ÌâÄ¿ÄÚÈÝ
A¡¢B¡¢C¡¢D¡¢EΪÎåÖÖ¶ÌÖÜÆÚÔªËØ¡£AÊÇÒÑ·¢ÏÖÔªËØÖÐÔ×Ó°ë¾¶×îСµÄ£»EÊǶÌÖÜÆÚÖÐÔ×Ó°ë¾¶×î´óµÄÖ÷×åÔªËØ£»BÓëEͬÖÜÆÚÇÒBµÄ×îÍâ²ãµç×ÓÊý±È´ÎÍâ²ãµç×ÓÊýÉÙÒ»¸ö£»C¡¢DΪͬÖÜÆÚµÄÏàÁÚÔªËØÇÒCµÄ°ë¾¶Ð¡ÓÚD£¬ËüÃǵĺËÍâµç×ÓÊýÖ®ºÍΪ15£¬ÊԻشð£º
¢ÅBµÄÃû³ÆÎª£º £»DµÄÔ×ӽṹʾÒâͼΪ£º
¢ÆÓÉA¡¢C¹¹³ÉµÄ4Ô×Ó·Ö×ӵĵç×ÓʽΪ £¬¸Ã»¯ºÏÎïÊôÓÚ »¯ºÏÎÌîÀë×Ó»ò¹²¼Û£©
¢ÇAµ¥ÖÊÔÚBµ¥ÖÊÖÐȼÉÕʱµÄ»ðÑæÎª É«
¢ÈEµ¥ÖÊÔÚCµ¥ÖÊÖÐȼÉյĻ¯Ñ§·´Ó¦·½³ÌʽΪ£º
¢ÉÓÉA¡¢C¡¢DÈýÖÖÔªËØ×é³ÉµÄÀë×Ó»¯ºÏÎïµÄ»¯Ñ§Ê½Îª £¨ÈÎдһÖÖ£©
¡¾´ð°¸¡¿
£¨1£©Cl
£¨2£©¹²¼Û
£¨3£©²Ô°×
£¨4£©2Na+O2=Na2O2
(5)NH4NO3
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
¡¾»¯Ñ§--Ñ¡ÐÞ3£ºÎïÖʽṹÓëÐÔÖÊ¡¿
ÏÖÓÐÆßÖÖÔªËØ£¬ÆäÖÐA¡¢B¡¢C¡¢D¡¢EΪ¶ÌÖÜÆÚÖ÷×åÔªËØ£¬F¡¢GΪµÚËÄÖÜÆÚÔªËØ£¬ËüÃǵÄÔ×ÓÐòÊýÒÀ´ÎÔö´ó£®Çë¸ù¾ÝÏÂÁÐÏà¹ØÐÅÏ¢£¬»Ø´ðÎÊÌ⣮
£¨1£©GλÓÚ ×å Çø£¬¼Ûµç×ÓÅŲ¼Ê½Îª £®
£¨2£©B»ù̬Ô×ÓÖÐÄÜÁ¿×î¸ßµÄµç×Ó£¬Æäµç×ÓÔÆÔÚ¿Õ¼äÓÐ ¸ö·½Ïò£¬Ô×Ó¹ìµÀ³Ê
ÐΣ®
£¨3£©»³öCÔ×ӵĵç×ÓÅŲ¼Í¼ £®
£¨4£©ÒÑÖªBA5ΪÀë×Ó»¯ºÏÎд³öÆäµç×Óʽ £®
£¨5£©DE3ÖÐÐÄÔ×ÓµÄÔÓ»¯·½Ê½Îª £¬Óü۲ãµç×Ó¶Ô»¥³âÀíÂÛÍÆ²âÆä¿Õ¼ä¹¹ÐÍΪ £®
£¨6£©Óõç×Óʽ±íʾFÔªËØÓëEÔªËØÐγɻ¯ºÏÎïµÄÐγɹý³Ì £®
ÏÖÓÐÆßÖÖÔªËØ£¬ÆäÖÐA¡¢B¡¢C¡¢D¡¢EΪ¶ÌÖÜÆÚÖ÷×åÔªËØ£¬F¡¢GΪµÚËÄÖÜÆÚÔªËØ£¬ËüÃǵÄÔ×ÓÐòÊýÒÀ´ÎÔö´ó£®Çë¸ù¾ÝÏÂÁÐÏà¹ØÐÅÏ¢£¬»Ø´ðÎÊÌ⣮
| AÔªËØµÄºËÍâµç×ÓÊýºÍµç×Ó²ãÊýÏàµÈ£¬Ò²ÊÇÓîÖæÖÐ×î·á¸»µÄÔªËØ |
| BÔªËØÔ×ӵĺËÍâpµç×ÓÊý±Èsµç×ÓÊýÉÙ1 |
| CÔ×ӵĵÚÒ»ÖÁµÚËĵçÀëÄÜ·Ö±ðÊÇ£º I1=738kJ/mol I2=1451kJ/mol I3=7733kJ/mol I4=10540kJ/mol |
| DÔ×ÓºËÍâËùÓÐp¹ìµÀÈ«Âú»ò°ëÂú |
| EÔªËØµÄÖ÷×åÐòÊýÓëÖÜÆÚÊýµÄ²îΪ4 |
| FÊÇǰËÄÖÜÆÚÖе縺ÐÔ×îСµÄÔªËØ |
| GÔÚÖÜÆÚ±íµÄµÚÆßÁÐ |
£¨2£©B»ù̬Ô×ÓÖÐÄÜÁ¿×î¸ßµÄµç×Ó£¬Æäµç×ÓÔÆÔÚ¿Õ¼äÓÐ
ÐΣ®
£¨3£©»³öCÔ×ӵĵç×ÓÅŲ¼Í¼
£¨4£©ÒÑÖªBA5ΪÀë×Ó»¯ºÏÎд³öÆäµç×Óʽ
£¨5£©DE3ÖÐÐÄÔ×ÓµÄÔÓ»¯·½Ê½Îª
£¨6£©Óõç×Óʽ±íʾFÔªËØÓëEÔªËØÐγɻ¯ºÏÎïµÄÐγɹý³Ì