ÌâÄ¿ÄÚÈÝ

£¨14·Ö£©A¡¢B¡¢C¡¢D¡¢EÎåÖÖ¶ÌÖÜÆÚÖ÷×åÔªËØ·ÖÕ¼Èý¸öÖÜÆÚ£¬A¡¢B¡¢CΪͬһÖÜÆÚÒÀ´ÎÏàÁÚµÄ3ÖÖÔªËØ£¬AºÍCµÄÔ­×ÓÐòÊýÖ®±ÈΪ3¡Ã4£¬EÔ­×ӵĵç×Ó²ãÊýµÈÓÚ×îÍâ²ãµç×ÓÊý£¬DµÄÔ­×ÓÐòÊýСÓÚE¡£ÇëÓû¯Ñ§ÓÃÓï»Ø´ðÏà¹ØÎÊÌ⣺

£¨1£©AÔªËØÔÚÖÜÆÚ±íÖеÄλÖà                 ¡£

£¨2£©±È½ÏCºÍE¼òµ¥Àë×Ӱ뾶´óС£º            ¡£

£¨3£©ÔªËØEµÄÒ»ÖÖ³£¼ûµÄ¿ÉÈÜÐÔÑÎÈÜÒº³Ê¼îÐÔ£¬ÆäÔ­ÒòÊÇ£¨ÓÃÀë×Ó·½³Ìʽ±íʾ£©£º      £»

£¨4£©X¡¢Y¡¢Z¡¢¼×¡¢ÒÒ¡¢±ûÊÇÓÉA¡¢B¡¢C·Ö±ðÓëDÐγɵĻ¯ºÏÎÁùÖÖ»¯ºÏÎï¿ÉÒÔÅųÉÏÂ±í£¬ÆäÖÐͬһºáÐеķÖ×ÓÖеç×ÓÊýÏàͬ£¬Í¬Ò»×ÝÐеÄÎïÖÊËùº¬ÔªËØÖÖÀàÏàͬ£¬ÆäÖÐX¡¢Y¡¢¼×³£Î³£Ñ¹ÏÂΪÆøÌ壬Z¡¢ÒÒ¡¢±û³£Î³£Ñ¹ÏÂΪҺÌå¡£

¢Ù¼×µÄ·Ö×ÓʽΪ               £»±ûµÄµç×ÓʽΪ          ¡£

¢ÚÒҵķÖ×ÓʽΪB2D4£¬ÒҺͱû³£×÷»ð¼ýÍƽøÆ÷µÄȼÁÏ£¬·´Ó¦ºóµÄ²úÎïÎÞÎÛȾ¡£ÒÑÖª8gҺ̬ÒÒÓë×ãÁ¿ÒºÌ¬±ûÍêÈ«·´Ó¦£¬²úÎï¾ùΪÆøÌåʱ£¬·Å³öÈÈÁ¿Îª160.35kJ£¬ÊÔд³ö¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ£º                               ¡£

 

£¨1£©µÚ¶þÖÜÆÚ¢ôA×å¡££¨2·Ö£©

   £¨2£©O2->Al3+£¨2·Ö£©¡£

   £¨3£©A1O2¡ª+2H2OAl(OH)3+OH¡ª£¨3·Ö£©

   £¨4£©¢ÙC2H6£¬    £¨¸÷2·Ö£©

       ¢ÚN2H4(1)+2H2O2(1)=N2(g)+4H2O(g)  ¡÷H=-641.40kJ/mol£¨3·Ö£©

½âÎö:ÂÔ

 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø