ÌâÄ¿ÄÚÈÝ
£¨6·Ö£©¡°æÏ¶ðÈýºÅ¡±ÎÀÐÇÓÚ2013Äê12ÔÂ2ÈÕ1ʱ30·Ö57ÃëЯ´øÔÂÇò³µ¡°ÓñÍúš±ÔÚÎ÷²ýÎÀÐÇ·¢ÉäÖÐÐÄ·¢ÉäÉý¿Õ£¬²¢ÇÒ»ñµÃÁËÔ²Âú³É¹¦¡£»ð¼ýÍÆ½øÆ÷ÖÐ×°ÓÐÇ¿»¹Ô¼ÁҺ̬ëÂ(N2H4)ºÍÇ¿Ñõ»¯¼ÁҺ̬H2O2£¬µ±ËüÃÇ»ìºÏ·´Ó¦Ê±£¬²úÉú´óÁ¿µªÆøºÍË®ÕôÆø£¬²¢·Å³ö´óÁ¿µÄÈÈ£®ÒÑÖª0.4 molҺ̬ëÂÓë×ãÁ¿µÄҺ̬H2O2·´Ó¦Éú³ÉµªÆøºÍË®ÕôÆø£¬·Å³ö256.652 kJµÄÈÈÁ¿¡£
£¨1£©·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪ_________________________________________________¡£
£¨2£©ÓÖÖªH2O(l)===H2O(g£©¦¤H£½£«44 kJ¡¤mol£1£¬Ôò16 gҺ̬ëÂÓë×ãÁ¿ÒºÌ¬¹ýÑõ»¯Çâ·´Ó¦Éú³ÉҺ̬ˮʱ·Å³öµÄÈÈÁ¿ÊÇ_______ kJ¡£
£¨3£©´Ë·´Ó¦ÓÃÓÚ»ð¼ýÍÆ½ø£¬³ýÊͷųö´óÁ¿µÄÈȺͿìËÙ²úÉú´óÁ¿ÆøÌåÍ⣬»¹ÓÐÒ»¸öºÜ´óµÄÓŵãÊÇ________________¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿