ÈçͼËùʾ£¬¼×¡¢ÒÒÁ½³Øµç¼«²ÄÁ϶¼ÊÇÌú°ôºÍʯī°ô£¬Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©ÈôÁ½³ØÖоùΪCuSO4ÈÜÒº£¬·´Ó¦Ò»¶Îʱ¼äºó£º
¢ÙÓкìÉ«ÎïÖÊÎö³öµÄÊǼ׳ØÖеÄ________°ô£¬ÒÒ³ØÖеÄ________°ô¡£
¢ÚÒÒ³ØÖÐÑô¼«µÄµç¼«·´Ó¦Ê½ÊÇ__________________________________¡£
£¨2£©ÈôÁ½³ØÖоùΪ±¥ºÍNaClÈÜÒº£º
¢Ùд³öÒÒ³ØÖÐ×Ü·´Ó¦µÄÀë×Ó·½³Ìʽ______________________________¡£
¢Ú¼×³ØÖÐʯīµç¼«Éϵĵ缫·´Ó¦ÊôÓÚ______(Ìî¡°Ñõ»¯·´Ó¦¡±»ò¡°»¹Ô­·´Ó¦¡±)¡£
¢Û½«ÊªÈóµÄKIµí·ÛÊÔÖ½·ÅÔÚÒÒ³Øʯīµç¼«¸½½ü£¬·¢ÏÖÊÔÖ½±äÀ¶£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ__________________________________¡£
¢ÜÈôÒÒ³ØתÒÆ0.02 mol e£­ºóֹͣʵÑ飬³ØÖÐÈÜÒºÌå»ýÊÇ200 mL£¬ÔòÈÜÒº»ìºÏÔȺóµÄpH£½_______¡£

£¨1£©¢Ùʯī  Ìú ¢Ú
(2)¢Ù
¢Ú»¹Ô­·´Ó¦
¢Û
¢Ü 13

½âÎö

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÈçͼËùʾ£¬¼×¡¢ÒÒÁ½³Øµç¼«²ÄÁ϶¼ÊÇÌú°ôºÍʯī°ô£¬Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÈôÁ½³ØÖоùΪCuSO4ÈÜÒº£¬·´Ó¦Ò»¶Îʱ¼äºó£º
¢ÙÓкìÉ«ÎïÖÊÎö³öµÄÊǼ׳ØÖеÄ
̼
̼
°ô£¬ÒÒ³ØÖеÄ
Ìú
Ìú
°ô£®
¢ÚÒÒ³ØÖÐÑô¼«µÄµç¼«·´Ó¦Ê½ÊÇ
4OH--4e-=2H2O+O2¡ü
4OH--4e-=2H2O+O2¡ü
£®
£¨2£©ÈôÁ½³ØÖоùΪ±¥ºÍNaClÈÜÒº£º
¢Ùд³öÒÒ³ØÖÐ×Ü·´Ó¦µÄÀë×Ó·½³Ìʽ
2Cl-+2H2O
 µç½â 
.
 
Cl2¡ü+H2¡ü+2OH-
2Cl-+2H2O
 µç½â 
.
 
Cl2¡ü+H2¡ü+2OH-
£®
¢Ú¼×³ØÖÐʯīµç¼«Éϵĵ缫·´Ó¦ÊôÓÚ
»¹Ô­·´Ó¦
»¹Ô­·´Ó¦
£¨Ìî¡°Ñõ»¯·´Ó¦¡±»ò¡°»¹Ô­·´Ó¦¡±£©£®
¢Û½«ÊªÈóµÄKIµí·ÛÊÔÖ½·ÅÔÚÒÒ³Øʯīµç¼«¸½½ü£¬·¢ÏÖÊÔÖ½±äÀ¶£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ
KI+Cl2=I2+2KCl
KI+Cl2=I2+2KCl
£®
¢ÜÈôÒÒ³ØתÒÆ0.02mol e-ºóֹͣʵÑ飬³ØÖÐÈÜÒºÌå»ýÊÇ200mL£¬ÔòÈÜÒº»ìºÏÔȺóµÄpH=
13
13
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÈçͼËùʾ£¬¼×¡¢ÒÒÁ½³Øµç¼«²ÄÁ϶¼ÊÇÌú°ôºÍ̼°ô£¬Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÈôÁ½³ØÖоùΪCuSO4ÈÜÒº£¬·´Ó¦Ò»¶Îʱ¼äºó£º
¢ÙÓкìÉ«ÎïÖÊÎö³öµÄÊǼ׳ØÖеÄ
̼
̼
°ô£¬ÒÒ³ØÖеÄ
Ìú
Ìú
°ô£®
¢ÚÒÒ³ØÖÐÑô¼«µÄµç¼«·´Ó¦Ê½ÊÇ
4OH--4e-=2H2O+O2¡ü
4OH--4e-=2H2O+O2¡ü
£®
£¨2£©ÈôÁ½³ØÖоùΪ±¥ºÍNaClÈÜÒº£º
¢Ùд³öÒÒ³ØÖÐ×Ü·´Ó¦µÄÀë×Ó·½³Ìʽ
2NaCl+2H2O
 Í¨µç 
.
 
Cl2¡ü+H2¡ü+2NaOH
2NaCl+2H2O
 Í¨µç 
.
 
Cl2¡ü+H2¡ü+2NaOH
£®
¢Ú¼×³ØÖÐ̼¼«Éϵ缫·´Ó¦Ê½ÊÇ
O2+2H2O+4e-=4OH-
O2+2H2O+4e-=4OH-
£¬ÒÒ³Ø̼¼«Éϵ缫·´Ó¦ÊôÓÚ
Ñõ»¯·´Ó¦
Ñõ»¯·´Ó¦
£¨Ìî¡°Ñõ»¯·´Ó¦¡±»ò¡°»¹Ô­·´Ó¦¡±£©£®
¢ÛÈôÒÒ³ØתÒÆ0.02mol e-ºóֹͣʵÑ飬³ØÖÐÈÜÒºÌå»ýÊÇ200mL£¬ÔòÈÜÒº»ìºÏ¾ùÔȺóµÄËùµÃNaOHÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈΪ
0.1mol/L
0.1mol/L
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

(ÿ¿Õ3·Ö,¹²21·Ö)ÈçͼËùʾ£¬¼×¡¢ÒÒÁ½³Øµç¼«²ÄÁ϶¼ÊÇÌú°ôÓë̼°ô£¬Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©ÈôÁ½³ØÖоùΪCuSO4ÈÜÒº£¬·´Ó¦Ò»¶Îʱ¼äºó£º

¢ÙÓкìÉ«ÎïÖÊÎö³öµÄÊǼ׳ØÖеÄ______°ô£¬ÒÒ³ØÖеÄ______°ô¡£

¢ÚÒÒ³ØÖÐÑô¼«µÄµç¼«·´Ó¦Ê½ÊÇ____________________¡£

£¨2£©ÈôÁ½³ØÖоùΪ±¥ºÍNaClÈÜÒº£º

¢Ùд³öÒÒ³ØÖÐ×Ü·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ___________________¡£

¢Ú¼×³ØÖÐ̼¼«µÄµç¼«·´Ó¦Ê½ÊÇ______________¡£

¢Û½«ÊªÈóµÄµí·Û¡ªKIÊÔÖ½·ÅÔÚÒÒ³Ø̼¼«¸½½ü£¬·¢ÏÖÊÔÖ½±äÀ¶£¬¹ýÒ»¶Îʱ¼äºóÓÖ·¢ÏÖÀ¶É«ÍËÈ¥¡£ÕâÊÇÒòΪ¹ýÁ¿µÄCl2ÓÖ½«Éú³ÉµÄI2Ñõ»¯¡£Èô·´Ó¦µÄCl2ºÍI2ÎïÖʵÄÁ¿Ö®±ÈΪ5¡Ã1£¬ÇÒÉú³ÉÁ½ÖÖËᣬ¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ_______________________________¡£

¢ÜÈôÒÒ³ØתÒÆ0.02mol e£­ºóֹͣʵÑ飬³ØÖÐÈÜÒºÌå»ýÊÇ200mL£¬Ôò·´Ó¦ºóÈÜÒºµÄpH£½               ¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÈçͼËùʾ£¬¼×¡¢ÒÒÁ½³Øµç¼«²ÄÁ϶¼ÊÇÌú°ôºÍʯī°ô£¬Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©ÈôÁ½³ØÖоùΪCuSO4ÈÜÒº£¬·´Ó¦Ò»¶Îʱ¼äºó£º

¢ÙÓкìÉ«ÎïÖÊÎö³öµÄÊǼ׳ØÖеÄ________°ô£¬ÒÒ³ØÖеÄ________°ô¡£

¢ÚÒÒ³ØÖÐÑô¼«µÄµç¼«·´Ó¦Ê½ÊÇ__________________________________¡£

£¨2£©ÈôÁ½³ØÖоùΪ±¥ºÍNaClÈÜÒº£º

¢Ùд³öÒÒ³ØÖÐ×Ü·´Ó¦µÄÀë×Ó·½³Ìʽ______________________________¡£

¢Ú¼×³ØÖÐʯīµç¼«Éϵĵ缫·´Ó¦ÊôÓÚ______(Ìî¡°Ñõ»¯·´Ó¦¡±»ò¡°»¹Ô­·´Ó¦¡±)¡£

¢Û½«ÊªÈóµÄKIµí·ÛÊÔÖ½·ÅÔÚÒÒ³Øʯīµç¼«¸½½ü£¬·¢ÏÖÊÔÖ½±äÀ¶£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ__________________________________¡£

¢ÜÈôÒÒ³ØתÒÆ0.02 mol e£­ºóֹͣʵÑ飬³ØÖÐÈÜÒºÌå»ýÊÇ200 mL£¬ÔòÈÜÒº»ìºÏÔȺóµÄpH£½_______¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2012½ìºþ±±Ê¡Î人Êи߶þÉÏѧÆÚÆÚÄ©¿¼ÊÔ»¯Ñ§ÊÔ¾í ÌâÐÍ£ºÌî¿ÕÌâ

ÈçͼËùʾ£¬¼×¡¢ÒÒÁ½³Øµç¼«²ÄÁ϶¼ÊÇÌú°ôºÍʯī°ô£¬Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©ÈôÁ½³ØÖоùΪCuSO4ÈÜÒº£¬·´Ó¦Ò»¶Îʱ¼äºó£º

¢ÙÓкìÉ«ÎïÖÊÎö³öµÄÊǼ׳ØÖеÄ________°ô£¬ÒÒ³ØÖеÄ________°ô¡£

¢ÚÒÒ³ØÖÐÑô¼«µÄµç¼«·´Ó¦Ê½ÊÇ__________________________________¡£

£¨2£©ÈôÁ½³ØÖоùΪ±¥ºÍNaClÈÜÒº£º

¢Ùд³öÒÒ³ØÖÐ×Ü·´Ó¦µÄÀë×Ó·½³Ìʽ______________________________¡£

¢Ú¼×³ØÖÐʯīµç¼«Éϵĵ缫·´Ó¦ÊôÓÚ______(Ìî¡°Ñõ»¯·´Ó¦¡±»ò¡°»¹Ô­·´Ó¦¡±)¡£

¢Û½«ÊªÈóµÄKIµí·ÛÊÔÖ½·ÅÔÚÒÒ³Øʯīµç¼«¸½½ü£¬·¢ÏÖÊÔÖ½±äÀ¶£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ__________________________________¡£

¢ÜÈôÒÒ³ØתÒÆ0.02 mol e£­ºóֹͣʵÑ飬³ØÖÐÈÜÒºÌå»ýÊÇ200 mL£¬ÔòÈÜÒº»ìºÏÔȺóµÄpH£½_______¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸