ÌâÄ¿ÄÚÈÝ

(17·Ö)ijͬѧÀûÓÃÏÂͼËùʾµÄʵÑé×°ÖýøÐÐÌú¸úË®ÕôÆø·´Ó¦µÄʵÑé,²¢¼ÌÐøÑо¿Ìú¼°Æ仯ºÏÎïµÄ²¿·ÖÐÔÖÊ¡£

ÒÑÖª:¢ÙFeO + 2H= Fe2+ + H2O¢ÚFe2O3 + 6H+ = 2Fe3+ +3 H2O ¢ÛFe3O4 + 8H+ = Fe2+ +2Fe3+ +4 H2O
Çë»Ø´ðÏÂÁÐÎÊÌâ:
£¨1£©Ó²ÖÊÊÔ¹ÜÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ                                       ¡£
£¨2£©¸ÃͬѧÓûÈ·¶¨·´Ó¦Ò»¶Îʱ¼äºóÓ²ÖÊÊÔ¹ÜÖйÌÌåÎïÖʵijɷÖ,Éè¼ÆÁËÈçÏÂʵÑé·½°¸:
¢Ù´ýÓ²ÖÊÊÔ¹ÜÀäÈ´ºó,È¡ÉÙÐíÆäÖеĹÌÌåÎïÖÊÈÜÓÚÏ¡ÁòËáµÃÈÜÒºB;
¢ÚÈ¡ÉÙÁ¿ÈÜÒºBµÎ¼ÓKSCNÈÜÒº,ÈôÈÜÒº±äºìÉ«Ôò˵Ã÷Ó²ÖÊÊÔ¹ÜÖйÌÌåÎïÖʵijɷÖÊÇ(Ö»ÓÐÒ»¸öÑ¡Ïî·ûºÏÌâÒâ)         ,ÈôÈÜҺδ±äºìÉ«Ôò˵Ã÷Ó²ÖÊÊÔ¹ÜÖйÌÌåÎïÖʵijɷÖÊÇ(Ö»ÓÐÒ»¸öÑ¡Ïî·ûºÏÌâÒâ)         ¡£

A£®Ò»¶¨ÓÐFe3O4,¿ÉÄÜÓÐFeB£®Ö»ÓÐFe(OH)3C£®Ò»¶¨ÓÐFe3O4ºÍFe
D£®Ò»¶¨ÓÐFe(OH)3,¿ÉÄÜÓÐFe E.Ö»ÓÐFe3O4
£¨3£©¸Ãͬѧ°´ÉÏÊöʵÑé·½°¸½øÐÐÁËʵÑé,½á¹ûÈÜҺδ±äºìÉ«,Ô­ÒòÊÇ                                   
                                     £¨ÓÃÀë×Ó·½³Ìʽ±íʾ£©¡£
£¨4£©¸ÃͬѧÂíÉÏÁíÈ¡ÉÙÁ¿ÈÜÒºB,ʹÆä¸úNaOHÈÜÒº·´Ó¦¡£Èô°´ÏÂͼËùʾµÄ²Ù×÷,¿É¹Û²ìµ½Éú³É°×É«³Áµí,ѸËÙ±ä³É»ÒÂÌÉ«,×îºó±ä³ÉºìºÖÉ«µÄÏÖÏó,Çëд³öÓëÉÏÊöÏÖÏóÏà¹ØµÄ·´Ó¦µÄ»¯Ñ§·½³Ìʽ                                                 ¡£

£¨5£©Ò»¶Îʱ¼äºó,¸Ãͬѧ·¢ÏÖ£¨3£©ÖÐδ±äºìµÄÈÜÒº±ä³ÉºìÉ«,˵Ã÷Fe2+ ¾ßÓР  ÐÔ¡£ÓÉ´Ë¿ÉÖª,ʵÑéÊÒÖк¬Fe2+µÄÑÎÈÜÒºÐèÏÖÓÃÏÖÅäÖƵÄÔ­ÒòÊÇ                           £¬²¢ÇÒÅäÖƺ¬Fe2+µÄÑÎÈÜҺʱӦ¼ÓÈëÉÙÁ¿         ¡£
£¨6£©ÒÒͬѧΪÁË»ñµÃ³Ö¾Ã°×É«µÄFe(OH)2³Áµí£¬×¼±¸ÓÃÓÒͼËùʾװÖã¬Óò»º¬O2µÄÕôÁóË®ÅäÖƵÄNaOHÈÜÒºÓëÐÂÖƵÄFeSO4ÈÜÒº·´Ó¦¡£»ñµÃ²»º¬O2µÄÕôÁóË®µÄ·½·¨ÊÇ______________¡£·´Ó¦¿ªÊ¼Ê±£¬´ò¿ªÖ¹Ë®¼ÐµÄÄ¿µÄÊÇ___________________________________£»Ò»¶Îʱ¼äºó£¬¹Ø±Õֹˮ¼Ð£¬ÔÚÊÔ¹Ü_______£¨Ìî¡°A¡±»ò¡°B¡±£©Öй۲쵽°×É«µÄFe(OH)2

£¨1£©3Fe + 4H2O(g) »òFe3O4 + 4H2 (2·Ö)        £¨2£©A,C(2·Ö)
£¨3£©Fe + 2Fe3+£½ 3Fe2+(2·Ö)(³öÏÖ´Ë·½³Ìʽ²Å¸ø·Ö,ֻдFe3O4 + 8H+£½ Fe2+ +2Fe3+ +4 H2O²»¸ø·Ö)
£¨4£©FeSO4 + 2NaOH £½ Fe(OH)2¡ý+ Na2SO4 (2·Ö) 4Fe(OH)2 + O2 + 2H2O £½ 4Fe(OH)3 (2·Ö)
£¨5£©»¹Ô­£¨1·Ö£©,Fe2+Ò×±»¿ÕÆøÖеÄÑõÆøÑõ»¯¶ø±äÖÊ£¨1·Ö£©,Ìú·Û£¨1·Ö£©
(6)°ÑÕôÁóË®Öó·Ð£¨1·Ö£©£¬ÀûÓÃÌú·ÛºÍÏ¡ÁòËá·´Ó¦²úÉúµÄÇâÆøÅųöÊÔ¹ÜA¡¢BÖеĿÕÆø£¬·ÀÖ¹Fe(OH)2±»¿ÕÆøÖÐÑõÆøÑõ»¯£¨2·Ö£©£¬B£¨1·Ö£©¡£

½âÎöÊÔÌâ·ÖÎö£º1£©Ó²ÖÊÊÔ¹ÜÖз¢Éú·´Ó¦ÊÇÌúÓëË®ÕôÆøµÄ·´Ó¦£¬×¢ÒâÌúÖ»ÄÜÉú³ÉËÄÑõ»¯ÈýÌúºÍÇâÆø£¬²»ÄÜд³öÈýÑõ»¯¶þÌú£»
£¨2£©ÈôÈÜÒº±äºìÉ«Ôò˵Ã÷ÈÜÒºÖÐÓÐÈý¼ÛÌúÀë×Ó£¬²»ÄÜÅųýÈÜÒºÖÐÊÇ·ñÓÐÑÇÌúÀë×Ó£¬ËùÒÔÓ²ÖÊÊÔ¹ÜÖйÌÌåÎïÖʵijɷÖÒ»¶¨ÓÐFe3O4,¿ÉÄÜÓÐFe£¬ÈôÈÜҺδ±äºìÉ«Ôò˵Ã÷ÈÜÒºÖÐÎÞÈý¼ÛÌúÀë×Ó£¬ÔòÒ»¶¨ÓйýÁ¿µÄÌú°ÑÈý¼ÛÌú»¹Ô­Îª¶þ¼ÛÌúÀë×Ó£¬Ó²ÖÊÊÔ¹ÜÖйÌÌåÎïÖʵijɷÖÒ»¶¨ÓÐFe3O4ºÍFe¡£
£¨3£©ËµÃ÷¹ýÁ¿µÄÌú°ÑÈý¼ÛÌú»¹Ô­Îª¶þ¼ÛÌú£¬Fe + 2Fe3+£½ 3Fe2+¡£
£¨4£©Éú³É°×É«³ÁµíÊÇÇâÑõ»¯ÑÇÌú,ѸËÙ±ä³É»ÒÂÌÉ«,×îºó±ä³ÉºìºÖÉ«ÊÇÇâÑõ»¯ÑÇÌú±»ÑõÆøÑõ»¯ÎªÇâÑõ»¯Ìú, FeSO4 + 2NaOH £½ Fe(OH)2¡ý+ Na2SO4£¬ 4Fe(OH)2 + O2 + 2H2O £½ 4Fe(OH)3¡£
£¨5£©Î´±äºìµÄÈÜÒº±ä³ÉºìÉ«£¬Fe2+ ±»Ñõ»¯ÎªÈý¼ÛÌúÀë×Ó£¬ËµÃ÷¾ßÓл¹Ô­ÐÔ¡£ÊµÑéÊÒÖк¬Fe2+µÄÑÎÈÜÒºÐèÏÖÓÃÏÖÅäÖƵÄÔ­ÒòÊÇÒ×±»¿ÕÆøÖеÄÑõÆøÑõ»¯¶ø±äÖÊ£¬²¢ÇÒÅäÖƺ¬Fe2+µÄÑÎÈÜҺʱӦ¼ÓÈëÉÙÁ¿Ìú·Û£¬¿ÉÒÔ°ÑÑõ»¯ºóµÄÈý¼ÛÌúÔÙ»¹Ô­³ÉÑÇÌúÀë×Ó¡£
£¨6£©»ñµÃ²»º¬O2µÄÕôÁóË®µÄ·½·¨Ò»°ãÊÇÖó·ÐÈÜÒº£¬¸Ï×ßÆøÌå¡£·´Ó¦¿ªÊ¼Ê±£¬´ò¿ªÖ¹Ë®¼ÐµÄÄ¿µÄÊÇÀûÓÃÌú·ÛºÍÏ¡ÁòËá·´Ó¦²úÉúµÄÇâÆøÅųöÊÔ¹ÜA¡¢BÖеĿÕÆø£¬·ÀÖ¹Fe(OH)2±»¿ÕÆøÖÐÑõÆøÑõ»¯£»Ò»¶Îʱ¼äºó£¬¹Ø±Õֹˮ¼Ð£¬AÖеÄÇâÆø²»ÄÜÅųö£¬ÔÚÊÔ¹ÜÄÚʹѹǿԽÀ´Ô½´ó£¬°ÑAÖеÄÁòËáÑÇÌú¸ÏÈëµ½BÖУ¬ÔÚBÖлá³öÏÖ°×É«³ÁµíFe(OH)2¡£
¿¼µã£º¿¼²éÌúÓëË®ÕôÆø·´Ó¦µÄʵÑ飬¼ìÑé¶þ¼ÛÌú¼°Èý¼ÛÌúµÄ·½·¨£¬ÇâÑõ»¯ÑÇÌúµÄÖƱ¸£¬ÌúÈý½ÇµÄת»¯¡£
µãÆÀ£º¶ÔÓÚÌúµ¥Öʼ°Æ仯ºÏÎïµÄת»¯ÊǸßÖеÄÖصãºÍÄѵ㣬¾­³£ÒÔÀë×ӵļìÑé¡¢³ýÔÓ¡¢ÎïÖʵÄÖƱ¸µÈÌâÐͳöÏÖ£¬±È½Ï×ۺϣ¬¿¼²éѧÉúµÄ×ÛºÏ˼άÄÜÁ¦¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

(17·Ö)ijѧÉú¿ÎÍâѧϰ»î¶¯Ð¡×éÕë¶Ô½Ì²ÄÖÐÍ­ÓëŨÁòËá·´Ó¦£¬Ìá³öÁËÑо¿¡°Äܹ»ÓëÍ­·´Ó¦µÄÁòËáµÄ×îµÍŨ¶ÈÊǶàÉÙ£¿¡±µÄÎÊÌ⣬²¢Éè¼ÆÁËÈçÏ·½°¸½øÐÐʵÑ飺

ʵÑéÊÔ¼Á£º18mol/LÁòËá20mL£¬´¿Í­·Û×ãÁ¿£¬×ãÁ¿2mol/LNaOHÈÜÒº

Çë¸ù¾ÝʵÑé»Ø´ðÎÊÌ⣺

¢ÅÊ×Ïȸù¾ÝÉÏͼËùʾ£¬×éװʵÑé×°Ö㬲¢ÔÚ¼ÓÈëÊÔ¼ÁÇ°ÏȽøÐР           ²Ù×÷¡£

¢ÆÉÕ±­ÖÐÓÃNaOHÈÜÒºÎüÊÕµÄÎïÖÊÊÇ£º      (Ìѧʽ)£¬ÀûÓõ¹ÖõÄ©¶·¶ø²»Êǽ«µ¼Æø¹ÜÖ±½ÓÉîÈëÉÕ±­ÖеÄÄ¿µÄÊÇ£º                            ¡£

¢Ç¼ÓÈÈÉÕÆ¿20·ÖÖÓ£¬ÉÕÆ¿Öз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ£º                     ¡£´ýÉÕÆ¿Öз´Ó¦»ù±¾½áÊø£¬³·È¥¾Æ¾«µÆ£¬ÀûÓÃÉÕÆ¿ÖеÄÓàÈÈʹ·´Ó¦½øÐÐÍêÈ«¡£È»ºóÓɵ¼¹ÜaͨÈë×ãÁ¿µÄ¿ÕÆø£¬ÒÔÈ·±£ÉÕÆ¿ÖеÄSO2ÆøÌåÈ«²¿½øÈëÉÕ±­ÖС£ÔÚ¸ÃʵÑé×°ÖÃÖеÄ

          (ÌîÒÇÆ÷Ãû³Æ)Æðµ½ÁËÈ·±£ÁòËáÌå»ý±£³Ö²»±äµÄ×÷Óá£

¢È½«³ä·Ö·´Ó¦ºóµÄÉÕ±­È¡Ï£¬ÏòÆäÖмÓÈë×ãÁ¿µÄËữµÄË«ÑõË®£¬ÔÙ¼ÓÈë×ãÁ¿µÄBaCl2ÈÜÒº£¬ÔÙ½øÐР        ¡¢          ¡¢          ºó³ÆÁ¿ÁòËá±µµÄÖÊÁ¿Îª13.98g£¬Çë¼ÆËãÄÜÓëÍ­·´Ó¦µÄÁòËáµÄŨ¶È×îµÍÊÇ         ¡£

¢ÉÓеÄͬѧÌá³öÔÚÉÏÃæ¢ÈÖпÉÒÔ²»±Ø¼ÓÈëËữµÄË«ÑõË®£¬Ö±½Ó½øÐкóÃæµÄʵÑ飬ҲÄܵõ½×¼È·µÄÊý¾Ý£¬Çë½áºÏÄãµÄÀí½â·ÖÎöÊÇ·ñÐèÒª¼ÓÈëË«ÑõË®¼°Ô­Òò£º               

                                                                         ¡£

 

(17·Ö)ʵÑéÊÒÖÐÓÐÈýƿʧȥ±êÇ©µÄËᣬ·Ö±ðÊÇŨÁòËᡢŨÑÎËáºÍŨÏõËá¡£
£¨1£©ÓÐͬѧÈÏΪ£¬ÓýðÊôÍ­×öÊÔ¼Á¿É¼ø±ðÉÏÊöÈýÖÖËᣬÇëÓû¯Ñ§·½³ÌʽºÍ±ØÒªµÄÎÄ×Ö¼ÓÒÔ˵Ã÷¡£____________________________________________________________________¡£
£¨2£©ÇëÉè¼ÆÒ»¸öʵÑé×°Öã¬Ê¹Í­ÓëÏ¡ÁòËá·´Ó¦Éú³ÉÁòËáÍ­ÓëÇâÆø¡£ÔÚÏÂÃæµÄ·½¿òÄÚ»­³ö×°ÖÃͼ£¬²¢±êÃ÷µç¼«²ÄÁÏÃû³ÆºÍµç½âÖÊÈÜÒºÃû³Æ¡£

£¨3£©ÊµÑéÊÒÖÐÓûÓÃŨÁòËáÅäÖÆ2.0mol¡¤L-1µÄÏ¡ÁòËá500mL¡£
¢ÙʵÑé¹ý³ÌÖÐËùÐèÒÇÆ÷³ýÁËÁ¿Í²¡¢ÉÕ±­¡¢²£Á§°ôÍ⣬»¹ÐèÒªµÄÒÇÆ÷ÓÐ____________________£»
¢ÚÏÂÁвÙ×÷ÖÐÔì³ÉËùÅäÈÜҺŨ¶ÈÆ«µÍµÄÊÇ_____________¡£
a£®ÓÃÁ¿Í²Á¿È¡Å¨ÁòËáʱÑöÊӿ̶ÈÏß
b£®¶¨ÈÝʱ¸©Êӿ̶ÈÏß
c£®×ªÒÆÈÜÒººóδϴµÓÉÕ±­ºÍ²£Á§°ô¾ÍÖ±½Ó¶¨ÈÝ
d£®¶¨ÈݺóÒ¡ÔÈ£¬·¢ÏÖÒºÃæµÍÓڿ̶ȣ¬Î´²¹¼ÓÕôÁóË®ÖÁ¿Ì¶È
£¨4£©ÏÂͼËùʾΪʵÑéÊÒÄ£Ä⹤ҵÉÏÀûÓð±´ß»¯Ñõ»¯·¨ÖƱ¸ÏõËáµÄʵÑé

ÓÃ×°ÖÃAÖƱ¸¸ÉÔï°±Æø£¬×°ÖÃBÖƱ¸¸ÉÔïÑõÆø¡£
¢Ù×°ÖÃAµÄÊÔ¹ÜÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ_______________________________¡£
×°ÖÃBµÄÉÕÆ¿ÄÚ·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ__________________________________¡£
¢ÚÏÂÁйØÓÚA¡¢B×°ÖõÄÐðÊö²»ÕýÈ·µÄÊÇ__________
a£®ÈôÑ¡ÔñºÏÊʵÄÊÔ¼Á£¬ÓÃB×°ÖÃÒ²¿ÉÖƱ¸°±Æø
b£®ÊµÑé¹ý³ÌÖУ¬A¡¢B×°ÖÃÖÐÒ»¶¨ÓÐÒ»¸ö·¢ÉúÑõ»¯»¹Ô­·´Ó¦
c£®UÐιÜÖеÄÊÔ¼Á¿ÉÒÔÏàͬ£¬µ«×÷Óò»Ïàͬ
¢Û°´ÕÕa½Óc£¬b½ÓdµÄ˳ÐòÁ¬½Ó×°ÖýøÐÐʵÑé¡£
²£Á§¹ÜÄÚ·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ___________________________________¡£
ʵÑé½áÊøºó£¬Ä³Í¬Ñ§²âµÃ×°ÖÃCÖÐÊÔ¹ÜÄÚËùµÃÈÜÒºµÄpH£¼7£¬µÃ³öµÄ½áÂÛΪ¸ÃÈÜÒºÒ»¶¨ÊÇÏõËá¡£¸Ã½áÂÛ__________£¨Ìî¡°ÑÏÃÜ¡±»ò¡°²»ÑÏÃÜ¡±£©£¬ÄãµÄÀíÓÉÊÇ_____________________________¡£

(17·Ö)ijѧÉú¿ÎÍâѧϰ»î¶¯Ð¡×éÕë¶Ô½Ì²ÄÖÐÍ­ÓëŨÁòËá·´Ó¦£¬Ìá³öÁËÑо¿¡°Äܹ»ÓëÍ­·´Ó¦µÄÁòËáµÄ×îµÍŨ¶ÈÊǶàÉÙ£¿¡±µÄÎÊÌ⣬²¢Éè¼ÆÁËÈçÏ·½°¸½øÐÐʵÑ飺
ʵÑéÊÔ¼Á£º18mol/LÁòËá20mL£¬´¿Í­·Û×ãÁ¿£¬×ãÁ¿2mol/LNaOHÈÜÒº
Çë¸ù¾ÝʵÑé»Ø´ðÎÊÌ⣺

¢ÅÊ×Ïȸù¾ÝÉÏͼËùʾ£¬×éװʵÑé×°Ö㬲¢ÔÚ¼ÓÈëÊÔ¼ÁÇ°ÏȽøÐР           ²Ù×÷¡£
¢ÆÉÕ±­ÖÐÓÃNaOHÈÜÒºÎüÊÕµÄÎïÖÊÊÇ£º      (Ìѧʽ)£¬ÀûÓõ¹ÖõÄ©¶·¶ø²»Êǽ«µ¼Æø¹ÜÖ±½ÓÉîÈëÉÕ±­ÖеÄÄ¿µÄÊÇ£º                            ¡£
¢Ç¼ÓÈÈÉÕÆ¿20·ÖÖÓ£¬ÉÕÆ¿Öз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ£º                     ¡£´ýÉÕÆ¿Öз´Ó¦»ù±¾½áÊø£¬³·È¥¾Æ¾«µÆ£¬ÀûÓÃÉÕÆ¿ÖеÄÓàÈÈʹ·´Ó¦½øÐÐÍêÈ«¡£È»ºóÓɵ¼¹ÜaͨÈë×ãÁ¿µÄ¿ÕÆø£¬ÒÔÈ·±£ÉÕÆ¿ÖеÄSO2ÆøÌåÈ«²¿½øÈëÉÕ±­ÖС£ÔÚ¸ÃʵÑé×°ÖÃÖеÄ
          (ÌîÒÇÆ÷Ãû³Æ)Æðµ½ÁËÈ·±£ÁòËáÌå»ý±£³Ö²»±äµÄ×÷Óá£
¢È½«³ä·Ö·´Ó¦ºóµÄÉÕ±­È¡Ï£¬ÏòÆäÖмÓÈë×ãÁ¿µÄËữµÄË«ÑõË®£¬ÔÙ¼ÓÈë×ãÁ¿µÄBaCl2ÈÜÒº£¬ÔÙ½øÐР        ¡¢          ¡¢          ºó³ÆÁ¿ÁòËá±µµÄÖÊÁ¿Îª13.98g£¬Çë¼ÆËãÄÜÓëÍ­·´Ó¦µÄÁòËáµÄŨ¶È×îµÍÊÇ         ¡£
¢ÉÓеÄͬѧÌá³öÔÚÉÏÃæ¢ÈÖпÉÒÔ²»±Ø¼ÓÈëËữµÄË«ÑõË®£¬Ö±½Ó½øÐкóÃæµÄʵÑ飬ҲÄܵõ½×¼È·µÄÊý¾Ý£¬Çë½áºÏÄãµÄÀí½â·ÖÎöÊÇ·ñÐèÒª¼ÓÈëË«ÑõË®¼°Ô­Òò£º               
                                                                         ¡£

(17·Ö)ijͬѧÀûÓÃÏÂͼËùʾµÄʵÑé×°ÖýøÐÐÌú¸úË®ÕôÆø·´Ó¦µÄʵÑé,²¢¼ÌÐøÑо¿Ìú¼°Æ仯ºÏÎïµÄ²¿·ÖÐÔÖÊ¡£

ÒÑÖª:¢ÙFeO + 2H= Fe2+ + H2O¢ÚFe2O3 + 6H+ = 2Fe3+ +3 H2O ¢ÛFe3O4 + 8H+ = Fe2+ +2Fe3+ +4 H2O

Çë»Ø´ðÏÂÁÐÎÊÌâ:

£¨1£©Ó²ÖÊÊÔ¹ÜÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ                                       ¡£

£¨2£©¸ÃͬѧÓûÈ·¶¨·´Ó¦Ò»¶Îʱ¼äºóÓ²ÖÊÊÔ¹ÜÖйÌÌåÎïÖʵijɷÖ,Éè¼ÆÁËÈçÏÂʵÑé·½°¸:

¢Ù´ýÓ²ÖÊÊÔ¹ÜÀäÈ´ºó,È¡ÉÙÐíÆäÖеĹÌÌåÎïÖÊÈÜÓÚÏ¡ÁòËáµÃÈÜÒºB;

¢ÚÈ¡ÉÙÁ¿ÈÜÒºBµÎ¼ÓKSCNÈÜÒº,ÈôÈÜÒº±äºìÉ«Ôò˵Ã÷Ó²ÖÊÊÔ¹ÜÖйÌÌåÎïÖʵijɷÖÊÇ(Ö»ÓÐÒ»¸öÑ¡Ïî·ûºÏÌâÒâ)         ,ÈôÈÜҺδ±äºìÉ«Ôò˵Ã÷Ó²ÖÊÊÔ¹ÜÖйÌÌåÎïÖʵijɷÖÊÇ(Ö»ÓÐÒ»¸öÑ¡Ïî·ûºÏÌâÒâ)         ¡£

A£®Ò»¶¨ÓÐFe3O4,¿ÉÄÜÓÐFe               B£®Ö»ÓÐFe(OH)3                        C£®Ò»¶¨ÓÐFe3O4ºÍFe

D£®Ò»¶¨ÓÐFe(OH)3,¿ÉÄÜÓÐFe E.Ö»ÓÐFe3O4                                                                                         

£¨3£©¸Ãͬѧ°´ÉÏÊöʵÑé·½°¸½øÐÐÁËʵÑé,½á¹ûÈÜҺδ±äºìÉ«,Ô­ÒòÊÇ                                   

                                     £¨ÓÃÀë×Ó·½³Ìʽ±íʾ£©¡£

£¨4£©¸ÃͬѧÂíÉÏÁíÈ¡ÉÙÁ¿ÈÜÒºB,ʹÆä¸úNaOHÈÜÒº·´Ó¦¡£Èô°´ÏÂͼËùʾµÄ²Ù×÷,¿É¹Û²ìµ½Éú³É°×É«³Áµí,ѸËÙ±ä³É»ÒÂÌÉ«,×îºó±ä³ÉºìºÖÉ«µÄÏÖÏó,Çëд³öÓëÉÏÊöÏÖÏóÏà¹ØµÄ·´Ó¦µÄ»¯Ñ§·½³Ìʽ                                                 ¡£

£¨5£©Ò»¶Îʱ¼äºó,¸Ãͬѧ·¢ÏÖ£¨3£©ÖÐδ±äºìµÄÈÜÒº±ä³ÉºìÉ«,˵Ã÷Fe2+ ¾ßÓР  ÐÔ¡£ÓÉ´Ë¿ÉÖª,ʵÑéÊÒÖк¬Fe2+µÄÑÎÈÜÒºÐèÏÖÓÃÏÖÅäÖƵÄÔ­ÒòÊÇ                           £¬²¢ÇÒÅäÖƺ¬Fe2+µÄÑÎÈÜҺʱӦ¼ÓÈëÉÙÁ¿         ¡£

£¨6£©ÒÒͬѧΪÁË»ñµÃ³Ö¾Ã°×É«µÄFe(OH)2³Áµí£¬×¼±¸ÓÃÓÒͼËùʾװÖã¬Óò»º¬O2µÄÕôÁóË®ÅäÖƵÄNaOHÈÜÒºÓëÐÂÖƵÄFeSO4ÈÜÒº·´Ó¦¡£»ñµÃ²»º¬O2µÄÕôÁóË®µÄ·½·¨ÊÇ______________¡£·´Ó¦¿ªÊ¼Ê±£¬´ò¿ªÖ¹Ë®¼ÐµÄÄ¿µÄÊÇ___________________________________£»Ò»¶Îʱ¼äºó£¬¹Ø±Õֹˮ¼Ð£¬ÔÚÊÔ¹Ü_______£¨Ìî¡°A¡±»ò¡°B¡±£©Öй۲쵽°×É«µÄFe(OH)2

 

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

¾«Ó¢¼Ò½ÌÍø