ÌâÄ¿ÄÚÈÝ


¹ýÑõ»¯ÇâÊÇÖØÒªµÄÑõ»¯¼Á¡¢»¹Ô­¼Á£¬ËüµÄË®ÈÜÒºÓÖ³ÆÎªË«ÑõË®£¬³£ÓÃÀ´Ïû

    ¶¾¡¢É±¾ú¡¢Æ¯°×µÈ¡£Ä³»¯Ñ§ÐËȤС×éȡһ¶¨Á¿µÄ¹ýÑõ»¯ÇâÈÜÒº£¬Óû׼ȷ²â¶¨¹ýÑõ»¯

    ÇâµÄº¬Á¿¡£ÇëÌîдÏÂÁпհףº

   £¨1£©È¡10.00 mLÃܶÈΪ¦Ñg/mLµÄ¹ýÑõ»¯ÇâÈÜҺϡÊÍÖÁ250mL¡£ÒÆÈ¡Ï¡ÊͺóµÄ¹ýÑõ»¯

        ÇâÈÜÒº25.00mLÖÁ×¶ÐÎÆ¿ÖУ¬¼ÓÈëÏ¡ÁòËáËữ£¬ÓÃÕôÁóˮϡÊÍ£¬×÷±»²âÊÔÑù¡£

   £¨2£©ÓøßÃÌËá¼Ø±ê×¼ÈÜÒºµÎ¶¨±»²âÊÔÑù£¬Æä·´Ó¦µÄÀë×Ó·½³ÌʽΪ               ¡£

   £¨3£©µÎ¶¨Ê±£¬½«¸ßÃÌËá¼Ø±ê×¼ÈÜҺעÈë        £¨Ìî¡°Ëáʽ¡±»ò¡°¼îʽ¡±£©µÎ¶¨¹Ü

        ÖУ¬µÎ¶¨µ½´ïÖÕµãµÄÏÖÏóÊÇ                                 ¡£

   £¨4£©Öظ´µÎ¶¨Èý´Î£¬Æ½¾ùºÄÓÃc mol/L KMnO4±ê×¼ÈÜÒºV mL£¬ÔòÔ­¹ýÑõ»¯ÇâÈÜÒºÖÐ

        ¹ýÑõ»¯ÇâµÄÖÊÁ¿·ÖÊýΪ                  ¡£

   £¨5£©ÈôµÎ¶¨Ç°µÎ¶¨¹Ü¼â×ìÖÐÓÐÆøÅÝ£¬µÎ¶¨ºóÆøÅÝÏûʧ£¬Ôò²â¶¨½á¹û          £¨Ìî

        ¡°Æ«¸ß¡±»ò¡°Æ«µÍ¡±»ò¡°²»±ä¡±£©¡£


  £¨2£©2MnO£«5H2O2£«6H+£½2Mn2+£«8H2O£«5O2¡ü£¨2·Ö£©
  £¨3£©Ëáʽ£¨1·Ö£©

       µÎÈë×îºóÒ»µÎ¸ßÃÌËá¼ØÈÜÒº£¬ÈÜÒº³Ê×ϺìÉ«£¬ÇÒ30ÃëÄÚ²»ÍÊÉ«£¨2·Ö£©

£¨4£©0.085cV/¦Ñ  [»ò(8.5cV/¦Ñ)%] £¨2·Ö£© 

  £¨5£©Æ«¸ß£¨2·Ö£©


Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

½¹ÑÇÁòËáÄÆ(Na2S2O5)Êdz£ÓõÄʳƷ¿¹Ñõ»¯¼ÁÖ®Ò»¡£Ä³Ñо¿Ð¡×é½øÐÐÈçÏÂʵÑ飺

ʵÑéÒ»¡¡½¹ÑÇÁòËáÄÆµÄÖÆÈ¡

²ÉÓÃÈçͼװÖÃ(ʵÑéǰÒѳý¾¡×°ÖÃÄÚµÄ¿ÕÆø)ÖÆÈ¡Na2S2O5¡£×°ÖâòÖÐÓÐNa2S2O5¾§ÌåÎö³ö£¬·¢ÉúµÄ·´Ó¦ÎªNa2SO3£«SO2===Na2S2O5¡£

(1)×°ÖâñÖвúÉúÆøÌåµÄ»¯Ñ§·½³ÌʽΪ______________________________________¡£

(2)Òª´Ó×°ÖâòÖлñµÃÒÑÎö³öµÄ¾§Ì壬¿É²ÉÈ¡µÄ·ÖÀë·½·¨ÊÇ______¡£

(3)×°ÖâóÓÃÓÚ´¦ÀíÎ²Æø£¬¿ÉÑ¡ÓõÄ×îºÏÀí×°ÖÃ(¼Ð³ÖÒÇÆ÷ÒÑÂÔÈ¥)Ϊ________(ÌîÐòºÅ)¡£

ʵÑé¶þ¡¡½¹ÑÇÁòËáÄÆµÄÐÔÖÊ

Na2S2O5ÈÜÓÚË®¼´Éú³ÉNaHSO3¡£

(4)Ö¤Ã÷NaHSO3ÈÜÒºÖÐHSOµÄµçÀë³Ì¶È´óÓÚË®½â³Ì¶È£¬¿É²ÉÓõÄʵÑé·½·¨ÊÇ________(ÌîÐòºÅ)¡£

a£®²â¶¨ÈÜÒºµÄpH¡¡¡¡b£®¼ÓÈëBa(OH)2ÈÜÒº

c£®¼ÓÈëÑÎËá  d£®¼ÓÈëÆ·ºìÈÜÒº

e£®ÓÃÀ¶É«Ê¯ÈïÊÔÖ½¼ì²â

(5)¼ì²âNa2S2O5¾§ÌåÔÚ¿ÕÆøÖÐÒѱ»Ñõ»¯µÄʵÑé·½°¸ÊÇ

________________________________________________________________________

________________________________________________________________________¡£

ʵÑéÈý¡¡ÆÏÌѾÆÖп¹Ñõ»¯¼Á²ÐÁôÁ¿µÄ²â¶¨

(6)ÆÏÌѾƳ£ÓÃNa2S2O5×÷¿¹Ñõ»¯¼Á¡£²â¶¨Ä³ÆÏÌѾÆÖп¹Ñõ»¯¼ÁµÄ²ÐÁôÁ¿(ÒÔÓÎÀëSO2¼ÆËã)µÄ·½°¸ÈçÏ£º

 (ÒÑÖª£ºµÎ¶¨Ê±·´Ó¦µÄ»¯Ñ§·½³ÌʽΪSO2£«I2£«2H2O===H2SO4£«2HI)

¢Ù°´ÉÏÊö·½°¸ÊµÑ飬ÏûºÄ±ê×¼I2ÈÜÒº25.00 mL£¬¸Ã´ÎʵÑé²âµÃÑùÆ·Öп¹Ñõ»¯¼ÁµÄ²ÐÁôÁ¿(ÒÔÓÎÀëSO2¼ÆËã)Ϊ________g¡¤L£­1¡£

¢ÚÔÚÉÏÊöʵÑé¹ý³ÌÖУ¬ÈôÓв¿·ÖHI±»¿ÕÆøÑõ»¯£¬Ôò²â¶¨½á¹û________(Ìî¡°Æ«¸ß¡±¡°Æ«µÍ¡±»ò¡°²»±ä¡±)¡£


Ñõ»¯Ð¿Îª°×É«·ÛÄ©£¬¿ÉÓÃÓÚʪÕѢµÈƤ·ô²¡µÄÖÎÁÆ¡£´¿»¯¹¤Òµ¼¶Ñõ»¯Ð¿(º¬ÓÐFe(¢ò), Mn(¢ò), Ni(¢ò)µÈÔÓÖÊ)µÄÁ÷³ÌÈçÏÂ:

 


¹¤ÒµZnO              ½þ³öÒº                                ÂËÒº             

 


           ÂËÒº                        Â˱ý             ZnO

Ìáʾ:ÔÚ±¾ÊµÁ³Ìõ¼þÏ£¬Ni(¢ò)²»Äܱ»Ñõ»¯:¸ßÃÌËá¼ØµÄ»¹Ô­²úÎïÊÇMnO2

»Ø´ðÏÂÁÐÎÊÌâ:

(1)·´Ó¦¢ÚÖгýµôµÄÔÓÖÊÀë×ÓÊÇ        £¬·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ

                                                                      £»

¼Ó¸ßÃÌËá¼ØÈÜҺǰ£¬ÈôpH½ÏµÍ£¬¶Ô³ýÔÓµÄÓ°ÏìÊÇ                                 £»

(2)·´Ó¦¢ÛµÄ·´Ó¦ÀàÐÍΪ         .¹ýÂ˵õ½µÄÂËÔüÖУ¬³ýÁ˹ýÁ¿µÄпÍ⻹ÓР          £»

(3)·´Ó¦¢ÜÐγɵijÁµíÒªÓÃˮϴ£¬¼ìÑé³ÁµíÊÇ·ñÏ´µÓ¸É¾»µÄ·½·¨ÊÇ

                                                                         ¡£

(4)·´Ó¦¢ÜÖвúÎïµÄ³É·Ö¿ÉÄÜÊÇZnCO3¡¤xZn(OH)2 .È¡¸É²ÙºóµÄÂ˱ý11.2g£¬ìÑÉÕºó¿ÉµÃµ½²úÆ·8.1 g. ÔòxµÈÓÚ                ¡£

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

¾«Ó¢¼Ò½ÌÍø