ÌâÄ¿ÄÚÈÝ

ijͭ¿óʯÖ÷Òªº¬Cu2(OH)2CO3£¬»¹º¬ÉÙÁ¿Fe¡¢SiµÄ»¯ºÏÎʵÑéÊÒÒÔ´ËÍ­¿óʯΪԭÁÏÖƱ¸CuSO4¡¤5H2O¼°CaCO3£¬²¿·Ö²½ÖèÈçÏ£º

Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÈÜÒºA³ýº¬ÓÐCu2+Í⣬»¹¿ÉÄܺ¬ÓеĽðÊôÀë×ÓÓÐ________(ÌîÀë×Ó·ûºÅ)£»ÔÚÈÜÒºAÖмÓÈëH2O2·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ______________¡£
£¨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_____________¡£
£¨1£©Fe2+¡¢Fe3+£¨2·Ö£©£»2Fe2++H2O2 +2H+£½2Fe3++2H2O £¨2·Ö£©
£¨2£©¢Ùa¡úg¡¢h¡úe¡¢d¡ú i £¨2·Ö£©£¬·ÀÖ¹µ¹Îü£¨1·Ö£©£»¢ÚA£¨2·Ö£© £¨3£©88.8%£¨2·Ö£©

ÊÔÌâ·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+£»Ë«ÑõË®¾ßÓÐÇ¿Ñõ»¯ÐÔ£¬ÄÜ°ÑÑÇÌúÀë×ÓÑõ»¯Éú³ÉÌúÀë×Ó£¬ËùÒÔÔÚÈÜÒºAÖмÓÈëH2O2·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ2Fe2++H2O2 +2H+£½2Fe3++2H2O¡£
£¨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%¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
£¨Ò»£©(4·Ö)À¨ºÅÖеÄÎïÖÊÊÇÔÓÖÊ£¬Ð´³ö³ýÈ¥ÕâЩÔÓÖʵÄÊÔ¼Á£º
£¨1£©MgO (Al2O3)                £¨2£©Cl2(HCl)           
£¨3£©FeCl3(FeCl2)               £¨4£©NaHCO3ÈÜÒº(Na2CO3)          
£¨¶þ£©(6·Ö)º£Ë®Öк¬ÓдóÁ¿µÄÂÈ»¯Ã¾£¬´Óº£Ë®ÖÐÌáȡþµÄÉú²úÁ÷³ÌÈçÏÂͼËùʾ£º

»Ø´ðÏÂÁÐÎÊÌ⣺
д³öÔÚº£Ë®ÖмÓÈëÑõ»¯¸ÆÉú³ÉÇâÑõ»¯Ã¾µÄ»¯Ñ§·½³Ìʽ                    £»
²Ù×÷¢ÙÖ÷ÒªÊÇÖ¸       £»ÊÔ¼Á¢Ù¿ÉÑ¡Óà        £»
²Ù×÷¢ÚÊÇÖ¸          £»¾­²Ù×÷¢Û×îÖտɵýðÊôþ¡£
£¨Èý£©£¨8·Ö£©ÊµÑéÊÒÅäÖÆ480ml 0£®1mol¡¤L-1µÄNa2CO3ÈÜÒº£¬»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Ó¦ÓÃÍÐÅÌÌìƽ³Æȡʮˮ̼ËáÄƾ§Ìå        g¡£
£¨2£©ÈçͼËùʾµÄÒÇÆ÷ÅäÖÆÈÜÒº¿Ï¶¨²»ÐèÒªµÄÊÇ      (ÌîÐòºÅ)£¬±¾ÊµÑéËùÐè²£Á§ÒÇÆ÷E¹æ¸ñΪ       mL¡£

£¨3£©ÈÝÁ¿Æ¿ÉϱêÓУº¢Ùζȡ¢¢ÚŨ¶È¡¢¢ÛÈÝÁ¿¡¢¢Üѹǿ¡¢¢Ý¿Ì¶ÈÏß¡¢¢ÞËáʽ»ò¼îʽÕâÁùÏîÖеĠ       ¡££¨ÌîÊý×Ö·ûºÅ£©
£¨4£©ÅäÖÆËùÐèµÄÖ÷ÒªÒÇÆ÷ÊÇ£ºaÈÝÁ¿Æ¿¡¢bÉÕ±­¡¢c½ºÍ·µÎ¹Ü¡¢dÍÐÅÌÌìƽ£¬ËüÃÇÔÚ²Ù×÷¹ý³ÌÖÐʹÓõÄÇ°ºó˳ÐòÊÇ       ¡££¨Ìîд×Öĸ£¬Ã¿ÖÖÒÇÆ÷Ö»ÄÜÑ¡ÔñÒ»´Î£©
£¨5£©²£Á§°ôÊÇ»¯Ñ§ÊµÑéÖг£ÓõÄÒ»ÖÖ²£Á§¹¤¾ß£¬ÔòÔÚÅäÖÆÈÜÒºµÄ¹ý³ÌÖв£Á§°ô¹²Æðµ½ÁË      ÖÖÓÃ;¡££¨ÌîдÊý×Ö£©
£¨6£©ÈôʵÑéʱÓöµ½ÏÂÁÐÇé¿ö£¬½«Ê¹ÈÜÒºµÄŨ¶ÈÆ«µÍµÄÊÇ       ¡£
A£®ÅäÖÆǰûÓн«ÈÝÁ¿Æ¿ÖеÄË®³ý¾¡£»
B£®Ì¼ËáÄÆʧȥÁ˲¿·Ö½á¾§Ë®£»
C£®Ì¼ËáÄƾ§Ìå²»´¿£¬ÆäÖлìÓÐÂÈ»¯ÄÆ£»
D£®³ÆÁ¿Ì¼ËáÄƾ§ÌåʱËùÓÃíÀÂëÉúÐ⣻
E£®¶¨ÈÝʱÑöÊӿ̶ÈÏß

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

¾«Ó¢¼Ò½ÌÍø