ÌâÄ¿ÄÚÈÝ


À¬»øÊÇ·Å´íµØ·½µÄ×ÊÔ´£¬¹¤Òµ·ÏÁÏÒ²¿ÉÒÔÔÙÀûÓá£Ä³»¯Ñ§ÐËȤС×éÔÚʵÑéÊÒÖÐÓÃ·ÏÆúµÄº¬ÂÁ¡¢Ìú¡¢Í­µÄºÏ½ðÖÆÈ¡ÁòËáÂÁÈÜÒº¡¢ÏõËáÍ­¾§ÌåºÍÌúºì(Fe2O3)¡£ÊµÑé·½°¸ÈçÏ£º

(1)д³öÂËÒºAÖмÓÈë×ãÁ¿ÁòËáºóËù·¢Éú·´Ó¦µÄÀë×Ó·½³Ìʽ£º________________________________________________________________________¡£

(2)ÒÑÖªFe(OH)3³ÁµíµÄpHÊÇ3¡«4£¬ÈÜÒºCͨ¹ýµ÷½ÚpH¿ÉÒÔʹFe3£«³ÁµíÍêÈ«¡£ÏÂÁÐÎïÖÊÖпÉÓÃ×÷µ÷ÕûÈÜÒºCµÄpHµÄÊÔ¼ÁÊÇ________(ÌîÐòºÅ)¡£

A£®Í­·Û¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡                     B£®°±Ë®

C£®ÇâÑõ»¯Í­                                 D£®¼îʽ̼ËáÍ­

(3)³£Î£¬ÈôÈÜÒºCÖнðÊôÀë×Ó¾ùΪ1 mol¡¤L£­1£¬Ksp[Fe(OH)3]£½4.0¡Á10£­38£¬Ksp[Cu(OH)2]£½2.2¡Á10£­20¡£¿ØÖÆpH£½4£¬ÈÜÒºÖÐc(Fe3£«)£½_____________________________________£¬

´Ëʱ________Cu(OH)2³ÁµíÉú³É(Ìî¡°ÓС±»ò¡°ÎÞ¡±)¡£

(4)½«20 mL Al2(SO4)3ÈÜÒºÓëµÈÎïÖʵÄÁ¿Å¨¶ÈµÄBa(OH)2ÈÜÒº70 mL»ìºÏ£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪ__________________________________________________________________¡£

(5)ÔÚ0.1 LµÄ»ìºÏËáÈÜÒºÖУ¬c(HNO3)£½2 mol¡¤L£­1£¬c(H2SO4)£½3 mol¡¤L£­1£¬½«0.3 molµÄÍ­·ÅÈë¼ÓÈȳä·Ö·´Ó¦ºó£¬±»»¹Ô­µÄHNO3µÄÎïÖʵÄÁ¿Îª________¡£


½âÎö¡¡(3)c(Fe3£«)£½£½4.0¡Á10£­8 mol¡¤L£­1£¬c(Cu2£«)¡¤c2(OH£­)£½1¡Á(10£­10)3£½1.0¡Á10£­30£¼Ksp[Cu(OH)2]£¬ËùÒÔÎÞCu(OH)2³ÁµíÉú³É¡£

(5)3Cu¡¡£«¡¡8H£«¡¡£«¡¡2NO===3Cu2£«£«2NO¡ü£«4H2O

  0.3 mol   0.8 mol    0.2 mol

ËùÒÔ±»»¹Ô­µÄHNO3Ϊ0.2 mol¡£

´ð°¸¡¡(1)AlO£«4H£«===Al3£«£«2H2O¡¡(2)CD¡¡(3)4.0¡Á10£­8¡¡ÎÞ¡¡

(4)2Al3£«£«3SO£«3Ba2£«£«7OH£­===3BaSO4¡ý£«Al(OH)3¡ý£«AlO£«2H2O¡¡(5)0.2 mol


Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

ÐÂÐͲÄÁÏÄÉÃ×¼¶Ìú·ÛÓëÆÕͨÌú·Û¾ßÓв»Í¬µÄÐÔÖÊ¡£ÒÑÖª£ºÔÚ²»Í¬Î¶ÈÏ£¬ÄÉÃ×¼¶Ìú·ÛÓëË®ÕôÆø·´Ó¦µÄ¹ÌÌå²úÎﲻͬ£¬Î¶ȵÍÓÚ570¡æÊ±£¬Éú³ÉFeO£»¸ßÓÚ570¡æÊ±£¬Éú³ÉFe3O4¡£

£¨1£©Ð´³öζȸßÓÚ570¡æÊ±·´Ó¦µÄ»¯Ñ§·½³Ìʽ       

                             ¡£

£¨2£©ÖÐѧ½Ì²ÄÖÐÓÃÓÒͼËùʾʵÑé×°Öã¬Íê³ÉÌú·ÛÓëË®ÕôÆø·´Ó¦µÄÑÝʾʵÑ顣ʵÑéÖÐʹÓ÷ÊÔíÒºµÄ×÷ÓÃÊÇ          

                      ¡£

£¨3£©Èç¹ûÌṩ¸øÄã3Ö§ÊԹܡ¢Ë®²Û¡¢Õô·¢Ãó¡¢½ºÈû¡¢µ¼¹Ü¡¢¾Æ¾«µÆ¼°Æä±ØÒªµÄÒÇÆ÷ºÍÎïÆ·£¬ÇëÔÚ´ðÌâ¾íµÄ·½¿òÖл­³öÄãÉè¼ÆµÄʵÑé×°ÖÃʾÒâͼ£¨°üÀ¨·´Ó¦Ê±ÈÝÆ÷ÖеÄÎïÖÊ£©¡£

˵Ã÷£º¢Ù±¾Ìâ×°ÖÃʾÒâͼÖеÄÒÇÆ÷¿ÉÒÔÓÃÏÂÃæµÄ·½Ê½±íʾ¡£

Ë®²Û£º ÊԹܣº  Õô·¢Ã󣺠 ²£Á§µ¼¹Ü£º»ò£¨µ«Ó¦±êʾ³öÔÚÒºÃæÉÏ»òÒºÃæÏ£©

¢ÚÌú¼Ų̈¡¢Ê¯ÃÞÍø¡¢¾Æ¾«µÆ¡¢²£Á§µ¼¹ÜÖ®¼äµÄÁª½Ó½º¹ÜµÈ£¬ÔÚʾÒâͼÖв»±Ø»­³ö¡£ÈçÐè¼ÓÈÈ£¬ÔÚÐè¼ÓÈȵÄÒÇÆ÷Ï·½£¬±êÒÔ¡°¡÷¡±±íʾ¡£

£¨4£©¼×ͬѧ¶ÔʵÑéºóµÄºÚÉ«¹ÌÌå²úÎïXº¬ÓÐÄÄЩÎïÖʽøÐÐÁËÈçÏÂʵÑ飺

ʵÑé²½Öè

ʵÑé²Ù×÷

ʵÑéÏÖÏó

¢ñ

È¡ÉÙÁ¿ºÚÉ«¹ÌÌå²úÎïX£¨¼Ù¶¨³É·Ö·Ö²¼¾ùÔÈ£©·ÅÈëÊÔ¹ÜÖУ¬¼ÓÈëÑÎËᣬ΢ÈÈ¡£

ºÚÉ«·ÛÄ©Öð½¥Èܽ⣬ÓÐÉÙÁ¿ÆøÅݲúÉú¡£

¢ò

ȡʵÑé¢ñ·´Ó¦ºóµÄÈÜÒºÉÙÐí£¬µÎ¼Ó¼¸µÎKSCNÈÜÒº£¬Õñµ´¡£

ÈÜҺûÓгöÏÖѪºìÉ«¡£

¢ó

ȡʵÑé¢ñ·´Ó¦ºóµÄÈÜÒºÉÙÐí£¬µÎ¼Ó¼¸µÎËáÐÔKMnO4ÈÜÒº£¬Õñµ´¡£

¸ßÃÌËá¼ØÈÜÒºÍÊÉ«¡£

¸ù¾ÝÒÔÉÏʵÑéÊÂʵ£¬ÄÜÈ·¶¨¹ÌÌåXÖдæÔÚµÄÎïÖÊÊÇ           £¬µ«²»ÄÜÈ·¶¨XµÄ³É·Ö£¬Ô­ÒòÊÇ                                                               ¡£

£¨5£©ÒÒͬѧΪÁË̽¾¿ÊµÑéºóµÄ¹ÌÌå²úÎïÊÇ·ñ´æÔÚFe3O4Éè¼ÆÏÂÁÐʵÑé·½°¸£º

¢ÙÈ¡¹ÌÌåÑùÆ·m1 g£¬ÈÜÓÚ×ãÁ¿µÄÏ¡ÑÎË᣻

¢ÚÏò¢Ù·´Ó¦ºóµÄÈÜÒºÖмÓÈë×ãÁ¿H2O2µÄºÍ°±Ë®£¬³ä·Ö·´Ó¦ºó£¬¹ýÂË¡¢Ï´µÓ¡¢¸ÉÔ

¢Û½«¢ÚÖÐËùµÃ¹ÌÌå½øÐÐׯÉÕ£¬µÃµ½ºìרɫ¹ÌÌåm2 g¡£

ÈÜÒºXÖз¢ÉúÑõ»¯»¹Ô­·´Ó¦µÄÀë×Ó·½³ÌʽΪ                                  £»m1Óëm2µÄ±ÈÖµÔÚ         ·¶Î§ÄÚʱ£¬²ÅÄÜÈ·¶¨¹ÌÌåÑùÆ·ÖÐÒ»¶¨´æÔÚFe3O4¡£

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

¾«Ó¢¼Ò½ÌÍø