ÌâÄ¿ÄÚÈÝ

±¥ºÍÑÎÈÜÒºWµÄµç½â²úÎï¿É·¢ÉúÏÂÁеÄϵÁз´Ó¦¡£Í¼ÖеÄÿһ·½¸ñ±íʾÓйصÄÒ»ÖÖÖ÷Òª·´Ó¦Îï»òÉú³ÉÎ·´Ó¦ÖмÓÈë»òÉú³ÉµÄË®ÒÔ¼°Éú³ÉµÄÆäËû²úÎïÒÑÂÔÈ¥£©¡£ÆäÖÐA¡¢B¡¢C¡¢D¡¢EÔÚ³£Î³£Ñ¹Ï¾ùÊÇÆøÌ¬ÎïÖÊ¡£
°´ÕÕ·´Ó¦¹ý³Ì£¬·ÖÎöÏÂÁÐÎÊÌ⣺
(1)È·¶¨ÎïÖÊ»¯Ñ§Ê½£ºW_________£¬C__________¡£
(2)Êéд±¥ºÍÑÎÈÜÒºWµÄµç½â»¯Ñ§·½³Ìʽ£º________________¡£
(3)Êéд·´Ó¦¢Ù¢ÚµÄ»¯Ñ§·½³Ìʽ£¬ÈôÊÇÑõ»¯»¹Ô­·´Ó¦£¬±ê³öµç×Ó×ªÒÆµÄ·½ÏòºÍÊýÄ¿£º
¢Ù___________________£¬¢Ú____________________¡£
(4)Êéд·´Ó¦¢Û(HÖÐͨÈëA)µÄÀë×Ó·½³Ìʽ£º____________________¡£
(5)ÎïÖÊIÊÇÒ»ÖÖÀë×Ó»¯ºÏÎÓõç×Óʽ±íʾÆä½á¹¹£º____________________¡£
(I)NH4Cl£»H2
(2)2NH4ClCl2¡ü+H2¡ü+NH3¡ü£»
(3)¢Ù£»¢ÚMg3N2+6H2O=3Mg(OH)2¡ý+2NH3¡ü
(4)Mg2++2NH3+2H2O=Mg(OH)2¡ý+2NH4+
(5)
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø