ÌâÄ¿ÄÚÈÝ

ijͭ¿óʯÖ÷Òªº¬Cu2(OH)2CO3£¬»¹º¬ÉÙÁ¿Fe¡¢SiµÄ»¯ºÏÎʵÑéÊÒÒÔ´ËÍ­¿óʯΪԭÁÏÖƱ¸CuSO4¡¤5H2O ¼°CaCO3£¬²¿·Ö²½ÖèÈçÏ£º

Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÈÜÒºA³ýº¬ÓÐCu2+Í⣬»¹¿ÉÄܺ¬ÓеĽðÊôÀë×ÓÓÐ________(ÌîÀë×Ó·ûºÅ)£¬ÑéÖ¤Ëùº¬Àë×ÓËùÓõÄÊÔ¼ÁÊÇ______________¡£
£¨2£©¿ÉÓÃÉú³ÉµÄCO2ÖÆÈ¡ÓÅÖÊ̼Ëá¸Æ¡£ÖƱ¸Ê±£¬ÏÈÏòÂÈ»¯¸ÆÈÜÒºÖÐͨÈë°±Æø£¬ÔÙͨÈëCO2¡£
¢ÙʵÑéÊÒͨ³£²ÉÓüÓÈÈÂÈ»¯ï§ºÍÇâÑõ»¯¸Æ»ìºÏÎïµÄ·½·¨ÖÆÈ¡°±Æø¡£Ä³Ñ§Ï°Ð¡×éÑ¡È¡ÏÂͼËù¸ø²¿·Ö×°ÖÃÖÆÈ¡²¢ÊÕ¼¯´¿¾»µÄ°±Æø¡£

Èç¹û°´ÆøÁ÷·½ÏòÁ¬½Ó¸÷ÒÇÆ÷½Ó¿Ú£¬ÄãÈÏΪÕýÈ·µÄ˳ÐòΪa¡ú______¡¢______¡ú______¡¢______¡ú i¡£ÆäÖÐÓëiÏàÁ¬Â©¶·µÄ×÷ÓÃÊÇ______________¡£
¢ÚʵÑéÊÒÖл¹¿ÉÓùÌÌåÇâÑõ»¯ÄƺÍŨ°±Ë®ÖÆÈ¡ÉÙÁ¿°±Æø£¬ÏÂÁÐ×îÊʺÏÍê³É¸ÃʵÑéµÄ¼òÒ××°ÖÃÊÇ_________(Ìî±àºÅ)

£¨3£©²â¶¨Í­¿óʯÖÐCu2(OH)2CO3ÖÊÁ¿°Ù·Öº¬Á¿µÄ·½·¨ÊÇ£ºa£®½«1.25gÍ­¿óʯÖÆÈ¡µÄCuSO4¡¤5H2OÓÚ׶ÐÎÆ¿ÖУ¬¼ÓÈëÊÊÁ¿Ë®ÍêÈ«Èܽ⣻b£®ÏòÈÜÒºÖмÓÈë100mL0.25mol/LµÄÇâÑõ»¯ÄÆÈÜҺʹCu2+ÍêÈ«³Áµí£»c£®¹ýÂË£»d£®ÂËÒºÖеÄÇâÑõ»¯ÄÆÈÜÒºÓÃ0.5mol/LÑÎËáµÎ¶¨ÖÁÖյ㣬ºÄÓÃ10mLÑÎËá¡£ÔòÍ­¿óʯÖÐCu2(OH)2CO3ÖÊÁ¿·ÖÊýΪ_____________¡£

£¨11·Ö£©£¨1£©Fe2+¡¢Fe3+£»ËáÐÔ¸ßÃÌËá¼ØÈÜÒº¡¢KSCNÈÜÒº¡££¨4·Ö£¬¸÷1·Ö£¬ÆäËûºÏÀí´ð°¸Ò²µÃ·Ö£©
£¨2£©¢Ùa¡úg¡¢h¡úe¡¢d¡ú i £¨2·Ö£©£¬·ÀÖ¹µ¹Îü£¨1·Ö£©£»¢ÚA£¨2·Ö£© £¨3£©88.8%£¨2·Ö£©

½âÎöÊÔÌâ·ÖÎö£º£¨1£©Cu2(OH)2CO3ÒÔ¼°Fe¡¢SiµÄ»¯ºÏÎïÓëÏ¡ÁòËá·´Ó¦Éú³ÉÁòËáÍ­¡¢ÁòËáÑÇÌú¡¢ÁòËáÌú¡£¶þÑõ»¯¹èÓëÏ¡ÁòËá²»·´Ó¦£¬ËùÒÔÈÜÒºÖÐAÖгýº¬ÓÐCu2+Í⣬»¹¿ÉÄܺ¬ÓеĽðÊôÀë×ÓÓÐFe2+¡¢Fe3+£»ÑÇÌúÀë×Ó¾ßÓл¹Ô­ÐÔ£¬ÄÜʹËáÐÔ¸ßÃÌËá¼ØÈÜÒºÍÊÉ«£¬Òò´Ë¿ÉÒÔ¼ìÑéÑÇÌúÀë×Ó¡£¼ìÑéÌúÀë×Ó¿ÉÒÔÓÃKSCNÈÜÒº¼ìÑé¡£
£¨2£©¢Ù¸ù¾Ý×°ÖÃͼ¿ÉÖª£¬A×°ÖÃÊÇÖƱ¸°±ÆøµÄ¡£ÓÉÓÚÉú³ÉµÄ°±ÆøÖк¬ÓÐË®ÕôÆø£¬ËùÒÔÐèÒª¸ÉÔѡÓüîʯ»Ò¸ÉÔï¡£°±ÆøµÄÃܶÈСÓÚ¿ÕÆøµÄ£¬ÇÒ°±Æø¼«Ò×ÈÜÓÚË®£¬ËùÒÔÓ¦¸ÃÓÃÏòÏÂÅÅ¿ÕÆø·¨ÊÕ¼¯£¬ÇÒÐèÒª½«¶àÓàµÄ°±Æø½øÐÐÎüÊÕ£¬Òò´ËÕýÈ·µÄ²Ù×÷˳ÐòÊÇa¡úg¡¢h¡úe¡¢d¡ú i£»°±Æø¼«Ò×ÈÜÓÚË®£¬Òò´ËÓëiÏàÁ¬µÄ©¶·µÄ×÷ÓÃÊÇ·ÀÖ¹µ¹Îü¡£
¢ÚÓùÌÌåÇâÑõ»¯ÄƺÍŨ°±Ë®ÖÆÈ¡ÉÙÁ¿°±Æø£¬Òò´ËÐèÒª·ÖҺ©¶·¡£·´Ó¦²»ÐèÒª¼ÓÈÈ£¬ÇÒ°±ÆøÓÃÏòÏÂÅÅ¿ÕÆø·¨ÊÕ¼¯£¬ËùÒÔÕýÈ·µÄ´ð°¸Ñ¡A¡£
£¨3£©ÏûºÄÑÎËáµÄÎïÖʵÄÁ¿ÊÇ0.01L¡Á0.5mol/L£½0.005mol£¬Ôò¸ù¾Ý·½³ÌʽNaOH£«HCl£½NaCl£«H2O¿ÉÖª£¬ÓëÑÎËá·´Ó¦µÄÇâÑõ»¯ÄÆÊÇ0.005mol¡£ÇâÑõ»¯ÄƵÄÎïÖʵÄÁ¿ÊÇ0.1L¡Á0.25mol/L£½0.025mol£¬ÔòÓëÁòËáÍ­·´Ó¦µÄÇâÑõ»¯ÄÆÊÇ0.025mol£­0.005mol£½0.020mol¡£Ôò¸ù¾Ý·½³Ìʽ2NaOH£«CuSO4£½Cu(OH)2¡ý£«Na2SO4¿ÉÖª£¬ÁòËáÍ­µÄÎïÖʵÄÁ¿ÊÇ0.020mol¡Â2£½0.010mol£¬ËùÒÔ¸ù¾ÝÔ­×ÓÊغã¿ÉÖª£¬Í­¿óʯÖÐCu2(OH)2CO3µÄÎïÖʵÄÁ¿ÊÇ0.010mol¡Â2£½0.005mol£¬ËùÒÔCu2(OH)2CO3ÖÊÁ¿·ÖÊýΪ¡Á100%£½88.8%¡£
¿¼µã£º¿¼²éÀë×Ó¼ìÑé¡¢°±ÆøÖƱ¸¡¢ÎïÖʺ¬Á¿µÄ²â¶¨ÒÔ¼°ÊµÑé·½°¸µÄÉè¼ÆÓëÆÀ¼ÛµÈ

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¸õÌú¿óµÄÖ÷Òª³É·Ö¿É±íʾΪFeO¡¤Cr2O3£¬»¹º¬ÓÐMgO¡¢Al2O3¡¢Fe2O3µÈÔÓÖÊ£¬ÒÔÏÂÊÇÒÔ¸õÌú¿óΪԭÁÏÖƱ¸ÖظõËá¼Ø£¨K2Cr2O7£©µÄÁ÷³Ìͼ£º

ÒÑÖª£º¢Ù4FeO¡¤Cr2O3+ 8Na2CO3+ 7O28Na2CrO4 + 2 Fe2O3 + 8CO2¡ü£»
¢ÚNa2CO3 + Al2O32NaAlO2 + CO2¡ü£»¢Û Cr2O72£­+ H2O2CrO42£­ + 2H+
¸ù¾ÝÌâÒâ»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©¹ÌÌåXÖÐÖ÷Òªº¬ÓÐ_________£¨Ìîд»¯Ñ§Ê½£©£»Òª¼ì²âËữ²Ù×÷ÖÐÈÜÒºµÄpHÊÇ·ñµÈÓÚ4.5£¬Ó¦¸ÃʹÓÃ__________£¨ÌîдÒÇÆ÷»òÊÔ¼ÁÃû³Æ£©¡£
£¨2£©Ëữ²½ÖèÓô×Ëáµ÷½ÚÈÜÒºpH<5£¬ÆäÄ¿µÄÊÇ_________________________________¡£
£¨3£©²Ù×÷¢óÓжಽ×é³É£¬»ñµÃK2Cr2O7¾§ÌåµÄ²Ù×÷ÒÀ´ÎÊÇ£º¼ÓÈëKCl¹ÌÌå¡¢Õô·¢Å¨Ëõ¡¢      ¡¢¹ýÂË¡¢_______¡¢¸ÉÔï¡£
£¨4£©Ï±íÊÇÏà¹ØÎïÖʵÄÈܽâ¶ÈÊý¾Ý£¬²Ù×÷¢ó·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ£ºNa2Cr2O7+2KCl ¡úK2Cr2O7¡ý+2NaCl¡£¸Ã·´Ó¦ÔÚÈÜÒºÖÐÄÜ·¢ÉúµÄÀíÓÉÊÇ_______________¡£

ÎïÖÊ
Èܽâ¶È/(g/100gË®)
0¡ãC
40¡ãC
80¡ãC
KCl
28
40.1
51.3
NaCl
35.7
36.4
38
K2Cr2O7
4.7
26.3
73
Na2Cr2O7
163
215
376
 
£¨5£©¸±²úÆ·YÖ÷Òªº¬ÇâÑõ»¯ÂÁ£¬»¹º¬ÉÙÁ¿Ã¾¡¢ÌúµÄÄÑÈÜ»¯ºÏÎï¼°¿ÉÈÜÐÔÔÓÖÊ£¬¾«È··ÖÎöYÖÐÇâÑõ»¯ÂÁº¬Á¿ÊdzÆÈ¡n gÑùÆ·£¬¼ÓÈë¹ýÁ¿______£¨ÌîдÊÔ¼Á£©¡¢Èܽ⡢¹ýÂË¡¢ÔÙ______£¨ÌîдÊÔ¼Á£©¡¢¡­¡­×ÆÉÕ¡¢ÀäÈ´¡¢³ÆÁ¿£¬µÃ¸ÉÔï¹ÌÌåm g ¡£¼ÆËãÑùÆ·ÖÐÇâÑõ»¯ÂÁµÄÖÊÁ¿·ÖÊýΪ_______£¨Óú¬m¡¢nµÄ´úÊýʽ±íʾ£©¡£
£¨6£©Áù¼Û¸õÓж¾£¬¶øCr3+Ïà¶Ô°²È«¡£¹¤Òµº¬¸õ£¨CrO3£©·ÏÔüÎÞº¦»¯´¦ÀíµÄ·½·¨Ö®Ò»ÊǸɷ¨½â¶¾£¬ÓÃú²»ÍêȫȼÉÕÉú³ÉµÄCO»¹Ô­CrO3¡£ÔÚʵÑéÊÒÖÐÄ£ÄâÕâÒ»¹ý³ÌµÄ×°ÖÃÈçÏ£º

COÓɼ×ËáÍÑË®ÖƵã»ÊµÑé½áÊøʱϨÃð¾Æ¾«µÆµÄ˳ÐòÊÇ_________________________¡£

ÅðþÄàÊÇÒ»ÖÖ¹¤Òµ·ÏÁÏ£¬Ö÷Òª³É·ÝÊÇMgO(Õ¼40%)£¬»¹ÓÐCaO¡¢MnO¡¢Fe2O3¡¢FeO¡¢Al2O3¡¢SiO2µÈÔÓÖÊ£¬ÒÔ´ËΪԭÁÏÖÆÈ¡µÄÁòËáþ£¬¿ÉÓÃÓÚӡȾ¡¢ÔìÖ½¡¢Ò½Ò©µÈ¹¤Òµ¡£´ÓÅðþÄàÖÐÌáÈ¡MgSO4¡¤7H2OµÄ¹¤ÒÕÁ÷³ÌÈçÏ£º

ÒÑÖª£ºNaClOÓëMn2+·´Ó¦²úÉúMnO2³Áµí¡£

³ÁµíÎï
Fe(OH)3
Al(OH)3
Fe(OH)2
¿ªÊ¼³ÁµípH
2.3
4.0
7.6
ÍêÈ«³ÁµípH
4.1
5.2
9.6
 
¸ù¾ÝÌâÒâ»Ø´ðÏÂÁÐÎÊÌ⣺
(1)ʵÑéÖÐÐèÓÃ1.00mol/LµÄÁòËá80.0mL£¬ÈôÓÃ98%µÄŨÁòËáÅäÖÆ£¬³ýÁ¿Í²¡¢²£Á§°ô¡¢½ºÍ·µÎ¹ÜÍ⣬»¹ÐèÒªµÄ²£Á§ÒÇÆ÷ÓР           ¡¢              ¡£
(2)ÂËÔüµÄÖ÷Òª³É·Ý³ýº¬ÓÐFe(OH)3¡¢Al(OH)3Í⣬»¹ÓР         ¡¢          ¡£
(3)¼ÓÈëµÄNaClO¿ÉÓëMn2+·´Ó¦²úÉúMnO2³Áµí£¬¸Ã·´Ó¦µÄÀë×Ó·½³Ìʽ£º                         ¡£
ÔÚµ÷½ÚpH£½5¡«6֮ǰ£¬»¹ÓÐÒ»ÖÖÀë×ÓÒ²»á±»NaClOÑõ»¯£¬¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽΪ£º
                                                                                     ¡£
(4)ΪÁ˼ìÑéÂËÒºÖÐFe3+ÊÇ·ñ±»³ý¾¡£¬¿ÉÑ¡ÓõÄÊÔ¼ÁÊÇ        ¡£
A£®KSCNÈÜÒº    B£®µí·ÛKIÈÜÒº   C£®H2O2     D£®KMnO4Ï¡ÈÜÒº
(5)ÒÑÖªMgSO4¡¢CaSO4µÄÈܽâ¶ÈÈçÏÂ±í£º
ζȣ¨¡æ£©
40
50
60
70
MgSO4
30.9
33.4
35.6
36.9
CaSO4
0.210
0.207
0.201
0.193
¡°³ý¸Æ¡±Êǽ«MgSO4ºÍCaSO4»ìºÏÈÜÒºÖеÄCaSO4³ýÈ¥£¬¸ù¾ÝÉϱíÊý¾Ý£¬¼òҪ˵Ã÷²Ù×÷²½Öè      ¡¢                    ¡£
(6)Èç¹û²âµÃÌṩµÄÅðþÄàµÄÖÊÁ¿Îª100.0g£¬µÃµ½µÄMgSO4¡¤7H2O196.8g£¬ÔòMgSO4¡¤7H2OµÄ²úÂÊΪ                    (Ïà¶Ô·Ö×ÓÖÊÁ¿£ºMgSO4¡¤7H2O£­246  MgO£­40)¡£

¸õÌú¿óµÄÖ÷Òª³É·Ö¿É±íʾΪFeO¡¤Cr2O3£¬»¹º¬ÓÐMgO¡¢Al2O3¡¢Fe2O3µÈÔÓÖÊ£¬ÒÔÏÂÊÇÒÔ¸õÌú¿óΪԭÁÏÖƱ¸ÖظõËá¼Ø£¨K2Cr2O7£©µÄÁ÷³Ìͼ£º

¸ù¾ÝÌâÒâ»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©¹ÌÌåXÖÐÖ÷Òªº¬ÓР     £¨Ìîд»¯Ñ§Ê½£©£»Òª¼ì²âËữ²Ù×÷ÖÐÈÜÒºµÄpHÊÇ·ñµÈÓÚ4£®5£¬Ó¦¸ÃʹÓà     £¨ÌîдÒÇÆ÷»òÊÔ¼ÁÃû³Æ£©¡£
£¨2£©Ëữ²½ÖèÓô×Ëáµ÷½ÚÈÜÒºpH <5£¬ÆäÄ¿µÄÊÇ      ¡£
£¨3£©²Ù×÷¢óÓжಽ×é³É£¬»ñµÃK2Cr2O7¾§ÌåµÄ²Ù×÷ÒÀ´ÎÊÇ£º¼ÓÈëKCl¹ÌÌå¡¢Õô·¢Å¨Ëõ¡¢
         ¡¢¹ýÂË¡¢       ¡¢¸ÉÔï¡£
£¨4£©Ï±íÊÇÏà¹ØÎïÖʵÄÈܽâ¶ÈÊý¾Ý£¬²Ù×÷III·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ¡£¸Ã·´Ó¦ÔÚÈÜÒºÖÐÄÜ·¢ÉúµÄÀíÓÉÊÇ£º                ¡£

£¨5£©¸±²úÆ·YÖ÷Òªº¬ÇâÑõ»¯ÂÁ£¬»¹º¬ÉÙÁ¿Ã¾¡¢ÌúµÄÄÑÈÜ»¯ºÏÎï¼°¿ÉÈÜÐÔÔÓÖÊ£¬¾«È··ÖÎöYÖÐÇâÑõ»¯ÂÁº¬Á¿µÄ·½·¨ÊdzÆÈ¡ngÑùÆ·£¬¼ÓÈë¹ýÁ¿         £¨ÌîдÊÔ¼Á£©¡¢Èܽ⡢¹ýÂË¡¢ÔÙ¼ÓÈë¹ýÁ¿             £¨ÌîдÊÔ¼Á£©¡¢¡­¡­×ÆÉÕ¡¢ÀäÈ´¡¢³ÆÁ¿£¬µÃ¸ÉÔï¹ÌÌåmg¡£¼ÆËãÏéÆ·ÖÐÇâÑõ»¯ÂÁµÄÖÊÁ¿·ÖÊýΪ            £¨Óú¬m¡¢nµÄ´úÊýʽ±íʾ£©¡£

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

¾«Ó¢¼Ò½ÌÍø