Ïû¶¾¼ÁÔÚÉú²úÉú»îÖÐÓм«ÆäÖØÒªµÄ×÷Ó㬿ª·¢¾ßÓйãÆÕ¡¢¸ßЧ¡¢µÍ¶¾µÄɱ¾ú¼ÁºÍÏû¶¾¼ÁÊǽñºó·¢Õ¹µÄÇ÷ÊÆ¡£
(1)Cl2¡¢H2O2¡¢ClO2(»¹Ô­²úÎïΪCl£­)¡¢O3(1 mol O3ת»¯Îª1 mol O2ºÍ1 mol H2O)µÈÎïÖʳ£±»ÓÃ×÷Ïû¶¾¼Á¡£µÈÎïÖʵÄÁ¿µÄÉÏÊöÎïÖÊÏû¶¾Ð§ÂÊ×î¸ßµÄÊÇ________(ÌîÐòºÅ)¡£

A£®Cl2B£®H2O2C£®ClO2D£®O3
(2)H2O2ÓÐʱ¿É×÷Ϊ¿óÒµ·ÏÒºÏû¶¾¼Á£¬ÓС°ÂÌÉ«Ñõ»¯¼Á¡±µÄÃÀ³Æ¡£ÈçÏû³ý²É¿óÒµ½ºÒºÖеÄÇ軯Îï(ÈçKCN)£¬¾­ÒÔÏ·´Ó¦ÊµÏÖ£ºKCN£«H2O2£«H2O=A£«NH3¡ü£¬ÔòÉú³ÉÎïAµÄ»¯Ñ§Ê½Îª________£¬H2O2±»³ÆΪ¡°ÂÌÉ«Ñõ»¯¼Á¡±µÄÀíÓÉÊÇ____________________________________
(3)Ư°×¼ÁÑÇÂÈËáÄÆ(NaClO2)ÔÚ³£ÎÂÓëºÚ°µ´¦¿É±£´æÒ»Äê¡£ÑÇÂÈËá²»Îȶ¨¿É·Ö½â£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪHClO2¨D¡úClO2¡ü£«H£«£«Cl£­£«H2O(δÅäƽ)¡£Ôڸ÷´Ó¦ÖУ¬µ±ÓÐ1 mol ClO2Éú³ÉʱתÒƵĵç×Ó¸öÊýԼΪ________¡£
(4)¡°84¡±Ïû¶¾Òº(Ö÷Òª³É·ÖÊÇNaClO)ºÍ½à²Þ¼Á(Ö÷Òª³É·ÖÊÇŨÑÎËá)²»ÄÜ»ìÓã¬Ô­ÒòÊÇ__________________(ÓÃÀë×Ó·½³Ìʽ±íʾ)¡£ÀûÓÃÂȼҵµÄ²úÎï¿ÉÒÔÉú²ú¡°84¡±Ïû¶¾Òº£¬Ð´³öÓйط´Ó¦µÄ»¯Ñ§·½³Ìʽ£º____________¡£

(1)C
(2)KHCO3¡¡H2O2ÊÇÑõ»¯¼Á£¬Æä²úÎïÊÇH2O£¬Ã»ÓÐÎÛȾ
(3)6.02¡Á1023
(4)ClO£­£«Cl£­£«2H£«=Cl2¡ü£«H2O¡¡Cl2£«2NaOH=NaClO£«NaCl£«H2O

½âÎö

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ijǿËáÐÔÈÜÒºX¿ÉÄܺ¬ÓÐBa2£«¡¢Al3£«¡¢NH¡¢Fe2£«¡¢Fe3£«¡¢CO¡¢SO¡¢SO¡¢Cl£­¡¢NOÖеÄÒ»ÖÖ»ò¼¸ÖÖ£¬È¡¸ÃÈÜÒº½øÐÐÁ¬ÐøʵÑ飬ʵÑé¹ý³ÌÈçÏ£º

¸ù¾ÝÒÔÉÏÐÅÏ¢£¬»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÉÏÊöÀë×ÓÖУ¬ÈÜÒºXÖгýH+Í⻹¿Ï¶¨º¬ÓеÄÀë×ÓÊÇ______£¬²»ÄÜÈ·¶¨ÊÇ·ñº¬ÓеÄÀë×Ó£¨M£©ÊÇ______£¬ÈôҪȷ¶¨¸ÃM£¨Èô²»Ö¹Ò»ÖÖ£¬¿ÉÈÎÑ¡Ò»ÖÖ£©ÔÚÈÜÒºXÖв»´æÔÚ£¬×î¿É¿¿µÄ»¯Ñ§·½·¨ÊÇ______£®
£¨2£©Ð´³ö·´Ó¦¢ÚµÄÀë×Ó·½³Ìʽ£º______£®
£¨3£©Í¨³£¿ÉÒÔÀûÓÃKClOÔÚÒ»¶¨Ìõ¼þÏÂÑõ»¯GÀ´ÖƱ¸Ò»ÖÖÐÂÐÍ¡¢¸ßЧ¡¢¶à¹¦ÄÜË®´¦Àí¼ÁK2FeO4£®Çëд³öÖƱ¸¹ý³ÌÖеÄÀë×Ó·½³Ìʽ______£®
£¨4£©¼ÙÉè²â¶¨A¡¢F¡¢I¾ùΪ0£®1mol£¬10mL XÈÜÒºÖÐn£¨H+£©=0£®4mol£¬µ±³ÁµíCÎïÖʵÄÁ¿´óÓÚ0£®7molʱ£¬ÈÜÒºXÖл¹Ò»¶¨º¬ÓÐ______£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

(16·Ö)ʵÑéÊÒÖÐÒÔ´ÖÍ­£¨º¬ÔÓÖÊ£©ÎªÔ­ÁÏ£¬Ä³ÖÖÖƱ¸Í­µÄÂÈ»¯ÎïµÄÁ÷³ÌÈçÏ£º

°´ÒªÇó»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©²Ù×÷¢ÙµÄËùÓõ½µÄ²£Á§ÒÇÆ÷ÓÐ______________¡£
£¨2£©ÉÏÊöÁ÷³ÌÖУ¬ËùµÃ¹ÌÌå1ÐèÒª¼ÓÏ¡ÑÎËáÈܽ⣬ÆäÀíÓÉÊÇ                       £»
ÈÜÒº1¿É¼ÓÊÔ¼ÁXÓÃÓÚµ÷½ÚpHÒÔ³ýÈ¥ÔÓÖÊ£¬ X¿ÉÑ¡ÓÃÏÂÁÐÊÔ¼ÁÖеģ¨ÌîÐòºÅ£©___________¡£
A£®NaOH     B£®NH3¡¤H2O       C£®CuO      D£®CuSO4   
£¨3£©·´Ó¦¢ÚÊÇÏòÈÜÒº2ÖÐͨÈëÒ»¶¨Á¿µÄSO2£¬¼ÓÈÈÒ»¶Îʱ¼äºóÉú³ÉCuCl°×É«³Áµí¡£Ð´³ö   
ÖƱ¸CuClµÄÀë×Ó·½³Ìʽ£º                                                   ¡£
£¨4£©ÏÖÓÃÈçͼËùʾµÄʵÑéÒÇÆ÷¼°Ò©Æ·À´ÖƱ¸´¿¾»¡¢¸ÉÔïµÄÂÈÆø²¢Óë´ÖÍ­·´Ó¦£¨Ìú¼Ų̈¡¢Ìú¼ÐÊ¡ÂÔ£©¡£

¢Ù°´ÆøÁ÷·½ÏòÁ¬½Ó¸÷ÒÇÆ÷½Ó¿Ú˳ÐòÊÇ£ºa¡ú  ¡¢¡¡¡¡¡ú¡¡¡¡¡¢¡¡¡¡¡ú¡¡¡¡¡¢¡¡¡¡¡ú¡¡¡¡¡¡¡£ÊµÑéÖдóÊԹܼÓÈÈÇ°Òª½øÐÐÒ»²½ÖØÒª²Ù×÷£¬Æä²Ù×÷ÊÇ                          ¡£
¢Ú·´Ó¦Ê±£¬Ê¢´ÖÍ­·ÛµÄÊÔ¹ÜÖеÄÏÖÏóÊÇ                              ¡£
£¨5£©ÔÚÈÜÒº2ת»¯ÎªCuCl2¡¤2H2OµÄ²Ù×÷¹ý³ÌÖУ¬·¢ÏÖÈÜÒºÑÕÉ«ÓÉÀ¶É«±äΪÂÌÉ«¡£Ð¡×éͬѧÓû̽¾¿ÆäÔ­Òò¡£ÒÑÖª£ºÂÈ»¯Í­ÈÜÒºÖÐÓÐÈçÏÂת»¯¹Øϵ£º
Cu(H2O)42+(aq) + 4Cl-(aq)  CuCl42-(aq) + 4 H2O (l)
À¶É«                      ÂÌÉ«
¸ÃͬѧȡÂÈ»¯Í­¾§ÌåÅä³ÉÀ¶ÂÌÉ«ÈÜÒºY£¬½øÐÐÈçÏÂʵÑ飬ÆäÖÐÄÜÖ¤Ã÷ÈÜÒºÖÐÓÐÉÏÊöת»¯¹ØϵµÄÊÇ             £¨ÌîÐòºÅ£©¡£
A£®½«YÏ¡ÊÍ£¬·¢ÏÖÈÜÒº³ÊÀ¶É«            B£®ÔÚYÖмÓÈëCuCl2¾§Ì壬ÈÜÒº±äΪÂÌÉ«
C£®ÔÚYÖмÓÈëNaCl¹ÌÌ壬ÈÜÒº±äΪÂÌÉ«  D£®È¡Y½øÐеç½â£¬ÈÜÒºÑÕÉ«×îÖÕÏûʧ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

Àë×Ó·´Ó¦ÊÇÖÐѧ»¯Ñ§ÖÐÖØÒªµÄ·´Ó¦ÀàÐÍ¡£»Ø´ðÏÂÁÐÎÊÌ⣺
(1)ÔÚ·¢ÉúÀë×Ó·´Ó¦µÄ·´Ó¦Îï»òÉú³ÉÎïÖУ¬Ò»¶¨´æÔÚÓР      (ÌîÐòºÅ)¡£
¢Ùµ¥ÖÊ£»¢ÚÑõ»¯Î¢Ûµç½âÖÊ£»¢ÜÑΣ»¢Ý»¯ºÏÎï
(2)¿ÉÓÃͼʾµÄ·½·¨±íʾ²»Í¬·´Ó¦ÀàÐÍÖ®¼äµÄ¹Øϵ¡£Èç·Ö½â·´Ó¦ºÍÑõ»¯»¹Ô­·´Ó¦¿É±íʾΪÏÂͼ¡£ÇëÔÚÏÂÃæµÄ·½¿òÖл­³öÀë×Ó·´Ó¦¡¢Öû»·´Ó¦ºÍÑõ»¯»¹Ô­·´Ó¦ÈýÕßÖ®¼äµÄ¹Øϵ¡£


(3)Àë×Ó·½³ÌʽÊÇÖØÒªµÄ»¯Ñ§ÓÃÓï¡£ÏÂÁÐÊÇÓйØÀë×Ó·½³ÌʽµÄһЩ´íÎó¹Ûµã£¬ÇëÔÚÏÂÁбí¸ñÖÐÓÃÏàÓ¦µÄ¡°Àë×Ó·½³Ìʽ¡±·ñ¶¨ÕâЩ¹Ûµã¡£

¢ÙËùÓеÄÀë×Ó·½³Ìʽ¾ù¿ÉÒÔ±íʾһÀà·´Ó¦
 
¢ÚËá¼îÖкͷ´Ó¦¾ù¿É±íʾΪH£«£«OH£­=H2O
 
¢ÛÀë×Ó·½³ÌʽÖз²ÊÇÄÑÈÜÐÔËá¡¢¼î¡¢ÑξùÒª±ê¡°¡ý¡±·ûºÅ
 
 
(4)ÊÔÁоٳöÈýÖÖ²»Í¬Àà±ðµÄÎïÖÊ(Ëá¡¢¼î¡¢ÑÎ)Ö®¼äµÄ·´Ó¦£¬ËüÃǶÔÓ¦µÄÀë×Ó·½³Ìʽ¶¼¿ÉÓá°Ba2£«£«SO42£­=BaSO4¡ý¡±À´±íʾ£¬Çëд³öÓйط´Ó¦µÄ»¯Ñ§·½³Ìʽ(3¸ö)£º
¢Ù                                                             £»
¢Ú                                                             £»
¢Û                                                            ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ÓÐÒ»ÖÖ°×É«·ÛÄ©£¬º¬ÓÐÏÂÁÐÒõÀë×ÓºÍÑôÀë×ÓÖеļ¸ÖÖ¡£
ÒõÀë×Ó£ºS2£­¡¢Cl£­¡¢NO¡¢¡¢¡¢¡¢¡£
ÑôÀë×Ó£ºNa£«¡¢Mg2£«¡¢Al3£«¡¢Ba2£«¡¢Fe2£«¡¢Fe3£«¡¢Cu2£«¡¢¡£
½«¸Ã°×É«·ÛÄ©½øÐÐÏÂÁÐʵÑ飬¹Û²ìµ½µÄÏÖÏóÈçÏ£º

ʵÑé²Ù×÷
ÏÖÏó
a.È¡ÉÙÁ¿·ÛÄ©£¬¼ÓË®¡¢Õñµ´
È«²¿Èܽ⡢
ÈÜÒºÎÞɫ͸Ã÷
 
b.ÏòËùµÃÈÜÒºÖÐÂýÂýµÎÈë¿ÁÐÔÄÆÈÜÒº£¬²¢¼ÓÈÈ
ÎÞÃ÷ÏÔÏÖÏó
c.È¡ÉÙÁ¿·ÛÄ©£¬¼ÓÑÎËá
ÎÞÃ÷ÏÔÏÖÏó
d.È¡ÉÙÁ¿·ÛÄ©£¬¼ÓÏ¡H2SO4ºÍÏ¡HNO3µÄ»ìºÏÒº
Óа×É«³ÁµíÉú³É
¸ù¾ÝʵÑéÍƶϣº
(1)´ÓaʵÑéÖУ¬¿ÉÍƶϷÛÄ©Öв»¿ÉÄÜÓР            (ÌîÀë×Ó·ûºÅ£¬ÏÂͬ)¡£
(2)´ÓbʵÑéÖУ¬¿ÉÍƶϷÛÄ©Öв»¿ÉÄÜÓР                        ¡£
(3)´ÓcʵÑéÖУ¬¿ÉÍƶϷÛÄ©Öв»¿ÉÄÜÓР                        ¡£
(4)´ÓdʵÑéÖУ¬¿ÉÍƶϷÛÄ©Öв»¿ÉÄÜÓР      £¬Ò»¶¨º¬ÓР      ¡£
(5)ÒÔÉϸ÷ʵÑéÈÔÎÞ·¨È·¶¨ÊÇ·ñ´æÔÚµÄÀë×ÓÊÇ           ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ij¹¤³§ÅųöµÄÎÛË®Öк¬ÓдóÁ¿µÄFe2£«¡¢Zn2£«¡¢Hg2£«ÈýÖÖ½ðÊôÀë×Ó¡£ÒÔÏÂÊÇij»¯Ñ§Ñо¿ÐÔѧϰС×éµÄͬѧÉè¼Æ³ýÈ¥ÎÛË®ÖеĽðÊôÀë×Ó£¬²¢»ØÊÕÂÌ·¯¡¢ð©·¯(ZnSO4¡¤7H2O)ºÍ¹¯µÄ·½°¸¡£
¡¾Ò©Æ·¡¿NaOHÈÜÒº¡¢Áò»¯ÄÆÈÜÒº¡¢Áò»¯ÑÇÌú¡¢Ï¡ÁòËá¡¢Ìú·Û
¡¾ÊµÑé·½°¸¡¿

¡¾ÎÊÌâ̽¾¿¡¿
(1)²½Öè¢òËù·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ                                ¡£
(2)²½Öè¢óÖгéÂ˵ÄÄ¿µÄÊÇ           £¬¸Ã²½Öè²úÉúFe(OH)3µÄ·´Ó¦µÄÀë×Ó·½³ÌʽΪ                                         ¡£
(3)²½Öè¢öÖеõ½ÁòËáпÈÜÒºµÄÀë×Ó·½³ÌʽΪ                                                               ¡£
(4)ÓûʵÏÖ²½Öè¢õ£¬Ðè¼ÓÈëµÄÊÔ¼ÁÓР      ¡¢       £¬ËùÉæ¼°µÄÖ÷Òª²Ù×÷ÒÀ´ÎΪ                                                         ¡£
(5)²½Öè¢ô³£Óõķ½·¨ÊÇ       £¬¸Ã²½ÖèÊÇ·ñ¶Ô»·¾³ÓÐÓ°Ï죿       (Ìî¡°ÊÇ¡±»ò¡°·ñ¡±)£¬ÈçÓÐÓ°Ï죬ÇëÄãÉè¼ÆÒ»¸öÂÌÉ«»·±£·½°¸À´ÊµÏÖ²½Öè¢ôµÄ·´Ó¦£º                                                            ¡£
(6)¸ÃÑо¿Ð¡×éµÄͬѧÔÚÇ¿¼îÈÜÒºÖУ¬ÓôÎÂÈËáÄÆÓëFe(OH)3·´Ó¦»ñµÃÁ˸ßЧ¾»Ë®¼ÁNa2FeO4£¬¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽΪ                             ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

¸ù¾ÝÒªÇóÌî¿Õ¡£
£¨1£©ÏÂÁÐÎïÖÊÖУ¬ÊôÓÚµç½âÖʵÄÊÇ            £¬ÊôÓڷǵç½âÖʵÄÊÇ        ¡££¨ÌîÐòºÅ£©
¢Ù̼ËáÇâÄÆ     ¢ÚCO2       ¢ÛÒÒ´¼     ¢ÜÌú
£¨2£©ÒûÓÃË®ÖеÄNO3-¶ÔÈËÀཡ¿µ»á²úÉúΣº¦£¬ÎªÁ˽µµÍÒûÓÃË®ÖÐNO3-µÄŨ¶È£¬¿ÉÒÔÔÚ¼îÐÔÌõ¼þÏÂÓÃÂÁ·Û½«NO3-»¹Ô­ÎªN2£¬Æ仯ѧ·½³ÌʽΪÈçÏ£º
10 Al £« 6 NaNO3 £« 4 NaOH £«18 H2O =" 10" NaAl£¨OH£©4 £« 3 N2¡ü
ÇëÓá°µ¥ÏßÇÅ·¨¡±±íʾÉÏÊö·´Ó¦Öеç×ÓתÒƵķ½ÏòºÍÊýÄ¿¡£
£¨3£©ÒÑÖª°±ÆøÓëÂÈÆøÔÚ³£ÎÂÌõ¼þÏ·¢ÉúÈçÏ·´Ó¦£º8 NH3£«3 Cl2£½ 6 NH4Cl £«N2£¬¸Ã·´Ó¦Öл¹Ô­¼ÁÓëÑõ»¯¼ÁµÄÎïÖʵÄÁ¿Ö®±ÈΪ             ¡£
£¨4£©½ñÓÐK2SO4ºÍAl2£¨SO4£©3»ìºÏÈÜÒº£¬ÒÑÖªÆäÖÐc£¨K+£©=" 0.2" mol¡¤L£­1£¬c£¨SO42£­£©= 0.7mol¡¤L£­1¡£ÔòÈÜÒºÖÐc£¨Al3+£©=          mol¡¤L£­1¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ÌúÊÇÓ¦ÓÃ×î¹ã·ºµÄ½ðÊô£¬ÌúµÄ±»¯Îï¡¢Ñõ»¯ÎïÒÔ¼°¸ß¼ÛÌúµÄº¬ÑõËáÑξùΪÖØÒª»¯ºÏÎï¡£
(1)Ҫȷ¶¨ÌúµÄijÂÈ»¯ÎïFeClxµÄ»¯Ñ§Ê½£¬¿ÉÓÃÀë×Ó½»»»ºÍµÎ¶¨µÄ·½·¨¡£ÊµÑéÖгÆÈ¡0.54 gµÄFeClxÑùÆ·£¬ÈܽâºóÏȽøÐÐÑôÀë×Ó½»»»Ô¤´¦Àí£¬ÔÙͨ¹ýº¬Óб¥ºÍOH£­µÄÒõÀë×Ó½»»»Öù£¬Ê¹Cl£­ºÍOH£­·¢Éú½»»»¡£½»»»Íê³Éºó£¬Á÷³öÈÜÒºµÄOH£­ÓÃ0.40 mol¡¤L£­1µÄÑÎËáµÎ¶¨£¬µÎÖÁÖÕµãʱÏûºÄÑÎËá25.0 mL¡£¼ÆËã¸ÃÑùÆ·ÖÐÂȵÄÎïÖʵÄÁ¿£¬²¢Çó³öFeClxÖÐxÖµ£º________(Áгö¼ÆËã¹ý³Ì)£»
(2)ÏÖÓÐÒ»º¬ÓÐFeCl2ºÍFeCl3µÄ»ìºÏÎïÑùÆ·£¬²ÉÓÃÉÏÊö·½·¨²âµÃn(Fe)£ºn(Cl)£½1£º2.1£¬Ôò¸ÃÑùÆ·ÖÐFeCl3µÄÎïÖʵÄÁ¿·ÖÊýΪ________¡£ÔÚʵÑéÊÒÖУ¬FeCl2¿ÉÓÃÌú·ÛºÍ________·´Ó¦ÖƱ¸£¬FeCl3¿ÉÓÃÌú·ÛºÍ________·´Ó¦ÖƱ¸£»
(3)FeCl3ÓëÇâµâËᷴӦʱ¿ÉÉú³É×ØÉ«ÎïÖÊ£¬¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽΪ______£»
(4)¸ßÌúËá¼Ø(K2FeO4)ÊÇÒ»ÖÖÇ¿Ñõ»¯¼Á£¬¿É×÷Ϊˮ´¦Àí¼ÁºÍ¸ßÈÝÁ¿µç³Ø²ÄÁÏ¡£FeCl3ÓëKClOÔÚÇ¿¼îÐÔÌõ¼þÏ·´Ó¦¿ÉÖÆÈ¡K2FeO4£¬Æä·´Ó¦µÄÀë×Ó·½³ÌʽΪ________£»ÓëMnO2£­Znµç³ØÀàËÆ£¬K2FeO4£­ZnÒ²¿ÉÒÔ×é³É¼îÐÔµç³Ø£¬K2FeO4ÔÚµç³ØÖÐ×÷ΪÕý¼«²ÄÁÏ£¬Æäµç¼«·´Ó¦Ê½Îª________£¬¸Ãµç³Ø×Ü·´Ó¦µÄÀë×Ó·½³ÌʽΪ________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

¹¤ÒµÉÏÓ÷ÛËéµÄúí·Ê¯(Ö÷Òªº¬Al2O3¡¢SiO2¼°ÌúµÄÑõ»¯Îï)ÖƱ¸¾»Ë®¼ÁBAC[Al2(OH)nCl6£­n]µÄÁ÷³ÌÈçÏ£º
 
(1)·ÛËéúí·Ê¯µÄÄ¿µÄÊÇ__________________________________________________£»
ÂËÔü¢ñµÄÖ÷Òª³É·ÖÊÇ________(Ìѧʽ)¡£
(2)²½Öè¢ÙÔÚÖó·ÐµÄ¹ý³ÌÖУ¬ÈÜÒºÖð½¥ÓÉÎÞÉ«±äΪÂÌÉ«£¬´ËʱÈÜÒºÖеÄÓÐÉ«Àë×ÓΪ_____
________(Ìѧʽ)£»ËæºóÈÜÒºÓÖ±äΪ×Ø»ÆÉ«£¬Ïà¹Ø·´Ó¦µÄÀë×Ó·½³ÌʽΪ_________£»²½Öè¢ÙµÄÖó·Ð×°ÖÃÉÏ·½Ðè°²×°Ò»³¤µ¼¹Ü£¬³¤µ¼¹ÜµÄ×÷ÓÃÊÇ__________________________¡£
(3)²½Öè¢ÚÖмÓÈëÊÊÁ¿µÄCa(OH)2²¢¿ØÖÆpH£¬ÆäÄ¿µÄ£ºÒ»ÊÇÉú³ÉBAC£»¶þÊÇ________
_______________________£»ÒÑÖªBACµÄ·ÖÉ¢ÖÊÁ£×Ó´óСÔÚ1¡«100 nmÖ®¼ä£¬ÓÉ´ËÇø±ðÂËÒº¢ñÓëBACÁ½ÖÖÒºÌåµÄÎïÀí·½·¨ÊÇ________________________£»ÈôCa(OH)2ÈÜÒº¹ýÁ¿£¬Ôò²½Öè¢ÛµÃµ½µÄBAC²úÂÊÆ«µÍ£¬Ð´³ö¸Ã·´Ó¦µÄÀë×Ó·½³Ìʽ£º_______________________
(4)Èô0.1 mol AlCl3ÔÚijζÈÏÂÈÜÓÚÕôÁóË®£¬µ±ÓÐ5%Ë®½âÉú³ÉAl(OH)3ÈÜҺʱ£¬ÎüÊÕÈÈÁ¿a kJ¡£Ð´³ö¸Ã¹ý³ÌµÄÈÈ»¯Ñ§·½³Ìʽ£º____________________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸