£¨10·Ö£©¡°Î÷Æø¶«Ê䡱ÊÇÎ÷²¿¿ª·¢µÄÖص㹤³Ì£¬ÕâÀïµÄÆøÌåÊÇÖ¸ÌìÈ»Æø£¬ÆäÖ÷Òª³É·ÖÊǼ×Íé¡£¹¤ÒµÉϽ«Ì¼ÓëË®ÔÚ¸ßÎÂÏ·´Ó¦ÖƵÃˮúÆø£¬Ë®ÃºÆøµÄÖ÷Òª³É·ÖÊÇCOºÍH2£¬Á½ÕßµÄÌå»ý±ÈԼΪ1:1¡£ÒÑÖª1mol COÆøÌåÍêȫȼÉÕÉú³ÉCO2ÆøÌå·Å³ö283kJÈÈÁ¿£¬1molH2ÍêȫȼÉÕÉú³ÉҺ̬ˮ·Å³ö286kJÈÈÁ¿£»1molCH4 ÆøÌåÍêȫȼÉÕÉú³ÉCO2ÆøÌåºÍҺ̬ˮ·Å³ö890kJÈÈÁ¿¡£
£¨1£©Ð´³öH2ÍêȫȼÉÕÉú³ÉҺ̬ˮµÄÈÈ»¯Ñ§·´Ó¦·½³Ìʽ£º_________________________£¬Èô1molCH4ÆøÌåÍêȫȼÉÕÉú³ÉCO2ÆøÌåºÍË®ÕôÆø£¬·Å³öµÄÈÈÁ¿________890kJ£¨Ìî¡°£¾¡±¡¢¡°£½¡±»ò¡°£¼¡±£©
£¨2£©ºöÂÔˮúÆøÖÐÆäËü³É·Ö£¬Ïàͬ״¿öÏÂÈôµÃµ½ÏàµÈµÄÈÈÁ¿£¬ËùÐèˮúÆøÓë¼×ÍéµÄÌå»ý±ÈԼΪ________£»È¼ÉÕÉú³ÉµÄCO2 µÄÖÊÁ¿±ÈԼΪ________________¡£
£¨3£©ÒÔÉÏÊý¾ÝºÍ¼ÆËã˵Ã÷£¬ÒÔÌìÈ»Æø´úÌæˮúÆø×÷ÃñÓÃȼÁÏ£¬Í»³öµÄÓŵãÊÇ________________¡£
£¨1£©2H2(g)+O2(g) == 2H2O(l)£»¡÷H=£572kJ•mol-1 <
£¨2£©25£º8 3£º1 £¨3£©È¼ÉÕÖµ¸ß£¬¼õÉÙCO2µÄÅÅ·ÅÁ¿£¬ÓÐÀûÓÚ±£»¤»·¾³
¡¾½âÎö¡¿£¨1£©1molH2ÍêȫȼÉÕÉú³ÉҺ̬ˮ·Å³ö286kJÈÈÁ¿£¬ËùÒÔH2ÍêȫȼÉÕÉú³ÉҺ̬ˮµÄÈÈ»¯Ñ§·´Ó¦·½³Ìʽ2H2(g)+O2(g) == 2H2O(l)£»¡÷H=£572kJ•mol-1¡£ÓÉÓÚË®ÕôÆøµÄÄÜÁ¿¸ßÓÚҺ̬ˮµÄÄÜÁ¿£¬ËùÒÔ1molCH4ÆøÌåÍêȫȼÉÕÉú³ÉCO2ÆøÌåºÍË®ÕôÆø£¬·Å³öµÄÈÈÁ¿Ð¡ÓÚ890kJ¡£
£¨2£©2molˮúÆøÍêȫȼÉշųöµÄÈÈÁ¿ÊÇ283kJ£«286kJ£½569kJ£¬Òª·Å³ö569kJµÄÄÜÁ¿ÐèÒª¼×ÍéÊÇ569¡Â890£½0.64mol£¬ÆäÌå»ý±ÈÔ½ÊÇ25©U8¡£0.64mol¼×ÍéÖÊÁ¿ÊÇ0.64mol¡Á16g/mol£½10.23g£¬ËùÒÔˮúÆøÓë¼×ÍéµÄÖÊÁ¿Ô¼Îª£¨28£«2£©¡Â10.23£½3©U1.
£¨3£©¸ù¾Ý£¨2£©¿ÉÖª·Å³öÏàͬµÄÈÈÁ¿£¬ÏûºÄµÄ¼×ÍéÖÊÁ¿ÉÙ£¬ËùÒÔȼÉÕÖµ¸ß£¬ÇÒÄܼõÉÙCO2µÄÅÅ·ÅÁ¿£¬ÓÐÀûÓÚ±£»¤»·¾³¡£
Ä꼶 | ¸ßÖÐ¿Î³Ì | Ä꼶 | ³õÖÐ¿Î³Ì |
¸ßÒ» | ¸ßÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ | ³õÒ» | ³õÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ |
¸ß¶þ | ¸ß¶þÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õ¶þ | ³õ¶þÃâ·Ñ¿Î³ÌÍƼö£¡ |
¸ßÈý | ¸ßÈýÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õÈý | ³õÈýÃâ·Ñ¿Î³ÌÍƼö£¡ |
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡°Î÷Æø¶«Ê䡱ÊÇÎ÷²¿¿ª·¢µÄÖص㹤³Ì£¬ÕâÀïµÄÆøÊÇÖ¸ÌìÈ»Æø£¬ÆäÖ÷Òª³É·ÖÊǼ×Íé¡£¹¤ÒµÉϽ«Ì¼ÓëË®ÔÚ¸ßÎÂÏ·´Ó¦ÖƵÃˮúÆø£¬Ë®ÃºÆøµÄÖ÷Òª³É·ÖÊÇCOºÍH2£¬Á½ÕßµÄÌå»ý±ÈԼΪ1¡Ã1¡£ÒÑÖª1 mol COÆøÌåÍêȫȼÉÕÉú³ÉCO2ÆøÌå·Å³ö283 kJÈÈÁ¿£»1molH2ÍêȫȼÉÕÉú³ÉҺ̬ˮ·Å³ö286 kJÈÈÁ¿£»1molCH4ÆøÌåÍêȫȼÉÕÉú³ÉCO2ÆøÌåºÍҺ̬ˮ·Å³ö890kJÈÈÁ¿¡£
(1)д³öH2ÍêȫȼÉÕÉú³ÉҺ̬ˮµÄÈÈ»¯Ñ§·´Ó¦·½³Ìʽ£º__________________£¬Èô1molCH4ÆøÌåÍêȫȼÉÕÉú³ÉCO2ÆøÌåºÍË®ÕôÆø£¬·Å³öµÄÈÈÁ¿____________890kJ(Ìî¡°£¾ ¡±¡°=¡±»ò¡°£¼¡±)¡£
(2)ºöÂÔˮúÆøÖÐÆäËû³É·Ö£¬Ïàͬ״¿öÏÂÈôµÃµ½ÏàµÈµÄÈÈÁ¿£¬ËùÐèˮúÆøÓë¼×ÍéµÄÌå»ý±ÈԼΪ____________£»È¼ÉÕÉú³ÉµÄCO2µÄÖÊÁ¿±ÈԼΪ____________¡£
(3)ÒÔÉÏÊý¾ÝºÍ¼ÆËã˵Ã÷£¬ÒÔÌìÈ»Æø´úÌæˮúÆø×÷ÃñÓÃȼÁÏ£¬Í»³öµÄÓŵãÊÇ________________________________________________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
£¨9·Ö£©¡°Î÷Æø¶«Ê䡱ÊÇÎ÷²¿¿ª·¢µÄÖص㹤³Ì£¬ÕâÀïµÄÆøÊÇÖ¸ÌìÈ»Æø£¬ÆäÖ÷Òª³É·ÖÊǼ×Íé¡£¹¤ÒµÉϽ«Ì¼ÓëË®ÔÚ¸ßÎÂÏ·´Ó¦ÖƵÃˮúÆø£¬Ë®ÃºÆøµÄÖ÷Òª³É·ÖÊÇCOºÍH2£¬Á½ÕßµÄÌå»ý±ÈԼΪ1¡Ã1¡£ÒÑÖª1 mol COÆøÌåÍêȫȼÉÕÉú³ÉCO2ÆøÌå·Å³ö283 kJÈÈÁ¿£»1 molÇâÆøÍêȫȼÉÕÉú³ÉҺ̬ˮ·Å³ö286 kJÈÈÁ¿£»1 mol CH4ÆøÌåÍêȫȼÉÕÉú³ÉCO2ÆøÌåºÍҺ̬ˮ·Å³ö890 kJÈÈÁ¿¡£
£¨1£©Ð´³öÇâÆøÍêȫȼÉÕÉú³ÉҺ̬ˮµÄÈÈ»¯Ñ§·´Ó¦·½³Ìʽ£º_____________________ £¬
Èô1 mol CH4ÆøÌåÍêȫȼÉÕÉú³ÉCO2ÆøÌåºÍË®ÕôÆø£¬·Å³öµÄÈÈÁ¿_______890 kJ£¨Ìî¡°>¡±¡¢¡°=¡±»ò¡°<¡±)¡£
£¨2£©ºöÂÔˮúÆøÖÐÆäËû³É·Ö£¬Ïàͬ״¿öÏÂÈôµÃµ½ÏàµÈµÄÈÈÁ¿£¬ËùÐèˮúÆøÓë¼×ÍéµÄÌå»ý±ÈԼΪ_______£»È¼ÉÕÉú³ÉµÄCO2µÄÖÊÁ¿±ÈԼΪ_______¡£
£¨3£©ÒÔÉÏÊý¾ÝºÍ¼ÆËã˵Ã÷£¬ÒÔÌìÈ»Æø´úÌæˮúÆø×÷ÃñÓÃȼÁÏ£¬Í»³öµÄÓŵãÊÇ___________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2010-2011Õã½Ê¡¸ßÒ»ÏÂѧÆÚÆÚÖп¼ÊÔ»¯Ñ§ÊÔ¾í ÌâÐÍ£ºÌî¿ÕÌâ
£¨9·Ö£©¡°Î÷Æø¶«Ê䡱ÊÇÎ÷²¿¿ª·¢µÄÖص㹤³Ì£¬ÕâÀïµÄÆøÊÇÖ¸ÌìÈ»Æø£¬ÆäÖ÷Òª³É·ÖÊǼ×Íé¡£¹¤ÒµÉϽ«Ì¼ÓëË®ÔÚ¸ßÎÂÏ·´Ó¦ÖƵÃˮúÆø£¬Ë®ÃºÆøµÄÖ÷Òª³É·ÖÊÇCOºÍH2£¬Á½ÕßµÄÌå»ý±ÈԼΪ1¡Ã1¡£ÒÑÖª1 mol COÆøÌåÍêȫȼÉÕÉú³ÉCO2ÆøÌå·Å³ö283 kJÈÈÁ¿£»1 molÇâÆøÍêȫȼÉÕÉú³ÉҺ̬ˮ·Å³ö286 kJÈÈÁ¿£»1 mol CH4ÆøÌåÍêȫȼÉÕÉú³ÉCO2ÆøÌåºÍҺ̬ˮ·Å³ö890 kJÈÈÁ¿¡£
£¨1£©Ð´³öÇâÆøÍêȫȼÉÕÉú³ÉҺ̬ˮµÄÈÈ»¯Ñ§·´Ó¦·½³Ìʽ£º_____________________ £¬
Èô1 mol CH4ÆøÌåÍêȫȼÉÕÉú³ÉCO2ÆøÌåºÍË®ÕôÆø£¬·Å³öµÄÈÈÁ¿_______890 kJ£¨Ìî¡°>¡±¡¢¡°=¡±»ò¡°<¡±)¡£
£¨2£©ºöÂÔˮúÆøÖÐÆäËû³É·Ö£¬Ïàͬ״¿öÏÂÈôµÃµ½ÏàµÈµÄÈÈÁ¿£¬ËùÐèˮúÆøÓë¼×ÍéµÄÌå»ý±ÈԼΪ_______£»È¼ÉÕÉú³ÉµÄCO2µÄÖÊÁ¿±ÈԼΪ_______¡£
£¨3£©ÒÔÉÏÊý¾ÝºÍ¼ÆËã˵Ã÷£¬ÒÔÌìÈ»Æø´úÌæˮúÆø×÷ÃñÓÃȼÁÏ£¬Í»³öµÄÓŵãÊÇ___________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2012½ìÔÆÄÏÊ¡¸ß¶þÏÂѧÆÚÆÚÖп¼ÊÔ»¯Ñ§ÊÔ¾í ÌâÐÍ£ºÌî¿ÕÌâ
£¨6·Ö£©±ûÍéȼÉÕ¿ÉÒÔͨ¹ýÒÔÏÂÁ½ÖÖ;¾¶£º
;¾¶I£ºC3H8(g) + 5O2(g) == 3CO2(g) +4H2O(l) ¦¤H£½-a kJ¡¤mol-1
;¾¶II£ºC3H8(g) ==C3H6(g)+ H2(g) ¦¤H£½+b kJ¡¤mol-1
2C3H6(g)+ 9O2(g) == 6CO2(g) +6H2O(l) ¦¤H£½-c kJ¡¤mol-1
2H2(g)+O2 (g) == 2H2O(l) ¦¤H£½-d kJ¡¤mol-1 ¡¡¡¡£¨abcd¾ùΪÕýÖµ£©
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÅжϵÈÁ¿µÄ±ûÍéͨ¹ýÁ½ÖÖ;¾¶·Å³öµÄÈÈÁ¿£¬Í¾¾¶I·Å³öµÄÈÈÁ¿ ______£¨Ìî¡°´óÓÚ¡±¡¢¡°µÈÓÚ¡±»ò¡°Ð¡ÓÚ¡±£©Í¾¾¶II·Å³öµÄÈÈÁ¿
£¨2£©ÓÉÓÚC3H8(g) ==C3H6(g)+ H2(g) µÄ·´Ó¦ÖУ¬·´Ó¦Îï¾ßÓеÄ×ÜÄÜÁ¿______£¨Ìî¡°´óÓÚ¡±¡¢¡°µÈÓÚ¡±»ò¡°Ð¡ÓÚ¡±£©Éú³ÉÎï¾ßÓеÄ×ÜÄÜÁ¿£¬ÄÇôÔÚ»¯Ñ§·´Ó¦Ê±¡£·´Ó¦Îï¾ÍÐèÒª______£¨Ìî¡°·Å³ö¡±¡¢»ò¡°ÎüÊÕ¡±£©ÄÜÁ¿²ÅÄÜת»¯ÎªÉú³ÉÎÒò´ËÆä·´Ó¦Ìõ¼þÊÇ______________
£¨3£©¡°Î÷Æø¶«Ê䡱ÊÇÎ÷²¿¿ª·¢µÄÖص㹤³Ì£¬ÕâÀïµÄÆøÌåÊÇÖ¸ÌìÈ»Æø£¬ÆäÖ÷Òª³É·ÖÊǼ×Í顣ʹÓùܵÀúÆøÓû§¸ÄÓÃÌìÈ»Æø£¬Ó¦µ÷ÕûÔî¾ß½øÆøÁ¿·§ÃÅ£¬¼´Ôö´ó_____£¨Ìî¡°¿ÕÆø¡±»ò¡°ÌìÈ»Æø¡±£©µÄ½øÈëÁ¿¡£¾Ý±¨µÀ£¬Í¬Ê±È¼Æø¼Û¸ñÒ²½«ÔÚÏÖÓеÄ0.95Ôª/m3µÄ»ù´¡Éϵ÷Õûµ½1.31Ôª/m3£¬Çëͨ¹ý¼ÆËã˵Ã÷Èô¾ÓÃñ±£³ÖÉú»îˮƽ²»±ä£¬ÔÚÏûºÄȼÆø·½ÃæµÄÏû·Ñ´óÔ¼ÊÇÏÖÔÚµÄ____±¶
²é¿´´ð°¸ºÍ½âÎö>>
°Ù¶ÈÖÂÐÅ - Á·Ï°²áÁбí - ÊÔÌâÁбí
ºþ±±Ê¡»¥ÁªÍøÎ¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨Æ½Ì¨ | ÍøÉÏÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | µçÐÅթƾٱ¨×¨Çø | ÉæÀúÊ·ÐéÎÞÖ÷ÒåÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | ÉæÆóÇÖȨ¾Ù±¨×¨Çø
Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com