ÒÑ֪ǦÐîµç³ØµÄ¹¤×÷Ô­ÀíΪPb+PbO2+2H2SO4

2PbSO4+2H2O,ÏÖÓÃÈçͼװÖýøÐеç½â(µç½âÒº×ãÁ¿),²âµÃµ±Ç¦Ðîµç³ØÖÐתÒÆ

0.4 molµç×ÓʱÌúµç¼«µÄÖÊÁ¿¼õÉÙ11.2 g¡£Çë»Ø´ðÏÂÁÐÎÊÌâ:

(1)AÊÇǦÐîµç³ØµÄ¡¡¡¡¡¡¡¡¼«,ǦÐîµç³ØÕý¼«·´Ó¦Ê½Îª¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡,·Åµç¹ý³ÌÖеç½âÒºµÄÃܶȡ¡¡¡¡¡¡¡(Ìî¡°¼õС¡±¡°Ôö´ó¡±»ò¡°²»±ä¡±)¡£

(2)Agµç¼«µÄµç¼«·´Ó¦Ê½ÊÇ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡,¸Ãµç¼«µÄµç¼«²úÎï¹²¡¡¡¡¡¡¡¡g¡£

(3)Cuµç¼«µÄµç¼«·´Ó¦Ê½ÊÇ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡,CuSO4ÈÜÒºµÄŨ¶È¡¡¡¡¡¡¡¡(Ìî¡°¼õС¡±¡°Ôö´ó¡±»ò¡°²»±ä¡±)¡£

(4)Èçͼ±íʾµç½â½øÐйý³ÌÖÐij¸öÁ¿(×Ý×ø±êx)Ëæʱ¼äµÄ±ä»¯ÇúÏß,Ôòx±íʾ¡¡¡¡¡¡¡¡¡£

a.¸÷UÐιÜÖвúÉúµÄÆøÌåµÄÌå»ý

b.¸÷UÐιÜÖÐÑô¼«ÖÊÁ¿µÄ¼õÉÙÁ¿

c.¸÷UÐιÜÖÐÒõ¼«ÖÊÁ¿µÄÔö¼ÓÁ¿


¸ù¾ÝÔÚµç½â¹ý³ÌÖÐÌúµç¼«ÖÊÁ¿µÄ¼õÉÙ¿ÉÅжÏAÊǵçÔ´µÄ¸º¼«,BÊǵçÔ´µÄÕý¼«,µç½âʱAg¼«×÷Òõ¼«,µç¼«·´Ó¦Ê½Îª2H++2e-====H2¡ü,Fe×÷Ñô¼«,µç¼«·´Ó¦Ê½ÎªFe-2e-====Fe2+,×ó²àUÐιÜÖÐ×Ü·´Ó¦Ê½ÎªFe+2H+====Fe2++H2¡ü¡£ÓÒ²àUÐιÜÏ൱ÓÚµç¶Æ×°ÖÃ,Znµç¼«×÷Òõ¼«,µç¼«·´Ó¦Ê½ÎªCu2++2e-====Cu,Í­µç¼«×÷Ñô¼«,µç¼«·´Ó¦Ê½ÎªCu-2e-====Cu2+,µç¶Æ¹ý³ÌÖÐCuSO4ÈÜÒºµÄŨ¶È±£³Ö²»±ä,¸ù¾ÝÉÏÊö·ÖÎö¿ÉµÃ´ð°¸¡£

´ð°¸:(1)¸º¡¡PbO2+4H++S+2e-====PbSO4+2H2O¡¡¼õС

(2)2H++2e-====H2¡ü¡¡0.4

(3)Cu-2e-====Cu2+¡¡²»±ä¡¡(4)b

¡¾·½·¨¼¼ÇÉ¡¿µç½â¼ÆËãµÄÒ»°ã½âÌâ˼·

(1)ÏÈ¿´Ñô¼«²ÄÁÏÊÇ»îÐԵ缫»¹ÊǶèÐԵ缫¡£

(2)ÕÒÈ«Àë×Ó,½«ÈÜÒºÖеÄËùÓÐÀë×Ó°´ÒõÑôÀë×Ó·Ö¿ª¡£

(3)¸ù¾Ýµç×ÓÊغã,´®Áªµç·ÖÐתÒƵĵç×ÓÊýÄ¿Ïàͬ,¼´ÒõÀë×Ó(»îÆõ缫)ʧȥµÄµç×ÓÊýµÈÓÚÑôÀë×ӵõ½µÄµç×ÓÊý,Áгö¹Øϵʽ½øÐмÆËã¡£³£ÓõļÆËã¹ØϵʽÓÐ:4H+¡«4OH-¡«2H2¡«O2¡«2Cu2+¡«2Cu¡«4Ag¡«4Ag+¡«4Cl-¡«2Cl2µÈ¡£


Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÓÃÈçͼװÖýøÐÐÌú·ÛÔÚ¸ßÎÂÏÂÓëË®ÕôÆø·´Ó¦µÄʵÑ飬²¢Óüòµ¥µÄ·½·¨ÊÕ¼¯¡¢¼ìÑéÉú³ÉµÄÇâÆø¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©Ð´³öÌúÔÚ¸ßÎÂÏÂÓëË®ÕôÆø·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º _______________________________¡£

£¨2£©¸ÉÔï¹ÜCÄÚÊ¢·ÅµÄÒ©Æ·ÊÇ________¡£¸ÉÔï¹ÜµÄ________(Ìî¡°m¡±»ò¡°n¡±)¶ËÓëµ¼¹Ü¿Ú g ÏàÁ¬½Ó¡£

£¨3£©ÈôÊÕ¼¯µ½±ê×¼×´¿öϵÄH2Ϊ22.4 L£¬Ôò²Î¼Ó·´Ó¦µÄÌú·ÛµÄÖÊÁ¿Îª________g¡£

£¨4£©µ±¹ÌÌåÖÊÁ¿Ôö¼Ó32 gʱ£¬Éú³ÉH2µÄÖÊÁ¿Îª________g¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


½«CuÓëCuOµÄ»ìºÏÎï20.8g¼ÓÈëµ½50mL 18.4mol/LŨH2SO4ÖУ¬¼ÓÈȳä·Ö·´Ó¦ÖÁ¹ÌÌåÎïÖÊÍêÈ«Èܽâ(²úÉúÆøÌåÈ«²¿Òݳö)£¬ÀäÈ´ºó½«ÈÜҺϡÊÍÖÁ1000ml£¬²âµÃc(H+)=0.84mol/L£»ÈôҪʹϡÊͺóÈÜÒºÖеÄCu2+³ÁµíÍêÈ«£¬Ó¦¼ÓÈë6.0mol/LµÄNaOHÈÜÒºµÄÌå»ýΪ

A£® 100mL    B£® 160mL    C£® 240mL     D£® 307mL

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÏÖÓÐÈçϸ÷»¯ºÏÎ¢Ù¾Æ¾«£¬¢ÚÂÈ»¯ï§£¬¢ÛÇâÑõ»¯±µ£¬¢Ü°±Ë®£¬¢ÝÕáÌÇ£¬¢Þ¸ßÂÈËᣬ¢ßÇâÁòËᣬ¢àÁòËáÇâ¼Ø£¬¢áÁ×Ëᣬ¢âÁòËá¡£

ÇëÓÃÎïÖʵÄÐòºÅÌîдÏÂÁпհףº

(1)ÊôÓÚµç½âÖʵÄÓÐ

________________________________________________________________________¡£

(2)ÊôÓÚÇ¿µç½âÖʵÄÓÐ

________________________________________________________________________¡£

(3)ÊôÓÚÈõµç½âÖʵÄÓÐ

________________________________________________________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ijУ»î¶¯Ð¡×éΪ̽¾¿½ðÊô¸¯Ê´µÄÏà¹ØÔ­Àí,Éè¼ÆÁËÈçͼaËùʾװÖÃ,ͼaµÄÌú°ôÄ©¶Î·Ö±ðÁ¬ÉÏÒ»¿éZnƬºÍCuƬ,²¢ÖÃÓÚº¬ÓÐK3Fe(CN)6¼°·Ó̪µÄ»ìºÏÄý½ºÉÏ¡£Ò»¶Îʱ¼äºó·¢ÏÖÄý½ºµÄijЩÇøÓò(ÈçͼbËùʾ)·¢ÉúÁ˱仯,ÒÑÖªFe2+¿ÉÓÃK3Fe(CN)6À´¼ìÑé(³ÊÀ¶É«)¡£ÔòÏÂÁÐ˵·¨²»ÕýÈ·µÄÊÇ(¡¡¡¡)

A.¼×Çø·¢ÉúµÄµç¼«·´Ó¦Ê½:Fe-2e-====Fe2+

B.ÒÒÇø²úÉúZn2+

C.±ûÇø³ÊÏÖºìÉ«

D.¶¡Çø³ÊÏÖÀ¶É«

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


¹¤ÒµºÏ³É°±µÄ·´Ó¦ÊÇÔÚ500¡æ×óÓÒ½øÐеģ¬ÕâÖ÷ÒªÊÇÒòΪ(¡¡¡¡)

A£®500¡æʱ´Ë·´Ó¦ËÙÂÊ×î¿ì

B£®500¡æʱNH3µÄƽºâŨ¶È×î´ó

C£®500¡æʱN2µÄת»¯ÂÊ×î¸ß

D£®500¡æʱ¸Ã·´Ó¦µÄ´ß»¯¼Á»îÐÔ×î´ó

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


¸ù¾ÝÅöײÀíÂÛ£¬·Ö×ÓÔÚ·¢Éú·´Ó¦Ê±±ØÐëÒª½øÐÐÓÐЧÅöײ¡£ÄÇЩ¾ßÓÐ×ã¹»¸ßÄÜÁ¿£¬ÄÜ·¢ÉúÓÐЧÅöײµÄ·Ö×Ó³ÆΪ»î»¯·Ö×Ó£¬ÒªÊ¹ÆÕͨ·Ö×Ó³ÉΪ»î»¯·Ö×ÓËùÐè×îСÄÜÁ¿³ÆΪ»î»¯ÄÜ(Ea)¡£Ò»¶¨Î¶ÈÏÂÆøÌå·Ö×ÓÖеĻ·Ö×Ó°Ù·ÖÊýÊÇÒ»¶¨µÄ£¬¶ø´ß»¯¼Á¿ÉÒԸıä»î»¯ÄܵĴóС¡£Èçͼ±íʾ298.15 Kʱ£¬N2¡¢H2ÓëNH3µÄƽ¾ùÄÜÁ¿ÓëºÏ³É°±·´Ó¦µÄ»î»¯ÄܵÄÇúÏßͼ£¬¾Ýͼ»Ø´ð£º

                    

(1)Èô·´Ó¦ÖÐÉú³É2 mol°±£¬Ôò·´Ó¦________(Ìî¡°ÎüÈÈ¡±»ò¡°·ÅÈÈ¡±)________kJ¡£

(2)ÔÚͼÖÐÇúÏß________(Ìî¡°a¡±»ò¡°b¡±)±íʾ¼ÓÈëÌú´¥Ã½µÄÄÜÁ¿±ä»¯ÇúÏߣ¬Ìú´¥Ã½Äܼӿ췴ӦËÙÂʵÄÔ­ÀíÊÇ_____________________________________________________________

________________________________________________________________________¡£

(3)Ä¿Ç°ºÏ³É°±¹¤Òµ¹ã·º²ÉÓõķ´Ó¦Ìõ¼þ500¡æ¡¢20 MPa¡«50 MPa¡¢Ìú´¥Ã½£¬·´Ó¦×ª»¯Âʲ»³¬¹ý50%£¬¹¤ÒµÉÏΪÁ˽øÒ»²½Ìá¸ß°±Æø²úÂÊ£¬ÄãÈÏΪÏÂÁдëÊ©×î¾­¼Ã¿ÉÐеÄÊÇ

________________________________________________________________________¡£

A£®½µµÍ·´Ó¦Î¶ȣ¬È÷´Ó¦Ïò×ÅÓÐÀûÓÚ°±ÆøÉú³ÉµÄ·½Ïò½øÐÐ

B£®Éý¸ßζȣ¬Èøü¶àµÄ·Ö×Ó±ä³É»î»¯·Ö×Ó

C£®Ñ°ÇóÄÜÔÚ¸üµÍµÄζÈÏÂÓкÜÇ¿´ß»¯»îÐÔµÄÐÂÐÍ´ß»¯¼Á

D£®Ñ°ÇóÐÂÐÍÄ͸ßѹ²ÄÁÏ£¬½«Ñ¹Ç¿Ôö´óÒ»±¶

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


Mg(OH)2ÄÑÈÜÓÚË®£¬µ«ËüÈܽâµÄ²¿·ÖÈ«²¿µçÀë¡£ÊÒÎÂÏÂʱ£¬±¥ºÍMg(OH)2ÈÜÒºµÄpH£½11£¬Èô²»¿¼ÂÇKWµÄ±ä»¯£¬Ôò¸ÃζÈÏÂMg(OH)2µÄÈܽâ¶ÈÊǶàÉÙ£¿(ÈÜÒºÃܶÈΪ1.0 g¡¤cm£­3)

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÏÂÁи÷ÖÖÇé¿öÏÂÒ»¶¨ÄÜ´óÁ¿¹²´æµÄÀë×Ó×éΪ(¡¡¡¡)

A£®pH£½7µÄÈÜÒºÖУºFe3£«¡¢Cl£­¡¢Na£«¡¢NO

B£®Ë®µçÀë³öµÄ[H£«]£½1¡Á10£­3 mol¡¤L£­1µÄË®ÈÜÒºÖУºNa£«¡¢CO¡¢Cl£­¡¢K£«

C£®pH£½1µÄË®ÈÜÒºÖУºNH¡¢Cl£­¡¢Mg2£«¡¢SO

D£®Al3£«¡¢HCO¡¢I£­¡¢Ca2£«

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸