¸ù¾ÝÒªÇóд³öÏÂÁз´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ
(1)Ò»¶¨Á¿ÇâÆøÓëÂÈÆø·´Ó¦Éú³ÉÂÈ»¯ÇâÆøÌå,µ±Éú³É1molÇâÂȼüʱ·Å³ö91.5kJµÄÈÈÁ¿________________________________________________________.
(2)ij»¯Ñ§·´Ó¦µÄÄÜÁ¿±ä»¯ÈçͼËùʾ,¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽÊÇ(·´Ó¦ÈÈÓÃabc±íʾ)
_____________________________________________.
(3)ij·´Ó¦µÄƽºâ³£ÊýÈç¹ûÓÐ1molN2 ÍêÈ«·´Ó¦,ÒªÎüÊÕÈÈÁ¿68kJ.д³ö¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ______________________________________________.
(4)ʵÑéÖв»ÄÜÖ±½Ó²â³öʯīºÍÇâÆøÉú³É¼×Íé·´Ó¦µÄ·´Ó¦ÈÈ,µ«¿É²â³ö¼×Í顢ʯī¡¢ÇâÆøȼÉյķ´Ó¦ÈÈ:
CH4(g)+2O2(g)=CO2(g)+2H2O(l)£»¡÷H1=-890.3kJ/mol
C(ʯī,s)+O2(g)=CO2(g)£»¦¤H2=-393.5 kJ/mol
H2(g)+1/2O2(g)=H2O(l)£»¡÷H3=-285.8kJ/moL
ÔòÓÉʯīÓëÇâÆø·´Ó¦Éú³É¼×ÍéµÄÈÈ»¯Ñ§·´Ó¦·½³ÌʽΪ__________________________________________.
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
ÎïÖÊÀà±ð | Ëá | ¼î | ÑÎ | Ñõ»¯Îï | Ç⻯Îï |
»¯Ñ§Ê½ | ¢ÙHCl ¢Ú H2SO4»òHNO3 H2SO4»òHNO3 |
¢Û NaOH»òKOH NaOH»òKOH ¢ÜBa£¨OH£©2 |
¢ÝNa2CO3 ¢Þ NaNO3»òKNO3»òK2SO4»òNa2SO4 NaNO3»òKNO3»òK2SO4»òNa2SO4 |
¢ßCO2 ¢àNa2O |
¢áNH3 ¢âH2O |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2012½ì±±¾©Êи߶þÏÂÆÚ3Ô·ÝÔ¿¼»¯Ñ§ÊÔ¾í ÌâÐÍ£ºÌî¿ÕÌâ
¸ù¾ÝÒªÇóд³öÏÂÁз´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ
(1)Ò»¶¨Á¿ÇâÆøÓëÂÈÆø·´Ó¦Éú³ÉÂÈ»¯ÇâÆøÌå,µ±Éú³É1molÇâÂȼüʱ·Å³ö91.5kJµÄÈÈÁ¿________________________________________________________.
(2)ij»¯Ñ§·´Ó¦µÄÄÜÁ¿±ä»¯ÈçͼËùʾ,¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽÊÇ(·´Ó¦ÈÈÓÃabc±íʾ)
_____________________________________________.
(3)ij·´Ó¦µÄƽºâ³£ÊýÈç¹ûÓÐ1molN2 ÍêÈ«·´Ó¦,ÒªÎüÊÕÈÈÁ¿68kJ.д³ö¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ______________________________________________.
(4)ʵÑéÖв»ÄÜÖ±½Ó²â³öʯīºÍÇâÆøÉú³É¼×Íé·´Ó¦µÄ·´Ó¦ÈÈ,µ«¿É²â³ö¼×Í顢ʯī¡¢ÇâÆøȼÉյķ´Ó¦ÈÈ:
CH4(g)+2O2(g)=CO2(g)+2H2O(l)£»¡÷H1=-890.3kJ/mol
C(ʯī,s)+O2(g)=CO2(g) £»¦¤H2=-393.5 kJ/mol
H2(g)+1/2O2(g)=H2O(l) £»¡÷H3=-285.8 kJ/moL
ÔòÓÉʯīÓëÇâÆø·´Ó¦Éú³É¼×ÍéµÄÈÈ»¯Ñ§·´Ó¦·½³ÌʽΪ__________________________________________.
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2012½ìÕã½Ê¡´ÈϪÊиßÈýÉÏѧÆÚÆÚÖп¼ÊÔ»¯Ñ§ÊÔ¾í ÌâÐÍ£ºÌî¿ÕÌâ
£¨12·Ö£©¸ù¾ÝÒªÇóÌî¿Õ£º
£¨1£©¸ù¾ÝÒªÇóд³öÏÂÁз´Ó¦µÄ»¯Ñ§·½³Ìʽ£º
¢ÙÓÐË®²Î¼ÓµÄÖû»·´Ó¦£¬³Ð×÷Ñõ»¯¼Á
¢ÚÓÐË®²Î¼ÓµÄÑõ»¯»¹Ô·´Ó¦£¬Ë®¼ÈÊÇÑõ»¯¼ÁÓÖÊÇ»¹Ô¼Á ¡£
¢ÛÓÐË®²Î¼ÓµÄÑõ»¯»¹Ô·´Ó¦£®µ«Ë®¼È²»ÊÇÑõ»¯¼ÁÓÖ²»ÊÇ»¹Ô¼Á ¡£
¢ÜÓÐË®²Î¼ÓµÄ¸´·Ö½â·´Ó¦ ¡£
£¨2£©¡°ÂÌÉ«ÊÔ¼Á¡±Ë«ÑõË®¿É×÷Ϊ¿óÒµ·ÏÔüÏû¶¾¼Á£¬Ïû³ý²É¿óÒµËÒÒºÖеÄÇ軯ÎÈçKCN£©µÄ»¯Ñ§·½³ÌʽΪ£ºKCN+H2O2+H2O=A+NH3¡ü
¢ÙÉú³ÉÎïAµÄ»¯Ñ§Ê½Îª ¡£
¢ÚÔÚ±ê×¼×´¿öÏÂÓÐ0.448L°±ÆøÉú³É£®ÔòתÒƵĵç×ÓÊýΪ ¡£
¢Û·´Ó¦Öб»Ñõ»¯µÄÔªËØΪ ¡£
£¨3£©Í¬ÎÂͬѹÏ£¬Í¬Ìå»ýµÄNH3ºÍH2SµÄÖÊÁ¿±ÈÊÇ £»Í¬ÖÊÁ¿µÄNH3ºÍH2SµÄÌå»ý±ÈÊÇ £»ÈôÁ½ÕßËù¼ªÇâÔÓÚ¸öÊýÏàµÈ£¬NH3ºÍH2SµÄÎïÖʵÄÁ¿±ÈÊÇ ¡£
£¨4£©½«±ê×¼×´¿öϵÄNH3£¨g£©VLÈÜÓÚË®ÖУ¬µÃµ½ÃܶÈΪb g¡¤cm-3µÄ°±Ë®a g£¬´ËʱÎïÖʵÄÁ¿Å¨¶ÈΪc mol¡¤L-1£¬ÔòÈÜÓÚË®ÖеÄNH3£¨g£©µÄÌå»ýVÊÇ L£¨ÓÃa¡¢b¡¢cµÈ±íʾ£©¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2010-2011ѧÄê±±¾©ÊÐÃÜÔƶþÖи߶þÏÂÆÚ3Ô·ÝÔ¿¼»¯Ñ§ÊÔ¾í ÌâÐÍ£ºÌî¿ÕÌâ
¸ù¾ÝÒªÇóд³öÏÂÁз´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ
(1)Ò»¶¨Á¿ÇâÆøÓëÂÈÆø·´Ó¦Éú³ÉÂÈ»¯ÇâÆøÌå,µ±Éú³É1molÇâÂȼüʱ·Å³ö91.5kJµÄÈÈÁ¿________________________________________________________.
(2)ij»¯Ñ§·´Ó¦µÄÄÜÁ¿±ä»¯ÈçͼËùʾ,¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽÊÇ(·´Ó¦ÈÈÓÃabc±íʾ)
_____________________________________________.
(3)ij·´Ó¦µÄƽºâ³£ÊýÈç¹ûÓÐ1molN2ÍêÈ«·´Ó¦,ÒªÎüÊÕÈÈÁ¿68kJ.д³ö¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ______________________________________________.
(4)ʵÑéÖв»ÄÜÖ±½Ó²â³öʯīºÍÇâÆøÉú³É¼×Íé·´Ó¦µÄ·´Ó¦ÈÈ,µ«¿É²â³ö¼×Í顢ʯī¡¢ÇâÆøȼÉյķ´Ó¦ÈÈ:
CH4(g)+2O2(g)=CO2(g)+2H2O(l)£»¡÷H1=-890.3kJ/mol
C(ʯī,s)+O2(g)=CO2(g) £»¦¤H2="-393.5" kJ/mol
H2(g)+1/2O2(g)=H2O(l) £»¡÷H3="-285.8" kJ/moL
ÔòÓÉʯīÓëÇâÆø·´Ó¦Éú³É¼×ÍéµÄÈÈ»¯Ñ§·´Ó¦·½³ÌʽΪ__________________________________________.
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