ÂÈ»¯Í­ÊÇÒ»Öֹ㷺ÓÃÓÚÉú²úÑÕÁÏ¡¢ÓÃľ²Ä·À¸¯¼ÁµÄ»¯¹¤²úÆ·¡£Ä³Ñо¿ÐÔѧϰС×éÓôÖÍ­£¨º¬ÔÓÖÊFe£©°´ÏÂÊöÁ÷³ÌÖƱ¸ÂÈ»¯Í­¾§Ìå¡£

£¨1£©¹ÌÌåAÓÃÏ¡ÑÎËáÈܽâ¶ø²»ÓÃË®ÈܽâµÄÔ­ÒòÊÇ________¡£
£¨2£©¼ÓÊÔ¼ÁXÓÃÓÚµ÷½ÚpHÒÔ³ýÈ¥ÔÓÖÊ£¬X¿ÉÑ¡ÓÃÏÂÁÐÊÔ¼ÁÖеÄ____£¨ÌîÐòºÅ£©¡£

A£®NaOHB£®NH3.H2OC£®CuOD£®Cu(OH)2E. CuSO4
£¨3£©ÂËÒºB¾­Ò»ÏµÁвÙ×÷¿ÉµÃÂÈ»¯Í­¾§Ì壬²Ù×÷µÄ³ÌÐòÒÀ´ÎΪ_________ ¡¢_________¡¢ ¹ýÂË¡¢×ÔÈ»¸ÉÔï¡£
£¨4£©ÊµÑéÊÒ²ÉÓÃÈçÏÂͼËùʾװÖ㬿Éʹ´ÖÍ­ÓëCl2·´Ó¦×ª»¯Îª¹ÌÌåA£¨²¿·Ö¼ÓÈÈÒÇÆ÷ºÍ¼Ð³Ö×°ÖÃÒÑÂÔÈ¥£©¡£

¢Ù¸Ã×°ÖÃÖÐÒÇÆ÷aµÄÃû³ÆÊÇ____£¬ÆäÖз¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ____________¡£
¢ÚÓÐͬѧÈÏΪӦÔÚŨÁòËáÏ´ÆøÆ¿Ç°Ôö¼ÓÎüÊÕHCIµÄ×°Öã¬ÄãÈÏΪÊÇ·ñ±ØÒª£¿____________£¨Ìî¡°ÊÇ¡±»ò¡°·ñ¡±£©¡£
¢Û¸Ã×°ÖôæÔÚÒ»¶¨µÄ°²È«Òþ»¼£¬Ïû³ý¸Ã°²È«Òþ»¼µÄ´ëÊ©ÊÇ_______________________¡£

£¨1£©ÒÖÖÆCu2£«¡¢Fe3£«µÈÀë×Ó·¢ÉúË®½â·´Ó¦    £¨2£©c d  £¨3£©Õô·¢Å¨Ëõ¡¢ÀäÈ´½á¾§
£¨4£©¢Ù Ô²µ×ÉÕÆ¿ MnO2 + 4H++2Cl£­ Mn2++ Cl2¡ü + 2H2O  
¢Ú ·ñ   ¢Û ÔÚ×°ÖÃCºÍDÖ®¼äÁ¬½ÓÒ»¸ö·Àµ¹Îü×°ÖÃ

½âÎöÊÔÌâ·ÖÎö£º£¨1£©´ÖÍ­Öк¬ÓÐCu¡¢Fe£¬ÓëÂÈÆø·´Ó¦Ê±´ËʱCuCl2¡¢FeCl3£¬ÓÉÓÚ¶þÕ߶¼ÊÇÇ¿ËáÈõ¼îÑΣ¬ÈÝÒ×·¢ÉúË®½â·´Ó¦²úÉúCu(OH)2¡¢Fe(OH)3¡£ÎªÁËÒÖÖÆCu2£«¡¢Fe3£«µÈÀë×Ó·¢ÉúË®½â·´Ó¦Í¨³£ÓÃÏ¡ÑÎËáÀ´Èܽ⡣
£¨2£©¼ÓÊÔ¼ÁXÓÃÓÚµ÷½ÚpHÒÔ³ýÈ¥ÔÓÖÊ£¬£¬µ«²»ÄÜÒýÈëеÄÔÓÖÊÀë×Ó¡£¸ù¾ÝÌâÄ¿¸ø¶¨µÄÎïÖÊ£¬¿ÉÑ¡ÓÃCuO¡¢Cu(OH)2¡£Ñ¡ÏîΪCD
£¨3£©½«³ýÈ¥Fe(OH)3³ÁµíµÄº¬ÓÐCuCl2ÂËÒº¾­Ò»ÏµÁвÙ×÷¿ÉµÃÂÈ»¯Í­¾§Ìå¡£CuCl2ÊÇÇ¿ËáÈõ¼îÑΣ¬Ë®½â²úÉúCu(OH)2ºÍHCl¡£ÑÎËáÓлӷ¢ÐÔ£¬ÈÝÒ×»Ó·¢Òݳö¡£µÃµ½µÄ¹ÌÌåÊÇCu(OH)2¡£ÓÉÓÚËüµÄÈܽâ¶ÈÊÜζȵÄÓ°Ïì±ä»¯½Ï´ó£¬ËùÒԿɲÉÓõIJÙ×÷µÄ³ÌÐòÒÀ´ÎΪÕô·¢Å¨Ëõ¡¢ÀäÈ´½á¾§¡¢¹ýÂË¡¢×ÔÈ»¸ÉÔï¡££¨4£©ÊµÑéÊÒÖÆÈ¡Cl2ÊÇÔÚÕôÁóÉÕÆ¿ÖÐÓÃŨÑÎËáÓë¶þÑõ»¯Ã̹²Èȵķ½·¨ÖÆÈ¡µÄ¡£·´Ó¦µÄÂËÔü·½³ÌʽΪ£ºMnO2 + 4H++2Cl£­ Mn2++ Cl2¡ü + 2H2O¡£¢ÚÓÐͬѧÈÏΪӦÔÚŨÁòËáÏ´ÆøÆ¿Ç°Ôö¼ÓÎüÊÕHCIµÄ×°Öã¬ÄãÈÏΪûÓбØÒª¡£ÒòΪCuÓëHCl²»»á·¢Éú·´Ó¦£¬Ö»ÓÐCl2·¢Éú·´Ó¦¡£¢ÛÓÉÓÚCl2ÈÝÒ×ÓëNaOH·¢Éú·´Ó¦£¬µ¼Öµ¼Æø¹ÜÖеÄѹǿ¼õС£¬ÕâʱÉÕ±­ÖеÄÈÜÒºÈÝÒ×µ¹ÎüÖÁ¼ÓÈÈ×°Ö㬹ʸÃ×°ÖôæÔÚÒ»¶¨µÄ°²È«Òþ»¼£¬Ïû³ý¸Ã°²È«Òþ»¼µÄ´ëÊ©ÊÇÔÚ×°ÖÃCºÍDÖ®¼äÁ¬½ÓÒ»¸ö·Àµ¹Îü×°Öá£
¿¼µã£º¿¼²éÆøÌåµÄÖƱ¸¡¢»ìºÏÎïµÄ·ÖÀë¡¢ÑεÄË®½âµÈ֪ʶ¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

A¡¢B¡¢C¡¢D¡¢E¡¢FÁùÖÖÎïÖʵÄÑæÉ«·´Ó¦¾ùΪ»ÆÉ«¡£ A¡¢B¡¢C¡¢DÓëÑÎËá·´Ó¦¾ùÉú³ÉE£¬´ËÍâB»¹Éú³É¡ªÖÖ¿ÉȼÐÔÆøÌ壻¶øC¡¢D»¹Éú³É¡ªÖÖÎÞÉ«ÎÞζµÄÆøÌåH£¬¸ÃÆøÌåÄÜʹ³ÎÇåµÄʯ»ÒË®±ä»ë×Ç¡£DºÍA¿É·´Ó¦Éú³ÉC£¬FºÍHÒ²¿É·´Ó¦Éú³ÉCºÍÁí¡ªÖÖÎÞÉ«ÎÞζÆøÌå¡£
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Ð´³öB¡¢CµÄ»¯Ñ§Ê½£ºB£º___________________; C£º___________________;
£¨2£©Ð´³öFºÍH2O·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º__________________________________;
£¨3£©Ð´³öÏÂÁз´Ó¦µÄÀë×Ó·½³Ìʽ
¢ÙDÈÜÒº+ÑÎË᣺_____________________________________________________;
¢ÚDÈÜÒº+AÈÜÒº£º___________________________________________________;

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ij»ìºÏÎïA£¬º¬ÓÐKAl£¨SO4£©2¡¢Al2O3ºÍFe2O3£¬ÔÚÒ»¶¨Ìõ¼þÏ¿ÉʵÏÖÏÂͼËùʾµÄÎïÖÊÖ®¼äµÄ±ä»¯£º

¾Ý´Ë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Í¼ÖÐÉæ¼°·ÖÀëÈÜÒºÓë³ÁµíµÄ·½·¨ÊÇ__________________¡£
£¨2£©B¡¢C¡¢D ÈýÖÖÎïÖʵĻ¯Ñ§Ê½Îª£ºB_________     C_________     D_________
£¨3£©³ÁµíEÓëÏ¡ÁòËá·´Ó¦µÄÀë×Ó·½³ÌʽΪ________________________________________¡£
£¨4£©½«³ÁµíFÖдæÔÚµÄÁ½ÖÖ½ðÊôÔªËØ×é³ÉµÄºÏ½ðÈÜÓÚ100 mL 4mol/LHClÈÜÒºÖУ¬È»ºóÔٵμÓ1 mol/L NaOHÈÜÒº£¬³ÁµíÖÊÁ¿mËæ¼ÓÈëNaOHÈÜÒºµÄÌå»ýV±ä»¯ÈçÏÂͼËùʾ¡£

ÒÑÖªV1£½160mL¡£¸ù¾ÝÒÔÉÏÐÅÏ¢»Ø´ð£º
¢Ù_________£¨Ìî¡°ÄÜ¡±»ò¡°²»ÄÜ¡±£©¼ÆËã³öV3
¢ÚV2Ϊ_________mL£¨ÈôÄÜËã³ö¾ßÌåÊý×Ö£¬ÇëÔÚºáÏßÉÏÌîд¾ßÌåÊý×Ö£»Èô²»ÄÜÇëÔÚºáÏßÉÏÌî¡°²»ÄÜÈ·¶¨¡±£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

¹¤ÒµÉϳ£ÓÃÌúÖÊÈÝÆ÷Ê¢×°ÀäŨÁòËᡣΪÑо¿ÌúÖʲÄÁÏÓëÈÈŨÁòËáµÄ·´Ó¦£¬Ä³Ñ§Ï°Ð¡×é½øÐÐÁËÒÔÏÂ̽¾¿»î¶¯£º
£¨1£©½«ÒÑÈ¥³ý±íÃæÑõ»¯ÎïµÄÌú¶¤£¨Ì¼Ëظ֣©·ÅÈëÀäŨÁòËáÖУ¬10·ÖÖÓºóÒÆÈëÁòËáÍ­ÈÜÒºÖУ¬Æ¬¿ÌºóÈ¡³ö¹Û²ì£¬Ìú¶¤±íÃæÎÞÃ÷ÏԱ仯£¬ÆäÔ­ÒòÊÇ                                                            ¡£
£¨2£©Áí³ÆÈ¡Ìú¶¤6.0g·ÅÈë15.0mlŨÁòËáÖУ¬¼ÓÈÈ£¬³ä·ÖÓ¦ºóµÃµ½ÈÜÒºX²¢ÊÕ¼¯µ½ÆøÌåY¡£
¢Ù¼×ͬѧÈÏΪXÖгýFe3+Í⻹¿ÉÄܺ¬ÓÐFe2+£¬ÈôҪȷÈÏÆäÖеÄFe2+£¬Ó¦Ñ¡Óà     £¨Ñ¡ÌîÐòºÅ£©¡£
a£®KSCNÈÜÒººÍÂÈË®                   b£®NaOHÈÜÒº
c£®Å¨°±Ë®                            d£®ËáÐÔKMnO4ÈÜÒº
¢ÚÒÒͬѧȡ336ml£¨±ê×¼×´¿ö£©ÆøÌåYͨÈë×ãÁ¿äåË®ÖУ¬·¢Éú·´Ó¦£º
SO2+Br2+2H2O=2HBr+H2SO4È»ºó¼ÓÈë×ãÁ¿BaCl2ÈÜÒº£¬¾­Êʵ±²Ù×÷ºóµÃ¸ÉÔï¹ÌÌå2.33g¡£ÓÉ´ËÍÆÖªÆøÌåYÖÐSO2µÄÌå»ý·ÖÊýΪ           ¡£
·ÖÎöÉÏÊöʵÑéÖÐSO2Ìå»ý·ÖÊýµÄ½á¹û£¬±ûͬѧÈÏΪÆøÌåYÖл¹¿ÉÄܺ¬ÓÐH2ºÍQÆøÌ塣Ϊ´ËÉè¼ÆÁËÏÂÁÐ̽¾¿ÊµÑé×°Öã¨Í¼ÖмгÖÒÇÆ÷Ê¡ÂÔ£©¡£

£¨3£©×°ÖÃBÖÐÊÔ¼ÁµÄ×÷ÓÃÊÇ                                             ¡£
£¨4£©ÈÏΪÆøÌåYÖл¹º¬ÓÐQµÄÀíÓÉÊÇ                                         £¨Óû¯Ñ§·½³Ìʽ±íʾ£©¡£
£¨5£©ÎªÈ·ÈÏQµÄ´æÔÚ£¬ÐèÔÚ×°ÖÃÖÐÌí¼ÓMÓÚ            £¨Ñ¡ÌîÐòºÅ£©¡£
a£®A֮ǰ          b£®A-B¼ä          c£®B-C¼ä        d£®C-D¼ä
£¨6£©Èç¹ûÆøÌåYÖк¬ÓУ¬Ô¤¼ÆʵÑéÏÖÏóÓ¦ÊÇ                                  ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ÒÑÖªA¡¢B¡¢C¡¢DΪ³£¼ûµ¥ÖÊ£¬ÆäÖÐB¡¢C¡¢D³£Î³£Ñ¹ÏÂΪÆøÌ壬¼×¡¢ÒÒ¡¢±û¡¢¶¡Îª»¯ºÏÎÒÒ³£ÎÂÏÂΪҺÌ壬±ûµÄÑæÉ«·´Ó¦Îª»ÆÉ«£¬ÏÂͼΪ¸÷ÖÖÎïÖÊÖ®¼äµÄÏ໥·´Ó¦

£¨1£©Ð´³öÏÂÁÐÎïÖʵĻ¯Ñ§Ê½£º
A          £¬B          £¬D          £¬±û         ¡£
£¨2£©±ûµç×ÓʽΪ             £¬·´Ó¦¢ÚÖÐÈôÓÐ11£®2L£¨±ê×¼×´¿öÏ£©BÉú³É£¬Ôò·¢ÉúתÒƵĵç
×ÓµÄÎïÖʵÄÁ¿Îª                         ¡£
£¨3£©Ð´³ö·´Ó¦¢ÛµÄ»¯Ñ§·½³Ìʽ£º                                              

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

´Ó»ÔÍ­¿ó£¨Cu2S£©ÖÐÀûÓûð·¨Á¶Í­¿ÉÒÔÌáÈ¡Í­£¬·¢ÉúÈçÏ·´Ó¦£º

ÏÂÃæÊÇÓÉCu2SÒ±Á¶Í­¼°ÖÆÈ¡CuSO4¡¤5H2OµÄÁ÷³Ìͼ£º

¢ÅCu2SÖÐÍ­ÔªËصĻ¯ºÏ¼ÛΪ  £¬µ±ÓÐ1molCu2SÓëO2·´Ó¦Éú³É2molCuʱ£¬×ªÒƵç×ÓµÄÎïÖʵÄÁ¿ÊÇ         ¡£
¢ÆCu2O¡¢CuOÖмÓÈë×ãÁ¿Ï¡ÁòËáµÃµ½µÄÌåϵAÖп´µ½ÈÜÒº³ÊÀ¶É«£¬ÇÒÓкìÉ«ÎïÖÊÉú³É£¬Çëд³öÉú³ÉºìÉ«ÎïÖʵÄÀë×Ó·½³Ìʽ                                              ¡£
¢ÇÈôʹAÖе¥ÖÊÈܽ⣬²Ù×÷¢ñÖмÓÈëµÄÊÔ¼Á×îºÃÊÇ       ¡££¨Ñ¡Ìî×Öĸ´úºÅ£©
A£®ÊÊÁ¿µÄHNO3     B£®ÊÊÁ¿µÄNaOH     C£®ÊÊÁ¿µÄH2O2
¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽΪ                                 ¡£
¢ÈÈ¡5£®00 gµ¨·¯ÑùÆ·Öð½¥Éý¸ßζÈʹÆä·Ö½â£¬·Ö½â¹ý³ÌµÄÈÈÖØÇúÏߣ¨ÑùÆ·ÖÊÁ¿Ëæζȱ仯µÄÇúÏߣ©ÈçÏÂͼËùʾ£º

¢ÙÓÉͼÖпÉÒÔ¿´³ö£¬µ¨·¯·Ö½âµÄ×îµÍζÈÊÇ       ¡£
¢Úͨ¹ý¼ÆËãÈ·¶¨258¡æʱ·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ                                  £¬eµã¶ÔÓ¦µÄ»¯Ñ§Ê½Îª         £¨¼ÆËã¹ý³ÌÂÔÈ¥£©¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

Óú¬ÓÐA12O3¡¢SiO2ºÍÉÙÁ¿FeO·xFe2O3µÄÂÁ»ÒÖƱ¸A12(SO4)3·18H2O¡££¬¹¤ÒÕÁ÷³ÌÈçÏÂ(²¿·Ö²Ù×÷ºÍÌõ¼þÂÔ):
¢ñ.ÏòÂÁ»ÒÖмÓÈë¹ýÁ¿Ï¡H2SO4£¬¹ýÂË;
¢ò.ÏòÂËÒºÖмÓÈë¹ýÁ¿KMnO4ÈÜÒº£¬µ÷½ÚÈÜÒºµÄpHԼΪ3£»
¢ó.¼ÓÈÈ£¬²úÉú´óÁ¿×ØÉ«³Áµí£¬¾²Öã¬ÉϲãÈÜÒº³Ê×ϺìÉ«;
¢ô.¼ÓÈëMnSO4ÖÁ×ϺìÉ«Ïûʧ£¬¹ýÂË£»
¢õ.ŨËõ¡¢½á¾§¡¢·ÖÀ룬µÃµ½²úÆ·¡£
H2SO4ÈܽâA12O3µÄÀë×Ó·½³ÌʽÊÇ                                
½«KMnO4 Ñõ»¯Fe2+µÄÀë×Ó·½³Ìʽ²¹³äÍêÕû£º
MnO4-+¡õFe2++¡õ      =Mn2++¡õFe3+ +¡õ      
ÉÏʽÖÐÑõ»¯¼ÁÊÇ              ,Ñõ»¯²úÎïÊÇ              ¡£
£¨3£©ÒÑÖª£ºÉú³ÉÇâÑõ»¯Îï³ÁµíµÄpH ks5u

 
Al£¨OH£©3
Fe£¨OH£©2
Fe£¨OH£©3
¿ªÊ¼³Áµíʱ
3.4
6.3
1.5
ÍêÈ«³Áµíʱ
4.7
8.3
2.8
×¢£º½ðÊôÀë×ÓµÄÆðʼŨ¶ÈΪ0.1mol·L-1
¸ù¾Ý±íÖÐÊý¾Ý½âÊͲ½Öè¢òµÄÄ¿µÄ£º                              ¡£
¼ºÖª:Ò»¶¨Ìõ¼þÏ£¬MnO4 - ¿ÉÓëMn2+·´Ó¦Éú³ÉMnO2¡£
¢Ù Ïò ¢ó µÄ³ÁµíÖмÓÈëŨHCI²¢¼ÓÈÈ£¬ÄÜ˵Ã÷³ÁµíÖдæÔÚMnO2µÄÏÖÏóÊÇ             ¡£
¢Ú¢ô ÖмÓÈëMnSO4µÄÄ¿µÄÊÇ                      ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

Fe2O3¾ßÓй㷺µÄÓÃ;¡£¼×ͬѧÔĶÁÓйØ×ÊÁϵÃÖª£ºÔÚ¸ßÎÂÏÂìÑÉÕFeCO3¿ÉÒԵõ½Fe2O3¡£ÎªÁ˽øÒ»²½ÑéÖ¤´Ë½áÂÛ£¬Ëû×öÁËÈçÏÂʵÑ飺

ʵÑé²½Öè
ʵÑé²Ù×÷
¢ñ
È¡Ò»¶¨ÖÊÁ¿µÄFeCO3¹ÌÌåÖÃÓÚÛáÛöÖУ¬¸ßÎÂìÑÉÕÖÁÖÊÁ¿²»ÔÙ¼õÇᣬÀäÈ´ÖÁÊÒÎÂ
¢ò
È¡ÉÙÁ¿ÊµÑé²½ÖèIËùµÃ¹ÌÌå·ÅÓÚÒ»½à¾»µÄÊÔ¹ÜÖУ¬ÓÃ×ãÁ¿µÄÏ¡ÁòËáÈܽâ
¢ó
ÏòʵÑé²½Öè¢òËùµÃÈÜÒºÖеμÓKSCNÈÜÒº£¬ÈÜÒº±äºì
 
Óɴ˼×ͬѧµÃ³ö½áÂÛ£º4FeCO3£«O22Fe2O3£«4CO2
(1)д³öʵÑé²½Öè¢óÖз¢Éú·´Ó¦µÄÀë×Ó·½³Ìʽ__________________________________¡£
(2)ÒÒͬѧÌá³öÁ˲»Í¬µÄ¿´·¨£ºìÑÉÕ²úÎï¿ÉÄÜÊÇFe3O4£¬ÒòΪFe3O4Ò²¿ÉÒÔÈÜÓÚÁòËᣬÇÒËùµÃÈÜÒºÖÐÒ²º¬ÓÐFe3£«¡£ÓÚÊÇÒÒͬѧ¶Ô¼×ͬѧµÄʵÑé²½Öè¢ó½øÐÐÁ˲¹³ä¸Ä½ø£º¼ìÑéʵÑé²½Öè¢òËùµÃÈÜÒºÖÐÊÇ·ñº¬ÓÐFe2£«¡£ËûÐèҪѡÔñµÄÊÔ¼ÁÊÇ____________(ÌîÐòºÅ)¡£
a£®ÂÈË®        b£®ÂÈË®£«KSCNÈÜÒº         c£®K3[Fe(CN)6](ÌúÇ軯¼ØÈÜÒº)
(3)±ûͬѧÈÏΪ¼´Ê¹µÃµ½ÁËÒÒͬѧԤÆÚµÄʵÑéÏÖÏó£¬Ò²²»ÄÜÈ·¶¨ìÑÉÕ²úÎïµÄ³É·Ö¡£ÄãÈÏΪ±ûͬѧ³Ö´Ë¿´·¨µÄÀíÓÉÊÇ____________¡£
(4)±ûͬѧ½øÒ»²½²éÔÄ×ÊÁϵÃÖª£¬ìÑÉÕFeCO3µÄ²úÎïÖеÄÈ·º¬ÓУ«2¼ÛÌúÔªËØ¡£ÓÚÊÇËûÉè¼ÆÁËÁíÒ»ÖÖÓÉFeCO3ÖÆÈ¡Fe2O3µÄ·½·¨£ºÏÈÏòFeCO3ÖÐÒÀ´Î¼ÓÈëÊÔ¼Á£ºÏ¡ÁòËá¡¢________ (ÌîÊÔ¼ÁÃû³Æ)ºÍ°±Ë®£»ÔÙ_________(Ìî²Ù×÷Ãû³Æ)£¬×ÆÉÕ£¬¼´¿ÉµÃµ½Fe2O3¡£
(5)¹¤ÒµÉÏÓÃÑõ»¯»¹Ô­µÎ¶¨·¨²âÁâÌú¿óÖÐFeCO3µÄÖÊÁ¿·ÖÊý£¬Í¨¹ý¿ØÖÆÑùÆ·µÄÖÊÁ¿£¬Ê¹µÎ¶¨Ê±ÏûºÄKMnO4ÈÜÒºÌå»ýΪc mL£¬¶ÔÓ¦ÁâÌú¿óÖÐFeCO3µÄÖÊÁ¿·ÖÊýΪc%£¬¿ÉÒÔ¼ò»¯¼ÆË㡣ijͬѧȡº¬FeCO3 c%µÄÁâÌú¿óa g£¬ÓÃ×ãÁ¿Ï¡ÁòËáÈܽâºó£¬ÔÙÓÃ0.0 200 mol¡¤L£­1µÄËáÐÔKMnO4ÈÜÒºµÎ¶¨(KMnO4±»»¹Ô­³ÉMn2£«)£¬×îÖÕÏûºÄKMnO4ÈÜÒºc mL¡£¼Ù¶¨¿óʯÖÐÎÞÆäËû»¹Ô­ÐÔÎïÖÊ£¬ÔòËùÈ¡ÁâÌú¿óµÄÖÊÁ¿a£½__________g¡£(FeCO3µÄĦ¶ûÖÊÁ¿Îª116 g¡¤mol£­1)

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ij»¯Ñ§ÐËȤС×éÓÃÖ»º¬ÓÐÂÁ¡¢Ìú¡¢Í­µÄ¹¤Òµ·ÏÁÏÖÆÈ¡´¿¾»µÄÂÈ»¯ÂÁÈÜÒº¡¢ÂÌ·¯¾§Ìå(FeSO4¡¤7H2O)ºÍµ¨·¯¾§Ì壬ÒÔ̽Ë÷¹¤Òµ·ÏÁϵÄÔÙÀûÓá£ÆäʵÑé·½°¸ÈçÏ£º

£¨1£©Ð´³öºÏ½ðÓëÉÕ¼îÈÜÒº·´Ó¦µÄÀë×Ó·½³Ìʽ£º                                                 ¡£
£¨2£©ÓÉÂËÒºAÖÆAlCl3ÈÜÒºµÄ;¾¶ÓТٺ͢ÚÁ½ÖÖ£¬Í¾¾¶¢ÚÖÐͨÈëµÄijÆøÌ壨¹Ì̬ʱ¿ÉÓÃÓÚÈ˹¤½µÓ꣩£¬Ð´³ö¸ÃÆøÌåµÄµç×Óʽ     ¡£ÄãÈÏΪ½ÏºÏÀíµÄ;¾¶ÊÇ    £¨Ìî¢Ù»ò¢Ú£©£¬ÀíÓÉÊÇ£º                          ¡£
£¨3£©ÂËÒºEÈô·ÅÖÃÔÚ¿ÕÆøÖÐÒ»¶Îʱ¼äºó£¬ÈÜÒºÖеÄÑôÀë×Ó³ýÁ˺ÍÍ⣬»¹¿ÉÄÜ´æÔÚ           £¨ÓÃÀë×Ó·ûºÅ±íʾ£©£¬¼ì²â¸ÃÀë×ӵķ½·¨ÊÇ                               ¡£
£¨4£©ÓÃÂËÔüFͨ¹ýÁ½ÖÖ;¾¶ÖÆÈ¡µ¨·¯£¬Óë;¾¶¢ÛÏà±È£¬Í¾¾¶¢ÜÃ÷ÏÔ¾ßÓеÄÁ½¸öÓŵãÊÇ£º                          ¡¢                               ¡£
£¨5£©Í¾¾¶¢Ü·¢ÉúµÄ·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º                                                       ¡£
£¨6£©ÊµÑéÊÒ´ÓCuSO4ÈÜÒºÖÆÈ¡µ¨·¯£¬²Ù×÷²½ÖèÓÐÕô·¢Å¨Ëõ¡¢ÀäÈ´½á¾§¡¢          ¡¢×ÔÈ»¸ÉÔï¡£

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸