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
1 |
2 |
1 |
2 |
1 |
2 |
1 |
2 |
13 |
2 |
13 |
2 |
1 |
2 |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
(6·Ö)°´ÒªÇóдÈÈ»¯Ñ§·½³Ìʽ£º
(1)ÒÑ֪ϡÈÜÒºÖУ¬1mol H2SO4ÓëNaOHÈÜҺǡºÃÍêÈ«·´Ó¦Éú³ÉÕýÑÎʱ£¬·Å³ö114.6 kJÈÈÁ¿£¬Ð´³ö±íʾH2SO4ÓëNaOH·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ_____________________¡£
(2)25¡æ¡¢101 kPaÌõ¼þϳä·ÖȼÉÕÒ»¶¨Á¿µÄ¶¡ÍéÆøÌå·Å³öÈÈÁ¿ÎªQ kJ£¬¾²â¶¨£¬½«Éú³ÉµÄCO2ͨÈë×ãÁ¿³ÎÇåʯ»ÒË®ÖвúÉú25 g°×É«³Áµí£¬Ð´³ö±íʾ¶¡ÍéȼÉÕÈȵÄÈÈ»¯Ñ§·½³Ìʽ ¡£
(3)ÒÑÖªÏÂÁÐÈÈ»¯Ñ§·½³Ìʽ£º
¢ÙCH3COOH(l)£«2O2(g)===2CO2(g)£«2H2O(l) ¦¤H1£½£870.3 kJ/mol
¢ÚC(s)£«O2(g)===CO2(g) ¡¡ ¦¤H2£½£393.5 kJ/mol
¢ÛH2(g)£«1/2O2(g)===H2O(l) ¦¤H3£½£285.8 kJ/mol
д³öÓÉC(s)¡¢H2(g)ºÍO2(g)»¯ºÏÉú³ÉCH3COOH(l)µÄÈÈ»¯Ñ§·½³Ìʽ ¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2011-2012ѧÄêɽÎ÷Ê¡´óͬһÖи߶þÉÏѧÆÚÆÚÖп¼ÊÔ»¯Ñ§ÊÔ¾í ÌâÐÍ£ºµ¥Ñ¡Ìâ
(8·Ö)°´ÒªÇóдÈÈ»¯Ñ§·½³Ìʽ£º
ÒÑ֪ϡÈÜÒºÖУ¬1 mol H2SO4ÓëNaOHÈÜÒº
Ç¡ºÃÍêÈ«·´Ó¦Ê±£¬·Å³ö114.6 kJÈÈÁ¿£¬
д³ö±íʾH2SO4ÓëNaOH·´Ó¦µÄÖкÍÈȵÄÈÈ»¯Ñ§·½³Ìʽ
_________________________________________________________________.
(2) 25¡æ¡¢101 kPaÌõ¼þϳä·ÖȼÉÕÒ»¶¨Á¿µÄ¶¡ÍéÆøÌå·Å³öÈÈÁ¿ÎªQ kJ£¬¾²â¶¨£¬½«Éú³ÉµÄCO2ͨÈë×ãÁ¿³ÎÇåʯ»ÒË®ÖвúÉú25 g°×É«³Áµí£¬Ð´³ö±íʾ¶¡ÍéȼÉÕÈȵÄÈÈ»¯Ñ§·½³Ìʽ______________________________________________________£®
(3) ÒÑÖªÏÂÁÐÈÈ»¯Ñ§·½³Ìʽ£º
¢ÙCH3COOH(l)£«2O2(g)===2CO2(g)£«2H2O(l) ¦¤H1£½£870.3 kJ/mol
¢ÚC(s)£«O2(g)===CO2(g)¡¡¦¤H2£½£393.5 kJ/mol
¢ÛH2(g)£«O2(g)===H2O(l)¦¤H3£½£285.8 kJ/mol
д³öÓÉC(s)¡¢H2(g)ºÍO2(g)»¯ºÏÉú³ÉCH3COOH(l)µÄÈÈ»¯Ñ§·½³Ìʽ_________________________________________£®
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2011-2012Äêн®ÎÚ³ľÆëÊеڰËÖÐѧ¸ßÒ»ÏÂѧÆÚÆÚÄ©¿¼ÊÔ»¯Ñ§¾í ÌâÐÍ£ºÌî¿ÕÌâ
(9·Ö)°´ÒªÇóдÈÈ»¯Ñ§·½³Ìʽ£º
(1)ÒÑ֪ϡÈÜÒºÖУ¬1 mol H2SO4ÓëNaOHÈÜҺǡºÃÍêÈ«·´Ó¦Ê±£¬·Å³ö114.6 kJÈÈÁ¿£¬Ð´³ö±íʾH2SO4ÓëNaOH·´Ó¦µÄÖкÍÈÈ»¯Ñ§·½³Ìʽ______________________________________.
(2)25¡æ¡¢101 kPaÌõ¼þϳä·ÖȼÉÕÒ»¶¨Á¿µÄ¶¡ÍéÆøÌå·Å³öÈÈÁ¿ÎªQ kJ£¬¾²â¶¨£¬½«Éú³ÉµÄCO2ͨÈë×ãÁ¿³ÎÇåʯ»ÒË®ÖвúÉú25 g°×É«³Áµí£¬Ð´³ö±íʾ¶¡ÍéȼÉÕÈÈ»¯Ñ§·½³Ìʽ___________________£®
(3)ÒÑÖªÏÂÁÐÈÈ»¯Ñ§·½³Ìʽ£º
¢ÙCH3COOH(l)£«2O2(g)===2CO2(g)£«2H2O(l) ¦¤H1£½£870.3 kJ/mol
¢ÚC(s)£«O2(g)===CO2(g) ¡¡ ¦¤H2£½£393.5 kJ/mol
¢ÛH2(g)£«O2(g)===H2O(l ) ¦¤H3£½£285.8 kJ/mol
д³öÓÉC(s)¡¢H2(g)ºÍO2(g)»¯ºÏÉú³ÉCH3COOH(l)µÄÈÈ»¯Ñ§·½³Ìʽ_______________________£®
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£ºÉ½Î÷Ê¡2010½ì¸ßÈý5ÔÂÄ©¿¼ÊÔÊÔÌâÀí×Û»¯Ñ§ ÌâÐÍ£ºÌî¿ÕÌâ
(6·Ö)°´ÒªÇóдÈÈ»¯Ñ§·½³Ìʽ£º
(1)ÒÑ֪ϡÈÜÒºÖУ¬1 mol H2SO4ÓëNaOHÈÜҺǡºÃÍêÈ«·´Ó¦Éú³ÉÕýÑÎʱ£¬·Å³ö114.6 kJÈÈÁ¿£¬Ð´³ö±íʾH2SO4ÓëNaOH·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ_____________________¡£
(2)25¡æ¡¢101 kPaÌõ¼þϳä·ÖȼÉÕÒ»¶¨Á¿µÄ¶¡ÍéÆøÌå·Å³öÈÈÁ¿ÎªQ kJ£¬¾²â¶¨£¬½«Éú³ÉµÄCO2ͨÈë×ãÁ¿³ÎÇåʯ»ÒË®ÖвúÉú25 g°×É«³Áµí£¬Ð´³ö±íʾ¶¡ÍéȼÉÕÈȵÄÈÈ»¯Ñ§·½³Ìʽ ¡£
(3)ÒÑÖªÏÂÁÐÈÈ»¯Ñ§·½³Ìʽ£º
¢ÙCH3COOH(l)£«2O2(g)===2CO2(g)£«2H2O(l) ¦¤H1£½£870.3 kJ/mol
¢ÚC(s)£«O2(g)===CO2(g) ¡¡ ¦¤H2£½£393.5 kJ/mol
¢ÛH2(g)£«1/2O2(g)===H2O(l ) ¦¤H3£½£285.8 kJ/mol
д³öÓÉC(s)¡¢H2(g)ºÍO2(g)»¯ºÏÉú³ÉCH3COOH(l)µÄÈÈ»¯Ñ§·½³Ìʽ ¡£
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