ÌâÄ¿ÄÚÈÝ

ÈçͼËùʾ£¬A-H¶¼ÊdzõÖл¯Ñ§Öг£¼ûµÄÎïÖÊ£¬ÒÑÖª£º³£ÎÂÏ£¬AÊǺÚÉ«¹ÌÌåµ¥ÖÊ£¬BÊǺÚÉ«¹ÌÌ廯ºÏÎDΪºìÉ«¹ÌÌåµ¥ÖÊ£¬FΪFe2O3£¬ËüÃǵÄת»¯¹ØϵÈçͼËùʾ¡£°´ÒªÇóÍê³ÉÏÂÁи÷Ì⣺

+F ¸ßÎÂ

 
   A                 C                  E

¸ßÎÂ

1

 

+H

 3

 
 

B                 D                   G

£¨1£©Ð´³öÏÂÁÐÎïÖʵĻ¯Ñ§Ê½£º C        £¬E         ¡£

£¨2£©Ð´³öÉÏͼÖТ٠¢Ú ¢Û·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º

  ¢Ù                                                      £»

  ¢Ú                                                      £»

  ¢Û                                                      ¡£

£¨3£©·´Ó¦µÄ¢Ù¢Û»ù±¾·´Ó¦ÀàÐͶ¼ÊÇ          £¬·´Ó¦¢ÚµÄ»ù±¾·´Ó¦ÀàÐÍÊÇ           ¡£

£¨1£©   C£ºCO2    E: CO

(2)    C+2CuO ¸ßΠ 2Cu+CO2  ¡ü (2·Ö)

CO2 +C  ¸ßΠ  2CO  £¨2·Ö£©

Fe+CuSO4       FeSO4+Cu    £¨2·Ö£©

£¨3£©Öû»·´Ó¦    »¯ºÏ·´Ó¦

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø