ÌâÄ¿ÄÚÈÝ

ij»¯Ñ§ÐËȤС×éÀûÓÃij·ÏÆúµÄÑõ»¯Í­Ð¿¿óÖÆÈ¡»îÐÔZnOʵÑéÁ÷³ÌÈçÏ£º

Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©¼ÓÈëÌú·Ûºó·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ_________________________________¡£
£¨2£©¼×¡¢ÒÒÁ½Í¬Ñ§Ñ¡ÓÃÏÂÁÐÒÇÆ÷£¬²ÉÓò»Í¬µÄ·½·¨À´ÖÆÈ¡°±Æø¡£

A             B
¢Ù¼×ͬѧʹÓõÄÒ©Æ·ÊÇÊìʯ»ÒÓëÂÈ»¯ï§£¬ÔòӦѡÓÃ×°ÖÃ_______£¨ÌîдװÖôúºÅ£©£¬Éú³É°±ÆøµÄ»¯Ñ§·½³ÌʽΪ_______________________________________£»
¢ÚÒÒͬѧѡÓÃÁË×°ÖÃB£¬ÔòʹÓõÄÁ½ÖÖÒ©Æ·µÄÃû³ÆΪ_______________¡£
£¨3£©H2O2µÄ×÷ÓÃÊÇ____________________________________________________¡£
£¨4£©³ýÌú¹ý³ÌÖеõ½µÄFe(OH)3¿ÉÓÃKClOÈÜÒºÔÚ¼îÐÔ»·¾³½«ÆäÑõ»¯µÃµ½Ò»ÖÖ¸ßЧµÄ¶à¹¦ÄÜË®´¦Àí¼Á£¨K2FeO4£©£¬¸Ã·´Ó¦ÖÐÑõ»¯¼ÁÓ뻹ԭ¼ÁµÄÎïÖʵÄÁ¿Ö®±ÈΪ____________¡£
£¨5£©ÒÒÖªÈÜÒºaÖк¬ÓÐCO32-¡¢SO42-Á½ÖÖËá¸ùÒõÀë×Ó£¬ÈôÖ»ÔÊÐíÈ¡ÓÃÒ»´ÎÑùÆ·£¬¼ìÑéÕâÖÖÀë×Ó´æÔÚµÄʵÑé²Ù×÷¹ý³ÌΪ________________________________________________________________________________¡£

£¨1£©Fe+2H+=Fe2+  +H2¡ü    Fe+Cu2+=Fe2++Cu £¨2£©¢ÙA   2NH4Cl+Ca£¨OH£©2 =CaCl2+2NH3¡ü +2H2O ¢ÚŨ°±Ë®ºÍ¼îʯ»Ò»òÉúʯ»Ò»òNaOH¹ÌÌ壨3£©½«ÈÜÒºÖÐFe2+Ñõ»¯ÎªFe3+  £¨4£©3:2 £¨5£©È¡ÉÙÁ¿ÈÜÒºaÓÚÊÔ¹ÜÖУ¬µÎ¼Ó¹ýÁ¿µÄÏ¡ÑÎËᣬÓÐÎÞÉ«ÆøÅݲúÉú¡£ËµÃ÷CO32-£¬¼ÌÐøµÎ¼ÓÂÈ»¯±µ»òÕßÇâÑõ»¯±µÈÜÒº£¬²úÉú°×É«³Áµí£¬ËµÃ÷ÓÐSO42-¡£

½âÎöÊÔÌâ·ÖÎö£º£¨1£©²ÎÕÕÁ÷³Ì£¬·´Ó¦µÄÎïÖÊÓÐÌú·Û£¬¹ýÁ¿µÄÏ¡ÁòËáºÍÑõ»¯Í­£¬¿ÉÒÔÍƳö¶ÔÓ¦µÄÀë×Ó·´Ó¦¡££¨2£©¢Ù¹ÌÌå¼ÓÈÈÑ¡ÔñA×°ÖᣢÚB×°Öò»ÐèÒª¼ÓÈÈ£¬ËùÒÔÓÃŨ°±Ë®ºÍ¼îʯ»Ò»òÉúʯ»Ò»òNaOH¹ÌÌå¡££¨3£©ÒòΪ²úÎïÓÐFe£¨OH£©3£¬ËùÒÔH2O2½«Fe2+Ñõ»¯ÎªFe3+  £¬Ë®½â³ÉFe£¨OH£©¡££¨4£©²ÎÕÕµç×ÓµÃʧÊغ㣬Fe(OH)3תÒÆ3¸öµç×Ó£¬KClOתÒÆ2¸öµç×Ó¡£ËùÒÔÎïÖʵÄÁ¿Ö®±ÈΪ3:2¡££¨5£©CO32-ÓëËá·´Ó¦Éú³ÉCO2 ,ÔÚ¼ÓÈë±µÊÔ¼ÁÉú³ÉBaSO4³Áµí¡£µ«µÚÒ»²½ÖеÄËáÊǹýÁ¿µÄ¡£
¿¼µã£ºÒÔÁ÷³ÌΪ±³¾°¿¼²ìѧÉú¶ÔÓÚ·ÖÎöʵÑé²Ù×÷ÄÜÁ¦£¬Àí½âÔªËØFe¡¢N¡¢CuµÈÐÔÖÊ¡£ÔÚÁ÷³ÌÌâÖÐÎÒÃÇӦעÒâÓÐЧµÄ·ÖÎöÁ÷³ÌÖеĻù±¾·´Ó¦ºÍÖØÒªµÄʵÑé²Ù×÷£¨Èç½á¾§£¬Õô·¢£¬ÖƱ¸£¬³ýÔӵȣ©

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

Èçͼ1ËùʾÊÇʵÑéÊÒÖÐÖÆÈ¡ÆøÌåµÄÒ»ÖÖ¼òÒ××°Öá£

£¨1£©Çë¼òÊö¼ìÑéËüµÄÆøÃÜÐԵķ½·¨_______________________¡£
£¨2£©ÀûÓÃÈçͼ1ËùʾװÖÿÉÒÔÖÆÈ¡£¨Ìî·´Ó¦Îï×´¿ö¼°·¢Éú·´Ó¦ÊÇ·ñÐèÒªµÄÌõ¼þ£©________________________ÆøÌå¡£
ijͬѧÉè¼ÆÈçͼ2ËùʾװÖã¬ÓôÖÌúÁ£Óë16.9%Ï¡ÏõËá·´Ó¦ÖÆÈ¡NOÆøÌ岢̽¾¿²úÎïÖÐÌúµÄ¼Û̬¡£Çë»Ø´ðÓйØÎÊÌâ¡£

£¨3£©ÒÑÖª16.9%Ï¡ÏõËáµÄÃܶÈΪ1.10g/cm3£¬ÔòÆäÎïÖʵÄÁ¿Å¨¶ÈΪ____________£¨¼ÆËã½á¹û±£ÁôÁ½Î»Ð¡Êý£©¡£ÓÃÈôÓÃ63%µÄÏõËáÅäÖÆ16.9%Ï¡ÏõËá500mL£¬ËùÐèµÄ²£Á§ÒÇÆ÷Óв£Á§°ô¡¢ÉÕ±­ºÍ            ¡£
£¨4£©µ±´ò¿ªÖ¹Ë®¼Ða¡¢¹Ø±Õֹˮ¼Ðbʱ£¬A×°ÖõĸÉÔï¹ÜÖй۲쵽µÄÏÖÏóÊÇ_______________________¡£B×°ÖÃÉÕ±­ÖÐÒºÌåµÄ×÷ÓÃÊÇ__________________________________¡£µ±A×°ÖÃÖÐÆøÌ弸ºõÎÞɫʱ£¬´ò¿ªÖ¹Ë®¼Ðb£¬¹Ø±Õֹˮ¼Ða£¬¿ÉÓÃC×°ÖÃÊÕ¼¯NOÆøÌå¡£
£¨5£©ÒÑÖªÏÂÁз´Ó¦¿ÉÒÔ·¢Éú£ºFe2O3+3KNO3+4KOH2K2FeO4+3KNO2+2H2O£¬¶øA×°ÖÃÖеÄÏ¡ÏõËἴʹ»»³ÉŨÏõËᣬҲ²»ÄÜÉú³É+6¼ÛµÄÌúµÄ»¯ºÏÎÆäÔ­ÒòÊÇ________¡£
a£®HNO3µÄÑõ»¯ÐÔ±ÈKNO3Èõ
b£®·´Ó¦µÄζȲ»¹»
c£®HNO3µÄÈÈÎȶ¨ÐÔ±ÈKNO3²î
d£®FeO42£­²»ÄÜ´æÔÚÓÚËáÐÔÈÜÒºÖÐ
£¨6£©ÏÖÓÐÒÇÆ÷ºÍÒ©Æ·£ºÊԹܺͽºÍ·µÎ¹Ü£¬0.1mol/LKSCNÈÜÒº¡¢0.2mol/LËáÐÔKMnO4ÈÜÒº¡¢0.1mol/LKIÈÜÒº¡¢ÂÈË®µÈ¡£ÇëÄãÉè¼ÆÒ»¸ö¼òµ¥ÊµÑ飬̽¾¿A×°ÖÃÉÕ±­ÀïÍêÈ«·´Ó¦ºóº¬Ìú¿ÉÄܵļÛ̬£¬ÌîдÏÂÁÐʵÑ鱨¸æ£º

ʵÑé²½Öè
²Ù×÷
ÏÖÏóÓë½áÂÛ
µÚÒ»²½
È¡ÉÙÁ¿ÒºÌå×°ÓÚÊԹܣ¬ÏòÊÔ
¹ÜÖеÎÈ뼸µÎKSCNÈÜÒº¡£
 
µÚ¶þ²½
 
ÈôÈÜÒº×ÏÉ«ÍÊÈ¥£¬Ôò˵Ã÷º¬ÓÐFe2+£»
ÈôÎÞÃ÷ÏԱ仯£¬Ôò˵Ã÷²»º¬Fe2+¡£
 

º£ÑóÖ²ÎïÈ纣´ø¡¢º£ÔåÖк¬ÓзḻµÄµâÔªËØ£¬µâÔªËØÒÔµâÀë×ÓÐÎʽ´æÔÚ¡£ÊµÑéÊÒ´Óº£ÔåÖÐÌáÈ¡µâµÄÁ÷³ÌÈçÏ£º

£¨1£©Ö¸³öÌáÈ¡µâµÄ¹ý³ÌÖÐÓйØʵÑé²Ù×÷µÄÃû³Æ£º
¢ÙΪ                £¬¢ÛΪ               £»¹ý³Ì¢ÚÖÐÓйط´Ó¦µÄÀë×Ó·½³ÌʽÊÇ                                            ¡£
£¨2£©ÌáÈ¡µâµÄ¹ý³ÌÖпɹ©Ñ¡ÔñµÄÓлúÈܼÁÊÇ(       ) 

A£®¼×±½¡¢¾Æ¾« B£®ËÄÂÈ»¯Ì¼¡¢±½ C£®ÆûÓÍ¡¢ÒÒËá D£®ÆûÓÍ¡¢¸ÊÓÍ
£¨3£©ÎªÊ¹º£Ôå»ÒÖеâÀë×Óת»¯ÎªµâµÄÓлúÈÜÒº£¬ÊµÑéÊÒÀïÓÐÉÕ±­¡¢²£Á§°ô¡¢ ¼¯ÆøÆ¿¡¢¾Æ¾«µÆ¡¢µ¼¹Ü¡¢Ô²µ×ÉÕÆ¿¡¢Ê¯ÃÞÍø£¬ÒÔ¼°±ØÒªµÄ¼Ð³Ö×°Öá¢ÎïÆ·£¬ÉÐȱÉٵIJ£Á§ÒÇÆ÷ÊÇ                      ¡£
£¨4£©´Óº¬µâµÄÓлúÈܼÁÖÐÌáÈ¡µâºÍ»ØÊÕÓлúÈܼÁ£¬»¹ÐèÒª¾­¹ýÕôÁó¡£Ö¸³öÏÂͼËùʾµÄʵÑé×°ÖÃÖеĴíÎóÖ®´¦£º ¢Ù                        £¬¢Ú                           ¡£

£¨5£©½øÐÐÉÏÊöÕôÁó²Ù×÷ʱ£¬Ê¹ÓÃˮԡ¼ÓÈȵÄÔ­ÒòÊÇ            £¬×îºó¾§Ì¬µâÔÚ            Àï¾Û¼¯¡£

´×ËáÑǸõË®ºÏÎï([Cr(CH3COO)2)]2¡¤2H2O£¬ÉîºìÉ«¾§Ì壩ÊÇÒ»ÖÖÑõÆøÎüÊÕ¼Á£¬Í¨³£ÒÔ¶þ¾ÛÌå·Ö×Ó´æÔÚ£¬²»ÈÜÓÚÀäË®ºÍÃÑ£¬Î¢ÈÜÓÚ´¼£¬Ò×ÈÜÓÚÑÎËᡣʵÑéÊÒÖƱ¸´×ËáÑǸõË®ºÏÎïµÄ×°ÖÃÈçͼËùʾ£¬Éæ¼°µÄ»¯Ñ§·½³ÌʽÈçÏ£º
Zn(s) + 2HCl(aq) = ZnCl2(aq) + H2(g);  2CrCl3(aq) + Zn(s)= 2CrCl2 (aq) + ZnCl2(aq)
2Cr2+(aq) + 4CH3COO-(aq) + 2H2O(l) = [Cr(CH3COO)2]2¡¤2H2O (s)

Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©¼ì²éÐé¿òÄÚ×°ÖÃÆøÃÜÐԵķ½·¨ÊÇ     ¡£
£¨2£©´×ËáÄÆÈÜÒºÓ¦·ÅÔÚ×°Öà    ÖУ¨ÌîдװÖñàºÅ£¬ÏÂͬ£©£»ÑÎËáÓ¦·ÅÔÚ×°Öà    ÖУ»
×°ÖÃ4µÄ×÷ÓÃÊÇ     ¡£
£¨3£©±¾ÊµÑéÖÐËùÓÐÅäÖÆÈÜÒºµÄË®ÐèÖó·Ð£¬ÆäÔ­ÒòÊÇ     ¡£
£¨4£©½«Éú³ÉµÄCrCl2ÈÜÒºÓëCH3COONaÈÜÒº»ìºÏʱµÄ²Ù×÷ÊÇ     ·§ÃÅA¡¢     ·§ÃÅB £¨Ìî¡°´ò¿ª¡±»ò¡°¹Ø±Õ¡±£©¡£
£¨5£©±¾ÊµÑéÖÐпÁ£Ðë¹ýÁ¿£¬ÆäÔ­ÒòÊÇ     ¡£
£¨6£©ÎªÏ´µÓ[Cr(CH3COO)2)]2¡¤2H2O²úÆ·£¬ÏÂÁз½·¨ÖÐ×îÊʺϵÄÊÇ     ¡£
A£®ÏÈÓÃÑÎËáÏ´£¬ºóÓÃÀäˮϴ        B£®ÏÈÓÃÀäˮϴ£¬ºóÓÃÒÒ´¼Ï´
C£®ÏÈÓÃÀäˮϴ£¬ºóÓÃÒÒÃÑÏ´        D£®ÏÈÓÃÒÒ´¼Ï´µÓ£¬ºóÓÃÒÒÃÑÏ´

ijÑо¿ÐÔѧϰС×齫һ¶¨Å¨¶ÈµÄNa2CO3ÈÜÒºµÎÈëMgSO4ÈÜÒºÖеõ½°×É«³Áµí¡£¼×ͬѧÈÏΪÁ½Õß·´Ó¦Ö»Éú³ÉMgCO3Ò»ÖÖ³Áµí£»ÒÒͬѧÈÏΪÕâÁ½ÕßÏ໥´Ù½øË®½â£¬Ö»Éú³ÉMg(OH)2Ò»ÖÖ³Áµí£»±ûͬѧÈÏΪÉú³ÉMgCO3ºÍMg(OH)2Á½ÖÖ³Áµí¡£(²éÔÄ×ÊÁÏÖª£ºMgCO3ºÍMg(OH)2¾ù²»´ø½á¾§Ë®)
(1)°´ÕÕÒÒͬѧµÄÀí½âNa2CO3ÈÜÒººÍMgSO4ÈÜÒº·´Ó¦µÄ»¯Ñ§·´Ó¦·½³ÌʽΪ___________________£»
ÔÚ̽¾¿³ÁµíÎï³É·ÖÇ°£¬Ð뽫³Áµí´ÓÈÜÒºÖзÖÀë²¢¾»»¯¡£¾ßÌå²Ù×÷Ϊ¢Ù¹ýÂË¢ÚÏ´µÓ¢Û¸ÉÔï¡£
(2)ÇëÓÃÏÂͼËùʾװÖã¬Ñ¡ÔñºÏÊʵÄʵÑé×°ÖúͱØÒªµÄÊÔ¼Á£¬Ö¤Ã÷³ÁµíÎïÖÐÖ»ÓÐ̼Ëáþ¡£

¢Ù¸÷×°ÖÃÁ¬½Ó˳ÐòΪ__________________(ÓÃ×°ÖñàºÅ±íʾ)£»
¢Ú×°ÖÃCÖÐ×°ÓÐÊÔ¼ÁµÄÃû³ÆÊÇ______________£»
¢ÛÄÜÖ¤Ã÷Éú³ÉÎïÖÐÖ»ÓÐMgCO3µÄʵÑéÏÖÏóÊÇ_______¡£
(3)ÈôMg(OH)2ºÍMgCO3Á½Õ߶¼ÓУ¬¿Éͨ¹ýÏÂÁÐËùʾװÖõÄÁ¬½Ó£¬½øÐж¨Á¿·ÖÎöÀ´²â¶¨Æä×é³É¡£

ʵÑé½áÊøʱͨÈë¹ýÁ¿µÄ¿ÕÆøµÄ×÷ÓÃÊÇ______________________________¡£
A×°ÖúÍE×°ÖõĹ²Í¬×÷ÓÃÊÇ__________________________¡£
¢ÛʵÑé¹ý³ÌÖвⶨµÄÊý¾ÝÓУºW1£ºÑùÆ·µÄÖÊÁ¿£¬W2£º·´Ó¦ºó×°ÖÃBÖвÐÔüµÄÖÊÁ¿£¬W3£º·´Ó¦Éú³ÉË®µÄÖÊÁ¿£¬W4£º·´Ó¦Éú³É¶þÑõ»¯Ì¼µÄÖÊÁ¿£¬ÎªÁ˲ⶨÆä×é³É£¬ÖÁÉÙÐèÒªÆäÖÐ____¸öÊý¾Ý£¬Ç뽫¿ÉÄܵÄ×éºÏÌîÈëϱíÖÐ(ÿ¸ö¿Õ¸ñÖÐÌîÒ»ÖÖ×éºÏ)¡£

´ïÖÝÊÐÇþÏؾ³ÄÚº¬ÓзḻµÄÉî²ãµØÏÂʳÑÎ×ÊÔ´£¬Ê³ÑÎÊÇÈÕ³£Éú»îÖеıØÐèÆ·£¬Ò²ÊÇÖØÒªµÄ»¯¹¤Ô­ÁÏ¡£
£¨1£©¾­²â¶¨¸ÃÏØÉî²ãµØÏ´ÖÑÎÖк¬ÓÐÉÙÁ¿K+¡¢Ca2+¡¢Mg2+¡¢Fe3+µÈÔÓÖÊÀë×Ó£¬Ä³Ñо¿ÐÔѧϰС×éÔÚʵÑéÊÒÌá´¿NaClµÄÁ÷³ÌÈçÏ£º

ËùÌṩµÄÊÔ¼Á£º±¥ºÍNa2CO3ÈÜÒº¡¢±¥ºÍK2CO3ÈÜÒº¡¢NaOHÈÜÒº¡¢BaCl2ÈÜÒº¡¢Ba(NO3)2ÈÜÒº¡¢75%ÒÒ´¼ÈÜÒº¡¢CCl4£¬ÒÇÆ÷¼°ÓÃÆ·×ÔÑ¡¡£
¢ÙÓû³ýÈ¥ÈÜÒºÖеÄCa2+¡¢Mg2+¡¢Fe3+¡¢SO42-£¬Ñ¡³öa²Ù×÷ÖÐËù´ú±íµÄÊÔ¼Á£¬°´µÎ¼Ó˳ÐòÒÀ´ÎΪ
                    £¨Ö»Ìѧʽ£©£¬b²½²Ù×÷µÄÃû³ÆÊÇ                    ¡£
¢ÚÏ´µÓ³ýÈ¥NaCl¾§Ìå±íÃ渽´øµÄÉÙÁ¿KCl£¬Ó¦Ñ¡ÓÃÊÔ¼ÁÊÇ                    £¬ÓÃPHÊÔÖ½²â¶¨ÂËÒº¢òPHÖµµÄ·½·¨ÊÇ                                         ¡£
£¨2£©ÓÃÌá´¿µÄNaClÅäÖÆ500mL£¬2.5mol¡¤L-1µÄNaClÈÜÒº£¬ËùÐèÒÇÆ÷³ýÉÕ±­£¬ÍÐÅÌÌìƽ£¨íÀÂëºÍÄ÷×Ó£©£¬Ò©³×£¬²£Á§°ôÍ⣬»¹ÐèÒª           £¨ÌîÒÇÆ÷Ãû³Æ£©£¬Ó¦³ÆÈ¡NaCl          g
(3)ÏÂÁвÙ×÷»áµ¼ÖÂËùÅäNaClÈÜҺŨ¶ÈÆ«¸ßµÄÊÇ             

A£®¶¨ÈÝÍê±Ïºó£¬¸ÇÈû£¬Ò¡ÔÈ£¬ÔÙ½«ÈÝÁ¿Æ¿ÖÃÓÚʵÑę́ÉÏ£¬·¢ÏÖÒºÃæµÍÓڿ̶ÈÏߣ¬ÔÙÌí¼ÓÕôÁóË®ÖÁ¿Ì¶ÈÏß¡£
B£®Î´½«Ï´µÓÉÕ±­ÄÚ±ÚµÄÈÜҺתÈëÈÝÁ¿Æ¿¡£ C£®¶¨ÈÝʱ£¬¸©Êӿ̶ÈÏß¡£
D£®×ªÒÆÈÜҺ֮ǰ£¬ÈÝÁ¿Æ¿ÄÚÓÐÉÙÁ¿ÕôÁóË®¡£ E¡¢³ÆÁ¿Ê±£¬ÌìƽָÕëÖ¸Ïò×óÅÌ¡£

²ÝËáÑÇÌú£¨FeC2O4?2H2O£©ÓÃ×÷·ÖÎöÊÔ¼Á¼°ÏÔÓ°¼ÁºÍÐÂÐ͵ç³Ø²ÄÁÏÁ×ËáÑÇÌú﮵ÄÉú²ú¡£ÒÑÖª£ºCOÄÜÓëÂÈ»¯îÙ£¨PdCl2£©ÈÜÒº·´Ó¦Éú³ÉºÚÉ«µÄîÙ·Û¡£»Ø´ðÏÂÁÐÎÊÌ⣺
I£®ÐËȤС×é¶Ô²ÝËáÑÇÌúµÄ·Ö½â²úÎï½øÐÐʵÑéºÍ̽¾¿¡£
£¨1£©½«ÆøÌå²úÎïÒÀ´Îͨ¹ýA¡¢³ÎÇåʯ»ÒË®£¬B¡¢ÂÈ»¯îÙ£¬¹Û²ìµ½AÖгÎÇåʯ»ÒË®¶¼±ä»ë×Ç£¬BÖгöÏÖºÚÉ«ÎïÖÊÉú³É£¬ÔòÉÏÊöÏÖÏó˵Ã÷ÆøÌå²úÎïÖÐÓР                      ¡£
£¨2£©Ì½¾¿·Ö½âµÃµ½µÄ¹ÌÌå²úÎïÖÐÌúÔªËصĴæÔÚÐÎʽ¡£     
¢ÙÌá³ö¼ÙÉè
¼ÙÉè1£º________£» ¼ÙÉè2£ºFeO£» ¼ÙÉè3£ºFeOºÍFe»ìºÏÎï
¢ÚÉè¼ÆʵÑé·½°¸Ö¤Ã÷¼ÙÉè3¡£
ÏÞÑ¡ÊÔ¼Á£º  1.0 mol?L£­1ÑÎËá¡¢3% H2O2¡¢0.1 mol?L£­1CuSO4¡¢20% KSCN¡¢ÕôÁóË®¡£

ʵÑé²½Öè
ÏÖÏóÓë½áÂÛ
²½Öè1 £ºÏòÊÔ¹ÜÖмÓÈëÉÙÁ¿¹ÌÌå²úÎÔÙ¼ÓÈë×ãÁ¿_________________£¬³ä·ÖÕñµ´
ÈôÈÜÒºÑÕÉ«Ã÷ÏԸı䣬ÇÒÓÐ_______Éú³É£¬ÔòÖ¤Ã÷ÓÐÌúµ¥ÖÊ´æÔÚ
²½Öè2£º ½«²½Öè1Öеõ½µÄ×ÇÒº¹ýÂË£¬²¢ÓÃÕôÁóˮϴµÓÖÁÏ´µÓÒºÎÞÉ«
 
²½Öè3£ºÈ¡²½Öè2µÃµ½µÄÉÙÁ¿¹ÌÌåÓëÊÔ¹ÜÖУ¬ µÎ¼Ó___________________________________
_______________________________________
 
__________________________________
___________________________________
 
II£®Ä³²ÝËáÑÇÌúÑùÆ·Öк¬ÓÐÉÙÁ¿²ÝËᣨΪ·½±ãÓÚ¼ÆË㣬²ÝËáÑÇÌúÖвÝËá¸ùºÍ²ÝËá·Ö×Ó¾ùÓÃC2O42£­´úÌ棩¡£ÏÖÓõζ¨·¨²â¶¨¸ÃÑùÆ·ÖÐFeC2O4µÄº¬Á¿¡£µÎ¶¨·´Ó¦·Ö±ðÊÇ£º5Fe2++MnO4£­+8H+=5Fe3+ +Mn2++4H2O¡¢5C2O42£­+2MnO4£­+16H+=10CO2¡ü+2Mn2++8H2O¡£
£¨3£©ÊµÑé·½°¸Éè¼ÆΪ£º
¢Ù½«×¼È·³ÆÁ¿µÄ0.20g²ÝËáÑÇÌúÑùÆ·ÖÃÓÚ250 mL׶ÐÎÆ¿ÄÚ£¬¼ÓÈëÊÊÁ¿2 mol/LµÄH2SO4ÈÜÒº£¬Ê¹ÑùÆ·Èܽ⣬¼ÓÈÈÖÁ70¡æ×óÓÒ£¬Á¢¼´ÓÃŨ¶ÈΪ0.02000 mol/LµÄ¸ßÃÌËá¼Ø±ê×¼ÈÜÒºµÎ¶¨ÖÁÖյ㣬¼Ç¼Êý¾Ý¡£Öظ´µÎ¶¨2´Î¡£Æ½¾ùÏûºÄV1 mL¡£
¢ÚÏòÉÏÊöµÎ¶¨»ìºÏÒºÖмÓÈëÊÊÁ¿µÄZn·ÛºÍ¹ýÁ¿µÄ2 mol/LµÄH2SO4ÈÜÒº£¬Öó·Ð5¡«8min£¬ÓÃKSCNÈÜÒºÔÚµãµÎ°åÉϼìÑéµãµÎÒº£¬Ö±ÖÁÈÜÒº²»Á¢¿Ì±äºì¡£½«ÂËÒº¹ýÂËÖÁÁíÒ»¸ö׶ÐÎÆ¿ÖУ¬¼ÌÐøÓÃ0.02000 mol/LµÄ¸ßÃÌËá¼Ø±ê×¼ÈÜÒºµÎ¶¨ÖÁÖյ㣬¼Ç¼Êý¾Ý¡£Öظ´µÎ¶¨2´Î¡£Æ½¾ùÏûºÄV2 mL¡£
¢ÛÈôijС×éµÄÒ»´Î²â¶¨Êý¾Ý¼Ç¼ÈçÏ£º V1= 18.90mL£¬V2=6.20mL¡£¸ù¾ÝÊý¾Ý¼ÆËã0.20gÑùÆ·ÖУºn£¨Fe2+£©=      £» n£¨C2O42£­£©=    £»FeC2O4µÄÖÊÁ¿·ÖÊýΪ               £¨¾«È·µ½0.01%£¬FeC2O4µÄʽÁ¿Îª144£©

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

¾«Ó¢¼Ò½ÌÍø