ÂÁ·ÛºÍÑõ»¯ÌúµÄ»ìºÏÎï½Ð×öÂÁÈȼÁ£¬ÂÁÈȼÁ·¢ÉúÂÁÈÈ·´Ó¦Ê±ÓÐÕâÑùµÄʵÑéÏÖÏ󣺡°·´Ó¦·Å³ö´óÁ¿µÄÈÈ£¬²¢·¢³öÒ«Ñ۵Ĺâ⡱£¬¡°Ö½Â©¶·µÄϲ¿±»ÉÕ´©£¬ÓÐÈÛÈÚÎïÂäÈëɳÖС±£®ÒÑÖª£ºAl¡¢FeµÄÈ۵㡢·ÐµãÊý¾ÝÈçÏ£º
ÎïÖÊ Al Fe
È۵㣨¡æ£© 660 1535
·Ðµã£¨¡æ£© 2467 2750
£¨1£©Ä³Í¬Ñ§²Â²â£¬ÂÁÈÈ·´Ó¦ËùµÃµ½µÄÈÛÈÚÎïÊÇÌúÂÁºÏ½ð£®ÀíÓÉÊÇ£º¸Ã·´Ó¦·ÅÈÈÄÜʹÌúÈÛ»¯£¬¶øÂÁµÄÈÛµã±ÈÌúµÍ£¬ËùÒÔÌúºÍÂÁÄÜÐγɺϽð£®ÄãÈÏΪËûµÄ½âÊÍÊÇ·ñºÏÀí£¿
 
£¨Ìî¡°ºÏÀí¡±»ò¡°²»ºÏÀí¡±£©£®
Ìú¡¢ÂÁ¡¢ÌúÂÁºÏ½ðÈýÖÖÎïÖÊÖÐÓ²¶È×î´óµÄÊÇ
 

£¨2£©¸ù¾ÝÒÑÓÐ֪ʶÕÒ³öÒ»ÖÖÑéÖ¤²úÎïÖÐÓÐFeµÄ×î¼òµ¥·½·¨£º
 
£®
£¨3£©Éè¼ÆÒ»¸ö¼òµ¥µÄʵÑé·½°¸£¬Ö¤Ã÷ÉÏÊöËùµÃµÄÈÛÈÚÎïÖк¬ÓнðÊôÂÁ£®ÇëÌîд±í¸ñ£º
ËùÓÃÊÔ¼Á NaOHÈÜÒº
ʵÑé²Ù×÷¼°ÏÖÏó È¡ÉÙÐí
 
£¬µÎ¼Ó
 
£¬Õñµ´£¬ÓÐÆøÅÝ
Óйط´Ó¦µÄ»¯Ñ§·½³Ìʽ 2Al+2NaOH+2H2O¡ú2NaAlO2+3H2¡ü
¿¼µã£ºÌ½¾¿ÂÁÈÈ·´Ó¦
רÌ⣺¼¸ÖÖÖØÒªµÄ½ðÊô¼°Æ仯ºÏÎï
·ÖÎö£º£¨1£©ÂÁµÄÈÛµã±ÈÌúµÍ£¬Éú³ÉÌúºÍҺ̬ÂÁÒ»ÆðµÎÂäÖÁÊ¢ÓÐɳ×ÓµÄÈÝÆ÷ÖÐÐγɺϽ𣻸ù¾ÝºÏ½ðµÄÎïÀíÐÔÖʽøÐÐÅжϣ»
£¨2£©¿ÉÒÔÓôÅÌú¼ìÑé²úÎïÖÐÊÇ·ñº¬ÓÐÌú£»
£¨3£©½ðÊôÂÁÄܺÍÇâÑõ»¯ÄÆ·´Ó¦·Å³öÇâÆø£¬¶ø½ðÊôÌúºÍÇâÑõ»¯ÄƲ»·´Ó¦£¬¾Ý´ËÉè¼ÆʵÑé·½°¸¼ìÑéºÏ½ðÖеÄÂÁ£®
½â´ð£º ½â£º£¨1£©ÂÁµÄÈÛµã±ÈÌúµÍ£¬Éú³ÉÌúºÍҺ̬ÂÁÒ»ÆðµÎÂäÖÁÊ¢ÓÐɳ×ÓµÄÈÝÆ÷ÖÐÐγɺϽð£¬ËùÒÔÂÁÈÈ·´Ó¦ËùµÃµ½µÄÈÛÈÚÎïÓ¦ÊÇÌúÂÁºÏ½ð£»ºÏ½ðµÄÓ²¶ÈÒ»°ã±È¸÷³É·Ö½ðÊôµÄÓ²¶È´ó£¬ËùÒÔÌú¡¢ÂÁ¡¢ÌúÂÁºÏ½ðÈýÖÖÎïÖÊÖÐÓ²¶È×î´óµÄÊÇÌúÂÁºÏ½ð£¬
¹Ê´ð°¸Îª£ººÏÀí£»ÌúÂÁºÏ½ð£»
£¨2£©ÑéÖ¤²úÎïÖÐÓÐFe£¬×î¼òµ¥µÄ·½·¨Îª£ºÓôÅÌú£¬ÈôÄÜÎüÒýÔò¿ÉÑéÖ¤²úÎïÖÐÓÐFe£¬
¹Ê´ð°¸Îª£ºÓôÅÌú£¬ÈôÄÜÎüÒýÔò¿ÉÑéÖ¤²úÎïÖÐÓÐFe£»
£¨3£©½ðÊôÂÁÄܺÍÇâÑõ»¯ÄÆ·´Ó¦·Å³öÇâÆø£º2Al+2OH-+2H2O=2AlO2-+3H2¡ü£¬¶ø½ðÊôÌúºÍÇâÑõ»¯ÄƲ»·´Ó¦£¬¿ÉÒÔÓÃÇâÑõ»¯ÄÆÈÜÒºÖ¤Ã÷ÉÏÊöËùµÃµÄ¿é×´ÈÛÈÚÎïÖк¬ÓнðÊôÂÁ£¬²Ù×÷·½·¨Îª£ºÈ¡ÉÙÐíÈÛÈÚÎµÎ¼ÓÇâÑõ»¯ÄÆÈÜÒº£¬Õñµ´£¬ÓÐÆøÅݲúÉú£¬Ö¤Ã÷ÈÛÈÚÎïÖÐÓнðÊôÂÁ£¬
¹Ê´ð°¸Îª£ºÈÛÈÚÎNaOHÈÜÒº£®
µãÆÀ£º±¾Ì⿼²éÂÁÈÈ·´Ó¦Ô­Àí¡¢½ðÊôÂÁµÄ»¯Ñ§ÐÔÖÊ£¬ÌâÄ¿ÄѶÈÖеȣ¬×¢ÒâÕÆÎÕÂÁÈÈ·´Ó¦·¢ÉúÌõ¼þ£¬Ã÷È·ÂÁµÄ»¯Ñ§ÐÔÖʼ°ºÏ½ðµÄÎïÀíÐÔÖÊ£¬£¨1£©ÎªÒ×´íµã£¬×¢Òâ·ÖÎö±íÖÐÊý¾Ý£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¶ÔÓÚ·´Ó¦aA+bB=dD+eE£¬¸Ã»¯Ñ§·´Ó¦ËÙÂʶ¨ÒåΪv=
v(A)
a
=
v(B)
b
=
v(D)
d
=
v(E)
e
£®Ê½ÖÐv£¨X£©Ö¸ÎïÖÊX=£¨X=A£¬B£¬D£¬E£©µÄ·´Ó¦ËÙÂÊ£¬a¡¢b¡¢d¡¢eÊÇ»¯Ñ§¼ÆÁ¿Êý£®298kʱ£¬²âµÃÈÜÒºÖеķ´Ó¦H2O2+2HI¨T2H2O+I2ÔÚ²»Í¬Å¨¶Èʱ»¯Ñ§·´Ó¦ËÙÂÊv¼ûÏÂ±í£ºÒÔÏÂ˵·¨ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
ʵÑé±àºÅ1234
c£¨HI£©/mol?L-10.1000.2000.3000.100
c£¨H2O2£©/mol?L-10.1000.1000.1000.200
v/mol?L-1?s-10.007600.01530.02270.0151
A¡¢ÊµÑé1¡¢2ÖУ¬v£¨H2O2£©ÏàµÈ
B¡¢½«Å¨¶È¾ùΪ0.200mol?L-1 H2O2ºÍHIÈÜÒºµÈÌå»ý»ìºÏ£¬·´Ó¦¿ªÊ¼Ê±v=0.0304mol?L-1?s-1
C¡¢vÓë¡°HIºÍH2O2Ũ¶ÈµÄ³Ë»ý¡±µÄ±ÈֵΪ³£Êý
D¡¢ÊµÑé4£¬·´Ó¦5ÃëºóH2O2Ũ¶È¼õÉÙÁË0.0755mol?L-1

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

µç½âÂÁ¼¼ÊõµÄ³öÏÖÓë³ÉÊìÈÃÂÁ´Ó»Ê¼ÒÕäÆ·±ä³ÉÆû³µ¡¢ÂÖ´¬¡¢º½Ì캽¿ÕÖÆÔì¡¢»¯¹¤Éú²úµÈÐÐÒµµÄÖØÒª²ÄÁÏ£®¹¤ÒµÉÏÓÃÂÁÍÁ¿ó£¨Ö÷Òª³É·ÖΪAl2O3£¬Fe2O3¡¢SiO2µÈ£©ÌáÈ¡´¿Al2O3×öÒ±Á¶ÂÁµÄÔ­ÁÏ£¬Ä³Ñо¿ÐÔѧϰС×éÉè¼ÆÁËÈçÏÂÌáÈ¡Á÷³Ìͼ

£¨1£©¹ÌÌå¢òµÄ»¯Ñ§Ê½Îª_
 
£¬¸Ã¹ÌÌåµÄÑÕɫΪ
 
£»
£¨2£©ÔÚʵ¼Ê¹¤ÒµÉú²úÁ÷³Ì¢ÝÖÐÐè¼ÓÈë±ù¾§Ê¯£¬Ä¿µÄÊÇ
 
£»
£¨3£©Ð´³öÁ÷³Ì¢ÛµÄÀë×Ó·½³Ìʽ
 
£»
£¨4£©ÂÁ·ÛÓëÑõ»¯Ìú·ÛÄ©ÔÚÒýȼÌõ¼þϳ£ÓÃÀ´º¸½Ó¸Ö¹ì£¬Ö÷ÒªÊÇÀûÓø÷´Ó¦
 
£»
£¨5£©´ÓÂÁÍÁ¿óÖÐÌáÈ¡ÂÁµÄ¹ý³Ì²»Éæ¼°µÄ»¯Ñ§·´Ó¦ÀàÐÍÊÇ
 
£»
A£®¸´·Ö½â·´Ó¦     B£®Ñõ»¯»¹Ô­·´Ó¦      C£®Öû»·´Ó¦      D£®·Ö½â·´Ó¦
£¨6£©Éú²ú¹ý³ÌÖгýNaOH¡¢H2O¿ÉÒÔÑ­»·Ê¹ÓÃÍ⣬»¹¿ÉÒÔÑ­»·Ê¹ÓõÄÎïÖÊÓÐ_
 
_£»
£¨7£©ÈôÏòÂËÒº¢ñÖÐÖðµÎµÎÈëNaOHÈÜÒºÖÁ¹ýÁ¿£¬²úÉú³ÁµíËæNaOHµÎÈë¹ØϵÕýÈ·µÄÊÇ
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÔÂÇòº¬ÓÐH¡¢He¡¢N¡¢Na¡¢Mg¡¢SiµÈÔªËØ£¬ÊÇÈËÀàδÀ´µÄ×ÊÔ´±¦¿â£®
£¨1£©3HeÊǸßЧÄÜÔ­ÁÏ£¬ÆäÔ­×ÓºËÄÚµÄÖÐ×ÓÊýΪ
 

£¨2£©NaµÄÔ­×ӽṹʾÒâͼΪ
 
£¬NaÔÚÑõÆøÖÐÍêȫȼÉÕËùµÃ²úÎïµÄµç×ÓʽΪ
 

£¨3£©MgCl2ÔÚ¹¤ÒµÉÏÓ¦Óù㷺£¬¿ÉÓÉMgOÖƱ¸£®
¢ÙMgOµÄÈÛµã±ÈBaOµÄÈÛµã
 
£¨Ìî¡°¸ß¡±»ò¡°µÍ¡±£®£©
¢ÚÔÂÇòÉÏij¿óʯ¾­´¦ÀíµÃµ½µÄMgOÖк¬ÓÐÉÙÁ¿SiO2£¬³ýÈ¥SiO2µÄÀë×Ó·½³ÌʽΪ
 
£»SiO2µÄ¾§ÌåÀàÐÍΪ
 
£®
¢ÛMgOÓë̼·ÛºÍÂÈÆøÔÚÒ»¶¨Ìõ¼þÏ·´Ó¦¿ÉÖƱ¸MgCl2£®ÈôβÆø¿ÉÓÃ×ãÁ¿NaOHÈÜÒºÍêÈ«ÎüÊÕ£¬ÔòÉú³ÉµÄÑÎΪ
 
£¨Ð´»¯Ñ§Ê½£©£®
£¨4£©ÔÂÈÀÖк¬ÓзḻµÄ3He£¬´ÓÔÂÈÀÖÐÌáÁ¶1kg 3He£¬Í¬Ê±¿ÉµÃ6000kg H2ºÍ700kg N2£¬ÈôÒԵõ½H2ºÍN2ΪԭÁϾ­Ò»ÏµÁз´Ó¦×î¶à¿ÉÉú²ú̼ËáÇâï§
 
kg£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨1£©³£ÎÂÏÂÈôÈÜÒºÖÐÓÉË®µçÀë²úÉúµÄc£¨OH-£©=1¡Á10-14mol?L-1£¬Âú×ã´ËÌõ¼þµÄÈÜÒºÖÐÒ»¶¨¿ÉÒÔ´óÁ¿¹²´æµÄÀë×Ó×éÊÇ
 
£®
A£®Al3+  Na+  NO3-  Cl-B£®K+  Na+  Cl-  NO3-
C£®K+  Na+  Cl-  CO32-D£®K+  NH4+  I-  NO3-
£¨2£©Ä³Î¶ȣ¨t¡æ£©Ê±£¬Ë®µÄÀë×Ó»ýΪKW=1¡Á10-13£¬Èô½«´ËζÈÏÂpH=11µÄ¿ÁÐÔÄÆÈÜÒºa LÓëpH=1µÄÏ¡ÁòËáb L»ìºÏ£¨Éè»ìºÏºóÈÜÒºÌå»ýµÄ΢С±ä»¯ºöÂÔ²»¼Æ£©£¬ËùµÃ»ìºÏҺΪÖÐÐÔ£¬Ôòa£ºbµÈÓÚ
 
£®
£¨3£©25¡æʱ£¬Èç¹ûÈ¡0.5mol/L HAÈÜÒºÓë0.5mol/L NaOHÈÜÒºµÈÌå»ý»ìºÏ£¨ºöÂÔ»ìºÏºóÈÜÒºÌå»ýµÄ±ä»¯£©£¬²âµÃpH=10£®ÊԻشðÒÔÏÂÎÊÌ⣺
¢Ù»ìºÏÈÜÒºÖÐÓÉË®µçÀë³öµÄc£¨H+£©
 
0.1mol/L NaOHÈÜÒºÖÐÓÉË®µçÀë³öµÄc£¨H+£©£¨Ìî¡°£¾¡±¡¢¡°£¼¡±¡¢»ò¡°=¡±£©£®
¢ÚÓ÷½³Ìʽ½âÊÍΪʲô»ìºÏºóÈÜÒºµÄpH£¾7
 
£®
¢Û»ìºÏÈÜÒºÖÐÏÂÁÐËãʽµÄ¼ÆËã½á¹û£¨Ìî¾ßÌåÊý×Ö£©£ºc£¨OH-£©-c£¨HA£©=
 
mol/L£®
£¨4£©ÒÑÖªNH4AÈÜҺΪÖÐÐÔ£¬ÓÖÖª½«HAÈÜÒº¼Óµ½Na2CO3ÈÜÒºÖÐÓÐÆøÌå·Å³ö£¬ÊÔÍƶϣ¨NH4£©2CO3ÈÜÒºµÄpH
 
7£¨Ìî¡°£¼¡±¡¢¡°£¾¡±»ò¡°=¡±£©£»ÏàͬζÈÏ£¬µÈÎïÖʵÄÁ¿Å¨¶ÈµÄÏÂÁÐËÄÖÖÑÎÈÜÒº°´pHÓÉ´óµ½Ð¡µÄÅÅÁÐ˳ÐòΪ
 
£¨ÌîÐòºÅ£©£®
a£®NH4HCO3     b£®NH4A     c£®£¨NH4£©2CO3     d£®NH4Cl£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

Na¡¢Cu¡¢O¡¢Si¡¢S¡¢ClÊdz£¼ûµÄÁùÖÖÔªËØ£®
£¨1£©NaλÓÚÔªËØÖÜÆÚ±íµÚ
 
ÖÜÆÚµÚ
 
×壻SµÄ»ù̬ԭ×ÓºËÍâÓÐ
 
¸öδ³É¶Ôµç×Ó£»SiµÄ»ù̬ԭ×ÓºËÍâµç×ÓÅŲ¼Ê½Îª
 
£®
£¨2£©Óá°£¾¡±»ò¡°£¼¡±Ìî¿Õ£º
µÚÒ»µçÀëÄÜÀë×Ӱ뾶ÈÛµãËáÐÔ
Si
 
S
O2-
 
Na+
NaCl
 
Si
H2SO4
 
HClO4
£¨3£©CuCl£¨s£©ÓëO2·´Ó¦Éú³ÉCuCl2£¨s£©ºÍÒ»ÖÖºÚÉ«¹ÌÌ壮ÔÚ25¡æ¡¢101kPaÏ£¬ÒÑÖª¸Ã·´Ó¦Ã¿ÏûºÄ1mol CuCl£¨s£©£¬·ÅÈÈ44.4kJ£¬¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽÊÇ
 
£®
£¨4£©ClO2³£ÓÃË®µÄ¾»»¯£¬¹¤ÒµÉÏ¿ÉÓÃCl2Ñõ»¯NaClO2ÈÜÒºÖÆÈ¡ClO2£®Ð´³ö¸Ã·´Ó¦µÄÀë×Ó·½³Ìʽ£¬²¢±ê³öµç×ÓתÒƵķ½ÏòºÍÊýÄ¿
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ϱíÊÇÔªËØÖÜÆÚ±íÖжÌÖÜÆÚÔªËصÄÒ»²¿·Ö£¬±íÖÐËùÁÐ×Öĸ·Ö±ð´ú±íÒ»ÖÖÔªËØ£®
£¨1£©ÉÏÊöÔªËØÖÐÐγɻ¯ºÏÎïÖÖÀà×î¶àµÄÊÇ
 
£¬Ô­×Ӱ뾶×î´óµÄÊÇ
 
£¬Àë×Ӱ뾶×î´óµÄÊÇ
 
£¨ÌîÃû³Æ£©£®×î¸ß¼ÛÑõ»¯Îï¶ÔӦˮ»¯ÎïËáÐÔ×îÇ¿µÄÊÇ
 
£¨Ìѧʽ£©
£¨2£©DµÄÇ⻯Îï±ÈGµÄÇ⻯ÎïÎȶ¨£¬ÆäÔ­ÒòÊÇ
 
£®
£¨3£©ÔÚÒ»¶¨Ìõ¼þÏ£¬AÓëEÐγɵÄÆø̬»¯ºÏÎïÈÜÓÚË®ºóÈÜÒº³Ê
 
 ÐÔ£¨Ìî¡°Ëᡱ¡¢¡°¼î¡±»ò¡°ÖС±£©£®
£¨4£©ÏÖÓÐÁíÒ»ÖÖÔªËØX£¬ÆäÔ­×ӵõç×ÓÄÜÁ¦×îÇ¿£¬ÔòXÔÚÖÜÆÚ±íÖеÄλÖÃ
 

£¨5£©¡°ÉñÖÛÁùºÅ¡±ÔØÈË·É´¬ÄÚÐèÒªÓÐÒ»ÖÖ»¯ºÏÎïÀ´ÎüÊÕº½ÌìÔ±ºô³öµÄCO2£¬ÄãÈÏΪ¸ÃÎïÖÊÓ¦¸ÃÊÇÓÉÉϱíÖеÄ
 
 £¨Ìî×Öĸ£©ÔªËØ×é³ÉµÄ£¬Óû¯Ñ§·½³Ìʽ±íʾÎüÊÕÔ­Àí£º
 
£®
£¨6£©Ð´³öÓÉÒÔÉÏÈÎÒâÁ½ÖÖÔªËØ×é³ÉµÄÖ»º¬¹²¼Û¼üµÄÎïÖʵĻ¯Ñ§Ê½
 
£¬Ö»º¬Àë×Ó¼üµÄÎïÖʵĻ¯Ñ§Ê½
 
£¬¼Èº¬Àë×Ó¼üÓÖº¬¹²¼Û¼üµÄÎïÖÊ
 
²»º¬»¯Ñ§¼üµÄÎïÖʵĻ¯Ñ§Ê½
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ijÈÜÒºÖпÉÄܺ¬ÓÐH+¡¢K+¡¢NH4+¡¢Mg2+¡¢Fe3+¡¢Al3+¡¢Cu2+¡¢SO42-¡¢I-¡¢CO32-µÈÀë×Ó£¬µ±Ïò¸ÃÈÜÒºÖмÓÈëijŨ¶ÈµÄNaOHÈÜҺʱ£¬·¢ÏÖ³ÁµíµÄÎïÖʵÄÁ¿ËæNaOHÈÜÒºµÄÌå»ý±ä»¯ÈçͼËùʾ£¬ÓÉ´Ë¿ÉÖª£º
£¨1£©¸ÃÈÜÒºÖп϶¨º¬ÓеÄÑôÀë×ÓÊÇ
 
£¬ÇÒ¸÷Àë×ÓÎïÖʵÄÁ¿Ö®±ÈΪ
 
£»¿Ï¶¨²»º¬µÄÑôÀë×ÓÊÇ
 
£®
£¨2£©È¡²¿·ÖÔ­ÈÜÒº£¬¼ÓÈëBaCl2ÈÜÒº£¬Óа×É«³ÁµíÉú³É£®¹ýÂ˳ýÈ¥³Áµíºó£¬ÏòÂËÒºÖеÎÈ뼸µÎÂÈË®£¬ÔÙ¼ÓÈëÉÙÁ¿ËÄÂÈ»¯Ì¼£¬Õñµ´ºó¾²Öã¬ËÄÂÈ»¯Ì¼³Ê×ÏÉ«£®ÔòÔ­ÈÜÒºÖп϶¨´æÔÚµÄÒõÀë×ÓÊÇ
 
£®
£¨3£©Ð´³öÔÚÏ߶Îcdʱ·´Ó¦µÄÀë×Ó·½³Ìʽ
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÄÜÕýÈ·±íʾÏÂÁз´Ó¦µÄÀë×Ó·½³ÌʽÊÇ£¨¡¡¡¡£©
A¡¢ÏòÇâÑõ»¯ÄÆÈÜÒºÖмÓÈëÉÙÁ¿ÂÁ·Û£º2Al+2OH-+2H2O¨T2AlO2-+3H2¡ü
B¡¢NH4HCO3ÈÜÒºÓë×ãÁ¿Ba£¨OH£©2ÈÜÒº»ìºÏ£ºHCO3-+Ba2++OH-¨TBaCO3¡ý+H2O
C¡¢Áò»¯ÄÆÈÜÒºÏÔ¼îÐÔµÄÔ­Òò£ºS2-+H2O¨TH2S+2OH-
D¡¢ÏòFeBr2ÈÜÒºÖÐͨÈë×ãÁ¿ÂÈÆø£º2Fe2++2Br-+2Cl2¨T2Fe3++Br2+4Cl-

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸