ÈçͼËùʾ3Ì×ʵÑé×°Ö㬷ֱð»Ø´ðÏÂÁÐÎÊÌâ¡£

£¨1£©×°ÖÃ1ΪÌúµÄÎüÑõ¸¯Ê´ÊµÑé¡£Ò»¶Îʱ¼äºó£¬ Ïò²åÈë̼°ôµÄ²£Á§Í²ÄÚµÎÈë·Ó̪ÈÜÒº£¬¿É¹Û²ìµ½Ì¼°ô¸½½üµÄÈÜÒº±äºì£¬¸Ãµç¼«·´Ó¦Îª                                  ¡£
£¨2£©×°ÖÃ2ÖеÄʯīÊÇ     ¼«£¨Ìî¡°Õý¡±»ò¡°¸º¡± £©£¬¸Ã×°Ö÷¢ÉúµÄ×Ü·´Ó¦µÄÀë×Ó·½³ÌʽΪ                                      ¡£
£¨3£©×°ÖÃ3Öм×ÉÕ±­Ê¢·Å200 mL 0.2 mol/LµÄNaClÈÜÒº£¬ÒÒÉÕ±­Ê¢·Å200 mL 1.0 mol/LµÄCuSO4ÈÜÒº¡£·´Ó¦Ò»¶Îʱ¼äºó£¬Í£Ö¹Í¨µç¡£Ïò¼×ÉÕ±­ÖеÎÈ뼸µÎ·Ó̪ÈÜÒº£¬¹Û²ìµ½Ê¯Ä«µç¼«¸½½üÊ×Ïȱäºì¡£

 

 
¢Ù µçÔ´µÄM¶ËΪ     ¼«£¬¼×ÉÕ±­ÖÐÌúµç¼«µÄµç¼«·´Ó¦Îª                   ¢ÚÍ£Ö¹µç½â£¬È¡³öCuµç¼«£¬Ï´µÓ¡¢¸ÉÔï¡¢³ÆÁ¿¡¢µç¼«ÔöÖØ 1.28 g£¬¼×ÉÕ±­ÖвúÉúµÄÆøÌå±ê×¼×´¿öÏÂÌå»ýΪ      mL ¡£

£¨12·Ö£©£¨1£©O2+ 4e- + 2H2O = 4OH-    £¨2£©Õý£¬ 2Fe3+ + Cu = 2Fe2+ + Cu2+
£¨3£©¢Ù Õý£¬ Fe - 2e-= Fe2+    ¢Ú  448

½âÎöÊÔÌâ·ÖÎö£º£¨1£©×°ÖÃ1ΪÌúµÄÎüÑõ¸¯Ê´£¬Ôò̼°ôÊÇÕý¼«£¬ÑõÆøµÃµ½µç×Ó£¬Éú³ÉOH£­£¬ËùÒÔ̼°ôÖÜΧÈÜÒºÏÔ¼îÐÔ£¬ÈÜÒº±äºì£¬¸Ãµç¼«·´Ó¦ÎªO2+ 4e- + 2H2O = 4OH-¡£
£¨2£©Í­ÊǽðÊô£¬Ê¯Ä«ÊǷǽðÊô£¬ËùÒÔʯīÊÇÕý¼«£¬ÈÜÒºÖеÄÌúÀë×ӵõ½µç×Ó¡£Í­ÊǸº¼«£¬Ê§È¥µç×Ó£¬ËùÒÔ×°ÖÃ2ÖеÄ×Ü·´Ó¦Ê½ÊÇ2Fe3+ + Cu = 2Fe2+ + Cu2+¡£
£¨3£©¢ÙÏò¼×ÉÕ±­ÖеÎÈ뼸µÎ·Ó̪ÈÜÒº£¬¹Û²ìµ½Ê¯Ä«µç¼«¸½½üÊ×Ïȱäºì£¬ËµÃ÷ÔÚʯīµç¼«ÉÏÉú³ÉOH£­Àë×Ó£¬¼´Ê¯Ä«ÊÇÒõ¼«£¬ÔòÌúÊÇÑô¼«£¬ËùÒÔµçÔ´µÄM¶ËΪÕý¼«£¬ÔòÌúµç¼«µÄµç¼«·´Ó¦ÎªFe - 2e-= Fe2+¡£
¢ÚCuµç¼«£¬Ï´µÓ¡¢¸ÉÔï¡¢³ÆÁ¿¡¢µç¼«ÔöÖØ 1.28 g£¬Õâ˵Ã÷ÒÒÖÐÎö³öÍ­µÄÖÊÁ¿ÊÇ1.28g£¬ÎïÖʵÄÁ¿ÊÇ0.02mol£¬×ªÒÆ0.04mol£¬ËùÒÔ¸ù¾Ýµç×ÓµÃʧÊغã¿ÉÖª£¬Éú³ÉÇâÆøµÄÎïÖʵÄÁ¿ÊÇ0.04mol¡Â2£½0.02mol£¬ÔÚ±ê×¼×´¿öϵÄÌå»ýÊÇ448ml¡£
¿¼µã£º¿¼²éµç»¯Ñ§Ó¦ÓõÄÓйØÅжϡ¢µç¼«·´Ó¦Ê½µÄÊéдÒÔ¼°ÓйؼÆËã
µãÆÀ£º¸ÃÌâÊǸ߿¼Öг£¼ûÌâÐÍ£¬ÊôÓÚÖеÈÄѶȵÄÊÔÌâ¡£ÊÔÌâ»ù´¡ÐÔÇ¿£¬²àÖضÔѧÉúÄÜÁ¦µÄÅàÑøÓë½âÌâ·½·¨µÄÔö´ó¡£×öÌâʱÐèҪעÒâµç¼«µÄÅжϺ͵缫·´Ó¦µÄÊéд£¬×¢Òâ´®Áªµç·Öи÷µç¼«×ªÒƵĵç×ÓÊýÄ¿ÏàµÈ£¬ÀûÓ÷´Ó¦µÄ·½³Ìʽ¼ÆËã¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÈçͼËùʾ3Ì×ʵÑé×°Ö㬷ֱð»Ø´ðÏÂÁÐÎÊÌ⣮

£¨1£©×°ÖÃ1ΪÌúµÄÎüÑõ¸¯Ê´ÊµÑ飮Ïò²åÈë̼°ôµÄ²£Á§Í²ÄÚµÎÈë·Ó̪ÈÜÒº£¬¿É¹Û²ìµ½Ì¼°ô¸½½üµÄÈÜÒº±äºì£¬¸Ãµç¼«·´Ó¦Îª
O2+4e-+2H2O¨T4OH-
O2+4e-+2H2O¨T4OH-
£®
£¨2£©×°ÖÃ2ÖеÄʯīÊÇ
Õý
Õý
¼«£¨Ìî¡°Õý¡±»ò¡°¸º¡±£©£¬¸Ã×°Ö÷¢ÉúµÄ×Ü·´Ó¦µÄÀë×Ó·½³ÌʽΪ
2Fe3++Cu¨T2Fe2++Cu2+
2Fe3++Cu¨T2Fe2++Cu2+
£®
£¨3£©×°ÖÃ3Öм×ÉÕ±­Ê¢·Å100mL 0.2mol/LµÄNaClÈÜÒº£¬ÒÒÉÕ±­Ê¢·Å100mL 0.5mol/LµÄCuSO4ÈÜÒº£®·´Ó¦Ò»¶Îʱ¼äºó£¬Í£Ö¹Í¨µç£®Ïò¼×ÉÕ±­ÖеÎÈ뼸µÎ·Ó̪£¬¹Û²ìµ½Ê¯Ä«µç¼«¸½½üÊ×Ïȱäºì£®
¢ÙµçÔ´µÄM¶ËΪ
Õý
Õý
¼«£»¼×ÉÕ±­ÖÐÌúµç¼«µÄµç¼«·´Ó¦Îª
Fe-2e-¨TFe2+
Fe-2e-¨TFe2+
£®
¢ÚÒÒÉÕ±­Öеç½â·´Ó¦µÄÀë×Ó·½³ÌʽΪ
2Cu2++2H2O
 µç½â 
.
 
2Cu+O2¡ü+4H+
2Cu2++2H2O
 µç½â 
.
 
2Cu+O2¡ü+4H+
£®
¢ÛÍ£Ö¹µç½â£¬È¡³öCuµç¼«£¬Ï´µÓ¡¢¸ÉÔï¡¢³ÆÁ¿¡¢µç¼«ÔöÖØ0.64g£¬¼×ÉÕ±­ÖвúÉúµÄÆøÌå±ê×¼×´¿öÏÂÌå»ýΪ
224
224
mL£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨12·Ö£©ÈçͼËùʾ3Ì×ʵÑé×°Ö㬷ֱð»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©×°ÖÃ1ΪÌúµÄÎüÑõ¸¯Ê´ÊµÑé¡£Ò»¶Îʱ¼äºó£¬Ïò²åÈë̼°ôµÄ²£Á§Í²ÄÚµÎÈë·Ó̪ÈÜÒº£¬¿É¹Û²ìµ½Ì¼°ô¸½½üµÄÈÜÒº±äºì£¬¸Ãµç¼«·´Ó¦Îª                             ¡£

£¨2£©×°ÖÃ2ÖеÄʯīÊÇ    ¼«£¨Ìî¡°Õý¡±»ò¡°¸º¡±£©£¬¸Ã×°Ö÷¢ÉúµÄ×Ü·´Ó¦µÄÀë×Ó·½³ÌʽΪ                            ¡£

£¨3£©×°ÖÃ3Öм×ÉÕ±­Ê¢·Å100 mL 0¡¢2 mol/LµÄNaClÈÜÒº£¬ÒÒÉÕ±­Ê¢·Å100 mL 0¡¢5 mol/LµÄCuSO4ÈÜÒº¡£·´Ó¦Ò»¶Îʱ¼äºó£¬Í£Ö¹Í¨µç¡£È¡³öCuµç¼«£¬Ï´µÓ¡¢¸ÉÔï¡¢³ÆÁ¿¡¢µç¼«ÔöÖØ 0¡¢64 g¡£

¢Ù µçÔ´µÄM¶ËΪ       ¼«£¬¼×ÉÕ±­ÖÐÌúµç¼«µÄµç¼«·´Ó¦Îª                    £»

¢Ú ÒÒÉÕ±­Öеç½â·´Ó¦µÄÀë×Ó·½³ÌʽΪ                                      £»

¢Û ¼×¡¢ÒÒÁ½ÉÕ±­ÖÐÉú³ÉµÄÆøÌå±ê×¼×´¿öϹ²     mL¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨12·Ö£©ÈçͼËùʾ3Ì×ʵÑé×°Ö㬷ֱð»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©×°ÖÃ1ΪÌúµÄÎüÑõ¸¯Ê´ÊµÑé¡£Ò»¶Îʱ¼äºó£¬Ïò²åÈë̼°ôµÄ²£Á§Í²ÄÚµÎÈë·Ó̪ÈÜÒº£¬¿É¹Û²ìµ½Ì¼°ô¸½½üµÄÈÜÒº±äºì£¬¸Ãµç¼«·´Ó¦Îª                             ¡£

£¨2£©×°ÖÃ2ÖеÄʯīÊÇ     ¼«£¨Ìî¡°Õý¡±»ò¡°¸º¡±£©£¬¸Ã×°Ö÷¢ÉúµÄ×Ü·´Ó¦µÄÀë×Ó·½³ÌʽΪ                            ¡£

£¨3£©×°ÖÃ3Öм×ÉÕ±­Ê¢·Å100 mL 0.2 mol/LµÄNaClÈÜÒº£¬ÒÒÉÕ±­Ê¢·Å100 mL 0.5 mol/LµÄCuSO4ÈÜÒº¡£·´Ó¦Ò»¶Îʱ¼äºó£¬Í£Ö¹Í¨µç¡£È¡³öCuµç¼«£¬Ï´µÓ¡¢¸ÉÔï¡¢³ÆÁ¿¡¢µç¼«ÔöÖØ 0.64 g¡£

¢Ù µçÔ´µÄM¶ËΪ       ¼«£¬¼×ÉÕ±­ÖÐÌúµç¼«µÄµç¼«·´Ó¦Îª                    £»

¢Ú ÒÒÉÕ±­Öеç½â·´Ó¦µÄÀë×Ó·½³ÌʽΪ                                      £»

¢Û ¼×¡¢ÒÒÁ½ÉÕ±­ÖÐÉú³ÉµÄÆøÌå±ê×¼×´¿öϹ²      mL¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2011-2012ѧÄ긣½¨Ê¡ÄÏ°²ÊÐÇȹâÖÐѧ¸ß¶þÉÏѧÆÚÆÚÄ©¿¼ÊÔ»¯Ñ§ÊÔ¾í ÌâÐÍ£ºÊµÑéÌâ

ÈçͼËùʾ3Ì×ʵÑé×°Ö㬷ֱð»Ø´ðÏÂÁÐÎÊÌâ¡£

£¨1£©×°ÖÃ1ΪÌúµÄÎüÑõ¸¯Ê´ÊµÑé¡£Ò»¶Îʱ¼äºó£¬ÈÜÒºÂÔÏÔdzÂÌÉ«£¬±íÃ÷Ìú±»     £¨Ìî¡°Ñõ»¯¡±»ò¡°»¹Ô­¡±£©£»Ïò²åÈë̼°ôµÄ²£Á§Í²ÄÚµÎÈë·Ó̪ÈÜÒº£¬¿É¹Û²ìµ½Ì¼°ô¸½½üµÄÈÜÒº±äºì£¬¸Ãµç¼«·´Ó¦Îª                   ¡£
£¨2£©×°ÖÃ2ÖеÄʯīÊÇ      ¼«£¨Ìî¡°Õý¡±»ò¡°¸º¡±£©£¬¸Ã×°Ö÷¢ÉúµÄ×Ü·´Ó¦µÄÀë×Ó·½³ÌΪ                             ¡£
£¨3£©×°ÖÃ3Öм×ÉÕ±­Ê¢·Å100 mL 0.2 mol/LµÄNaClÈÜÒº£¬ÒÒÉÕ±­Ê¢·Å100 mL 0.5 mol/LµÄCuSO4ÈÜÒº¡£·´Ó¦Ò»¶Îʱ¼äºó£¬Í£Ö¹Í¨µç¡£Ïò¼×ÉÕ±­ÖеÎÈ뼸µÎ·Ó̪£¬¹Û²ìµ½Ê¯Ä«µç¼«¸½½üÊ×Ïȱäºì¡£
¢ÙµçÔ´µÄM¶ËΪ     ¼«£»¼×ÉÕ±­ÖÐÌúµç¼«µÄµç¼«·´Ó¦Îª                ¡£
¢ÚÒÒÉÕ±­Öеç½â·´Ó¦µÄÀë×Ó·½³ÌʽΪ                          ¡£
¢ÛÍ£Ö¹µç½â£¬È¡³öCuµç¼«£¬Ï´µÓ¡¢¸ÉÔï¡¢³ÆÁ¿¡¢µç¼«ÔöÖØ0.64 g£¬¼×ÉÕ±­ÖвúÉúµÄÆøÌå±ê×¼×´¿öÏÂÌå»ýΪ         mL¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2011-2012ѧÄê°²»ÕÊ¡¸ßÈýµÚËÄ´ÎÔ¿¼»¯Ñ§ÊÔ¾í ÌâÐÍ£ºÌî¿ÕÌâ

£¨12·Ö£©ÈçͼËùʾ3Ì×ʵÑé×°Ö㬷ֱð»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©×°ÖÃ1ΪÌúµÄÎüÑõ¸¯Ê´ÊµÑé¡£Ò»¶Îʱ¼äºó£¬Ïò²åÈë̼°ôµÄ²£Á§Í²ÄÚµÎÈë·Ó̪ÈÜÒº£¬¿É¹Û²ìµ½Ì¼°ô¸½½üµÄÈÜÒº±äºì£¬¸Ãµç¼«·´Ó¦Îª                              ¡£

£¨2£©×°ÖÃ2ÖеÄʯīÊÇ     ¼«£¨Ìî¡°Õý¡±»ò¡°¸º¡±£©£¬¸Ã×°Ö÷¢ÉúµÄ×Ü·´Ó¦µÄÀë×Ó·½³ÌʽΪ                             ¡£

£¨3£©×°ÖÃ3Öм×ÉÕ±­Ê¢·Å100 mL 0¡¢2 mol/LµÄNaClÈÜÒº£¬ÒÒÉÕ±­Ê¢·Å100 mL 0¡¢5 mol/LµÄCuSO4ÈÜÒº¡£·´Ó¦Ò»¶Îʱ¼äºó£¬Í£Ö¹Í¨µç¡£È¡³öCuµç¼«£¬Ï´µÓ¡¢¸ÉÔï¡¢³ÆÁ¿¡¢µç¼«ÔöÖØ 0¡¢64 g¡£

¢Ù µçÔ´µÄM¶ËΪ       ¼«£¬¼×ÉÕ±­ÖÐÌúµç¼«µÄµç¼«·´Ó¦Îª                     £»

¢Ú ÒÒÉÕ±­Öеç½â·´Ó¦µÄÀë×Ó·½³ÌʽΪ                                       £»

¢Û ¼×¡¢ÒÒÁ½ÉÕ±­ÖÐÉú³ÉµÄÆøÌå±ê×¼×´¿öϹ²      mL¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸