ÌâÄ¿ÄÚÈÝ

úµÄÆø»¯ÊǸßЧ¡¢Çå½àµØÀûÓÃú̿µÄÖØҪ;¾¶Ö®Ò»¡£
(1)ÔÚ250C 101kPaʱ£¬H2ÓëO2»¯ºÏÉú³É1mol H2O(g)·Å³ö241.8kJµÄÈÈÁ¿£¬ÆäÈÈ»¯Ñ§·½³ÌʽΪ
___________
ÓÖÖª: ¢ÙC(s)£«O2(g)¨TCO2(g) ¡÷H£½£­393.5kJ/mol
¢ÚCO(g)£«O2(g)¨TCO2(g) ¡÷H£½£­283.0kJ/mol
½¹Ì¿ÓëË®ÕôÆø·´Ó¦Êǽ«¹ÌÌåú±äΪÆøÌåȼÁϵķ½·¨£¬C(s)£«H2O(g)¨TCO(g)£«H2(g) ¡÷H=____kJ/mol
(2) CO¿ÉÒÔÓëH2O(g)½øÒ»²½·¢Éú·´Ó¦: CO(g)£«H2O(g)CO2(g)£«H2(g) ¡÷H£¼0ÔÚºãÈÝÃܱÕÈÝÆ÷ÖУ¬Æðʼʱn(H2O)=0.20mol£¬n(CO)£½0.10 mol,ÔÚ8000Cʱ´ïµ½Æ½ºâ״̬£¬K£½1.0£¬Ôòƽºâʱ£¬ÈÝÆ÷ÖÐCOµÄת»¯ÂÊÊÇ_____________(¼ÆËã½á¹û±£ÁôһλСÊý)¡£
(3) ¹¤ÒµÉÏ´ÓúÆø»¯ºóµÄ»ìºÏÎïÖзÖÀë³öH2£¬½øÐа±µÄºÏ³É£¬ÒÑÖª·´Ó¦·´Ó¦N2(g)£«3H2(g2NH3(g)£¨¡÷H£¼0£©ÔÚµÈÈÝÌõ¼þϽøÐУ¬¸Ä±äÆäËû·´Ó¦Ìõ¼þ£¬ÔÚI¡¢II¡¢III½×¶ÎÌåϵÖи÷ÎïÖÊŨ¶ÈËæʱ¼ä±ä»¯µÄÇúÏßÈçÏÂͼËùʾ£º

¢ÙN2µÄƽ¾ù·´Ó¦ËÙÂÊv1(N2)¡¢vII(N2)¡¢vIII(N2)´Ó´óµ½Ð¡ÅÅÁдÎÐòΪ________£»
¢ÚÓɵÚÒ»´Îƽºâµ½µÚ¶þ´Îƽºâ£¬Æ½ºâÒƶ¯µÄ·½Ïò ÊÇ________£¬²ÉÈ¡µÄ´ëÊ©ÊÇ________¡£
¢Û±È½ÏµÚII½×¶Î·´Ó¦Î¶È(T2)ºÍµÚIII½×¶Î·´Ó¦Ëٶȣ¨T3)µÄ¸ßµÍ£ºT2________T3Ìî¡°¡µ¡¢=¡¢<¡±ÅжϵÄÀíÓÉÊÇ________________¡£

£¨14·Ö£©£¨1£©H2(g)+O2(g)¨TlH2O(g) ¡÷H£½£­241.8kJ/mol £¨2·Ö£©£»£«131.3kJ £¨1·Ö£©
£¨2£©66.7% £¨2·Ö£© £¨3£©¢Ùv1(N2)£¾vII(N2)£¾vIII(N2)£¨3·Ö£©
¢ÚÏòÕý·´Ó¦·½Ïò ´Ó·´Ó¦ÌåϵÖÐÒƳö²úÎïNH3£¨3·Ö£©
¢Û£¾ ´Ë·´Ó¦Îª·ÅÈÈ·´Ó¦£¬½µµÍζȣ¬Æ½ºâÏòÕý·´Ó¦·½ÏòÒƶ¯ £¨3·Ö£©

½âÎöÊÔÌâ·ÖÎö£º£¨1£©ÔÚ25¡æ¡¢101kPaʱ£¬H2ÓëO2»¯ºÏÉú³É1molH2O(g)·Å³ö241.8kJµÄÈÈÁ¿£¬ËùÒÔÆäÈÈ»¯Ñ§·½³ÌʽΪH2(g)+O2(g)¨TlH2O(g) ¡÷H£½£­241.8kJ/mol¡£
ÒÑÖª¢ÙC(s)£«O2(g)¨TCO2(g) ¡÷H£½£­393.5kJ/mol
¢ÚCO(g)£«O2(g)¨TCO2(g) ¡÷H£½£­283.0kJ/mol
¢ÛH2(g)+O2(g)¨TlH2O(g) ¡÷H£½£­241.8kJ/mol
ËùÒÔ¸ù¾Ý¸Ç˹¶¨ÂÉ£¬¢Ù£­¢Û£­¢Ú¼´µÃµ½C(s)£«H2O(g)¨TCO(g)£«H2(g)£¬Ôò·´Ó¦ÈÈ¡÷H£½£­393.5kJ/mol£«241.8kJ/mol£«283.0kJ/mol£½£«131.3kJ/mol¡£
£¨2£©            CO(g)£«H2O(g)CO2(g)£«H2(g)
ÆðʼÁ¿£¨mol£©     0.1     0.2         0      0
ת»¯Á¿£¨mol£©      x       x          x       x
ƽºâÁ¿£¨mol£©   0.1£­x  0.2£­x        x       x
ÓÉÓÚ·´Ó¦Ç°ºóÆøÌåµÄÌå»ý²»±ä£¬¿ÉÒÔÓÃÎïÖʵÄÁ¿´úÌæŨ¶È¼ÆËãƽºâ³£Êý
¼´£½1.0
½âµÃx£½
ËùÒÔƽºâʱ£¬ÈÝÆ÷ÖÐCOµÄת»¯ÂÊÊÇ¡Á100%£½66.7%
£¨3£©¢Ù¸ù¾ÝͼÏñ¿ÉÖª£¬ÐéÏß±íʾµªÆøµÄŨ¶È±ä»¯£¬Ôòv1(N2)£½£¨2mol/L£­1mol/L£©¡Â20min£½0.05mol/£¨L?min£©£¬vII(N2)£½£¨1mol/L£­0.62mol/L£©¡Â15min£½0.0253mol/£¨L?min£©£¬vIII(N2)£½£¨0.62mol/L£­0.5mol/L£©¡Â10min£½0.012mol/£¨L?min£©£¬¹ÊN2µÄƽ¾ù·´Ó¦ËÙÂÊv1(N2)£¾vII(N2)£¾vIII(N2).
¢Ú¸ù¾ÝͼÏñ¿ÉÖª£¬µÚ¢ò½×¶Î°±ÆøÊÇ´Ó0¿ªÊ¼µÄ£¬Ë²¼ä·´Ó¦ÎﵪÆøºÍÇâÆøŨ¶È²»±ä£¬Òò´Ë¿ÉÒÔÈ·¶¨µÚÒ»´Îƽºâºó´ÓÌåϵÖÐÒƳöÁË°±Æø£¬¼´¼õÉÙÉú³ÉÎïŨ¶È£¬Æ½ºâÕýÏòÒƶ¯¡£
¢ÛµÚ¢ó½×¶ÎµÄ¿ªÊ¼ÓëµÚ¢ò½×¶ÎµÄƽºâ¸÷ÎïÖʵÄÁ¿¾ùÏàµÈ£¬¸ù¾Ý°±ÆøºÍÇâÆøµÄÁ¿¼õÉÙ£¬°±ÆøµÄÁ¿Ôö¼Ó¿ÉÅжÏƽºâÊÇÕýÏòÒƶ¯µÄ¡£¸ù¾Ýƽºâ¿ªÊ¼Ê±Å¨¶ÈÈ·¶¨´ËƽºâÒƶ¯²»¿ÉÄÜÊÇÓÉŨ¶ÈµÄ±ä»¯ÒýÆðµÄ£¬ÁíÍâÌâÄ¿Ëù¸øÌõ¼þÈÝÆ÷µÄÌå»ý²»±ä£¬Ôò¸Ä±äѹǿҲ²»¿ÉÄÜ£¬Òò´ËÒ»¶¨ÎªÎ¶ȵÄÓ°Ïì¡£´Ë·´Ó¦ÕýÏòΪ·ÅÈÈ·´Ó¦£¬¿ÉÒÔÍƲâΪ½µµÍζȣ¬Òò´Ë´ïµ½Æ½ºâºóζÈÒ»¶¨±ÈµÚ¢ò½×¶ÎƽºâʱµÄζȵͣ¬¼´T2£¾T3¡£
¿¼µã£º¿¼²éÈÈ»¯Ñ§·½³ÌʽµÄÊéд£»¸Ç˹¶¨ÂɵÄÓ¦Óã»·´Ó¦ËÙÂʺÍƽºâ³£ÊýµÄ¼ÆËãÒÔ¼°Íâ½çÌõ¼þ¶Ô·´Ó¦ËÙÂʺÍƽºâ״̬µÄÓ°ÏìµÈ

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

½ðÊôÎÙÓÃ;¹ã·º£¬Ö÷ÒªÓÃÓÚÖÆÔìÓ²ÖÊ»òÄ͸ßεĺϽð£¬ÒÔ¼°µÆÅݵĵÆË¿¡£¸ßÎÂÏ£¬ÔÚÃܱÕÈÝÆ÷ÖÐÓÃH2»¹Ô­WO3¿ÉµÃµ½½ðÊôÎÙ£¬Æä×Ü·´Ó¦Îª£º
WO3 (s) + 3H2 (g)W (s) + 3H2O (g)
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
¢ÅÉÏÊö·´Ó¦µÄ»¯Ñ§Æ½ºâ³£Êý±í´ïʽΪ___________________________¡£
¢ÆijζÈÏ·´Ó¦´ïƽºâʱ£¬H2ÓëË®ÕôÆøµÄÌå»ý±ÈΪ2:3£¬ÔòH2µÄƽºâת»¯ÂÊΪ_____________________£»ËæζȵÄÉý¸ß£¬H2ÓëË®ÕôÆøµÄÌå»ý±È¼õС£¬Ôò¸Ã·´Ó¦Îª·´Ó¦_____________________£¨Ìî¡°ÎüÈÈ¡±»ò¡°·ÅÈÈ¡±£©¡£
¢ÇÉÏÊö×Ü·´Ó¦¹ý³Ì´óÖ·ÖΪÈý¸ö½×¶Î£¬¸÷½×¶ÎÖ÷Òª³É·ÖÓëζȵĹØϵÈçϱíËùʾ£º

ζÈ
25¡æ  ~  550¡æ  ~  600¡æ  ~  700¡æ
Ö÷Òª³É·Ý
WO3      W2O5      WO2        W
 
µÚÒ»½×¶Î·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ___________________________£»580¡æʱ£¬¹ÌÌåÎïÖʵÄÖ÷Òª³É·ÖΪ________£»¼ÙÉèWO3Íêȫת»¯ÎªW£¬ÔòÈý¸ö½×¶ÎÏûºÄH2ÎïÖʵÄÁ¿Ö®±ÈΪ____________________________________¡£
¢È ÒÑÖª£ºÎ¶ȹý¸ßʱ£¬WO2 (s)ת±äΪWO2 (g)£»
WO2 (s) + 2H2 (g)  W (s) + 2H2O (g)£»¦¤H £½ +66.0 kJ¡¤mol£­1
WO2 (g) + 2H2(g)  W (s) + 2H2O (g)£»¦¤H £½ £­137.9 kJ¡¤mol£­1
ÔòWO2 (s)  WO2 (g) µÄ¦¤H £½ ______________________¡£
¢ÉÎÙË¿µÆ¹ÜÖеÄWÔÚʹÓùý³ÌÖлºÂý»Ó·¢£¬Ê¹µÆË¿±äϸ£¬¼ÓÈëI2¿ÉÑÓ³¤µÆ¹ÜµÄʹÓÃÊÙÃü£¬Æ乤×÷Ô­ÀíΪ£ºW (s) +2I2 (g)WI4 (g)¡£ÏÂÁÐ˵·¨ÕýÈ·µÄÓÐ____________¡£
a£®µÆ¹ÜÄÚµÄI2¿ÉÑ­»·Ê¹ÓÃ
b£®WI4ÔÚµÆË¿ÉϷֽ⣬²úÉúµÄWÓÖ³Á»ýÔÚµÆË¿ÉÏ
c£®WI4ÔڵƹܱÚÉϷֽ⣬ʹµÆ¹ÜµÄÊÙÃüÑÓ³¤   
d£®Î¶ÈÉý¸ßʱ£¬WI4µÄ·Ö½âËÙÂʼӿ죬WºÍI2µÄ»¯ºÏËÙÂʼõÂý

ÄòËØ(H2NCONH2)ÊÇÒ»Öַdz£ÖØÒªµÄ¸ßµª»¯·Ê£¬ÔÚ¹¤Å©ÒµÉú²úÖÐÓÐ×ŷdz£ÖØÒªµÄµØλ¡£
£¨1£©¹¤ÒµÉϺϳÉÄòËصķ´Ó¦ÈçÏ£º
2NH3(l)+CO2(g)H2O(l)+H2NCONH2(l)  ¡÷H=-103£®7 kJ¡¤mol-1
ÏÂÁдëÊ©ÖÐÓÐÀûÓÚÌá¸ßÄòËصÄÉú³ÉËÙÂʵÄÊÇ           

A£®²ÉÓøßÎÂ
B£®²ÉÓøßѹ
C£®Ñ°ÕÒ¸ü¸ßЧµÄ´ß»¯¼Á
D£®¼õСÌåϵÄÚCO2Ũ¶È
£¨2£©ºÏ³ÉÄòËصķ´Ó¦ÔÚ½øÐÐʱ·ÖΪÈçÏÂÁ½²½£º
µÚÒ»²½£º2NH3(l)+CO2(g) H2NCOONH4(°±»ù¼×Ëáï§)(l) ¡÷H1
µÚ¶þ²½£ºH2NCOONH4(l) H2O(l)+H2NCONH2(l)  ¡÷H2£®
ijʵÑéС×éÄ£Ä⹤ҵÉϺϳÉÄòËصÄÌõ¼þ£¬ÔÚÒ»Ìå»ýΪ0£®5 LÃܱÕÈÝÆ÷ÖÐͶÈë4 mol°±ºÍl mol¶þÑõ»¯Ì¼£¬ÊµÑé²âµÃ·´Ó¦Öи÷×é·ÖËæʱ¼äµÄ±ä»¯ÈçÏÂͼIËùʾ£º

¢ÙÒÑÖª×Ü·´Ó¦µÄ¿ìÂýÓÉÂýµÄÒ»²½¾ö¶¨£¬ÔòºÏ³ÉÄòËØ×Ü·´Ó¦µÄ¿ìÂýÓɵڠ   ²½·´Ó¦¾ö¶¨£¬×Ü·´Ó¦½øÐе½   minʱµ½´ïƽºâ¡£
¢ÚµÚ¶þ²½·´Ó¦µÄƽºâ³£ÊýKËæζȵı仯ÈçÉÏͼIIËùʾ£¬Ôò¡÷H2    0(Ìî¡°>¡± ¡°<¡± »ò ¡°=¡±)
£¨3£©ÔÚζÈ70-95¡æʱ£¬¹¤ÒµÎ²ÆøÖеÄNO¡¢NO2¿ÉÒÔÓÃÄòËØÈÜÒºÎüÊÕ£¬½«Æäת»¯ÎªN2
¢ÙÄòËØÓëNO¡¢NO2ÈýÕßµÈÎïÖʵÄÁ¿·´Ó¦£¬»¯Ñ§·½³ÌʽΪ
                                                            
¢ÚÒÑÖª£ºN2(g)+O2(g)=2NO(g)£®¡÷H=180£®6 kJ¡¤mol-1
N2(g)+3H2(g)=2NH3(g) ¡÷H=-92£®4 kJ¡¤mol-1
2H2(g)+O2(g)=2H2O(g)  ¡÷H=-483£®6 kJ¡¤mol-1
Ôò4NO(g)+4NH3(g)+O2(g)=4N2(g)+6H2O(g)  ¡÷H=    kJ¡¤mol-1
£¨4£©ÄòËØȼÁϵç³Ø½á¹¹ÈçÉÏͼIIIËùʾ¡£Æ乤×÷ʱ¸º¼«µç¼«·´Ó¦Ê½
¿É±íʾΪ                                                   ¡£

£¨1£©¡¢¢ÙÓÃëÂ(N2H4)ΪȼÁÏ£¬ËÄÑõ»¯¶þµª×öÑõ»¯¼Á£¬Á½Õß·´Ó¦Éú³ÉµªÆøºÍÆø̬ˮ¡£
ÒÑÖª£ºN2(g)£«2O2(g)£½N2O4(g)  ¦¤H£½+10.7kJ¡¤mol-1
N2H4(g)£«O2(g)£½N2(g)£«2H2O(g)  ¦¤H£½-543kJ¡¤mol-1
д³öÆø̬ëºÍN2O4·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ£º                                ¡£
¢ÚÒÑÖªËÄÑõ»¯¶þµªÔÚ´óÆøÖлòÔڽϸßζÈϺÜÄÑÎȶ¨´æÔÚ£¬ÆäºÜÈÝÒ×ת»¯Îª¶þÑõ»¯µª¡£ÊÔÍƶÏÓɶþÑõ»¯µªÖÆÈ¡ËÄÑõ»¯¶þµªµÄ·´Ó¦Ìõ¼þ(»ò´ëÊ©)£º                         ¡£
£¨2£©¿Æѧ¼ÒÖÆÔì³öÒ»ÖÖʹÓùÌÌåµç½âÖʵÄȼÁϵç³Ø£¬ÆäЧÂʸü¸ß£¬¿ÉÓÃÓÚº½Ì캽¿Õ¡£

ͼ¼×ËùʾװÖÃÖУ¬ÒÔÏ¡ÍÁ½ðÊô²ÄÁÏΪ¶èÐԵ缫£¬ÔÚÁ½¼«ÉÏ·Ö±ðͨÈëCH4ºÍ¿ÕÆø£¬ÆäÖйÌÌåµç½âÖÊÊDzôÔÓÁËY2O3µÄZrO2¹ÌÌ壬ËüÔÚ¸ßÎÂÏÂÄÜ´«µ¼Ñô¼«Éú³ÉµÄO2-(O2+4e ¡ú2O2-)
¢Ùcµç¼«Îª       £¬dµç¼«Éϵĵ缫·´Ó¦Ê½Îª                            ¡£
¢ÚͼÒÒÊǵç½â100mL 0.5mol¡¤L-1 CuSO4ÈÜÒº£¬aµç¼«Éϵĵ缫·´Ó¦Ê½Îª          ¡£Èôaµç¼«²úÉú56mL(±ê×¼×´¿ö)ÆøÌ壬ÔòËùµÃÈÜÒºµÄpH=     (²»¿¼ÂÇÈÜÒºÌå»ý±ä»¯)£¬ÈôҪʹµç½âÖÊÈÜÒº»Ö¸´µ½µç½âÇ°µÄ״̬£¬¿É¼ÓÈë       (Ñ¡Ìî×ÖĸÐòºÅ)
a.CuO    b.Cu(OH)2     c.CuCO3     d.Cu2(OH)2CO3

ÔËÓ÷´Ó¦Ô­ÀíÑо¿µª¡¢Áò¡¢ÂÈ¡¢µâ¼°Æ仯ºÏÎïµÄ·´Ó¦ÓÐÖØÒªÒâÒå¡£
£¨1£©ÔÚ·´Ó¦£º2SO2(g)+O2(g)2SO3(g)µÄ»ìºÏÌåϵÖУ¬SO3µÄ°Ù·Öº¬Á¿ºÍζȵĹØϵÈçÏÂͼ(ÇúÏßÉÏÈκÎÒ»µã¶¼±íʾƽºâ״̬£©£º

¢Ù2SO2(g)+O2(g)2SO3(g)µÄ¡÷H        0£¨Ìî¡°>¡±»ò¡°<¡±£©£»ÈôÔÚºãΡ¢ºãѹʱ£¬Ïò¸ÃƽºâÌåϵÖÐͨÈ뺤Æøƽºâ½«        Òƶ¯£¨Ìî¡°Ïò×󡱡¢¡°ÏòÓÒ¡±»ò¡°²»¡±£©£»
¢Úµ±Î¶ÈΪT1£¬·´Ó¦½øÐе½×´Ì¬Dʱ£¬VÕý        VÄ棨Ìî¡°>¡±¡¢¡°<¡±»ò¡°=¡±£©¡£
£¨2£©¢ÙÏÂͼÊÇÒ»¶¨Ìõ¼þÏ£¬N2ºÍH2·¢Éú¿ÉÄæ·´Ó¦Éú³É1mol NH3µÄÄÜÁ¿±ä»¯Í¼£¬¸Ã·´Ó¦µÄÈÈ»¯Ñ§·´Ó¦·½³Ìʽ                 ¡£(¡÷HÓú¬Q1¡¢Q2µÄ´úÊýʽ±íʾ£©

¢Ú25¡ãCʱ£¬½«a mol ? L¨D1µÄ°±Ë®Óëb mol ? L¨D1µÄÑÎËáµÈÌå»ý»ìºÏ£¬ËùµÃÈÜÒºµÄpH=7£¬Ôòc (NH4+)         c(Cl¨D)£¬a        b£¨Ìî¡°>¡±¡¢¡°<¡±»ò¡°=¡±£©£»
£¨3£©º£Ë®Öк¬ÓдóÁ¿ÒÔ»¯ºÏ̬ÐÎʽ´æÔÚµÄÂÈ¡¢µâÔªËØ¡£ÒÑÖª£º250Cʱ£¬Ksp(AgCl)=1.6¡Á10¨D10mol2?L¨D2¡¢Ksp(AgI)=1.5¡Á10¨D16mol2?L¨D2¡£
ÔÚ 250Cʱ£¬Ïò 10mL0.002mol?L¨D1µÄ NaClÈÜÒºÖеÎÈë 10mL0.002mol?L¨D1AgNO3ÈÜÒº£¬Óа×É«³ÁµíÉú³É£¬ÏòËùµÃ×ÇÒºÖмÌÐøµÎÈË0.1mol ?L¨D1µÄNaIÈÜÒº£¬°×É«³ÁµíÖð½¥×ª»¯Îª»ÆÉ«³Áµí£¬ÆäÔ­ÒòÊÇ        £¬¸Ã·´Ó¦µÄÀë×Ó·½³Ìʽ                 ¡£

£¨14·Ö£©Áò»¯ÎïÔÚ×ÔÈ»½çÖеIJ¿·ÖÑ­»·¹ØϵÈçÏ¡£

£¨1£©H2SÔÚ¿ÕÆøÖпÉÒÔȼÉÕ¡£
ÒÑÖª£º 2H2S(g) + O2(g) 2S(s) + 2H2O(g)  ¦¤H= £­442.38 kJ/mol    ¢Ù
S(s) + O2(g)  SO2(g)             ¦¤H=£­297.04 kJ/mol    ¢Ú
H2S(g)ÓëO2(g)·´Ó¦²úÉúSO2(g)ºÍH2O(g)µÄÈÈ»¯Ñ§·½³ÌʽÊÇ     ¡£
£¨2£©SO2ÊÇ´óÆøÎÛȾÎº£Ë®¾ßÓÐÁ¼ºÃµÄÎüÊÕSO2µÄÄÜÁ¦£¬Æä¹ý³ÌÈçÏ¡£
¢Ù SO2ÈÜÓÚº£Ë®Éú³ÉH2SO3£¬H2SO3×îÖÕ»áµçÀë³öSO32¡ª£¬ÆäµçÀë·½³ÌʽÊÇ     ¡£
¢Ú SO32¡ª¿ÉÒÔ±»º£Ë®ÖеÄÈܽâÑõÑõ»¯ÎªSO42¡ª¡£º£Ë®µÄpH»á     £¨Ìî¡°Éý¸ß¡± ¡¢¡°²»±ä¡±»ò¡°½µµÍ¡±£©¡£
¢Û Ϊµ÷Õûº£Ë®µÄpH£¬¿É¼ÓÈëÐÂÏʵĺ£Ë®£¬Ê¹ÆäÖеÄHCO3¡ª²ÎÓë·´Ó¦£¬Æä·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ     ¡£
¢Ü ÔÚÉÏÊö·´Ó¦µÄͬʱÐèÒª´óÁ¿¹ÄÈë¿ÕÆø£¬ÆäÔ­ÒòÊÇ     ¡£
£¨3£©×ÔÈ»½çµØ±í²ãÔ­ÉúÍ­µÄÁò»¯Îï¾­Ñõ»¯¡¢ÁÜÂË×÷Óúó±ä³ÉCuSO4ÈÜÒº£¬ÏòµØÏÂÉî²ãÉø͸£¬Óöµ½ÄÑÈܵÄZnS£¬ÂýÂýת±äΪͭÀ¶£¨CuS£©£¬Óû¯Ñ§ÓÃÓï±íʾÓÉZnSת±äΪCuSµÄ¹ý³Ì£º     ¡£
£¨4£©SO2ºÍO2ÔÚH2SO4ÈÜÒºÖпÉÒÔ¹¹³ÉÔ­µç³Ø£¬Æ为¼«·´Ó¦Ê½ÊÇ     ¡£

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø