ÌâÄ¿ÄÚÈÝ

A¡¢B¡¢C¡¢DËÄÖÖ¶ÌÖÜÆÚÔªËØ£¬A¡¢B¡¢CͬÖÜÆÚ£¬AµÄÔ­×Ó°ë¾¶ÊÇͬÖÜÆÚÖÐ×î´óµÄ£»B¡¢DͬÖ÷×å¡£¼ºÖªDÔªËØµÄÒ»ÖÖµ¥ÖÊÊÇÈÕ³£Éú»îÖÐÒûË®»ú³£ÓõÄÏû¶¾¼Á£¬CÔªËØµÄµ¥ÖÊ¿ÉÒÔ´ÓA¡¢BÁ½ÔªËØ×é³ÉµÄ»¯ºÏÎïµÄË®ÈÜÒºÖÐÖû»³öBÔªËØµÄµ¥ÖÊ¡£

£¨1£©CÔªËØÔÚÖÜÆÚ±íÖеÄλÖà              ¡£

£¨2£©AÔªËØÓëË®·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ                              ¡£

£¨3£©Ð´³öCÔªËØµÄµ¥ÖÊ´ÓA¡¢BÁ½ÔªËØ×é³ÉµÄ»¯ºÏÎïµÄË®ÈÜÒºÖÐÖû»³öBÔªËØµÄµ¥ÖʵĻ¯Ñ§·½³Ìʽ                               ¡£

£¨4£©ºÍ¾ù¾ßÓÐÆ¯°×ÐÔ£¬¶þÕߵᝰ×Ô­Àí       ¡££¨Ìî¡°Ïàͬ¡±»ò¡°²»Í¬¡±£©

£¨5£©ÇâÆøÊǺϳɰ±µÄÖØÒªÔ­ÁÏ£¬ºÏ³É°±·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽÈçÏ£º

       3H2£«N2  2NH3 ¡÷H=£­92£®4kJ¡¤mol£­1

¢Ùµ±ºÏ³É°±·´Ó¦´ïµ½Æ½ºâºó£¬¸Ä±äijһÍâ½çÌõ¼þ £¨²»

¸Ä±ä¡¢ºÍµÄÁ¿£©£¬·´Ó¦ËÙÂÊÓëʱ¼äµÄ¹Ø

ϵÈçÓÒͼËùʾ¡£Í¼ÖÐʱÒýÆðƽºâÒÆ¶¯µÄÌõ¼þ¿ÉÄÜ

ÊÇ        £¬ÆäÖбíʾƽºâ»ìºÏÎïÖеĺ¬Á¿×î

¸ßµÄÒ»¶Îʱ¼äÊÇ       ¡£

¢ÚζÈΪT¡æÊ±£¬½«2a molºÍa mol·ÅÈë0£®5 L  ÃܱÕÈÝÆ÷ÖУ¬³ä·Ö·´Ó¦ºó²âµÃµÄת»¯ÂÊΪ50£¥¡£Ôò¸Ã·´Ó¦µÄƽºâ³£ÊýΪ           ¡£

 

£¨1£©µÚÈýÖÜÆÚ£¬µÚVIIA×å   £¨2£© 2Na+2H2O=2Na++2OH-+H2¡ü

£¨3£© Cl2+Na2S=2NaCl+S¡ý    £¨4£©²»Í¬    £¨5£©¼Óѹ  t2-t3  4/a2

½âÎö:ÂÔ

 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø