ÌâÄ¿ÄÚÈÝ

ÒÑÖªH2(g)£«1/2O2(g)===H2O(g) ¦¤H£½£­241.8 kJ¡¤mol£­1ÏÂÁÐ˵·¨Öв»ÕýÈ·µÄÊÇ( )

A£®H2µÄȼÉÕÈÈΪ241.8 kJ¡¤mol£­1
B£®2H2(g)£«O2(g)===2H2O(g)¦¤H£½£­483.6 kJ¡¤mol£­1
C£®1 mol H2ÍêȫȼÉÕÉú³ÉҺ̬ˮ·Å³öµÄÈÈÁ¿´óÓÚ241.8 kJ
D£®¶ÏÁÑ1 mol H2OµÄ»¯Ñ§¼üÎüÊÕµÄ×ÜÄÜÁ¿´óÓÚ¶ÏÁÑ1 mol H2ºÍ0.5 mol O2µÄ»¯Ñ§¼üËùÎüÊÕµÄ×ÜÄÜÁ¿

A

½âÎöÊÔÌâ·ÖÎö£ºÈ¼ÉÕÈÈÊÇÔÚÒ»¶¨Ìõ¼þÏ£¬1mol¿ÉȼÎïÍêȫȼÉÕÉú³ÉÎȶ¨µÄÑõ»¯ÎïʱËù·Å³öµÄÈÈÁ¿£¬ËùÒÔA²»ÕýÈ·£¬Ë®µÄÎȶ¨×´Ì¬ÊÇҺ̬£»·´Ó¦ÈȺͻ¯Ñ§¼ÆÁ¿Êý³ÉÕý±È£¬BÕýÈ·£»Æø̬ˮµÄÄÜÁ¿¸ßÓÚҺ̬ˮµÄÄÜÁ¿£¬ËùÒÔ1 mol H2ÍêȫȼÉÕÉú³ÉҺ̬ˮ·Å³öµÄÈÈÁ¿´óÓÚ241.8 kJ£¬CÕýÈ·£»·´Ó¦ÈÈ»¹µÈÓڶϼüÎüÊÕµÄÄÜÁ¿ºÍÐγɻ¯Ñ§¼üËù·Å³öµÄÄÜÁ¿µÄ²îÖµ£¬DÕýÈ·£¬´ð°¸Ñ¡A¡£
¿¼µã£»¿¼²é·´Ó¦ÈȵÄÓйØÅжÏ
µãÆÀ£º´Óºê¹ÛµÄ½Ç¶È·ÖÎö£¬·´Ó¦ÈȾÍÊÇ·´Ó¦ÎïºÍÉú³ÉÎï×ÜÄÜÁ¿µÄÏà¶Ô´óС£¬¶ø´Ó΢¹ÛµÄ½Ç¶È·ÖÎö£¬·´Ó¦ÈȾÍÊǶϼüÎüÊÕµÄÄÜÁ¿ºÍÐγɻ¯Ñ§¼üËù·Å³öµÄÄÜÁ¿µÄ²îÖµ¡£ÁíÍ⻹ÐèҪעÒâµÄÊÇ·´Ó¦ÈÈ»¹ÓëÎïÖʵÄ״̬ÓйØϵ£¬ÔÚÅжÏʱÐèҪעÒâ¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø