ÌâÄ¿ÄÚÈÝ

(1)»ð¼ýÍƽøÆ÷ÖÐÊ¢ÓÐÇ¿»¹Ô­¼ÁҺ̬ëÂ(N2H4)ºÍÇ¿Ñõ»¯¼ÁҺ̬˫ÑõË®£¬µ±ËüÃÇ»ìºÏ·´Ó¦Ê±£¬¼´²úÉú´óÁ¿µªÆøºÍË®ÕôÆø£¬²¢·Å³ö´óÁ¿ÈÈ£®ÒÑÖª0.4 molҺ̬ëÂÓë×ãÁ¿ÒºÌ¬Ë«ÑõË®·´Ó¦£¬Éú³ÉµªÆøºÍË®ÕôÆø£¬·Å³ö256 kJµÄÈÈÁ¿£®

¢Ùд³ö¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ________£®

¢ÚÓÖÒÑÖªH2O(l)£½H2O(g)£¬¦¤H£½44 kJ¡¤mol£­1£¬Ôò16 gҺ̬ëÂÓëҺ̬˫ÑõË®·´Ó¦Éú³ÉҺ̬ˮʱ·Å³öµÄÈÈÁ¿ÊÇ________kJ£®

(2)ʵÑéÊÒÓÃ×ãÁ¿µÄ½ðÊôпÓëÌå»ý¡¢Å¨¶ÈÒ»¶¨µÄÏ¡ÁòËá·´Ó¦ÖÆÈ¡ÇâÆøʱ(²»¿¼ÂÇζȱ仯)£¬ÒªÏëʹ·´Ó¦ËÙÂʼõÂý£¬ÓÖ²»Ó°Ïì²úÉúÇâÆøµÄ×ÜÁ¿£¬ÏÂÁдëÊ©ÖÐÄܴﵽĿµÄµÄÊÇ________£®

¢Ù¼ÓÈëÉÙÁ¿¹ÌÌå̼ËáÄÆ£»

¢Ú¼ÓÈëÉÙÁ¿¹ÌÌå´×ËáÄÆ£»

¢Û¼ÓÈëÊÊÁ¿µÄÁòËáÄÆÈÜÒº£»

¢Ü¼ÓÈëÊÊÁ¿µÄÏõËáÄÆÈÜÒº£»

¢ÝµÎ¼ÓÉÙÁ¿CuSO4ÈÜÒº£»

¢Þ¼ÓÈÈ

´ð°¸£º
½âÎö£º

¡¡¡¡(1)¢ÙN2H4(l)£«2H2O2(l)£½N2(g)£«4H2O(g)¡¡¦¤H£½£­640 kJ¡¤mol£­1¡¡(2·Ö)

¡¡¡¡¢Ú408¡¡(2·Ö)

¡¡¡¡(2)¢Ú¢Û(3·Ö)


Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
»ð¼ýÍƽøÆ÷ÖÐÊ¢ÓÐÇ¿»¹Ô­¼ÁҺ̬루N2H4£©ºÍÇ¿Ñõ»¯¼ÁҺ̬˫ÑõË®£®µ±ËüÃÇ»ìºÏ·´Ó¦Ê±£¬¼´²úÉú´óÁ¿µªÆøºÍË®ÕôÆø£¬²¢·Å³ö´óÁ¿µÄÈÈ£®ÒÑÖª0.4molҺ̬ëÂÓë×ãÁ¿µÄҺ̬˫ÑõË®·´Ó¦£¬Éú³ÉµªÆøºÍË®ÕôÆø£¬·Å³öQKJµÄÈÈÁ¿£®
£¨1£©·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪ
N2H4£¨1£©+2H2O2£¨1£©=N2£¨g£©+4H2O£¨g£©¡÷H=-2.5QkJ?mol-1
N2H4£¨1£©+2H2O2£¨1£©=N2£¨g£©+4H2O£¨g£©¡÷H=-2.5QkJ?mol-1

£¨2£©´Ë·´Ó¦ÓÃÓÚ»ð¼ýÍƽø£¬³ýÊÍ·Å´óÁ¿ÈȺͿìËÙ²úÉú´óÁ¿ÆøÌåÍ⣬»¹ÓÐÒ»¸öºÜ´óµÄÓŵãÊÇ
Éú³ÉÎïΪµªÆøºÍË®£¬²»ÎÛȾ¿ÕÆø
Éú³ÉÎïΪµªÆøºÍË®£¬²»ÎÛȾ¿ÕÆø

£¨3£©ÔÚÎ÷²¿´ó¿ª·¢µÄÊ®´ó¹¤ºÍÖУ¬¡±Î÷Æø¶«ÊäÊÇÆäÖÐÖ®Ò»£®¡°Î÷Æø¶«Ê䡱Öеġ±Æø¡±ÊÇÖ¸
C
C

A£®Ë®ÃºÆø    B£®ÁѽâÆø    C£®ÌìÈ»Æø    D£®µçʯÆø
£¨4£©2001Äê2ÔÂ24ÈÕ£¬ÖÐÑëµçÊǪ́±¨µÀ£º½üÄêÀ´±±¾©µÄ¿ÕÆøÖÊÁ¿Ìá¸ß£¬ÒÑʯ¼Òׯ¢ò¼¶±ê×¼£®ÕâÖ÷ÒªµÃÁ¦ÓÚ°áǨһ´ËÆø³¬±êÅŷŵŤ³§ºÍ´óÁ¿Ê¹ÓÃȼÆøȼÁÏ£¬¼õÉÙÁËSO2ºÍ·Û³¾µÄÅÅ·Å£¬ÌرðÊÇʹÓÃÁËÐí¶àµÄ»·±£ÐÍÆû³µ£¬ÅÅ·ÅÎÛȾֵ´ó´ó½µµÍ£®ÒÑÖª1g¼×ÍéÍêȫȼÉÕÉú³ÉÆø̬ˮʱ·Å³öÈÈÁ¿50.125KJ£®ÊÔд³ö¼×ÍéȼÉÕµÄÈÈ»¯Ñ§·½³Ìʽ
CH4£¨g£©+2O2£¨g£©¨TCO2£¨g£©+2H2O£¨l£©¡÷H=-802KJ/mol
CH4£¨g£©+2O2£¨g£©¨TCO2£¨g£©+2H2O£¨l£©¡÷H=-802KJ/mol
£®

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

¾«Ó¢¼Ò½ÌÍø