ÌâÄ¿ÄÚÈÝ

ÖÐÑëµçÊǪ́±¨µÀ£¬Ò»ÖÖÃû½Ð¡°ÅŶ¾»ùÇ¿Àë×ÓÅŶ¾ÒÇ¡±µÄ²úÆ·ÕýÔÚÊг¡ÈÈÏú¡£ÏòÅŶ¾ÅèÄÚµ¹ÈëÁË´¿¾»µÄÎÂË®£¬Ë«½Å·ÅÈëÅèÖУ¬Æô¶¯µçÔ´¿ª¹Ø£¬¼ÓÈëÁËÊÊÁ¿¾«ÑΡ£¹ýÒ»¶Îʱ¼ä£¬ÅèÖпªÊ¼³öÏÖÂÌÉ«¡¢ºìºÖÉ«µÄÐõ×´Îï¡£ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ


  1. A.
    ´Ë¡°ÅŶ¾»ùÇ¿Àë×ÓÅŶ¾ÒÇ¡±Ò»¶¨ÊÇÓöèÐÔµç¼«ÖÆ³ÉµÄ
  2. B.
    ¼ÓһЩ¾«ÑεÄÖ÷ҪĿµÄÊÇÆðµ½ÏûÑ×ɱ¾úµÄ×÷ÓÃ
  3. C.
    ÂÌÉ«¡¢ºìºÖÉ«µÄÐõ×´Îï¾ÍÊǴӽŵ×ÅųöµÄÌåÄÚ¶¾ËØ
  4. D.
    ÂÌÉ«¡¢ºìºÖÉ«µÄÐõ×´ÎïÊÇÇâÑõ»¯ÑÇÌú¡¢ÇâÑõ»¯ÌúÐγɵĻìºÏÎï
D

ÕýÈ·´ð°¸£ºD
¸ÃÒÇÆ÷ÓÃÌú×÷µç¼«£¬µç½âNaClÈÜÒº£¬Ñô¼«:Fe¨D2e-=Fe2£« £¬Òõ¼«£º2H2O£«2e-=H2¡ü£«2OH¨D
4Fe(OH)2£«O2£«2H2O="4Fe" (OH)3 , Fe(OH)2ÓɻҰף¬±ä³É»ÒÂÌÔÙ±ä³ÉºìºÖÉ«¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø