ÌâÄ¿ÄÚÈÝ

ÅðËá¾§Ìå³ÉƬ״£¬Óл¬Äå¸Ð£¬¿É×÷È󻬼Á£¬ÅðËá·Ö×ӽṹ¿É±íʾΪ¡£ÅðËá¶ÔÈËÌåµÄÊÜÉË×éÖ¯ÓлººÍµÄ·À¸¯×÷Ó㬹ʿÉÓÃÓÚÒ½Ò©ºÍʳƷ·À¸¯µÈ·½Ãæ

   £¨1£©¸ù¾ÝÒÔÉÏËùÊö¿ÉÖªÅðËáÓ¦ÊôÓÚ_____________¡£

A£®Ç¿Ëá                      B£®ÖÐÇ¿Ëá         C£®ÈõËá

   £¨2£©Ñо¿±íÃ÷£ºÔÚ´ó¶àÊýÇé¿öÏ£¬ÔªËصÄÔ­×ÓÔÚÐγɷÖ×Ó»òÀë×Óʱ£¬Æä×îÍâµç×Ó²ã¾ßÓдﵽ8µç×ÓÎȶ¨½á¹¹µÄÇ÷ÊÆ¡£ÔÚÅðËá·Ö×ÓÖУ¬×îÍâ²ã´ïµ½8¸öµç×ÓÎȶ¨½á¹¹µÄÔ­×ÓÓÐ_______¸ö¡£

   £¨3£©ÅðËáºÍ¼×´¼ÔÚŨÁòËá´æÔÚµÄÌõ¼þÏ£¬¿ÉÉú³É»Ó·¢ÐÔÅðËáÈý¼×õ¥£¬ÊÔд³öÅðËáÍêÈ«õ¥»¯µÄ»¯Ñ§·½³Ìʽ£¨×¢Ã÷·´Ó¦Ìõ¼þ£©_______________________________________¡£

   £¨4£©ÒÑÖªÅðËá0.01mol¿É±»20mL0.5mol¡¤L-1NaOHÈÜҺǡºÃÍêÈ«Öкͣ¬¾Ý´ËÍÆ²â£ºÅðËáÔÚË®ÖÐÏÔËáÐÔµÄÔ­ÒòÊÇ£¨Ð´µçÀë·½³Ìʽ£©_________________________________¡£

 

£¨1£©C

£¨2£©3

£¨3£©

£¨4£©H3BO3 +H2OB(OH)4-+H+

½âÎö:

 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
ÅðËá¾§Ìå³ÊƬ״£¬Óл¬Äå¸Ð£¬¿É×öÈ󻬼Á£®ÅðËá¶ÔÈËÌåµÄÊÜÉË×éÖ¯ÓлººÍºÍ·À¸¯×÷Ó㬹ʿÉÒÔÓÃÓÚÒ½Ò©ºÍʳƷ·À¸¯µÈ·½Ã森ÊԻشðÏÂÁÐÎÊÌ⣺
£¨1£©´ÓÌâÄ¿ÐÅÏ¢¿ÉÖªÅðËáÓ¦ÊôÓÚ
Èõ
Èõ
ËᣨÌî¡°Èõ¡±»ò¡°Ç¿¡±£©
£¨2£©ÅðËáµÄ·Ö×ÓʽΪH3BO3£¬ÒÑÖªHÓëO³É¼ü£¬ÔòÆä·Ö×ӽṹʽΪ£º
£®
£¨3£©Ñо¿±íÃ÷£ºÔÚ´ó¶àÊýÇé¿öÏ£¬ÔªËصÄÔ­×ÓÔÚÐγɷÖ×Ó»òÀë×Óʱ£¬Æä×îÍâ²ãÓдﵽ8µç×ÓÎȶ¨½á¹¹µÄÇ÷Ïò£®ÒÑÖª0.01molÅðËá¿ÉÒÔ±»20mL 0.5mol?L-1NaOHÈÜҺǡºÃÍêÈ«Öкͣ¬¾Ý´ËÍÆ²âÅðËáÔÚË®ÈÜÒºÖгÊËáÐÔµÄÔ­ÒòÊÇ£¨ÓÃÀë×Ó·½³Ìʽ±íʾ£©
B£¨OH£©3+H2O=B£¨OH£©4-+H+
B£¨OH£©3+H2O=B£¨OH£©4-+H+
£»Ð´³öÅðËáÓëNaOHÈÜÒº·´Ó¦µÄÀë×Ó·½³Ìʽ£º
B£¨OH£©3+OH-=B£¨OH£©4-
B£¨OH£©3+OH-=B£¨OH£©4-
£®
£¨4£©ÅðËáºÍ¼×´¼ÔÚŨÁòËá´æÔÚµÄÌõ¼þÏ£¬¿ÉÉú³É»Ó·¢ÐÔÅðËáõ¥£¬ÊÔд³öÅðËáÍêÈ«õ¥»¯µÄ·½³Ìʽ£º
B£¨OH£©3+3CH3OH¨TB£¨OCH3£©3+3H2O
B£¨OH£©3+3CH3OH¨TB£¨OCH3£©3+3H2O
£®
£¨5£©¿ÆÑ§¼Ò·¢ÏÖÅð»¯Ã¾ÔÚ39Kʱ³Ê³¬µ¼ÐÔ£¬ÔÚÅð»¯Ã¾¾§ÌåµÄÀíÏëÄ£ÐÍÖУ¬Ã¾Ô­×ÓºÍÅðÔ­×ÓÊÇ·Ö²ãÅŲ¼µÄ£¬Ò»²ãþһ²ãÅðÏà¼äÅÅÁУ®

ÈçͼÊǸþ§Ìå΢¹Û¿Õ¼äÖÐÈ¡³öµÄ²¿·ÖÔ­×ÓÑØZÖá·½ÏòµÄͶӰ£¬°×ÇòÊÇþԭ×ÓͶӰ£¬ºÚÇòÊÇÅðÔ­×ÓͶӰ£¬Í¼ÖеÄÅðÔ­×ÓºÍþԭ×ÓͶӰÔÚÍ¬Ò»Æ½ÃæÉÏ£®¸ù¾ÝÏÂͼȷ¶¨Åð»¯Ã¾µÄ»¯Ñ§Ê½Îª
MgB2
MgB2
£®

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

¾«Ó¢¼Ò½ÌÍø