ÌâÄ¿ÄÚÈÝ

£¨12·Ö£©ÂÒÈӷϾɵç³Ø»áµ¼ÖÂÑÏÖصĻ·¾³ÎÛȾ£¬Ò»½Ú·Ïµç³Ø¾ÍÊÇÒ»¿Å¡°Õ¨µ¯¡±¡£Ä³»¯Ñ§ÐËȤС×éµÄͬѧÔÚ̽¾¿·Ï¸Éµç³ØÄڵĺÚÉ«¹ÌÌå»ØÊÕÀûÓÃʱ£¬½øÐÐÁËÈçͼËùʾµÄʵÑ飺

²éÔĽ̲Ŀɵõ½ÏÂÁÐÐÅÏ¢£º
¢ÙÆÕͨпÃ̵ç³ØµÄºÚÉ«ÎïÖÊÖ÷Òª³É·ÖΪMnO2£®NH4Cl£®ZnCl2µÈÎïÖÊ¡£
¢ÚZn£¨OH£©2ÄÜÈܽâÓÚ¹ýÁ¿µÄ°±Ë®ÖС£
Çë»Ø´ðÒÔÏÂÎÊÌ⣺
£¨1£©²Ù×÷¢Û×ÆÉÕÂËÔüʱËùÓõ½µÄÖ÷ÒªÒÇÆ÷Óоƾ«µÆ£®²£Á§°ô£®            £®ÄàÈý½ÇºÍÈý½Å¼Ü£»×ÆÉÕÂËÔüÖеĺÚÉ«¹ÌÌåʱ£¬²úÉúÒ»ÖÖʹ³ÎÇåµÄʯ»ÒË®±ä»ë×ǵÄÆøÌ壬ÓÉ´ËÍƲâÂËÔüÖл¹´æÔÚ̼¡£
£¨2£©²Ù×÷¢ÜµÄÊÔ¹ÜÖмÓÈë¢ÛÖÐËùµÃºÚÉ«ÂËÔü£¬ÊÔ¹ÜÖÐѸËÙ²úÉúÄÜʹ´ø»ðÐǵÄľÌõ¸´È¼µÄÆøÌ壬¾Ý´Ë¿É³õ²½È϶¨ºÚÉ«¹ÌÌåΪ               ¡£
£¨3£©¸Ãͬѧ¶ÔÂËÒºµÄ³É·Ö½øÐмìÑ飬ÒÔÈ·ÈÏÊÇ·ñº¬ÓÐNH4ClºÍZnCl2£¬ÏÂÃæÊÇËû×öÍêʵÑéºóËùдµÄʵÑ鱨¸æ£¬ÇëÄãд³öÆä¿Õ°×´¦µÄÄÚÈÝ£º
ʵÑéÄ¿µÄ
²Ù×÷
ʵÑéÏÖÏó
½áÂÛ
¼ìÑéCl£­
È¡ÉÙÐíÂËÒºÓÚÊÔ¹ÜÖУ¬             
                                 
                    
                             
º¬ÓÐCl£­
¼ìÑéNH4£«
È¡ÉÙÐíÂËÒºÓÚÊÔ¹ÜÖУ¬            
                                 
                    
                    
º¬ÓÐNH4£«
¼ìÑéZn2£«
È¡ÉÙÐíÂËÒºÓÚÊÔ¹ÜÖУ¬             
                                 
                    
                    
º¬ÓÐZn2£«
£¨4£©¸ù¾ÝÒÔÉÏʵÑ鱨¸æ£¬¹ØÓÚÂËÒºµÄ³É·Ö£¬Í¬Ñ§ÃǵĽáÂÛÊÇ£ºÂËÒºÖк¬ÓÐÂÈ»¯ï§ºÍÂÈ»¯Ð¿£¬ÈôÏë´ÓÂËÒºÖеõ½ÈÜÖʹÌÌ壬»¹Ó¦½øÐеòÙ×÷ÊÇ                 ¡£ÈôÒª½«ËùµÃµÄÈÜÖʹÌÌåÖеÄÎïÖʼÓÒÔ·ÖÀ룬¿ÉÓà             ·¨¡£
£¨1£©ÛáÛö £¨1·Ö£©           £¨2£©¶þÑõ»¯ÃÌ£¨»òMnO2£© £¨2·Ö£©
£¨3£©£¨Ã¿¿Õ1·Ö£©
²Ù×÷
ʵÑéÏÖÏó
¼ÓÈëÏõËáËữµÄÏõËáÒøÈÜÒº
Óа×É«³Áµí²úÉú
¼ÓÈëŨÇâÑõ»¯ÄÆÈÜÒº²¢¼ÓÈÈ£¬½«ÈóʪµÄºìɫʯÈïÊÔÖ½ÖÃÓÚÊԹܿڸ½½ü
ºìɫʯÈïÊÔÖ½±ä³ÉÀ¶É«
Ï¡°±Ë®
ÏȲúÉú°×É«³Áµí£¬¼ÌÐø¼ÒÈ¥°±Ë®£¬³ÁµíÈܽâ
      £¨4£©ÔÚ»ìºÏÒºÖмÓÈëÉÙÁ¿ÑÎËᣬ¼ÓÈÈÕô·¢£¨¼ÓÈÈŨËõ£©£¬ÀäÈ´½á¾§£¬¹ýÂË£¨2·Ö£©
¼ÓÈÈ£¨1·Ö£©
ÂÔ
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
Èý²ÝËáºÏÌúËá¼Ø¾§ÌåK3[Fe(C2O4)3]¡¤3H2O¿ÉÓÃÓÚÉãÓ°ºÍÀ¶É«Ó¡Ë¢£»ÒÑÖª²ÝËáÊÜÈÈÒ×·Ö½â²úÉúCO2ºÍCO¡£Ä³Ð¡×齫ÎÞË®Èý²ÝËáºÏÌúËá¼Ø°´ÈçͼËùʾװÖýøÐÐʵÑ飺

£¨1£©ÈôʵÑéÖУ¬¹Û²ìµ½B¡¢EÖÐÈÜÒº¾ù±ä»ë×Ç£¬ÔòDÖеÄÏÖÏóÊÇ_______________________¡£
£¨2£©ÄãÈÏΪ¸ÃʵÑéÉè¼Æ»¹ÐèÒªÈçºÎ¸Ä½ø£¨ÇëÖÁÉÙд³öÁ½´¦£©¢Ù            £»¢Ú              ¡£
£¨3£©¸ÃС×éͬѧ²éÔÄ×ÊÁϺóÍÆÖª£¬¹ÌÌå²úÎïÖУ¬ÌúÔªËز»¿ÉÄÜÒÔÈý¼ÛÐÎʽ´æÔÚ£¬¶øÑÎÖ»ÓÐK2CO3¡£ÑéÖ¤¹ÌÌå²úÎïÖмØÔªËØ´æÔڵķ½·¨ÊÇ        £¬ÏÖÏóÊÇ           ¡£
£¨4£©¹ÌÌå²úÎïÖÐÌúÔªËØ´æÔÚÐÎʽµÄ̽¾¿¡£ ¢ÙÌá³öºÏÀí¼ÙÉ裺
¼ÙÉè1£º         £»    ¼ÙÉè2£º              £»¼ÙÉè3£º         ¡£
Éè¼ÆʵÑé·½°¸Ö¤Ã÷ÉÏÊöµÄ¼ÙÉ裨ÏÞÑ¡ÊÔ¼Á£ºÅ¨ÁòËá¡¢1.0 mol¡¤L£­1 HNO3¡¢1.0 mol¡¤L£­1ÑÎËá¡¢1.0 mol¡¤L£­1 NaOH¡¢3% H2O2¡¢0.1 mol¡¤L£­1 KI¡¢0.1 mol¡¤L£­1 CuSO4¡¢20% KSCN¡¢³ÎÇåʯ»ÒË®¡¢ÕôÁóË®£©£º¢ÚÈôijѧÉúÉè¼ÆʵÑéÀ´Ì½¾¿ÉÏÊö¼ÙÉèÖеÄÆäÖÐÖ®Ò»³ÉÁ¢£¬ÇëÄã°ïÖúÑ¡ÔñÒ»ÏÂʵÑé¹ý³ÌÖÐÐèÒªÉÏÊöÊÔ¼ÁÖеÄ3% H2O2¡¢³ÎÇåʯ»ÒË®­                          ¡£
(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               
                                                                         ¡£
²â¶¨º£Ë®ÖÐÈܽâÑõµÄ²½ÖèÈçÏ£º
£¨1£©Ë®ÑùµÄ¹Ì¶¨¡£È¡amLº£Ë®ÑùѸËÙ¼ÓÈë¹Ì¶¨¼ÁMnSO4ÈÜÒººÍ¼îÐÔKIÈÜÒº£¨º¬KOH£©£¬Á¢¼´ÈûºÃÆ¿Èû£¬²¢Õñµ´Ò¡ÔÈ£¬Ê¹Ö®³ä·Ö·´Ó¦¡£
£¨2£©Ëữ¡£¿ªÈûºóѸËÙ¼ÓÊÊÁ¿1£º1µÄÁòËᣬÔÙѸËÙÈûºÃÆ¿Èû¡£·´¸´Õñµ´ÖÁ³ÁµíÍêÈ«Èܽ⡣
£¨3£©µÎ¶¨¡£ÏòÈÜÒºÖмÓÈë1mL0.5%µí·ÛÈÜÒº£¬ÔÙÓÃbmol/LNa2S2O3±ê×¼ÈÜÒºµÎ¶¨ÖÁÈÜÒº¸ÕºÃÍÊÉ«¡£
ÓйصÄÀë×Ó·´Ó¦ÈçÏ£º
2Mn2+ +O2 + 4OH- = 2MnO(OH)2  (¸Ã·´Ó¦¼«¿ì) 
MnO(OH)2 +2I- + 4H+ = Mn2+ + I2 + 3H2O
I2 + 2S2O32- = 2I- + S4O62-   Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÔÚ½øÐУ¨1£©¡¢£¨2£©²Ù×÷ʱ£¬Òª×¢Ò⡰ѸËÙ¡±£¬Èç¹û²Ù×÷»ºÂý£¬»áʹ²â¶¨½á¹û     £¨Ìî¡°Æ«¸ß¡±»ò¡°Æ«µÍ¡±»ò¡°ÎÞÓ°Ï족£©Ô­ÒòÊÇ                                 ¡£
£¨2£©µÎ¶¨Ç°£¬ÓÃÕôÁóˮϴ¾»µÎ¶¨¹Ü¼´¼ÓÈëNa2S2O3±ê×¼ÈÜÒº£¬»áʹ²â¶¨½á¹û      £¨Ìî¡°Æ«¸ß¡±»ò¡°Æ«µÍ¡±»ò¡°ÎÞÓ°Ï족£©Ô­ÒòÊÇ                                 ¡£
£¨3£©µÎ¶¨Ç°£¬Î´ÅųýµÎ¶¨¹ÜÖеÄÆøÅÝ¡£Ôڵζ¨¹ý³ÌÖÐÆøÅݵÄÅųý£¬»áʹ²â¶¨½á¹û   £¨Ìî¡°Æ«¸ß¡±»ò¡°Æ«µÍ¡±»ò¡°ÎÞÓ°Ï족£©Ô­ÒòÊÇ                                 ¡£
I£®ÏÂͼÊÇÖÐѧ½Ì²ÄÖÐÍ­ÓëÏ¡ÏõËá·´Ó¦µÄʵÑé×°Öã¬Çë¾Í´Ë»Ø´ðÏÂÁÐÎÊÌ⣺
(1)¹ÜAÖз¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ              ¡£
¢ÆÊÔ¹ÜAÖй۲쵽µÄÏÖÏóÊÇ                                 ¡£

¢ÇÊÔ¹ÜBÖÐÊ¢×°NaOHÈÜÒºµÄÄ¿µÄÊÇ                        ¡£
II£®ÉÏÊö×°ÖôæÔÚÃ÷ÏԵIJ»×㣬Ϊ¿Ë·þÉÏÊöȱµã£¬Ä³¿ÎÍâС×éµÄѧÉúÉè¼ÆÁËÈçͼËùʾµÄ×°Öá£

ʵÑé²½ÖèÈçÏ£º¢Ù¼ì²é×°ÖõÄÆøÃÜÐÔ£»¢ÚÏòÊÔ¹ÜÖмÓÈëÒ»¶¨Á¿Ì¼Ëá
¸Æ¹ÌÌ壻¢ÛÏòÊÔ¹ÜÖмÓÈë¹ýÁ¿Ï¡ÏõËᣬ²¢Ñ¸ËÙ¸ÇÉÏ´øÍ­Ë¿ºÍµ¼¹ÜµÄ
ÏðƤÈû£»¢Ü´ý·´Ó¦ÍêÈ«ºó£¬½«µ¼¹Ü²åÈëÊÔ¹ÜÄÚ½Ó½üÒºÃ棨Èçͼ£©£»
¢Ý½«Í­Ë¿²åÈëµ½ÈÜÒºÖУ»¢Þ·´Ó¦Ò»¶Îʱ¼äºó£¬ÓÃ×¢ÉäÆ÷ÏòÊÔ¹ÜÄÚ
ÍÆÈëÑõÆø£¨»ò¿ÕÆø£©¡£
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
¢ÈʵÑéÖмÓÈëCaCO3µÄ×÷ÓÃÊÇ                                      ¡£
¢ÉʵÑé¢ÜµÄʵÑéÄ¿µÄÊÇ                                       ¡£
¢ÊʵÑé¢ÞÖÐÍÆÈëÑõÆø»ò¿ÕÆøµÄÄ¿µÄÊÇ                               ¡£
ÒÔ»ÆÍ­¿ó£¨CuFeS2£©ÎªÔ­ÁÏÒ±Á¶Í­µÄ·´Ó¦Îª£º
8CuFeS2+21O28Cu+4FeO3+2Fe2O3+16SO2

£¨1£©ÈôCuΪ+2¼Û£¬Ôò·´Ó¦ÖÐÿÉú³É1molCuÐèҪתÒÆ             molµç×Ó¡£
£¨2£©Îª×ÛºÏÀûÓïÆøÖеĴóÁ¿SO2£¬ÏÂÁз½°¸ºÏÀíµÄÊÇ           ¡£
a£®ÓÃŨÁòËáÎüÊÕ         b£®ÓÃÓÚÖƱ¸ÁòËá
c£®¸ß¿ÕÅÅ·Å             d£®Óô¿¼îÒºÎüÊÕÖÆÑÇÁòËáÄÆ
£¨3£©Îª¼ìÑéËùµÃ¿óÔüÊÇÖÐÊÇ·ñº¬ÓÐFeO¡¢Fe2O3¡¢CuO¡¢A12O3¡¢SiO2µÈÑõ»¯Î½øÐÐÁËÒÔÏÂʵÑ飻
¢ÙÈ¡Ò»¶¨Á¿¿óÔü·ÛÄ©£¬¼ÓÈëÊÊÁ¿Å¨ÁòËáºó¼ÓÈÈ£¬¹ÌÌåÈ«²¿Èܽ⣬µÃÈÜÒºA£»½«²úÉúµÄÆøÌåͨÈëÆ·ºìÈÜÒºÖУ¬ÈÜÒºÍÊÉ«¡£ÓÉ´ËÅжϿóÔüÖÐÒ»¶¨º¬ÓР        £¬Ò»¶¨Ã»ÓР     ¡£
¢Ú½«ÈÜÒºAÏ¡Êͺó·Ö³ÉÁ½·Ý£¬È¡ÆäÖÐÒ»·Ý£¬¼ÓÈë¹ýÁ¿NaOHÈÜÒº£¬ÓгÁµí²úÉú£¬¾²ÖúóÈ¡ÉϲãÇåÒº£¬Í¨Èë¹ýÁ¿µÄCO2ÈÜÒº±ä»ë×Ç¡£ÓÉ´ËÅжϳö¿óÔüÖк¬ÓР        £¬Ð´³öͨÈë¹ýÁ¿µÄCO2Ëù·¢Éú·´Ó¦µÄÀë×Ó·½³Ìʽ£º              ¡£
¢ÛÈ¡ÁíÒ»·ÝÈÜÒºA¼ÓÈë¹ýÁ¿µÄÌú·Û³ä·Ö·´Ó¦ºó¹ýÂË£¬µÃµ½µÄ¹ÌÌåÖÐÓкìÉ«ÎïÖÊ£¬ÓÉ´ËÅжϿóÔüÖÐÓР            £¬Ð´³ö´Ë²½²Ù×÷Öеõ½ºìÉ«ÎïÖÊËù·¢Éú·´Ó¦µÄÀë×Ó·½³Ìʽ            ¡£

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

¾«Ó¢¼Ò½ÌÍø