C¡¢D¡¢G¡¢I¾ùΪ¶ÌÖÜÆÚÔªËØÐγɵĵ¥ÖÊ£¬D¡¢G¡¢IΪ³£¼û·Ç½ðÊôÆø̬µ¥ÖÊ¡£DÔªËصÄÔ­×ÓµÄ×îÍâ²ãµç×ÓÊýÊÇ´ÎÍâ²ãµç×ÓÊýµÄ3±¶£¬C¡¢GͬÖÜÆÚ£¬ÇÒÔ­×Ó×îÍâ²ãµç×ÓÊýÏà²î4£¬ËüÃǵļòµ¥Àë×Óµç×Ó²ã½á¹¹²»Í¬¡£Ï໥¼äÓÐÈçÏÂת»¯¹Øϵ£º

ÇëÌî¿Õ£º

(1)  DÓëIÄÜÐγÉÔ­×Ó¸öÊý±ÈΪ1£º1µÄ¹²¼Û»¯ºÏÎÇëд³öÆä·Ö×Óʽ£º            £»

£¨2£©LÊÇÄ¿Ç°Ó¦ÓÃ×î¹ã·ºµÄ½ðÊô£¬ÓÃ̼°ô×÷Ñô¼«£¬L×÷Òõ¼«£¬Ð´³öµç½âEË®ÈÜÒºµÄ»¯Ñ§Ê½£º                                   ¡£

£¨3£©»¯ºÏÎïK Öк¬ÓÐ×é³Éµ¥ÖÊLµÄÔªËØ£¬ÇÒ¸ÃÔªËصÄÖÊÁ¿·ÖÊýΪ70%¡£·´Ó¦¢ÙµÄ»¯Ñ§·½³ÌʽÊÇ                                                                  £¬Òý·¢¸Ã·´Ó¦µÄ²Ù×÷ÊÇ                      

£¨4£©Ð´³öA+F ¡úJµÄÀë×Ó·½³Ìʽ£º                                     ¡£

 

£¨1£©H2O2 £¨2·Ö£©

(2) 2KCl+2H2O2KOH+H2¡ü+Cl2¡ü £¨3·Ö£©

(3) Fe2O3 +2Al2Fe +Al2O3 £¨3·Ö£©

ÔÚ»ìºÏÎïÉϼÓÉÙÁ¿KClO3£¬²åÉÏMgÌõ²¢½«Æäµãȼ£¨3·Ö£©

(4)Al2O3+2OH-===2AlO2-+H2O£¨3·Ö£©

½âÎö:ÂÔ

 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

C¡¢D¡¢G¡¢I¾ùΪ¶ÌÖÜÆÚÔªËØÐγɵĵ¥ÖÊ£¬D¡¢G¡¢IΪ³£¼û·Ç½ðÊôÆø̬µ¥ÖÊ£®DµÄ×îÍâ²ãµç×ÓÊýÊÇ´ÎÍâ²ãµç×ÓÊýµÄ3±¶£¬C¡¢GͬÖÜÆÚ£¬ÇÒ×îÍâ²ãµç×ÓÊýÏà²î4£¬ËüÃǵļòµ¥Àë×Óµç×Ó²ã½á¹¹²»Í¬£®Ï໥¼äÓÐÈçÏÂת»¯¹Øϵ£º

ÇëÌî¿Õ£º
£¨1£©DÓëIÄÜÐγÉÔ­×Ó¸öÊý±ÈΪ1£º1µÄ¹²¼Û»¯ºÏÎÇëд³öÆä·Ö×Óʽ£º
H2O2
H2O2
£»
GÓëIÒ²ÄÜÐγÉÔ­×Ó¸öÊý±ÈΪ1£º1µÄ¹²¼Û»¯ºÏÎÇëд³öÆäµç×Óʽ£º
£®
£¨2£©LÊÇÄ¿Ç°Ó¦ÓÃ×î¹ã·ºµÄ½ðÊô£¬ÓÃ̼°ô×÷Ñô¼«£¬L×÷Òõ¼«£¬Ð´³öµç½â1L1mol?L-1µÄEË®ÈÜÒºµÄ»¯Ñ§·½³Ìʽ£º
2KCl+2H2O
 µç½â 
.
 
2KOH+Cl2¡ü+H2¡ü
2KCl+2H2O
 µç½â 
.
 
2KOH+Cl2¡ü+H2¡ü
£¬µ±ÈÜÒºµÄpH=13ʱ£¬ÀíÂÛÉÏÉú³ÉGµÄµ¥ÖÊÔÚ±ê×¼×´¿öϵÄÌå»ýΪ
1.12L
1.12L
£»Èô½«ÉÏÊöÁ½µç¼«²ÄÁϵ÷»»£¬ÔòÑô¼«Éϵĵ缫·´Ó¦Îª
Fe-2e-=Fe2+
Fe-2e-=Fe2+
£®
£¨3£©»¯ºÏÎïK Öк¬ÓÐ×é³Éµ¥ÖÊLµÄÔªËØ£¬ÇÒ¸ÃÔªËصÄÖÊÁ¿·ÖÊýΪ70%£®·´Ó¦C¡úLµÄ»¯Ñ§·½³ÌʽÊÇ
Fe2O3+2Al
 ¸ßΠ
.
 
Al2O3+2Fe
Fe2O3+2Al
 ¸ßΠ
.
 
Al2O3+2Fe

£¨4£©Ð´³öA+F¡úJµÄÀë×Ó·½³Ìʽ£º
Al2O3+2OH-=2AlO2-+H2O
Al2O3+2OH-=2AlO2-+H2O
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

A¡«LΪÖÐѧ»¯Ñ§ÒÑѧ¹ýµÄÎïÖÊ£¬Ï໥¼äÓÐÈçÏÂת»¯¹Øϵ£¨²¿·Ö²úÎïÈçH2OÒÑÂÔÈ¥£©£®ÆäÖÐC¡¢D¡¢G¡¢I¾ùΪ¶ÌÖÜÆÚÔªËØÐγɵĵ¥ÖÊ£¬D¡¢G¡¢IΪ³£¼ûÆø̬µ¥ÖÊ£»DµÄÔ­×Ó×îÍâ²ãµç×ÓÊýÊÇ´ÎÍâ²ãµç×ÓÊýµÄ3±¶£»C¡¢GµÄÔªËØͬÖÜÆÚ£¬ÇÒÔ­×Ó×îÍâ²ãµç×ÓÊýÏà²î4£»KΪºì×ØÉ«¹ÌÌå·ÛÄ©£®
¾«Ó¢¼Ò½ÌÍø
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©DµÄÔªËØÓëIµÄÔªËØÄÜÐγÉÔ­×Ó¸öÊý±ÈΪ1£º1µÄ¹²¼Û»¯ºÏÎÆäµç×ÓʽΪ
 
£®
£¨2£©Ð´³öÓÃʯī×÷µç¼«£¬µç½âEµÄË®ÈÜÒºµÄÀë×Ó·½³Ìʽ£º
 
£®ÆøÌåGÔÚ
 
¼«Òݳö£¨Ìî¡°Ñô¡±»ò¡°Òõ¡±£©£®
£¨3£©Ð´³öC+K¡úL+AµÄ»¯Ñ§·½³Ìʽ£º
 
£»¸Ã·´Ó¦ÔÚ¹¤ÒµÉϱ»³ÆΪ
 
·´Ó¦£®
£¨4£©Ð´³öA+F¡úJµÄÀë×Ó·½³Ìʽ£º
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

A¡«LΪÖÐѧ»¯Ñ§ÒÑѧ¹ýµÄÎïÖÊ,Ï໥¼äÓÐÈçÏÂת»¯¹Øϵ(²¿·Ö²úÎïÈçH2OÒÑÂÔÈ¥)¡£ÆäÖÐC¡¢D¡¢G¡¢I¾ùΪ¶ÌÖÜÆÚÔªËØÐγɵĵ¥ÖÊ£¬D¡¢G¡¢IΪ³£¼ûÆø̬µ¥ÖÊ¡£DµÄÔ­×Ó×îÍâ²ãµç×ÓÊýÊÇ´ÎÍâ²ãµç×ÓÊýµÄ3±¶£¬C¡¢GµÄÔªËØͬÖÜÆÚ£¬ÇÒÔ­×Ó×îÍâ²ãµç×ÓÊýÏà²î4£¬ËüÃǵļòµ¥Àë×Óµç×Ó²ã½á¹¹²»Í¬¡£KΪºì×ØÉ«¹ÌÌå·ÛÄ©¡£?

ÇëÌî¿Õ£º?

(1)DµÄÔªËØÓëIµÄÔªËØÄÜÐγÉÔ­×Ó¸öÊý±ÈΪ1¡Ã1µÄ¹²¼Û»¯ºÏÎÆäµç×ÓʽΪ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡£?

(2)д³öÓÃʯī×÷µç¼«£¬µç½âEË®ÈÜÒºµÄÀë×Ó·½³Ìʽ¡¡¡¡¡¡¡¡¡¡¡£ÆøÌåGÔÚ¡¡¡¡¡¡¼«Òݳö(Ìî¡°Ñô¡±»ò¡°Òõ¡±)¡£?

(3)д³öC+K¡úL+AµÄ»¯Ñ§·½³Ìʽ¡¡¡¡¡¡¡¡¡¡¡¡¡¡£¬Ö¸³öÒý·¢´Ë·´Ó¦µÄ·½·¨£º¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡£

(4)д³öA+F¡úJµÄÀë×Ó·½³Ìʽ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡£?

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2010½ìÁÉÄþÊ¡´óÁ¬Ð­×÷Ìå¸ß¶þÉÏѧÆÚ¾ºÈü£¨Àí×Û£©»¯Ñ§²¿·Ö ÌâÐÍ£ºÌî¿ÕÌâ

C¡¢D¡¢G¡¢I¾ùΪ¶ÌÖÜÆÚÔªËØÐγɵĵ¥ÖÊ£¬D¡¢G¡¢IΪ³£¼û·Ç½ðÊôÆø̬µ¥ÖÊ¡£DÔªËصÄÔ­×ÓµÄ×îÍâ²ãµç×ÓÊýÊÇ´ÎÍâ²ãµç×ÓÊýµÄ3±¶£¬C¡¢GͬÖÜÆÚ£¬ÇÒÔ­×Ó×îÍâ²ãµç×ÓÊýÏà²î4£¬ËüÃǵļòµ¥Àë×Óµç×Ó²ã½á¹¹²»Í¬¡£Ï໥¼äÓÐÈçÏÂת»¯¹Øϵ£º

ÇëÌî¿Õ£º
(1)  DÓëIÄÜÐγÉÔ­×Ó¸öÊý±ÈΪ1£º1µÄ¹²¼Û»¯ºÏÎÇëд³öÆä·Ö×Óʽ£º            £»
£¨2£©LÊÇÄ¿Ç°Ó¦ÓÃ×î¹ã·ºµÄ½ðÊô£¬ÓÃ̼°ô×÷Ñô¼«£¬L×÷Òõ¼«£¬Ð´³öµç½âEË®ÈÜÒºµÄ»¯Ñ§Ê½£º                                   ¡£
£¨3£©»¯ºÏÎïK Öк¬ÓÐ×é³Éµ¥ÖÊLµÄÔªËØ£¬ÇÒ¸ÃÔªËصÄÖÊÁ¿·ÖÊýΪ70%¡£·´Ó¦¢ÙµÄ»¯Ñ§·½³ÌʽÊÇ                                                                 £¬Òý·¢¸Ã·´Ó¦µÄ²Ù×÷ÊÇ                      
£¨4£©Ð´³öA+F ¡ú JµÄÀë×Ó·½³Ìʽ£º                                     ¡£

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸