16£®Ä³Í¸Ã÷ÈÜÒºÖнö¿ÉÄܺ¬ÓÐÏÂÁÐÀë×ÓÖеļ¸ÖÖ£ºFe2+¡¢${NH}_{4}^{+}$¡¢K+¡¢Al3+¡¢Ba2+¡¢Cl-¡¢${NO}_{2}^{-}$¡¢${CO}_{3}^{2-}$¡¢${SO}_{4}^{2-}$¡¢${MnO}_{4}^{-}$½«¸ÃÈÜÒº·Ö³ÉÈýµÈ·Ý£¬½øÐÐÈçÏÂʵÑ飺
¢ÙÔÚµÚÒ»·ÝÈÜÒºÖмÓÈë×ãÁ¿Ï¡ÑÎËᣬÁ¢¼´³öÏÖ´óÁ¿µÄºì×ØÉ«ÆøÌ壬ͬʱµÃµ½ÎÞÉ«ÈÜÒº£®
¢ÚÔÚµÚ¶þ·ÝÈÜÒºÖмÓÈë×ãÁ¿µÄÇâÑõ»¯ÄÆÈÜÒº£¬³ä·Ö¼ÓÈȺóÊÕ¼¯µ½2.24LÆøÌ壨±ê×¼×´¿öÏ£©£¬ÔÙÔÚËùµÃµÄÈÜÒºÖÐͨÈë¹ýÁ¿µÄ¶þÑõ»¯Ì¼£¬×îÖյõ½7.8g°×É«³Áµí£®
¢ÛÔÚµÚÈý·ÝÈÜÒºÖмÓÈë×ãÁ¿µÄÂÈ»¯±µÈÜÒº£¬×îÖյõ½46.6g°×É«³Áµí£®
ÊԻشðÏÂÁÐÎÊÌ⣺
£¨1£©Ð´³öʵÑé¢ÙÖвúÉúºì×ØÉ«ÆøÌåµÄÀë×Ó·½³Ìʽ£º2H++2NO2-¨TNO¡ü+NO2¡ü+H2O£»
£¨2£©Ð´³öʵÑé¢ÚÖвúÉú°×É«³ÁµíµÄÀë×Ó·½³Ìʽ£ºAlO2-+CO2+2H2O=Al£¨OH£©3¡ý+HCO3-£»
£¨3£©Ô­ÈÜÒºÖп϶¨´æÔÚµÄÀë×ÓÊÇNO2-¡¢Al3+¡¢SO42-¡¢K+¡¢NH4+£»
£¨4£©Ô­ÈÜÒºÖв»ÄÜÈ·¶¨ÊÇ·ñ´æÔÚµÄÀë×ÓÊÇCl-£¬ÊÔÉè¼ÆʵÑé·½°¸ÑéÖ¤¸ÃÀë×ÓÊÇ·ñ´æÔÚ£º¿ÉÈ¡ÉÙÁ¿Ô­ÈÜÒºÓÚÊÔ¹ÜÖУ¬µÎ¼Ó×ãÁ¿µÄÏõËá±µÈÜÒº£¬¹ýÂË£¬ÔÙÔÚÂËÒºÖеμÓÏõËáËữµÄÏõËáÒøÈÜÒº£¬ÈçÉú³É³Áµí£¬ËµÃ÷º¬ÓÐCl-£¬ÈçÎÞ³ÁµíÉú³É£¬ËµÃ÷²»º¬Cl-£®

·ÖÎö ¢ÙÔÚµÚÒ»·ÝÈÜÒºÖмÓÈë×ãÁ¿Ï¡ÑÎËᣬÁ¢¼´³öÏÖ´óÁ¿µÄºì×ØÉ«ÆøÌ壬˵Ã÷Éú³ÉNO2£¬ÔòÓ¦´æÔÚNO2-£¬ÔòÈÜÒºÖв»´æÔÚFe2+¡¢MnO4-µÈÀë×Ó£»
¢ÚÔÚµÚ¶þ·ÝÈÜÒºÖмÓÈë×ãÁ¿µÄÇâÑõ»¯ÄÆÈÜÒº£¬³ä·Ö¼ÓÈȺóÊÕ¼¯µ½2.24LÆøÌ壨±ê×¼×´¿öÏ£©£¬Ó¦Éú³É°±Æø£¬ËµÃ÷º¬ÓÐNH4+£¬ÇÒΪ0.1mol£¬ÔÙÔÚËùµÃµÄÈÜÒºÖÐͨÈë¹ýÁ¿µÄ¶þÑõ»¯Ì¼£¬×îÖյõ½7.8g°×É«³Áµí£¬Ó¦ÎªAl£¨OH£©3³Áµí£¬Îª0.1mol£¬ËµÃ÷º¬ÓÐAl3+£¬ÔòÒ»¶¨²»º¬ÓÐCO32-µÈÀë×Ó£»
¢ÛÔÚµÚÈý·ÝÈÜÒºÖмÓÈë×ãÁ¿µÄÂÈ»¯±µÈÜÒº£¬×îÖյõ½46.6g°×É«³Áµí£¬Ó¦ÎªBaSO4£¬Îª0.2mol£¬ËµÃ÷º¬ÓÐSO42-£¬Ôò²»º¬ÓÐBa2+£¬
´æÔÚ3n£¨Al3+£©+n£¨NH4+£©=2n£¨SO42-£©£¬¼´3¡Á0.1mol+0.2mol=2¡Á0.2mol£¬µ«Òò´æÔÚNO2-£¬ÔòÓɵçºÉÊغã¿ÉÖªÒ»¶¨º¬ÓÐK+£¬ÒԴ˽â´ð¸ÃÌ⣮

½â´ð ½â£º¢ÙÔÚµÚÒ»·ÝÈÜÒºÖмÓÈë×ãÁ¿Ï¡ÑÎËᣬÁ¢¼´³öÏÖ´óÁ¿µÄºì×ØÉ«ÆøÌ壬˵Ã÷Éú³ÉNO2£¬ÔòÓ¦´æÔÚNO2-£¬ÔòÈÜÒºÖв»´æÔÚFe2+¡¢MnO4-µÈÀë×Ó£»
¢ÚÔÚµÚ¶þ·ÝÈÜÒºÖмÓÈë×ãÁ¿µÄÇâÑõ»¯ÄÆÈÜÒº£¬³ä·Ö¼ÓÈȺóÊÕ¼¯µ½2.24LÆøÌ壨±ê×¼×´¿öÏ£©£¬Ó¦Éú³É°±Æø£¬ËµÃ÷º¬ÓÐNH4+£¬ÇÒΪ0.1mol£¬ÔÙÔÚËùµÃµÄÈÜÒºÖÐͨÈë¹ýÁ¿µÄ¶þÑõ»¯Ì¼£¬×îÖյõ½7.8g°×É«³Áµí£¬Ó¦ÎªAl£¨OH£©3³Áµí£¬Îª0.1mol£¬ËµÃ÷º¬ÓÐAl3+£¬ÔòÒ»¶¨²»º¬ÓÐCO32-µÈÀë×Ó£»
¢ÛÔÚµÚÈý·ÝÈÜÒºÖмÓÈë×ãÁ¿µÄÂÈ»¯±µÈÜÒº£¬×îÖյõ½46.6g°×É«³Áµí£¬Ó¦ÎªBaSO4£¬Îª0.2mol£¬ËµÃ÷º¬ÓÐSO42-£¬Ôò²»º¬ÓÐBa2+£¬
´æÔÚ3n£¨Al3+£©+n£¨NH4+£©=2n£¨SO42-£©£¬¼´3¡Á0.1mol+0.2mol=2¡Á0.2mol£¬µ«Òò´æÔÚNO2-£¬ÔòÓɵçºÉÊغã¿ÉÖªÒ»¶¨º¬ÓÐK+£¬
£¨1£©´æÔÚNO2-£¬¼ÓÈëÑÎËá·¢Éú2H++2NO2-¨TNO¡ü+NO2¡ü+H2O£¬¹Ê´ð°¸Îª£º2H++2NO2-¨TNO¡ü+NO2¡ü+H2O£»
£¨2£©ÊµÑé¢ÚÖвúÉú°×É«³ÁµíΪAl£¨OH£©3£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪAlO2-+CO2+2H2O=Al£¨OH£©3¡ý+HCO3-£¬¹Ê´ð°¸Îª£ºAlO2-+CO2+2H2O=Al£¨OH£©3¡ý+HCO3-£»
£¨3£©ÓÉÒÔÉÏ·ÖÎö¿ÉÖª£¬¿Ï¶¨´æÔÚµÄÀë×ÓÓÐNO2-¡¢Al3+¡¢SO42-¡¢K+¡¢NH4+£¬¹Ê´ð°¸Îª£ºNO2-¡¢Al3+¡¢SO42-¡¢K+¡¢NH4+£»
£¨4£©Òò²»ÄÜÈ·¶¨NO2-µÄÁ¿£¬Ôò²»ÄÜÈ·¶¨ÊÇ·ñº¬ÓÐCl-£¬¿ÉÈ¡ÉÙÁ¿Ô­ÈÜÒºÓÚÊÔ¹ÜÖУ¬µÎ¼Ó×ãÁ¿µÄÏõËá±µÈÜÒº£¬¹ýÂË£¬ÔÙÔÚÂËÒºÖеμÓÏõËáËữµÄÏõËáÒøÈÜÒº£¬ÈçÉú³É³Áµí£¬ËµÃ÷º¬ÓÐCl-£¬ÈçÎÞ³ÁµíÉú³É£¬ËµÃ÷²»º¬Cl-£¬
¹Ê´ð°¸Îª£ºCl-£»¿ÉÈ¡ÉÙÁ¿Ô­ÈÜÒºÓÚÊÔ¹ÜÖУ¬µÎ¼Ó×ãÁ¿µÄÏõËá±µÈÜÒº£¬¹ýÂË£¬ÔÙÔÚÂËÒºÖеμÓÏõËáËữµÄÏõËáÒøÈÜÒº£¬ÈçÉú³É³Áµí£¬ËµÃ÷º¬ÓÐCl-£¬ÈçÎÞ³ÁµíÉú³É£¬ËµÃ÷²»º¬Cl-£®

µãÆÀ ±¾Ì⿼²éÁ˳£¼ûÀë×ӵļìÑé¡¢Àë×Ó¹²´æµÄÅжϣ¬ÌâÄ¿ÄѶÈÖеȣ¬×¢ÒâÕÆÎÕ³£¼ûÀë×ӵĻ¯Ñ§ÐÔÖʼ°¼ìÑé·½·¨£¬Äܹ»¸ù¾ÝµçºÉÊغãƽºâÈÜÒºÖÐδ֪Àë×ӵĴæÔÚÇé¿ö£¬ÕýÈ·Àí½âÌâ¸ÉÐÅÏ¢Êǽâ´ð±¾ÌâµÄ¹Ø¼ü£¬Ò×´íµãΪ¼ØÀë×ÓµÄÅжϣ¬´ðÌâʱעÒ⣮

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

6£®ÏÂÁÐÎïÖʲ»ÊôÓڸ߷Ö×Ó»¯ºÏÎïµÄÊÇ£¨¡¡¡¡£©
A£®ÓÍÖ¬B£®ÑòëC£®µí·ÛD£®Ïð½º

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

7£®ÏÂÁз´Ó¦ÖУ¬ÏõËá¼È±íÏÖËáÐÔ£¬ÓÖ±íÏÖÑõ»¯ÐÔµÄÊÇ£¨¡¡¡¡£©
A£®Fe2O3¸úÏ¡ÏõËá·´Ó¦B£®Al£¨OH£©3¸úÏ¡ÏõËá·´Ó¦
C£®CuO¸úÏ¡ÏõËá·´Ó¦D£®Fe£¨OH£©2¸úÏ¡ÏõËá·´Ó¦

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

4£®»¯Ñ§ÓëÈËÀàµÄÉú²ú¡¢Éú»îÃÜÇÐÏà¹Ø£¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®ÔÚ¿ÕÆøÖУ¬¸ÖÌúÖÆÆ·ºÍÂÁÖÆÆ·¶¼ÈÝÒ×·¢ÉúÎüÑõ¸¯Ê´
B£®¾ÛÒÒÏ©ËÜÁÏÒ×ÀÏ»¯£¬ÊÇÒòΪ·¢ÉúÁ˼ӳɷ´Ó¦
C£®Ê³Æ·°ü×°´üÖг£·ÅÈëС´üµÄÉúʯ»Ò£¬Ä¿µÄÊÇ·ÀֹʳƷÑõ»¯±äÖÊ
D£®¿ÕÆøÖÊÁ¿Ö¸±êÖÐÓÐÒ»ÏîÖ¸±êÃû³ÆΪPM2.5£¬Ö¸µÄÊÇ¿ÕÆøÖÐÖ±¾¶¡Ü2.5΢Ã׵ĹÌÌå»òÒºÌåµÄ×ܳÆ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

11£®ÓÃCaSO4´úÌæO2ÓëȼÁÏCO·´Ó¦£¬¼È¿ÉÒÔÌá¸ßȼÉÕЧÂÊ£¬ÓÖÄܵõ½¸ß´¿CO2£¬ÊÇÒ»ÖÖ¸ßЧ¡¢Çå½à¡¢¾­¼ÃµÄÐÂÐÍȼÉÕ¼¼Êõ£¬·´Ó¦¢ÙΪÖ÷·´Ó¦£¬·´Ó¦¢ÚºÍ¢ÛΪ¸±·´Ó¦£®
¢Ù$\frac{1}{4}$CaSO4£¨s£©+CO£¨g£©?$\frac{1}{4}$CaS£¨s£©+CO2£¨g£©¡÷H1=-47.3kJ/mol
¢ÚCaSO4£¨s£©+CO£¨g£©?CaO£¨s£©+CO2£¨g£©+SO2£¨g£©¡÷H2=+210.5kJ/mol
¢ÛCO£¨g£©?$\frac{1}{2}$C£¨s£©+$\frac{1}{2}$CO2£¨g£©¡÷H3=-86.2kJ/mol
£¨1£©·´Ó¦2CaSO4£¨s£©+7CO£¨g£©?£¨s£©+CaO£¨s£©+C£¨s£©+6CO2£¨g£©+SO2£¨g£©µÄ¡÷H=4¡÷H1+¡÷H2+2¡÷H3£¨Óá÷H1¡÷H2¡÷H3±íʾ£©£®
£¨2£©·´Ó¦¢Ù¡«¢ÛµÄƽºâ³£ÊýµÄ¶ÔÊýlgKË淴ӦζÈTµÄ±ä»¯ÇúÏß¼ûͼ18£®½áºÏ¸÷·´Ó¦µÄ¡÷H£¬¹éÄÉlgK¡«TÇúÏ߱仯¹æÂÉ£º
a£©·ÅÈÈ·´Ó¦µÄlgKËæζÈÉý¸ß¶øϽµ
b£©·Å³ö»òÎüÊÕÈÈÁ¿Ô½´óµÄ·´Ó¦£¬ÆälgKÊÜζÈÓ°ÏìÔ½´ó
£¨3£©ÏòÊ¢ÓÐCaSO4µÄÕæ¿ÕºãÈÝÈÝÆ÷ÖгäÈëCO£¬·´Ó¦¢ÙÓÚ900¡ãC´ïµ½Æ½ºâ£¬cƽºâ£¨CO£©=8.0¡Á10-5mol•L-1£¬¼ÆËãCOµÄת»¯ÂÊ99%£¨ºöÂÔ¸±·´Ó¦£¬½á¹û±£Áô2λÓÐЧÊý×Ö£©£®
£¨4£©Îª¼õÉÙ¸±²úÎ»ñµÃ¸ü´¿¾»µÄCO2£¬¿ÉÔÚ³õʼȼÁÏÖÐÊÊÁ¿¼ÓÈëO2£®
£¨5£©ÒÔ·´Ó¦¢ÙÖÐÉú³ÉµÄCaSΪԭÁÏ£¬ÔÚÒ»¶¨Ìõ¼þϾ­Ô­×ÓÀûÓÃÂÊ100%µÄ¸ßη´Ó¦£¬¿ÉÔÙÉú³ÉCaSO4£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪCaS+2O2$\frac{\underline{\;Ò»¶¨Ìõ¼þ\;}}{\;}$ CaSO4£»ÔÚÒ»¶¨Ìõ¼þÏÂCO2¿ÉÓë¶Ô¶þ¼×±½·´Ó¦£¬ÔÚÆä±½»·ÉÏÒýÈëÒ»¸öôÈ»ù£¬²úÎïµÄ½á¹¹¼òʽΪ£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

1£®Í¼ÖÐA¡¢BΪ¶à¿×¶èÐԵ缫£¬C¡¢DΪ¼ÐÔÚ½þ¹ýNa2SO4ÈÜÒººÍ·Ó̪µÄʪÂËÖ½ÌõµÄ²¬¼Ð£¬a¡¢bΪµçÔ´Á½¼«£®ÊµÑéÇ°£¬ÍùÁ½Ö§ÊÔ¹ÜÖгäÂúKOHÈÜÒº²¢µ¹Á¢ÓÚÊ¢ÓÐKOHÈÜÒºµÄË®²ÛÖУ®ÊµÑéʱ£¬¶Ï¿ªK1£¬±ÕºÏK2¡¢K3£¬Í¨µçÒ»¶Îʱ¼ä£¬Á½Ö§ÊÔ¹ÜÊÕ¼¯µÄÆøÌåÌå»ýÈçͼËùʾ£®¸ù¾ÝÒªÇó»Ø´ðÎÊÌ⣺

£¨1£©µçÔ´µÄa¼«Îª¸º¼«£¨Ìî¡°Õý¡±»ò¡°¸º¡±£©£»
£¨2£©²¬¼ÐCÉÏ·´Ó¦µÄµç¼«·´Ó¦Ê½Îª4H++4e=2H2¡ü£»
£¨3£©×°ÖÃֱͨÁ÷µçʱ£¬ÏÂÁÐÅжϲ»ÕýÈ·µÄÊÇ¢Ú
¢Ùµç¼«AÉÏ·¢ÉúÑõ»¯·´Ó¦            ¢Úµç×ÓÑØa¡úC¡úD¡úb·¾¶Á÷¶¯
¢ÛƬ¿ÌºóDµã¸½½üc£¨SO42-£©Ôö´ó     ¢ÜƬ¿Ìºó¿É¹Û²ìµ½ÂËÖ½Cµã±äºìÉ«£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

8£®ÔÚ10LÃܱÕÈÝÆ÷ÖзÅÈë0.50 mol X£¬ÔÚÒ»¶¨Î¶ÈÏ·¢Éú·´Ó¦£ºX£¨g£©?2Y£¨g£©+Z£¨s£©¡÷H£¾0£¬ÈÝÆ÷ÄÚÆøÌå×ÜѹǿpË淴Ӧʱ¼ätµÄ±ä»¯¹ØϵÈçͼËùʾ£®ÒÔÏ·ÖÎöÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®¸ÃζÈÏ´˷´Ó¦µÄƽºâ³£ÊýK=0.64mol/L
B£®´Ó·´Ó¦¿ªÊ¼µ½t1ʱµÄƽ¾ù·´Ó¦ËÙÂÊv£¨X£©=0.008mol/£¨L•min£©
C£®ÓûÌá¸ßƽºâÌåϵÖÐYµÄÌå»ý·ÖÊý£¬¿ÉÉý¸ßÌåϵζȻò¼õÉÙZµÄÁ¿
D£®ÆäËûÌõ¼þ²»±ä£¬ÔÙ³äÈë0.1 mol ÆøÌåX£¬Æ½ºâÕýÏòÒƶ¯£¬XµÄת»¯ÂÊÔö´ó

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

13£®ÏÂÁÐÎïÖʵÄÀà±ðÓëËùº¬ÄÜÍŶ¼ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
¢Ù·ÓÀà-OH
¢ÚôÈËáÀà-COOH
¢ÛÈ©Àà
¢ÜCH3-O-CH3 ÃÑÀà
¢Ýõ¥Àà
A£®¢Ù¢Ú¢Û¢Ü¢ÝB£®¢Ú¢ÜC£®¢Ú¢Û¢ÜD£®¢Ú¢Ü¢Ý

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

14£®ÌúÓëÌúºÏ½ðÊÇÉú»îÖг£¼ûµÄ²ÄÁÏ£¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®ÌúÓëÑÎËá·´Ó¦£¬ÌúºÏ½ð²»ÓëÑÎËá·´Ó¦
B£®´¿ÌúµÄÓ²¶ÈºÍÇ¿¶È¾ù¸ßÓÚÌúºÏ½ð
C£®²»Ðâ¸ÖÊÇÌúºÏ½ð£¬Ö»º¬½ðÊôÔªËØ
D£®Ò»¶¨Ìõ¼þÏ£¬Ìú·Û¿ÉÓëË®ÕôÆø·´Ó¦

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸