9£®Ä³Í­¿óʯº¬Ñõ»¯Í­¡¢Ñõ»¯ÑÇÍ­¡¢ÈýÑõ»¯¶þÌúºÍ´óÁ¿Âöʯ£¨SiO2£©£¬ÏÖ²ÉÓÃËá½þ·¨´Ó¿óʯÖÐÌáÈ¡Í­£¬ÆäÁ÷³ÌͼÈçÏ£º

ÒÑÖª£º¢Ùµ±¿óʯÖÐÈýÑõ»¯¶þÌúº¬Á¿Ì«µÍʱ£¬¿ÉÓÃÁòËáºÍÁòËáÌúµÄ»ìºÏÒº½þ³öÍ­£»¢Ú·´ÝÍÈ¡ºóµÄË®²ãÊÇÁòËáÍ­ÈÜÒº£¬Cu2+Ũ¶ÈԼΪ50g/L£®»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©¿óʯÓÃÏ¡ÁòËá½þ³ö¹ý³ÌÖÐÑõ»¯ÑÇÍ­·¢ÉúµÄ·´Ó¦Îª£ºCu2O+2H+¨TCu2++Cu+H2O£»Çëд³ö¸Ã¹ý³ÌÖз¢ÉúµÄÁíÒ»¸öÑõ»¯»¹Ô­·´Ó¦µÄÀë×Ó·½³Ìʽ£ºCu+2Fe3+¨T2Fe2++Cu2+£®
£¨2£©Ð´³öÓöèÐԵ缫µç½âË®²ãµÄµç½â×Ü·´Ó¦·½³Ìʽ£º2CuSO4+2H2O$\frac{\underline{\;µç½â\;}}{\;}$O2¡ü+2Cu+2H2SO4£®
£¨3£©Ñ­»·Öз´ÝÍÈ¡¼ÁBµÄÖ÷Òª³É·ÖÊÇH2SO4£®
£¨4£©Ä³Í­¿óʯÑùÆ·ÖУ¬Èô½öº¬Ñõ»¯ÑÇÍ­¡¢ÈýÑõ»¯¶þÌúºÍÂöʯÈýÖÖÎïÖÊ£®È¡¸Ã¿óʯÑùÆ·200.0g£¬ÓÃ100mL1.0mol•L-1H2SO4ÈÜÒº½þÈ¡ºó£¬»¹Ðè¼ÓÈë10mL 1.0mol•L-1 Fe2£¨SO4£©3ÈÜÒº²ÅÄÜʹͭȫ²¿½þ³ö£¬½þÈ¡Òº¾­³ä·Öµç½âºó¿ÉµÃµ½ 6.4gCu£®ÇóÍ­¿óʯÑùÆ·ÖÐÑõ»¯ÑÇÍ­ºÍÑõ»¯ÌúµÄÖÊÁ¿·ÖÊý£¿

·ÖÎö £¨1£©Ñõ»¯Í­¡¢ÈýÑõ»¯¶þÌú¾ù¿ÉÒÔºÍÇ¿Ëá·¢Éú·´Ó¦Éú³ÉÑκÍË®£¬½ðÊôÍ­ºÍÈý¼ÛÌúÖ®¼ä·¢ÉúÑõ»¯»¹Ô­·´Ó¦£»
£¨2£©Ë®²ãΪÁòËáÍ­ÈÜÒº£¬µç½âÁòËáÍ­ÈÜÒºÉú³ÉÍ­¡¢ÁòËáºÍÑõÆø£»
£¨3£©·´ÝÍÈ¡ºóµÄË®²ãÊÇÁòËáÍ­ÈÜÒº£¬Óɵç½âÖªÉú³ÉÁòËᣬËùÒÔÑ­»·Öз´ÝÍÈ¡¼ÁBÊÇÁòË᣻
£¨4£©ÒÀ¾Ý·´Ó¦¹ý³ÌÖÐÍ­ÔªËØÊغã¼ÆËãÑõ»¯ÑÇÍ­º¬Á¿£»ÒÀ¾ÝÑõ»¯»¹Ô­·´Ó¦µÄµç×ÓÊغã¼ÆËãÑõ»¯ÌúµÄÖÊÁ¿·ÖÊý£®

½â´ð ½â£º£¨1£©¿óʯÓÃÏ¡ÁòËá½þ³ö¹ý³ÌÖÐÑõ»¯ÑÇÍ­·¢ÉúµÄ·´Ó¦Îª£¬Cu2O+2H+¨TCu2++Cu+H2O£»½ðÊôÍ­ºÍÈý¼ÛÌúÖ®¼ä·¢ÉúÑõ»¯»¹Ô­·´Ó¦£»·´Ó¦µÄÀë×Ó·½³ÌʽΪ£ºCu+2Fe3+¨T2Fe2++Cu2+£»
¹Ê´ð°¸Îª£ºCu+2Fe3+¨T2Fe2++Cu2+£»
£¨2£©Ë®²ãΪÁòËáÍ­ÈÜÒº£¬µç½âÁòËáÍ­ÈÜÒºÉú³ÉÍ­¡¢ÁòËáºÍÑõÆø£¬·¢Éú·´Ó¦µÄ·½³ÌʽΪ£º2CuSO4+2H2O $\frac{\underline{\;µç½â\;}}{\;}$O2¡ü+2Cu+2H2SO4£»
¹Ê´ð°¸Îª£º2CuSO4+2H2O $\frac{\underline{\;µç½â\;}}{\;}$O2¡ü+2Cu+2H2SO4 £»
£¨3£©·´ÝÍÈ¡ºóµÄË®²ãÊÇÁòËáÍ­ÈÜÒº£¬Óɵç½âÖªÉú³ÉÁòËᣬËùÒÔÑ­»·Öз´ÝÍÈ¡¼ÁBÊÇÁòË᣻
¹Ê´ð°¸Îª£ºH2SO4£»
£¨4£©È¡¸Ã¿óʯÑùÆ·200.0g£¬ÓÃ100mL1.0mol•L-1H2SO4ÈÜÒº½þÈ¡ºó£¬»¹Ðè¼ÓÈë10mL 1.0mol•L-1 Fe2£¨SO4£©3ÈÜÒº²ÅÄÜʹͭȫ²¿½þ³ö£¬½þÈ¡Òº¾­³ä·Öµç½âºó¿ÉµÃµ½ 6.4gCu£»ÒÀ¾ÝÍ­ÔªËØÊغãÑõ»¯ÑÇÍ­ÖеÄÍ­ÔªËØ×îºóÉú³ÉÍ­£¬Í­CuÎïÖʵÄÁ¿=$\frac{6.4g}{64g/mol}$=0.1mol£»Ñõ»¯ÑÇÍ­ÎïÖʵÄÁ¿Îª0.05mol£»ÖÊÁ¿Îª0.05mol¡Á144g/mol=7.2g£»Cu2O%=$\frac{7.2g}{200g}$¡Á100%=3.6%£»ÒÀ¾ÝÑõ»¯»¹Ô­·´Ó¦µç×ÓÊغãFe3+¡«Cu+¡«Cu¼ÆËãµÃµ½£ºn£¨Fe3+£©=0.1mol£»Ô­ÑùÆ·ÖÐÌúÔªËØÎïÖʵÄÁ¿Îª=0.1mol-0.02mol=0.08mol£»
ËùÒÔFe2O3ÎïÖʵÄÁ¿Îª£º0.04mol£¬ÖÊÁ¿·ÖÊý=$\frac{0.04mol¡Á160g/mol}{200g}$¡Á100%=3.2%£¬
´ð£ºCu2OµÄÖÊÁ¿·ÖÊý£º3.6%£» Fe2O3µÄÖÊÁ¿·ÖÊý£º3.2%£®

µãÆÀ ±¾Ìâ¿´ËÆ¿¼²ìËá½þ·¨´Ó¿óʯÖÐÌáÈ¡Í­ÕâÑùÒ»¸ö¸´ÔӵŤÒÕÁ÷³Ì£¬Êµ¼ÊÉÏ¿¼²ìÁ˳£¼û½ðÊô¼°Æ仯ºÏÎïµÄÐÔÖÊ¡¢µç½â¾«Á¶Í­£¬¼°¶ÔÓлúÎï½á¹¹µÄ½á¹¹·ÖÎö£¬ÌâÄ¿ÄѶȽϴó£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

13£®²â֪ijÖÖ±´úÌþÔªËØ×é³ÉÖк¬Ì¼ºÍäåµÄÖÊÁ¿·ÖÊý·Ö±ðΪ12.8%ºÍ85.0%£¬ÒÑÖª¸Ã±´úÌþÏà¶Ô·Ö×ÓÖÊÁ¿Îª188£¬Çó¸Ã±´úÌþµÄ·Ö×Óʽ£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

14£®AÔªËصÄÑôÀë×ÓÓëBÔªËصÄÒõÀë×Ó¾ßÓÐÏàͬµÄµç×Ó²ã½á¹¹£®ÓйØÁ½ÔªËصÄÐðÊö£¬ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
¢ÙÔ­×Ӱ뾶A£¼B    
¢ÚÀë×Ӱ뾶A£¾B    
¢ÛÔ­×ÓÐòÊýA£¾B
¢ÜÔ­×Ó×îÍâ²ãµç×ÓÊýA£¼B    
¢ÝAµÄ×î¸ßÕý¼ÛÓëBµÄ¸º¼Û¾ø¶ÔÖµÒ»¶¨ÏàµÈ£®
A£®¢Ù¢ÚB£®¢Û¢ÜC£®¢Û¢ÝD£®¢Û¢Ü¢Ý

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

11£®Ä³»¯ºÏÎïX¿É¿´×÷ÓÉCO2ºÍNaOH·´Ó¦ËùµÃ£®ÎªÈ·¶¨Æä×é³É£¬Ïà¹ØʵÑé¼°Êý¾ÝÈç±íËùʾ£º
 ¢ñ¢ò¢ó¢ô
ÑÎËáÒºµÄÌå»ý
£¨mL£©
30303030
ÑùÆ·£¨g£©3.324.155.817.47
¶þÑõ»¯Ì¼µÄÌå»ý
£¨mL£©
672840896672
£¨1£©ÓÃ2.49gÑùÆ·½øÐÐͬÑùµÄʵÑéʱ£¬²úÉúCO2504mL£®
£¨2£©ÁìÈ¡3.32g»¯ºÏÎïXÑùÆ·ÓÚ300¡æ¼ÓÈÈ·Ö½âÖÁÍêÈ«£¨Na2CO3²»·Ö½â£©£¬²úÉúCO2112mL£¨±ê×¼×´¿ö£©ºÍË®0.45g£¬»¯ºÏÎïXµÄ»¯Ñ§Ê½2Na2CO3•NaHCO3•2H2O£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

4£®ÊµÑéÓú¬ÓÐÔÓÖÊ£¨FeO¡¢Fe2O3£©µÄ·ÏCuOÖƱ¸µ¨·¯¾§Ìå¾­ÀúÏÂÁйý³Ì£¬Fe3+ÔÚPH=5ʱÒÑÍêÈ«³Áµí£¬Æä·ÖÎö´íÎóµÄÊÇ£º£¨¡¡¡¡£©
ÑùÆ·$¡ú_{¢Ù¡÷}^{×ãÁ¿ÁòËá}$ÈÜÒº¢ñ$¡ú_{¢Ú}^{H_{2}O_{2}}$ÈÜÒº¢ò$¡ú_{µ÷PH=5}^{¢ÛCuO}$×ÇÒº$¡ú_{¹ýÂË}^{¢Ü}$ÂËÒº¢ó$¡ú_{ÀäÈ´½á¾§}^{Õô·¢Å¨Ëõ}$ $¡ú_{Ï´µÓ}^{¹ýÂË}$CuSO4•5H2O¾§Ì壮
A£®ÓÃ18mol/LŨÁòËáÅäÖÆÈܽâËùÓÃ4mol/LµÄÏ¡ÁòËᣬ²£Á§ÒÇÆ÷Ò»°ãÖ»ÓÐ3ÖÖ
B£®ÀûÓÃCu£¨OH£©2Ìæ´úCuOÒ²¿Éµ÷ÊÔÈÜÒºpH²»Ó°ÏìʵÑé½á¹û
C£®Ï´µÓ¾§Ì壺ÏòÂ˳ö¾§ÌåµÄ©¶·ÖмÓÉÙÁ¿Ë®½þû¾§Ì壬×ÔÈ»Á÷Ï£¬Öظ´2-3´Î
D£®ÈôÒªÓÃʵÑé²â¶¨ËùµÃÁòËáÍ­¾§ÌåÖнᾧˮµÄÊýÄ¿ÒªÓõ½ÛáÛöºÍÛáÛöǯ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

14£®Ìú¡¢ÂÁ¼°Æ仯ºÏÎïÔÚÉú²úºÍÉú»îÖÐÓÐ׏㷺µÄÓ¦Óã®
£¨1£©Ä³Ñо¿ÐÔѧϰС×éÉè¼ÆÁËÈçͼ1ËùʾװÖÃ̽¾¿¸ÖÌúµÄ¸¯Ê´Óë·À»¤£®ÔÚÏàͬÌõ¼þÏ£¬Èý×é×°ÖÃÖÐÌúµç¼«¸¯Ê´×î¿ìµÄÊÇ¢Ù £¨Ìî×°ÖÃÐòºÅ£©£¬¸Ã×°ÖÃÖÐÕý¼«µç¼«·´Ó¦Ê½ÎªO2+2H2O+4e-=4OH-£»Îª·ÀÖ¹½ðÊôFe±»¸¯Ê´£¬¿ÉÒÔ²ÉÓÃÉÏÊö¢Ú¢Û£¨Ìî×°ÖÃÐòºÅ£©×°ÖÃÔ­Àí½øÐзÀ»¤£»×°ÖâÛÖÐ×Ü·´Ó¦µÄÀëÓè·½³ÌʽΪ2Cl-+2H2O$\frac{\underline{\;µç½â\;}}{\;}$H2¡ü+Cl2¡ü+2OH-£®
£¨2£©ÐÂÐ͹ÌÌåLiFePO4¸ôĤµç³Ø¹ã·ºÓ¦ÓÃÓڵ綯Æû³µ£®
µç³Ø·´Ó¦ÎªFePO4+Li$?_{³äµç}^{·Åµç}$LiFePO4£¬µç½âÖÊΪº¬Li+µÄµ¼µç¹ÌÌ壬Çҳ䡢·Åµçʱµç³ØÄÚÁ½¼«¼äµÄ¸ôĤֻÔÊÐíLi+×ÔÓÉͨ¹ý¶øµ¼µç£®¸Ãµç³Ø·ÅµçʱLi+ÏòÕý¼«Òƶ¯£¨Ìî¡°Õý¡±»ò¡°¸º¡±£©£¬¸º¼«·´Ó¦ÎªLi-e-=Li+£¬ÔòÕý¼«·´Ó¦Ê½ÎªFePO4+Li++e-=LiFePO4£®
£¨3£©Ñõ»¯ÌúÊÇÖØÒª¹¤ÒµÑÕÁÏ£¬Ó÷ÏÌúмÖƱ¸ËüµÄÁ÷³ÌÈçͼ2£º»Ø´ðÏÂÁÐÎÊÌ⣺
¢Ù²Ù×÷I¡¢¢òµÄÃû³Æ·Ö±ðÊǹýÂË¡¢Ï´µÓ£®
¢Úд³öÔÚ¿ÕÆøÖжÍÉÕFeCO3µÄ»¯Ñ§·½³Ìʽ4FeCO3+O2$\frac{\underline{\;¸ßÎÂ\;}}{\;}$2Fe2O3+4CO2£»
£¨4£©ÓÐЩͬѧÈÏΪKMnO4ÈÜÒºµÎ¶¨Ò²ÄܽøÐÐÌúÔªËغ¬Á¿µÄ²â¶¨£¨5Fe2++MnO${\;}_{4}^{-}$+8H+¨T5Fe3++Mn2++4H2O£©£®
a£®³ÆÈ¡2.850gÂÌ·¯ £¨FeSO4•7H2O£©²úÆ·£¬Èܽ⣬ÔÚ250mLÈÝÁ¿Æ¿Öж¨ÈÝ£»
b£®Á¿È¡25.00mL´ý²âÈÜÒºÓÚ׶ÐÎÆ¿ÖУ»
c£®ÓÃÁòËáËữµÄ0.01000mol/KMnO4ÈÜÒºµÎ¶¨ÖÁÖյ㣬ÏûºÄKMnO4ÈÜÒºÌå»ýµÄƽ¾ùֵΪ 20.00mL£®¼ÆËãÉÏÊöÑùÆ·ÖÐFeSO4•7H2OµÄÖÊÁ¿·ÖÊýΪ0.9754£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

1£®ÊµÑéÊÒÀォ°×Á×·ÅÔÚË®Àï±£´æ£¬¾Ý´Ë¶Ô°×Á×ÐÔÖÊ×÷³öÈçÏÂÍƲ⣬²»ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®ËüÒ×ÓëÑõÆø·´Ó¦B£®ËüÓöË®²»·´Ó¦C£®ËüÄÑÈÜÓÚË®D£®Ëü±ÈË®ÖØ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

18£®ÔÚ50mL 4mol/LµÄÇâÑõ»¯ÄÆÈÜÒºÖУ¬Í¨ÈëÒ»¶¨Á¿µÄÁò»¯Ç⣬·´Ó¦Íê±Ï£¬ÔÚ³£Î¼õѹÌõ¼þÏ£¬ÓõªÆø°ÑÈÜÒº´µ¸É£¬µÃµ½°×É«¹ÌÌå7.92g£¬Í¨¹ý¼ÆËãÈ·¶¨°×É«¹ÌÌåµÄ×é³É¼°¸÷×é·ÖµÄÖÊÁ¿£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

19£®ÏÂÁÐÓйØÎïÖʽṹµÄ˵·¨ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®PCl3ºÍH2O·Ö×ÓÖÐËùÓÐÔ­×ÓµÄ×îÍâ²ã¶¼´ïµ½8µç×ÓÎȶ¨½á¹¹
B£®HBrµÄµç×ÓʽΪ 
C£®3.4 g°±ÆøÖк¬ÓÐ0.6NA¸öN-H¼ü
D£®78 g Na2O2¾§ÌåÖÐËùº¬Òõ¡¢ÑôÀë×Ó¸öÊý¾ùΪ4NA

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸