒµÉÏÓÃÂÈÆøºÍÏûʯ»Ò·´Ó¦ÖÆȡƯ°×·ÛµÄ»¯Ñ§·´Ó¦·½³Ìʽ£º        ¡£
£¨3£©ÇëÄãд³öθÊæƽ£¨º¬ÓÐÇâÑõ»¯ÂÁ£©ÖÎÁÆθËᣨÑÎËᣩ¹ý¶àµÄÀë×Ó·´Ó¦·½³Ìʽ£º    ¡£
£¨4£©Ó¡Ë¢µç·°åÊÇÓɸ߷Ö×Ó²ÄÁϺÍÍ­²­¸´ºÏ¶ø³É£¬¿ÌÖÆÓ¡Ë¢µç·°åʱ£¬ÒªÓÃFeCl3ÈÜÒº×÷Ϊ¡°¸¯Ê´Òº¡±¡£
¢Ùд³öFeCl3ÈÜÒº¡°¸¯Ê´¡±Í­²­µÄÀë×Ó·½³Ìʽ     ¡£
¢Úij»¯Ñ§ÐËȤС×éΪ·ÖÎöij³§¼Ò¸¯Ê´ÒÔºóËùµÃ»ìºÏÈÜÒº×é³É£¬ÊÔ̽ÐÔµØÏò100 mL¸¯Ê´ºóµÄ»ìºÏÈÜÒºÖмÓÈë2.80gÌú·Û£¬½á¹ûÈ«²¿ÈܽâÇÒδ¼û¹ÌÌåÎö³ö¡£Ôò»ìºÏÈÜÒºµÄ×é³ÉΪ£¨Ìîд»¯Ñ§Ê½£©£º     
£¨1£©Ê¯»Òʯ£¬Á¶Ìú¸ß¯        £¨2£©2Cl2+2Ca(OH)2= Ca(ClO)2 +CaCl2£«2H 2O
£¨3£©Al(OH)3+3H+= Al3+£«3H 2  £¨4£©¢Ù2Fe3++Cu=2Fe2++Cu2+  ¢ÚFeCl3¡¢FeCl2¡¢CuCl2

ÊÔÌâ·ÖÎö£º£¨1£©Éú²ú²£Á§µÄÖ÷ÒªÔ­ÁÏÊÇ´¿¼î¡¢Ê¯»ÒʯºÍʯӢ£¬Éú²úË®ÄàµÄÖ÷ÒªÔ­ÁÏÊÇð¤ÍÁ¡¢Ê¯»Òʯ£¬Á¶ÌúµÄÖ÷ÒªÔ­ÁÏÊÇÌú¿óʯ¡¢½¹Ì¿¡¢Ê¯»ÒʯºÍ¿ÕÆø£¬ËùÒÔ¶¼ÒªÓõ½µÄÔ­ÁÏÊÇʯ»Òʯ¡£¹¤ÒµÁ¶ÌúµÄÖ÷ÒªÉ豸ÊǸ߯¡£
£¨2£©ÂÈÆøºÍÏûʯ»ÒÖÆƯ°×·ÛµÄ·½³ÌʽΪ2Cl2+2Ca(OH)2= Ca(ClO)2 +CaCl2£«2H 2
£¨3£©Î¸ÊæƽÖÎÁÆθËá¹ý¶àµÄÀë×Ó·½³ÌʽΪAl(OH)3+3H+= Al3+£«3H 2
£¨4£©¢ÙFeCl3ÈÜÒº¡°¸¯Ê´¡±Í­²­µÄÀë×Ó·½³Ìʽ2Fe3++Cu=2Fe2++Cu2+¡£Ïò¸¯Ê´ºóµÄÈÜÒº×ܼÓÈë2.80gÌú·Û£¬Ã»ÓйÌÌåÎö³ö£¬ËµÃ÷Cu2+ûÓвÎÓë·´Ó¦£¬ËùÒÔÈÜÒºÖдæÔÚFe3+£¬Ôò»ìºÏÈÜÒºµÄ×é³ÉΪFeCl3¡¢FeCl2¡¢CuCl2¡£
µãÆÀ£º±¾Ìâ·Ç³£»ù´¡¼òµ¥£¬Ö÷Òª¿¼²é·½³ÌʽµÄÊéд¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø