ΪÑé֤±Ëص¥ÖÊÑõ»¯ÐÔµÄÏà¶ÔÇ¿Èõ,ijС×éÓÃÏÂͼËùʾװÖýøÐÐʵÑé(¼Ð³ÖÒÇÆ÷ÒÑÂÔÈ¥,ÆøÃÜÐÔÒѼìÑé)¡£

ʵÑé¹ý³Ì:
¢ñ.´ò¿ªµ¯»É¼Ð,´ò¿ª»îÈûa,µÎ¼ÓŨÑÎËá¡£
¢ò.µ±BºÍCÖеÄÈÜÒº¶¼±äΪ»Æɫʱ,¼Ð½ôµ¯»É¼Ð¡£
¢ó.µ±BÖÐÈÜÒºÓÉ»ÆÉ«±äΪ×غìɫʱ,¹Ø±Õ»îÈûa¡£
¢ô.¡­¡­
(1)AÖвúÉú»ÆÂÌÉ«ÆøÌå,Æ仯ѧ·½³ÌʽÊÇ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡£ 
(2)ÑéÖ¤ÂÈÆøµÄÑõ»¯ÐÔÇ¿ÓÚµâµÄʵÑéÏÖÏóÊÇ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡£ 
(3)BÖÐÈÜÒº·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ¡¡¡¡¡¡¡¡¡¡¡¡¡£ 
(4)ΪÑéÖ¤äåµÄÑõ»¯ÐÔÇ¿ÓÚµâ,¹ý³Ì¢ôµÄ²Ù×÷ºÍÏÖÏóÊÇ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡£ 
(5)¹ý³Ì¢óʵÑéµÄÄ¿µÄÊÇ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡£ 
(6)ÂÈ¡¢äå¡¢µâµ¥ÖʵÄÑõ»¯ÐÔÖð½¥¼õÈõµÄÔ­Òò:ͬÖ÷×åÔªËØ´ÓÉϵ½Ï¡¡¡¡¡¡¡¡,µÃµç×ÓÄÜÁ¦Öð½¥¼õÈõ¡£ 

(1)16HCl(Ũ)+2KMnO4=2MnCl2+2KCl+5Cl2¡ü+8H2O
(2)µí·ÛKIÊÔÖ½±äÀ¶
(3)Cl2+2Br-=Br2+2Cl-
(4)´ò¿ª»îÈûb,½«ÉÙÁ¿CÖÐÈÜÒºµÎÈëDÖÐ,¹Ø±Õ»îÈûb,È¡ÏÂDÕñµ´¡£¾²ÖúóCCl4²ãÈÜÒº±äΪ×ϺìÉ«
(5)È·ÈÏCµÄ»ÆÉ«ÈÜÒºÖÐÎÞCl2,ÅųýCl2¶ÔäåÖû»µâʵÑéµÄ¸ÉÈÅ
(6)Ô­×Ӱ뾶Öð½¥Ôö´ó

½âÎö

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

CuSO4ÈÜÒºÓëK2C2O4ÈÜÒº»ìºÏÉú³ÉÒ»ÖÖÀ¶É«¼ØÑÎË®ºÏÎïKaCub(C2O4)c¡¤nH2O¡£Í¨¹ýÏÂÊöʵÑéÈ·¶¨¸Ã¾§ÌåµÄ×é³É¡£
²½Öèa£º³ÆÈ¡0.6720 gÑùÆ·£¬·ÅÈë׶ÐÎÆ¿£¬¼ÓÈëÊÊÁ¿2mol¡¤L£­1Ï¡ÁòËᣬ΢ÈÈʹÑùÆ·Èܽ⡣ÔÙ¼ÓÈë30 mLË®¼ÓÈÈ£¬ÓÃ0.2000mol¡¤L£­1 KMnO4ÈÜÒºµÎ¶¨ÖÁÖյ㣬ÏûºÄ8.00 mL KMnO4ÈÜÒº¡£Óйط´Ó¦£º2MnO4£­£«5C2O42£­£«16H£«=2Mn2+£«8H2O£«10CO2¡ü
²½Öèb£º½Ó׎«ÈÜÒº³ä·Ö¼ÓÈÈ¡£ÀäÈ´ºó£¬µ÷½ÚpH²¢¼ÓÈë¹ýÁ¿µÄKI¹ÌÌ壬ÈÜÒº±äΪ×ØÉ«²¢²úÉú°×É«³ÁµíCuI¡£ÓÃ0.2500mol¡¤L£­1 Na2S2O3±ê×¼ÈÜÒºµÎ¶¨ÖÁÖյ㣬ÏûºÄ8.00 mL Na2S2O3ÈÜÒº¡£µÎ¶¨Ê±·´Ó¦Îª£ºI2£«2S2O32£­=2I£­£«S4O62£­
£¨1£©²½ÖèbÖÐÉú³É°×É«³ÁµíµÄÀë×Ó·½³Ìʽ£º                   ¡£
£¨2£©²½ÖèbÖС°½«ÈÜÒº³ä·Ö¼ÓÈÈ¡±µÄÄ¿µÄÊÇ                     ¡£
£¨3£©ÇëÍê³ÉÏÂÁÐÈ·¶¨ÑùÆ·×é³ÉµÄ¼ÆËã¹ý³Ì¡£
¢Ù¼ÆËãÑùÆ·ÖÐn (C2O42£­)£¨Ð´³ö¼ÆËã¹ý³Ì£©
¢Ú¼ÆËãÑùÆ·ÖÐn (Cu2+) £¨Ð´³ö¼ÆËã¹ý³Ì£©
¢Û¸ù¾Ý         Ô­Àí¿ÉÇó³ön (K+)£¬ÔÙÒÀ¾Ý         Ô­ÀíÇóµÃn (H2O)
¢Ü¸ÃÑùÆ·¾§ÌåµÄ»¯Ñ§Ê½Îª        

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º¼ÆËãÌâ

°Ñ19£®2 g µÄCu·ÅÈë×ãÁ¿µÄÏ¡ÏõËáÖУ¬Î¢ÈÈÖÁCuÍêÈ«·´Ó¦¡£
ÒÑÖª£º3Cu + 8HNO3(Ï¡) = 3Cu(NO3)2 +2NO¡ü+ 4H2OÇó£º
£¨1£©²Î¼Ó·´Ó¦µÄÏõËáµÄÎïÖʵÄÁ¿£»
£¨2£©±»»¹Ô­µÄÏõËáµÄÖÊÁ¿£»
£¨3£©Éú³ÉµÄNOÔÚ±ê×¼×´¿öϵÄÌå»ý¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÊµÑéÌâ

£¨14·Ö£©2013Äê6Ô£¬ÎÒ¹ú¡°òÔÁú¡±ºÅÔÙ´ÎˢС°ÖйúÉî¶È¡±¡ª¡ªÏÂDZ7062Ã×£¬ÎªÎÒ¹úÉ¿óÎï×ÊÔ´µÄ¿ª·¢µì¶¨ÁË»ù´¡¡£º£ÑóÉî´¦ÓзḻµÄÃ̽áºË¿ó£¬Ã̽áºËµÄÖ÷Òª³É·ÖÊÇMnO2£¬Í¬Ê±»¹º¬ÓлÆÍ­¿ó¡£
¢ñ¡¢¡°òÔÁú¡±ºÅÍâ¿ÇÊÇÓÃÌØÊâµÄîѺϽð²ÄÁÏÖƳɣ¬Ëü¿ÉÒÔÔÚ7000mµÄÉÖгÐÊÜÖØѹ£¬TiÊÇÒÔîÑ°×·Û£¨TiO2£©ÎªÔ­ÁϽøÐÐÉú²ú£¬îÑ°×·ÛÊÇÀûÓÃTiO2£«·¢ÉúË®½âÉú³ÉîÑËᣨH2TiO3£©³Áµí£¬ÔÙìÑÉÕ³ÁµíÖƵõġ£TiO2£«·¢ÉúË®½âµÄÀë×Ó·½³ÌʽΪ____________________________¡£
¢ò¡¢MnO2ÊÇÒ»ÖÖÖØÒªµÄÎÞ»ú¹¦ÄܲÄÁÏ£¬¹¤ÒµÉÏ´ÓÃ̽áºËÖÐÖÆÈ¡´¿¾»µÄMnO2¹¤ÒÕÁ÷³ÌÈçÏÂͼËùʾ£º

£¨1£©²½Öè¢òÖÐÒÔNaClO3ΪÑõ»¯¼Á£¬µ±Éú³É0.05molMnO2ʱ£¬ÏûºÄ0.1mol/LµÄNaClO3ÈÜÒº200ml£¬¸Ã·´Ó¦Àë×Ó·½³ÌʽΪ_______________________________¡£
£¨2£©ÒÑÖªÈÜÒºBµÄÈÜÖÊÖ®Ò»¿ÉÑ­»·ÓÃÓÚÉÏÊöÉú²ú£¬´ËÎïÖʵÄÃû³ÆÊÇ____________¡£
¢ó¡¢ÀûÓûÆÍ­¿óÁ¶Í­²úÉúµÄ¯Ôü£¨º¬Fe2O3¡¢FeO¡¢SiO2¡¢Al2O3µÈ£©¿ÉÖƱ¸Fe2O3£¬·½·¨Îª£º
£¨A£©ÓùýÁ¿µÄÏ¡ÑÎËá½þȡ¯Ôü¡¢¹ýÂË£»£¨B£©ÏòÂËÒºÖмÓÈë5£¥µÄH2O2£¬ÔÙÏòÆäÖмÓÈë¹ýÁ¿µÄNaOHÈÜÒº£¬¹ýÂË£¬½«³ÁµíÏ´µÓ¡¢¸ÉÔï¡¢ìÑÉյõ½Fe2O3¡£¸ù¾ÝÒÔÉÏÐÅÏ¢»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©£¨B£©ÖÐÏòÂËÒºÖмÓÈë5%µÄH2O2£¬ÆäÄ¿µÄÊÇ_________________________________¡£
£¨2£©Éè¼ÆʵÑéÖ¤Ã÷¯ÔüÖк¬ÓÐFeO___________________________________________¡£
£¨3£©½«ìÑÉյõ½µÄFe2O3»¹Ô­ÎªFeµ¥ÖÊ£¬ÔÙ½«ÖÊÁ¿Îªm gµÄFeµ¥ÖÊ·Ö³ÉÏàµÈµÄËÄ·Ý£¬·Ö±ðÓë50mL¡¢100mL¡¢150mL¡¢200mLµÄµÈŨ¶ÈµÄÏ¡ÏõËá·´Ó¦£¬·´Ó¦²úÎïNOÔÚ±ê¿öϵÄÌå»ý¼û¸½±í£º

ʵÑé
¢Ù
¢Ú
¢Û
¢Ü
V(HNO3)/ml
50
100
150
200
V(NO)/L
1.344
2.688
3.36
3.36
Ôò£º¢Ùm£½________g¡£
¢Úд³öʵÑé¢Ú·¢Éú·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º_____________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÊµÑéÌâ

»¯ºÏÎïKxFe(C2O4)y¡¤3H2O(FeΪ£«3¼Û)ÊÇÒ»ÖÖ¹âÃô²ÄÁÏ£¬ÊµÑéÊÒ¿ÉÒÔÓÃÈçÏ·½·¨À´ÖƱ¸ÕâÖÖ²ÄÁϲ¢²â¶¨ÕâÖÖ²ÄÁϵÄ×é³É£º

£¨1£©½á¾§Ê±Ó¦½«ÈÜÒºÓñùË®ÀäÈ´ÖÃÓÚÔÚºÚ°µ´¦µÈ´ý¾§ÌåµÄÎö³ö£¬ÕâÑù²Ù×÷µÄÔ­ÒòÊÇ£º________________________________________________________________________¡£
£¨2£©²Ù×÷¢ñµÄÃû³ÆÊÇ______________________¡£
£¨3£©³ÆÈ¡Ò»¶¨ÖÊÁ¿µÄ¾§ÌåÖÃÓÚ׶ÐÎÆ¿ÖУ¬¼ÓÈë×ãÁ¿µÄÕôÁóË®ºÍÏ¡H2SO4£¬½«C2O42-ת»¯ÎªH2C2O4ºóÓÃ0.100 0 mol¡¤L£­1KMnO4ÈÜÒºµÎ¶¨£¬µ±ÏûºÄKMnO4ÈÜÒº24.00 mLʱǡºÃÍêÈ«·´Ó¦£¬H2C2O4ÓëËáÐÔKMnO4ÈÜÒº·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ£º__________________________¡£ÔÙÏòÈÜÒºÖмÓÈëÊÊÁ¿µÄ»¹Ô­¼Á£¬Ç¡ºÃ½«Fe3£«Íêȫת»¯ÎªFe2£«£¬ÓÃKMnO4ÈÜÒº¼ÌÐøµÎ¶¨¡£µ±Fe2£«ÍêÈ«Ñõ»¯Ê±£¬ÓÃÈ¥KMnO4ÈÜÒº4.00 mL£¬´ËµÎ¶¨·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ____________¡£
£¨4£©ÅäÖÆ100 mL 0.100 0 mol¡¤L£­1KMnO4ÈÜÒº¼°µÎ¶¨ÊµÑéÖÐËùÐèµÄ²£Á§ÒÇÆ÷³ýÉÕ±­¡¢²£Á§°ô¡¢½ºÍ·µÎ¹Ü¡¢Á¿Í²¡¢×¶ÐÎÆ¿Í⻹ÓÐ________(ÌîÒÇÆ÷Ãû³Æ)¡£µÎ¶¨ÖÕµãʱÈÜÒºÑÕɫΪ________É«£¬ÇÒ30ÃëÄÚ²»±äÉ«¡£
£¨5£©¾­¼ÆË㣬»¯ºÏÎïKxFe(C2O4)y¡¤3H2OÖУ¬x£½________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÊµÑéÌâ

Ö¤Ã÷±×åÔªËصķǽðÊôÐÔÇ¿Èõ£¬Ä³Ð¡×éÓÃÏÂͼËùʾװÖýøÐÐʵÑé(¼Ð³ÖÒÇÆ÷ÒÑÂÔÈ¥£¬ÆøÃÜÐÔÒѼì²é)¡£

ʵÑé¹ý³Ì£º
¢ñ£®´ò¿ªµ¯»É¼Ð£¬´ò¿ª»îÈûa£¬µÎ¼ÓŨÑÎËá¡£
¢ò£®µ±BºÍCÖеÄÈÜÒº¶¼±äΪ»Æɫʱ£¬¼Ð½ôµ¯»É¼Ð¡£
¢ó£®µ±BÖÐÈÜÒºÓÉ»ÆÉ«±äΪ×غìɫʱ£¬¹Ø±Õ»îÈûa¡£
¢ô£®¡­¡­
£¨1£©ÑéÖ¤ÂÈÆøµÄÑõ»¯ÐÔÇ¿ÓÚµâµÄʵÑéÏÖÏóÊÇ______________________________¡£
£¨2£©BÖÐÈÜÒº·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ______________________________¡£
£¨3£©ÎªÑéÖ¤äåµÄÑõ»¯ÐÔÇ¿Óڵ⣬¹ý³Ì ¢ô µÄ²Ù×÷ºÍÏÖÏóÊÇ________________________________________¡£
£¨4£©¹ý³Ì¢óʵÑéµÄÄ¿µÄÊÇ__________________________________________________¡£
£¨5£©ÂÈ¡¢äå¡¢µâµ¥ÖʵÄÑõ»¯ÐÔÖð½¥¼õÈõµÄÔ­Òò£ºÍ¬Ö÷×åÔªËØ´ÓÉϵ½ÏÂ,Ô­×Ӱ뾶Öð½¥        £¬µÃµç×ÓÄÜÁ¦Öð½¥______¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÊµÑéÌâ

ij»¯Ñ§Ð¡×éÔÚѧϰԪËØÖÜÆÚÂÉ֪ʶºó£¬¶Ô½Ì²ÄÖÐCl2½«Fe2£«Ñõ»¯³ÉFe3£«µÄʵÑé½øÒ»²½Ë¼¿¼£¬²¢Ìá³öÎÊÌ⣺Cl2Äܽ«Fe2£«Ñõ»¯³ÉFe3£«£¬ÄÇôBr2ºÍI2ÄÜ·ñ½«Fe2£«Ñõ»¯³ÉFe3£«£¿
»·½ÚÒ»£ºÀíÂÛÍƲ⡣
²¿·ÖͬѧÈÏΪBr2ºÍI2¶¼Äܽ«Fe2£«Ñõ»¯³ÉFe3£«£¬ÒÀ¾ÝÊÇ____________________
_________________________________________________________________¡£
²¿·ÖͬѧÈÏΪBr2ºÍI2¶¼²»Äܽ«Fe2£«Ñõ»¯³ÉFe3£«¡£»¹ÓÐͬѧÈÏΪBr2Äܽ«
Fe2£«Ñõ»¯³ÉFe3£«£¬¶øI2²»ÄÜ£¬ÒÀ¾ÝÊÇͬһÖ÷×å´ÓÉϵ½Ï±Ëص¥ÖʵÄÑõ»¯ÐÔÖð½¥¼õÈõ¡£
»·½Ú¶þ£ºÉè¼ÆʵÑé½øÐÐÑéÖ¤¡£
Ïò´óÊÔ¹ÜÖмÓÈëÊÊÁ¿Ìú·Û£¬¼ÓÈë10 mLÏ¡ÑÎËᣬÕñµ´ÊԹܣ¬³ä·Ö·´Ó¦ºó£¬Ìú·ÛÓÐÊ£Ó࣬ȡÉϲãÇåÒº½øÐÐÏÂÁÐʵÑé¡£
ʵÑé1£º

ÊÔ¹Ü
²Ù×÷
ÏÖÏó
¢Ù
ÏÈÏòÊÔ¹ÜÖмÓÈë2 mLÐÂÖƵÄFeCl2ÈÜÒº£¬ÔÙÏòÊÔ¹ÜÖеμÓÉÙÁ¿ºì×ØÉ«µÄäåË®£¬Õñµ´ÊÔ¹Ü
ÈÜҺΪ
»ÆÉ«
¢Ú
ÏÈÏòÊÔ¹ÜÖмÓÈë2 mLÐÂÖƵÄFeCl2ÈÜÒº£¬ÔÙÏòÊÔ¹ÜÖеμÓÉÙÁ¿×Ø»ÆÉ«µÄµâË®£¬Õñµ´ÊÔ¹Ü
ÈÜҺΪ
»ÆÉ«
 
»·½ÚÈý£ºÊµÑéÏÖÏóµÄ·ÖÎöÓë½âÊÍ¡£
£¨1£©¼×ͬѧÈÏΪÊԹܢÙÖеÄÏÖÏó˵Ã÷äåË®Äܽ«Fe2£«Ñõ»¯£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪ_________________________________________________________________¡£
ÒÒͬѧÈÏΪӦ¸Ã²¹×öʵÑ飬²ÅÄܵóö¼×ͬѧµÄ½áÂÛ¡£ÇëÄã°ïÖúÒÒͬѧÍê³ÉʵÑ飺
ʵÑé2£º
²Ù×÷
ÏÖÏó
 
 
 
£¨2£©¸ÃС×éͬѧ¶ÔÊԹܢÚÖÐËùµÃµÄÈÜÒº³Ê»ÆÉ«µÄÔ­ÒòÕ¹¿ªÌÖÂÛ£¬²¢Ìá³öÁËÁ½ÖÖ¼ÙÉ裺
¼ÙÉè1£ºµâË®ÓëFeCl2ÈÜÒº²»·´Ó¦£¬»ÆÉ«ÊǵâˮϡÊͺóµÄÑÕÉ«¡£
¼ÙÉè2.________________¡£
ʵÑé3£º½øÐÐʵÑéÒÔÅжϼÙÉèÊÇ·ñ³ÉÁ¢¡£
²Ù×÷
ÏÖÏó
ÏòÊԹܢÚËùµÃµÄÈÜÒºÖмÌÐø¼ÓÈë0.5 mL CCl4£¬³ä·ÖÕñµ´£¬¾²ÖÃÒ»¶Îʱ¼ä¡£È¡³öÉϲãÈÜÒº£¬µÎ¼ÓKSCNÈÜÒº
¾²Öúó£¬ÉϲãÈÜÒº¼¸ºõÎÞÉ«£¬Ï²ãÈÜҺΪ×ÏÉ«£»È¡ÉϲãÈÜÒºµÎ¼ÓKSCNÈÜÒººó£¬ÈÜÒº³ÊdzºìÉ«
 
±ûͬѧÈÏΪʵÑé3µÄÏÖÏó¿ÉÒÔ˵Ã÷¼ÙÉè2³ÉÁ¢£¬¶¡Í¬Ñ§ÈÏΪ²»ÑϽ÷£¬ÓÚÊÇÉè¼ÆÁËʵÑé4¼ÌÐø̽¾¿¡£
ʵÑé4£º
²Ù×÷
ÏÖÏó
ÏòÁíÒ»Ö§ÊÔ¹ÜÖмÓÈë2 mLÐÂÖƵÄFeCl2ÈÜÒº£¬µÎ¼Ó0.5 mLµâË®ºó£¬ÔÙ¼ÓÈë0.5 mLÒÒËáÒÒõ¥£¬³ä·ÖÕñµ´£¬¾²ÖÃÒ»¶Îʱ¼ä¡£È¡³öϲãÈÜÒº£¬µÎ¼ÓKSCNÈÜÒº
¾²Öúó£¬ÉϲãÈÜҺΪ×ÏÉ«£¬Ï²ãÈÜÒº¼¸ºõÎÞÉ«£»È¡Ï²ãÈÜÒº£¬ÏòÆäÖеμÓKSCNÈÜÒººó£¬ÈÜҺûÓгÊdzºìÉ«
 
ÄãÈÏΪʵÑé4ÖмÓÈëÒÒËáÒÒõ¥µÄÖ÷ҪĿµÄÊÇ___________________________¡£
¶¡Í¬Ñ§¸ù¾ÝʵÑé4µÄÏÖÏóµÃ³ö½áÂÛ£ºÔÚ±¾´ÎʵÑéÌõ¼þÏ£¬µâË®ÓëFeCl2ÈÜÒº·´Ó¦µÄ³Ì¶ÈºÜС¡£
£¨3£©Cl2¡¢Br2¡¢I2Ñõ»¯Fe2£«µÄÄÜÁ¦Öð½¥¼õÈõ£¬´ÓÔ­×ӽṹ½Ç¶È½âÊÍÔ­Òò£º______________________________________________________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÊµÑéÌâ

µâÊÇÈËÌå±ØÐèµÄ΢Á¿ÔªËØÖ®Ò»£¬ÓС°ÖÇÁ¦ÔªËØ¡±Ö®³Æ¡£Ê³ÓüӵâʳÑοÉÔ¤·Àµâȱ·¦²¡¡£²éÔÄ×ÊÁÏÖª£º¢ÙÔÚËáÐÔÌõ¼þÏ£¬I¡ªÄܱ»NO3-Àë×ÓÑõ»¯³ÉIO3-Àë×Ó£¬±»H2O2»òO2µÈÑõ»¯³ÉI2£»¢ÚIO3-Àë×ÓÄܱ»HSO3-Àë×Ó»¹Ô­³ÉI2¡£
ÇëÄãÀûÓÃʵÑéÊÒ³£ÓÃÒÇÆ÷ºÍÏÂÁÐÏÞÑ¡ÊÔ¼Á£¬ÒÀ´ÎÑо¿Ä³Ê³ÑÎÑùÆ·ÖÐËù¼ÓµâµÄ´æÔÚÐÎʽÊÇI2¡¢I-¡¢IO3-ÖеÄÄÄÒ»ÖÖ¡£
ÏÞÑ¡ÊÔ¼ÁÈçÏ£º1.0 mol?L-1HNO3ÈÜÒº¡¢1.0 mol?L-1 H2SO4ÈÜÒº¡¢1.0 mol?L-1NaHSO3ÈÜÒº¡¢3% H2O2ÈÜÒº¡¢1%µí·ÛÈÜÒº¡¢ÕôÁóË®
¢ñ£®Ìá³ö¼ÙÉè
¼ÙÉè1£º¸ÃʳÑÎÑùÆ·Öк¬I2 £»      
¼ÙÉè2£º¸ÃʳÑÎÑùÆ·Öк¬I¡ª
¼ÙÉè3£º                         ¡£
¢ò£®Éè¼Æ·½°¸ÓëʵÑé̽¾¿
½«ÊÊÁ¿Ê³ÑÎÑùÆ·ÈÜÓÚÕôÁóË®ÖƳÉÈÜÒº£¬Çë°´ÒªÇóÌîдϱí

ʵÑé²½Öè
Ô¤ÆÚÏÖÏóÓë½áÂÛ
²½Öè1£ºÈ¡ÉÙÁ¿ÉÏÊöÈÜҺעÈëÊÔ¹ÜÖУ¬µÎÈ뼸µÎµí·ÛÈÜÒºÕñµ´¡£
ÈôÈÜÒºÏÔ          £¬Ôò¼ÙÉè1³ÉÁ¢£»·ñÔò£¬¼ÙÉè1²»³ÉÁ¢£¬ÔÙ½øÐв½Öè2
²½Öè2£º                               
                                      
                                     
ÈôÈÜÒºÏÔÀ¶É«£¬Ôò¼ÙÉè2³ÉÁ¢£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪ               £»·ñÔò£¬¼ÙÉè2²»³ÉÁ¢£¬ÔÙ½øÐв½Öè3
²½Öè3£º                              
                                      
                                      
                                     
                                     
 
¢ó£®ÎÊÌâÓë˼¿¼
´¿µÄKIO3»òKI¶¼¿ÉÒÔ×÷ΪʳÓüӵâÑÎÖеâµÄÀ´Ô´¡£´Ó»¯Ñ§½Ç¶ÈÀ´¿´£¬ÊÔÑéÖÐÌí¼Ó            £¨Ìî¡°KIO3¡±»ò¡°KI¡±£©¸üºÃ£»ÀíÓÉÊÇ                             ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºµ¥Ñ¡Ìâ

þ¡¢ÂÁ»ìºÏ·ÛÄ©10.2g£¬½«ËüÈÜÓÚ500mL4mol/LµÄÑÎËáÀÈôҪʹ³ÁµíÖÊÁ¿´ïµ½×î´óÖµ£¬ÔòÐè¼ÓÈë2mol¡¤l£­1µÄÇâÑõ»¯ÄÆÈÜÒºµÄÌå»ýΪ

A£®1000mLB£®500mLC£®100mLD£®1500mL

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸