ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿£¨1£©¢Ùд³ö2-¼×»ù¶¡ÍéµÄ½á¹¹¼òʽ£º_____ ¢Úд³öСËÕ´òµÄ»¯Ñ§Ê½£º_____

£¨2£©Ð´³öÍ­ÓëÂÈ»¯ÌúÈÜÒº·´Ó¦µÄÀë×Ó·½³Ìʽ£º_______________________________

£¨3£©ÌúƬ¡¢Í­Æ¬Á´½Óºó½þÈëÏ¡ÁòËáÐγÉÔ­µç³Ø£¬Õý¼«·´Ó¦Ê½ÊÇ______________

£¨4£©Ð´³ö±½ºÍÒºäåÔÚ´ß»¯¼Á×÷ÓÃÏÂÉú³Éäå±½µÄ»¯Ñ§·½³Ìʽ____________________

£¨5£©ÒÑÖª0.25mol N2H4(g)ÍêȫȼÉÕÉú³ÉµªÆøºÍÆø̬ˮʱ·Å³ö133.5KJÈÈÁ¿¡£Ð´³ö¸Ã·´Ó¦µÄȼÉÕÈÈ»¯Ñ§·½³Ìʽ____________

¡¾´ð°¸¡¿CH3CH(CH3)CH2CH3 NaHCO3 Cu+2Fe3+=Cu2++2Fe2+ 2H£«£«2e£­=H2¡ü +Br2 + HBr N2H4(g)+O2(g)=N2(g)+2H2O(g) ¡÷H=-534kJ¡¤mol£­1

¡¾½âÎö¡¿

£¨1£©2-¼×»ù¶¡ÍéµÄ½á¹¹¼òʽΪCH3CH(CH3)CH2CH3£¬Ð¡ËÕ´òµÄ»¯Ñ§Ê½ÎªNaHCO3£»

£¨2£©Í­ÓëÂÈ»¯ÌúÈÜÒº·´Ó¦µÄÀë×Ó·½³ÌʽΪCu+2Fe3+=Cu2++2Fe2+£»

£¨3£©ÌúƬ¡¢Í­Æ¬Á´½Óºó½þÈëÏ¡ÁòËáÐγÉÔ­µç³Ø£¬ÇâÀë×ÓÔÚÕý¼«·¢Éú·´Ó¦£¬µç¼«·´Ó¦Ê½Îª2H£«£«2e£­=H2¡ü£»

£¨4£©±½ºÍÒºäåÔÚ´ß»¯¼Á×÷ÓÃÏÂÉú³Éäå±½µÄ»¯Ñ§·½³ÌʽΪ+Br2 + HBr£»

£¨5£©0.25mol N2H4(g)ÍêȫȼÉÕÉú³ÉµªÆøºÍÆø̬ˮʱ·Å³ö133.5KJÈÈÁ¿£¬Ôò1mol N2H4(g)ÍêȫȼÉÕÉú³ÉµªÆøºÍÆø̬ˮʱ·Å³ö534KJÈÈÁ¿£¬ÈÈ»¯Ñ§·½³ÌʽΪN2H4(g)+O2(g)=N2(g)+2H2O(g) ¡÷H=-534kJ¡¤mol£­1¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø