ÏÂÁÐÓйØÈÈ»¯Ñ§µÄÐðÊöÕýÈ·µÄÊÇ | |
[¡¡¡¡] | |
A£® |
ÒÑÖª2H2(g)£«O2(g)2H2O(g)£»¡÷H£½£483.6kJ/mol£¬ÔòÇâÆøµÄȼÉÕÈÈΪ241.8kJ |
B£® |
ÈκηÅÈÈ·´Ó¦ÔÚ³£ÎÂÌõ¼þÏÂÒ»¶¨ÄÜ·¢Éú·´Ó¦ |
C£® |
º¬20.0gNaOHµÄÏ¡ÈÜÒºÓëÏ¡ÑÎËáÍêÈ«Öкͣ¬·Å³ö28.7kJµÄÈÈÁ¿£¬Ôò±íʾ¸Ã·´Ó¦ÖкÍÈȵÄÈÈ»¯Ñ§·½³ÌʽΪ£ºNaOH(aq)£«HCl(aq)NaCl(aq)£«H2O(l)£»¡÷H£½£57.4kJ/mol |
D£® |
CaCO3(s)£½CaO(s)£«CO2(g);¡÷H1CaO(s)£«H2O(l)£½Ca(OH)2(s);¡÷H2£¬Ôò¡÷H2>¡÷H1 |
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
A¡¢ÇâÆøµÄȼÉÕÈÈΪ285.5kJ/mo1£¬ÔòË®µç½âµÄÈÈ»¯Ñ§·½³ÌʽΪ£º2H2O£¨l£©¨T2H2£¨g£©+O2£¨g£©£»¡÷H=+285.5kJ/mol | ||||
B¡¢1mol¼×ÍéÍêȫȼÉÕÉú³ÉCO2ºÍH2O£¨l£©Ê±·Å³ö890kJÈÈÁ¿£¬ËüµÄÈÈ»¯Ñ§·½³ÌʽΪ
| ||||
C¡¢ÒÑÖª2C£¨s£©+O2£¨g£©¨T2CO£¨g£©£»¡÷H=-221kJ?mol-1£¬ÔòCµÄȼÉÕÈÈΪ110.5kJ/mol | ||||
D¡¢HFÓëNaOHÈÜÒº·´Ó¦£ºH+£¨aq£©+OH-£¨aq£©¨TH2O£¨l£©£»¡÷H=-57.3kJ/mol |
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