ÌâÄ¿ÄÚÈÝ
3£®ÒÑÖªAÊǵ¥ÖÊ£¬A¡¢B¡¢C¡¢D¡¢E 5ÖÖÎïÖʾùº¬Í¬Ò»ÖÖÔªËØ£¬XÊǵؿÇÖк¬Á¿×î¶àµÄÔªËØÐγɵĵ¥ÖÊ£¬ËüÃǵÄÏ໥ת»¯¹ØϵÈçͼËùʾ£®ÊԻشðÏÂÁÐÎÊÌ⣮£¨1£©Í¨³£Çé¿öÏ£¬ÈôAΪÆøÌ壬C¡¢D¶¼ÊÇ´óÆøÎÛȾÎ
¢Ùд³öÏÂÁз´Ó¦µÄ»¯Ñ§·½³Ìʽ£ºD¡úE3NO2+H2O=NO+2HNO3£®
¢Ú¹¤ÒµÉÏÖÆÈ¡BµÄ»¯Ñ§·½³Ìʽ£ºN2+3H2$?_{¸ßθßѹ}^{´ß»¯¼Á}$2NH3£®
¢ÛʵÑéÊÒÖмìÑéBµÄ²Ù×÷·½·¨ÓÃʪÈóµÄºìɫʯÈïÊÔÖ½¿¿½üÊԹܿڣ¬ÊÔÖ½±äÀ¶£®
¢Ü±ê×¼×´¿öÏ£¬½«Ê¢ÂúDµÄÊԹܵ¹¿ÛÔÚÊ¢ÂúË®µÄË®²ÛÖУ¬Ò»¶Îʱ¼äºó£¬¼Ù¶¨ÈÜÖʲ»À©É¢£¬ÔòÊÔ¹ÜÖÐËùµÃÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈΪ0.045mol/L£®
£¨2£©Í¨³£Çé¿öÏ£¬ÈôAΪµ»ÆÉ«¹ÌÌ壺
¢Ùд³öBµÄÈÜÒºÓëC·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º2H2S+SO2=3S¡ý+2H2O£®
¢Ú½«CͨÈëäåË®ÖеÄÏÖÏóÊÇäåË®ÍÊÉ«£¬·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽÊÇSO2+Br2+2H2O=4H++SO42-+2Br-£®
·ÖÎö XÊǵؿÇÖк¬Á¿×î¶àµÄÔªËØÐγɵĵ¥ÖÊ£¬Ó¦ÎªO2£¬·ûºÏת»¯¹ØϵµÄӦΪS»òN2£¬
£¨1£©Í¨³£Çé¿öÏ£¬ÈôAΪÆøÌ壬C¡¢D¶¼ÊÇ´óÆøÎÛȾÎÔòAΪN2£¬BΪNH3£¬CΪNO£¬DΪNO2£¬EΪHNO3£»
£¨2£©Í¨³£Çé¿öÏ£¬ÈôAΪµ»ÆÉ«¹ÌÌ壬ӦΪS£¬ÔòBΪH2S£¬CΪSO2£¬DΪSO3¡¢EΪH2SO4£¬½áºÏ¶ÔÓ¦ÎïÖʵÄÐÔÖÊÒÔ¼°ÌâÄ¿ÒªÇó½â´ð¸ÃÌ⣮
½â´ð ½â£ºXÊǵؿÇÖк¬Á¿×î¶àµÄÔªËØÐγɵĵ¥ÖÊ£¬Ó¦ÎªO2£¬·ûºÏת»¯¹ØϵµÄӦΪS»òN2£¬
£¨1£©Í¨³£Çé¿öÏ£¬ÈôAΪÆøÌ壬C¡¢D¶¼ÊÇ´óÆøÎÛȾÎÔòAΪN2£¬BΪNH3£¬CΪNO£¬DΪNO2£¬EΪHNO3£¬
¢ÙD¡úEΪ¶þÑõ»¯µªÓëË®ÖÆÏõËáµÄ·´Ó¦£¬·½³ÌʽΪ3NO2+H2O=NO+2HNO3£¬
¹Ê´ð°¸Îª£º3NO2+H2O=NO+2HNO3£»
¢Ú¹¤ÒµÉÏÓõªÆøºÍÇâÆøºÏ³É°±£¬ÖÆÈ¡BµÄ»¯Ñ§·½³ÌʽΪN2+3H2$?_{¸ßθßѹ}^{´ß»¯¼Á}$2NH3£¬
¹Ê´ð°¸Îª£ºN2+3H2$?_{¸ßθßѹ}^{´ß»¯¼Á}$2NH3£»
¢Û°±ÆøΪ¼îÐÔÆøÌ壬ÓëË®·´Ó¦Éú³ÉNH3•H2O£¬µçÀë×Ó³öOH-Àë×Ó£¬ÈÜÒº³Ê¼îÐÔ£¬¿ÉÓÃʪÈóµÄºìɫʯÈïÊÔÖ½¼ìÑ飬ÈçÊÔÖ½±äÀ¶É«£¬Ôò˵Ã÷Óа±ÆøÉú³É£¬
¹Ê´ð°¸£ºÓÃʪÈóµÄºìɫʯÈïÊÔÖ½¿¿½üÊԹܿڣ¬ÊÔÖ½±äÀ¶£»
¢Ü¼ÙÉè¸ÃÈÝÆ÷µÄÈÝ»ýΪ3L£¬Ôò¶þÑõ»¯µªµÄÌå»ýΪ3L£®¶þÑõ»¯µªºÍË®·´Ó¦µÄ·½³ÌʽΪ£º
3NO2+H2O=2HNO3+NO£¬
3¡Á22.4L 1¡Á22.4L
3L 1L
·´Ó¦Ç°ºóÆøÌåµÄÌå»ýÓÉ3L±äΪ1L£¬ËùÒÔÈÜÒºµÄÌå»ýΪ2L£»¸ÃÈÜÒºµÄÈÜÖÊΪÏõËᣬ
3NO2+H2O=2HNO3+NO£¬
3¡Á22.4L 2mol
3L $\frac{5}{56}$mol
ËùÒÔÏõËáµÄÎïÖʵÄÁ¿Å¨¶ÈΪC=$\frac{n}{V}$=0.045mol/L£®
¹Ê´ð°¸Îª£º0.045mol/L£»
£¨2£©Í¨³£Çé¿öÏ£¬ÈôAΪµ»ÆÉ«¹ÌÌ壬ӦΪS£¬ÔòBΪH2S£¬CΪSO2£¬DΪSO3¡¢EΪH2SO4£¬
¢ÙBÓëCΪH2SºÍSO2µÄ·´Ó¦£¬·½³ÌʽΪ2H2S+SO2=3S¡ý+2H2O£¬
¹Ê´ð°¸Îª£º2H2S+SO2=3S¡ý+2H2O£»
¢Ú¶þÑõ»¯Áò¾ßÓл¹ÔÐÔ£¬¿ÉÓëäåË®·¢ÉúÑõ»¯»¹Ô·´Ó¦£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪSO2+Br2+2H2O=4H++SO42-+2Br-£¬¿É¹Û²ìµ½äåË®ÍÊÉ«£¬
¹Ê´ð°¸Îª£ºäåË®ÍÊÉ«£»SO2+Br2+2H2O=4H++SO42-+2Br-£®
µãÆÀ ±¾Ì⿼²éÁËÎÞ»úÎïÍƶϣ¬Îª¸ßƵ¿¼µã£¬²àÖØÓÚѧÉúµÄ·ÖÎöÄÜÁ¦µÄ¿¼²é£¬¸ù¾ÝÎïÖʵÄÑÕÉ«¡¢ÎïÖʵÄÐÔÖʽøÐÐÍƶϣ¬AÄÜÁ¬Ðø±»Ñõ»¯£¬ËµÃ÷AÖдæÔÚµÄijÖÖÔªËØÓжàÖÖ»¯ºÏ¼Û£¬ÔÙ½áºÏEµÄÐÔÖÊ·ÖÎö½â´ð£¬ÌâÄ¿ÄѶȲ»´ó£®
A£® | ÓëË®·´Ó¦Éú³ÉÒÒ´¼ | B£® | ÓëäåË®·´Ó¦Ê¹Ö®ÍÊÉ« | ||
C£® | ÓëÇâÆø·´Ó¦Éú³ÉÒÒÍé | D£® | ÓëÑõÆø·´Ó¦Éú³É¶þÑõ»¯Ì¼ºÍË® |
A£® | H2O2×÷Ñõ»¯¼Á | B£® | H2O2ʧȥµç×Ó | C£® | µâÔªËر»»¹Ô | D£® | I-±íÏÖÑõ»¯ÐÔ |
A£® | »ìºÏÆøÌåÖÐÒ»¶¨Óм×Íé | B£® | »ìºÏÆøÌåÖпÉÄÜÊǼ×ÍéºÍÒÒÍé | ||
C£® | »ìºÏÆøÌåÖÐÒ»¶¨Ã»ÓбûÍé | D£® | »ìºÏÆøÌåÖпÉÄÜÓÐÒÒÏ© |
A£® | ½ðÊôµ¥ÖÊÔÚÑõ»¯»¹Ô·´Ó¦ÖÐ×ÜÊÇ×÷»¹Ô¼Á | |
B£® | ·Ç½ðÊôµ¥ÖÊÔÚÑõ»¯»¹Ô·´Ó¦ÖÐ×ÜÊÇ×÷Ñõ»¯¼Á | |
C£® | ÄÆÔ×ÓÔÚÑõ»¯»¹Ô·´Ó¦ÖÐʧȥ1¸öµç×Ó£¬¶øÂÁÔ×Óʧȥ3¸öµç×Ó£¬ËùÒÔÄƵĻ¹ÔÐÔСÓÚÂÁ | |
D£® | ½ðÊôÔªËر»»¹Ôʱһ¶¨Éú³É½ðÊôµ¥ÖÊ |
A£® | v£¨NH3£©=0.003 mol•£¨L•s£©-1 | B£® | v£¨NO£©=0.08 mol•£¨L•s£©-1 | ||
C£® | v£¨H2O£©=0.003 mol•£¨L•s£©-1 | D£® | v£¨O2£©=0.01 mol•£¨L•s£©-1 |