ÏÂÃæÊdz£¼ûµÄµç»¯Ñ§×°ÖÃͼ£º
 
ͼ1                 Í¼2                 Í¼3
Çë»Ø´ðÒÔÏÂÎÊÌ⣺£¨1£©Í¼1ÖÐÍ­°åÉÏÌúí¶¤´¦ÈÝÒ×ÉúÐ⣬³ÆΪ              ¸¯Ê´£¬±»¸¯Ê´µÄ½ðÊôÊÇ         £¬Ô­µç³ØµÄ×Ü·´Ó¦·½³ÌʽÊÇ                          ¡£
£¨2£©Í¼2ÖÐa¡¢bÊǶà¿×ʯīµç¼«£¬¶Ï¿ªK2£¬±ÕºÏK1Ò»¶Îʱ¼ä£¬¹Û²ìµ½Á½Ö»²£Á§¹ÜÄÚ¶¼ÓÐÆøÅݽ«µç¼«°üΧ£¬b¼«Éϵĵ缫·´Ó¦Ê½Îª                   £¬OH-Ïò          £¨Ìîa»òb£©¼«Òƶ¯¡£È»ºó¶Ï¿ªK1£¬±ÕºÏK2£¬¹Û²ìµ½µçÁ÷¼ÆAµÄÖ¸ÕëÓÐƫת¡£b¼«Éϵĵ缫·´Ó¦Ê½Îª                         £¬OH-Ïò               £¨Ìîa»òb£©¼«Òƶ¯¡£
£¨3£©Í¼3ÖУ¬X¡¢Y¶¼ÎªÊ¯Ä«µç¼«£¬ÔÚUÐ͹ÜÁ½²à·Ö±ð¼ÓÈëÒ»µÎ×ÏɫʯÈïÊÔÒº£¬ÔòͨµçºóX¸½½üÏÔ______É«£¬Y¸½½üÏÔ_______É«¡£

£¨1£©ÎüÑõ     Fe      2Fe+O2+2H2O=2Fe(OH)2 
£¨2£©     2H++2e£­=H2      a     H2+2OH¡ª£«2e£­=2H2O    b  £¨3£©À¶  ºì

½âÎö

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÏÂÃæÊdz£¼ûµÄµç»¯Ñ§×°ÖÃͼ£¬¢Ù¢Û¢ÜÖоùΪ¶èÐԵ缫£¬ÏÂÁÐ˵·¨²»ÕýÈ·µÄÊÇ£¨¡¡¡¡£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÏÂÃæÊdz£¼ûµÄµç»¯Ñ§×°ÖÃͼ£¬¢Ù¢Û¢ÜÖоùΪ¶èÐԵ缫£¬Çë»Ø´ðÏÂÁÐÎÊÌâ:

            

¢Ùµç½âCuCl2ÈÜҺװÖà                           ¢Úµç¶ÆͭʵÑé×°ÖÃ

           

¢ÛÇâÑõȼÁϵç³ØʾÒâͼ                            ¢ÜÀë×Ó½»»»Ä¤·¨µç½âÔ­ÀíʾÒâͼ

(1)д³ö×°ÖâÙÖз¢Éú·´Ó¦µÄ×Ü»¯Ñ§·½³Ìʽ__________________________________¡£

(2)×°ÖâÚÖÐͭƬΪ______¼«£¬ÈôͭƬºÍÌúÖÆÆ·µÄÖÊÁ¿ÏàµÈ£¬µç½âÒ»¶Îʱ¼äºó£¬µç·ÖÐÓÐ2 molµç×ÓתÒÆ£¬´ËʱͭƬºÍÌúÖÆÆ·µÄÖÊÁ¿²îΪ____________g¡£

(3)×°ÖâÛÖÐb¼«Îª______¼«£¬¸Ã¼«µÄµç¼«·´Ó¦·½³ÌʽΪ_________________¡£

(4)×°ÖâÜΪ¹¤ÒµÉϵç½âʳÑÎË®ÖÆÈ¡NaOHµÄÔ­ÀíʾÒâͼ£¬ÆäÖÐÑôÀë×Ó½»»»Ä¤½«µç½â²Û¸ô³ÉÑô¼«ÊÒºÍÒõ¼«ÊÒ£¬Ñô¼«ÊÒ¼ÓÈ뾫ÖƵı¥ºÍʳÑÎË®£¬Òõ¼«ÊÒ¼ÓÈë´¿Ë®£¨º¬Ò»¶¨Á¿µÄNaOH£©¡£

Çë½âÊÍÒõ¼«µÄ´¿Ë®ÖмÓÈëNaOHµÄÔ­Òò£º__________________________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÏÂÃæÊdz£¼ûµÄµç»¯Ñ§×°ÖÃͼ£¬¢Ù¢Û¢ÜÖоùΪ¶èÐԵ缫£¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ

A. ×°ÖâÙÖеç½âÒ»¶Îʱ¼äºó£¬ÈÜÒºµÄPH²»±ä¡£

B. ×°ÖâÚÖÐb¼«ÎªÕý¼«£¬¸Ã¼«µÄµç¼«·´Ó¦·½³ÌʽΪO2+4H++ 4e-=2H2O

C. ×°ÖâÛÖÐͭƬΪÑô¼«£¬ÈôͭƬºÍÌúÖÆÆ·µÄÖÊÁ¿ÏàµÈ£¬µç½âÒ»¶Îʱ¼äºó£¬µç·ÖÐÓÐ2mol

µç×ÓתÒÆ£¬´ËʱͭƬºÍÌúÖÆÆ·µÄÖÊÁ¿²îΪ128g

D. ×°ÖâÜÖÐÑôÀë×Ó½»»»Ä¤ÉÏÿ͸¹ý1mol Na+ʱ£¬ÔòÒõ¼«ÉϲúÉú11.2L H2

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÏÂÃæÊdz£¼ûµÄµç»¯Ñ§×°ÖÃͼ£º

 

ͼ1                 Í¼2                 Í¼3

Çë»Ø´ðÒÔÏÂÎÊÌ⣺£¨1£©Í¼1ÖÐÍ­°åÉÏÌúí¶¤´¦ÈÝÒ×ÉúÐ⣬³ÆΪ               ¸¯Ê´£¬±»¸¯Ê´µÄ½ðÊôÊÇ         £¬Ô­µç³ØµÄ×Ü·´Ó¦·½³ÌʽÊÇ                          ¡£

£¨2£©Í¼2ÖÐa¡¢bÊǶà¿×ʯīµç¼«£¬¶Ï¿ªK2£¬±ÕºÏK1Ò»¶Îʱ¼ä£¬¹Û²ìµ½Á½Ö»²£Á§¹ÜÄÚ¶¼ÓÐÆøÅݽ«µç¼«°üΧ£¬b¼«Éϵĵ缫·´Ó¦Ê½Îª                   £¬OH-Ïò          £¨Ìîa»òb£©¼«Òƶ¯¡£È»ºó¶Ï¿ªK1£¬±ÕºÏK2£¬¹Û²ìµ½µçÁ÷¼ÆAµÄÖ¸ÕëÓÐƫת¡£b¼«Éϵĵ缫·´Ó¦Ê½Îª                         £¬OH-Ïò                £¨Ìîa»òb£©¼«Òƶ¯¡£

£¨3£©Í¼3ÖУ¬X¡¢Y¶¼ÎªÊ¯Ä«µç¼«£¬ÔÚUÐ͹ÜÁ½²à·Ö±ð¼ÓÈëÒ»µÎ×ÏɫʯÈïÊÔÒº£¬ÔòͨµçºóX¸½½üÏÔ______É«£¬Y¸½½üÏÔ_______É«¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸