ÌâÄ¿ÄÚÈÝ

пºÍÂÁ¶¼ÊÇ»îÆÃ½ðÊô£¬ÆäÇâÑõ»¯Îï¼ÈÄÜÈÜÓÚÇ¿ËᣬÓÖÄÜÈÜÓÚÇ¿¼î£®µ«ÇâÑõ»¯ÂÁ²»ÈÜÓÚ°±Ë®£¬¶øÇâÑõ»¯Ð¿ÄÜÈÜÓÚ°±Ë®ÖÐÉú³ÉZn(NH3)£®»Ø´ðÏÂÁÐÎÊÌ⣺

(1)µ¥ÖÊÂÁÈÜÓÚÇâÑõ»¯ÄÆÈÜÒººó£¬ÈÜÒºÖÐÂÁÔªËØµÄ´æÔÚÐÎʽΪ________£®(Óû¯Ñ§Ê½±íʾ)

(2)д³öпÓëÇâÑõ»¯ÄÆÈÜÒº·´Ó¦µÄ»¯Ñ§·½³Ìʽ________£®

(3)ÏÂÁи÷×éÖеÄÁ½ÖÖÈÜÒº£¬ÓÃÏ໥µÎ¼ÓµÄʵÑé·½·¨¼´¿É¼ø±ðµÄÊÇ________£®

¢ÙÁòËáÂÁºÍÇâÑõ»¯ÄÆ¡¡¢ÚÁòËáÂÁºÍ°±Ë®

¢ÛÁòËáпºÍÇâÑõ»¯ÄÆ¡¡¢ÜÁòËáпºÍ°±Ë®

(4)д³ö¿ÉÈÜÐÔÂÁÑÎÓ백ˮ·´Ó¦µÄÀë×Ó·½³Ìʽ£º________£®

ÊÔ½âÊÍʵÑéÊÒ²»ÊÊÒËÓÿÉÈÜÐÔпÑÎÓ백ˮ·´Ó¦ÖƱ¸ÇâÑõ»¯Ð¿µÄÔ­Òò________£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
пºÍÂÁ¶¼ÊÇ»îÆÃ½ðÊô£¬ÆäÇâÑõ»¯Îï¼ÈÄÜÈÜÓÚÇ¿ËᣬÓÖÄÜÈÜÓÚÇ¿¼î£®µ«ÊÇÇâÑõ»¯ÂÁ²»ÈÜÓÚ°±Ë®£¬¶øÇâÑõ»¯Ð¿ÄÜÈÜÓÚ°±Ë®£¬Éú³ÉZn£¨NH3£©42+£®»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©µ¥ÖÊÂÁÈÜÓÚÇâÑõ»¯ÄÆÈÜÒººó£¬ÈÜÒºÖÐÂÁÔªËØµÄ´æÔÚÐÎʽΪ
Na[Al£¨0H£©4]
Na[Al£¨0H£©4]
£¨Óû¯Ñ§Ê½±íʾ£©£®
£¨2£©Ð´³öпºÍÇâÑõ»¯ÄÆÈÜÒº·´Ó¦µÄ»¯Ñ§·½³Ìʽ
Zn+2NaOH+2H2O=Na2[Zn£¨0H£©4]+H2¡ü
Zn+2NaOH+2H2O=Na2[Zn£¨0H£©4]+H2¡ü
£®
£¨3£©ÏÂÁи÷×éÖеÄÁ½ÖÖÈÜÒº£¬ÓÃÏ໥µÎ¼ÓµÄʵÑé·½·¨¼´¿É¼ø±ðµÄÊÇ
¢Ù¢Û¢Ü
¢Ù¢Û¢Ü
£®
¢ÙÁòËáÂÁºÍÇâÑõ»¯ÄÆ ¢ÚÁòËáÂÁºÍ°±Ë®¢ÛÁòËáпºÍÇâÑõ»¯ÄÆ¢ÜÁòËáпºÍ°±Ë®
£¨4£©Ð´³ö¿ÉÈÜÐÔÂÁÑÎÓ백ˮ·´Ó¦µÄÀë×Ó·½³Ìʽ
Al3++3NH3?H2O=Al£¨OH£©3¡ý+3NH4+
Al3++3NH3?H2O=Al£¨OH£©3¡ý+3NH4+
£®ÊÔ½âÊÍÔÚʵÑéÊÒ²»ÊÊÒËÓÿÉÈÜÐÔпÑÎÓ백ˮ·´Ó¦ÖƱ¸ÇâÑõ»¯Ð¿µÄÔ­Òò
¿ÉÈÜÐÔпÑÎÓ백ˮ·´Ó¦²úÉúµÄÇâÑõ»¯Ð¿¿ÉÈÜÓÚ¹ýÁ¿µÄ°±Ë®ÖУ¬Éú³ÉZn£¨NH3£©42+£¬°±Ë®µÄÁ¿²»Ò׿ØÖÆ
¿ÉÈÜÐÔпÑÎÓ백ˮ·´Ó¦²úÉúµÄÇâÑõ»¯Ð¿¿ÉÈÜÓÚ¹ýÁ¿µÄ°±Ë®ÖУ¬Éú³ÉZn£¨NH3£©42+£¬°±Ë®µÄÁ¿²»Ò׿ØÖÆ
£®

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

¾«Ó¢¼Ò½ÌÍø