ÌâÄ¿ÄÚÈÝ

(14·Ö)

ϱíÊÇÔªËØÖÜÆÚ±íµÄÒ»²¿·Ö£¬Çë»Ø´ðÓйØÎÊÌ⣺
    Ö÷×å
ÖÜÆÚ
IA
IIA
IIIA
IVA
VA
VIA
VIIA
0
2
 
 
 
¢Ù
¢Ú
 
¢Û
 
3
 
 
¢Ý
 
 
¢Þ
¢ß
¢à
4
¢á
¢Ü
 
 
 
 
¢â
 
£¨1£©±íÖл¯Ñ§ÐÔÖÊ×î²»»îÆõÄÔªËØ£¬ÆäÔ­×ӽṹʾÒâͼΪ          ¡£
£¨2£©±íÖÐÄÜÐγÉÁ½ÐÔÇâÑõ»¯ÎïµÄÔªËØÊÇ            (ÓÃÔªËØ·ûºÅ±íʾ)£¬Ð´³ö¸ÃÔªËصÄÖÊÓë¢á×î¸ß¼ÛÑõ»¯ÎïµÄË®»¯Îï·´Ó¦µÄ»¯Ñ§·½³Ìʽ              
£¨3£©¢ÜÔªËØÓë¢ßÔªËØÐγɻ¯ºÏÎïµÄµç×Óʽ                    
£¨4£©¢Ù¡¢¢Ú¡¢¢Þ¡¢¢ßËÄÖÖÔªËصÄ×î¸ß¼ÛÑõ»¯ÎïµÄË®»¯ÎïÖÐËáÐÔ×îÇ¿µÄÊÇ              
                    (Ìѧʽ)¡£
£¨5£©¢ÛÔªËØÓë¢âÔªËØÁ½Õߺ˵çºÉÊýÖ®²îÊÇ           ¡£
£¨6£©Éè¼ÆʵÑé·½°¸£º±È½Ï¢ßÓë¢âµ¥ÖÊÑõ»¯ÐÔµÄÇ¿Èõ£¬Ç뽫·½°¸ÌîÈëÏÂ±í¡£
ʵÑé²½Öè
ʵÑéÏÖÏóÓë½áÂÛ
 
 






£¨1£©ÂÔ £¨2£©Al   2Al + 2KOH + 2H2O = 2KAlO2 + 3H2 ¡ü 
£¨3£©ÂÔ  £¨4£©HClO4  £¨5£©26  
£¨6£©£¨ÆäËûºÏÀí´ð°¸Ò²¸ø·Ö£©
ʵÑé²½Öè
ʵÑéÏÖÏóÓë½áÂÛ
Íùһ֧СÊÔ¹ÜÖмÓÈëÉÙÁ¿NaBrÈÜÒº£¬µÎ¼ÓÂÈË®
Èç¹ûÈÜÒºÓÉÎÞÉ«±äΪ³ÈÉ«£¬ËµÃ÷Cl2µÄ·Ç½ðÊôÐÔ±ÈBr2Ç¿
 
ÂÔ
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø