ÌâÄ¿ÄÚÈÝ

ÂÌ·¯£¨FeSO4?7H2O£©ÊÇÖÎÁÆȱÌúÐÔƶѪµÄÌØЧҩ¡£Ä³Ñ§Ð£µÄ»¯Ñ§ÐËȤС×éµÄͬѧ¶ÔÂÌ·¯½øÐÐÁËÈçϵÄ̽¾¿£º

FeSO4?7H2OµÄÖƱ¸

¸Ã»¯Ñ§ÐËȤС×éµÄͬѧÔÚʵÑéÊÒͨ¹ýÈçÏÂʵÑéÓÉ·ÏÌúм£¨º¬ÉÙÁ¿Ñõ»¯Í­¡¢Ñõ»¯ÌúµÈÔÓÖÊ£©ÖƱ¸FeSO4¡¤7H2O¾§Ì壺

¢Ù½«5%Na2CO3ÈÜÒº¼ÓÈ뵽ʢÓÐÒ»¶¨Á¿·ÏÌúмµÄÉÕ±­ÖУ¬¼ÓÈÈÊý·ÖÖÓ£¬ÓÃÇãÎö·¨³ýÈ¥

Na2CO3ÈÜÒº£¬È»ºó½«·ÏÌúмÓÃˮϴµÓ2¡«3±é¡£

¢ÚÏòÏ´µÓ¹ýµÄ·ÏÌúмÖмÓÈë¹ýÁ¿µÄÏ¡ÁòËᣬ¿ØÖÆζÈÔÚ50¡«80¡æÖ®¼äÖÁÌúмºÄ¾¡£»

¢Û³ÃÈȹýÂË£¬½«ÂËҺתÈëµ½ÃܱÕÈÝÆ÷ÖУ¬¾²Öá¢ÀäÈ´½á¾§£»

¢Ü´ý½á¾§Íê±Ïºó£¬Â˳ö¾§Ì壬ÓÃÉÙÁ¿±ùˮϴµÓ2¡«3´Î£¬ÔÙÓÃÂËÖ½½«¾§ÌåÎü¸É£»

¢Ý½«ÖƵõÄFeSO4¡¤7H2O¾§Ìå·ÅÔÚÒ»¸öС¹ã¿ÚÆ¿ÖУ¬Ãܱձ£´æ¡£

Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©ÊµÑé²½Öè¢ÙµÄÄ¿µÄÊÇ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡£

£¨2£©ÊµÑé²½Öè¢ÚÃ÷ÏÔ²»ºÏÀí£¬ÀíÓÉÊÇ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡£

£¨3£©ÎªÁËÏ´µÓ³ýÈ¥¾§Ìå±íÃ渽×ŵÄÁòËáµÈÔÓÖÊ£¬ÊµÑé²½Öè¢ÜÖÐÓÃÉÙÁ¿±ùˮϴµÓ¾§Ì壬ԭÒòÊÇ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡ ¡£

£¨¶þ£©Ì½¾¿ÂÌ·¯£¨FeSO4¡¤7H2O£©ÈÈ·Ö½âµÄ²úÎï

ÒÑÖªSO3µÄÈÛµãÊÇ16.8¡ãC£¬·ÐµãÊÇ44.8¡ãC£¬¸ÃС×éÉè¼ÆÈçÏÂͼËùʾµÄʵÑé×°Öã¨Í¼ÖмÓÈÈ¡¢¼Ð³ÖÒÇÆ÷µÈ¾ùÊ¡ÂÔ£©£º

¡¾ÊµÑé¹ý³Ì¡¿

¢ÙÒÇÆ÷Á¬½Óºó£¬¼ì²é×°ÖÃAÓëBÆøÃÜÐÔ£»

¢ÚÈ¡Ò»¶¨Á¿ÂÌ·¯¹ÌÌåÖÃÓÚAÖУ¬Í¨ÈëN2ÒÔÇý¾¡×°ÖÃÄڵĿÕÆø£¬¹Ø±Õk£¬Óþƾ«µÆ¼ÓÈÈÓ²Öʲ£Á§¹Ü£»

¢Û¹Û²ìµ½A ÖйÌÌåÖð½¥±äºì×ØÉ«£¬BÖÐÊÔ¹ÜÊÕ¼¯µ½ÎÞÉ«ÒºÌ壬CÖÐÈÜÒºÍÊÉ«£»

¢Ü´ýAÖз´Ó¦ÍêÈ«²¢ÀäÈ´ÖÁÊÒκó£¬È¡ÉÙÁ¿·´Ó¦ºó¹ÌÌåÓÚÊÔ¹ÜÖУ¬¼ÓÈëÁòËáÈܽ⣬ȡÉÙÁ¿µÎÈ뼸µÎKSCNÈÜÒº£¬ÈÜÒº±äºìÉ«£»

¢ÝÍùB×°ÖõÄÊÔ¹ÜÖеÎÈ뼸µÎBaCl2ÈÜÒº£¬ÈÜÒº±ä»ë×Ç¡£

(4£©ÊµÑé½á¹û·ÖÎö

½áÂÛ1£ºBÖÐÊÕ¼¯µ½µÄÒºÌåÊÇ?????????????????? £»

½áÂÛ2£ºCÖÐÈÜÒºÍÊÉ«£¬¿ÉÍÆÖª²úÎïÖÐÓÐ???? ?????????????? £»

½áÂÛ3£º×ۺϷÖÎöÉÏÊöʵÑé¢ÛºÍ¢Ü¿ÉÍÆÖª¹ÌÌå²úÎïÒ»¶¨ÓÐFe2O3¡£

¡¾ÊµÑ鷴˼¡¿

£¨5£©ÇëÖ¸³ö¸ÃС×éÉè¼ÆµÄʵÑé×°ÖõÄÃ÷ÏÔ²»×㣺??????????????????????????? ¡£

£¨6£©·Ö½âºóµÄ¹ÌÌåÖпÉÄܺ¬ÓÐÉÙÁ¿FeO£¬È¡ÉÏÊöʵÑé¢ÜÖÐÑÎËáÈܽâºóµÄÈÜÒºÉÙÐíÓÚÊÔ¹ÜÖУ¬Ñ¡ÓÃÒ»ÖÖÊÔ¼Á¼ø±ð£¬¸ÃÊÔ¼Á×îºÏÊʵÄÊÇ?????????? ¡£

a£®ÂÈË®ºÍKSCNÈÜÒº???? b£®ËáÐÔKMnO4ÈÜÒº????? c£®H2O2???? d£®NaOHÈÜÒº

 

¡¾´ð°¸¡¿

£¨£±£©³ýÓÍÎÛ£»

£¨2£©Ó¦¸ÃÌúм¹ýÁ¿£¨»ò·´Ó¦ºóÈÜÒºÖбØÐëÓÐÌúÊ£Óࣩ£¬·ñÔòÈÜÒºÖпÉÄÜÓÐFe3+´æÔÚ£»

£¨3£©ÓñùˮϴµÓ¿É½µµÍÏ´µÓ¹ý³ÌÖÐFeSO4¡¤7H2OµÄËðºÄ£»

£¨4£©H2SO4ÈÜÒº¡¢SO2£»

£¨5£©ÔÚC×°ÖúóÔö¼ÓÒ»Ì×βÆø´¦Àí×°Öã»?

£¨6£©b¡£

¡¾½âÎö¡¿

ÊÔÌâ·ÖÎö£º£¨1£©Ì¼ËáÄÆÈÜÒºÓë·ÏÌúмµÄ³É·Ö¾ù²»ÄÜ·´Ó¦£¬¹ÊÖ»ÆðÈ¥ÓÍÎÛµÄ×÷Ó㻣¨2£©½áºÏ·ÏÌúмµÄ³É·ÖºÍºó±ßµÄ²½Öè¿ÉÖª£¬±ØÐë±£Ö¤ÌúÓÐÊ£Ó࣬²ÅÄܱ£Ö¤ÈÜÒºÖÐûÓÐÌúÀë×Ó£»£¨3£©½µµÍζȣ¬¼õСÂÌ·¯µÄÈܽâ¶È£»£¨4£©ÁòËáÑÇÌú¾§ÌåÊÜÈÈÓÐË®ÕôÆûÉú³É£¬¸ù¾ÝÈýÑõ»¯ÁòµÄÈÛµã¿ÉÖª£¬±ùË®Äܽ«ÈýÑõ»¯ÁòºÍˮҺ»¯£¬ÔÚBÖз´Ó¦Éú³ÉÁòË᣻CÖÐÊÇÉú³ÉµÄ¶þÑõ»¯ÁòʹƷºìÈÜÒºÍÊÉ«£»£¨5£©¶þÑõ»¯ÁòÒª½øÐÐβÆø´¦Àí£»£¨6£©¹ÌÌåÖÐÓÐÑõ»¯Ìú£¬ÈÜÓÚÑÎËáºóÓÐÈý¼ÛÌúÉú³É£¬ÒªÏë¼ìÑé¶þ¼ÛÌúµÄ´æÔÚ£¬ÐèÀûÓÃÑÇÌúÀë×ӵĻ¹Ô­ÐÔ£¬Ê¹ËáÐÔ¸ßÃÌËá¼ØÈÜÒºÍÊÉ«£¬¹ÊÑ¡b¡£

¿¼µã£º¿¼²éÑÇÌúÀë×ÓºÍÌúÀë×ӵļìÑé¡¢¶þÑõ»¯ÁòµÄƯ°××÷Óá¢ÎïÖʵÄÖƱ¸ºÍÓ¦ÓÃÌâÄ¿ÐÅÏ¢½âÌâµÄÄÜÁ¦

 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
£¨2010?ÄÏͨģÄ⣩ij»¯Ñ§Ñо¿ÐÔѧϰС×éͨ¹ý²éÔÄ×ÊÁÏ£¬Éè¼ÆÁËÈçͼËùʾµÄ·½·¨ÒÔº¬Äø·Ï´ß»¯¼ÁΪԭÁÏÀ´ÖƱ¸NiSO4?7H2O£®ÒÑ֪ij»¯¹¤³§µÄº¬Äø´ß»¯¼ÁÖ÷Òªº¬ÓÐNi£¬»¹º¬ÓÐAl£¨31%£©¡¢Fe£¨1.3%£©µÄµ¥Öʼ°Ñõ»¯ÎÆäËû²»ÈÜÔÓÖÊ£¨3.3%£©£®

²¿·ÖÑôÀë×ÓÒÔÇâÑõ»¯ÎïÐÎʽÍêÈ«³ÁµíʱµÄpHÈçÏ£º
³ÁµíÎï Al£¨OH£©3 Fe£¨OH£©3 Fe£¨OH£©2 Ni£¨OH£©2
pH 5.2 3.2 9.7 9.2
»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©²Ù×÷a¡¢cÖÐÐèʹÓõÄÒÇÆ÷³ýÌú¼Ų̈£¨´øÌúȦ£©¡¢¾Æ¾«µÆ¡¢ÉÕ±­¡¢²£Á§°ôÍ⻹ÐèÒªµÄÖ÷ÒªÒÇÆ÷Ϊ
©¶·¡¢Õô·¢Ãó
©¶·¡¢Õô·¢Ãó
£®
£¨2£©¡°¼î½þ¡±¹ý³ÌÖз¢ÉúµÄÀë×Ó·½³ÌʽÊÇ
2Al+2OH-+2H2O¨T2AlO2-+3H2¡ü¡¢Al2O3+2OH-¨T2AlO2-+3H2O
2Al+2OH-+2H2O¨T2AlO2-+3H2¡ü¡¢Al2O3+2OH-¨T2AlO2-+3H2O
£®
£¨3£©¡°Ëá½þ¡±Ê±Ëù¼ÓÈëµÄËáÊÇ
H2SO4
H2SO4
£¨Ìѧʽ£©£®Ëá½þºó£¬¾­²Ù×÷a·ÖÀë³ö¹ÌÌå¢Ùºó£¬ÈÜÒºÖпÉÄܺ¬ÓеĽðÊôÀë×ÓÊÇ
Ni2+¡¢Fe2+
Ni2+¡¢Fe2+
£®
£¨4£©²Ù×÷bΪµ÷½ÚÈÜÒºµÄpH£¬ÄãÈÏΪpHµÄ×î¼Ñµ÷¿Ø·¶Î§ÊÇ
3.2-9.2
3.2-9.2
£®
£¨5£©¡°µ÷pHΪ2¡«3¡±µÄÄ¿µÄÊÇ
·ÀÖ¹ÔÚŨËõ½á¾§¹ý³ÌÖÐNi2+Ë®½â
·ÀÖ¹ÔÚŨËõ½á¾§¹ý³ÌÖÐNi2+Ë®½â
£®
£¨6£©²úÆ·¾§ÌåÖÐÓÐʱ»á»ìÓÐÉÙÁ¿ÂÌ·¯£¨FeSO4?7H2O£©£¬ÆäÔ­Òò¿ÉÄÜÊÇ
H2O2µÄÓÃÁ¿²»×㣨»òH2O2ʧЧ£©¡¢±£ÎÂʱ¼ä²»×ãµ¼ÖÂFe2+δ±»ÍêÈ«Ñõ»¯Ôì³ÉµÄ
H2O2µÄÓÃÁ¿²»×㣨»òH2O2ʧЧ£©¡¢±£ÎÂʱ¼ä²»×ãµ¼ÖÂFe2+δ±»ÍêÈ«Ñõ»¯Ôì³ÉµÄ
£®
´Óº¬Äø·Ï´ß»¯¼ÁÖпɻØÊÕÄø£¬ÆäÁ÷³ÌÈçÏ£º

ijÓÍÖ¬»¯¹¤³§µÄº¬Äø´ß»¯¼ÁÖ÷Òªº¬ÓÐNi£¬»¹º¬ÓÐAl£¨31%£©¡¢Fe£¨1.3%£©µÄµ¥Öʼ°Ñõ»¯ÎÆäËû²»ÈÜÔÓÖÊ£¨3.3%£©£®²¿·ÖÑôÀë×ÓÒÔÇâÑõ»¯ÎïÐÎʽÍêÈ«³ÁµíʱµÄpHÈçÏ£º
³ÁµíÎï Al£¨OH£©3 Fe£¨OH£©3 Fe£¨OH£©2 Ni£¨OH£©2
pH 5.2 3.2 9.7 9.2
»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©¡°¼î½þ¡±µÄÄ¿µÄÊdzýÈ¥
Al¡¢Al2O3¡¢ÓÍÖ¬
Al¡¢Al2O3¡¢ÓÍÖ¬
£®
£¨2£©¡°Ëá½þ¡±Ê±Ëù¼ÓÈëµÄËáÊÇ
H2SO4
H2SO4
£¨Ìѧʽ£©£®Ëá½þºó£¬ÂËÒº¢ÚÖпÉÄܺ¬ÓеĽðÊôÀë×ÓÊÇ
Ni2+¡¢Fe2+
Ni2+¡¢Fe2+
£®
£¨3£©¡°µ÷pHΪ2¡«3¡±µÄÄ¿µÄÊÇ
·ÀÖ¹ÔÚŨËõ½á¾§¹ý³ÌÖÐNi2+Ë®½â
·ÀÖ¹ÔÚŨËõ½á¾§¹ý³ÌÖÐNi2+Ë®½â
£®
£¨4£©²úÆ·¾§ÌåÖÐÓÐʱ»á»ìÓÐÉÙÁ¿ÂÌ·¯£¨FeSO4?7H2O£©£¬¿ÉÄÜÊÇÓÉÓÚÉú²ú¹ý³ÌÖÐ
H2O2µÄÓÃÁ¿²»×㣨»òH2O2ʧЧ£©¡¢±£ÎÂʱ¼ä²»×ãµ¼ÖÂFe2+δ±»ÍêÈ«Ñõ»¯Ôì³ÉµÄ
H2O2µÄÓÃÁ¿²»×㣨»òH2O2ʧЧ£©¡¢±£ÎÂʱ¼ä²»×ãµ¼ÖÂFe2+δ±»ÍêÈ«Ñõ»¯Ôì³ÉµÄ
£®
£¨5£©NiSO4ÔÚÇ¿¼îÈÜÒºÖÐÓÃNaClOÑõ»¯£¬¿ÉÖƵüîÐÔÄøïÓµç³Øµç¼«²ÄÁÏ--NiOOH£®¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ
2Ni2++ClO-+4OH-=2NiOOH¡ý+Cl-+H2O
2Ni2++ClO-+4OH-=2NiOOH¡ý+Cl-+H2O
£®
Èý²ÝËáºÏÌú£¨¢ó£©Ëá¼Ø¾§Ì壨K3[Fe£¨C2O4£©3]?3H2O£©ÓкÜÖØÒªµÄÓÃ;£®¿ÉÓÃÈçͼÁ÷³ÌÀ´ÖƱ¸£®¸ù¾ÝÌâÒâÍê³ÉÏÂÁи÷Ì⣺

£¨1£©ÈôÓÃÌúºÍÏ¡ÁòËáÖƱ¸ÂÌ·¯£¨FeSO4?7H2O£©¹ý³ÌÖУ¬ÆäÖÐ
Ìú
Ìú
£¨ÌîÎïÖÊÃû³Æ£©ÍùÍùÒª¹ýÁ¿£¬ÀíÓÉÊÇ
·ÀÖ¹Fe2+±»Ñõ»¯
·ÀÖ¹Fe2+±»Ñõ»¯
£®
£¨2£©Òª´ÓÈÜÒºÖеõ½ÂÌ·¯£¬±ØÐë½øÐеÄʵÑé²Ù×÷ÊÇ
bcae
bcae
£®£¨°´Ç°ºó˳ÐòÌ
a£®¹ýÂËÏ´µÓ     b£®Õô·¢Å¨Ëõ     c£®ÀäÈ´½á¾§    d£®×ÆÉÕ    e£®¸ÉÔï
ijÐËȤС×éΪ²â¶¨Èý²ÝËáºÏÌúËá¼Ø¾§Ì壨K3[Fe£¨C2O4£©3]?3H2O£©ÖÐÌúÔªËغ¬Á¿£¬×öÁËÈçÏÂʵÑ飺
²½Öè1£º³ÆÁ¿5.000gÈý²ÝËáºÏÌúËá¼Ø¾§Ì壬ÅäÖƳÉ250mlÈÜÒº£®
²½Öè2£ºÈ¡ËùÅäÈÜÒº25.00mlÓÚ׶ÐÎÆ¿ÖУ¬¼ÓÏ¡H2SO4Ëữ£¬µÎ¼ÓKMnO4ÈÜÒºÖÁ²ÝËá¸ùÇ¡ºÃÈ«²¿±»Ñõ»¯³É¶þÑõ»¯Ì¼£¬Í¬Ê±£¬MnO4-±»»¹Ô­³ÉMn2+£®Ïò·´Ó¦ºóµÄÈÜÒºÖмÓÈëÒ»¶¨Á¿Ð¿·Û£¬¼ÓÈÈÖÁ»ÆÉ«¸ÕºÃÏûʧ£¬¹ýÂË£¬Ï´µÓ£¬½«¹ýÂ˼°Ï´µÓËùµÃÈÜÒºÊÕ¼¯µ½×¶ÐÎÆ¿ÖУ¬´Ëʱ£¬ÈÜÒºÈÔÀïËáÐÔ£®
²½Öè3£ºÔÚËáÐÔÌõ¼þÏ£¬ÓÃ0.010mol/L KMnO4ÈÜÒºµÎ¶¨²½Öè¶þËùµÃÈÜÒºÖÁÖյ㣬¹²×öÈý´ÎʵÑ飬ƽ¾ùÏûºÄKMnO4ÈÜÒº20.00ml£¬µÎ¶¨ÖÐMnO4-£¬±»»¹Ô­³ÉMn2+£®
£¨3£©²½Öè1ÖУ¬ÅäÖÆÈý²ÝËáºÏÌúËá¼ØÈÜÒºÐèҪʹÓõIJ£Á§ÒÇÆ÷³ýÉÕ±­¡¢²£Á§°ôÒÔÍ⻹ÓÐ
250mlÈÝÁ¿Æ¿
250mlÈÝÁ¿Æ¿
£»Ö÷Òª²Ù×÷²½ÖèÒÀ´ÎÊÇ£º³ÆÁ¿¡¢Èܽ⡢תÒÆ¡¢
Ï´µÓ
Ï´µÓ
¡¢¶¨ÈÝ¡¢Ò¡ÔÈ£®
£¨4£©²½Öè2ÖУ¬¼ÓÈëп·ÛµÄÄ¿µÄÊÇ
»¹Ô­Fe3+ΪFe2+
»¹Ô­Fe3+ΪFe2+
£®
£¨5£©²½Öè3ÖУ¬·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ£º
MnO4-+5Fe2++8H+¨TMn2++5Fe3++4H2O
MnO4-+5Fe2++8H+¨TMn2++5Fe3++4H2O
£®
£¨6£©²½Öè2ÖУ¬Èô¼ÓÈëµÄKMnO4µÄÈÜÒºµÄÁ¿²»¹»£¬Ôò²âµÃµÄÌúº¬Á¿
Æ«¸ß
Æ«¸ß
£®£¨Ñ¡Ìî¡°Æ«µÍ¡±¡¢¡°Æ«¸ß¡±¡¢¡°²»±ä¡±£©
£¨7£©Ä³Í¬Ñ§½«8.74gÎÞË®Èý²ÝËáºÏÌúËá¼Ø£¨K3[Fe£¨C2O4£©3]£©ÔÚÒ»¶¨Ìõ¼þϼÓÈȷֽ⣬ËùµÃ¹ÌÌåµÄÖÊÁ¿Îª5.42g£¬Í¬Ê±µÃµ½ÃܶÈΪ1.647g/L£¨ÒÑÕۺϳɱê×¼×´¿öÏ£©ÆøÌ壮Ñо¿¹ÌÌå²úÎïµÃÖª£¬ÌúÔªËز»¿ÉÄÜÒÔÈý¼ÛÐÎʽ´æÔÚ£¬¶øÑÎÖ»ÓÐK2CO3£®Ð´³ö¸Ã·Ö½â·´Ó¦µÄ»¯Ñ§·½³Ìʽ
2K3[Fe£¨C2O4£©3]¨T3K2CO3+Fe+FeO+4CO+5CO2
2K3[Fe£¨C2O4£©3]¨T3K2CO3+Fe+FeO+4CO+5CO2
£®
ʵÑéÊÒÀûÓÃÁòË᳧ÉÕÔü£¨Ö÷Òª³É·ÖΪÌúµÄÑõ»¯Îï¼°ÉÙÁ¿FeS¡¢SiO2µÈ£©ÖƱ¸¾ÛÌú£¨¼îʽÁòËáÌúµÄ¾ÛºÏÎºÍÂÌ·¯£¨FeSO4?7H2O£©£¬¹ý³ÌÈçÏ£º

£¨1£©½«¹ý³Ì¢ÚÖеIJúÉúµÄÆøÌåͨÈëÏÂÁÐÈÜÒºÖУ¬ÈÜÒº»áÍÊÉ«µÄÊÇ
ACD
ACD
£»
A£®Æ·ºìÈÜÒº     B£®×ÏɫʯÈïÈÜÒº     C£®ËáÐÔKMnO4ÈÜÒº     D£®äåË®
£¨2£©¹ý³Ì¢ÙÖУ¬FeSºÍO2¡¢H2SO4·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º
4FeS+3O2+6H2SO4=2Fe2£¨SO4£©3+6H2O+4S
4FeS+3O2+6H2SO4=2Fe2£¨SO4£©3+6H2O+4S
£»
£¨3£©¹ý³Ì¢ÛÖУ¬Ðè¼ÓÈëµÄÎïÖÊÊÇ
Fe
Fe
£»
£¨4£©¹ý³Ì¢ÜÖУ¬Õô·¢½á¾§ÐèҪʹÓþƾ«µÆ¡¢Èý½Ç¼Ü£¬»¹ÐèÒªµÄÒÇÆ÷ÓÐ
Õô·¢Ãó¡¢²£Á§°ô
Õô·¢Ãó¡¢²£Á§°ô
£»
£¨5£©¹ý³Ì¢Ýµ÷½ÚpH¿ÉÑ¡ÓÃÏÂÁÐÊÔ¼ÁÖеÄ
C
C
£¨ÌîÑ¡ÏîÐòºÅ£©£»
A£®Ï¡ÁòËá         B£®CaCO3       C£®NaOHÈÜÒº
£¨6£©¹ý³Ì¢ÞÖУ¬½«ÈÜÒºZ¼ÓÈȵ½70-80¡æ£¬Ä¿µÄÊÇ
´Ù½øFe3+µÄË®½â
´Ù½øFe3+µÄË®½â
£»
£¨7£©ÊµÑéÊÒΪ²âÁ¿ËùµÃµ½µÄ¾ÛÌúÑùÆ·ÖÐÌúÔªËصÄÖÊÁ¿·ÖÊý£¬½øÐÐÏÂÁÐʵÑ飮
¢ÙÓ÷ÖÎöÌìƽ³ÆÈ¡2.700gÑùÆ·£»
¢Ú½«ÑùÆ·ÈÜÓÚ×ãÁ¿µÄÑÎËáºó£¬¼ÓÈë¹ýÁ¿µÄÂÈ»¯±µÈÜÒº£»
¢Û¹ýÂË¡¢Ï´µÓ¡¢¸ÉÔ³ÆÁ¿£¬µÃ¹ÌÌåÖÊÁ¿Îª3.495g£®Èô¸Ã¾ÛÌúÖ÷Òª³É·ÖΪ[£¨Fe£¨OH£©£¨SO4£©]n£»
Ôò¸Ã¾ÛÌúÑùÆ·ÖÐÌúÔªËصÄÖÊÁ¿·ÖÊýΪ
31.1%
31.1%
£®£¨¼ÙÉèÔÓÖÊÖв»º¬ÌúÔªËغÍÁòÔªËØ£©

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

¾«Ó¢¼Ò½ÌÍø