(10·Ö)Na2S2O3¡¤5H2O(Ë׳ƺ£²¨)ÊÇÕÕÏàÒµ³£ÓõÄÒ»ÖÖ¶¨Ó°¼Á£¬³£²ÉÓÃÏ·¨ÖƱ¸£º½«ÑÇÁòËáÄÆÈÜÒºÓëÁò·Û»ìºÏ¹²ÈÈ£¬Éú³ÉÁò´úÁòËáÄÆNa2SO3+S£½Na2S2O3£¬ÂËÈ¥Áò·Û£¬ÔÙ½«ÂËҺŨËõ¡¢ÀäÈ´£¬¼´ÓÐNa2S2O3¡¤5H2O¾§ÌåÎö³ö¡£¸Ã·¨ÖƵõľ§ÌåÖг£»ìÓÐÉÙÁ¿Na2SO3ºÍNa2SO4µÄÔÓÖÊ.

Ϊ²â¶¨Ò»ÖÖº£²¨¾§ÌåÑùÆ·µÄ³É·Ö£¬Ä³Í¬Ñ§³ÆÈ¡Èý·ÝÖÊÁ¿²»Í¬µÄ¸ÃÑùÆ·£¬·Ö±ð¼ÓÈëÏàͬŨ¶ÈµÄÏ¡H2SO4ÈÜÒº20mL£¬³ä·Ö·´Ó¦ºóÂ˳öÁò£¬²¢½«ÂËҺ΢ÈÈ(¼Ù¶¨Éú³ÉµÄSO2È«²¿Òݳö)£¬²âµÃÓйØʵÑéÊý¾ÝÈçÏÂ(±ê×¼×´¿ö)£º

 

µÚÒ»·Ý

µÚ¶þ·Ý

µÚÈý·Ý

ÑùÆ·µÄÖÊÁ¿/g

12.60

18.90

28.00

¶þÑõ»¯ÁòµÄÌå»ý/L

1.12

1.68

2.24

ÁòµÄÖÊÁ¿/g

1.28

1.92

2.56

(1)ÑùÆ·ÓëÏ¡ÁòËá¿ÉÄÜ·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ__________________________________£»

_____________________________________________¡£

(2)¸ù¾ÝÉϱíÊý¾Ý·ÖÎö£¬¸ÃÑùÆ·_________________________(ÌîÑ¡Ïî×Öĸ)£»

A£®º¬ÓÐNa2S2O3¡¤5H2OºÍNa2SO4Á½ÖֳɷÖ

B£®º¬ÓÐNa2S2O3¡¤5H2OºÍNa2SO3Á½ÖֳɷÖ

C£®º¬ÓÐNa2S2O3¡¤5H2O¡¢Na2SO3ºÍNa2SO4ÈýÖֳɷÖ

(3)¸ÃÑùÆ·Öи÷³É·ÖµÄÎïÖʵÄÁ¿Ö®±ÈΪ___________________________£»

(4)Ëù¼ÓÁòËáÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈΪ___________________________mol/L¡£

 

¡¾´ð°¸¡¿

(10·Ö)

(1)£¨¸÷2·Ö£¬¹²4·Ö£©S2O32-+2H+£½S¡ý+SO2¡ü+H2O   SO32-+2H+£½SO2¡ü+H2O

(2)£¨2·Ö£©C

(3)£¨2·Ö£©n(Na2S2O3):n(Na2SO3):n(Na2SO4) = 4:1:1

(4)£¨2·Ö£©5mol¡¤L-1

¡¾½âÎö¡¿ÂÔ

 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

(10·Ö)Na2S2O3¡¤5H2O(Ë׳ƺ£²¨)ÊÇÕÕÏàÒµ³£ÓõÄÒ»ÖÖ¶¨Ó°¼Á£¬³£²ÉÓÃÏ·¨ÖƱ¸£º½«ÑÇÁòËáÄÆÈÜÒºÓëÁò·Û»ìºÏ¹²ÈÈ£¬Éú³ÉÁò´úÁòËáÄÆNa2SO3+S£½Na2S2O3£¬ÂËÈ¥Áò·Û£¬ÔÙ½«ÂËҺŨËõ¡¢ÀäÈ´£¬¼´ÓÐNa2S2O3¡¤5H2O¾§ÌåÎö³ö¡£¸Ã·¨ÖƵõľ§ÌåÖг£»ìÓÐÉÙÁ¿Na2SO3ºÍNa2SO4µÄÔÓÖÊ.

Ϊ²â¶¨Ò»ÖÖº£²¨¾§ÌåÑùÆ·µÄ³É·Ö£¬Ä³Í¬Ñ§³ÆÈ¡Èý·ÝÖÊÁ¿²»Í¬µÄ¸ÃÑùÆ·£¬·Ö±ð¼ÓÈëÏàͬŨ¶ÈµÄÏ¡H2SO4ÈÜÒº20mL£¬³ä·Ö·´Ó¦ºóÂ˳öÁò£¬²¢½«ÂËҺ΢ÈÈ(¼Ù¶¨Éú³ÉµÄSO2È«²¿Òݳö)£¬²âµÃÓйØʵÑéÊý¾ÝÈçÏÂ(±ê×¼×´¿ö)£º

 

µÚÒ»·Ý

µÚ¶þ·Ý

µÚÈý·Ý

ÑùÆ·µÄÖÊÁ¿/g

12.60

18.90

28.00

¶þÑõ»¯ÁòµÄÌå»ý/L

1.12

1.68

2.24

ÁòµÄÖÊÁ¿/g

1.28

1.92

2.56

(1)ÑùÆ·ÓëÏ¡ÁòËá¿ÉÄÜ·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ__________________________________£»

_____________________________________________¡£

(2)¸ù¾ÝÉϱíÊý¾Ý·ÖÎö£¬¸ÃÑùÆ·_________________________(ÌîÑ¡Ïî×Öĸ)£»

A£®º¬ÓÐNa2S2O3¡¤5H2OºÍNa2SO4Á½ÖֳɷÖ

B£®º¬ÓÐNa2S2O3¡¤5H2OºÍNa2SO3Á½ÖֳɷÖ

C£®º¬ÓÐNa2S2O3¡¤5H2O¡¢Na2SO3ºÍNa2SO4ÈýÖֳɷÖ

(3)¸ÃÑùÆ·Öи÷³É·ÖµÄÎïÖʵÄÁ¿Ö®±ÈΪ___________________________£»

(4)Ëù¼ÓÁòËáÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈΪ___________________________mol/L¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2012½ìºÓ±±Ê¡ÈýºÓÒ»ÖиßÈýÉÏѧÆÚµÚ¶þ´ÎÔ¿¼»¯Ñ§ÊÔ¾í ÌâÐÍ£º¼ÆËãÌâ

(10·Ö)Na2S2O3¡¤5H2O(Ë׳ƺ£²¨)ÊÇÕÕÏàÒµ³£ÓõÄÒ»ÖÖ¶¨Ó°¼Á£¬³£²ÉÓÃÏ·¨ÖƱ¸£º½«ÑÇÁòËáÄÆÈÜÒºÓëÁò·Û»ìºÏ¹²ÈÈ£¬Éú³ÉÁò´úÁòËáÄÆNa2SO3+S£½Na2S2O3£¬ÂËÈ¥Áò·Û£¬ÔÙ½«ÂËҺŨËõ¡¢ÀäÈ´£¬¼´ÓÐNa2S2O3¡¤5H2O¾§ÌåÎö³ö¡£¸Ã·¨ÖƵõľ§ÌåÖг£»ìÓÐÉÙÁ¿Na2SO3ºÍNa2SO4µÄÔÓÖÊ.
Ϊ²â¶¨Ò»ÖÖº£²¨¾§ÌåÑùÆ·µÄ³É·Ö£¬Ä³Í¬Ñ§³ÆÈ¡Èý·ÝÖÊÁ¿²»Í¬µÄ¸ÃÑùÆ·£¬·Ö±ð¼ÓÈëÏàͬŨ¶ÈµÄÏ¡H2SO4ÈÜÒº20mL£¬³ä·Ö·´Ó¦ºóÂ˳öÁò£¬²¢½«ÂËҺ΢ÈÈ(¼Ù¶¨Éú³ÉµÄSO2È«²¿Òݳö)£¬²âµÃÓйØʵÑéÊý¾ÝÈçÏÂ(±ê×¼×´¿ö)£º

 
µÚÒ»·Ý
µÚ¶þ·Ý
µÚÈý·Ý
ÑùÆ·µÄÖÊÁ¿/g
12.60
18.90
28.00
¶þÑõ»¯ÁòµÄÌå»ý/L
1.12
1.68
2.24
ÁòµÄÖÊÁ¿/g
1.28
1.92
2.56
(1)ÑùÆ·ÓëÏ¡ÁòËá¿ÉÄÜ·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ__________________________________£»
_____________________________________________¡£
(2)¸ù¾ÝÉϱíÊý¾Ý·ÖÎö£¬¸ÃÑùÆ·_________________________(ÌîÑ¡Ïî×Öĸ)£»
A£®º¬ÓÐNa2S2O3¡¤5H2OºÍNa2SO4Á½ÖֳɷÖ
B£®º¬ÓÐNa2S2O3¡¤5H2OºÍNa2SO3Á½ÖֳɷÖ
C£®º¬ÓÐNa2S2O3¡¤5H2O¡¢Na2SO3ºÍNa2SO4ÈýÖֳɷÖ
(3)¸ÃÑùÆ·Öи÷³É·ÖµÄÎïÖʵÄÁ¿Ö®±ÈΪ___________________________£»
(4)Ëù¼ÓÁòËáÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈΪ___________________________mol/L¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2013½ìºÓÄÏÊ¡Î÷»ªÒ»¸ß¸ßÈýÉÏѧÆÚµÚÒ»´ÎÔ¿¼»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÌî¿ÕÌâ

£¨10·Ö)£©

£¨1£©¹¤ÒµÉϽ«ÂÈÆøͨÈëʯ»ÒÈé[Ca(OH)2]ÖÆȡƯ°×·Û£¬Àë×Ó·´Ó¦·½³ÌʽΪ                       ¡£

£¨2£©³£ÎÂÏÂ,Ïò20 mL 1.0 µÄÈÜÒºÖÐÖðµÎ¼ÓÈëµÈÎïÖʵÄÁ¿Å¨¶ÈµÄ

ÈÜÒº,Éú³É³ÁµíµÄÁ¿Óë¼ÓÈëÇâÑõ»¯±µÈÜÒºµÄÌå»ý¹ØϵÈçͼËùʾ¡£a¡¢b¡¢c¡¢d·Ö±ð±íʾʵÑéʱ²»Í¬½×¶ÎµÄÈÜÒº¡£ÆäÖÐbµãËùʾÈÜÒº³Ê?         ?(ÌËáÐÔ¡±¡°ÖÐÐÔ¡±»ò¡±¼îÐÔ¡±),b- cÖ®¼ä·´Ó¦µÄÀë×Ó·½³ÌʽΪ                   ¡£

£¨3£©µâËá¼ØÊÇʳÑÎÖеÄÌí¼Ó¼Á¡£¼ìÑéµâÑÎÊÇ·ñ¼ÓÓеâËá¼Ø£¬¶¨ÐÔ¼ìÑé·½·¨ÊÇÔÚËáÐÔ½éÖÊÖмӻ¹Ô­¼ÁKCNS£¬Æä·´Ó¦ÈçÏ£º6IO3£­+5CNS£­+H++2H2O=3I2+5HCN+5SO42£­£¬¼ìÑéʱ£¬³ýÐèÓÃKCNSÍ⣬»¹ÐèÒªµÄÒ»ÖÖÊÔ¼Á×îºÃÊÇ       ¡£¶¨Á¿²â¶¨²úÆ·ÖÐKIO3µÄº¬Á¿Ê±£¬¿ÉÏÈÓÃË®ÈܽâÔÙͨÈëSO2£¬È»ºóÓÃNa2S2O3±ê×¼ÈÜÒºµÎ¶¨Îö³öµÄµâ£¬Í¨SO2µÄÀë×Ó·½³ÌʽΪ£º                          ¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

(10·Ö) Na2S2O3?5H2O£¨Ë׳ƺ£²¨£©ÊÇÕÕÏàÒµ³£ÓõÄÒ»ÖÖ¶¨Ó°¼Á¡£¹¤ÒµÉÏÖƵõÄNa2S2O3?5H2O¾§ÌåÖпÉÄܺ¬ÓÐNa2SO3¡£Îª²â¶¨Ä³º£²¨ÑùÆ·µÄ³É·Ö£¬³ÆÈ¡Èý·ÝÖÊÁ¿²»Í¬µÄ¸ÃÑùÆ·£¬·Ö±ð¼ÓÈëÏàͬŨ¶ÈµÄÁòËáÈÜÒº30 mL£¬³ä·Ö·´Ó¦£¨Na2S2O3£«H2SO4=Na2SO4£«SO2¡ü£«S¡ý£«H2O£©ºóÂ˳öÁò£¬Î¢ÈÈÂËҺʹSO2È«²¿Òݳö¡£²âµÃÓйØʵÑéÊý¾ÝÈçÏÂ±í£¨ÆøÌåÌå»ýÒÑ»»ËãΪ±ê×¼×´¿ö£©¡£

 

µÚÒ»·Ý

µÚ¶þ·Ý

µÚÈý·Ý

ÑùÆ·µÄÖÊÁ¿/g

6.830

13.660

30.000

¶þÑõ»¯ÁòÆøÌåµÄÌå»ý/L

0.672

1.344

2.688

ÁòµÄÖÊÁ¿/g

0.800

1.600

3.200

ÊÔ¼ÆË㣺

£¨1£©ËùÓÃÁòËáÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈΪ            ¡£

£¨2£©ÑùÆ·ÖÐn(Na2S2O3?5H2O)£ºn(Na2SO3)=            ¡£

£¨3£©Ä³»·¾³¼à²âС×éÓÃÉÏÊöº£²¨ÑùÆ·ÅäÖƺ¬Na2S2O3 0.100 mol?L-1µÄº£²¨ÈÜÒº£¬²¢ÀûÓÃËü²â¶¨Ä³¹¤³§·ÏË®ÖÐBa2+µÄŨ¶È¡£ËûÃÇÈ¡·ÏË®50.00 mL£¬¿ØÖÆÊʵ±µÄËá¶È¼ÓÈë×ãÁ¿µÄK2Cr2O7ÈÜÒº£¬µÃBaCrO4³Áµí£»³Áµí¾­Ï´µÓ¡¢¹ýÂ˺ó£¬ÓÃÊÊÁ¿µÄÏ¡ÑÎËáÈܽ⣬´ËʱCrO42-È«²¿×ª»¯ÎªCr2O72-£»ÔÙ¼Ó¹ýÁ¿KIÈÜÒº£¬³ä·Ö·´Ó¦ºó£¬ÓÃÉÏÊöÅäÖƵĺ£²¨ÈÜÒº½øÐе樣¬·´Ó¦Íêȫʱ£¬²âµÃÏûºÄº£²¨ÈÜÒºµÄÌå»ýΪ36.00 mL¡£ÒÑÖªÓйط´Ó¦µÄÀë×Ó·½³ÌʽΪ£º

¢ÙCr2O72- + 6I- + 14H+      2Cr3+ + 3I2 + 7H2O

¢ÚI2 + 2S2O32-      2I- + S4O62-

¢ÛI2 + SO32- + H2O      2I- + SO42- + 2H+

ÔòµÎ¶¨¹ý³ÌÖпÉÓà         ×÷ָʾ¼Á¡£¼ÆËã¸Ã¹¤³§·ÏË®ÖÐBa2+µÄÎïÖʵÄÁ¿Å¨¶È¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸