ÌâÄ¿ÄÚÈÝ

£¨¹²22·Ö£©

¢ñ£®£¨4·Ö£©ÏÂÁÐʵÑé²Ù×÷ÖУ¬ºÏÀíµÄÊÇ             

A£®ÊµÑéÊÒÖÆÒÒϩʱ£¬Ôھƾ«ºÍŨÁòËáµÄ»ìºÏÒºÖУ¬·ÅÈ뼸ƬËé´ÉƬ£¬¼ÓÈÈ»ìºÏÎʹҺÌåζÈѸËÙÉýµ½170¡æ

B£®ÑéÖ¤äåÒÒÍéË®½â²úÎïʱ£¬½«äåÒÒÍéºÍÇâÑõ»¯ÄÆÈÜÒº»ìºÏ£¬³ä·ÖÕñµ´ÈÜÒº¡¢¾²Ö㬴ýÒºÌå·Ö²ãºó£¬µÎ¼ÓÏõËáÒøÈÜÒº

C£®½«Í­Ë¿Íä³ÉÂÝÐý×´£¬Ôھƾ«µÆÉϼÓÈȱäºÚºó£¬Á¢¼´ÉìÈëÎÞË®ÒÒ´¼ÖУ¬Íê³ÉÒÒ´¼Ñõ»¯ÎªÒÒÈ©µÄʵÑé

D£®±½·ÓÖеμÓÉÙÁ¿Ï¡äåË®£¬¿ÉÓÃÀ´¶¨Á¿¼ìÑé±½·Ó

E£®¹¤Òµ¾Æ¾«ÖÆÈ¡ÎÞË®¾Æ¾«Ê±£¬ÏȼÓÉúʯ»ÒÈ»ºóÕôÁó£¬ÕôÁó±ØÐ뽫ζȼƵÄË®ÒøÇò²åÈë·´Ó¦ÒºÖУ¬²â¶¨·´Ó¦ÒºÎ¶È

¢ò£®£¨10·Ö£©ÊµÑéÊÒÓÃÏÂͼËùʾװÖÃÖƱ¸äå±½£¬²¢ÑéÖ¤¸Ã·´Ó¦ÊÇÈ¡´ú·´Ó¦¡£

(1) ¹Ø±ÕF»îÈû£¬´ò¿ªC»îÈû£¬ÔÚ×°ÓÐÉÙÁ¿±½µÄÈý¾±Æ¿ÖÐÓÉA¿Ú¼ÓÈëÉÙÁ¿ä壬ÔÙ¼ÓÈëÉÙÁ¿Ìúм£¬ÈûסA¿Ú£¬ÔòÈý¾±Æ¿Öз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º                             ¡£

(2) DÊÔ¹ÜÄÚ×°µÄÊÇ                £¬Æä×÷ÓÃÊÇ                         ¡£

(3) EÊÔ¹ÜÄÚ×°µÄÊÇ                £¬EÊÔ¹ÜÄÚ³öÏÖµÄÏÖÏóΪ                                    ¡£

(4) ´ýÈý¿ÚÉÕÆ¿Öеķ´Ó¦¼´½«½áÊøʱ(´ËʱÆøÌåÃ÷ÏÔ¼õÉÙ)£¬´ò¿ªF»îÈû£¬¹Ø±ÕC»îÈû£¬¿ÉÒÔ¿´µ½µÄÏÖÏóÊÇ                                                                                 ¡£

(5) ÉÏÒ»²½µÃµ½´Öäå±½ºó£¬ÒªÓÃÈçϲÙ×÷¾«ÖÆ£º

aÕôÁó£» bˮϴ£» cÓøÉÔï¼Á¸ÉÔ d 10%NaOHÈÜҺϴµÓ£» eˮϴ

ÕýÈ·µÄ²Ù×÷˳ÐòÊÇ                            

¢ó£®£¨8·Ö£©ÓÃÈçÓÒͼËùʾװÖýøÐÐʵÑ飬½«AÖðµÎ¼ÓÈëBÖУº

£¨1£© ÈôBΪNa2CO3·ÛÄ©£¬CΪC6H5ONaÈÜÒº£¬ÊµÑéÖй۲쵽СÊÔ¹ÜÄÚÈÜÒºÓɳÎÇå±ä»ë×Ç£¬ÔòÊÔ¹ÜCÖл¯Ñ§·´Ó¦µÄÀë×Ó·½³Ìʽ£º________________           ¡£È»ºóÍùÉÕ±­ÖмÓÈë·ÐË®£¬¿É¹Û²ìµ½ÊÔ¹ÜCÖеÄÏÖÏ󣺠               ¡£

£¨2£© ÈôBÊÇÉúʯ»Ò£¬¹Û²ìµ½CÈÜÒºÖÐÏÈÐγɳÁµí£¬È»ºó³ÁµíÈܽâ.µ±³ÁµíÍêÈ«Èܽ⣬ǡºÃ±ä³ÎÇåʱ£¬¹Ø±ÕE.È»ºóÍùСÊÔ¹ÜÖмÓÈëÉÙÁ¿ÒÒÈ©ÈÜÒº£¬ÔÙÍùÉÕ±­ÖмÓÈëÈÈË®£¬¾²ÖÃƬ¿Ì£¬¹Û²ìµ½ÊԹܱڳöÏÖ¹âÁÁµÄÒø¾µ£¬ÔòAÊÇ         £¨ÌîÃû³Æ£©£¬CÊÇ        £¨Ìѧʽ£©£¬ÓëÒÒÈ©µÄ»ìºÏºó£¬¸ÃÈÜÒºÖз´Ó¦µÄ»¯Ñ§·½³Ìʽ£º        _____________¡£ÒÇÆ÷DÔÚ´ËʵÑéÖеÄ×÷ÓÃÊÇ                         ¡£

 

¡¾´ð°¸¡¿

¢ñ£®AC£¨ÍêÈ«ÕýÈ·µÃ4·Ö£¬´ð¶ÔÒ»¸öµÃ2·Ö£¬ÓдíÎó´ð°¸µÃ0·Ö£©

¢ò£®£¨1£©  (2·Ö)

(2)  CCl4£»£¨1·Ö£©ÎüÊÕ»Ó·¢³öµÄ±½ÕôÆûºÍäå¡££¨1·Ö£©

(3)  AgNO3ÈÜÒº(1·Ö)£»µ¼¹Ü¿Ú²úÉú´óÁ¿°×Îí£¬£¨1·Ö£©Éú³Éµ­»ÆÉ«³Áµí¡££¨1·Ö£©

(4)  Ë®µ¹ÎüÖÁÈý¿ÚÉÕÆ¿ÖУ¨1·Ö£©

(5)  bdeca (»òedbca)£¨2·Ö£©

¢ó£®£¨1£©  (2·Ö)ÈÜÒºÓÉ»ë×DZä³ÎÇå¡££¨1·Ö£©

£¨2£© Ũ°±Ë®£¨1·Ö£©£¬AgNO3(1·Ö)£¬CH3CHO£«2Ag(NH3)2OHCH3COONH4£«H2O£«2Ag¡ý£«3NH3  (2·Ö)·ÀÖ¹µ¹Îü£¨1·Ö£©.

¡¾½âÎö¡¿ÂÔ

 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

£¨Ã¿¿Õ2·Ö£¬¹²22·Ö£©

I.ʵÑé²âµÃ0.01mol/LµÄKMnO4µÄÁòËáÈÜÒººÍ0.1mol/LµÄH2C2O4ÈÜÒºµÈÌå»ý»ìºÏºó£¬·´Ó¦ËÙÂʦÔ[mol/(L ¡¤ s)]Ó뷴Ӧʱ¼ät£¨s£©µÄ¹ØϵÈçͼËùʾ¡£»Ø´ðÈçÏÂÎÊÌ⣺

£¨1£©¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º                           

£¨2£©0¡út2ʱ¼ä¶ÎÄÚ·´Ó¦ËÙÂÊÔö´óµÄÔ­ÒòÊÇ£º               £¬

£¨3£©t2¡útʱ¼ä¶ÎÄÚ·´Ó¦ËÙÂʼõСµÄÔ­ÒòÊÇ£º                                     £¬

£¨4£©Í¼ÖÐÒõÓ°²¿·Ö¡°Ãæ»ý¡±±íʾt1¡út3ʱ¼äÀï                    ¡£

£Á£®Mn2+ÎïÖʵÄÁ¿Å¨¶ÈµÄÔö´ó           B£®Mn2+ÎïÖʵÄÁ¿µÄÔö¼Ó 

C£®SO42-ÎïÖʵÄÁ¿Å¨¶È                  D£®MnO4-ÎïÖʵÄÁ¿Å¨¶ÈµÄ¼õС

II. Ϊ±È½ÏFe3+ºÍCu2+¶ÔH2O2·Ö½âµÄ´ß»¯Ð§¹û£¬Ä³»¯Ñ§Ñо¿Ð¡×éµÄͬѧ·Ö±ðÉè¼ÆÁËÈçÏÂͼ¼×¡¢ÒÒËùʾµÄʵÑé¡£Çë»Ø´ðÏà¹ØÎÊÌ⣺

£¨1£©¶¨ÐÔ·ÖÎö£ºÈçÉÏͼ¼×¿É¹Û²ì                  £¬¶¨ÐԱȽϵóö½áÂÛ¡£ÓÐͬѧÌá³ö½«FeCl3¸ÄΪFe2(SO4)3¸üΪºÏÀí£¬ÆäÀíÓÉÊÇ                ¡£

£¨2£©¶¨Á¿·ÖÎö£ºÈçÉÏͼÒÒËùʾ£¬ÊµÑéʱ¾ùÒÔÉú³É40mLÆøÌåΪ׼£¬ÆäËû¿ÉÄÜÓ°ÏìʵÑéµÄÒòËؾùÒѺöÂÔ¡£Í¼ÖÐÒÇÆ÷AµÄÃû³ÆΪ           £¬ÊµÑéÖÐÐèÒª²âÁ¿µÄÊý¾ÝÊÇ                                 

£¨3£©¼ÓÈë0.01mol MnO2·ÛÄ©ÓÚ60mL H2O2ÈÜÒºÖУ¬ÔÚ±ê×¼×´¿öÏ·ųöÆøÌåÌå»ýºÍʱ¼äµÄ¹ØϵÈçͼËùʾ¡£Éè·Å³öÆøÌåµÄ×ÜÌå»ýΪV mL¡£

¢Ù·Å³öV/3 mLÆøÌåʱËùÐèʱ¼äΪ           min¡£

¢Ú ¸ÃH2O2ÈÜÒºµÄŨ¶ÈΪ                    

¢ÛA¡¢B¡¢C¡¢D¸÷µã·´Ó¦ËÙÂÊ¿ìÂýµÄ˳ÐòΪ        >     >      >     

 

£¨Ã¿¿Õ2·Ö£¬¹²22·Ö£©
I.ʵÑé²âµÃ0.01mol/LµÄKMnO4µÄÁòËáÈÜÒººÍ0.1mol/LµÄH2C2O4ÈÜÒºµÈÌå»ý»ìºÏºó£¬·´Ó¦ËÙÂʦÔ[mol/(L ¡¤ s)]Ó뷴Ӧʱ¼ät£¨s£©µÄ¹ØϵÈçͼËùʾ¡£»Ø´ðÈçÏÂÎÊÌ⣺

£¨1£©¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º                           
£¨2£©0¡út2ʱ¼ä¶ÎÄÚ·´Ó¦ËÙÂÊÔö´óµÄÔ­ÒòÊÇ£º              £¬
£¨3£©t2¡útʱ¼ä¶ÎÄÚ·´Ó¦ËÙÂʼõСµÄÔ­ÒòÊÇ£º                                     £¬
£¨4£©Í¼ÖÐÒõÓ°²¿·Ö¡°Ãæ»ý¡±±íʾt1¡út3ʱ¼äÀï                    ¡£
£Á£®Mn2+ÎïÖʵÄÁ¿Å¨¶ÈµÄÔö´ó           B£®Mn2+ÎïÖʵÄÁ¿µÄÔö¼Ó 
C£®SO42-ÎïÖʵÄÁ¿Å¨¶È                  D£®MnO4-ÎïÖʵÄÁ¿Å¨¶ÈµÄ¼õС
II.Ϊ±È½ÏFe3+ºÍCu2+¶ÔH2O2·Ö½âµÄ´ß»¯Ð§¹û£¬Ä³»¯Ñ§Ñо¿Ð¡×éµÄͬѧ·Ö±ðÉè¼ÆÁËÈçÏÂͼ¼×¡¢ÒÒËùʾµÄʵÑé¡£Çë»Ø´ðÏà¹ØÎÊÌ⣺

£¨1£©¶¨ÐÔ·ÖÎö£ºÈçÉÏͼ¼×¿É¹Û²ì                  £¬¶¨ÐԱȽϵóö½áÂÛ¡£ÓÐͬѧÌá³ö½«FeCl3¸ÄΪFe2(SO4)3¸üΪºÏÀí£¬ÆäÀíÓÉÊÇ                ¡£
£¨2£©¶¨Á¿·ÖÎö£ºÈçÉÏͼÒÒËùʾ£¬ÊµÑéʱ¾ùÒÔÉú³É40mLÆøÌåΪ׼£¬ÆäËû¿ÉÄÜÓ°ÏìʵÑéµÄÒòËؾùÒѺöÂÔ¡£Í¼ÖÐÒÇÆ÷AµÄÃû³ÆΪ           £¬ÊµÑéÖÐÐèÒª²âÁ¿µÄÊý¾ÝÊÇ                                 
£¨3£©¼ÓÈë0.01mol MnO2·ÛÄ©ÓÚ60mL H2O2ÈÜÒºÖУ¬ÔÚ±ê×¼×´¿öÏ·ųöÆøÌåÌå»ýºÍʱ¼äµÄ¹ØϵÈçͼËùʾ¡£Éè·Å³öÆøÌåµÄ×ÜÌå»ýΪV mL¡£

¢Ù·Å³öV/3 mLÆøÌåʱËùÐèʱ¼äΪ           min¡£
¢Ú¸ÃH2O2ÈÜÒºµÄŨ¶ÈΪ                    
¢ÛA¡¢B¡¢C¡¢D¸÷µã·´Ó¦ËÙÂÊ¿ìÂýµÄ˳ÐòΪ       >     >     >     

£¨¹²22·Ö£©Ä³ÊµÑéС×éÓÃÏÂÁÐ×°ÖýøÐÐÈçϵÄʵÑé¡£ 
£¨1£©ÊµÑé¹ý³ÌÖÐÍ­Íø³öÏÖºìÉ«ºÍºÚÉ«½»ÌæµÄÏÖÏó£¬Çëд³öÏàÓ¦µÄ»¯Ñ§·½³Ìʽ
                                ¡¢                                  ¡£
ʵÑéС×éÔÚ²»¶Ï¹ÄÈë¿ÕÆøµÄÇé¿öÏ£¬Ï¨Ãð¾Æ¾«µÆ£¬·¢ÏÖ·´Ó¦ÈÔÄܼÌÐø½øÐУ¬ËµÃ÷¸Ã·´Ó¦ÊÇ  ¡¡  ¡¡¡¡ ·´Ó¦¡£

£¨2£©¼×ºÍÒÒÁ½¸öˮԡ×÷Óò»Ïàͬ¡£¼×µÄ×÷ÓÃÊÇ                              £»
ÒÒµÄ×÷ÓÃÊÇ                             ¡£
£¨3£©·´Ó¦½øÐÐÒ»¶Îʱ¼äºó£¬¸ÉÔïµÄÔ‡¹ÜaÖÐÄÜÊÕ¼¯µ½²»Í¬µÄÎïÖÊ£¬ÄÇôÊÕ¼¯µ½µÄÓлúÎïÊÇ£¨ÇëÌîдÆä½á¹¹¼òʽ£©         £»¼¯ÆøÆ¿ÖÐÊÕ¼¯µ½µÄÆøÌåµÄ×îÖ÷Òª³É·ÖÊÇ     ¡££¨ÌîдÆøÌåµÄ·Ö×Óʽ£©
£¨4£©ÈôÊÔ¹ÜaÖÐÊÕ¼¯µ½µÄÒºÌåÓÃ×ÏɫʯÈïÊÔÖ½¼ìÑ飬ÊÔÖ½ÏÔºìÉ«£¬ÊµÑéС×éµÄͬѧÈÏΪÕâ¿ÉÄÜÊÇ´æÔÚ¸±²úÎïÒÒËá¡£³ýÈ¥¸ÃÎïÖÊ£¬¿ÉÒÔʹÓóø·¿Öг£¼ûµÄÒ»ÖÖ»¯ºÏÎ¸ÃÎïÖÊΪ    ¡££¨Ìîд»¯Ñ§Ê½»ò·Ö×Óʽ£©
£¨5£©Çëд³öÒÔÉÏʵÑéÉæ¼°µÄÓлúÎïÒÒ´¼ÔÚÉú»îÉú²úÖеÄÒ»ÖÖ¾ßÌåÓ¦Ó㺠           ¡£
£¨6£©¸ÃʵÑéµÄÄ¿µÄÊÇ£º                                     

£¨Ã¿¿Õ2·Ö£¬¹²22·Ö£©

I.ʵÑé²âµÃ0.01mol/LµÄKMnO4µÄÁòËáÈÜÒººÍ0.1mol/LµÄH2C2O4ÈÜÒºµÈÌå»ý»ìºÏºó£¬·´Ó¦ËÙÂʦÔ[mol/(L ¡¤ s)]Ó뷴Ӧʱ¼ät£¨s£©µÄ¹ØϵÈçͼËùʾ¡£»Ø´ðÈçÏÂÎÊÌ⣺

£¨1£©¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º                           

£¨2£©0¡út2ʱ¼ä¶ÎÄÚ·´Ó¦ËÙÂÊÔö´óµÄÔ­ÒòÊÇ£º               £¬

£¨3£©t2¡útʱ¼ä¶ÎÄÚ·´Ó¦ËÙÂʼõСµÄÔ­ÒòÊÇ£º                                      £¬

£¨4£©Í¼ÖÐÒõÓ°²¿·Ö¡°Ãæ»ý¡±±íʾt1¡út3ʱ¼äÀï                     ¡£

£Á£®Mn2+ÎïÖʵÄÁ¿Å¨¶ÈµÄÔö´ó            B£®Mn2+ÎïÖʵÄÁ¿µÄÔö¼Ó 

C£®SO42-ÎïÖʵÄÁ¿Å¨¶È                   D£®MnO4-ÎïÖʵÄÁ¿Å¨¶ÈµÄ¼õС

II. Ϊ±È½ÏFe3+ºÍCu2+¶ÔH2O2·Ö½âµÄ´ß»¯Ð§¹û£¬Ä³»¯Ñ§Ñо¿Ð¡×éµÄͬѧ·Ö±ðÉè¼ÆÁËÈçÏÂͼ¼×¡¢ÒÒËùʾµÄʵÑé¡£Çë»Ø´ðÏà¹ØÎÊÌ⣺

£¨1£©¶¨ÐÔ·ÖÎö£ºÈçÉÏͼ¼×¿É¹Û²ì                   £¬¶¨ÐԱȽϵóö½áÂÛ¡£ÓÐͬѧÌá³ö½«FeCl3¸ÄΪFe2(SO4)3¸üΪºÏÀí£¬ÆäÀíÓÉÊÇ                 ¡£

£¨2£©¶¨Á¿·ÖÎö£ºÈçÉÏͼÒÒËùʾ£¬ÊµÑéʱ¾ùÒÔÉú³É40mLÆøÌåΪ׼£¬ÆäËû¿ÉÄÜÓ°ÏìʵÑéµÄÒòËؾùÒѺöÂÔ¡£Í¼ÖÐÒÇÆ÷AµÄÃû³ÆΪ            £¬ÊµÑéÖÐÐèÒª²âÁ¿µÄÊý¾ÝÊÇ                                 

£¨3£©¼ÓÈë0.01mol MnO2·ÛÄ©ÓÚ60mL H2O2ÈÜÒºÖУ¬ÔÚ±ê×¼×´¿öÏ·ųöÆøÌåÌå»ýºÍʱ¼äµÄ¹ØϵÈçͼËùʾ¡£Éè·Å³öÆøÌåµÄ×ÜÌå»ýΪV mL¡£

¢Ù·Å³öV/3 mLÆøÌåʱËùÐèʱ¼äΪ            min¡£

¢Ú ¸ÃH2O2ÈÜÒºµÄŨ¶ÈΪ                    

¢ÛA¡¢B¡¢C¡¢D¸÷µã·´Ó¦ËÙÂÊ¿ìÂýµÄ˳ÐòΪ        >      >      >     

 

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

¾«Ó¢¼Ò½ÌÍø