ÌâÄ¿ÄÚÈÝ

£¨14·Ö£©A¡¢B¡¢C¡¢D¡¢E¡¢FÁùÖÖµÄת»¯¹ØϵÈçÓÒͼ£¬ÆäÖÐAΪӦÓÃ×î¹ã·ºµÄ½ðÊô£¬Ñõ»¯ÎïE¼ÈÄÜÓëÑÎËá·´Ó¦£¬ÓÖÄÜÓëNaOHÈÜÒº·´Ó¦¡£X¡¢YÊôÓÚͬһÖÜÆÚ£¬ÇҺ˵çºÉÊýÏà²î4¡£

ÉÏÊöת»¯ÖÐijЩ·´Ó¦Ìõ¼þδָ³ö£¬Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Ö¸³öCÖеĻ¯Ñ§¼üÀàÐÍ£º          £»DµÄ»¯Ñ§Ê½Îª            £»ÔªËØX¡¢YµÄÔ­×Ӱ뾶´óС˳ÐòÊÇ                     £¨ÓÃÔªËØ·ûºÅ±íʾ£©¡£
£¨2£©EÓëNaOHÈÜÒº·´Ó¦µÄÀë×Ó·½³ÌʽΪ                                   ¡£
£¨3£©·´Ó¦A+H2O¡úC+DµÄ»¯Ñ§·½³ÌʽΪ                                    ¡£
£¨4£©FÄÜ´Ù½øH2OµÄµçÀ룬ÆäÔ­ÒòÊÇ                                      £¨ÓÃÀë×Ó·½³Ìʽ±íʾ£©¡£
£¨5£©250Cʱ£¬½«pH£½1µÄÑÎËáÓëpH£½12µÄNaOHÈÜÒº°´Ìå»ý±È1£º9»ìºÏ£¨»ìºÏʱÈÜÒºÌå»ý±ä»¯ºöÂÔ²»¼Æ£©£¬»ìºÏ¾ùÔȺóËùµÃÈÜÒºµÄpH£½          ¡£
ÿ¿Õ2·Ö£¬¹²14·Ö
£¨1£©¹²¼Û¼ü£»   Fe3O4£»      Al£¾Cl     (¸÷2·Ö)
£¨2£©Al2O3+2OH-=2AlO2-+H2O       (2·Ö) 
(3)Fe + 4H2O(g) =¸ßÎÂ=Fe3O4 + 4H2         £¨2·Ö£©
(4) 3Fe3+ + 3H2O  Fe(OH)+ 3H+                         £¨2·Ö£©
AΪӦÓÃ×î¹ã·ºµÄ½ðÊôÔòΪÌú£¬ÓëÑÎËá·´Ó¦£¬Éú³ÉÂÈ»¯ÑÇÌúºÍÇâÆø£¬ÌúÓëË®·´Ó¦Éú³ÉFe3O4ºÍH2,¼´ Fe + 4H2O(g) ==Fe3O4 + 4H£¬ÄÇôCÊÇÇâÆø£¬´æÔڵĻ¯Ñ§¼üÊǹ²¼Û¼ü£¬BΪÂÈ»¯ÑÇÌú£¬·Ç½ðÊôµ¥ÖÊX2ΪÂÈÆø£¬ËùÒÔYΪAl £¬FΪÂÈ»¯Ìú£»Ñõ»¯ÎïE¼ÈÄÜÓëÑÎËá·´Ó¦£¬ÓÖÄÜÓëNaOHÈÜÒº·´Ó¦£¬EΪAl2O3£¬Al2O3+2OH-=2AlO2-+H2O¡£
FÄÜ´Ù½øH2OµÄµçÀëÊÇÒòΪFe3+·¢ÉúÁËË®½â,3Fe3+ + 3H2O  Fe(OH)+ 3H+£»£¨5£©ÖÐpH£½12µÄNaOHÈÜÒºÖÐOH-Ũ¶ÈΪ0.01mol/l,»ìºÏʱH+¹ýÁ¿£¬»ìºÏºóCH+=(0.1¡Á1£­0.01¡Á9)/10=0.001 mol/l,PH=3
(5)   3        £¨2·Ö£©
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
£¨15·Ö£© A¡¢B¡¢C¡¢D¶¼ÊÇÖÐѧ»¯Ñ§³£¼ûµÄÎïÖÊ£¬ÆäÖÐA¡¢B¡¢C¾ùº¬ÓÐͬһÖÖÔªËØ¡£ÔÚÒ»¶¨Ìõ¼þÏÂÏ໥ת»¯µÄ¹ØϵÈçͼËùʾ¡£

Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÈôDΪÓÃÁ¿×î´ó¡¢ÓÃ;×î¹ãµÄ½ðÊôµ¥ÖÊ£¬¼ÓÈÈÕô¸ÉBµÄÈÜҺûÓеõ½BµÄÑΣ¬ÔòBµÄ»¯Ñ§Ê½¿ÉÄÜΪ          ¡££¨Ð´³öÒ»ÖÖ¼´¿É£©
£¨2£©Èôͨ³£Çé¿öÏÂA¡¢B¡¢C¡¢D¶¼ÊÇÆøÌ壬ÇÒBºÍDΪ¿ÕÆøµÄÖ÷Òª³É·Ö£¬ÔòAµÄµç×ÓʽΪ            £¬BµÄ½á¹¹Ê½Îª             ¡£
£¨3£©ÈôDΪÂȼҵµÄÖØÒª²úÆ·£¬·´Ó¦£¨III£©µÄÀë×Ó·½³ÌʽΪ                ¡£
£¨4£©ÈôA¡¢B¡¢CµÄÈÜÒº¾ùÏÔ¼îÐÔ£¬CΪ±ºÖƸâµãµÄ·¢½Í·ÛµÄÖ÷Òª³É·ÖÖ®Ò»£¬Ò²¿É×÷ΪҽÁÆÉÏÖÎÁÆθËá¹ý¶àÖ¢µÄÒ©¼Á¡£
¢Ù25¡æʱ£¬pH¾ùΪ10µÄA¡¢BÁ½ÈÜÒºÖУ¬ÓÉË®µçÀë³öµÄÇâÑõ¸ùÀë×ÓŨ¶ÈÖ®±ÈΪ        ¡£
¢Ú25¡æʱ£¬0£®1mol¡¤L-1µÄA¡¢B¡¢CÈýÖÖÈÜÒº£¬·Ö±ðÓÃˮϡÊͲ»Í¬µÄ±¶Êýºó£¬ÈÜÒºµÄpHÏàͬ£¬ÔòÏ¡ÊͺóÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶È×î´óµÄÊÇ           ÈÜÒº£¨ÌîÈÜÖʵĻ¯Ñ§Ê½£©¡£
¢Û½«µÈÎïÖʵÄÁ¿µÄBºÍCÈÜÓÚË®ÐγɻìºÏÈÜÒº£¬ÈÜÒºÖи÷ÖÖÀë×ÓŨ¶ÈÓÉ´óµ½Ð¡Ë³ÐòΪ                                                       ¡£
£¨13·Ö£©ÓÐA¡¢B¡¢C¡¢DËÄÖÖÔªËØ£¬ÆäÖÐAÔªËغÍBÔªËصÄÔ­×Ó¶¼ÓÐ1¸öδ³É¶Ôµç×Ó£¬A+±ÈB¡ªÉÙÒ»¸öµç×Ӳ㣬BÔ­×ÓµÃÒ»¸öµç×ÓÌîÈë3p¹ìµÀºó£¬ 3p¹ìµÀÒѳäÂú£»CÔ­×ÓµÄp¹ìµÀÖÐÓÐ3¸öδ³É¶Ôµç×Ó£¬ÆäÆø̬Ç⻯ÎïÔÚË®ÖеÄÈܽâ¶ÈÔÚͬ×åÔªËØËùÐγɵÄÇ⻯ÎïÖÐ×î´ó£»DµÄ×î¸ß»¯ºÏ¼ÛºÍ×îµÍ»¯ºÏ¼ÛµÄ´úÊýºÍ4£¬Æä×î¸ß¼ÛÑõ»¯ÎïÖк¬DµÄÖÊÁ¿·ÖÊýΪ40%£¬ÇÒÆäºËÄÚÖÊ×ÓÊýµÈÓÚÖÐ×ÓÊý¡£RÊÇÓÉA¡¢DÁ½ÔªËØÐγɵÄÀë×Ó»¯ºÏÎÆäÖÐA+ÓëD2¡ªÀë×ÓÊýÖ®±ÈΪ2£º1¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©AÔªËØÐγɵľ§ÌåÊôÓÚÃÜÖöѻý·½Ê½£¬ÔòÆ侧Ì徧°ûÀàÐÍÊôÓÚ¡¡¡¡¡¡¡¡¡¡¡¡¡££¨Ìîд¡°Áù·½¡±¡¢¡°ÃæÐÄÁ¢·½¡±»ò¡°ÌåÐÄÁ¢·½¡±£©¡£
£¨2£©B¡ªµÄµç×ÓÅŲ¼Ê½¡¡¡¡¡¡£¬ÔÚCB3·Ö×ÓÖÐCÔªËØÔ­×ÓµÄÔ­×Ó¹ìµÀ·¢ÉúµÄÊÇ_______ÔÓ»¯¡£
£¨3£©CµÄÇ⻯Îï¿Õ¼ä¹¹ÐÍΪ¡¡¡¡¡¡¡¡¡¡¡¡£¬ÆäÇ⻯ÎïÔÚͬ×åÔªËØËùÐγɵÄÇ⻯ÎïÖзеã×î¸ßµÄÔ­ÒòÊÇ¡¡¡¡¡¡¡¡¡¡¡¡¡£
£¨4£©BÔªËصĵ縺ÐÔ¡¡¡¡¡¡¡¡¡¡¡¡DÔªËصĵ縺ÐÔ£¨Ìî¡°>¡±£¬¡°<¡±»ò¡°=¡±£©£»ÓÃÒ»¸ö»¯Ñ§·½³Ì
ʽ˵Ã÷B¡¢DÁ½ÔªËØÐγɵĵ¥ÖʵÄÑõ»¯ÐÔÇ¿Èõ£º¡¡¡¡¡¡¡¡¡¡¡¡¡£

£¨5£©ÈçÉÏͼËùʾÊÇRÐγɵľ§ÌåµÄ¾§°û£¬É辧°ûµÄ±ß
³¤Îªacm¡£ÔòR¾§ÌåµÄÃܶÈΪ¡¡¡¡¡¡¡¡¡¡¡¡¡££¨°¢·ü¼ÓµÂÊý³£ÊýÓÃNA±íʾ£©

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

¾«Ó¢¼Ò½ÌÍø