ÌâÄ¿ÄÚÈÝ

½¨ÖþÓõĺìשºÍÇàש£¬ÆäÑÕÉ«ÊÇÓÉÆäÖк¬ÓеIJ»Í¬¼Û̬µÄÌúÑõ»¯ÎïËùÖ¡£ÎÒ¹ú¹Å´úשÍß½¨ÖþÄܹ»±£³Öµ½ÏÖÔڵģ¬¼¸ºõÎÞÒ»ÀýÍâµÄÊÇÓÉÇàש½¨³É¡£ÓÐÈËÌá³öÒÔÏ¿ÉÄܵÄÔ­Òò£º
¢ÙÇàשÖꬵÄÊÇFeO£»¢ÚÇàשÖꬵÄÊÇFe3O4£»¢ÛFe2O3Óö³¤ÆÚËáÓêÇÖÊ´»á·ç»¯ÈܽâʹשÍßÇ¿¶È½µµÍ£»¢ÜFe3O4ÐÔÖÊÎȶ¨£»¢ÝFeOÐÔÖÊÎȶ¨¡£ÄãÈÏΪÓеÀÀíµÄÊÇ
A£®¢Ù¢Û¢ÝB£®¢Ú¢Û¢ÜC£®¢Û¢ÝD£®¢Ù¢Ú¢Û¢Ü¢Ý
A

ÊÔÌâ·ÖÎö£ºÇàשÊǼÓˮȱÑõÌõ¼þÏÂÉú³ÉµÄ»¹Ô­¼ÁÇâÆøºÍÒ»Ñõ»¯Ì¼°ÑÑõ»¯Ìú»¹Ô­ÎªµÍ¼ÛµÄÑõ»¯ÑÇÌúËùÖ£¬Æ俹Ñõ»¯ÐԺͿ¹¸¯Ê´ÐÔ¾ùÔöÇ¿¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
£¨15·Ö£©ÓÃFeCl3ËáÐÔÈÜÒºÍѳýH2SºóµÄ·ÏÒº£¬Í¨¹ý¿ØÖƵçѹµç½âµÃÒÔÔÙÉú¡£Ä³Í¬Ñ§Ê¹ÓÃʯīµç¼«£¬ÔÚ²»Í¬µçѹ£¨x£©Ïµç½âpH=1µÄ0.1mol/LFeCl2ÈÜÒº£¬Ñо¿·ÏÒºÔÙÉú»úÀí¡£¼Ç¼ÈçÏ£¨a¡¢b¡¢c´ú±íµçѹֵ£º£©
ÐòºÅ
µçѹ/V
Ñô¼«ÏÖÏó
¼ìÑéÑô¼«²úÎï
I
x¡Ýa
µç¼«¸½½ü³öÏÖ»ÆÉ«£¬ÓÐÆøÅݲúÉú
ÓÐFe3+¡¢ÓÐCl2
II
a£¾x¡Ýb
µç¼«¸½½ü³öÏÖ»ÆÉ«£¬ÎÞÆøÅݲúÉú
ÓÐFe3+¡¢ÎÞCl2
III
b£¾x£¾0
ÎÞÃ÷ÏԱ仯
ÎÞFe3+¡¢ÎÞCl2
£¨1£©ÓÃKSCNÈÜÒº¼ìÑé³öFe3+µÄÏÖÏóÊÇ_______¡£
£¨2£©IÖУ¬Fe2+²úÉúµÄÔ­Òò¿ÉÄÜÊÇCl-ÔÚÑô¼«·Åµç£¬Éú³ÉµÄCl2½«Fe2+Ñõ»¯¡£Ð´³öÓйط´Ó¦µÄ·½³Ìʽ_____¡£
£¨3£©ÓÉIIÍƲ⣬Fe3+²úÉúµÄÔ­Òò»¹¿ÉÄÜÊÇFe2+ÔÚÑô¼«·Åµç£¬Ô­ÒòÊÇFe2+¾ßÓÐ_____ÐÔ¡£
£¨4£©IIÖÐËäδ¼ì²â³öCl2£¬µ«Cl-ÔÚÑô¼«ÊÇ·ñ·ÅµçÈÔÐè½øÒ»²½ÑéÖ¤¡£µç½âpH=1µÄNaClÈÜÒº×ö¶ÔÕÕʵÑ飬¼Ç¼ÈçÏ£º
ÐòºÅ
µçѹ/V
Ñô¼«ÏÖÏó
¼ìÑéÑô¼«²úÎï
IV
a£¾x¡Ýc
ÎÞÃ÷ÏԱ仯
ÓÐCl2
V
c£¾x¡Ýb
ÎÞÃ÷ÏԱ仯
ÎÞCl2
¢ÙNaClÈÜÒºµÄŨ¶ÈÊÇ________mol/L¡£
¢ÚIVÖмì²âCl2µÄʵÑé·½·¨:____________________¡£
¢ÛÓëII¶Ô±È£¬µÃ³öµÄ½áÂÛ£¨Ð´³öÁ½µã£©£º___________________¡£
£¨15·Ö£©ÊµÑéÊÒÒÔ·ÏͭмΪԭÁÏÖÆÈ¡¼îʽ̼ËáÍ­¡¾Cu2(OH)2CO3¡¿µÄ²½ÖèÈçÏ£º
²½ÖèÒ»£º·ÏͭмÖÆÏõËáÍ­
Èçͼ£¬
ÓýºÍ·µÎ¹ÜÎüȡŨHNO3»ºÂý¼Óµ½×¶ÐÎÆ¿ÄڵķÏͭмÖÐ(·Ïͭм¹ýÁ¿)£¬³ä·Ö·´Ó¦ºó¹ýÂË£¬µÃµ½ÏõËáÍ­ÈÜÒº¡£
²½Öè¶þ£º¼îʽ̼ËáÍ­µÄÖƱ¸
Ïò´óÊÔ¹ÜÖмÓÈë̼ËáÄÆÈÜÒººÍÏõËáÍ­ÈÜÒº£¬Ë®Ô¡¼ÓÈÈÖÁ70¡æ×óÓÒ£¬ÓÃ0.4 mol/LµÄNaOHÈÜÒºµ÷½ÚpHÖÁ8.5£¬Õñµ´£¬¾²Ö㬹ýÂË£¬ÓÃÈÈˮϴµÓ£¬ºæ¸É£¬µÃµ½¼îʽ̼ËáÍ­²úÆ·¡£ 
²½ÖèÈý£º¼îʽ̼ËáÍ­µÄ×é³É²â¶¨
¼îʽ̼ËáÍ­¿É±íʾΪ£ºxCuCO3 ¡¤yCu (OH)2 ¡¤zH2O£¬¿É²ÉÓÃÇâÆø»¹Ô­·¨À´È·¶¨£¬Æä·´Ó¦Ô­ÀíΪ£º
xCuCO3 ¡¤yCu (OH)2 ¡¤zH2O +  H2¡ú Cu +  CO2  +  H2O£¨Î´Åäƽ£©
Íê³ÉÏÂÁÐÌî¿Õ£º
£¨1£©²½ÖèÒ»ÖУ¬·´Ó¦¿ªÊ¼Ê±£¬Æ¿ÄÚµÄÏÖÏóÊÇ                                       £¬
ÓøÃ×°ÖÃÖÆÈ¡ÏõËáÍ­£¬ºÃ´¦ÊÇ                                                  ¡£
£¨2£©²½Öè¶þÖУ¬Ë®Ô¡¼ÓÈÈËùÐèÒÇÆ÷ÓР               ¡¢                  (¼ÓÈÈ¡¢¼Ð³ÖÒÇÆ÷¡¢Ê¯ÃÞÍø³ýÍâ)£»Ï´µÓµÄÄ¿µÄÊÇ                                         ¡£
£¨3£© ²½ÖèÈýÖУ¬¢ÙÒÔ×Öx¡¢y¡¢zĸΪϵÊý£¬ÅäƽÇâÆø»¹Ô­·¨µÄ»¯Ñ§·½³Ìʽ£º
      xCuCO3 ¡¤yCu (OH)2 ¡¤zH2O+       H2¡ú     Cu+      CO2+      H2O
¢Ú³ÆÈ¡24.0gij¼îʽ̼ËáÍ­ÑùÆ·£¬³ä·Ö·´Ó¦ºóµÃµ½12.8 g²ÐÁôÎÉú³É4.4g¶þÑõ»¯Ì¼ºÍ7.2gË®¡£¸ÃÑùÆ·ÖнᾧˮÖÊÁ¿Îª        g£¬»¯Ñ§Ê½Îª                  ¡£
£¨10·Ö£©ÏÂͼÊÇÑо¿Í­ÓëŨÁòËáµÄ·´Ó¦×°Öãº

£¨1£©AÊÔ¹ÜÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ                                   ¡£
£¨2£©·´Ó¦Ò»¶Îʱ¼äºó£¬¿É¹Û²ìµ½BÊÔ¹ÜÖеÄÏÖÏóΪ                                    ¡£
£¨3£©CÊԹܿڽþÓÐNaOHÈÜÒºµÄÃÞÍÅ×÷ÓÃÊÇ                                           ¡£
£¨4£©ÊµÑé½áÊøºó£¬Ö¤Ã÷AÊÔ¹ÜÖз´Ó¦ËùµÃ²úÎïÊÇ·ñº¬ÓÐÍ­Àë×ӵIJÙ×÷·½·¨ÊÇ                                              ¡£
£¨5£©ÔÚÍ­ÓëŨÁòËá·´Ó¦µÄ¹ý³ÌÖУ¬·¢ÏÖÓкÚÉ«ÎïÖʳöÏÖ£¬¾­²éÔÄÎÄÏ×»ñµÃÏÂÁÐ×ÊÁÏ¡£
×ÊÁÏ1

¸½±íÍ­ÓëŨÁòËá·´Ó¦²úÉúºÚÉ«ÎïÖʵÄÏà¹ØÐÔÖÊ
×ÊÁÏ2
XÉäÏß¾§Ìå·ÖÎö±íÃ÷£¬Í­ÓëŨÁòËá·´Ó¦Éú³ÉµÄºÚÉ«ÎïÖÊΪCu2S¡¢CuS¡¢Cu7S4ÖеÄÒ»ÖÖ»ò¼¸ÖÖ¡£
 ½öÓÉÉÏÊö×ÊÁϿɵóöµÄÕýÈ·½áÂÛÊÇ            ¡£
A£®Í­ÓëŨÁòËᷴӦʱËùÉæ¼°µÄ·´Ó¦¿ÉÄܲ»Ö¹Ò»¸ö B£®ÁòËáŨ¶ÈÑ¡ÔñÊʵ±£¬¿É±ÜÃâ×îºó²úÎïÖгöÏÖºÚÉ«ÎïÖÊ£»C£®¸Ã·´Ó¦·¢ÉúµÄÌõ¼þÖ®Ò»ÊÇÁòËáŨ¶È¡Ý15 mol¡¤L D£®ÁòËáŨ¶ÈÔ½´ó£¬ºÚÉ«ÎïÖÊÔ½¿ì³öÏÖ¡¢Ô½ÄÑÏûʧ

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

¾«Ó¢¼Ò½ÌÍø