ÌâÄ¿ÄÚÈÝ
¡¾ÌâÄ¿¡¿(1)ÒÑÖª2 molÇâÆøȼÉÕÉú³ÉҺ̬ˮʱ·Å³ö572 kJµÄÈÈÁ¿£¬·´Ó¦·½³ÌʽÊÇ2H2(g)£«O2(g)===2H2O(l)¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺
¢Ù¸Ã·´Ó¦µÄÉú³ÉÎïÄÜÁ¿×ܺÍ________(Ìî¡°´óÓÚ¡±¡¢¡°Ð¡ÓÚ¡±»ò¡°µÈÓÚ¡±)·´Ó¦ÎïÄÜÁ¿×ܺ͡£
¢ÚÈô2 molÇâÆøÍêȫȼÉÕÉú³ÉË®ÕôÆø£¬Ôò·Å³öµÄÈÈÁ¿_____(Ìî¡°´óÓÚ¡±¡¢¡°Ð¡ÓÚ¡±»ò¡°µÈÓÚ¡±)572 kJ¡£
(2) 2.3gÓлúÎïC2H6OºÍÒ»¶¨Á¿µÄÑõÆø»ìºÏµãȼ,Ç¡ºÃÍêȫȼÉÕ,Éú³ÉCO2ºÍҺ̬ˮ£¬²¢·Å³ö68.35kJÈÈÁ¿,Ôò¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽÊÇ________________________________¡£
(3) FeS2±ºÉÕ²úÉúµÄSO2¿ÉÓÃÓÚÖÆÁòËá¡£ÒÑÖª25 ¡æ¡¢101 kPaʱ£º
2SO2(g)£«O2(g) 2SO3(g)¡¡¦¤H1£½£197 kJ¡¤mol£1
H2O(g)===H2O(l)¡¡¦¤H2£½£44 kJ¡¤mol£1
2SO2(g)£«O2(g)£«2H2O(g)===2H2SO4(l)¡¡¦¤H3£½£545 kJ¡¤mol£1
ÔòSO3(g)ÓëH2O(l)·´Ó¦Éú³ÉH2SO4(l)µÄÈÈ»¯Ñ§·½³ÌʽÊÇ_________________________________¡£
¡¾´ð°¸¡¿ СÓÚ Ð¡ÓÚ C2H6O£¨l£©+3O2£¨g£©=2CO2£¨g£©+3H2O(l)¡÷H=-1367kJ/mol SO3£¨g£©£«H2O£¨l£©=H2SO4 £¨l£©¦¤H£½£152 kJ/mol
¡¾½âÎö¡¿(1)¢ÙÇâÆøȼÉÕÉú³ÉҺ̬ˮʱ·ÅÈÈ£¬Ôò·´Ó¦ÎïµÄÄÜÁ¿¸ßÓÚÉú³ÉÎïµÄÄÜÁ¿£¬¹Ê´ð°¸Îª£ºÐ¡ÓÚ£»
¢ÚҺ̬ˮ±äΪˮÕôÆøÊÇÎüÈȵĹý³Ì£¬2molÇâÆøȼÉÕÉú³ÉҺ̬ˮʱ·Å³ö572kJÈÈÁ¿£¬Éú³ÉÆø̬ˮʱ·Å³öµÄÈÈÁ¿Ð¡ÓÚ572kJ£¬¹Ê´ð°¸Îª£ºÐ¡ÓÚ£»
(2) 2.3gÓлúÎïC2H6OµÄÎïÖʵÄÁ¿Îª=0.05mol£¬ºÍÒ»¶¨Á¿µÄÑõÆø»ìºÏµãȼ£¬Ç¡ºÃÍêȫȼÉÕ£¬Éú³ÉCO2ºÍҺ̬ˮ£¬²¢·Å³ö68.35kJÈÈÁ¿¡£Ôò1molC2H6OÍêȫȼÉշųöµÄÈÈÁ¿Îª68.35kJ¡Á
=1367kJ£¬Ôò¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪC2H6O(l)+3O2(g)=2CO2(g)+3H2O(l)¡÷H=-1367kJ/mol£¬¹Ê´ð°¸Îª£ºC2H6O(l)+3O2(g)=2CO2(g)+3H2O(l)¡÷H=-1367kJ/mol£»
(3)2SO2(g)+O2(g)2SO3(g)¡÷H1=Ò»197kJ/mol ¢Ù£¬2H2O(g)=2H2O(1)¡÷H2=-44kJ/mol ¢Ú£¬2SO2(g)+O2(g)+2H2O(g)=2H2SO4(l)¡÷H3=Ò»545kJ/mol¢Û£¬ÀûÓøÇ˹¶¨ÂÉ£º(¢Û-¢Ù-¢Ú)¡Á
µÃSO3 (g)+H2O(l)=H2SO4(l)¡÷H=-152kJ/mol£¬¹Ê´ð°¸Îª£ºSO3(g)+H2O(l)=H2SO4(l)¡÷H3=-152kJ/mol¡£
![](http://thumb2018.1010pic.com/images/loading.gif)
¡¾ÌâÄ¿¡¿ÀûÓÃÁòËáÔü£¨Ö÷Òªº¬Fe2O3¡¢SiO2¡¢Al2O3µÈÔÓÖÊ£©ÖƱ¸Ñõ»¯ÌúµÄ¹¤ÒÕÁ÷³ÌÈçÏ£º
ϱíÁгöÁËÏà¹Ø¸ÃʵÑéÌõ¼þϽðÊôÀë×ÓÉú³ÉÇâÑõ»¯Îï³ÁµíµÄpH¡£
³ÁµíÎï | Fe(OH)3 | Al(OH)3 | Fe(OH)2 |
¿ªÊ¼³Áµí | 2.7 | 3.8 | 7.5 |
ÍêÈ«³Áµí | 3.2 | 5.2 | 9.7 |
£¨1£©¡°Ëá½þ¡±Ê±ÁòËáÒªÊʵ±¹ýÁ¿£¬ÆäÄ¿µÄ³ýÁËÌá¸ßÌúµÄ½þ³öÂÊÍ⣬»¹¿ÉÒÔ_______¡£¡°Ëá½þ¡±²½ÖèºóÐèÒª²â¶¨ÈÜÒºÖÐFe3+µÄº¬Á¿£¬ÆäÔÒòÊÇ____________¡£
£¨2£©¡°»¹Ô¡±Êǽ«Fe3+ת»¯ÎªFe2+£¬Í¬Ê±FeS2±»Ñõ»¯ÎªSO42-£¬¸Ã·´Ó¦µÄÀë×Ó·½³ÌΪ________¡£
£¨3£©ÂËÔüµÄÖ÷Òª³É·ÖÊÇFeS2ºÍ_______(Ìѧʽ)¡£
£¨4£©¹ýÂ˺óµÄÂËÒºÖк¬ÓÐFe3+£¬²úÉúFe3+µÄÔÒòÊÇ________(ÓÃÀë×Ó·½³Ìʽ±íʾ£©¡£
£¨5£©ÇëÉè¼ÆÓá°¹ýÂË¡±ºóµÄÈÜÒºÖƱ¸Fe2O3µÄʵÑé·½°¸£º___________£¨ÊµÑéÖбØÐëʹÓõÄÊÔ¼ÁÓУº5% µÄH2O2ÈÜÒº£¬0.5mol/LNaOHÈÜÒº£©¡£
¡¾ÌâÄ¿¡¿ÏÂͼÊÇʵÑéÊÒÖƱ¸ÂÈÆø²¢½øÐÐһϵÁÐÏà¹ØʵÑéµÄ×°ÖÃ(¼Ð³ÖÉ豸ÒÑÂÔ)¡£
£¨1£©ÊµÑéÊÒÖÆÂÈÆøµÄ»¯Ñ§·½³ÌʽΪ____________________________¡£
£¨2£©×°ÖÃBÖб¥ºÍʳÑÎË®µÄ×÷ÓÃÊÇ______________£»Çëд³ö×°ÖÃBµÄÁíÒ»¸ö×÷ÓÃ_________________________________¡£
£¨3£©×°ÖÃCµÄʵÑéÄ¿µÄÊÇÑéÖ¤ÂÈÆøÊÇ·ñ¾ßÓÐƯ°×ÐÔ£¬ Ϊ´ËCÖТñ¡¢¢ò¡¢¢óÒÀ´Î·ÅÈëÎïÖʵÄ×éºÏÊÇ________(Ìî±àºÅ)¡£
񅧏 | a | b | c | d |
¢ñ | ¸ÉÔïµÄÓÐÉ«²¼Ìõ | ¸ÉÔïµÄÓÐÉ«²¼Ìõ | ʪÈóµÄÓÐÉ«²¼Ìõ | ʪÈóµÄÓÐÉ«²¼Ìõ |
¢ò | ¼îʯ»Ò | ¹è½º | ŨÁòËá | ÎÞË®ÂÈ»¯¸Æ |
¢ó | ʪÈóµÄÓÐÉ«²¼Ìõ | ʪÈóµÄÓÐÉ«²¼Ìõ | ¸ÉÔïµÄÓÐÉ«²¼Ìõ | ¸ÉÔïµÄÓÐÉ«²¼Ìõ |
£¨4£©Éè¼Æ×°ÖÃD¡¢EµÄÄ¿µÄÊDZȽÏÂÈ¡¢äå¡¢µâµÄ·Ç½ðÊôÐÔ¡£·´Ó¦Ò»¶Îʱ¼äºó£¬´ò¿ª»îÈû£¬½«×°ÖÃDÖÐÉÙÁ¿ÈÜÒº¼ÓÈë×°ÖÃEÖУ¬Õñµ´£¬¹Û²ìµ½µÄÏÖÏóÊÇ____________________£¬¸ÃÏÖÏó_______(Ìî¡°ÄÜ¡±»ò¡°²»ÄÜ¡±)˵Ã÷äåµÄ·Ç½ðÊôÐÔÇ¿Óڵ⣬ÔÒòÊÇ_______________¡£
£¨5£©ÓÐÈËÌá³ö£¬×°ÖÃFÖпɸÄÓÃ×ãÁ¿µÄNa2SO3ÈÜÒºÎüÊÕÓàÂÈ£¬ÊÔд³öÏàÓ¦µÄÀë×Ó·´Ó¦·½³Ìʽ£º__________________¡£