ÌâÄ¿ÄÚÈÝ

6£®ÏÂÁз´Ó¦µÄÀë×Ó·½³ÌʽÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®ÖÆ×÷Ó¡Ë¢µç·°å£ºFe3++Cu¨TCu2++Fe2+
B£®µÈÎïÖʵÄÁ¿µÄBa£¨OH£©2ÓëNH4HSO4ÔÚÏ¡ÈÜÒºÖз´Ó¦£ºBa2++2OH-+2H++SO42-¨TBaSO4¡ý+2H2O
C£®Ca£¨OH£©2ÈÜÒºÓë×ãÁ¿Ca£¨HCO3£©2ÈÜÒº·´Ó¦£ºCa2++2HCO3-+2OH-¨T2CaCO3¡ý+2H2O+CO32-
D£®ÐÂÖÆ±¥ºÍÂÈË®ÖмÓÈëʯ»Òʯ¿ÉÌá¸ßÈÜÒºÖÐHClOŨ¶È£ºCaCO3+2Cl2+H2O¨TCa2++2Cl-+CO2¡ü+2H2O

·ÖÎö A£®Àë×Ó·½³ÌʽÁ½±ßÕýµçºÉ²»ÏàµÈ£¬Î¥·´Á˵çºÉÊØºã£»
B£®¶þÕßÎïÖʵÄÁ¿ÏàµÈʱ£¬ï§¸ùÀë×ÓÒ²²ÎÓë·´Ó¦£»
C£®Ì¼ËáÇâ¸Æ×ãÁ¿£¬·´Ó¦²úÎïÖв»»á´æÔÚ̼Ëá¸ùÀë×Ó£»
D£®HClÓë̼Ëá¸Æ·´Ó¦£¬´Ù½øÁËÂÈÆøÓëË®µÄ·´Ó¦£¬ÓÐÀûÓÚHClOµÄÉú³É£®

½â´ð ½â£ºA£®ÌúÀë×ÓÓëÍ­·´Ó¦Éú³ÉÑÇÌúÀë×ÓºÍÍ­Àë×Ó£¬ÕýÈ·µÄÀë×Ó·½³ÌʽΪ£ºFe3++Cu¨TCu2++Fe2+£¬¹ÊA´íÎó£»
B£®µÈÎïÖʵÄÁ¿µÄBa£¨OH£©2ÓëNH4HSO4ÔÚÏ¡ÈÜÒºÖз´Ó¦£ºNH4++Ba2++2OH-+H++SO42-¨TBaSO4¡ý+H2O+NH3•H2O£¬¹ÊB´íÎó£»
C£®Ca£¨OH£©2ÈÜÒºÓë×ãÁ¿Ca£¨HCO3£©2ÈÜÒº·´Ó¦Éú³É̼Ëá¸Æ³ÁµíºÍË®£¬ÕýÈ·µÄÀë×Ó·½³ÌʽΪ£ºCa2++HCO3-+OH-¨TCaCO3¡ý+H2O+£¬¹ÊC´íÎó£»
D£®ÐÂÖÆ±¥ºÍÂÈË®ÖмÓÈëʯ»Òʯ¿ÉÌá¸ßÈÜÒºÖÐHClOŨ¶È£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪ£ºCaCO3+2Cl2+H2O¨TCa2++2Cl-+CO2¡ü+2H2O£¬¹ÊDÕýÈ·£»
¹ÊÑ¡D£®

µãÆÀ ±¾Ì⿼²éÁËÀë×Ó·½³ÌʽµÄÊéдÅжϣ¬Îª¸ß¿¼µÄ¸ßƵÌ⣬ÌâÄ¿ÄѶȲ»´ó£¬×¢ÒâÃ÷È·Àë×Ó·½³ÌʽÕýÎóÅжϳ£Ó÷½·¨£º¼ì²é·´Ó¦Îï¡¢Éú³ÉÎïÊÇ·ñÕýÈ·£¬¼ì²é¸÷ÎïÖʲð·ÖÊÇ·ñÕýÈ·£¬ÈçÄÑÈÜÎï¡¢Èõµç½âÖʵÈÐèÒª±£Áô»¯Ñ§Ê½£¬¼ì²éÊÇ·ñ·ûºÏÊØºã¹ØÏµ£¨È磺ÖÊÁ¿ÊغãºÍµçºÉÊØºãµÈ£©µÈ£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
19£®ÀûÓÃÄøîâ¿ó£¨Ö÷Òªº¬ÓÐMoS2¡¢NiS2¡¢NiS¡¢FeS2¡¢SiO2ºÍC£©Öк¬ÓеÄ̼×÷»¹Ô­¼Á£¬½øÐÐÑ¡ÔñÐÔ»¹Ô­ÈÛÁ¶£¬¿ÉÒÔÌáÈ¡ÄøÌúºÏ½ð£¬Í¬Ê±µÃµ½îâËáÄÆ¾§Ì壨Na2MoO4•2H2O£©£¬ÆäÖ÷ÒªÁ÷³ÌÈçͼ£º

£¨1£©Ñ¡ÔñÐÔ»¹Ô­ÈÛÁ¶µÄÓŵãÖ®Ò»Êǽ«SÔªËØ×ª»¯ÎªNa2S£¬±ÜÃâÁËÎÛȾÐÔÆøÌåSO2µÄ²úÉú£®¼ìÑéÂ¯ÆøÖÐÊÇ·ñº¬ÓÐSO2µÄ·½·¨Îª½«Â¯ÆøÍ¨ÈëÆ·ºìÖУ¬ÈôÆ·ºìÍÊÉ«£¬¼ÓÈȺóÓÖ»Ö¸´ºìÉ«£¬Ö¤Ã÷Â¯ÆøÖÐÓÐSO2£¬ÈôÆ·ºì²»ÍÊÉ«£¬ÔñÂ¯ÆøÖв»º¬ÓÐSO2
£¨2£©»¹Ô­ÈÛÁ¶Ê±£¬FeS2·¢ÉúµÄ·´Ó¦Îª3FeS2+6Na2CO3+11C+8O2 $\frac{\underline{\;¸ßÎÂ\;}}{\;}$3Fe+6Na2S+17CO2£¬¸Ã·´Ó¦ÖÐÑõ»¯²úÎïÊÇ£¬·´Ó¦Ã¿×ªÒÆ11mol e-£¬Éú³ÉÌúµÄÖÊÁ¿Îª42g£®
£¨3£©Ë®½þÒºÖÐͨÈëÊÊÁ¿CO2ºÍ¿ÕÆø£¬È»ºó¹ýÂË£¬ËùµÃÂËÔüµÄÖ÷Òª³É·ÖΪH2SiO3£®
£¨4£©ÖªîâËáÄÆÈÜÒºÖÐc£¨MoO42-£©=0.4 0mol•L-1£¬c£¨CO32-£©=0.10mol•L-1£®ÓÉîâËáÄÆÈÜÒºÖÆ±¸îâËáÄÆ¾§Ìåʱ£¬Ðè¼ÓÈëBa£¨OH£©2¹ÌÌåÒÔ³ýÈ¥CO32-£®µ±BaMoO4¿ªÊ¼³Áµíʱ£¬CO32-µÄÈ¥³ýÂÊÊÇ90%[¼ºÖªKsp£¨BaCO3£©=1¡Á10-9¡¢Ksp£¨BaMoO4£©=4.0¡Á10-8£¬ºöÂÔÈÜÒºµÄÌå»ý±ä»¯]
£¨5£©îâËáÄÆºÍÔ¹ðõ£¼¡°±ËáµÄ»ìºÏÒº³£×÷ÎªÌ¼ËØ¸ÖµÄ»ºÊ´¼Á£®³£ÎÂÏ£¬Ì¼ËظÖÔÚÈýÖÖ²»Í¬½éÖÊÖеĸ¯Ê´ËÙÂÊʵÑé½á¹ûÈçͼ£º

¢ÙÒªÊ¹Ì¼ËØ¸ÖµÄ»ºÊ´Ð§¹û×îÓÅ£¬îâËáÄÆºÍÔ¹ðõ£¼¡°±ËáµÄŨ¶È±ÈÓ¦1£º1
¢Úµ±ÁòËáµÄŨ¶È´óÓÚ90%ʱ£¬¸¯Ê´ËÙÂʼ¸ºõΪÁ㣬ԭÒòÊdz£ÎÂÏÂŨÁòËá¾ßÓÐÇ¿Ñõ»¯ÐÔ£¬»áʹÌú¶Û»¯
¢ÛËæ×ÅÑÎËáºÍÁòËáŨ¶ÈµÄÔö´ó£¬Ì¼ËظÖÔÚÑÎËáÖеĸ¯Ê´ËÙÂÊÃ÷ÏԼӿ죬ÆäÔ­Òò¿ÉÄÜÊÇ£¬ÇëÉè¼ÆÊµÑéÖ¤Ã÷ÉÏÊö½áÂÛ£ºÏòÉÏÊöÁòËáÈÜÒºÖмÓÈëÉÙÁ¿ÂÈ»¯ÄƵȿÉÈÜÐÔÂÈ»¯Îï£¬Ì¼ËØ¸ÖµÄ¸¯Ê´ËÙÂÊÃ÷ÏԼӿ죮
14£®¸ßÂ¯ÃºÆøÎªÁ¶Ìú¹ý³ÌÖвúÉúµÄ¸±²úÆ·£¬Ö÷Òª³É·ÖΪCO¡¢CO2¡¢N2µÈ£¬ÆäÖÐCOº¬Á¿Ô¼Õ¼25%×óÓÒ£®
£¨1£©Ñо¿±íÃ÷£¬ÓÉCO¿ÉÒÔÖ±½ÓºÏ³É¶àÖÖ»¯¹¤²úÆ·£¬ÈçÉú²ú¼×´¼£®
ÒÑÖª£º2CH3OH£¨l£©+3O2£¨g£©¨T2CO2£¨g£©+4H2O£¨l£©¡÷H=-1453kJ•mol-1
2H2£¨g£©+O2£¨g£©¨T2H2O£¨l£©¡÷H=-571.6kJ•mol-1
CO£¨g£©+2H2£¨g£©¨TCH3OH£¨l£©¡÷H=-128.1kJ•mol-1£®ÔòCO£¨g£©È¼ÉÕÈÈΪ283 kJ•mol-1£®
£¨2£©ÈËÌåÖм¡ºìµ°°×£¨Mb£©ÓëѪºìµ°°×£¨Hb£©µÄÖ÷Òª¹¦ÄÜΪÔËÊäÑõÆøÓë¶þÑõ»¯Ì¼£®¼¡ºìµ°°×£¨Mb£©¿ÉÒÔÓëС·Ö×ÓX£¨ÈçÑõÆø»òÒ»Ñõ»¯Ì¼£©½áºÏ£¬·´Ó¦·½³ÌʽΪ£ºMb£¨aq£©+X£¨g£©?MbX£¨aq£©
¢Ùͨ³£ÓÃp ±íʾ·Ö×ÓX µÄѹÁ¦£¬p0 ±íʾ±ê׼״̬´óÆøÑ¹£¬ÈôX ·Ö×ӵį½ºâŨ¶ÈΪ$\frac{p}{{p}_{0}}$£¬Ð´³öÉÏÊö·´Ó¦µÄƽºâ³£Êý±í´ïʽ£ºK=$\frac{c£¨MbX£©{P}_{0}}{c£¨Mb£©P}$£®ÇëÓÃ$\frac{p}{{p}_{0}}$¼°K ±íʾÎü¸½Ð¡·Ö×ӵļ¡ºìµ°°×£¨MbX£©Õ¼×ܼ¡ºìµ°°×µÄ±ÈÀý£®
¢ÚÔÚ³£ÎÂÏ£¬¼¡ºìµ°°×ÓëCO ½áºÏ·´Ó¦µÄƽºâ³£ÊýK£¨CO£©Ô¶´óÓÚÓëO2½áºÏµÄƽºâ³£ÊýK£¨O2£©£¬ÈçͼÄÜ´ú±í½áºÏÂÊ£¨f£©Óë´ËÁ½ÖÖÆøÌåѹÁ¦£¨p£©µÄ¹ØÏµµÄÊÇC

¢ÛÈËÌåÖеÄѪºìµ°°×£¨Hb£©Í¬ÑùÄÜÎü¸½O2¡¢CO2ºÍH+£¬Ïà¹Ø·´Ó¦·½³Ìʽ¼°Æä»¯Ñ§Æ½ºâ³£Êý·Ö±ðÊÇ£º
¢ñ£®Hb£¨aq£©+H+£¨aq£©?HbH+£¨aq£©£»K1
¢ò£®HbH+£¨aq£©+O2£¨g£©?HbHO2£¨aq£©+H+£»K2
¢ó£®Hb£¨aq£©+O2£¨g£©?HbO2£¨aq£©K3
¢ô£®HbO2£¨aq£©+H+£¨aq£©+CO2£¨g£©?Hb£¨H+£©CO2£¨aq£©+O2£¨g£©£»¡÷H4
ÔòK3=K1•K2£¨ÓÃK1¡¢K2±íʾ£©£¬ÈôµÍÎÂÏ·´Ó¦¢ôÄÜ×Ô·¢½øÐУ¬Ôò¸Ã·´Ó¦¡÷H£¼0£¬¡÷S£¼  0£¨Ìî¡°£¾¡±¡°£¼¡±»ò¡°=¡±£©
£¨3£©CO2µÄ´¦Àí ·½·¨ÓжàÖÖ£¬½«ÉÙÁ¿CO2ÆøÌåͨÈëʯ»ÒÈéÖгä·Ö·´Ó¦£¬´ïµ½Æ½ºâºó£¬²âµÃÈÜÒºÖÐc£¨OH-£©=cmol/L£¬Ôòc£¨CO32-£©=$\frac{a{c}^{2}}{b}$mol/L£¨Óú¬a¡¢b¡¢cµÄ´úÊýʽ±íʾ£©£®£¨ÒÑÖªKap£¨CaCO3£©=a£¬Kap[Ca£¨OH£©2]=b£©

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

¾«Ó¢¼Ò½ÌÍø