ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿ÄÜÔ´ÊÇÏÖ´úÎÄÃ÷µÄÔ­¶¯Á¦£¬Í¨¹ý»¯Ñ§·½·¨¿ÉÒÔʹÄÜÁ¿°´ÈËÃÇËùÆÚÍûµÄÐÎʽת»¯£¬´Ó¶ø¿ª±ÙÐÂÄÜÔ´ºÍÌá¸ßÄÜÔ´µÄÀûÓÃÂÊ¡£

£¨1£©ÇâÆøÔÚO2ÖÐȼÉյķ´Ó¦ÊÇ_____ÈÈ·´Ó¦£¨Ìî¡°·Å¡±»ò¡°Îü¡±£©£¬ÕâÊÇÓÉÓÚ·´Ó¦ÎïµÄ×ÜÄÜÁ¿_____Éú³ÉÎïµÄ×ÜÄÜÁ¿£¨Ìî¡°´óÓÚ¡±¡¢¡°Ð¡ÓÚ¡±»ò¡°µÈÓÚ¡±£¬ÏÂͬ£©£»

£¨2£©´Ó»¯Ñ§·´Ó¦µÄ±¾ÖʽǶÈÀ´¿´£¬ÊÇÓÉÓÚ¶ÏÁÑ·´Ó¦ÎïÖеĻ¯Ñ§¼üÎüÊÕµÄ×ÜÄÜÁ¿______ÐγɲúÎïµÄ»¯Ñ§¼ü·Å³öµÄ×ÜÄÜÁ¿¡£ÒÑÖªÆÆ»µ1 mol H¨CH¼ü¡¢1mol O=O¼ü¡¢1 mol H¨CO¼üʱ·Ö±ðÐèÒªÎüÊÕ436 kJ¡¢498 kJ¡¢463 kJµÄÄÜÁ¿¡£Ôò2 mol H2(g)ºÍ1mol O2(g)ת»¯Îª2mol H2O(g)ʱ·Å³öµÄÈÈÁ¿Îª_______¡£:

¡¾´ð°¸¡¿£¨1£©·Å£¬´óÓÚ £¨2£©Ð¡ÓÚ¡¢482 kJ

¡¾½âÎö¡¿

ÊÔÌâ·ÖÎö£º£¨1£©ÇâÆøÔÚO2ÖÐȼÉյķ´Ó¦ÊÇ·ÅÈÈ·´Ó¦£¬ÕâÊÇÓÉÓÚ·´Ó¦ÎïµÄ×ÜÄÜÁ¿´óÓÚÉú³ÉÎïµÄ×ÜÄÜÁ¿£»

£¨2£©´Ó»¯Ñ§·´Ó¦µÄ±¾ÖʽǶÈÀ´¿´£¬ÊÇÓÉÓÚ¶ÏÁÑ·´Ó¦ÎïÖеĻ¯Ñ§¼üÎüÊÕµÄ×ÜÄÜÁ¿Ð¡ÓÚÐγɲúÎïµÄ»¯Ñ§¼ü·Å³öµÄ×ÜÄÜÁ¿¡£·´Ó¦ÈÈ£½·´Ó¦ÎïÖÐ×ܼüÄÜÖ®ºÍ£­Éú³ÉÎïÖÐ×ܼüÄÜÖ®ºÍ£¬Ôò2 mol H2(g)ºÍ1mol O2(g)ת»¯Îª2mol H2O(g)ʱ·Å³öµÄÈÈÁ¿2¡Á2¡Á463 kJ£­2¡Á436 kJ£­498 kJ£½482kJ¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø