ʵÑéÊÒÀûÓÃÁòË᳧ÉÕÔü(Ö÷Òª³É·ÖΪÌúµÄÑõ»¯Îï¼°ÉÙÁ¿FeS¡¢SiO2µÈ)ÖƱ¸¾ÛÌú(¼îʽÁòËáÌúµÄ¾ÛºÏÎï)ºÍÂÌ·¯(FeSO4¡¤7H2O)£¬¹ý³ÌÈçÏ£º

(1)½«¹ý³Ì¢ÚÖвúÉúµÄÆøÌåͨÈëÏÂÁÐÈÜÒºÖУ¬ÈÜÒº»áÍÊÉ«µÄÊÇ________¡£
A£®Æ·ºìÈÜÒº¡¡¡¡¡¡  B£®×ÏɫʯÈïÈÜÒº
C£®ËáÐÔKMnO4ÈÜÒº¡¡¡¡¡¡  D£®äåË®
(2)¹ý³Ì¢ÙÖУ¬FeSºÍO2¡¢H2SO4·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ______________________________
(3)¹ý³Ì¢ÛÖУ¬Ðè¼ÓÈëµÄÎïÖÊÊÇ________¡£
(4)¹ý³Ì¢ÜÖУ¬Õô·¢½á¾§ÐèҪʹÓþƾ«µÆ¡¢Èý½Å¼Ü¡¢ÄàÈý½Ç£¬»¹ÐèÒªµÄÒÇÆ÷ÓÐ________________¡£
(5)¹ý³Ì¢Ýµ÷½ÚpH¿ÉÑ¡ÓÃÏÂÁÐÊÔ¼ÁÖеÄ________(ÌîÑ¡ÏîÐòºÅ)¡£
A£®Ï¡ÁòËá¡¡¡¡B£®CaCO3¡¡¡¡C£®NaOHÈÜÒº
(6)¹ý³Ì¢ÞÖУ¬½«ÈÜÒºZ¼ÓÈȵ½70¡«80 ¡æ£¬Ä¿µÄÊÇ______________________¡£

(1)ACD
(2)4FeS£«3O2£«6H2SO4=2Fe2(SO4)3£«6H2O£«4S
(3)Fe(»òÌú)¡¡(4)Õô·¢Ãó¡¢²£Á§°ô¡¡(5)B
(6)´Ù½øFe3£«µÄË®½â

½âÎö

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

(14·Ö)ÓÐÈýÖÖ½ðÊôµ¥ÖÊA¡¢B¡¢C£¬ÆäÖÐAµÄÑæÉ«·´Ó¦Îª»ÆÉ«£¬B¡¢CÊdz£¼û½ðÊô¡£ÈýÖÖ½ðÊôµ¥ÖÊA¡¢B¡¢CÄÜÓëÆøÌå¼×¡¢ÒÒ¡¢±û¼°ÎïÖÊD¡¢E¡¢F¡¢G¡¢HÖ®¼ä·¢ÉúÈçÏÂת»¯¹Øϵ£¨Í¼ÖÐÓÐЩ·´Ó¦µÄ²úÎïºÍ·´Ó¦µÄÌõ¼þûÓбê³ö£©¡£

Çë¸ù¾ÝÒÔÉÏÐÅÏ¢»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Ð´³öÏÂÁÐÎïÖʵĻ¯Ñ§Ê½£º
A_____________£»H ______________£» G___________£»ÒÒ____________¡£
£¨2£©Ð´³öÏÂÁз´Ó¦µÄ»¯Ñ§·½³Ìʽ£º
·´Ó¦¢Ù________________________________£»·´Ó¦¢Ú_________________________________¡£
£¨3£© ·´Ó¦¢ÚÖеç×ÓתÒƵÄÊýĿΪ6.02¡Á1023¸öʱÏûºÄµÄÆøÌåÒÒµÄÎïÖʵÄÁ¿Îª        mol

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ʵÑéÌâ(14·Ö)£º¢ñ¡¢HCOOHÊÇÒ»ÖÖÎÞÉ«¡¢Ò×»Ó·¢µÄÒºÌ壬ijѧϰС×é²ÎÕÕ£ºÔ­Àí£¬ÓÃÏÂÁÐÒÇÆ÷£¨¼ÓÈȼ°¹Ì¶¨×°ÖÃÊ¡ÂÔ£©ÖƱ¸¸ÉÔï¡¢´¿¾»µÄCO£¬²¢ÓÃCO»¹Ô­CuO·ÛÄ©¡£
£¨1£©ÈôËùÖÆÆøÌåÁ÷Ïò´Ó×óÏòÓÒʱ£¬ÉÏÊöÒÇÆ÷Á¬½ÓΪ£º A¡ú(   ) ¡ú(    )¡ú(    )¡ú(   ) ¡ú(   )
£¨2£©A×°ÖÃÖÐСÊԹܵÄ×÷Óã¨ÖÁÉÙ´ðÁ½Ìõ£©£º
¢Ù                               ¡£¢Ú                               ¡£
£¨3£©Ö¤Ã÷HCOOHÊÜÈȲúÎïÖÐÓÐCOµÄÏÖÏóΪ£º                            ¡£
£¨4£©±¾ÊµÑéÓÐ3´¦Óõ½¾Æ¾«µÆ£¬³ýA¡¢B´¦Í⣬»¹È±Ò»¸ö¾Æ¾«µÆ£¬Ó¦·ÅÔÚ       ´¦.
¢ò¡¢Ñ§Ï°Ð¡×é²éÔÄ×ÊÁÏÖª£º
CuµÄÑÕɫΪºìÉ«»ò×ϺìÉ«£¬¶øCu2OµÄÑÕɫҲΪºìÉ«»òשºìÉ«¡£¢Ú4CuO2 Cu2O + O2¡ü£»¢Û Cu2O +2H+
= Cu+Cu2+ + H2O £»Òò´Ë¶ÔCO³ä·Ö»¹Ô­CuOºóËùµÃºìÉ«¹ÌÌåÊÇ·ñº¬ÓÐCu2O½øÐÐÁËÈÏÕæµÄÑо¿£¬Ìá³öÏÂÁÐÉè¼Æ·½°¸£º
·½°¸¢Ù£ºÈ¡¸ÃºìÉ«¹ÌÌåÈÜÓÚ×ãÁ¿Ï¡ÏõËáÖУ¬¹Û²ìÈÜÒºÑÕÉ«µÄ±ä»¯¡£
·½°¸¢Ú£ºÈ¡¸ÃºìÉ«¹ÌÌåÈÜÓÚ×ãÁ¿Ï¡ÁòËáÖУ¬¹Û²ìÈÜÒºÑÕÉ«µÄ±ä»¯¡£
£¨5£©Ð´³öCu2OÓëÏ¡ÏõËá·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º                                 ¡£
£¨6£©ÇëÄãÆÀ¼Û·½°¸¢ÚµÄºÏÀíÐÔ£¬²¢¼òÊöÀíÓÉ£º·½°¸¢Ú£º            £¬ÀíÓÉ£º                        ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ÇпªµÄ½ðÊôNa±©Â¶ÔÚ¿ÕÆøÖУ¬Æä±ä»¯¹ý³ÌÈçÏ£º

£¨1£©·´Ó¦¢ñµÄ·´Ó¦¹ý³ÌÓëÄÜÁ¿±ä»¯µÄ¹ØϵÈçÏ£º

¢Ù ·´Ó¦¢ñ ÊÇ¡¡ ¡¡·´Ó¦£¨Ìî¡°·ÅÈÈ¡±»ò¡°ÎüÈÈ¡±£©£¬ÅжÏÒÀ¾ÝÊÇ¡¡¡¡¡¡¡£
¢Ú 1 mol Na(s)È«²¿Ñõ»¯³ÉNa2O(s)µÄÈÈ»¯Ñ§·½³ÌʽÊÇ¡¡¡¡¡¡¡£
£¨2£©·´Ó¦¢òÊÇNa2OÓëË®µÄ·´Ó¦£¬Æä²úÎïµÄµç×ÓʽÊÇ     ¡£
£¨3£©°×É«·ÛĩΪNa2CO3¡£½«ÆäÈÜÓÚË®ÅäÖÆΪ0.1 mol/L Na2CO3ÈÜÒº£¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ     £¨Ìî×Öĸ£©¡£

A£®Éý¸ßζȣ¬ÈÜÒºµÄpH½µµÍ
B£®c(OH£­)£­c (H£«)£½c (HCO3£­)£«2 c (H2CO3)
C£®¼ÓÈëÉÙÁ¿NaOH¹ÌÌ壬c (CO32¨D)Óëc (Na£«)¾ùÔö´ó
D£®c (Na£«) > c (CO32¨D) > c (HCO3¨D) > c(OH¨D) > c (H£«)
£¨4£© ÄƵç³ØµÄÑо¿¿ª·¢ÔÚÒ»¶¨³Ì¶ÈÉÏ¿É»ººÍÒòï®×ÊÔ´¶ÌȱÒý·¢µÄµç³Ø·¢Õ¹ÊÜÏÞÎÊÌâ¡£
¢Ù ÄƱÈï®»îÆã¬ÓÃÔ­×ӽṹ½âÊÍÔ­Òò_______¡£
¢ÚZEBRA µç³ØÊÇÒ»ÖÖÄƵç³Ø£¬×Ü·´Ó¦ÎªNiCl2 + 2Na  Ni + 2NaCl¡£ÆäÕý¼«·´Ó¦Ê½ÊÇ_____¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

´¿¼î£¨»¯Ñ§Ê½ÎªNa2CO3£©ÊÇÒ»ÖÖÖØÒªµÄ»¯¹¤Ô­ÁÏ¡£ÏÖ´ú»¯¹¤Éú²úÓÐÈýÖÖ¹¤ÒÕ£º
Ò»¡¢²¼À¼ÖƼ¡£ÒÔʳÑÎΪԭÁÏÖƼ¸Ã·¨·ÖÈý²½£º
¢ÙÓÃÂÈ»¯ÄÆÓëÁòËá·´Ó¦ÖÆÁòËáÄÆ£º2NaCl+H2SO4=Na2SO4+2HCl£»
¢ÚÓý¹Ì¿»¹Ô­ÁòËáÄƵÃÁò»¯ÄÆ£ºNa2SO4+4C=Na2S+4CO¡ü
¢ÛÓÃÁò»¯ÄÆÓëʯ»Òʯ·´Ó¦ÖÆ̼ËáÄÆ£ºNa2S+CaCO3=Na2CO3+CaS
¶þ¡¢°±¼î·¨¼´Ë÷¶ûάÖƼ¡£ÒÔʳÑΡ¢°±¡¢¶þÑõ»¯Ì¼ÎªÔ­ÁÏ£¬Æä·´Ó¦Ò²·ÖÈý²½½øÐУº
¢ÙNH3+CO2+H2O=NH4HCO3
¢ÚNH4HCO3+NaCl=NaHCO3+NH4Cl
¢Û2NaHCO3=Na2CO3+CO2¡ü+H2O
Èý¡¢ºòÊÏÖƼ¡£µÍÎÂÏÂÏÈÏò±¥ºÍʳÑÎË®ÖÐͨÈë°±Æø£¬ÔÙͨÈë¶þÑõ»¯Ì¼¿ÉÎö³ö̼ËáÇâÄÆ£¬ÔÙ¼ÓÈëϸÑÎÄ©£¬ÒòͬÀë×ÓЧӦ£¬µÍÎÂÂÈ»¯ï§Èܽâ¶ÈͻȻ½µµÍ£¬¶øʳÑεÄÈܽâ¶È±ä»¯²»´ó£¬ËùÒÔÂÈ»¯ï§Îö³ö¶øʳÑβ»Îö³ö£»ÔÙÓð±±¥ºÍºóͨ¶þÑõ»¯Ì¼£¬½á¹ûÍù·µÎö³öNaHCO3ºÍNH4Cl¡£¸Ã·¨Éú²úµÄ´¿¼îÖÊÁ¿ÓÅÁ¼£¬´¿°×ÈçÑ©¡£
£¨1£©Í¨¹ýÈýÖÖ·½·¨µÄ±È½Ï£¬²¼À¼ÖƼ¹¤ÒÕµÄȱµãÓР                                          (дÁ½µã)¡£
£¨2£©°±¼î·¨¹¤ÒÕµÄÖÐÑ­»·ÀûÓõÄÎïÖÊÊÇ        (Ìѧʽ)£»²úÆ·µÄ¸±²úÎïNH4Cl¼È¿ÉÒÔ×öµª·ÊÓÖ¿ÉÒÔÖØÐÂÉú³É°±Æø¡£Ð´³öNH4ClÓëÉúʯ»Ò·´Ó¦µÄ»¯Ñ§·½³Ìʽ          ¡£
£¨3£©ºòÊÏÖƼ·´Ó¦µÄ·½³ÌʽΪ                                           ¡£
£¨4£©ÎªÊ²Ã´ºòÊÏÖƼ¹¤ÒÕÖÐÏÈÏò±¥ºÍʳÑÎË®ÖÐͨÈë°±Æø£¬ÔÙͨÈë¶þÑõ»¯Ì¼¡£ÀíÓÉÊÇ                                                     (дÁ½µã)¡£
£¨5£©ºòÊÏÖƼ²úÆ·´¿¼îÖк¬ÓÐ̼ËáÇâÄÆ¡£Èç¹ûÓüÓÈÈ·Ö½âµÄ·½·¨²â¶¨´¿¼îÖÐ̼ËáÇâÄƵÄÖÊÁ¿·ÖÊý£¬ÓÃm1±íʾ¼ÓÈÈÇ°´¿¼îÑùÆ·µÄÖÊÁ¿£¬m2±íʾ¼ÓÈȺó¹ÌÌåµÄÖÊÁ¿¡£Ôò´¿¼îÖÐ̼ËáÇâÄƵÄÖÊÁ¿·ÖÊý¿É±íʾΪ£º                          ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

Ìú¡¢ÂÁ¡¢Í­µÈ½ðÊô¼°Æ仯ºÏÎïÔÚÈÕ³£Éú»îÖÐÓ¦Óù㷺£¬Çë¸ù¾ÝÏÂÁÐʵÑé»Ø´ðÎÊÌâ¡£
(1)ÉúÌúÖк¬ÓÐÒ»ÖÖÌú̼»¯ºÏÎïX(Fe3C)¡£XÔÚ×ãÁ¿µÄ¿ÕÆøÖиßÎÂìÑÉÕ£¬Éú³ÉÓдÅÐԵĹÌÌåY£¬½«YÈÜÓÚ¹ýÁ¿ÑÎËáºóÈÜÒºÖдóÁ¿´æÔÚµÄÑôÀë×ÓÊÇ             £»YÓë¹ýÁ¿Å¨ÏõËá·´Ó¦ºóÈÜÒºÖк¬ÓеÄÑεĻ¯Ñ§Ê½Îª                               ¡£
(2)ijÈÜÒºÖÐÓÐMg2£«¡¢Fe2£«¡¢Al3£«¡¢Cu2£«µÈÀë×Ó£¬ÏòÆäÖмÓÈë¹ýÁ¿µÄNaOHÈÜÒººó£¬¹ýÂË£¬½«ÂËÔü¸ßÎÂ×ÆÉÕ£¬²¢½«×ÆÉÕºóµÄ¹ÌÌåͶÈë¹ýÁ¿µÄÏ¡ÑÎËáÖУ¬ËùµÃÈÜÒºÓëÔ­ÈÜÒºÏà±È£¬ÈÜÒºÖдóÁ¿¼õÉÙµÄÑôÀë×ÓÊÇ       ¡£
A£®Mg2£«      B£®Fe2£«        C£®Al3£«       D£®Cu2£«
(3)Ñõ»¯ÌúÊÇÖØÒªµÄ¹¤ÒµÑÕÁÏ£¬Ó÷ÏÌúмÖƱ¸ËüµÄÁ÷³ÌÈçÏ£º

»Ø´ðÏÂÁÐÎÊÌ⣺
¢Ù²Ù×÷¢ñµÄÃû³ÆÊÇ       £»²Ù×÷¢òµÄÃû³ÆÊÇ       £»²Ù×÷¢òµÄ·½·¨Îª                               ¡£
¢ÚÇëд³öÉú³ÉFeCO3³ÁµíµÄÀë×Ó·½³Ìʽ£º                          ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

Ìú¼°ÌúµÄ»¯ºÏÎïÓ¦Óù㷺£¬ÈçFeCl3¿ÉÓÃ×÷´ß»¯¼Á¡¢Ó¡Ë¢µç·ͭ°å¸¯Ê´¼ÁºÍÍâÉËֹѪ¼ÁµÈ¡£
(1)д³öFeCl3ÈÜÒº¸¯Ê´Ó¡Ë¢µç·ͭ°åµÄÀë×Ó·½³Ìʽ_______________________
(2)Èô½«(1)Öеķ´Ó¦Éè¼Æ³ÉÔ­µç³Ø£¬Çë»­³öÔ­µç³ØµÄ×°ÖÃͼ£¬±ê³öÕý¡¢¸º¼«£¬²¢Ð´³öµç¼«·´Ó¦Ê½¡£

Õý¼«·´Ó¦_______________________________
¸º¼«·´Ó¦_______________________________

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

£¨1£©¸ßεç½â¼¼ÊõÄܸßЧʵÏÖCO2(g) + H2O(g) ="CO(g)" + H2(g) +O2(g) £¬¹¤×÷Ô­ÀíʾÒâͼÈçÏ£º

¢Ùµç¼«b·¢Éú       £¨Ìî¡°Ñõ»¯¡±»ò¡°»¹Ô­¡±£©·´Ó¦¡£
¢ÚCO2Ôڵ缫a·ÅµçµÄ·´Ó¦Ê½ÊÇ                              ¡£
£¨2£©¹¤ÒµÉÏÓÃij¿óÔü£¨º¬ÓÐCu2O¡¢Al2O3¡¢Fe2O3¡¢SiO2£©ÌáÈ¡Í­µÄ²Ù×÷Á÷³ÌÈçÏ£º

ÒÑÖª£º Cu2O + 2H=" Cu" + Cu2+ + H2O

³ÁµíÎï
Cu(OH)2
Al(OH)3
Fe(OH)3
Fe(OH)2
¿ªÊ¼³ÁµípH
5.4
4.0
1.1
5.8
³ÁµíÍêÈ«pH
6.7
5.2
3.2
8.8
 
¢Ù¹ÌÌå»ìºÏÎïAÖеijɷÖÊÇ            ¡£
¢Ú·´Ó¦¢ñÍê³Éºó£¬ÌúÔªËصĴæÔÚÐÎʽΪ           ¡££¨ÌîÀë×Ó·ûºÅ£©
Çëд³öÉú³É¸ÃÀë×ÓµÄÀë×Ó·½³Ìʽ                                       ¡£
¢ÛxµÄÊýÖµ·¶Î§ÊÇ3.2¡ÜpH£¼4.0£¬y¶ÔÓ¦µÄÊýÖµ·¶Î§ÊÇ             ¡£
¢ÜÏÂÁйØÓÚNaClOµ÷pHµÄ˵·¨ÕýÈ·µÄÊÇ        £¨ÌîÐòºÅ£©¡£
a£®¼ÓÈëNaClO¿ÉʹÈÜÒºµÄpH½µµÍ
b£®NaClOÄܵ÷½ÚpHµÄÖ÷ÒªÔ­ÒòÊÇÓÉÓÚ·¢Éú·´Ó¦ClO£­+ H+HClO£¬ClO£­ÏûºÄH+£¬´Ó¶ø´ïµ½µ÷½ÚpHµÄÄ¿µÄ
c£®NaClOÄܵ÷½ÚpHµÄÖ÷ÒªÔ­ÒòÊÇÓÉÓÚNaClOË®½âClO£­+ H2OHClO+OH£­£¬OH£­ÏûºÄH+ £¬´Ó¶ø´ïµ½µ÷½ÚpHµÄÄ¿µÄ
¢ÝʵÑéÊÒÅäÖÆÖÊÁ¿·ÖÊýΪ20.0%µÄCuSO4ÈÜÒº£¬ÅäÖƸÃÈÜÒºËùÐèµÄCuSO4¡¤5H2OÓëH2OµÄÖÊÁ¿Ö®±ÈΪ         ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

Ìú¼°ÌúµÄ»¯ºÏÎïÓ¦Óù㷺£¬ÈçFeCl3¿ÉÓÃ×÷´ß»¯¼Á¡¢Ó¡Ë¢µç·ͭ°å¸¯Ê´¼ÁºÍÍâÉËֹѪ¼ÁµÈ¡£
£¨1£©Ð´³öFeCl3ÈÜÒº¸¯Ê´Ó¡Ë¢µç·ͭ°åµÄÀë×Ó·½³Ìʽ£º                                    ¡£
£¨2£©Èô½«£¨1£©Öеķ´Ó¦Éè¼Æ³ÉÔ­µç³Ø£¬Çë»­³öÔ­µç³ØµÄ×°ÖÃͼ£¬±ê³öÕý¡¢
¸º¼«£¬²¢Ð´³öµç¼«·´Ó¦Ê½¡£
Õý¼«·´Ó¦£º                                                    ¡£
¸º¼«·´Ó¦£º                                                    ¡£
£¨3£©¸¯Ê´Í­°åºóµÄ»ìºÏÈÜÒºÖУ¬ÈôCu2£«¡¢Fe3£«ºÍFe2£«µÄŨ¶È¾ùΪ0.10 mol¡¤L£­1£¬Çë²ÎÕÕϱí¸ø³öµÄÊý¾ÝºÍÒ©Æ·£¬¼òÊö³ýÈ¥CuCl2ÈÜÒºÖÐFe3£«ºÍFe2£«µÄʵÑé²½Öè                                                      ¡£

 
ÇâÑõ»¯Î↑ʼ³ÁµíʱµÄpH
ÇâÑõ»¯Îï³Áµí
ÍêȫʱµÄpH
Fe3£«
Fe2£«
1.9
7.0
3.2
9.0
Cu2£«
4.7
6.7
ÌṩµÄÒ©Æ·£ºCl2¡¡Å¨H2SO4¡¡NaOHÈÜÒº¡¡CuO¡¡Cu
 
£¨4£©Ä³¿ÆÑÐÈËÔ±·¢ÏÖÁÓÖʲ»Ðâ¸ÖÔÚËáÖи¯Ê´»ºÂý£¬µ«ÔÚijЩÑÎÈÜÒºÖи¯Ê´ÏÖÏóÃ÷ÏÔ¡£Çë´ÓÉϱíÌṩµÄÒ©Æ·ÖÐÑ¡ÔñÁ½ÖÖ£¨Ë®¿ÉÈÎÑ¡£©£¬Éè¼Æ×î¼ÑʵÑ飬ÑéÖ¤ÁÓÖʲ»Ðâ¸ÖÒ×±»¸¯Ê´¡£Óйط´Ó¦µÄ»¯Ñ§·½³Ìʽ                            £»                              £»ÁÓÖʲ»Ðâ¸Ö¸¯Ê´µÄʵÑéÏÖÏó                                        ¡£

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸