ÌâÄ¿ÄÚÈÝ

ÌúÔÚÀäµÄŨÁòËáÖÐÄÜ·¢Éú¶Û»¯ÏÖÏó¡£Ä³ÐËȤС×éµÄͬѧ·¢ÏÖ½«Ò»¶¨Á¿µÄÌúÓëŨÁòËá¼ÓÈÈʱ£¬¹Û²ìµ½ÌúÍêÈ«Èܽ⣬²¢²úÉú´óÁ¿ÆøÌ塣Ϊ´Ë£¬ËûÃÇÉè¼ÆÁËÈçÏÂ×°ÖÃÑéÖ¤Ëù²úÉúµÄÆøÌå¡£

£¨1£©Ö¤Ã÷·´Ó¦Ëù²úÉúµÄÆøÌåÖÐÓÐSO2Éú³ÉµÄÏÖÏóÊÇ                           ¡£
£¨2£©Ö¤Ã÷ÆøÌåÖк¬ÓÐÇâÆøµÄʵÑéÏÖÏóÊÇ                                       ¡£
£¨3£©ÎªÁ˽øÒ»²½Ì½¾¿·´Ó¦ºóAÈÜÒºÖÐÌúÔªËصļÛ̬£¬ËûÃǽøÐÐÁËÈçϵļÙÉ裺
¼ÙÉè1£ºÈÜÒºÖÐÌúÔªËؼÈÓÐFe3+Ò²ÓÐFe2+
¼ÙÉè2£ºÈÜÒºÖÐÌúÔªËØÖ»ÓÐFe3+
¼ÙÉè3£ºÈÜÒºÖÐÌúÔªËØÖ»ÓÐ________________
»ùÓÚ¼ÙÉè1£¬ÏÖÓÐÊÔ¼Á£º0.01 mol/LËáÐÔKMnO4ÈÜÒº¡¢Ï¡äåË®ÈÜÒº¡¢0.1 mal/L KIÈÜÒº¡¢
µí·ÛÈÜÒº¡¢KSCNÈÜÒº£¬ÕôÁóË®¡£Çë̽¾¿ÆäËùµÃÈÜÒº¡£ÇëÍê³É±íÖÐÄÚÈÝ¡£
¡¾ÊµÑé̽¾¿¡¿
ʵÑé²Ù×÷
Ô¤ÆÚÏÖÏó
½áÂÛ
È¡·´Ó¦ºóµÄAÈÜÒº·Ö×°ÔÚa¡¢bÁ½ÊԹܣ¬²½Öè¢Ù£ºÍùaÊÔ¹ÜÖеÎÈë         
                            ¡£
                       
                       
                 
                 
²½Öè¢Ú£ºÍùbÊÔ¹ÜÖеÎÈë         
                            ¡£
                       
                       
ÈÜÒºº¬ÓÐFe3+
 
£¨16·Ö£©£¨1£©Æ·ºìÊÔÒº±ädz£¨»òÍÊÉ«£© £¨2·Ö£©
£¨2£©EÖкÚÉ«(CuO)±ä³ÉºìÉ«£¬FÖа×É«·ÛÄ©(CuSO4)±ä³ÉÀ¶É«£»£¨2·Ö£©
£¨3£©¼ÙÉè3£ºÈÜÒºÖÐÌúÔªËØÖ»ÓÐFe2+ £¨2·Ö£©
¡¾ÊµÑé̽¾¿¡¿
ʵÑé²Ù×÷
Ô¤ÆÚÏÖÏó
½áÂÛ
È¡·´Ó¦ºóµÄAÈÜÒº·Ö×°ÔÚa¡¢bÁ½ÊԹܣ¬²½Öè¢Ù£ºÍùaÊÔ¹ÜÖеÎÈëÉÙÁ¿£¨1·Ö£©0.01mol/LËáÐÔKMnO4ÈÜÒº£¨1·Ö£©¡£
KMnO4ÈÜÒºµÄ×ϺìÉ«ÍÊÈ¥£¨»ò±ädz£©£¨2·Ö£©
 ÈÜÒºº¬ÓÐFe2+£¨2·Ö£©
²½Öè¢Ú£ºÍùbÊÔ¹ÜÖеÎÈ뼸µÎ£¨1·Ö£©KSCNÈÜÒº£¬³ä·ÖÕñµ´£¨1·Ö£©[»ò¼¸µÎ£¨»òÉÙÁ¿£©0.1mol/LKIÈÜÒººÍµí·ÛÈÜÒº£¨1·Ö£©] ¡£
ÈÜÒº±äΪѪºìÉ«£¨2·Ö£©
[»òÈÜÒº±äΪÀ¶É«£¨2·Ö£©]
ÈÜÒºº¬ÓÐFe3+
 

ÊÔÌâ·ÖÎö£º£¨1£©SO2¾ßÓÐƯ°×ÐÔ¡¢ËáÐÔ¡¢»¹Ô­ÐÔºÍÑõ»¯ÐÔ£¬¶Áͼ¿ÉÖª£¬¸ÃʵÑéÓÃÆ·ºìÈÜÒº¼ìÑéÊÇ·ñÓÐSO2Éú³É£¬ÈôÆ·ºìÈÜÒºÍÊÉ«»ò±ädz£¬ËµÃ÷×°ÖÃAÖз´Ó¦²úÉúµÄÆøÌ庬ÓÐSO2£»£¨2£©ÇâÆø¾ßÓÐÇ¿»¹Ô­ÐÔ£¬ÔÚ¼ÓÈÈÌõ¼þÏÂÄÜʹºÚÉ«µÄÑõ»¯Í­»¹Ô­ÎªºìÉ«µÄµ¥ÖÊÍ­£¬H2±»Ñõ»¯ÎªÄÜʹ°×É«µÄÎÞË®CuSO4¹ÌÌå±äÀ¶µÄH2O£¬Èô×°ÖÃEÖкÚÉ«¹ÌÌå±äΪºìÉ«£¬FÖа×É«·ÛÄ©±äΪÀ¶É«£¬ËµÃ÷×°ÖÃAÖзųöµÄÆøÌåÖк¬ÓÐH2£»£¨3£©³£ÎÂÏÂÌú±»Å¨ÁòËá¶Û»¯£¬±íÃæÓÐÒ»²ãÖÂÃÜÑõ»¯ÎﱡĤ£¬¼ÓÈÈÏÂÇ¿Ñõ»¯ÐÔ¡¢Ç¿ËáÐÔµÄŨÁòËá¿ÉÒÔ½«ÌúÑõ»¯ÎªÌúÑΣ¬¶ø¹ýÁ¿µÄµ¥ÖÊÌú¿ÉÒÔ½«ÌúÑλ¹Ô­ÎªÑÇÌúÑΣ¬Òò´Ë·´Ó¦ºóAµÄÈÜÒºÖÐÌúÔªËØ¿ÉÄܼÈÓÐFe2+ÓÖÓÐFe3+£¬Ò²¿ÉÄÜÖ»ÓÐFe3+£¬»¹¿ÉÄÜÖ»ÓÐFe2+£¬¸ù¾ÝÒÑÖª¼ÙÉè1¡¢¼ÙÉè2µÄÐÅÏ¢ÍƶϼÙÉè3ΪÈÜÒºÖÐÌúÔªËØÖ»ÓÐFe2+£»¸ù¾ÝÌú¼°Æ仯ºÏÎïµÄÐÔÖÊ¿ÉÖª£¬Fe2+¾ßÓл¹Ô­ÐÔ£¬ÄÜʹËáÐÔKMnO4ÈÜÒºÍÊÉ«£¬Fe3+¾ßÓÐÑõ»¯ÐÔ£¬Äܽ«KIÈÜÒº»¹Ô­ÎªÄÜʹµí·ÛÈÜÒº±äÀ¶µÄI2£¬Fe2+ÓöKSCNÈÜÒº²»±äºì£¬¶øFe3+ÓöKSCNÈÜÒº±äºì£¬´óÁ¿º¬ÓÐFe3+µÄÈÜÒº³Ê»ÆÉ«£¬ÓëäåË®µÄÑÕÉ«Ïàͬ£¬ÓÉ´Ë¿ÉÒÔÑ¡ÔñÊʵ±µÄÊÔ¼ÁÉè¼ÆʵÑé·½°¸ÑéÖ¤AµÄÈÜÒºÖмÈÓÐFe2+ÓÖÓÐFe3+¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
ijͬѧÂÃÓÎʱ·¢ÏÖ£¬Ãç×åÈ˵ÄÒøÊÎÃÀÀö¶ø¸»ÓÐÃñ×åÎÄ»¯£¬ÖÆ×÷ÒøÊÎʱ¿ÉÒÔÑ¡ÓÃFe£¨NO3)3ÈÜÒº×öÊ´¿Ì¼Á¡£ÊÜ´ËÆô·¢£¬¸ÃͬѧËùÔڵĻ¯Ñ§ÐËȤС×éÔÚʵÑéÊÒÑ¡ÓÃFe(NO3)3ÈÜÒºÇåÏ´×ö¹ýÒø¾µ·´Ó¦µÄÊԹܣ¬·¢ÏÖ²»µ«Òø¾µÈܽ⣬¶øÇÒ½ÏÉÙ²úÉú´Ì¼¤ÐÔÆøÌå¡£
»¯Ñ§ÐËȤС×é¶ÔFe(NO3)3ÈÜÒºÈܽâÒøµÄÔ­Àí½øÐÐ̽¾¿£º
¡¾Ìá³ö¼ÙÉè¡¿¼ÙÉè1£º Fe(NO3)3ÈÜÒºÏÔËáÐÔ£¬ÔÚ´ËËáÐÔÌõ¼þÏÂNO3-ÄÜÑõ»¯Ag£»
¼ÙÉè2£ºFe3+¾ßÓÐÑõ»¯ÐÔ£¬ÄÜÑõ»¯Ag
¡¾ÑéÖ¤¼ÙÉè¡¿
£¨1£©¼×ͬѧÑéÖ¤¼ÙÉè1¡£
¢ÙËûÓõ­×ÏÉ«µÄFe(N03)3¡¤9H20¾§Ì壨·ÖÎö´¿£¬Mr=404)ÅäÖÆ1.5mol/LµÄFe(N03)3ÂäÒº100mL¡£ÐèÒª³ÆÈ¡_____g Fe(N03)3¡¤9H20¾§Ì壬ÅäÖƹý³ÌÖÐËùÓõ½µÄÒÇÆ÷³ýÉÕ±­¡¢²£Á§°ôÍ⻹±ØÐ裺__________
¢Ú²âµÃ1.5  mol/LµÄFe(NO3)3ÈÜÒºpHԼΪ1£¬ÆäÔ­ÒòÓû¯Ñ§ÓÃÓï±íʾΪ____¡£
¢Û½«pH=1µÄHN03ÈÜÒº¼ÓÈëµ½¶ÆÓÐÒø¾µµÄÊÔ¹ÜÖУ¬Õñµ´£¬¹Û²ìµ½Òø¾µÂýÂýÈܽ⣬²úÉúÎÞÉ«ÆøÌå²¢ÔÚÒºÃæÉÏ·½±äΪºì×ØÉ«£¬ÈÜÒºÖз¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽӦÊÇ_____
¢Ü½«1.5mol/LµÄFe(NO3)3ÈÜÒº¼ÓÈëµ½¶ÆÓÐÒø¾µµÄÊÔ¹ÜÖУ¬Õñµ´£¬¹Û²ìµ½Òø¾µºÜ¿ìÈܽ⣬²¢ÇÒÈÜÒºÑÕÉ«¼ÓÉî¡£
£¨2£©ÒÒͬѧÑéÖ¤¼ÙÉè2¡£·Ö±ðÓÃÈÜÖʵÄÖÊÁ¿·ÖÊýΪ2%¡¢10%µÄ×ãÁ¿FeCl3ÈÜÒº¼ÓÈëµ½¶ÆÓÐÒø¾µµÄÊÔ¹ÜÖУ¬Õñµ´£¬¶¼¿´²»³öÒø¾µÈܽ⡣ÒÒͬѧÓɴ˵óö½áÂÛ£¬¼ÙÉè2²»³ÉÁ¢¡£
ÄãÊÇ·ñͬÒâÒҵĽáÂÛ?_______,¼òÊöÀíÓÉ£º_______
¡¾Ë¼¿¼Óë½»Á÷¡¿I¼×ͬѧµÄʵÑé¢ÜÖУ¬ÈÜÒºÑÕɫΪʲô»á¼ÓÉ²éÔÄ×ÊÁϵÃÖª£¬Fe2+ÄÜÓëNOÐγÉÅäÀë×Ó£º£¨×ØÉ«)¡£ÒÑÖª£¬Í¬Å¨¶ÈµÄÏõËáÑõ»¯ÐÔ±ÈFe3+ÂÔÇ¿¡£
¸ù¾ÝÒÔÉÏÐÅÏ¢×ۺϷÖÎö£¬Å¨¡¢Ï¡Fe(N03)3ÈÜÒºÈܽâÒø¾µÊ±£¬·¢ÉúµÄ·´Ó¦Óкβ»Í¬£¿
__________________________________________
ÏÖÓÐһƿʵÑéÊÒ·ÅÖÃÒѾõĿÉÄܱ»Ñõ»¯µÄNa2SO3¹ÌÌ壬ΪÁËÑо¿ËüµÄ×é³É£¬ÇëÄã²ÎÓëͬѧÃǽøÐеÄÈçÏÂ̽¾¿»î¶¯£º
¿ÉÑ¡ÓÃÊÔ¼Á£ºÅ¨H2SO4¡¢Å¨HNO3¡¢10%ÑÎËá¡¢0.1mol/LH2SO4¡¢0.1mol/LHNO3¡¢0.1mol/LBaCl2¡¢0.1mol/LBa(NO3)2¡¢3%H2O2¡¢10%NaOHÈÜÒº¡¢ÕôÁóË®¡¢Æ·ºìÈÜÒº£»ÒÇÆ÷×ÔÑ¡¡£
£¨1£©Ìá³ö¼ÙÉè
¼ÙÉèÒ»£º¹ÌÌåÈ«²¿ÊÇNa2SO3£» ¼ÙÉè¶þ£º¹ÌÌåÈ«²¿ÊÇNa2SO4£»
¼ÙÉèÈý£º                                                  ¡£
£¨2£©Éè¼ÆʵÑé·½°¸(ÂÔ)£»Ñ¡ÓÃÏÂͼװÖýøÐÐʵÑ飬¸Ã×°ÖõÄÓŵãÊÇ                 ¡£

£¨3£©½øÐÐʵÑ飺ÇëÔÚϱíÖÐÓüòÒªÎÄ×Öд³öʵÑé²Ù×÷¡¢Ô¤ÆÚÏÖÏóºÍ½áÂÛ¡£
ʵÑé²½Öè
Ô¤ÆÚÏÖÏóºÍ½áÂÛ
²½Öè1£ºÈ¡ÊÊÁ¿¹ÌÌåÑùÆ·ÓÚ΢ÐÍÊÔ¹ÜÖУ»ÔÚW¹Üa´¦µÎÈë        ¡¢b´¦µÎÈë      £»Óýº¹Ü½«W¹ÜÓë΢ÐÍÊÔ¹ÜÁ¬½ÓºÃ
 
²½Öè2£ºÓÃÕëͲÎüÈë       £¬½«ÕëÍ·´©¹ý΢ÐÍÊԹܵĽºÈû£¬Ïò¹ÌÌåÑùÆ·ÖÐ×¢Èë¸ÃÈÜÒº¡£
                               
                               
                              ¡£
²½Öè3£º²¦³öÕëͲ£¬ÎüÈëÕôÁóˮϴ¾»£»ÔÙÎüÈë
               ×¢Èë΢ÐÍÊÔ¹ÜÖÐ
                               
                               
                              ¡£
 
£¨4£©Èô½«ÉÏÊö̽¾¿¹ý³ÌÖÐÉú³ÉµÄÆøÌåͨÈëµ½×ãÁ¿µÄÐÂÖÆƯ°×·ÛŨÈÜÒºÖУ¬Ôò¿ÉÉú³É°×É«³Áµí¡£ÊÔд³ö¸Ã·´Ó¦µÄÀë×Ó·½³Ìʽ£º                                        ¡£

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

¾«Ó¢¼Ò½ÌÍø