ÌâÄ¿ÄÚÈÝ

ij»¯Ñ§ÐËȤС×éÓÃÏÂͼװÖõç½âCuSO4ÈÜÒº£¬²â¶¨Í­µÄÏà¶Ô·Ö×ÓÖÊÁ¿¡£

£¨1£©ÈôʵÑéÖвⶨÔÚ±ê×¼×´¿öÏ·ųöµÄÑõÆøµÄÌå»ýVL£¬AÁ¬½ÓÖ±Á÷µçÔ´µÄ__________ (Ìî¡°Õý¼«¡±»ò¡°¸º¼«¡±£©¡£
£¨2£©µç½â¿ªÊ¼Ò»¶Îʱ¼äºó£¬ÔÚUÐιÜÖпɹ۲쵽µÄÏÖÏó£­____________________________¡£
µç½âµÄÀë×Ó·½³ÌʽΪ                                                           ¡£
£¨3£©ÊµÑéÖл¹Ðè²â¶¨µÄÊý¾ÝÊÇ_______________£¨ÌîдÐòºÅ£©¡£
¢ÙA¼«µÄÖÊÁ¿ÔöÖØm g      ¢ÚB¼«µÄÖÊÁ¿ÔöÖØm g
£¨4£©ÏÂÁÐʵÑé²Ù×÷ÖбØÒªµÄÊÇ____________£¨Ìî×Öĸ£©¡£

A£®³ÆÁ¿µç½âÇ°µç¼«µÄÖÊÁ¿
B£®µç½âºó£¬µç¼«ÔÚºæ¸É³ÆÖØÇ°£¬±ØÐëÓÃÕôÁóË®³åÏ´
C£®¹Îϵç½â¹ý³ÌÖе缫ÉÏÎö³öµÄÍ­£¬²¢ÇåÏ´¡¢³ÆÖØ
D£®µç¼«ÔÚºæ¸É³ÆÖصIJÙ×÷ÖбØÐë°´¡°ºæ¸É¡ª³ÆÖØÒ»ÔÙºæ¸ÉÒ»ÔÙ³ÆÖØ¡±½øÐÐ
E.ÔÚÓпÕÆø´æÔÚµÄÇé¿öÏ£¬ºæ¸Éµç¼«±ØÐë²ÉÓõÍκæ¸ÉµÄ·½·¨
£¨5£©Í­µÄÏà¶ÔÔ­×ÓÖÊÁ¿Îª£º_______________________£¨Óú¬ÓÐm¡¢VµÄ¼ÆËãʽ±íʾ£©¡£
£¨6£©Èç¹ûÓüîÐÔ£¨KOHΪµç½âÖÊ£©¼×´¼È¼Áϵç³Ø×÷ΪµçÔ´½øÐÐʵÑ飬·Åµçʱ¸º¼«µÄµç¼«·´Ó¦Ê½Îª                                                           ¡£

£¨1£©¸º £¨2£©A¼«±ä´Ö£¬B¼«ÓÐÆøÅݲúÉú£¬ÈÜÒºÑÕÉ«±ädz2Cu2£«£«2H2O2Cu£«O2¡ü£«4H£«
£¨3£©¢Ù  £¨4£©ABDE  £¨5£©11.2m/V £¨6£©CH3OH£«8OH£­£­6e£­=CO32£­£«6H2O

½âÎöÊÔÌâ·ÖÎö£º£¨1£©²âµÃO2µÄÌå»ýΪVL£¬A¼«Îª¸º¼«£¬
£¨2£©ÏÖÏóΪA¼«±ä´Ö£¬B¼«ÓÐÆøÅݲúÉú£¬ÈÜÒºÑÕÉ«±ädz£¬µç¼«·´Ó¦Ê½Îª2Cu2£«£«2H2O2Cu£«O2¡ü£«4H£«£¨4£©µç¼«ÉÏÎö³öµÄCuµÄÖÊÁ¿²»ÐèÒª¹ÒÏ£¬Ö»ÐèÒª³Æ³öA¼«ÖÊÁ¿µÄÔö¼ÓÁ¿£¬¹Ê²»ÐèÒªCÏîµÄ²Ù×÷¡£
£¨5£© ÉèÍ­µÄĦ¶ûÖÊÁ¿ÎªMmol/g  2Cu¡«O2
                                                           2M   1
m   V/22.4
½âµÃM=11.2m/V
¿¼µã£º¿¼²éµç½âÔ­Àí¡¢¼ÆËãµÈÏà¹Ø֪ʶ¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

°²»Õ°¼É½Ìú¿ó×øÂäÔÚÂí°°É½¾³ÄÚ£¬¿ó´²ÊôÓÚ½Ïà»ðɽһÇÖÈëÑÒÐÍÌú¿ó´²£¬Ë׳ơ°çãÑÒÐÍ¡±Ìú¿ó£¬ÊÇÂí¸ÖÖØÒªµÄ¿óÇøÖ®Ò»¡£Ìú¿óÖб»³ÆΪºìÁ±Ê¯µÄÌú¿óº¬ÃÌÁ¿¸ß£¬ÃÌÊÇÒ±Á¶Ã̸ֵÄÖØÒªÔ­ÁÏ¡£ºìÁ±Ê¯Ö÷Òª³É·ÖÓдÅÌú¿óFe3O4¡¢ÁâÌú¿óFeCO3¡¢ÃÌ¿ó£¨MnO2ºÍMnCO3£©Ê¯ÃÞMg3Si3O7(OH)4µÈ¡£¹¤ÒµÉϽ«ºìÁ±Ê¯´¦ÀíºóÔËÓÃÒõÀë×ÓĤµç½â·¨µÄм¼ÊõÌáÈ¡½ðÊôÃ̲¢ÖƵÄÂÌÉ«¸ßЧˮ´¦Àí¼Á£¨K2FeO4£©¡£¹¤ÒµÁ÷³ÌÈçÏ£º

£¨1£©¹¤ÒµÉÏΪÌá¸ßÏ¡ÁòËá½þȡЧÂÊÒ»°ã²ÉÈ¡µÄ´ëÊ©ÊÇ£¨ÈÎÒâдÁ½ÖÖ·½·¨£©
¢Ù                                 ¢Ú                                    
£¨2£©Ê¯ÃÞ»¯Ñ§Ê½ÎªMg3Si3O7(OH)4Ò²¿ÉÒÔ±íʾ³ÉÑõ»¯ÎïÐÎʽ£¬Ñõ»¯Îï±í´ïʽΪ        ¡£
£¨3£©ÒÑÖª²»Í¬½ðÊôÀë×ÓÉú³ÉÇâÑõ»¯Îï³ÁµíËùÐèµÄpHÈçÏÂ±í£º

Àë×Ó
Fe3+
Al3+
Fe2+
Mn2+
Mg2+
¿ªÊ¼³ÁµíµÄpH
2.7
3.7
7.0
7.8
9.3
ÍêÈ«³ÁµíµÄpH
3.7
4.7
9.6
9.8
10.8
 
¹ý³Ì¢ÚÖмӰ±Ë®µ÷½ÚÈÜÒºµÄpHµÈÓÚ6£¬ÔòÂËÔüBµÄ³É·Ö                  ¡£
£¨4£©½þ³öÒºÖÐÒÔMn2+ÐÎʽ´æÔÚ£¬ÇÒÂËÔüAÖÐÎÞMnO2Ô­Òò                         ¡£
£¨5£©µç½â×°ÖÃÖмýÍ·±íʾÈÜÒºÖÐÒõÀë×ÓÒƶ¯µÄ·½Ïò£»ÔòAµç¼«ÊÇÖ±Á÷µçÔ´µÄ       ¼«¡£Êµ¼ÊÉú²úÖУ¬Ñô¼«ÒÔÏ¡ÁòËáΪµç½âÒº£¬Ñô¼«µÄµç¼«·´Ó¦Ê½Îª                                         ¡£

£¨6£©ÂËÔü¾­·´Ó¦¢ÜÉú³ÉÂÌÉ«¸ßЧˮ´¦Àí¼ÁµÄÀë×Ó·½³Ìʽ                                          ¡£
K2FeO4±»ÓþΪÂÌÉ«¸ßЧˮ´¦Àí¼ÁµÄÔ­ÒòÊÇ                                          ¡£

ij·´Ó¦Öз´Ó¦ÎïÓëÉú³ÉÎïÓУºFeCl2¡¢FeCl3¡¢CuCl2¡¢Cu¡£
£¨1£©½«ÉÏÊö·´Ó¦Éè¼Æ³ÉµÄÔ­µç³ØÈçͼ¼×Ëùʾ£¬Çë»Ø´ðÏÂÁÐÎÊÌ⣺
 
¢ÙͼÖÐXÈÜÒºÊÇ_____________________________£»
¢Úʯīµç¼«ÉÏ·¢ÉúµÄµç¼«·´Ó¦Ê½Îª__________________________________________£»
¢ÛÔ­µç³Ø¹¤×÷ʱ£¬ÑÎÇÅÖеÄ____________(Ìî¡°K£«¡±»ò¡°Cl£­¡±)²»¶Ï½øÈëXÈÜÒºÖС£
£¨2£©½«ÉÏÊö·´Ó¦Éè¼Æ³ÉµÄµç½â³ØÈçͼÒÒËùʾ£¬ÒÒÉÕ±­ÖнðÊôÑôÀë×ÓµÄÎïÖʵÄÁ¿Óëµç×ÓתÒƵÄÎïÖʵÄÁ¿µÄ±ä»¯¹ØϵÈçͼ±û£¬Çë»Ø´ðÏÂÁÐÎÊÌ⣺
¢ÙMÊÇ__________¼«£»
¢Úͼ±ûÖеĢÚÏßÊÇ______________µÄ±ä»¯¡£
¢Ûµ±µç×ÓתÒÆΪ2 molʱ£¬ÏòÒÒÉÕ±­ÖмÓÈë________ L 5 mol¡¤L£­1 NaOHÈÜÒº²ÅÄÜʹËùÓеĽðÊôÑôÀë×Ó³ÁµíÍêÈ«¡£
£¨3£©ÌúµÄÖØÒª»¯ºÏÎï¸ßÌúËáÄÆ(Na2FeO4)ÊÇÒ»ÖÖÐÂÐÍÒûÓÃË®Ïû¶¾¼Á£¬¾ßÓкܶàÓŵ㡣
¢Ù¸ßÌúËáÄÆÉú²ú·½·¨Ö®Ò»Êǵç½â·¨£¬ÆäÔ­ÀíΪFe£«2NaOH£«2H2O=Na2FeO4£«3H2¡ü£¬Ôòµç½âʱÑô¼«µÄµç¼«·´Ó¦Ê½ÊÇ__________________________________¡£
¢Ú¸ßÌúËáÄÆÉú²ú·½·¨Ö®¶þÊÇÔÚÇ¿¼îÐÔ½éÖÊÖÐÓÃNaClOÑõ»¯Fe(OH)3Éú³É¸ßÌúËáÄÆ¡¢ÂÈ»¯ÄƺÍË®£¬¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽΪ_______________________________________¡£

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

¾«Ó¢¼Ò½ÌÍø