7£®ÂÈ»¯ÑÇÍ­£¨CuCl£©ÊÇÖØÒªµÄ»¯¹¤Ô­ÁÏ£®¹¤ÒµÉϳ£Í¨¹ýÏÂÁз´Ó¦ÖƱ¸CuCl£º
2CuSO4+Na2SO3+2NaCl+Na2CO3¡ú2CuCl¡ý+3Na2SO4+CO2¡ü
£¨1£©CuClÖƱ¸¹ý³ÌÖÐÐèÒªÅäÖÆÖÊÁ¿·ÖÊýΪ20.0%µÄCuSO4ÈÜÒº£¬ÅäÖƸÃÈÜÒºËùÐèµÄCuSO4•5H2OÓëH2OµÄÖÊÁ¿Ö®±ÈΪ5£º11£®
£¨2£©×¼È·³ÆÈ¡ËùÅäÖƵÄ0.2500g CuClÑùÆ·ÖÃÓÚÒ»¶¨Á¿µÄ0.5mol•L-1 FeCl3ÈÜÒºÖУ¬´ýÑùÆ·ÍêÈ«Èܽâºó£¬¼ÓË®20mL£¬ÓÃ0.1000mol•L-1µÄCe£¨SO4£©2ÈÜÒºµÎ¶¨µ½Öյ㣬ÏûºÄ24.60mLCe£¨SO4£©2ÈÜÒº£®Óйط´Ó¦µÄÀë×Ó·½³ÌʽΪ£ºFe3++CuCl¡úFe2++Cu2++Cl-Ce4++Fe2+¡úFe3++Ce3+
ÒÑÖª¹ú¼Ò±ê×¼¹æ¶¨ºÏ¸ñµÄCuCl²úÆ·µÄÖ÷ÒªÖÊÁ¿Ö¸±êΪCuClµÄÖÊÁ¿·ÖÊý´óÓÚ96.50%£®ÊÔͨ¹ý¼ÆËã˵Ã÷ÉÏÊöÑùÆ·ÖÐCuClµÄÖÊÁ¿·ÖÊýÊÇ·ñ·ûºÏ¹ú¼Ò±ê×¼£®

·ÖÎö £¨1£©ÉèÐèÒªCuSO4•5H2OµÄÖÊÁ¿Îªx£¬H2OµÄÖÊÁ¿Îªy£¬¸ù¾Ý»¯Ñ§Ê½¼ÆËã³öÁòËáÍ­µÄÖÊÁ¿m£¨CuSO4£©£¬ÈÜÒºÖÊÁ¿Îªx+y£¬ÀûÓÃÖÊÁ¿·ÖÊýÁз½³Ì£¬¼ÆËãx¡¢yµÄ±ÈÀý¹Øϵ£®
£¨2£©¸ù¾Ý¹Øϵʽ¼ÆËãn£¨CuCl£©£¬½ø¶ø¼ÆËãÑùÆ·ÖÐm£¨CuCl£©£¬¼ÆËã0.2500gºÏ¸ñµÄCuClÖк¬ÓÐCuClµÄÖÊÁ¿£¬½øÐбȽÏÅжϣ®

½â´ð ½â£º£¨1£©ÉèÐèÒªCuSO4•5H2OµÄÖÊÁ¿Îªx£¬H2OµÄÖÊÁ¿Îªy£¬Ôò£º$\frac{\frac{160}{250}x}{x+y}$¡Á100%=20.0%£¬¼´16x=5£¨x+y£©£¬ËùÒÔ x£ºy=5£º11£¬
´ð£ºËùÐèCuSO4•5H2OÓëH2OµÄÖÊÁ¿Ö®±ÈΪ5£º11£»
£¨2£©ÉèÑùÆ·ÖÐCuClµÄÖÊÁ¿Îªx£¬Ôò£º
ÓÉ»¯Ñ§·´Ó¦·½³Ìʽ¿ÉÖª£ºCuCl¡«¡«¡«Fe2+¡«¡«¡«Ce4+ 
                                         1                         1
                                   n£¨CuCl£©        24.60¡Á10-3L¡Á0.1000 mol/L
ËùÒÔ n£¨CuCl£©=24.60¡Á10-3L¡Á0.1000 mol/L=2.46¡Á10-3mol£®ËùÒÔ¸ÃÑùÆ·CuClµÄÖÊÁ¿Îª2.46¡Á10-3mol¡Á99.5g/mol=0.2448g£®
0.2500gºÏ¸ñµÄCuClÖк¬ÓÐCuClµÄÖÊÁ¿0.2500g¡Á96.5%=0.2413g£¬Ð¡ÓÚ0.2448g£¬¹Ê¸ÃÑùÆ·ÖÐCuClµÄÖÊÁ¿·ÖÊý·ûºÏ±ê×¼£®
´ð£ºÑùÆ·ÖÐCuClµÄÖÊÁ¿·ÖÊý·ûºÏ±ê×¼£®

µãÆÀ ±¾Ì⿼²éÎïÖʵÄÁ¿Å¨¶È¼ÆËã¡¢µÎ¶¨¼ÆËãµÈ£¬ÄѶÈÖеȣ¬×¢ÒâÔڵζ¨¼ÆËãÖÐÓÉÓÚÉæ¼°·´Ó¦½Ï¶à£¬Í¨³£ÀûÓùØϵʽ¼ÆË㣮

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

17£®¹¤ÒµÉÏÀûÓ÷ú̼îæ¿ó£¨Ö÷Òª³É·ÖCeCO3F£©ÌáÈ¡CeCl3µÄÒ»ÖÖ¹¤ÒÕÁ÷³ÌÈçÏ£º

Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©CeCO3FÖУ¬CeÔªËصĻ¯ºÏ¼ÛΪ+3£®
£¨2£©Ëá½þ¹ý³ÌÖÐÓÃÏ¡ÁòËáºÍH2O2Ìæ»»HCl²»»áÔì³É»·¾³ÎÛȾ£®Ð´³öÏ¡ÁòËá¡¢H2O2ÓëCeO2·´Ó¦µÄÀë×Ó·½³Ìʽ£ºH2O2+2CeO2+6H+=2Ce3++4H2O+O2¡ü£®
£¨3£©ÏòCe£¨BF4£©3ÖмÓÈëKClÈÜÒºµÄÄ¿µÄÊDZÜÃâÈý¼ÛîæÒÔCe£¨BF4£©3³ÁµíµÄÐÎʽËðʧ»ò³ýÈ¥BF4-»òÌá¸ßCeCl3µÄ²úÂÊ£®
£¨4£©ÈÜÒºÖеÄC£¨Ce3+£©µÈÓÚ1¡Á10-5mol£®l-1£¬¿ÉÈÏΪCe3+³ÁµíÍêÈ«£¬´ËʱÈÜÒºµÄPHΪ9£®£¨ÒÑÖªKSP[Ce£¨OH£©3]=1¡Á10-20£©
£¨5£©¼ÓÈÈCeCl3.6H2OºÍNH4ClµÄ¹ÌÌå»ìºÏÎï¿ÉµÃµ½ÎÞË®CeCl3£¬ÆäÖÐNH4ClµÄ×÷ÓÃÊÇNH4Cl¹ÌÌåÊÜÈÈ·Ö½â²úÉúHCl£¬ÒÖÖÆCeCl3Ë®½â£®
£¨6£©×¼È·³ÆÈ¡0.7500gCeCl3ÑùÆ·ÖÃÓÚ׶ÐÎÆ¿ÖУ¬¼ÓÈëÊÊÁ¿¹ýÁòËáï§ÈÜÒº½«Ce3+Ñõ»¯ÎªCe4+£¬È»ºóÓÃ0.1000mol£®l-1£¨NH4£©2Fe£¨SO4£©2±ê×¼ÈÜÒºµÎ¶¨ÖÁÖյ㣬ÏûºÄ25.00ml±ê×¼ÈÜÒº£®£¨ÒÑÖª£ºFe2+Ce4+=Ce3++Fe3+£©
¢Ù¸ÃÑùÆ·ÖÐCeCl3µÄÖÊÁ¿·ÖÊýΪ82.2%£®
¢ÚÈôʹÓþÃÖõģ¨NH4£©2Fe£¨SO4£©2±ê×¼ÈÜÒº½øÐе樣¬²âµÃ¸ÃCeCl3ÑùÆ·µÄÖÊÁ¿·ÖÊýÆ«´ó£¨Ìî¡°Æ«´ó¡±¡¢¡°Æ«Ð¡¡±»ò¡°ÎÞÓ°Ï족£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

18£®°¢Ë¾Æ¥ÁÖ£¨ÒÒõ£Ë®ÑîËᣬ£©ÊÇÊÀ½çÉÏÓ¦ÓÃ×î¹ã·ºµÄ½âÈÈ¡¢ÕòÍ´ºÍ¿¹Ñ×Ò©£®ÒÒõ£Ë®ÑîËáÊÜÈÈÒ׷ֽ⣬·Ö½âζÈΪ128¡æ¡«135¡æ£®Ä³Ñ§Ï°Ð¡×éÔÚʵÑéÊÒÒÔË®ÑîËᣨÁÚôÇ»ù±½¼×ËᣩÓë´×ËáôûΪÖ÷ÒªÔ­ÁϺϳɰ¢Ë¾Æ¥ÁÖ£¬·´Ó¦Ô­ÀíÈçÏ£º

ÖƱ¸»ù±¾²Ù×÷Á÷³ÌÈçÏ£º
´×Ëáôû+Ë®ÑîËá$\stackrel{ŨÁòËá}{¡ú}$$\stackrel{Ò¡ÔÈ}{¡ú}$$\stackrel{85¡æ-90¡æ¼ÓÈÈ}{¡ú}$$\stackrel{ÀäÈ´}{¡ú}$$¡ú_{Ï´µÓ}^{¼õѹ¹ýÂË}$´Ö²úÆ·
Ö÷ÒªÊÔ¼ÁºÍ²úÆ·µÄÎïÀí³£ÊýÈçϱíËùʾ£º
Ãû³ÆÏà¶Ô·Ö×ÓÖÊÁ¿ÈÛµã»ò·Ðµã£¨¡æ£©Ë®ÈÜÐÔ
Ë®ÑîËá138158£¨È۵㣩΢ÈÜ
´×Ëáôû102139.4£¨·Ðµã£©Ò×Ë®½â
ÒÒõ£Ë®ÑîËá180135£¨È۵㣩΢ÈÜ
Çë¸ù¾ÝÒÔÉÏÐÅÏ¢»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÖƱ¸°¢Ë¾Æ¥ÁÖʱ£¬ÒªÊ¹ÓøÉÔïµÄÒÇÆ÷µÄÔ­ÒòÊÇ´×ËáôûºÍË®Ò×·¢Éú·´Ó¦£®
£¨2£©ºÏ³É°¢Ë¾Æ¥ÁÖʱ£¬×îºÏÊʵļÓÈÈ·½·¨ÊÇˮԡ¼ÓÈÈ£®

£¨3£©Ìá´¿´Ö²úÆ·Á÷³ÌÈçÏ£¬¼ÓÈÈ»ØÁ÷×°ÖÃÈçͼ£º
´Ö²úÆ·$¡ú_{·Ðʯ}^{ÒÒËáÒÒõ¥}$$¡ú_{»ØÁ÷}^{¼ÓÈÈ}$$\stackrel{³ÃÈȹýÂË}{¡ú}$$¡ú_{¼õѹ¹ýÂË}^{ÀäÈ´}$$¡ú_{¸ÉÔï}^{Ï´µÓ}$ÒÒõ£Ë®ÑîËá
¢ÙʹÓÃζȼƵÄÄ¿µÄÊÇ¿ØÖƼÓÈȵÄζȣ¬·ÀÖ¹ÒÒõ£Ë®ÑîËáÊÜÈÈÒ׷ֽ⣮
¢ÚÀäÄýË®µÄÁ÷½ø·½ÏòÊÇa£¨Ìî¡°a¡±»ò¡°b¡±£©£»
¢Û³ÃÈȹýÂ˵ÄÔ­ÒòÊÇ·ÀÖ¹ÒÒõ£Ë®ÑîËá½á¾§Îö³ö£®
¢ÜÏÂÁÐ˵·¨ÕýÈ·µÄÊÇabc£¨ÌîÑ¡Ïî×Öĸ£©£®
a£®´ËÖÖÌá´¿·½·¨ÖÐÒÒËáÒÒõ¥µÄ×÷ÓÃÊÇ×öÈܼÁ
b£®´ËÖÖÌá´¿´Ö²úÆ·µÄ·½·¨½ÐÖؽᾧ
c£®¸ù¾ÝÒÔÉÏÌá´¿¹ý³Ì¿ÉÒԵóö°¢Ë¾Æ¥ÁÖÔÚÒÒËáÒÒõ¥ÖеÄÈܽâ¶ÈµÍÎÂʱ´ó
d£®¿ÉÒÔÓÃ×ÏɫʯÈïÈÜÒºÅжϲúÆ·ÖÐÊÇ·ñº¬ÓÐδ·´Ó¦ÍêµÄË®ÑîËá
£¨4£©ÔÚʵÑéÖÐÔ­ÁÏÓÃÁ¿£º2.0gË®ÑîËá¡¢5.0mL´×Ëáôû£¨¦Ñ=1.08/cm3£©£¬×îÖճƵòúÆ·ÖÊÁ¿Îª2.2g£¬ÔòËùµÃÒÒõ£Ë®ÑîËáµÄ²úÂÊΪ84.3%£¨ÓðٷÖÊý±íʾ£¬Ð¡Êýµãºóһ룩£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÊµÑéÌâ

15£®äå±½ÊÇÒ»ÖÖ³£ÓõĻ¯¹¤Ô­ÁÏ£®ÊµÑéÊÒÖƱ¸äå±½µÄʵÑé²½ÖèÈçÏ£º
²½Öè1£ºÔÚaÖмÓÈë15mL±½ºÍÉÙÁ¿Ìúм£¬ÔÙ½«bÖÐ4.0mLÒºäåÂýÂý¼ÓÈëµ½aÖУ¬³ä·Ö·´Ó¦£®
²½Öè2£ºÏòaÖмÓÈë10mLË®£¬È»ºó¹ýÂ˳ýȥδ·´Ó¦µÄÌúм£®
²½Öè3£ºÂËÒºÒÀ´ÎÓÃ10mLË®¡¢8mL 10%µÄNaOHÈÜÒº¡¢10mL Ë®Ï´µÓ£¬
·ÖÒºµÃ´Öäå±½£®
±½äåäå±½
ÃܶÈ/gcm-30.883.101.50
·Ðµã/¡æ8059156
ÔÚË®ÖеÄÈܽâ¶È΢ÈÜ΢ÈÜ΢ÈÜ
²½Öè4£ºÏò·Ö³öµÄ´Öäå±½ÖмÓÈëÉÙÁ¿µÄÎÞË®ÂÈ»¯¸Æ£¬¾²ÖᢹýÂ˼´µÃ´Ö²úÆ·£®
£¨1£©²½Öè1ÒÇÆ÷aÖз¢ÉúµÄÖ÷Òª·´Ó¦ÊÇ£»
£¨2£©ÒÇÆ÷cµÄÃû³ÆÊÇÀäÄý¹Ü£¬ÒÇÆ÷dµÄ×÷ÓÃÊÇÎüÊÕHBr·ÀÎÛȾ£¬·Àµ¹Îü£»
£¨3£©½«bÖеÄÒºäåÂýÂý¼ÓÈëµ½aÖУ¬¶ø²»ÄÜ¿ìËÙ¼ÓÈëµÄÔ­ÒòÊÇ·ÀÖ¹·´Ó¦·Å³öµÄÈÈʹC6H6¡¢Br2»Ó·¢¶øÓ°Ïì²úÂÊ£»
£¨4£©ÒÇÆ÷cµÄ×÷ÓÃÊÇÀäÄý»ØÁ÷£¬»ØÁ÷µÄÖ÷ÒªÎïÖÊÓÐC6H6¡¢Br2£¨Ìѧʽ£©£»
£¨5£©²½Öè4µÃµ½µÄ´Ö²úÆ·Öл¹º¬ÓÐÔÓÖʱ½£¬ÒÑÖª±½¡¢äå±½µÄÓйØÎïÀíÐÔÖÊÈçÉÏ±í£¬ÔòÒª½øÒ»²½Ìá´¿´Ö²úÆ·£¬»¹±ØÐë½øÐеÄʵÑé²Ù×÷Ãû³ÆÊÇÕôÁó£®
£¨6£©ÊµÑé½áÊøºó£¬ÊÔÉè¼ÆʵÑé·½°¸¼ìÑéd×°ÖÃËùµÃÈÜÒºÖк¬ÓÐBr-£º
ʵÑé²Ù×÷£ºÈ¡ÉÙÁ¿dÖÐÈÜÒºÓÚÊÔ¹ÜÖУ¬µÎ¼Ó¹ýÁ¿Ï¡HNO3£¬ÔÙ¼ÓÈëAgNO3ÈÜÒº£»
ʵÑéÏÖÏóºÍ½áÂÛ£ºÓе­»ÆÉ«³ÁµíÉú³É£¬ÈÜÒºÖк¬ÓÐBr-£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

2£®ÓÉÓÚË®ÌåÖк¬NµÄÎïÖÊ£¬º¬PµÄÎïÖʶøÒýÆðµÄÎÛȾ£¬½Ð×öË®ÌåµÄ¸»ÓªÑø»¯£®µ±Ë®Ìå·¢Éú¸»ÓªÑø»¯Ê±£¬ÓÉÓÚȱÑõ£¬ÖÂʹ´ó¶àÊýË®Éú¶¯Îï²»ÄÜÉú´æ£¬ÖÂËÀµÄ¶¯Ö²ÎïÒź¡ÔÚË®µ×³Á»ý¸¯Àã¬Ê¹Ë®Öʲ»¶Ï¶ñ»¯£¬Ó°ÏìÓæÒµÉú²ú£¬²¢Í¨¹ýʳÎïÁ´Î£º¦ÈËÌ彡¿µ£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

12£®Ä³»¯Ñ§Ñ§Ï°Ð¡×éµÄÑо¿¿ÎÌâÊÇ£ºÌ½¾¿²â¶¨²ÝËᾧÌ壨H2C2O4•xH2O£©ÖÐxµÄÖµ£®¸Ã×éͬѧͨ¹ý²éÔÄ×ÊÁϲéÑ°µÃ£¬²ÝËáÒ×ÈÜÓÚË®£¬Ë®ÈÜÒº¿ÉÒÔÓÃËáÐÔKMnO4ÈÜÒº½øÐе樣º
2MnO4-+5H2C2O4+6H+=2Mn2++10CO2¡ü+8H2O
ѧϰС×éµÄͬѧÉè¼ÆÁ˵ζ¨µÄ·½·¨²â¶¨xÖµ£®
¢Ù³ÆÈ¡1.260g´¿²ÝËᾧÌ壬½«ÆäÖƳÉ100.00mLË®ÈÜҺΪ´ý²âÒº£®
¢ÚÈ¡25.00mL´ý²âÒº·ÅÈë׶ÐÎÆ¿ÖУ¬ÔÙ¼ÓÈëÊÊÁ¿µÄÏ¡H2SO4
¢ÛÓÃŨ¶ÈΪ0.1000mol/LµÄËáÐÔKMnO4±ê×¼ÈÜÒº½øÐе樣¬´ïµ½ÖÕµãʱÏûºÄÁË10.00mL£®
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÅäÖÆ100.00mL´ý²âҺʱ£¬Ê¹ÓõIJ£Á§ÒÇÆ÷³ýÉÕ±­¡¢²£Á§°ô¼°½ºÍ·µÎ¹ÜÍ⣬»¹ÐèÒª100mLÈÝÁ¿Æ¿£®
£¨2£©µÎ¶¨Ê±£¬½«ËáÐÔKMnO4±ê׼ҺװÔÚÈçͼÖеļף¨Ìî¡°¼×¡±»ò¡°ÒÒ¡±£©µÎ¶¨¹ÜÖУ®±¾ÊµÑéµÎ¶¨´ïµ½ÖÕµãµÄ±êÖ¾Êǵ±µÎÈë×îºóÒ»µÎËáÐÔKMnO4ÈÜҺʱ£¬ÈÜÒºÓÉÎÞÉ«±äΪdz×ÏÉ«£¬ÇÒ°ë·ÖÖÓÄÚ²»ÍÊÉ«£®
£¨3£©Í¨¹ýÉÏÊöÊý¾Ý£¬ÇóµÃx=2£®
£¨4£©ÌÖÂÛ£º
¢ÙÈôµÎ¶¨ÖÕµãʱ¸©Êӵζ¨¹Ü£¬ÔòÓɴ˲âµÃµÄxÖµ»áÆ«´ó£¨Ìî¡°Æ«´ó¡±¡¢¡°Æ«Ð¡¡±»ò¡°²»±ä¡±£¬ÏÂͬ£©£®
¢ÚÈôµÎ¶¨Ê±ËùÓõÄËáÐÔKMnO4ÈÜÒºÒò¾ÃÖöøµ¼ÖÂŨ¶È±äС£¬ÔòÓɴ˲âµÃµÄxÖµ»áƫС£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

19£®ÓÃÒÑ֪Ũ¶ÈµÄNaOHÈÜÒº²â¶¨Î´ÖªÅ¨¶ÈµÄÑÎËᣨµÎ¶¨Ê±ÑÎËáÖÃÓÚËáʽµÎ¶¨¹ÜÖУ©£¬ÒÔϲÙ×÷»áÔì³ÉËù²âÑÎËáÈÜҺŨ¶ÈÆ«¸ßµÄÊÇ£¨¡¡¡¡£©
A£®ËáʽµÎ¶¨¹ÜδÓôý²âÈÜÒºÈóÏ´
B£®¼îʽµÎ¶¨¹ÜδÓôý×°ÈÜÒºÈóÏ´
C£®×¶ÐÎƿδÓôý×°ÈÜÒºÈóÏ´
D£®Ôڵζ¨Ç°µÎ¶¨¹Ü¼â×첿·ÖÓÐÆøÅÝ£¬µÎ¶¨ºóÆøÅÝÏûʧ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

16£®ÔÚÒ»¸öÈÝ»ý¹Ì¶¨µÄÃܱÕÈÝÆ÷ÖУ¬·¢Éú·´Ó¦£ºCO£¨g£©+2H2£¨g£©?CH3OH£¨g£©¡÷H£¼0£®µÚ2minʱֻ¸Ä±äÒ»¸öÌõ¼þ£¬·´Ó¦Çé¿öÈç±í£ºÏÂÁÐ˵·¨²»ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
ʱ¼äc£¨CO£©/mol•L-1c£¨H2£©/mol•L-1c£¨CH3OH£©/mol•L-1
Æðʼ130
µÚ2min0.82.60.2
µÚ4min0.41.80.6
µÚ6min0.41.80.6
A£®µÚ4 minÖÁµÚ6 min¸Ã»¯Ñ§·´Ó¦´¦ÓÚƽºâ״̬
B£®µÚ2 minʱ£¬Èç¹ûÖ»¸Ä±äijһÌõ¼þ£¬Ôò¸Ä±äµÄÌõ¼þ¿ÉÄܼÓÈëÁËH2
C£®µÚ2 minʱ£¬Èç¹ûÖ»¸Ä±äijһÌõ¼þ£¬Ôò¸Ä±äµÄÌõ¼þ¿ÉÄÜÊÇʹÓô߻¯¼Á
D£®µÚ6 minʱ£¬ÆäËûÌõ¼þ²»±ä£¬Èç¹ûÉý¸ßζȣ¬Õý·´Ó¦ËÙÂÊÔö´ó

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

17£®ÊÀ²©Ô°µØÇø¸ÄÔìÇ°£¬¹æ»®ÇøÄÚÓÐÒ»×ù´óÐ͸ÖÌú³§£¬¸½½ü¾ÓÃñÔø±¥Êܵ½¸Ã³§²úÉúµÄ×غìÉ«ÑÌÎíµÄÀ§ÈÅ£®¹À¼ÆÕâÒ»¿ÕÆøÎÛȾÎï¿ÉÄܺ¬ÓУ¨¡¡¡¡£©
A£®FeO·Û³¾B£®Fe3O4·Û³¾C£®Fe2O3·Û³¾D£®NO2ÆøÌå

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸