ij¹¤³§ÓÃÈíÃ̿󣨺¬MnO2Ô¼70£¥¼°Al2O3£©ºÍÉÁп¿ó£¨º¬ZnSÔ¼80£¥¼°FeS£©£¬¹²Í¬Éú²úMnO2ºÍZn£¨¸Éµç³ØÔ­ÁÏ£©¡£
ÒÑÖª£º¢ÙAÊÇMnSO4¡¢ZnSO4¡¢Fe2£¨SO4£©3¡¢Al2£¨SO4£©3µÄ»ìºÏÒº¡£
¢ÚIVÖеĵç½â·´Ó¦Ê½ÎªMnSO4+ZnSO4+2H2O MnO2+Zn+2H2SO4¡£

£¨1£©AÖÐÊôÓÚ»¹Ô­²úÎïµÄÊÇ   ¡£
£¨2£©MnCO3¡¢Zn2£¨OH£©2CO3µÄ×÷ÓÃÊÇ   £»¢òÐèÒª¼ÓÈȵÄÔ­ÒòÊÇ   £»CµÄ»¯Ñ§Ê½ÊÇ   ¡£
£¨3£©¢óÖз¢ÉúµÄÀë×Ó·½³ÌʽΪ   £¬   £»
£¨4£©Èç¹û²»¿¼ÂÇÉú²úÖеÄËðºÄ£¬³ý¿óʯÍ⣬Ð蹺ÂòµÄ»¯¹¤Ô­ÁÏÊÇ   ¡£

£¨1£©MnSO4    £¨2·Ö£©
£¨2£©Ôö´óÈÜÒºµÄpH£¬Ê¹Fe3+ºÍAl3+¾ùÉú³É³Áµí£¨2·Ö£©£»´Ù½øFe3+ºÍAl3+µÄË®½â£¨2·Ö£©£» H2SO4£¨2·Ö£©
£¨3£©Mn2+ + CO32- = MnCO3¡ý£¨2·Ö£© 2Zn2+ + 2CO32- + H2O = Zn2£¨OH£©2CO3 ¡ý+ CO2¡ü£¨2·Ö£©
£¨4£©´¿¼îºÍÁòËᣨ2·Ö£©

½âÎöÊÔÌâ·ÖÎö£º£¨1£©±È½ÏÐÅÏ¢¢ÙAÓëÈíÃÌ¿óÖÐÔªËØ»¯ºÏ¼ÛµÄ±ä»¯¿ÉÖª£¬MnÔªËØ»¯ºÏ¼ÛÓÉ+4¼Û½µµÍΪ+2¼Û£¬ËùÒÔAÖл¹Ô­²úÎïΪMnSO4¡£
£¨2£©Óɹ¤ÒÕÁ÷³Ì¿ÉÖª£¬MnCO3¡¢Zn2£¨OH£©2CO3µÄ×÷ÓþÍÊǵ÷½ÚpH£¬Ê¹Fe3+¡¢Al3+³ÁµíÍêÈ«£»Fe3+¡¢Al3+³ÁµíÈÝÒ×ÐγɽºÌ壬²»ÀûÓÚÇâÑõ»¯Ìú¡¢ÇâÑõ»¯ÂÁ³Á½µ£¬¢òÖмÓÈȵÄÄ¿µÄÊǼÓËÙ³ÁµíÉú³É£¬·ÀÖ¹½ºÌå³öÏÖ£¬²¢Ê¹ÐγɽºÌåµÄÇâÑõ»¯ÂÁºÍÇâÑõ»¯ÌúÒ²Éú³É³Áµí£»²Ù×÷¢ñ¼ÓÈÈ¡¢½þÈ¡ÐèÒªÁòËᣬÓÉ¢Ú¿ÉÖªCΪÁòËᣬѭ»·ÀûÓá£
£¨3£©ÈÜÒºBÖк¬ÓÐMn2+¡¢Zn2+£¬¸ù¾ÝÁ÷³Ìͼ£¬¼ÓÈëNa2CO3ºó£¬Éú³ÉMnCO3ºÍ Zn2£¨OH£©2CO3£¬Àë×Ó·½³ÌʽΪ£ºMn2+ + CO32- = MnCO3¡ý£» 2Zn2+ + 2CO32- + H2O = Zn2£¨OH£©2CO3 ¡ý+ CO2¡ü
£¨4£©ÓÉÁ÷³Ìͼ¿ÉÖª£¬Ðè¼ÓÈë̼ËáÄÆ¡¢ÁòËᣬËùÒÔ³ý¿óʯÍ⣬Ð蹺ÂòµÄ»¯¹¤Ô­ÁÏÊÇ´¿¼îºÍÁòËá¡£
¿¼µã£º±¾Ì⿼²é»¯Ñ§Á÷³ÌµÄ·ÖÎö¡¢ÊÔ¼ÁµÄ×÷Óá¢Àë×Ó·½³ÌʽµÄÊéд¡¢Ñõ»¯»¹Ô­·´Ó¦¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

Á×ËáÌúï®(LiFePO4)±»ÈÏΪÊÇ×îÓÐǰ;µÄï®Àë×Óµç³ØÕý¼«²ÄÁÏ¡£Ä³ÆóÒµÀûÓø»Ìú½þ³öÒºÉú³ÉÁ×ËáÌúï®,¿ª±ÙÁË´¦ÀíÁòËáÑÇÌú·ÏÒºÒ»ÌõÐÂ;¾¶¡£ÆäÖ÷ÒªÁ÷³ÌÈçÏÂ:

ÒÑÖª:H2TiO3ÊÇÖÖÄÑÈÜÓÚË®µÄÎïÖÊ¡£
(1)îÑÌú¿óÓÃŨÁòËá´¦Àí֮ǰ,ÐèÒª·ÛËé,ÆäÄ¿µÄÊÇ                             ¡¡¡£
(2)TiO2+Ë®½âÉú³ÉH2TiO3µÄÀë×Ó·½³ÌʽΪ                                  ¡¡¡£
(3)¼ÓÈëNaClO·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ                                     ¡¡¡£
(4)ÔÚʵÑéÖÐ,´ÓÈÜÒºÖйýÂ˳öH2TiO3ºó,ËùµÃÂËÒº»ë×Ç,Ó¦ÈçºÎ²Ù×÷¡¡            ¡£
(5)Ϊ²â¶¨îÑÌú¿óÖÐÌúµÄº¬Á¿,ijͬѧȡ¾­Å¨ÁòËáµÈ´¦ÀíµÄÈÜÒº(´ËʱîÑÌú¿óÖеÄÌúÒÑÈ«²¿×ª»¯Îª¶þ¼ÛÌúÀë×Ó),²ÉÈ¡KMnO4±ê×¼ÒºµÎ¶¨Fe2+µÄ·½·¨:(²»¿¼ÂÇKMnO4ÓëÆäËûÎïÖÊ·´Ó¦)Ôڵζ¨¹ý³ÌÖÐ,ÈôδÓñê×¼ÒºÈóÏ´µÎ¶¨¹Ü,Ôòʹ²â¶¨½á¹û¡¡¡¡¡¡¡¡(Ìî¡°Æ«¸ß¡±¡¢¡°Æ«µÍ¡±»ò¡°ÎÞÓ°Ï족),µÎ¶¨ÖÕµãµÄÏÖÏóÊÇ¡¡                                 ¡£µÎ¶¨·ÖÎöʱ,³ÆÈ¡a gîÑÌú¿ó,´¦Àíºó,ÓÃc mol/L KMnO4±ê×¼ÒºµÎ¶¨,ÏûºÄV mL,ÔòÌúÔªËصÄÖÊÁ¿·ÖÊýµÄ±í´ïʽΪ¡¡¡¡¡¡¡¡¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

»¯Ñ§ÊµÑéÊÒ²úÉúµÄ·ÏÒºÖк¬ÓдóÁ¿»áÎÛȾ»·¾³µÄÎïÖÊ£¬ÎªÁ˱£»¤»·¾³£¬ÕâЩ·ÏÒº±ØÐë¾­´¦Àíºó²ÅÄÜÅÅ·Å¡£Ä³»¯Ñ§ÊµÑéÊÒ²úÉúµÄ·ÏÒºÖк¬ÓÐÁ½ÖÖ½ðÊôÀë×Ó£ºFe3+¡¢Cu2+£¬»¯Ñ§Ð¡×éÉè¼ÆÁËÈçÏÂͼËùʾµÄ·½°¸¶Ô·ÏÒº½øÐд¦Àí£¬ÒÔ»ØÊÕ½ðÊô£¬±£»¤»·¾³¡£

£¨1£©²Ù×÷¢ÙµÄÃû³ÆÊÇ      ¡£
£¨2£©³ÁµíAÖк¬ÓеĽðÊôµ¥ÖÊÓР               ¡£
£¨3£©²Ù×÷¢ÚÖÐÔÚËáÐÔÏ·¢Éú·´Ó¦µÄÀë×Ó·½³Ìʽ                                    ¡£
£¨4£©¼ìÑéÈÜÒºBÖк¬ÓеĽðÊôÑôÀë×Ó³£ÓõÄÊÔ¼ÁÊÇ                   ¡£
£¨5£©²Ù×÷¢ÛÖз¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ                                          ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ÓÃϼʯÑÒ£¨»¯Ñ§Ê½Îª KNa3[AlSiO4]4£¬Ö÷Òª³É·ÝNa2O¡¢K2O¡¢Al2O3¡¢SiO2£©ÖÆ̼ËáÄÆ¡¢Ì¼Ëá¼ØºÍÑõ»¯ÂÁµÄ¹¤ÒÕÁ÷³ÌÈçÏ£º

ÒÑÖª£ºNaHCO3ÈÜÒºµÄpHԼΪ8¡«9£¬Na2CO3ÈÜÒºµÄpHԼΪ11¡«12¡£Èܽâ¹ýÂ˹¤Ðò²úÉúµÄÂËÒºÖк¬ÄÆ¡¢¼ØºÍÂÁµÄ¿ÉÈÜÐÔÑÎÀ࣬¸ÆºÍ¹èµÈÆäËûÔÓÖÊÔÚÂËÔüϼʯÄàÖС£²¿·ÖÎïÖʵÄÈܽâ¶È¼ûÓÒͼ¡£

ÊԻشðÏÂÁÐÎÊÌ⣺
£¨1£©×ÆÉյõ½¹ÌÌåMµÄ»¯Ñ§·½³ÌʽÊÇ________________________________¡£
£¨2£©XÎïÖÊÊÇ___________£¬ÂËÒºWÖÐÖ÷Òªº¬ÓеÄÀë×ÓÓÐ____________¡££¨Ð´ÈýÖÖ£©
£¨3£©²Ù×÷¢ñµÃµ½Ì¼ËáÄƾ§ÌåµÄ²Ù×÷Ϊ¡¡¡¡ ¡¢    ¡¢    ¡¢Ï´µÓ¡¢¸ÉÔï¡£
£¨4£©Ì¼Ëữ¢ñÖз¢ÉúÖ÷Òª·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ__________________________¡£
£¨5£©Ì¼Ëữ¢òµ÷ÕûpH£½8µÄÄ¿µÄÊÇ_______________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

Ϊ̽¾¿Ä³¿¹ËáÒ©XµÄ×é³É£¬½øÐÐÈçÏÂʵÑ飺
²éÔÄ×ÊÁÏ£º
¢Ù¿¹ËáÒ©X¿ÉÄܵÄ×é³É¿ÉÒÔ±íʾΪ£ºMgmAln(OH)p(CO3)q(SiO3)r£¨m¡¢n¡¢p¡¢q¡¢rΪ¡Ý0µÄÕûÊý£©¡£
¢ÚÔÚpH=5.0ʱ³ÁµíÍêÈ«£»ÔÚpH=8.8ʱ¿ªÊ¼³Áµí£¬pH=11.4ʱ³ÁµíÍêÈ«¡£
ʵÑé¹ý³Ì£º

²½Öè
ʵÑé²Ù×÷
ʵÑéÏÖÏó
I
ÏòXµÄ·ÛÄ©ÖмÓÈë¹ýÁ¿ÑÎËá
²úÉúÆøÌåA£¬µÃµ½ÎÞÉ«ÈÜÒº
II
Ïò¢ñËùµÃµÄÈÜÒºÖеμӰ±Ë®£¬µ÷½ÚpHÖÁ5~ 6£¬¹ýÂË
Éú³É°×É«³ÁµíB
III
Ïò³ÁµíBÖмӹýÁ¿NaOHÈÜÒº
³ÁµíÈ«²¿Èܽâ
IV
ÏòIIµÃµ½µÄÂËÒºÖеμÓNaOHÈÜÒº£¬µ÷½ÚpHÖÁ12
Éú³É°×É«³ÁµíC
 
£¨1£©¢ñÖÐÆøÌåA¿Éʹ³ÎÇåʯ»ÒË®±ä»ë×Ç£¬AµÄ»¯Ñ§Ê½ÊÇ           ¡£
£¨2£©IIÖÐÉú³ÉB·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ                                      ¡£
£¨3£©IIIÖÐBÈܽⷴӦµÄÀë×Ó·½³ÌʽÊÇ                                     ¡£
£¨4£©³ÁµíCµÄ»¯Ñ§Ê½ÊÇ             ¡£
£¨5£©ÈôÉÏÊön(A)©Un(B)©Un(C)=1©U2©U3£¬ÔòXµÄ»¯Ñ§Ê½ÊÇ                   ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

¹¤ÒµÉÏÓÃÉÁп¿ó£¨Ö÷Òª³É·ÖΪZnS£¬»¹º¬ÓÐFe2O3µÈÔÓÖÊ£©ÎªÔ­ÁÏÉú²úZnSO4¡¤7H2OµÄ¹¤ÒÕÁ÷³ÌÈçÏ£º

£¨1£©´ÓÂËÔüAÖпɻñµÃÒ»ÖÖµ­»ÆÉ«·Ç½ðÊôµ¥Öʵĸ±²úÆ·£¬Æ仯ѧʽΪ        ¡£
£¨2£©½þÈ¡¹ý³ÌÖÐFe2(SO4)3µÄ×÷ÓÃÊÇ               £¬½þȡʱFe2(SO4)3ÓëZnS·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ                                            ¡£
£¨3£©³ýÌú¹ý³Ì¿ØÖÆÈÜÒºµÄpHÔÚ5.4×óÓÒ£¬¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽΪ                     ¡£¸Ã¹ý³ÌÔÚ¿ÕÆøÈë¿Ú´¦Éè¼ÆÁËÒ»¸öÀàËÆÁÜÔ¡ÅçÍ·µÄ×°Öã¬ÆäÄ¿µÄÊÇ                                         ¡£
£¨4£©Öû»·¨³ýÖؽðÊôÀë×ÓËùÓÃÎïÖÊCΪ         ¡£
£¨5£©ÁòËáпµÄÈܽâ¶ÈÓëζÈÖ®¼äµÄ¹ØϵÈçÏÂ±í£º

ζÈ/¡æ
0
20
40
60
80
100
Èܽâ¶È/g
41.8
54.1
70.4
74.8
67.2
60.5
 
´Ó³ýÖؽðÊôºóµÄÁòËáпÈÜÒºÖлñµÃÁòËáп¾§ÌåµÄʵÑé²Ù×÷Ϊ           ¡¢            ¡¢¹ýÂË¡¢¸ÉÔï¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ä廯¸Æ(CaBr2¡¤2H2O)ÊÇÒ»ÖÖ°×É«¾§Ì壬Ò×ÈÜÓÚË®£¬ÓкÜÇ¿µÄÎüʪÐÔ£¬ÊǹâÃôÖ½¿óȪˮºÍÃð»ð¼ÁµÄÖØÒª³É·Ö£¬ÔÚÒ½Ò©ÉÏÓÃ×÷ÖÎÁÆÉñ¾­Ë¥ÈõµÈµÄÒ©ÎҲÓÃ×÷»¯Ñ§·ÖÎö¡£Óù¤Òµ´óÀíʯ£¨º¬ÓÐÉÙÁ¿Al3+¡¢Fe3+µÈÔÓÖÊ£©ÖƱ¸ä廯¸ÆµÄÖ÷ÒªÁ÷³ÌÈçÏÂ

»Ø´ðÏÂÁÐÎÊÌâ
£¨1£©Èܽâʱ·¢ÉúµÄÖ÷Òª·´Ó¦µÄÀë×Ó·½³ÌʽΪ               
£¨2£©³ýÔÓ²½Öè¿ØÖÆÈÜÒºµÄpHԼΪ8£®0µÄÄ¿µÄÊÇ               £¬               
£¨3£©ÂËÒºÓÃÇâäåËáËữµÄÄ¿µÄÊÇ           £¬²Ù×÷aÖ÷Òª°üÀ¨               £¬           ºÍ¹ýÂË
£¨4£©ÖƵõÄä廯¸Æ¾§Ìå¿ÉÒÔͨ¹ýÈçϲ½Öè²â¶¨Æä´¿¶È£º
¢Ù³ÆÈ¡5£®00gä廯¸Æ¾§ÌåÑùÆ·£¬¢ÚÈܽ⣻¢ÛµÎÈë×ãÁ¿Naa2CO3ÈÜÒº£¬³ä·Ö·´Ó¦ºó¹ýÂË£¬¢Üºæ¸É¡¢ÀäÈ´£»¢Ý³ÆÁ¿¡£ÈôµÃµ½2£® 00 g̼Ëá¸Æ£¬ÔòÑùÆ·µÄ´¿¶ÈΪ               
£¨5£©ä廯¸Æ¾§ÌåÖÐäåÀë×Ӻ͸ÆÀë×ӵļìÑé
¢Ù½«ÉÙÁ¿ä廯¸Æ¾§ÌåÈÜÓÚË®£¬¼ÓÈëÏõËáËữµÄAgNO3ÈÜÒº£¬ÊµÑéÏÖÏóΪ               £¬·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ               
¢Ú½«ÉÙÁ¿ä廯¸Æ¾§ÌåÈÜÓÚË®£¬µÎ¼Ó²ÝËáÄÆÈÜÒº£¬ÊµÑéÏÖÏóΪ               £¬·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ               

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

NiSO4¡¤6H2OÊÇÒ»ÖÖÂÌÉ«Ò×ÈÜÓÚË®µÄ¾§Ì壬¹ã·ºÓÃÓÚ»¯Ñ§¶ÆÄø¡¢Éú²úµç³ØµÈ£¬¿ÉÓɵç¶Æ·ÏÔü£¨³ýº¬ÄøÍ⣬»¹º¬ÓУºCu¡¢Zn¡¢Fe¡¢CrµÈÔÓÖÊ£©ÎªÔ­ÁÏ»ñµÃ¡£²Ù×÷²½ÖèÈçÏ£º

£¨1£©¼ÓNa2SµÄÄ¿µÄÊdzýȥͭ¡¢Ð¿µÈÔÓÖÊ£¬Çëд³ö³ýÈ¥Cu2+µÄÀë×Ó·½³Ìʽ__________ __________
£¨2£© ¼Ó6%µÄH2O2ʱ£¬Î¶Ȳ»Äܹý¸ß£¬ÊÇÒòΪ£º  _____         ________ ¡£
£¨3£© ³ýÌú·½·¨£ºÓÃH2O2³ä·ÖÑõ»¯ºó£¬ÔÙÓÃNaOH¿ØÖÆpHÖµ3¡«4·¶Î§ÄÚÉú³ÉÇâÑõ»¯Ìú³Áµí¡£ÔÚÉÏÊö·½·¨ÖУ¬Ñõ»¯¼Á¿ÉÓÃNaClO3´úÌ棬Çëд³öÓÃÂÈËáÄÆÑõ»¯Fe2+µÄÀë×Ó·½³ÌʽΪ£º___________________________________________________________________________
£¨4£©ÉÏÊöÁ÷³ÌÖÐÂËÒº¢óµÄÖ÷Òª³É·ÖÊÇ£º¡¡           ¡¡¡££¨Ìѧʽ£©
£¨5£©²Ù×÷¢ñ°üÀ¨ÒÔϹý³Ì£º¹ýÂË£¬Óà          £¨ÌîÊÔ¼Á»¯Ñ§Ê½£©Èܽ⣬         £¬Ï´µÓ»ñµÃ²úÆ·¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

Á×ËáÌúﮣ¨LiFePO4£©±»ÈÏΪÊÇ×îÓÐǰ;µÄï®Àë×Óµç³ØÕý¼«²ÄÁÏ¡£Ä³ÆóÒµÀûÓø»Ìú½þ³öÒºÉú³ÉÁ×ËáÌúﮣ¬¿ª±ÙÁË´¦ÀíÁòËáÑÇÌú·ÏÒºÒ»ÌõÐÂ;¾¶¡£ÆäÖ÷ÒªÁ÷³ÌÈçÏ£º

ÒÑÖª£ºH2LiO3ÊÇÖÖÄÑÈÜÓÚË®µÄÎïÖÊ¡£
(1)îÑÌú¿óÓÃŨÁòËá´¦Àí֮ǰ£¬ÐèÒª·ÛË飬ÆäÄ¿µÄ                                  
(2)TiO2+Ë®½âÉú³ÉH2TiO3µÄÀë×Ó·½³Ìʽ                                         
(3)¼ÓÈëNaClO·¢Éú·´Ó¦µÄÀë×Ó·½³Ìʽ                                
(4)ÔÚʵÑéÖУ¬´ÓÈÜÒºÖйýÂ˳öH2TiO3ºó£¬ËùµÃÂËÒº»ë×Ç£¬Ó¦ÈçºÎ²Ù×÷               ¡£
(5)Ϊ²â¶¨îÑÌú¿óÖÐÌúµÄº¬Á¿£¬Ä³Í¬Ñ§È¡¾­Å¨ÁòËáµÈ´¦ÀíµÄÈÜÒº£¨´ËʱîÑÌú¿óÖеÄÌú¼ºÈ«²¿×ª»¯Îª¶þ¼ÛÌúÀë×Ó£©£¬²ÉÈ¡KMnO4±ê×¼ÒºÀ춨Fe2+µÄ·½·¨£º(²»¿¼ÂÇKMnO4ÓëÆäËûÎïÖÊ·´Ó¦)Ôڵζ¨¹ý³ÌÖУ¬ÈôδÓñê×¼ÒºÈóÏ´µÎ¶¨¹Ü£¬Ôòʹ²â¶¨½á¹û        ¡£ (Ìî¡°Æ«¸ß¡¢Æ«µÍ¡¢ÎÞÓ°Ï족)£¬µÎ¶¨ÖÕµãµÄÏÖÏó                ¡£µÎ¶¨·ÖÎöʱ£¬³ÆÈ¡a gîÑÌú¿ó£¬´¦Àíºó£¬ÓÃcmol/LKMnO4±ê×¼ÒºµÎ¶¨£¬ÏûºÄVmL£¬ÔòÌúÔªËصÄÖÊÁ¿·ÖÊýµÄ±í´ïʽΪ            

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸