ÌâÄ¿ÄÚÈÝ
£¨10·Ö£©ÏÂÁпòͼËùʾµÄÎïÖÊת»¯¹ØϵÖУ¬¼×ÊÇÈÕ³£Éú»îÖг£¼ûµÄ½ðÊô£¬ÒÒ¡¢±û¡¢¶¡Êdz£¼ûµÄÆøÌåµ¥ÖÊ¡£ÆøÌåBÓëÆøÌåCÏàÓö²úÉú´óÁ¿µÄ°×ÑÌÉú³ÉE£¬AÊÇÇ¿¼î£¬DÊǺ£Ë®ÖÐŨ¶È×î¸ßµÄÑΣ¨²¿·Ö·´Ó¦ÎïºÍÉú³ÉÎï¼°Ë®ÒÑÂÔÈ¥£©¡£
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Ð´³öÒÒÓë±ûÉú³ÉBµÄ»¯Ñ§·½³Ìʽ£º
£¨2£©Ð´³ö¼×ºÍAÈÜÒº·´Ó¦µÄÀë×Ó·½³Ìʽ£º
£¨3£©¼ìÑéEÖÐÑôÀë×Óʱ£¬È¡ÉÙÁ¿EÓÚÊÔ¹ÜÖÐ ÔòÖ¤Ã÷EÖÐÓиÃÑôÀë×Ó¡£
£¨4£©Ð´³öʵÑéÊÒÖÆÈ¡BµÄ»¯Ñ§·½³Ìʽ£º
£¨5£©¹¤ÒµÉÏÓÃB×÷ÔÁÏÖƱ¸Ä³Ëᣬд³ö¸ÃËáµÄŨÈÜÒºÔÚ³£ÎÂÏÂÓëij½ðÊô·´Ó¦£¬Éú³ÉÂÌÉ«ÈÜÒºµÄ»¯Ñ§·½³Ìʽ£º
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Ð´³öÒÒÓë±ûÉú³ÉBµÄ»¯Ñ§·½³Ìʽ£º
£¨2£©Ð´³ö¼×ºÍAÈÜÒº·´Ó¦µÄÀë×Ó·½³Ìʽ£º
£¨3£©¼ìÑéEÖÐÑôÀë×Óʱ£¬È¡ÉÙÁ¿EÓÚÊÔ¹ÜÖÐ ÔòÖ¤Ã÷EÖÐÓиÃÑôÀë×Ó¡£
£¨4£©Ð´³öʵÑéÊÒÖÆÈ¡BµÄ»¯Ñ§·½³Ìʽ£º
£¨5£©¹¤ÒµÉÏÓÃB×÷ÔÁÏÖƱ¸Ä³Ëᣬд³ö¸ÃËáµÄŨÈÜÒºÔÚ³£ÎÂÏÂÓëij½ðÊô·´Ó¦£¬Éú³ÉÂÌÉ«ÈÜÒºµÄ»¯Ñ§·½³Ìʽ£º
£¨10·Ö£©£¨Ã¿¿Õ2·Ö£©£¨1£©N2+3H22NH3
£¨2£©2Al+2OH-+2H2O=2AlO2-+3H2¡ü
£¨3£©¼ÓÈëNaOHÈÜÒº£¬¼ÓÈÈ£¬ÓÃʪÈóµÄºìɫʯÈïÊÔÖ½¿¿½üÊԹܿÚÊÔÖ½±äÀ¶
£¨4£©2NH4Cl+Ca(OH)2CaCl2+2NH3¡ü+2H2O
£¨5£©Cu+4HNO3(Ũ)=Cu(NO3)2+2NO2¡ü+2H2O
£¨2£©2Al+2OH-+2H2O=2AlO2-+3H2¡ü
£¨3£©¼ÓÈëNaOHÈÜÒº£¬¼ÓÈÈ£¬ÓÃʪÈóµÄºìɫʯÈïÊÔÖ½¿¿½üÊԹܿÚÊÔÖ½±äÀ¶
£¨4£©2NH4Cl+Ca(OH)2CaCl2+2NH3¡ü+2H2O
£¨5£©Cu+4HNO3(Ũ)=Cu(NO3)2+2NO2¡ü+2H2O
±¾ÌâÊÇÎÞ»ú¿òͼÌ⣬¸ù¾ÝÕÒ׼ͻÆƵ㡣ÆøÌåBÓëÆøÌåCÏàÓö²úÉú´óÁ¿µÄ°×ÑÌÉú³ÉE£¬ËùÒÔEÊÇÂÈ»¯ï§£¬ÔòCÊÇÂÈ»¯Ç⣬BÊÇ°±Æø£¬ÒÒÊǵªÆø£¬±ûÊÇÇâÆø£¬¶¡ÊÇÂÈÆø¡£DÊǺ£Ë®ÖÐŨ¶È×î¸ßµÄÑΣ¬ËùÒÔDÊÇÂÈ»¯ÄÆ£¬ÔòAÊÇÇâÑõ»¯ÄÆ¡£½ðÊô¼×ÄܺÍÇâÑõ»¯ÄÆ·´Ó¦Éú³ÉÇâÆø£¬ËùÒÔ¼×ÊÇÂÁ¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿