£¨14·Ö£©Ä³»¯Ñ§Ñо¿ÐÔѧϰС×éͨ¹ý²éÔÄ×ÊÁÏ£¬Éè¼ÆÁËÈçÏÂͼËùʾµÄ·½·¨À´ÖƱ¸KMnO4¡£
ÒÑÖª£º¢Ù3MnO2+KClO3+6KOH3K2MnO4+KCl+3H2O
¢Ú¼¸ÖÖÎïÖʵÄÈܽâ¶È£º

ζÈ
Èܽâ¶È/g
K2CO3
KOH
KMnO4
20¡æ
111
112
6.38
60¡æ
127
154
22.1
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
¢ÅµÚÒ»²½¹ÌÌåÈÛÈÚʱ³ýÈý½Å¼Ü¡¢ÄàÈý½Ç¡¢Ï¸Ìú°ô¡¢ÛáÛöǯÍ⣬»¹ÐèÏÂÁÐÒÇÆ÷ÖеĠ ¡ø  ¡£
A£®Õô·¢ÃóB£®ÉÕ±­C£®¾Æ¾«µÆD£®ÌúÛáÛö
¢Æ·´Ó¦¢ñÖÐÑõ»¯²úÎïÓ뻹ԭ²úÎïµÄÎïÖʵÄÁ¿Ö®±ÈΪ  ¡ø  £»¸Ã·´Ó¦ÖпÉÒÔÑ­»·ÀûÓõIJúÎïÊÇ  ¡ø  ¡£
¢Ç·´Ó¦¢òµÄÌõ¼þΪµç½â£¬Ð´³ö·´Ó¦¢òµÄ»¯Ñ§·½³Ìʽ  ¡ø  ¡£
¢È;¾¶Ò»¡¢¶þÖеIJÙ×÷a¡¢bÏàͬ£¬¾ù°üÀ¨  ¡ø  ¡¢¹ýÂ˵È3²½¡£
¢Éͨ¹ýÓòÝËáµÎ¶¨KMnO4ÈÜÒºµÄ·½·¨¿É²â¶¨KMnO4´ÖÆ·µÄ´¿¶È£¨ÖÊÁ¿·ÖÊý£©¡£
¢ÙʵÑéʱҪ½«²ÝËᾧÌ壨H2C2O4¡¤2H2O£©Åä³É±ê×¼ÈÜÒº£¬ÔòÅäÖÆ100mL1.5mol¡¤L¡ª1µÄ²ÝËáÈÜÒº£¬ÐèÒª³ÆÈ¡²ÝËᾧÌåµÄÖÊÁ¿Îª  ¡ø  ¡£
¢Ú¸ÃʵÑéÖгý²ÝËᾧÌåµÄÖÊÁ¿Í⣬»¹ÐèÒª²É¼¯µÄÊý¾ÝÓР ¡ø  ¡£

¢ÅCD
¢Æ2:1   MnO2
¢Ç2K2MnO4£«2H2O2KOH£«2KMnO4£«H2¡ü
¢ÈÕô·¢Å¨Ëõ¡¢½µÎ½ᾧ
¢É¢Ù18.9g ¢ÚµÎ¶¨Ê±ÏûºÄ²ÝËáÈÜÒºµÄÌå»ý¡¢KMnO4´ÖÆ·µÄÖÊÁ¿

½âÎö

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ij»¯Ñ§Ñо¿ÐÔѧϰС×é¶ÔijÎÞÉ«Ë®ÑùµÄ³É·Ö½øÐмìÑ飬ÒÑÖª¸ÃË®ÑùÖÐÖ»¿ÉÄܺ¬ÓÐK+¡¢Mg2+¡¢Fe3+¡¢Cu2+¡¢Al3+¡¢Ag+¡¢Ca2+¡¢CO
 
2-
3
¡¢SO
 
2-
4
¡¢Cl-ÖеÄÈô¸ÉÖÖÀë×Ó£®¸ÃС×éͬѧȡ100mLË®Ñù½øÐÐʵÑ飬ÏòË®ÑùÖÐÏȵμÓÏõËá±µÈÜÒº£¬ÔٵμÓ1mol/LµÄÏõËᣬʵÑé¹ý³ÌÖгÁµíÖÊÁ¿µÄ±ä»¯Çé¿öÈçÓÒͼËùʾ£º
£¨1£©Ë®ÑùÖÐÒ»¶¨º¬ÓеÄÒõÀë×ÓÊÇ
SO42-£¬CO32-
SO42-£¬CO32-
£¬ÆäÎïÖʵÄÁ¿Å¨¶ÈÖ®±ÈΪ
1£º2
1£º2
£®
£¨2£©Ð´³öBC¶ÎËù±íʾ·´Ó¦µÄÀë×Ó·½³Ìʽ£º
BaCO3+2H+=Ba2++CO2¡ü+H2O
BaCO3+2H+=Ba2++CO2¡ü+H2O
£®
£¨3£©ÓÉBµ½Cµã±ä»¯¹ý³ÌÖÐÏûºÄÏõËáµÄÌå»ýΪ
40ml
40ml
£®
£¨4£©ÊÔ¸ù¾ÝʵÑé½á¹ûÍƲâK+ÊÇ·ñ´æÔÚ£¿
ÊÇ
ÊÇ
£¨Ìî¡°ÊÇ¡±»ò¡°·ñ¡±£©£»Èô´æÔÚ£¬K+µÄÎïÖʵÄÁ¿Å¨¶Èc£¨K+£©µÄ·¶Î§ÊÇ
¡Ý0.6mol/L
¡Ý0.6mol/L
£®£¨ÈôK+²»´æÔÚ£¬Ôò²»±Ø»Ø´ð¸ÃÎÊ£©£®
£¨5£©Éè¼Æ¼òµ¥ÊµÑéÑéÖ¤Ô­Ë®ÑùÖпÉÄÜ´æÔÚµÄÀë×Ó£º
È¡ÉÙÁ¿Ë®ÑùÓëÊÔ¹ÜÖУ¬ÏòÊÔ¹ÜÖмÓÈë¹ýÁ¿ÉýÎÂÏõËá±µÈÜÒººÍÏ¡ÏõËᣬ´ý³ÁµíÍêÈ«ºÍÎÞÆøÌåÉú³Éºó£¬ÏòÉϲãÇåÒºÖеμÓÊÊÁ¿µÄÏõËáÒøÈÜÒº£¬ÈôÉú³É°×É«³Áµí£¬ÔòÔ­Ë®ÑùÖÐ º¬ÓÐCl-£¬ÈôÎÞ°×É«³ÁµíÉú³É£¬Ö¤Ã÷ÎÞCl-´æÔÚ
È¡ÉÙÁ¿Ë®ÑùÓëÊÔ¹ÜÖУ¬ÏòÊÔ¹ÜÖмÓÈë¹ýÁ¿ÉýÎÂÏõËá±µÈÜÒººÍÏ¡ÏõËᣬ´ý³ÁµíÍêÈ«ºÍÎÞÆøÌåÉú³Éºó£¬ÏòÉϲãÇåÒºÖеμÓÊÊÁ¿µÄÏõËáÒøÈÜÒº£¬ÈôÉú³É°×É«³Áµí£¬ÔòÔ­Ë®ÑùÖÐ º¬ÓÐCl-£¬ÈôÎÞ°×É«³ÁµíÉú³É£¬Ö¤Ã÷ÎÞCl-´æÔÚ
£®£¨Ð´³öʵÑé²½Öè¡¢ÏÖÏóºÍ½áÂÛ£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨2010?ÄÏͨģÄ⣩ij»¯Ñ§Ñо¿ÐÔѧϰС×éͨ¹ý²éÔÄ×ÊÁÏ£¬Éè¼ÆÁËÈçͼËùʾµÄ·½·¨ÒÔº¬Äø·Ï´ß»¯¼ÁΪԭÁÏÀ´ÖƱ¸NiSO4?7H2O£®ÒÑ֪ij»¯¹¤³§µÄº¬Äø´ß»¯¼ÁÖ÷Òªº¬ÓÐNi£¬»¹º¬ÓÐAl£¨31%£©¡¢Fe£¨1.3%£©µÄµ¥Öʼ°Ñõ»¯ÎÆäËû²»ÈÜÔÓÖÊ£¨3.3%£©£®

²¿·ÖÑôÀë×ÓÒÔÇâÑõ»¯ÎïÐÎʽÍêÈ«³ÁµíʱµÄpHÈçÏ£º
³ÁµíÎï Al£¨OH£©3 Fe£¨OH£©3 Fe£¨OH£©2 Ni£¨OH£©2
pH 5.2 3.2 9.7 9.2
»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©²Ù×÷a¡¢cÖÐÐèʹÓõÄÒÇÆ÷³ýÌú¼Ų̈£¨´øÌúȦ£©¡¢¾Æ¾«µÆ¡¢ÉÕ±­¡¢²£Á§°ôÍ⻹ÐèÒªµÄÖ÷ÒªÒÇÆ÷Ϊ
©¶·¡¢Õô·¢Ãó
©¶·¡¢Õô·¢Ãó
£®
£¨2£©¡°¼î½þ¡±¹ý³ÌÖз¢ÉúµÄÀë×Ó·½³ÌʽÊÇ
2Al+2OH-+2H2O¨T2AlO2-+3H2¡ü¡¢Al2O3+2OH-¨T2AlO2-+3H2O
2Al+2OH-+2H2O¨T2AlO2-+3H2¡ü¡¢Al2O3+2OH-¨T2AlO2-+3H2O
£®
£¨3£©¡°Ëá½þ¡±Ê±Ëù¼ÓÈëµÄËáÊÇ
H2SO4
H2SO4
£¨Ìѧʽ£©£®Ëá½þºó£¬¾­²Ù×÷a·ÖÀë³ö¹ÌÌå¢Ùºó£¬ÈÜÒºÖпÉÄܺ¬ÓеĽðÊôÀë×ÓÊÇ
Ni2+¡¢Fe2+
Ni2+¡¢Fe2+
£®
£¨4£©²Ù×÷bΪµ÷½ÚÈÜÒºµÄpH£¬ÄãÈÏΪpHµÄ×î¼Ñµ÷¿Ø·¶Î§ÊÇ
3.2-9.2
3.2-9.2
£®
£¨5£©¡°µ÷pHΪ2¡«3¡±µÄÄ¿µÄÊÇ
·ÀÖ¹ÔÚŨËõ½á¾§¹ý³ÌÖÐNi2+Ë®½â
·ÀÖ¹ÔÚŨËõ½á¾§¹ý³ÌÖÐNi2+Ë®½â
£®
£¨6£©²úÆ·¾§ÌåÖÐÓÐʱ»á»ìÓÐÉÙÁ¿ÂÌ·¯£¨FeSO4?7H2O£©£¬ÆäÔ­Òò¿ÉÄÜÊÇ
H2O2µÄÓÃÁ¿²»×㣨»òH2O2ʧЧ£©¡¢±£ÎÂʱ¼ä²»×ãµ¼ÖÂFe2+δ±»ÍêÈ«Ñõ»¯Ôì³ÉµÄ
H2O2µÄÓÃÁ¿²»×㣨»òH2O2ʧЧ£©¡¢±£ÎÂʱ¼ä²»×ãµ¼ÖÂFe2+δ±»ÍêÈ«Ñõ»¯Ôì³ÉµÄ
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â

£¨2011?Ì©ÖÝһģ£©Ä³»¯Ñ§Ñо¿ÐÔѧϰС×éΪ̽¾¿Ä³Æ·ÅÆ»¨ÉúÓÍÖв»±¥ºÍÖ¬·¾ËáµÄº¬Á¿£¬½øÐÐÁËÈçÏÂʵÑ飺
²½Öè¢ñ£º³ÆÈ¡0.4g»¨ÉúÓÍÑùÆ·£¬ÖÃÓÚÁ½¸ö¸ÉÔïµÄµâÆ¿£¨Èçͼ£©ÄÚ£¬¼ÓÈë10mLËÄÂÈ»¯Ì¼£¬ÇáÇáÒ¡¶¯Ê¹ÓÍÈ«²¿Èܽ⣮ÏòµâÆ¿ÖмÓÈë25.00mLº¬0.01mol IBrµÄÎÞË®ÒÒËáÈÜÒº£¬¸ÇºÃÆ¿Èû£¬ÔÚ²£Á§ÈûÓëÆ¿¿ÚÖ®¼äµÎ¼ÓÊýµÎ10%µâ»¯¼ØÈÜÒº·â±Õ·ì϶£¬ÒÔÃâIBrµÄ»Ó·¢Ëðʧ£®
²½Öè¢ò£ºÔÚ°µ´¦·ÅÖÃ30min£¬²¢²»Ê±ÇáÇáÒ¡¶¯£®30minºó£¬Ð¡Ðĵشò¿ª²£Á§Èû£¬ÓÃÐÂÅäÖƵÄ10%µâ»¯¼Ø10mLºÍÕôÁóË®50mL°Ñ²£Á§ÈûºÍÆ¿¾±ÉϵÄÒºÌå³åÏ´ÈëÆ¿ÄÚ£®
²½Öè¢ó£º¼ÓÈëָʾ¼Á£¬ÓÃ0.1mol?L-1Áò´úÁòËáÄÆÈÜÒºµÎ¶¨£¬ÓÃÁ¦Õñµ´µâÆ¿£¬Ö±ÖÁÖյ㣮
²â¶¨¹ý³ÌÖз¢ÉúµÄÏà¹Ø·´Ó¦ÈçÏ£º
¢Ù
¢ÚIBr+KI=I2+KBr
¢ÛI2+2S2O32-=2I-+S4O62-
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÒÑ֪±ËØ»¥»¯ÎïIBrµÄÐÔÖÊÓë±Ëص¥ÖÊÀàËÆ£¬ÊµÑéÖÐ׼ȷÁ¿È¡IBrÈÜҺӦѡÓõÄÒÇÆ÷ÊÇ
ËáʽµÎ¶¨¹Ü£¨»òÒÆÒº¹Ü £©
ËáʽµÎ¶¨¹Ü£¨»òÒÆÒº¹Ü £©
£¬µâÆ¿²»¸ÉÔï»á·¢Éú·´Ó¦µÄ»¯Ñ§·½³Ìʽ
IBr+H2O=HIO+HBr
IBr+H2O=HIO+HBr
£®
£¨2£©²½Öè¢òÖеâÆ¿ÔÚ°µ´¦·ÅÖÃ30min£¬²¢²»Ê±ÇáÇáÒ¡¶¯µÄÔ­ÒòÊÇ
µâÆ¿ÖÃÓÚ°µ´¦¿É¼õÉÙä廯Çâ»Ó·¢£¬²»¶Ï½Á¶¯¿ÉÒÔÈÃÎïÖʼä³ä·Ö·´Ó¦
µâÆ¿ÖÃÓÚ°µ´¦¿É¼õÉÙä廯Çâ»Ó·¢£¬²»¶Ï½Á¶¯¿ÉÒÔÈÃÎïÖʼä³ä·Ö·´Ó¦
£®
£¨3£©²½Öè¢óÖÐËù¼Óָʾ¼ÁΪ
µí·ÛÈÜÒº
µí·ÛÈÜÒº
£¬µÎ¶¨ÖÕµãµÄÏÖÏó
ÈÜÒºÓÉÀ¶É«Ç¡ºÃ±äΪÎÞÉ«ÇÒ30ÃëÄÚ²»±ä»¯
ÈÜÒºÓÉÀ¶É«Ç¡ºÃ±äΪÎÞÉ«ÇÒ30ÃëÄÚ²»±ä»¯
£®
£¨4£©·´Ó¦½áÊøºó´ÓÒºÌå»ìºÏÎïÖлØÊÕËÄÂÈ»¯Ì¼£¬ËùÐè²Ù×÷ÓÐ
·ÖÒº¡¢ÕôÁó
·ÖÒº¡¢ÕôÁó
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

±µ£¨Ba£©ºÍïÈ£¨Sr£©¼°Æ仯ºÏÎïÔÚ¹¤ÒµÉÏÓÐ׏㷺µÄÓ¦Óã¬ËüÃÇÔڵؿÇÖг£ÒÔÁòËáÑεÄÐÎʽ´æÔÚ£¬BaSO4ºÍSrSO4¶¼ÊÇÄÑÈÜÐÔÑΣ®¹¤ÒµÉÏÌáÈ¡±µºÍïÈʱÊ×ÏȽ«BaSO4ºÍSrSO4ת»¯³ÉÄÑÈÜÈõËáÑΣ®
ÒÑÖª£ºSrSO4£¨s£©?Sr2+£¨aq£©+SO2-4£¨aq£©  KSP=2.5¡Á10-7
SrSO3£¨s£©?Sr2+£¨aq£©+CO2-4£¨aq£©   KSP=2.5¡Á10-9
£¨1£©½«SrSO4ת»¯³ÉSrCO3µÄÀë×Ó·½³ÌʽΪ
 
£¬¸Ã·´Ó¦µÄƽºâ³£Êý±í´ïʽΪ
 
£»¸Ã·´Ó¦ÄÜ·¢ÉúµÄÔ­ÒòÊÇ
 
£®£¨ÓóÁµíÈܽâÔÙƽºâµÄÓйØÀíÂÛ½âÊÍ£©
£¨2£©¶ÔÓÚÉÏÊö·´Ó¦£¬ÊµÑéÖ¤Ã÷Ôö´óCO2-3µÄŨ¶È»ò½µµÍζȶ¼ÓÐÀûÓÚÌá¸ßSrSO4µÄת»¯ÂÊ£®ÅжÏÔÚÏÂÁÐÁ½ÖÖÇé¿öÏ£¬Æ½ºâ³£ÊýKµÄ±ä»¯Çé¿ö£¨Ìî¡°Ôö´ó¡±¡¢¡°¼õС¡±»ò¡°²»±ä¡±£©£º
¢ÙÉý¸ßζȣ¬Æ½ºâ³£ÊýK½«
 
£»
¢ÚÔö´óCO2-3µÄŨ¶È£¬Æ½ºâ³£ÊýK½«
 
£®
£¨3£©ÒÑÖª£¬SrSO4ºÍSrCO3ÔÚËáÖеÄÈܽâÐÔÓëBaSO4ºÍBaCO3ÀàËÆ£¬Éè¼ÆʵÑéÖ¤Ã÷ÉÏÊö¹ý³ÌÖÐSrSO4ÊÇ·ñÍêȫת»¯³ÉSrCO3£®ÊµÑéËùÓÃÊÔ¼ÁΪ
 
£»ÊµÑéÏÖÏó¼°ÆäÏàÓ¦½áÂÛ
 
£®
£¨4£©BaCl2ÈÜÒººÍBa£¨NO3£©2ÈÜÒºÊÇʵÑéÖмìÑéSO2-4µÄ³£ÓÃÊÔ¼Á£®Ä³»¯Ñ§Ñо¿ÐÔѧϰС×é¼ìÑéijÈÜÒºÖдæÔÚSO2-4ʱ£¬Ê×ÏȼÓÈëBa£¨NO3£©2ÈÜÒº£¬²úÉú°×É«³Áµí£¬È»ºó¼ÓÈë¹ýÁ¿Ï¡ÏõËᣬ°×É«³Áµí²»Èܽ⣬Óɴ˵óö½áÂÛ£ºÈÜÒºÖÐÒ»¶¨º¬ÓÐSO2-4£®ÄãÈÏΪ¸ÃÍÆÀíÊÇ·ñÑÏÃÜ£¿ÊÔ˵Ã÷ÀíÓÉ£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ij»¯Ñ§Ñо¿ÐÔѧϰС×éΪÁ˽â´Ó¹¤ÒµäåÖÐÌá´¿äåµÄ·½·¨£¬²éÔÄÁËÓйØ×ÊÁÏ£¬Br2µÄ·ÐµãΪ59¡æ£®Î¢ÈÜÓÚË®£¬Óж¾ÐÔºÍÇ¿¸¯Ê´ÐÔ£®ËûÃDzιÛÉú²ú¹ý³Ìºó£¬»æÖÆÁËÈçͼװÖüòͼ£®
¾«Ó¢¼Ò½ÌÍø
ÇëÄã²ÎÓë·ÖÎöÌÖÂÛ£º
¢ÙͼÖÐÒÇÆ÷BµÄÃû³Æ£º
 
£¬ÀäÈ´Ë®´Ó
 
½ø£¬
 
³ö£»
¢ÚʵÑé×°ÖÃÆøÃÜÐÔÁ¼ºÃ£¬Òª´ïµ½Ìá´¿äåµÄÄ¿µÄ£¬²Ù×÷ÖÐÈçºÎ¿ØÖƹؼüÌõ¼þ£º
 
£®
¢ÛCÖÐÒºÌå²úÉúÑÕɫΪ
 
£»
¢ÜÉÕ±­AµÄ×÷ÓÃ
 
£¬DÖÐÒºÌåµÄ×÷ÓÃÊÇ
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸