ÌâÄ¿ÄÚÈÝ

̼ËáÄÆ¡ª¹ýÑõ»¯Çâ¼ÓºÏÎï(aNa2CO3¡¤bH2O2)¾ßÓÐƯ°×¡¢É±¾ú×÷Óá£ÊµÑéÊÒÓá°´¼Îö·¨¡±ÖƱ¸¸ÃÎïÖʵÄʵÑé²½ÖèÈçÏ£º
µÚ1²½£ºÈ¡ÊÊÁ¿Ì¼ËáÄÆÈܽâÓÚÒ»¶¨Á¿Ë®Àµ¹ÈëÉÕÆ¿ÖУ»ÔÙ¼ÓÈëÉÙÁ¿Îȶ¨¼Á(MgCl2ºÍNa2SiO3)£¬½Á°è¾ùÔÈ¡£
µÚ2²½£º½«ÊÊÁ¿30%µÄH2O2ÈÜÒºÔÚ½Á°è״̬ϵÎÈëÉÕÆ¿ÖУ¬ÓÚ15 ¡æ×óÓÒ·´Ó¦1 h¡£
µÚ3²½£º·´Ó¦Íê±ÏºóÔÙ¼ÓÈëÊÊÁ¿ÎÞË®ÒÒ´¼£¬¾²Öᢽᾧ£¬¹ýÂË¡¢¸ÉÔïµÃ²úÆ·¡£
(1)µÚ1²½ÖУ¬Îȶ¨¼ÁÓëË®·´Ó¦Éú³ÉÁ½ÖÖ³£¼ûµÄÄÑÈÜÎÆ仯ѧ·½³ÌʽΪ___________________________________________________________¡£
(2)µÚ2²½ÖУ¬·´Ó¦±£³ÖΪ15 ¡æ×óÓҿɲÉÈ¡µÄ´ëÊ©ÊÇ_____________________
___________________________________________________¡£
(3)µÚ3²½ÖУ¬ÎÞË®ÒÒ´¼µÄ×÷ÓÃÊÇ____________________________________¡£
(4)H2O2µÄº¬Á¿¿ÉºâÁ¿²úÆ·µÄÓÅÁÓ¡£ÏÖ³ÆÈ¡m g(Ô¼0.5 g)ÑùÆ·£¬ÓÃÐÂÖó·Ð¹ýµÄÕôÁóË®ÅäÖƳÉ250 mLÈÜÒº£¬È¡25.0 mLÓÚ׶ÐÎÆ¿ÖУ¬ÏÈÓÃÏ¡ÁòËáËữ£¬ÔÙÓÃc mol¡¤L£­1 KMnO4ÈÜÒºµÎ¶¨ÖÁÖյ㡣
¢ÙÅäÖÆ250 mLÈÜÒºËùÐèµÄ²£Á§ÒÇÆ÷ÓÐÉÕ±­¡¢²£Á§°ô¡¢Á¿Í²________¡¢________£»
¢ÚµÎ¶¨ÖÕµã¹Û²ìµ½µÄÏÖÏóÊÇ______________________________________¡£

(5)¿ÉÄ£ÄâÓÃÕôÁ󷨲ⶨÑùÆ·ÖÐ̼ËáÄƵĺ¬Á¿¡£×°ÖÃÈçÓÒͼËùʾ(¼ÓÈȺ͹̶¨×°ÖÃÒÑÂÔÈ¥)£¬ÊµÑé²½ÖèÈçÏ£º
²½Öè1£º°´ÓÒͼËùʾ×é×°ÒÇÆ÷£¬¼ì²é×°ÖÃÆøÃÜÐÔ¡£
²½Öè2£º×¼È·Á¿È¡(4)ÖÐËùÅäÈÜÒº50 mLÓÚÉÕÆ¿ÖС£
²½Öè3£º×¼È·Á¿È¡40.00 mLÔ¼0.2 mol¡¤L£­1 NaOHÈÜÒºÁ½·Ý£¬·Ö±ð×¢ÈëÉÕ±­ºÍ׶ÐÎÆ¿ÖС£
²½Öè4£º´ò¿ª»îÈûK1¡¢K2£¬¹Ø±Õ»îÈûK3»º»ºÍ¨È뵪ÆøÒ»¶Îʱ¼äºó£¬¹Ø±ÕK1¡¢K2£¬´ò¿ªK3£»¾­·ÖҺ©¶·ÏòÉÕÆ¿ÖмÓÈë10 mL 3 mol¡¤L£­1ÁòËáÈÜÒº¡£
²½Öè5£º¼ÓÈÈÖÁÉÕÆ¿ÖеÄÒºÌå·ÐÌÚ£¬ÕôÁ󣬲¢±£³ÖÒ»¶Îʱ¼ä¡£
²½Öè6£º¾­K1ÔÙ»º»ºÍ¨È뵪ÆøÒ»¶Îʱ¼ä¡£
²½Öè7£ºÏò׶ÐÎÆ¿ÖмÓÈëËá¼îָʾ¼Á£¬ÓÃc1 mol¡¤L£­1 H2SO4±ê×¼ÈÜÒºµÎ¶¨ÖÁÖյ㣬ÏûºÄH2SO4±ê×¼ÈÜÒºV1 mL¡£
²½Öè8£º½«ÊµÑé²½Öè1¡«7Öظ´Á½´Î¡£
¢Ù²½Öè3ÖУ¬×¼È·ÒÆÈ¡40.00 mL NaOHÈÜÒºËùÐèҪʹÓõÄÒÇÆ÷ÊÇ________£»
¢Ú²½Öè1¡«7ÖУ¬È·±£Éú³ÉµÄ¶þÑõ»¯Ì¼±»ÇâÑõ»¯ÄÆÈÜÒºÍêÈ«ÎüÊÕµÄʵÑé²½ÖèÊÇ________(ÌîÐòºÅ)£»
¢ÛΪ»ñµÃÑùÆ·ÖÐ̼ËáÄƵĺ¬Á¿£¬»¹Ðè²¹³äµÄʵÑéÊÇ______________________¡£
¡¡(1)MgCl2£«Na2SiO3£«2H2O===2NaCl£«Mg(OH)2¡ý£«H2SiO3¡ý¡¡(2)15 ¡æˮԡ»òÀäˮԡ¡¡(3)½µµÍ¹ý̼ËáÄƵÄÈܽâ¶È(ÓÐÀûÓÚ¾§ÌåÎö³ö)
(4)¢Ù250 mLÈÝÁ¿Æ¿¡¡½ºÍ·µÎ¹Ü¡¡¢ÚÈÜÒº³Ê·ÛºìÉ«ÇÒ30 s²»ÍÊÉ«
(5)¢Ù¼îʽµÎ¶¨¹Ü¡¡¢Ú1,5,6¡¡¢ÛÓÃH2SO4±ê×¼ÈÜÒºµÎ¶¨NaOHÈÜÒºµÄŨ¶È
¡¡(1)ÓÉÓÚÎȶ¨¼ÁÊÇÂÈ»¯Ã¾ºÍ¹èËáÄÆ£¬ÓÉÌâÒâ»ìºÏºóÉú³ÉÁ½ÖÖÄÑÈÜ»¯ºÏÎÒò´Ë¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪMgCl2£«Na2SiO3£«2H2O===Mg(OH)2¡ý£«H2SiO3¡ý£«2NaCl¡£(2)ÒªÏë½ÏÎȶ¨µÄ±£³ÖÈÜÒºµÄζÈΪ15 ¡æ£¬½ÏºÃµÄ·½·¨ÊÇÓÃ15 ¡æµÄˮԡ¡£(3)·´Ó¦Íê±ÏºóÏòÈÜÒºÖмÓÈëÎÞË®ÒÒ´¼£¬Ö÷Òª×÷ÓÃÊǽµµÍ¹ý̼ËáÄƵÄÈܽâ¶È£¬ÓÐÀûÓÚÆä½á¾§Îö³ö¡£(4)ÅäÖÆ250 mLÈÜÒºÐèÒªµÄ¹Ø¼üÒÇÆ÷ÊÇ250 mLµÄÈÝÁ¿Æ¿×÷ΪÁ¿Æ÷£¬×¼È·È·¶¨ÈÜÒºµÄÌå»ý£¬»¹ÐèÒªÉÕ±­¡¢Á¿Í²¡¢²£Á§°ô¡¢½ºÍ·µÎ¹ÜµÈ£»ÓÉÓÚÓÃËáÐÔ¸ßÃÌËá¼ØÈÜÒºµÎ¶¨Ë«ÑõË®£¬Òò´Ë´ïµ½ÖÕµãʱ£¬¹ýÁ¿µÄ¸ßÃÌËá¼Ø¿ÉÒÔ×öָʾ¼Á£¬ÖÕµãʱµÄÏÖÏóΪµÎÈë×îºóÒ»µÎËáÐÔ¸ßÃÌËá¼ØÈÜÒºÑÕÉ«³Ê·ÛºìÉ«£¬ÇÒ°ë·ÖÖÓ²»±äÉ«¡£(5)Ҫ׼ȷÁ¿È¡40.00 mL NaOHÈÜÒº£¬ÐèÒª¾«¶È¸ßµÄÁ¿Æ÷£¬²»ÄÜÓÃÁ¿Í²£¬Ó¦¸ÃÓüîʽµÎ¶¨¹Ü£»ÊµÑéʱÏÈͨÈ뵪ÆøÊÇΪÁË·ÀֹϵͳÖпÕÆøµÄCO2Ó°ÏìʵÑé½á¹û£¬×îºóͨµªÆøÊÇΪÁË°ÑÉú³ÉµÄCO2È«²¿±»ÉÕ¼îÎüÊÕ£¬¼ì²é×°ÖÃÆøÃÜÐÔÊÇΪÁË·ÀÖ¹Éú³ÉµÄCO2Òݳö£¬¼ÓÈÈÖó·ÐÉÕÆ¿ÖеÄÈÜÒºÊÇ°ÑÉú³É²¢ÈܽâÔÚÈÜÒºÖеÄCO2¸Ï³öÈ»ºó±»ÉÕ¼îÎüÊÕ£¬Òò´ËʵÑé²½Öè1¡¢5¡¢6ÊÇΪÁ˱£Ö¤Éú³ÉµÄCO2ÍêÈ«±»ÉÕ¼îÎüÊÕ£»ÓÉÓÚʵÑéÖÐÉÕ¼îÈÜÒºµÄŨ¶ÈÊÇ´óԼΪ0.2 mol¡¤L£­1£¬Òò´ËÒªÏë׼ȷ¼ÆËãʵÑé½á¹û£¬»¹ÐèÒªÖªµÀÉÕ¼îÈÜÒºµÄ׼ȷŨ¶È£¬Õâ¾ÍÐèÒªÓñê×¼ÁòËáÈÜÒºµÎ¶¨ÉÕ¼îÈÜÒºµÄŨ¶È¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
ÁòËáÍ­ÊÜÈÈ·Ö½âÉú³ÉÑõ»¯Í­ºÍÆøÌ壬¼ÓÈÈζȲ»Í¬£¬ÆøÌå³É·ÖÒ²²»Í¬¡£ÆøÌå³É·Ö¿ÉÄܺ¬SO2¡¢SO3ºÍO2ÖеÄÒ»ÖÖ¡¢Á½ÖÖ»òÈýÖÖ¡£Ä³»¯Ñ§¿ÎÍâ»î¶¯Ð¡×éͨ¹ýÉè¼Æ̽¾¿ÐÔʵÑ飬²â¶¨·´Ó¦²úÉúµÄSO2¡¢SO3ºÍO2µÄÎïÖʵÄÁ¿£¬²¢¼ÆËãÈ·¶¨¸÷ÎïÖʵĻ¯Ñ§¼ÆÁ¿Êý£¬´Ó¶øÈ·¶¨CuSO4·Ö½âµÄ»¯Ñ§·½³Ìʽ¡£ÊµÑéÓõ½µÄÒÇÆ÷ÈçÏÂͼËùʾ£º

[Ìá³ö²ÂÏë]
¢ñ.ËùµÃÆøÌåµÄ³É·Ö¿ÉÄÜÖ»º¬SO3Ò»ÖÖ£»
¢ò.ËùµÃÆøÌåµÄ³É·Ö¿ÉÄܺ¬ÓÐ________Á½ÖÖ£»
¢ó.ËùµÃÆøÌåµÄ³É·Ö¿ÉÄܺ¬ÓÐ________ÈýÖÖ¡£
[ʵÑé̽¾¿]
ʵÑé²Ù×÷¹ý³ÌÂÔ¡£ÒÑ֪ʵÑé½áÊøʱ£¬ÁòËáÍ­ÍêÈ«·Ö½â¡£
£¨1£©ÇëÄã×éװ̽¾¿ÊµÑéµÄ×°Ö㬰´´Ó×óÖÁÓҵķ½Ïò£¬¸÷ÒÇÆ÷½Ó¿ÚµÄÁ¬½Ó˳ÐòΪ¢Ù¡ú¢á¡ú¢â¡ú¢Þ¡ú¢Ý¡ú________¡ú________¡ú________¡ú________¡ú¢Ú£¨Ìî½Ó¿ÚÐòºÅ£©¡£
£¨2£©ÈôʵÑé½áÊøʱBÖÐÁ¿Í²Ã»ÓÐÊÕ¼¯µ½Ë®£¬ÔòÖ¤Ã÷²ÂÏë________ÕýÈ·¡£
£¨3£©ÓÐÁ½¸öʵÑéС×é½øÐиÃʵÑ飬ÓÉÓÚ¼ÓÈÈʱµÄζȲ»Í¬£¬ÊµÑé½áÊøºó²âµÃÏà¹ØÊý¾ÝÒ²²»Í¬£¬Êý¾ÝÈçÏ£º
ʵÑé
С×é
³ÆÈ¡CuSO4
µÄÖÊÁ¿/g
×°ÖÃCÔö¼Ó
µÄÖÊÁ¿/g
Á¿Í²ÖÐË®µÄÌå»ýÕÛËã³É±ê×¼×´¿öÏÂÆøÌåµÄÌå»ý/mL
Ò»
6.4
2.56
448
¶þ
6.4
2.56
224
 
Çëͨ¹ý¼ÆË㣬ÍƶϳöµÚһС×éºÍµÚ¶þС×éµÄʵÑéÌõ¼þÏÂCuSO4·Ö½âµÄ»¯Ñ§·½³Ìʽ¡£
µÚһС×飺________________________________________________________£»
µÚ¶þС×飺________________________________________________________¡£
ÁòËáÑÇÌúï§[(NH4)2SO4¡¤FeSO4¡¤6H2O]ΪdzÂÌÉ«¾§Ì壬Ò×ÈÜÓÚË®£¬²»ÈÜÓھƾ«£¬ÔÚË®ÖеÄÈܽâ¶È±ÈFeSO4»ò(NH4)2SO4¶¼ÒªÐ¡¡£ÊµÑéÊÒÖг£ÒÔ·ÏÌúмΪԭÁÏÀ´ÖƱ¸£¬Æä²½ÖèÈçÏ£º

ͼ1
²½Öè1£ºÌúмµÄ´¦Àí¡£½«·ÏÌúм·ÅÈëÈȵÄ̼ËáÄÆÈÜÒºÖнþÅݼ¸·ÖÖÓºó£¬ÓÃͼ1Ëùʾ·½·¨·ÖÀë³ö¹ÌÌ岢ϴµÓ¡¢¸ÉÔï¡£
²½Öè2£ºFeSO4ÈÜÒºµÄÖƱ¸¡£½«´¦ÀíºÃµÄÌúм·ÅÈë׶ÐÎÆ¿£¬¼ÓÈë¹ýÁ¿µÄ3 mol¡¤
L£­1H2SO4ÈÜÒº£¬¼ÓÈÈÖÁ³ä·Ö·´Ó¦ÎªÖ¹¡£³ÃÈȹýÂË(Èçͼ2Ëùʾ)£¬ÊÕ¼¯ÂËÒººÍÏ´µÓÒº¡£

ͼ2
²½Öè3£ºÁòËáÑÇÌú淋ÄÖƱ¸¡£ÏòËùµÃFeSO4ÈÜÒºÖмÓÈë±¥ºÍ(NH4)2SO4ÈÜÒº£¬¾­¹ý¼ÓÈÈŨËõ¡¢ÀäÈ´½á¾§¡¢¹ýÂË¡¢ÒÒ´¼Ï´µÓºóµÃµ½ÁòËáÑÇÌú茶§Ìå¡£
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
(1)²½Öè1ÖÐͼ1·ÖÀë·½·¨³ÆΪ________·¨¡£
(2)²½Öè2ÖÐÓÐÒ»´¦Ã÷ÏÔ²»ºÏÀíµÄÊÇ___________________________________¡£
³ÃÈȹýÂ˵ÄÀíÓÉÊÇ________________________________________________¡£
(3)²½Öè3¼ÓÈÈŨËõ¹ý³ÌÖУ¬µ±________ʱֹͣ¼ÓÈÈ¡£ÓÃÎÞË®ÒÒ´¼Ï´µÓ¾§ÌåµÄÔ­ÒòÊÇ______________________________________________________________¡£
(4)FeSO4¡¤7H2OÔÚ³±ÊªµÄ¿ÕÆøÖÐÒ×±»Ñõ»¯³ÉFe(OH)SO4¡¤3H2O£¬Ð´³ö¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ________________________________________________________¡£
¿ÕÆø´µ³ö·¨¹¤ÒÕ,ÊÇÄ¿Ç°¡°º£Ë®Ìáä塱µÄ×îÖ÷Òª·½·¨Ö®Ò»¡£Æ乤ÒÕÁ÷³ÌÈçÏÂ:

(1)äåÔÚÖÜÆÚ±íÖÐλÓÚ¡¡¡¡¡¡¡¡ÖÜÆÚ¡¡¡¡¡¡¡¡×å¡£ 
(2)²½Öè¢ÙÖÐÓÃÁòËáËữ¿ÉÌá¸ßCl2µÄÀûÓÃÂÊ,ÀíÓÉÊÇ                                                                            
(3)²½Öè¢ÜÀûÓÃÁËSO2µÄ»¹Ô­ÐÔ,·´Ó¦µÄÀë×Ó·½³ÌʽΪ                                                         
(4)²½Öè¢ÞµÄÕôÁó¹ý³ÌÖÐ,ζÈÓ¦¿ØÖÆÔÚ80~90 ¡æ,ζȹý¸ß»ò¹ýµÍ¶¼²»ÀûÓÚÉú²ú,Çë½âÊÍÔ­Òò¡¡                                                                                                        ¡£ 
(5)²½Öè¢àÖÐäåÕôÆøÀäÄýºóµÃµ½ÒºäåÓëäåË®µÄ»ìºÏÎï,¿ÉÀûÓÃËüÃǵÄÃܶÈÏà²îºÜ´óµÄÌصã½øÐзÖÀë¡£·ÖÀëÒÇÆ÷µÄÃû³ÆÊÇ¡¡                                                      ¡£ 
(6)²½Öè¢Ù¡¢¢ÚÖ®ºó²¢Î´Ö±½ÓÓá°º¬Br2µÄº£Ë®¡±½øÐÐÕôÁóµÃµ½Òºäå,¶øÊǾ­¹ý¡°¿ÕÆø´µ³ö¡±¡°SO2ÎüÊÕ¡±¡°Ñõ»¯¡±ºóÔÙÕôÁó,ÕâÑù²Ù×÷µÄÒâÒåÊÇ¡¡                                                                            ¡£ 

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

¾«Ó¢¼Ò½ÌÍø