ÌâÄ¿ÄÚÈÝ

£¨10·Ö£©Ç°20ºÅÔªËØA¡¢B¡¢C¡¢D£¬AÔªËØËù´¦µÄÖÜÆÚÊý¡¢Ö÷×åÐòÊý¡¢Ô­×ÓÐòÊý¾ùÏàµÈ£»BµÄÔ­×Ó°ë¾¶ÊÇÆäËùÔÚÖ÷×åÖÐ×îСµÄ£¬BµÄ×î¸ß¼ÛÑõ»¯Îï¶ÔӦˮ»¯ÎïµÄ»¯Ñ§Ê½ÎªHBO3£»CÔªËØÔ­×ÓµÄ×îÍâ²ãµç×ÓÊý±È´ÎÍâ²ãÉÙ2¸ö£»CµÄÒõÀë×ÓÓëDµÄÑôÀë×Ó¾ßÓÐÏàͬµÄµç×ÓÅŲ¼£¬Á½ÔªËØ¿ÉÐγɻ¯ºÏÎïD2C¡£

(1) BÔªËØµÄÔ­×ӽṹʾÒâͼ    £»

(2)CÔÚÖÜÆÚ±íÖеÄλÖà £»

(3)»¯ºÏÎïD2CµÄµç×Óʽ__£»

(4) BµÄÇ⻯ÎïµÄË®ÈÜÒº¼ÓÈëBµÄ×î¸ß¼ÛÑõ»¯Îï¶ÔӦˮ»¯ÎïµÄÏ¡ÈÜÒº£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪ£º                  £»

(5)ÓÐÈËÈÏΪ£¬Aµ¥ÖÊÊÇÒ»ÖÖÓÅÖÊÄÜÔ´£¬ÄãÈÏΪËü×÷ΪÄÜÔ´µÄÓŵãÊÇ£º                  

                                                

 

(1) £¨NµÄÔ­×ӽṹʾÒâͼ£©£¨2·Ö£©¡£

(2) µÚÈýÖÜÆÚµÚ¢öA×å £¨2·Ö£©£»(3) £¨2·Ö£©¡£

(4)   NH3¡¤H2O+ H+ = NH4+ + H2O  £¨2·Ö£©¡£

£¨5£©ÈÈÖµ¸ß£»À´Ô´¹ã£»ÎÞÎÛȾ¡££¨2·Ö£¬ÉÙÒ»¸ö¿Û1·Ö£©

½âÎö:ǰ20ºÅÔªËØÖÐÖ»ÓÐÇâÔªËØµÄÖÜÆÚÊý¡¢Ö÷×åÐòÊý¡¢Ô­×ÓÐòÊý¾ùÏàµÈ£¬AΪÇ⣻BÔªËØÔÚHBO3»¯ºÏ¼ÛÊÇ+5¼ÛËùÒÔÊǵÚÎåÖ÷×åÔªËØ£¬ÓÖËüÊÇËùÔÚ×å°ë¾¶×îСµÄ£¬BΪµªÔªËØ£»CÖ»ÄÜÊÇÁòÔªËØ£»ÔÙÓÉD2CÖªDµÄ»¯ºÏ¼ÛÊÇ+1¼Û£¬CµÄÒõÀë×ÓÓëDµÄÑôÀë×Ó¾ßÓÐÏàͬµÄµç×ÓÅŲ¼£¬ËùÒÔDÊǼØÔªËØ¡£ÕâÑùºÜÈÝÒ×½â¾öÒÔÉÏÎÊÌâ¡£

 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø