ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿ÈçͼΪʵÑéÊÒijŨÑÎËáÊÔ¼ÁÆ¿±êÇ©ÉϵÄÓйØÊý¾Ý£¬ÊÔ¸ù¾Ý±êÇ©ÉϵÄÓйØÊý¾Ý»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©¸ÃŨÑÎËáÖÐHClµÄÎïÖʵÄÁ¿Å¨¶ÈΪ____mol¡¤L-1¡£

£¨2£©È¡ÓÃÈÎÒâÌå»ýµÄ¸ÃÑÎËáÈÜҺʱ£¬ÏÂÁÐÎïÀíÁ¿Öв»ËæËùÈ¡Ìå»ýµÄ¶àÉÙ¶ø±ä»¯µÄÊÇ___

A.ÈÜÒºÖÐHClµÄÎïÖʵÄÁ¿ B.ÈÜÒºµÄŨ¶È

C.ÈÜÒºµÄÃÜ¶È D.ÈÜÒºÖÐCl-µÄÊýÄ¿

£¨3£©ÏÂÁÐÈÝÁ¿Æ¿µÄʹÓ÷½·¨ÖУ¬²»ÕýÈ·µÄÊÇ____

A.ʹÓÃÈÝÁ¿Æ¿Ç°¼ìÑéÊÇ·ñ©ˮ

B.ÈÝÁ¿Æ¿ÓÃˮϴ¾»ºó£¬ÔÙÓôýÅäÈÜҺϴµÓ

C.ÅäÖÆÈÜҺʱ£¬Èç¹ûÊÔÑùÊǹÌÌ壬°Ñ³ÆºÃµÄ¹ÌÌåÓÃÖ½ÌõСÐĵ¹ÈëÈÝÁ¿Æ¿ÖУ¬»ºÂý¼ÓË®ÖÁÀë¿Ì¶ÈÏß1¡«2cm´¦£¬ÓýºÍ·µÎ¹Ü¼ÓÕôÁóË®ÖÁ¿Ì¶ÈÏß¡£

D.ÅäÖÆÈÜҺʱ£¬ÈôÊÔÑùÊÇÒºÌ壬ÓÃÁ¿Í²È¡ÑùºóÓò£Á§°ôÒýÁ÷µ¹ÈëÈÝÁ¿Æ¿ÖУ¬»ºÂý¼ÓË®ÖÁÀë¿Ì¶ÈÏß1¡«2cm´¦£¬ÓýºÍ·µÎ¹Ü¼ÓÕôÁóË®ÖÁ¿Ì¶ÈÏß¡£

E.¸ÇºÃÆ¿Èû£¬ÓÃʳָ¶¥×¡Æ¿Èû£¬ÁíÒ»Ö»ÊÖÍÐסƿµ×£¬°ÑÈÝÁ¿Æ¿·´¸´µ¹×ª¶à´Î£¬Ò¡ÔÈ¡£

£¨4£©Ä³Ñ§ÉúÓûÓÃÉÏÊöŨÑÎËáºÍÕôÁóË®ÅäÖÆ480mLÎïÖʵÄÁ¿Å¨¶ÈΪ0.200 mol¡¤L-1µÄÏ¡ÑÎËá¡£

¢ÙËùÐè²£Á§ÒÇÆ÷³ýÉÕ±­¡¢Á¿Í²¡¢²£Á§°ô¡¢½ºÍ·µÎ¹ÜÍ⻹Ðè_____¡£

¢Ú¸ÃѧÉúÐèÒªÓÃÁ¿Í²Á¿È¡___mLÉÏÊöŨÑÎËá½øÐÐÅäÖÆ¡£ÔÚÅäÖÆ¹ý³ÌÖУ¬ÏÂÁÐʵÑé²Ù×÷»áʹËùÅäÖÆµÄÏ¡ÑÎËáµÄÎïÖʵÄÁ¿Å¨¶ÈÆ«´óµÄÓÐ____

A.×ªÒÆÈÜÒººóδϴµÓÉÕ±­ºÍ²£Á§°ô¾ÍÖ±½Ó¶¨ÈÝ

B.ÓÃÁ¿Í²Á¿È¡Å¨ÑÎËáʱ¸©ÊÓ¹Û²ì°¼ÒºÃæ

C.ÔÚÈÝÁ¿Æ¿Öж¨ÈÝʱ¸©Êӿ̶ÈÏß

D.¶¨Èݺó°ÑÈÝÁ¿Æ¿µ¹ÖÃÒ¡ÔÈ£¬·¢ÏÖÒºÃæµÍÓڿ̶ÈÏߣ¬ÓÖ¼ÓË®ÖÁ¿Ì¶ÈÏß

¡¾´ð°¸¡¿11.9 BC BCD 500mLÈÝÁ¿Æ¿ 8.4 C

¡¾½âÎö¡¿

£¨1£©¸ù¾Ý ¼ÆËã¸ÃŨÑÎËáÖÐHClµÄÎïÖʵÄÁ¿Å¨¶È£»

£¨2£©¸ù¾Ýn=cVÅжϣ»

£¨2£©A¡¢ÈÝÁ¿Æ¿ÓÐÆ¿Èû£¬Ê¹ÓÃǰ±ØÐë¼ì²éÈÝÁ¿Æ¿ÊÇ·ñ©ˮ£»
B¡¢ÈÝÁ¿Æ¿ÈôÓôýÅäÒºÈóÏ´£¬µ¼ÖÂÅäÖÆµÄÈÜÒºÖÐÈÜÖʵÄÎïÖʵÄÁ¿Æ«´ó£»
C¡¢ÈÝÁ¿Æ¿ÊÇÓÃÓÚÅäÖÆÒ»¶¨Å¨¶ÈµÄ¶¨Á¿ÒÇÆ÷£¬²»Äܹ»ÔÚÈÝÁ¿Æ¿ÖÐÈܽâ¹ÌÌ壻
D¡¢Ó¦¸ÃÔÚÉÕ±­ÖÐÏ¡ÊÍŨÈÜÒº£¬²»Äܹ»ÔÚÈÝÁ¿Æ¿ÖÐÖ±½ÓÏ¡ÊÍ

E.ΪËùÅäÈÜÒº¾ùÒ»£¬¸ÇºÃÆ¿Èû£¬ÓÃʳָ¶¥×¡Æ¿Èû£¬ÁíÒ»Ö»ÊÖÍÐסƿµ×£¬°ÑÈÝÁ¿Æ¿·´¸´µ¹×ª¶à´Î£¬Ò¡ÔÈ¡£

£¨4£©ÅäÖÆ480mLÎïÖʵÄÁ¿Å¨¶ÈΪ0.200 mol¡¤L-1µÄÏ¡ÑÎËáÐè500mLÈÝÁ¿Æ¿£»¸ù¾ÝÏ¡ÊÍǰºóÈÜÖʵÄÎïÖʵÄÁ¿²»±ä¼ÆËãÐèŨÑÎËáµÄÌå»ý£»¸ù¾Ý ·ÖÎöÎó²î¡£

£¨1£© mol¡¤L-1£»

(2)ÈÜҺΪ¾ùÒ»Îȶ¨·Öɢϵ£¬Å¨¶È¡¢ÃܶÈÓëÈÜÒºÌå»ýÎ޹أ»¸ù¾Ýn=cV£¬Ìå»ý²»Í¬Ê±£¬ÈÜÖʵÄÎïÖʵÄÁ¿ºÍÀë×ÓÊýÄ¿±ä»¯£¬¹ÊÑ¡BC£»

£¨3£©A¡¢ÈÝÁ¿Æ¿Ê¹ÓÃʱ£¬Ó¦Ïȼì²éÊÇ·ñ©ˮ£¬È»ºóÓÃÕôÁóˮϴµÓ¸É¾»¼´¿É£¬¹ÊAÕýÈ·£»
B¡¢ÈÝÁ¿Æ¿Ï´¾»ºó²»ÄÜÓÃËùÅäÖÆÈÜÒºÈóÏ´£¬·ñÔòÓ°ÏìÅäÖÆµÄÈÜÒºµÄŨ¶È£¬¹ÊB´íÎó£»
C¡¢ÈÝÁ¿Æ¿Ö»ÄÜÓÃÀ´ÅäÖÆÈÜÒº£¬²»ÄÜÔÚÈÝÁ¿Æ¿ÖÐÈܽ⣬Ӧ¸ÃÔÚÉÕ±­ÖÐÈܽ⣬¹ÊC´íÎó£»
D¡¢ÈÝÁ¿Æ¿Ö»ÄÜÓÃÀ´ÅäÖÆÈÜÒº£¬²»ÄÜÔÚÈÝÁ¿Æ¿ÖÐÏ¡ÊÍ£¬Ó¦¸ÃÔÚÉÕ±­ÖÐÏ¡ÊÍ£¬¹ÊD´íÎó£»
E¡¢Ò¡ÔÈʱ£¬¸ÇºÃÆ¿Èû£¬ÓÃʳָ¶¥×¡Æ¿Èû£¬ÁíÒ»Ö»ÊÖÍÐסƿµ×£¬°ÑÈÝÁ¿Æ¿·´¸´µ¹×ª£¬¹ÊEÕýÈ·£»Ñ¡BCD£»

£¨4£©ÅäÖÆ480mLÎïÖʵÄÁ¿Å¨¶ÈΪ0.200 mol¡¤L-1µÄÏ¡ÑÎËáÐèÓÃ500mLÈÝÁ¿Æ¿¶¨ÈÝ£¬¢ÙËùÐè²£Á§ÒÇÆ÷³ýÉÕ±­¡¢Á¿Í²¡¢²£Á§°ô¡¢½ºÍ·µÎ¹ÜÍ⻹Ðè500mLÈÝÁ¿Æ¿£»

ÉèÐèÒª11.9mol/LµÄÑÎËáµÄÌå»ýΪV£¬¸ù¾ÝÏ¡ÊͶ¨ÂÉÓУº0.2mol/L¡Á0.5L=11.9mol/L¡ÁV£¬½âµÃV=0.0084L=8.4mL£»

A.×ªÒÆÈÜÒººóδϴµÓÉÕ±­ºÍ²£Á§°ô¾ÍÖ±½Ó¶¨ÈÝ£¬ÈÜÖÊÆ«ÉÙ£¬ËùÅäÖÆµÄÏ¡ÑÎËáµÄÎïÖʵÄÁ¿Å¨¶ÈƫС£¬¹Ê²»Ñ¡A£»

B.ÓÃÁ¿Í²Á¿È¡Å¨ÑÎËáʱ¸©ÊÓ¹Û²ì°¼ÒºÃæ£¬Å¨ÑÎËáµÄÌå»ýƫС£¬ÈÜÖÊÆ«ÉÙ£¬ËùÅäÖÆµÄÏ¡ÑÎËáµÄÎïÖʵÄÁ¿Å¨¶ÈƫС£¬¹Ê²»Ñ¡B£»

C.ÔÚÈÝÁ¿Æ¿Öж¨ÈÝʱ¸©Êӿ̶ÈÏߣ¬ÈÜÒºÌå»ýƫС£¬ËùÅäÖÆµÄÏ¡ÑÎËáµÄÎïÖʵÄÁ¿Å¨¶ÈÆ«´ó£¬¹ÊÑ¡C£»

D.¶¨Èݺó°ÑÈÝÁ¿Æ¿µ¹ÖÃÒ¡ÔÈ£¬·¢ÏÖÒºÃæµÍÓڿ̶ÈÏߣ¬ÓÖ¼ÓË®ÖÁ¿Ì¶ÈÏߣ¬ÈÜÒºÌå»ýÆ«´ó£¬ËùÅäÖÆµÄÏ¡ÑÎËáµÄÎïÖʵÄÁ¿Å¨¶ÈƫС£¬¹Ê²»Ñ¡D¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿Ä³Ð¡×éÀûÓÃH2C2O4ÈÜÒººÍËáÐÔKMnO4ÈÜÒº·´Ó¦À´Ì½¾¿¡°Íâ½çÌõ¼þ¶Ô»¯Ñ§·´Ó¦ËÙÂʵÄÓ°Ï족¡£ÊµÑéʱ£¬ÏÈ·Ö±ðÁ¿È¡Á½ÖÖÈÜÒº£¬È»ºóµ¹ÈëÊÔ¹ÜÖÐѸËÙÕñµ´»ìºÏ¾ùÔÈ£¬¿ªÊ¼¼ÆÊ±£¬Í¨¹ý²â¶¨ÍÊÉ«ËùÐèʱ¼äÀ´ÅжϷ´Ó¦µÄ¿ìÂý¡£¸ÃС×éÉè¼ÆÁËÈçÏ·½°¸¡£

ʵÑé±àºÅ

ÊÒÎÂÏ£¬ÊÔ¹ÜÖÐËù¼ÓÊÔ¼Á¼°ÆäÓÃÁ¿/mL

ÊÒÎÂÏÂÈÜÒºÑÕÉ«ÍÊÖÁÎÞÉ«ËùÐèʱ¼ä/min

0.6mol/L

H2C2O4ÈÜÒº

H2O

0.2mol/LKMnO4ÈÜÒº

3mol/L

Ï¡ÁòËá

1

3.0

2.0

2.0

3.0

4.0

2

3.0

3.0

2.0

2.0

5.2

3

3.0

4.0

2.0

1.0

6.4

(1)¸ù¾ÝÉϱíÖеÄʵÑéÊý¾Ý£¬¿ÉÒԵõ½µÄ½áÂÛÊÇ______________________________¡£

(2)¸ÃС×éͬѧ¸ù¾Ý¾­Ñ黿֯ÁËn(Mn2+)ËæÊ±¼ä±ä»¯Ç÷ÊÆµÄʾÒâͼ£¬Èçͼ1Ëùʾ¡£µ«ÓÐͬѧ²éÔÄÒÑÓеÄʵÑé×ÊÁÏ·¢ÏÖ£¬¸ÃʵÑé¹ý³ÌÖÐn(Mn2+)ËæÊ±¼ä±ä»¯µÄÇ÷ÊÆÓ¦Èçͼ2Ëùʾ¡£¸ÃС×éͬѧ¸ù¾Ýͼ2ËùʾÐÅÏ¢Ìá³öÁËеļÙÉ裬²¢¼ÌÐø½øÐÐʵÑé̽¾¿¡£

¸ÃС×éͬѧÌá³öµÄ¼ÙÉèÊÇ___________________________________¡£

¢ÚÇëÄã°ïÖú¸ÃС×éͬѧÍê³ÉʵÑé·½°¸£¬²¢Ìîд±íÖпհס£

ʵÑé±àºÅ

ÊÒÎÂÏ£¬ÊÔ¹ÜÖÐËù¼ÓÊÔ¼Á¼°ÆäÓÃÁ¿/mL

ÔÙÏòÊÔ¹ÜÖмÓÈëÉÙÁ¿¹ÌÌåX

ÊÒÎÂÏÂÈÜÒºÑÕÉ«ÍÊÖÁÎÞÉ«ËùÐèʱ¼ä/min

0.6mol/L

H2C2O4ÈÜÒº

H2O

0.2mol/LKMnO4ÈÜÒº

3mol/L

Ï¡ÁòËá

4

3.0

2.0

2.0

3.0

t

¹ÌÌåXÊÇ__________¡£

¢ÛÈô¸ÃС×éͬѧÌá³öµÄ¼ÙÉè³ÉÁ¢£¬Ê±¼ät__________4.0min£¨Ìî>¡¢=»ò<£©¡£

(3)Ϊ̽¾¿Î¶ȶԻ¯Ñ§·´Ó¦ËÙÂʵÄÓ°Ï죬¸ÃС×éͬѧ׼±¸ÔÚÉÏÊöʵÑé»ù´¡ÉϼÌÐø½øÐÐʵÑ飬ÇëÄã°ïÖú¸ÃС×éͬѧÍê³É¸ÃʵÑé·½°¸Éè¼Æ____________________________

¡¾ÌâÄ¿¡¿ÇåÔ¶ÊÐijУµÄ»¯Ñ§ÐËȤС×é¾­³£×ö̽¾¿ÊµÑ飺

(Ò»)ΪÁË̽¾¿Ò»Ñõ»¯µªÄÜ·ñ±»Na2O2ÍêÈ«ÎüÊÕ£¬Éè¼ÆÁËÈçÏÂʵÑé¡£×°ÖÃÈçÏ£¨¼ÓÈÈ×°ÖÃÊ¡ÂÔ£©£º

²éÔÄ×ÊÁÏËùÖª£º¢Ù2NO+Na2O2=2NaNO2£»¢ÚËáÐÔÌõ¼þÏ£¬NO»òNO2¶¼ÄÜÓëKMnO4ÈÜÒº·´Ó¦Éú³ÉNO3-¡£»Ø´ðÏÂÁÐÎÊÌ⣺

(1)ÒÇÆ÷aÃû³Æ£º____________¡£

(2)BÆ¿ÄÚ×°µÄÎïÖÊÊÇ£º_________¡£

(3)ÈôNOÄܱ»Na2O2ÍêÈ«ÎüÊÕ£¬E×°ÖÃÖеÄÏÖÏóΪ__________________¡£

(4)Èý¾±ÉÕÆ¿AÖз´Ó¦µÄ»¯Ñ§·½³ÌʽΪ______________________¡£

(5)C×°ÖõÄ×÷ÓÃÊÇ_________________________¡£

(¶þ)ʵÑéÊÒ³£ÓÃNa2SO3¹ÌÌåÓëŨÁòËá·´Ó¦ÖÆÈ¡SO2

(6)ijͬѧ²â¶¨²¿·Ö±äÖʵÄNa2SO3ÑùÆ·ÖÐNa2SO3µÄº¬Á¿£¨ÒÑÖªÔÚËáÐÔÌõ¼þÏÂIO3-Äܽ«SO32£­Ñõ»¯ÎªSO42-£¬×ÔÉí»¹Ô­ÎªI-£©£º

¢ÙÓõç×ÓÌìÆ½³ÆÈ¡16.00gNa2SO3¹ÌÌåÅä³Él00mLÈÜÒº£¬È¡25.00mLÓÚ×¶ÐÎÆ¿ÖУ¬²¢¼ÓÈ뼸µÎµí·ÛÈÜÒº¡£

¢ÚÓÃ0.1000mol/LËáÐÔKIO3ÈÜÒº£¨ÁòËáËữ£©µÎ¶¨£¬Èý´ÎƽÐÐʵÑéËùÓñê×¼ÒºÌå»ýµÄƽ¾ùֵΪ24.00mL¡£ÔòµÎ¶¨ÖÕµãʱ׶ÐÎÆ¿ÖвúÉúµÄÏÖÏó______________£¬Ð´³öÓë²úÉúÖÕµãÏÖÏóÓйط´Ó¦µÄÀë×Ó·½³Ìʽ__________£¬ÑùÆ·ÖÐNa2SO3µÄÖÊÁ¿·ÖÊýΪ_________¡££¨¼ÆËã½á¹û±£ÁôËÄλÓÐЧÊý×Ö£©

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø