ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿I.ÏÖÓÐÎåÖÖÈÜÒº£¬·Ö±ðº¬ÏÂÁÐÀë×Ó£º¢ÙAg+£¬¢ÚMg2+£¬¢ÛFe2+£¬¢ÜAl3+£¬¢ÝFe3+¡£

(1)д³ö·ûºÏÏÂÁÐÌõ¼þµÄÀë×Ó·ûºÅ£º¼ÈÄܱ»Ñõ»¯ÓÖÄܱ»»¹Ô­µÄÀë×ÓÊÇ______£¬¼ÓÌú·ÛºóÈÜÒºÔöÖصÄÊÇ____£»

(2)ÏòFe2+µÄÈÜÒºÖеμÓNaOHÈÜÒº£¬ÏÖÏóÊÇ__________________¡£

(3)³ýÈ¥FeCl2ÖÐFeCl3ÖÐËùÉæ¼°µÄÀë×Ó·½³Ìʽ£º________________¡£

II.(1)ÈçͼËùʾ£¬½«ÂÈÆøÒÀ´Îͨ¹ýÊ¢ÓиÉÔïÓÐÉ«²¼ÌõµÄ¹ã¿ÚÆ¿ºÍÊ¢ÓÐʪÈóÓÐÉ«²¼ÌõµÄ¹ã¿ÚÆ¿£¬¿É¹Û²ìµ½µÄÏÖÏóÊÇ£º____________£¬¸ÃʵÑéÖ¤Ã÷ÆðƯ°××÷ÓõÄÊÇ______¡£(Ìѧʽ)

(2)¢ÙÂÈÆøÓж¾£¬ÊµÑéÊÒÎüÊÕ¶àÓàµÄÂÈÆøµÄÔ­ÀíÊÇ(ÓÃÀë×Ó·½³Ìʽ±íʾ)__________________£»

¢Ú¸ù¾ÝÕâÒ»Ô­Àí£¬¹¤ÒµÉϳ£ÓÃÁ®¼ÛµÄʯ»ÒÈéÎüÊÕ¹¤ÒµÂÈÆøβÆøÖƵÃƯ°×·Û£¬Æ¯°×·ÛµÄÓÐЧ³É·ÖÊÇ_________(Ìѧʽ)¡£

¢Û³¤ÆÚ¶ÖÃÓÚ¿ÕÆøÖеÄƯ°×·Û»áʧЧ£¬Ê§Ð§µÄÔ­ÒòÊÇ(Óû¯Ñ§·½³Ìʽ±íʾ)___________________________£¬________________________¡£

¢ÜƯ°×·ÛÊÇ·ñÍêȫʧЧ¿ÉÓÃÏ¡ÑÎËá¼ìÑ飬¼ÓÏ¡ÑÎËáºó²úÉúµÄÆøÌåÊÇ______(Ìî×Öĸ´úºÅ)¡£

A.O2 B.Cl2 C.CO2 D.HClO

¡¾´ð°¸¡¿Fe2+ Fe3+ Éú³É°×É«Ðõ×´³Áµí£¬³ÁµíѸËÙÓÉ°×É«±äΪ»ÒÂÌÉ«£¬×îºó±äΪºìºÖÉ« Fe+2Fe3+¨T3Fe2+ ¸ÉÔïµÄÓÐÉ«²¼ÌõÎÞÃ÷ÏÔÏÖÏ󣬳±ÊªµÄÓÐÉ«²¼ÌõÍÊÉ« HClO Cl2+2OH-=Cl-+ClO-+H2O Ca(ClO)2 Ca(ClO)2+CO2+H2O=CaCO3¡ý+2HClO 2HClO2HCl+O2¡ü C

¡¾½âÎö¡¿

I. (1)¾ÓÓÚÖмä¼Û̬µÄ½ðÊôÑôÀë×Ó¼ÈÄܱ»Ñõ»¯ÓÖÄܱ»»¹Ô­£¬ÌúÄܽ«Òø´ÓÈÜÒºÖÐÖû»³öÀ´£¬ÄܺÍFe3+·´Ó¦Éú³ÉFe2+£»

(2) ÏòFe2+µÄÈÜÒºÖеμÓNaOHÈÜÒº£¬·¢Éú·´Ó¦£ºFe2++2OH-=Fe(OH)2¡ý£¬°×É«ÇâÑõ»¯ÑÇÌú³Áµí£¬ºÜÈÝÒ×±»ÑõÆøÑõ»¯ÎªÇâÑõ»¯Ìú£»

(3) Fe3+Äܱ»Ìú»¹Ô­Éú³ÉFe2+£¬ÇÒ²»ÒýÈëÐÂÔÓÖÊ£»

II.(1)ÂÈÆøÎÞƯ°×ÐÔ£¬ÂÈÆøÓëË®·´Ó¦Éú³ÉµÄ´ÎÂÈËá¾ßÓÐƯ°×ÐÔ£»

(2)ÀûÓÃÂÈÆøºÍÇâÑõ»¯ÄÆÈÜÒº·´Ó¦À´´¦ÀíβÆø£»Ê¯»ÒÈéÎüÊÕ¹¤ÒµÂÈÆøβÆøÖƵÃƯ°×·ÛÉú³ÉÂÈ»¯¸Æ¡¢´ÎÂÈËá¸Æ£¬´ÎÂÈËá¸ÆΪƯ°×·ÛµÄÓÐЧ³É·Ö£»³¤ÆÚ¶ÖÃÓÚ¿ÕÆøÖеÄƯ°×·Û£¬´ÎÂÈËá¸Æ±äÖÊΪ̼Ëá¸Æ£¬¼ÓÏ¡ÑÎËáºó²úÉúµÄÆøÌåΪ¶þÑõ»¯Ì¼.

I.(1)Fe2+ÖÐFeÔªËصĻ¯ºÏ¼Û´¦ÓÚÖмä¼Û̬£¬¼ÈÄܱ»Ñõ»¯ÓÖÄܱ»»¹Ô­£¬ÌúÄÜÓë¢ÙµÄAg+ºÍ¢ÝÖÐFe3+·¢ÉúÀë×Ó·´Ó¦£¬¶Ô¢Ù·´Ó¦Îª£ºFe+2Ag+=2Ag+Fe2+£¬ÈÜÒºÖÊÁ¿¼õÇ᣻¶ÔÓڢݷ¢Éú·´Ó¦Îª£ºFe+2Fe3+=3Fe2+£¬ÈÜÒºÖÊÁ¿Ôö¼Ó£»¹Ê¼ÈÄܱ»Ñõ»¯ÓÖÄܱ»»¹Ô­µÄÀë×ÓÊÇFe2+£¬¼ÓÌú·ÛºóÈÜÒºÔöÖصÄÊÇFe3+£»

(2) Fe2+ºÍÇâÑõ»¯ÄÆÈÜÒºÖеÄOH-·´Ó¦Éú³ÉµÄÇâÑõ»¯ÑÇÌú³ÁµíFe2++2OH-=Fe(OH)2¡ý£¬ÇâÑõ»¯ÑÇÌúºÜÈÝÒ×±»ÑõÆøÑõ»¯ÎªÇâÑõ»¯Ìú£¬4Fe(OH)2+O2+2H2O=4Fe(OH)3£¬ËùÒÔ¿´µ½µÄÏÖÏóÊÇ£º³öÏÖ°×É«³Áµí£¬Ñ¸ËÙ±äΪ»ÒÂÌÉ«×îºó±äΪºìºÖÉ«£»

(3)³ýÈ¥FeCl2ÖÐFeCl3Ñ¡ÓÃÌú·Û£¬ FeCl3¿ÉÓëFe·´Ó¦Éú³ÉFeCl2ÈÜÒº£¬²»ÒýÈëÐÂÔÓÖÊ£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪ£ºFe+2Fe3+=3Fe2+£»

II.(1)ÂÈÆøÎÞƯ°×ÐÔ£¬ÂÈÆøÓëË®·´Ó¦Éú³ÉµÄ´ÎÂÈËá¾ßÓÐƯ°×ÐÔ£»Cl2+H2O=HCl+HClO£»Òò´Ë»á¿´µ½£º¸ÉÔïµÄÓÐÉ«²¼ÌõÎÞÃ÷ÏÔÏÖÏ󣬳±ÊªµÄÓÐÉ«²¼ÌõÍÊÉ«£»¸ÃʵÑéÖ¤Ã÷ÆðƯ°××÷ÓõÄÊÇHClO£»

(2) ¢ÙΪÁË·ÀÖ¹ÂÈÆøβÆøÎÛȾ¿ÕÆø£¬¿ÉÓÃNaOHÈÜÒºÎüÊÕ£¬¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽΪCl2+2OH--=Cl-+ClO-+H2O£»

¢ÚCl2Óëʯ»ÒÈé·¢Éú·´Ó¦£º2Cl2+2Ca(OH)2=CaCl2+Ca(ClO)2+2H2O£¬µÃµ½Æ¯°×·Û£¬Æ¯°×·ÛµÄÓÐЧ³É·ÖÊÇCa(ClO)2£»

¢Û³¤ÆÚ¶ÖÃÓÚ¿ÕÆøÖеÄƯ°×·Û£¬ÓÐЧ³É·ÖCa(ClO)2»áºÍ¿ÕÆøÖеĶþÑõ»¯Ì¼Ë®·´Ó¦Éú³É̼Ëá¸ÆºÍ´ÎÂÈËᣬ·½³ÌʽΪ£ºCa(ClO)2+CO2+H2O=CaCO3¡ý+2HClO£¬HClO²»Îȶ¨£¬¹âÕÕÈÝÒ׷ֽ⣬·Ö½â·´Ó¦·½³ÌʽΪ£º2HClO2HCl+O2¡ü£»

¢ÜƯ°×·Û³¤ÆÚ±©Â¶ÔÚ¿ÕÆøÖУ¬×îºóµÃµ½µÄÖ÷Òª³É·ÖÖк¬Ì¼Ëá¸Æ£¬¼ÓÏ¡ÑÎËáºó·¢Éú¸´·Ö½â·´Ó¦£¬CaCO3+2HCl=CaCl2+H2O+CO2¡ü£»¹ÊºÏÀíÑ¡ÏîÊÇC¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿ÊµÑéÊÒÓÃÈçÏÂ×°ÖÃÖÆÈ¡ÂÈÆø£¬²¢ÓÃÂÈÆø½øÐÐʵÑé¡£»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©AÖÐÊ¢ÓÐŨÑÎËᣬBÖÐÊ¢ÓÐMnO2£¬Ð´³ö·´Ó¦µÄÀë×Ó·½³Ìʽ___________________¡£

£¨2£©DÖзÅÈëŨH2SO4ÆäÄ¿µÄÊÇ_____________________________¡£

£¨3£©EÖÐΪºìÉ«¸É²¼Ìõ£¬FÖÐΪºìɫʪ²¼Ìõ£¬¿É¹Û²ìµ½µÄÏÖÏóÊÇ___________£¬¶Ô±ÈEºÍFÖÐÏÖÏóµÄ²îÒì¿ÉµÃ³öµÄ½áÂÛÊÇ________________________________¡£

£¨4£©G´¦µÄÏÖÏóÊÇ____________________________________¡£

£¨5£©Óû¯Ñ§·½³Ìʽд³öH´¦Î²ÆøÎüÊÕ×°ÖÃÖеķ´Ó¦Ô­Àí____________¡£

£¨6£©¼ÒÍ¥Öг£ÓÃÏû¶¾Òº£¨Ö÷Òª³É·ÖNaClO£©Óë½à²ÞÁ飨Ö÷Òª³É·ÖÑÎËᣩÇå½àÎÀÉú¡£Ä³Æ·ÅÆÏû¶¾Òº°ü×°ÉÏ˵Ã÷ÈçÏÂͼ¡£

¢ÙÏû¶¾ÒºÓë½à²ÞÁé²»ÄÜͬʱʹÓã¬Ô­ÒòÊÇ£¨ÓÃÀë×Ó·½³Ìʽ±íʾ£©____________¡£

¢ÚÐè¡°Ãܱձ£´æ¡±µÄÔ­Òò____________________________________________¡£

£¨7£©¹¤ÒµÉÏÓÃÑÇÂÈËáÄƺÍÏ¡ÑÎËáΪԭÁÏÖƱ¸ ClO2 £¬Ð´³ö·´Ó¦µÄ»¯Ñ§·½³Ìʽ____________¡£Cl2ºÍClO2¶¼ÊÇÇ¿Ñõ»¯¼Á,ÄÜɱËÀË®ÖеIJ¡¾ú¡£µÈÖÊÁ¿µÄClO2µÄÑõ»¯ÄÜÁ¦ÊÇCl2µÄ_______±¶¡£

£¨8£©Çè(CN)2¡¢ÁòÇè(SCN)2µÄ»¯Ñ§ÐÔÖʺͱËØ(X2)ºÜÏàËÆ£¬»¯Ñ§ÉϳÆΪÄâ±ËØ£¬È磺[(SCN)2£«H2O = HSCN£«HSCNO]¡£ËüÃÇÒõÀë×ӵĻ¹Ô­ÐÔÇ¿ÈõΪ Cl-<Br-<CN-<SCN-<I-¡£ÊÔд³ö£ºKBrºÍKSCNµÄ»ìºÏÈÜÒºÖмÓÈë(CN)2£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ_________________¡£

¡¾ÌâÄ¿¡¿»¯Ñ§ÊÇÒ»ÃÅÒÔʵÑéΪ»ù´¡µÄѧ¿Æ£¬ÊԻشðÒÔÏÂÎÊÌ⣺

IÓÃÖÊÁ¿·ÖÊýΪ36.5%µÄŨÑÎËá(ÃܶÈΪ1.16 g¡¤cm£­3)ÅäÖƳÉ1 mol¡¤L£­1µÄÏ¡ÑÎËá¡£ÏÖʵÑéÊÒ½öÐèÒªÕâÖÖÑÎËá220 mL£¬ÊԻشðÏÂÁÐÎÊÌ⣺

£¨1£©ÅäÖÆÏ¡ÑÎËáʱ£¬Ó¦Ñ¡ÓÃÈÝÁ¿Îª________mLµÄÈÝÁ¿Æ¿¡£

£¨2£©¾­¼ÆËãÐèÒª________mLŨÑÎËᣬÔÚÁ¿È¡Ê±ÒËÑ¡ÓÃÏÂÁÐÁ¿Í²ÖеÄ________¡£

A£®5 mL B£®10 mL C£®25 mL D£®50 mL

£¨3£©ÔÚÁ¿È¡Å¨ÑÎËáºó£¬½øÐÐÁËÏÂÁвÙ×÷£º

¢ÙµÈÏ¡Ê͵ÄÑÎËáµÄζÈÓëÊÒÎÂÒ»Öºó£¬Ñز£Á§°ô×¢Èë250 mLÈÝÁ¿Æ¿ÖС£

¢ÚÍùÈÝÁ¿Æ¿ÖÐСÐļÓÕôÁóË®ÖÁÒºÃæÀëÈÝÁ¿Æ¿¿Ì¶ÈÏß1¡«2 cmʱ£¬¸ÄÓýºÍ·µÎ¹Ü¼ÓÕôÁóË®£¬Ê¹ÈÜÒºµÄÒºÃæÓëÆ¿¾±µÄ¿Ì¶È±êÏßÏàÇС£

¢ÛÔÚÊ¢ÑÎËáµÄÉÕ±­ÖÐ×¢ÈëÕôÁóË®£¬²¢Óò£Á§°ô½Á¶¯£¬Ê¹Æä»ìºÏ¾ùÔÈ¡£

¢ÜÓÃÕôÁóˮϴµÓÉÕ±­ºÍ²£Á§°ô2ÖÁ3´Î£¬²¢½«Ï´µÓҺȫ²¿×¢ÈëÈÝÁ¿Æ¿¡£

ÉÏÊö²Ù×÷ÖУ¬ÕýÈ·µÄ˳ÐòÊÇ(ÌîÐòºÅ)____________¡£

£¨4£©ÔÚÉÏÊöÅäÖƹý³ÌÖУ¬ÓøոÕÏ´µÓ½à¾»µÄÁ¿Í²À´Á¿È¡Å¨ÑÎËᣬÆäÅäÖƵÄÏ¡ÑÎËáŨ¶È________(Ìî¡°Æ«¸ß¡±¡¢¡°Æ«µÍ¡±»ò¡°ÎÞÓ°Ï족)¡£ÈôδÓÃÕôÁóˮϴµÓÉÕ±­ÄÚ±Ú»òδ½«Ï´µÓҺעÈëÈÝÁ¿Æ¿£¬ÔòÅäÖƵÄÏ¡ÑÎËáŨ¶ÈÊÇ________(Ìî¡°Æ«¸ß¡±¡¢¡°Æ«µÍ¡±»ò¡°ÎÞÓ°Ï족)¡£

IIʵÑéÊÒÓÃÒÔÏÂ×°ÖýøÐÐʵÑé

£¨5£©×°Öü׺ÍÎìÓÃÓÚÖÆÈ¡²¢ÊÕ¼¯ÉÙÁ¿°±Æø£¬¼×Öз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º ______________________¡£ Îì×°ÖÃÊÕ¼¯°±ÆøÓ¦´ÓÊÕ¼¯×°ÖõÄ_____(Ìî×ÖĸÐòºÅ)µ¼¹Ü½øÆø¡£

£¨6£©Ñ¡ÓÃ×°ÖÃÒÒ¡¢±û¡¢ÎìÖƱ¸¡¢ÊÕ¼¯Ò»Ñõ»¯µªÆøÌ壬ÒÒÖÐËÜÁÏ°åÉÏÈô·ÅÖÃʯ»Òʯ£¬ÎìÖРʢÂúÏ¡NaOHÈÜÒº£¬ÒÇÆ÷ÕýÈ·µÄÁ¬½Ó˳ÐòΪ________________________ (ÓýӿÚ×Öĸ±íʾ)¡£

¡¾ÌâÄ¿¡¿²ÝËáÑÇÌú£¨FeC2O4¡¤2H2O£©¿ÉÓÃÓÚÖƱ¸Á×ËáÌú﮵ç³ØÕý¼«²ÄÁÏ¡£ÊµÑéÊÒÖƱ¸²ÝËáÑÇÌú²¢²â¶¨Æä×é³ÉµÄʵÑéÁ÷³ÌÈçÏ£º

£¨1£© ¡°Èܽ⡱ʱ¼ÓÈȵÄÄ¿µÄÊÇ_____________________£»¡°³ÁÌú¡±Ê±½«ÈÜÒºÖó·Ð²¢ÔÚ²»¶Ï½Á°èϼÓÈëH2C2O4ÈÜÒº£¬¡°²»¶Ï½Á°è¡±³ý¿Éʹ·´Ó¦Îï³ä·Ö½Ó´¥Í⣬ÁíһĿµÄÊÇ_____________________¡£

£¨2£© ¡°¹ýÂË¡¢Ï´µÓ¡±Ê±£¬ÄÜ˵Ã÷³ÁµíÒÑÏ´µÓ¸É¾»µÄÒÀ¾ÝÊÇ_____________________¡£

£¨3£©°´ÒÔϲ½Öè¿ÉÒԲⶨ²úÆ·ÖвÝËá¸ùµÄÖÊÁ¿·ÖÊý¡£

¢Ù³ÆÈ¡0.1600g²ÝËáÑÇÌúÓÚ׶ÐÎÆ¿ÖУ¬¼ÓÈë25mL2mol¡¤L£­1µÄH2SO4ÈÜÒº£¬¼ÓÈÈÖÁ40~50¡æ£¬Ê¹ÑùÆ·Èܽ⡣

¢ÚÓÃ0.02000 mol¡¤L£­1KMnO4ÈÜÒºµÎ¶¨ÖÁÖյ㣬ÏûºÄKMnO4ÈÜÒº23.80mL¡£

[5C2O+2MnO+16H+=10CO2¡ü+2Mn2++8H2O£»5Fe2++MnO+8H+=5Fe3++Mn2++4H2O]

¢ÛÔڢڵζ¨ºóµÄÈÜÒºÖмÓÈë×ãÁ¿Zn·Û£¨2Fe3++Zn = 2Fe2++Zn2+£©ºÍ5mL2mol¡¤L£­1 µÄH2SO4ÈÜÒº£¬Öó·ÐÔ¼10min¡£

¢Ü½«ÂËÒº¹ýÂËÖÁÁíÒ»¸ö׶ÐÎÆ¿ÖУ¬ÓÃ10mL1mol¡¤L£­1µÄH2SO4ÈÜҺϴµÓ׶ÐÎÆ¿£¬½«È«²¿Fe2+תÒÆÈë׶ÐÎÆ¿ÖУ¬ÔÙÓÃ0.02000 mol¡¤L£­1KMnO4ÈÜÒºµÎ¶¨ÖÁÖյ㣬ÏûºÄKMnO4ÈÜÒºÌå»ý8.00 mL¡£

£¨I£©²âµÃµÄ²úÆ·ÖÐn(Fe) ¡Ãn(C2O) _________1¡Ã1£¨Ìî¡°>¡± ¡°£½¡±»ò¡°<¡±£©¡£

£¨II£©Çë¼ÆËã²úÆ·ÖÐC2OµÄÖÊÁ¿·ÖÊý(д³ö¼ÆËã¹ý³Ì)¡£______________

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

¾«Ó¢¼Ò½ÌÍø