ÌâÄ¿ÄÚÈÝ

½üÄêÀ´£¬ËáÓêÎÛȾ½ÏΪÑÏÖØ£¬·ÀÖÎËáÓê³ÉÁËÆÈÔÚü½ÞµÄÎÊÌâ¡£
(1)ÓÐÈËÌá³öÁËÒ»ÖÖÀûÓÃÂȼҵ²úÆ·´¦Àíº¬¶þÑõ»¯Áò·ÏÆøµÄ·½·¨£¬Á÷³ÌÈçÏ£º
(¢ñ)½«º¬SO2µÄ·ÏÆøͨÈëµç½â±¥ºÍʳÑÎË®ºóËùµÃµ½µÄÈÜÒºÖУ¬µÃNaHSO3ÈÜÒº¡£
(¢ò)½«µç½â±¥ºÍʳÑÎË®ËùµÃÆøÌå·´Ó¦ºóÖƵÃÑÎËá¡£
(¢ó)½«ÑÎËá¼ÓÈëNaHSO3ÈÜÒºÖУ¬·´Ó¦ËùµÃµ½µÄSO2ÆøÌå»ØÊÕ£¬Éú³ÉµÄNaClÑ­»·ÀûÓá£
¢Ùд³ö²½Öè(¢ñ)·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º                                    ¡£
¢Úд³ö²½Öè(¢ò)Öеç½â±¥ºÍʳÑÎË®µÄ»¯Ñ§·½³Ìʽ£º                              ¡£
¢Ûд³ö²½Öè(¢ó)·´Ó¦µÄÀë×Ó·½³Ìʽ£º                                     ¡£
(2)»¹ÓÐѧÕßÌá³öÀûÓÃFe2£«¡¢Fe3£«µÈÀë×ӵĴ߻¯×÷Ó㬳£ÎÂϽ«SO2Ñõ»¯³ÉSO42-¶øʵÏÖSO2µÄ»ØÊÕÀûÓá£Ä³Ñо¿ÐÔѧϰС×é¾Ý´ËÉè¼ÆÁËÈçÏ·½°¸£¬ÔÚʵÑéÊÒÌõ¼þϲⶨת»¯Æ÷ÖÐSO2Ñõ»¯³ÉSO42-µÄת»¯ÂÊ¡£

¢Ù¸ÃС×é²ÉÓÃÏÂͼװÖÃÔÚʵÑéÊҲⶨģÄâÑÌÆøÖÐSO2µÄÌå»ý·ÖÊý£¬XÈÜÒº¿ÉÒÔÊÇ       ¡£(ÌîдÐòºÅ)
A£®µâµÄµí·ÛÈÜÒºB£®ËáÐÔ¸ßÃÌËá¼ØÈÜÒº
C£®ÇâÑõ»¯ÄÆÈÜÒºD£®ÂÈ»¯±µÈÜÒº
¢ÚÈôÉÏÊöʵÑéÊÇÔÚ±ê×¼×´¿öϽøÐеģ¬Óû²â¶¨×ª»¯Æ÷ÖÐSO2Ñõ»¯³ÉSOµÄת»¯ÂÊ£¬ÒÑÖªÆøÌåÁ÷ËÙ£¬»¹Ðè²â¶¨µÄÊý¾ÝÓР                 ¡¢                          ¡£
(1)¢ÙSO2£«NaOH=NaHSO3
¢Ú2NaCl£«2H2O2NaOH£«H2¡ü£«Cl2¡ü
¢ÛHSO3-£«H£«=SO2¡ü£«H2O
(2)¢ÙA¡¢B¡¡¢ÚʵÑéʱ¼ä¡¡¼ÓÈëÑÎËáËữµÄBaCl2ÈÜÒººóÉú³É³ÁµíµÄÖÊÁ¿
Çó½âSO2ÔÚ»ìºÏÆøÖеÄÌå»ý·ÖÊý£¬ÐèÒªÇó³öÁ½¸öÁ¿£¬Ò»¸öÊÇSO2µÄÌå»ý£¬Ò»¸öÊÇ»ìºÏÆøµÄ×ÜÌå»ý¡£ÀûÓÃÁ¿Æø×°Ö㬿ÉÇó³öÎüÊÕSO2ºóÓàÆøµÄÌå»ý£»ÀûÓÃÏ´ÆøÆ¿ÖеÄXÈÜÒº£¬Çó³öSO2µÄÌå»ý¡£ËùÒÔ¶ÔÓÚÏ´ÆøÆ¿ÖеÄÈÜÒº£¬±ØÐëÄÜÓëSO2·´Ó¦£¬ÇÒÄÜ·¢ÉúÃ÷ÏÔµÄÑÕÉ«±ä»¯£¬ÒÔ±ãÈ·¶¨·´Ó¦µÄÖյ㡣Çó½âSO2µÄת»¯ÂÊ£¬Ó¦Çó³öSO2ÆøÌåµÄ×ÜÁ¿ºÍÉú³ÉSOµÄÁ¿¡£ÒòΪÔÚÇ°Ò»ÎÊÖÐÒÑÇó³öSO2ÔÚ»ìºÏÆøÖеĺ¬Á¿£¬ËùÒÔÖ»ÐèÈ·¶¨»ìºÏÆøµÄ×ÜÁ¿¡£ÀûÓÃÁ÷ËÙÇó×ÜÁ¿£¬Ö»ÐèÖªµÀͨÈëʱ¼ä¡£ÒªÇóSO2µÄת»¯ÂÊ£¬Ö»ÐèÈ·¶¨³ÁµíÖÊÁ¿¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
·ÖijͬѧΪ̽¾¿Í­¸úŨÁòËáµÄ·´Ó¦Çé¿ö£¬ÓÃÏÂͼËùʾװÖýøÐÐÁËÓйØʵÑé

£¨1£©BÊÇÓÃÀ´ÊÕ¼¯ÊµÑéÖÐÉú³ÉµÄÆøÌåµÄ×°Ö㬵«Î´½«µ¼¹Ü»­È«£¬ÇëÔÚ´ðÌâ¾íÉϰѵ¼¹Ü²¹³ä
ÍêÕû        ¡£
£¨2£©Çëд³öÍ­¸úŨÁòËá·´Ó¦µÄ»¯Ñ§·½³Ìʽ         ¡£
£¨3£©ÊµÑéÖÐDÖеÄÏÖÏó         ¡£
£¨4£©ÊµÑéÖУ¬¸Ãͬѧȡ6.4gͭƬºÍ12mL 18mol¡¤L-1H2SO4ÈÜÒº·ÅÔÚÔ²µ×ÉÕÆ¿Öй²ÈÈ£¬Ö±µ½·´Ó¦Í£Ö¹£¬×îºó·¢ÏÖÉÕÆ¿Öл¹ÓÐͭƬʣÓࣻ¸Ãͬѧ¸ù¾ÝËùѧµÄ»¯Ñ§ÖªÊ¶Åжϻ¹»áÓÐÒ»¶¨Á¿H2SO4µÄÊ£Ó࣬ÏÂÁÐÒ©Æ·ÖÐÄܹ»ÓÃÀ´ÑéÖ¤·´Ó¦Í£Ö¹ºóµÄÉÕÆ¿ÖÐÈ·ÓÐÊ£ÓàµÄH2SO4µÄÊÇ         Ìî×Öĸ±àºÅ£©¡£
a£®BaCl2ÈÜÒº    b£®Ba(NO3)2ÈÜÒº     c£®Òø·Û     d£®Na2CO3·ÛÄ©
£¨5£©Èô½«16gÍ­Óë50mLH2SO4ÎïÖʵÄÁ¿Å¨¶ÈΪһ¶¨ÖµµÄŨÁòËá·´Ó¦£¬Í­ÍêÈ«Èܽ⡣Çë»Ø´ð£º
¢Ù·´Ó¦ÖвúÉúµÄÆøÌåÔÚ±ê×¼×´¿öϵÄÌå»ýΪ         L¡£
¢Ú¸Ã·´Ó¦Öб»»¹Ô­µÄH2SO4µÄÎïÖʵÄÁ¿Îª         mol¡£
¢Û´ý²úÉúµÄÆøÌåÈ«²¿Êͷźó£¬ÏòÈÜÒºÖеμÓVmL a mol¡¤L-1NaOHÈÜÒº£¬Ç¡ºÃʹÈÜÒºÖР
µÄCu2+È«²¿×ª»¯Îª³Áµí£¬ÔòԭŨÁòËáÖÐH2SO4µÄÎïÖʵÄÁ¿Å¨¶È=         mol¡¤L-1¡£
ʵÑéÊÒÓÃÍ­ÖƱ¸CuSO4ÈÜÒºÓжàÖÖ·½°¸£¬Ä³ÊµÑéС×é¸ø³öÁËÒÔÏÂÈýÖÖ·½°¸£º

Çë»Ø´ðÓйØÎÊÌ⣺
£¨1£©¼×·½°¸£º
¢Ùд³ö¸Ã·´Ó¦µÄÀë×Ó·½³Ìʽ                  £»
¢ÚΪÁ˽ÚÔ¼Ô­ÁÏ£¬ÁòËáºÍÏõËáµÄÎïÖʵÄÁ¿Ö®±È×î¼ÑΪ£¬n(H2SO4)£ºn(HNO3)=           ¡£
£¨2£©ÒÒ·½°¸£º½«6.4gÍ­Ë¿·Åµ½90mL 1.5mol¡¤L-1µÄÏ¡ÁòËáÖУ¬¿ØÎÂÔÚ50¡æ¡£¼ÓÈë40mL 10%µÄH2O2£¬·´Ó¦0.5Сʱ£¬Éýε½60¡æ£¬³ÖÐø·´Ó¦1Сʱºó£¬¾­Ò»ÏµÁвÙ×÷£¬µÃCuSO4¡¤5H2O 20.0g¡¾ÒÑÖªÓйØĦ¶ûÖÊ Á¿£ºM(Cu)=64g/mol£¬ M(CuSO4¡¤5H2O) =250g/mol¡¿¡£
¢Ù·´Ó¦Ê±Î¶ȿØÖÆÔÚ50¡æ~60¡æ£¬²»Ò˹ý¸ßµÄÔ­ÒòÊÇ                      £»
¢Ú±¾ÊµÑéCuSO4¡¤5H2OµÄ²úÂÊΪ                  ¡£
£¨3£©±û·½°¸£º½«¿ÕÆø»òÑõÆøÖ±½ÓͨÈ뵽ͭ·ÛÓëÏ¡ÁòËáµÄ»ìºÏÎïÖУ¬·¢ÏÖÔÚ³£ÎÂϼ¸ºõ²»·´Ó¦¡£Ïò·´Ó¦ÒºÖмÓÉÙÁ¿FeSO4£¬¼´·¢Éú·´Ó¦£¬Éú³ÉÁòËáÍ­¡£·´Ó¦ÍêÈ«ºó£¬¼ÓÎïÖÊAµ÷½ÚpHÖÁ4 £¬È»ºó¹ýÂË¡¢Å¨Ëõ¡¢½á¾§¡£
¢ÙÎïÖÊA¿ÉÑ¡ÓÃÒÔϵĠ         £¨ÌîÐòºÅ£©£»
A£®CaOB£®NaOHC£®CuCO3D£®Cu2(OH)2CO3 E£®Fe2(SO4)3
¢Ú·´Ó¦ÖмÓÈëÉÙÁ¿FeSO4¿É¼ÓËÙÍ­µÄÑõ»¯£¬FeSO4µÄ×÷ÓÃÊÇ                  £»
£¨4£©¶Ô±È¼×¡¢ÒÒ¡¢±ûÈýÖÖʵÑé·½°¸£¬±û·½°¸µÄÓŵãÓУ¨Ð´Á½Ìõ£©£º
                                  ¡¢                                 ¡£

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

¾«Ó¢¼Ò½ÌÍø