Çë¸ù¾Ý½Ì²ÄÉϵÄʵÑé¡°ÄÆÓëÂÈÆø·´Ó¦¡±£¬Íê³ÉÏÂÁÐÎÊÌâ¡£

£¨1£©È¡Ò»¿éÂ̶¹´óµÄ½ðÊôÄÆ£¨ÇÐÈ¥Ñõ»¯²ã£©£¬ÓÃÂËÖ½Îü¾»Æä±íÃæµÄúÓÍ£¬·ÅÔÚʯÃÞÍøÉÏ£¬Óþƾ«µÆ΢ÈÈ¡£´ýÄÆÈÛ³ÉÇò״ʱ£¬½«Ê¢ÓÐÂÈÆøµÄ¼¯ÆøƿѸËÙµ¹¿ÛÔÚÄƵÄÉÏ·½¡£¸ù¾ÝËùѧµÄ֪ʶ·ÖÎö¸ÃʵÑé´æÔÚÄÄЩȱµã£¿
¢Ù____________________________________________________£»
¢Ú____________________________________________________£»
¢Û________________¡££¨ÖÁÉÙÌî2Ìõ£©
£¨2£©Ä³Í¬Ñ§¸ù¾ÝÒÔÉÏ´æÔÚµÄȱµã¸Ä½øʵÑé×°Öã¬ÈçͼËùʾ£º

ʵÑé²½Ö裺
a£®È¡Â̶¹Á£´óµÄÄÆ£¬ÓÃÂËÖ½Îü¸É±íÃæµÄúÓÍ£¬ÇÐÈ¥Ñõ»¯²ã£¬·ÅÈë²£Á§¹ÜÖУ¬°´Í¼Ê¾°²×°ºÃÒÇÆ÷£»
b£®ÂýÂýµÎ¼ÓŨÑÎËᣬÁ¢¼´¾çÁÒ·´Ó¦²úÉúÂÈÆø£»
c£®µ±²£Á§¹ÜÖгäÂú»ÆÂÌÉ«ÆøÌåʱ£¬ÔÙ¼ÓÈÈÄÆ£¬ÄÆÈÛ»¯²¢È¼ÉÕ¡£
¢Ù¹Û²ìµ½µÄʵÑéÏÖÏóÓУºµ±µÎ¼ÓŨÑÎËáºóÊÔ¹ÜÖвúÉú________É«ÆøÌ壻ÄƾçÁÒȼÉÕ£¬»ðÑæ³Ê________É«ÇÒÓÐ________Éú³É£¬·´Ó¦½áÊøºó£¬¹Ü±ÚÉϹ۲쵽ÓÐ________Éú³É¡£
¢Ú¸Ä½øºóµÄʵÑéÓŵ㣺a.________£¬b.________£¬c.______£¨ÖÁÉٻشð2Ìõ£©¡£
¢Ûд³öNaÔÚCl2ÖÐȼÉյĻ¯Ñ§·½³Ìʽ________________£¬²¢Óõç×Óʽ±íʾ²úÎïÐγɹý³Ì________________¡£

£¨1£©¢ÙNaÔ¤ÏÈÔÚ¿ÕÆøÖмÓÈÈ£¬»áÉú³ÉÑõ»¯Î¿ÉÄÜÓ°ÏìNaÔÚCl2ÖÐȼÉÕ
¢ÚÔ¤ÏÈÊÕ¼¯µÄCl2ÔÚ½øÐÐʵÑéʱ¿ÉÄܲ»¹»»ò¹ýÁ¿
¢ÛʵÑé¹ý³ÌÖлáÔì³ÉÎÛȾ
£¨2£©¢Ù»ÆÂÌ¡¡»Æ¡¡°×ÑÌ¡¡°×É«¹ÌÌå
¢Úa.Õû¸öʵÑé¹ý³ÌÖÐÂÈÆø±£³ÖÒ»¶¨Å¨¶ÈºÍ´¿¶È£¬±ÜÃâ·¢Éú¸±·´Ó¦¡¡b£®°²È«¿É¿¿£¬ÎÛȾÉÙ¡¡c£®¿ÉÒÔ±ÜÃâÄÆÔÚ¼ÓÈÈʱÉú³ÉNa2O2
¢Û2Na£«Cl22NaCl    

½âÎö

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º¼ÆËãÌâ

½«Ò»Ð¡¿é½ðÊôÄÆͶÈ뵽ʢÓÐ100ml AlCl3ºÍMgCl2µÄ»ìºÏÈÜÒºµÄÉÕ±­ÖУ¬·¢ÏÖÉÕ±­ÖÐÓÐÆøÅݲúÉú£¬ÔÚÕâ¹ý³ÌÖл¹¿´µ½ÉÕ±­ÖÐÓа×É«³Áµí²úÉú£¬³ÁµíÎïÏȶàºóÉÙ¡£·´Ó¦Íê±Ïºó£¬ÊÕ¼¯µ½±ê×¼×´¿öÏÂÇâÆø13.44LͬʱµÃµ½21.4g³Áµí£¬½«³ÁµíÈÜÓÚ¹ýÁ¿µÄNaOHÈÜÒºÖУ¬·¢ÏÖ³Áµí¼õÉÙÁË15.6g£¬ÇóÔ­»ìºÏÈÜÒºÖÐMg2+¡¢Al3+¼°Cl-µÄÎïÖʵÄÁ¿Å¨¶È¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÊµÑéÌâ

ΪÁË¡°Ì½¾¿Ìú¼°Æ仯ºÏÎïµÄÑõ»¯ÐÔ»ò»¹Ô­ÐÔ¡±£®
£¨1£©ÊµÑéǰͬѧԤ²â£ºFe2+¿Ï¶¨¼ÈÓл¹Ô­ÐÔÓÖÓÐÑõ»¯ÐÔ£®ÄãÈÏΪËûÔ¤²âµÄÒÀ¾ÝÊÇ£º__         
£¨2£©¼×ͬѧÓûÓÃʵÑéÖ¤Ã÷ËûµÄÔ¤²â£®ÊµÑéÊÒÌṩÁËÏÂÁÐÊÔ¼Á£º3£¥µÄH2O2ÈÜÒº¡¢Ð¿Á£¡¢Í­Æ¬¡¢0£®1mol¡¤L -lFeCl2ÈÜÒº¡¢KSCNÈÜÒº¡¢ÐÂÖÆÂÈË®£®
¢ÙÈô¼Æ»®ÔÚ0£®1 mol¡¤L-l FeCl2ÈÜÒºÖеÎÈëÐÂÖÆÂÈË®£¬Ì½¾¿Fe2+µÄ»¹Ô­ÐÔ£¬ÄãÔ¤¼Æ¿ÉÄÜ·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ              
¢ÚʵÑéÖУ¬¼×ͬѧ·¢ÏÖÏÖÏó²»Ì«Ã÷ÏÔ£¬ÀÏʦ·ÖÎö¿ÉÄÜÊDzúÎïµÄº¬Á¿Ì«µÍ£¬½¨Òé¿ÉÒÔͨ¹ý¼ìÑéFe2+·´Ó¦µÄ²úÎïµÄ´æÔÚÒÔ»ñÈ¡Ö¤¾Ý£®ÄãÈÏΪ¿ÉÑ¡_____µÎÈëСÃ÷ËùµÃµÄ»ìºÏÒºÖУ¬²¢Í¨¹ýÈÜÒº³öÏÖ___É«µÄÏÖÏó£¬Ö¤Ã÷¸ÃͬѧµÄ¹ÛµãºÍʵÑé·½°¸¶¼ÊÇÕýÈ·µÄ£®
¢Û¶ÔÓÚÖ¤Ã÷Fe2+¾ßÓÐÑõ»¯ÐÔ£¬¸ÃͬѧÈÏΪ½ðÊôµ¥Öʶ¼¾ßÓл¹Ô­ÐÔ£¬²¢·Ö±ð½«Í­Æ¬¡¢Ð¿Á£Í¶ÈëFeCl2ÈÜÒºÖУ¬½á¹ûпÁ£Öð½¥±äС£®ÓÉ´Ë˵Ã÷ÈýÖÖ½ðÊôµÄ»¹Ô­ÐÔÓÉÇ¿ÖÁÈõµÄ˳ÐòΪ________£®
£¨3£©¼×ͬѧ·ÖÎöH2O2ÖÐÑõÔªËØÏÔ£­1¼Û(Öмä¼Û)£¬²¢Ìá³öÒÉÎÊ£ºH2O2ÓëFeCl2µÄ·´Ó¦Ê±£¬Fe2+»¹×÷Ñõ»¯¼ÁÂð£¿
¢ÙÇëÄãΪ¸ÃͬѧÊáÀíÏà¹ØÎÊÌ⣺ÈôFe2+ÔÚ·´Ó¦ÖбíÏÖ³öÑõ»¯ÐÔӦת»¯³É______(Ìî΢Á£·ûºÅ£¬ÏÂͬ)£¬ÈôFe2+ÔÚ·´Ó¦ÖбíÏÖ³ö»¹Ô­ÐÔӦת»¯³É_____£®
¢Úʵ¼ÊÉÏFe2+»¹Ô­ÐÔ½ÏÇ¿£¬ÊµÑéÊÒµÄFeCl2ÈÜÒº³£ÒòÑõ»¯¶ø±äÖÊ£®³ýÔӵķ½·¨ÊÇ£º                £¬Ïà¹Ø·´Ó¦µÄÀë×Ó·½³Ìʽ£º_____      £®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÊµÑéÌâ

ÔÚʵÑéÊҲⶨ̼ËáÄÆÓë̼ËáÇâÄƵĻìºÏÎïÖУ¬Ì¼ËáÄƵÄÖÊÁ¿·ÖÊý[Ó÷ûºÅw(Na2CO3)±íʾ]£¬³ÆÈ¡´Ë»ìºÏÎï5.lg£¬ÈÜÓÚË®ÖУ¬Åä³É250mLÈÜÒº¡£
a£®(10·Ö)·½°¸Ò»£º³Áµí·¨²âw(Na2CO3)ÀûÓû¯Ñ§·´Ó¦°ÑHCO3£­¡¢CO32£­Íêȫת»¯Îª³Áµí£¬³ÆÈ¡³ÁµíµÄÖÊÁ¿£¬Óɴ˼ÆËã»ìºÏÎïÖÐw (Na2CO3)¡£
£¨1£©Á¿È¡100 mLÅäÖƺõÄÈÜÒºÓÚÉÕ±­ÖУ¬µÎ¼Ó×ãÁ¿³Áµí¼Á£¬°ÑÈÜÒºÖÐHCO3£­¡¢CO32£­Íêȫת»¯Îª³Áµí£¬Ó¦Ñ¡µÄÊÔ¼ÁÊÇ___________ £¨Ìî±àºÅ£©¡£

A£®CaCl2 B£®MgSO4 C£®£®NaCI D£®Ba(OH)2
£¨2£©¼òÊöÖ¤Ã÷HCO3£­¡¢CO32£­ÒÑÍêÈ«³ÁµíµÄʵÑé²Ù×÷_________________________¡£
£¨3£©¹ýÂË£¬ÌáÈ¡³Áµí£¬Ôò¹ýÂ˲Ù×÷ËùÐèÒªµÄ²£Á§ÒÇÆ÷ÓÐ________________________¡£
£¨4£©Ï´µÓ³Áµí£¬¼òÊöÏ´µÓ³ÁµíµÄ²Ù×÷_____________________________¡£
£¨5£©¸ÉÔï³ä·Ö£¬³ÆÈ¡³ÁµíµÄÖÊÁ¿Îª9.8g£¬Óɴ˼ÆËãw(Na2CO3)¡£Èç¹û´Ë²½ÖУ¬³Áµíδ¸ÉÔï³ä·Ö¾Í³ÆÁ¿£¬Ôò²âµÃw (Na2CO3)________________£¨ÌîÆ«´ó»òƫС¡¢ÎÞÓ°Ï죩¡£
b£®·½°¸¶þ£ºµÎ¶¨·¨²âw(Na2CO3)Á¿È¡25.00 mLÅäÖƺõÄÈÜÒº¼ÓÈë׶ÐÎÆ¿ÖУ¬µÎ¼Ó2µÎ·Ó̪ÊÔ¼Á£¬Ò¡ÔÈ£¬ÓÃ0.2000 mol/LµÄÑÎËá½øÐе樵½Öյ㡣Öظ´´Ë²Ù×÷2´Î£¬ÏûºÄÑÎËáµÄÌå»ýƽ¾ùֵΪ20.00 mL¡£      [ÒÑÖª±¥ºÍµÄ̼ËáÈÜÒºPHΪ3.9]
£¨1£©Á¿È¡25.00 mLÅäÖƺõÄÈÜÒº£¬Ó¦Ñ¡Ôñ_______________ÒÇÆ÷À´Íê³É¡£
£¨2£©Åжϵζ¨ÖÕµãµÄÒÀ¾ÝÊÇ_____________________£¬´Ë¹ý³ÌÖз´Ó¦µÄÀë×Ó·½³ÌʽΪ__________________________________________________¡£
£¨3£©´Ë·¨²âµÃw(Na2CO3)=________%£¨±£ÁôÁ½Î»Ð¡Êý£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÊµÑéÌâ

ÈýÑõ»¯¶þÌúºÍÑõ»¯ÑÇÍ­¶¼ÊǺìÉ«·ÛÄ©£¬³£ÓÃ×÷ÑÕÁÏ¡£Ä³Ð£Ò»»¯Ñ§ÊµÑéС×éͨ¹ýʵÑéÀ´Ì½¾¿Ò»ºìÉ«·ÛÄ©ÊÇFe2O3¡¢Cu2O»ò¶þÕߵĻìºÏÎ̽¾¿¹ý³ÌÈçÏ£º
²éÔÄ×ÊÁÏ£ºCu2OÊÇÒ»ÖÖ¼îÐÔÑõ»¯ÎÈÜÓÚÏ¡ÁòËáÉú³ÉCuºÍCuSO4£¬ÔÚ¿ÕÆøÖмÓÈÈÉú³ÉCuO¡£
Ìá³ö¼ÙÉè
¼ÙÉè1£ººìÉ«·ÛÄ©ÊÇFe2O3
¼ÙÉè2£ººìÉ«·ÛÄ©ÊÇCu2O
¼ÙÉè3£ººìÉ«·ÛÄ©ÊÇFe2O3ºÍCu2OµÄ»ìºÏÎï
Éè¼Æ̽¾¿ÊµÑé
È¡ÉÙÁ¿·ÛÄ©·ÅÈë×ãÁ¿Ï¡ÁòËáÖУ¬ÔÚËùµÃÈÜÒºÖÐÔٵμÓKSCNÊÔ¼Á¡£
£¨1£©Èô¼ÙÉè1³ÉÁ¢£¬ÔòʵÑéÏÖÏóÊÇ_____________________________________________¡£
£¨2£©ÈôµÎ¼ÓKSCNÊÔ¼ÁºóÈÜÒº²»±äºìÉ«£¬ÔòÖ¤Ã÷Ô­¹ÌÌå·ÛÄ©ÖÐÒ»¶¨²»º¬ÈýÑõ»¯¶þÌú¡£ÄãÈÏΪÕâÖÖ˵·¨ºÏÀíÂð£¿________¡£¼òÊöÄãµÄÀíÓÉ(²»Ðèд³ö·´Ó¦·½³Ìʽ)____________
________________________________________________________________________¡£
£¨3£©Èô¹ÌÌå·ÛÄ©ÍêÈ«ÈܽâÎÞ¹ÌÌå´æÔÚ£¬µÎ¼ÓKSCNÊÔ¼ÁʱÈÜÒº²»±äºìÉ«£¬ÔòÖ¤Ã÷Ô­¹ÌÌå·ÛÄ©ÊÇ________£¬Ð´³ö·¢Éú·´Ó¦µÄÀë×Ó·½³Ìʽ________________________________¡£
̽¾¿ÑÓÉì
¾­ÊµÑé·ÖÎö£¬È·¶¨ºìÉ«·ÛĩΪFe2O3ºÍCu2OµÄ»ìºÏÎï¡£
£¨4£©ÊµÑéС×éÓûÓüÓÈÈ·¨²â¶¨Cu2OµÄÖÊÁ¿·ÖÊý£¬È¡a g¹ÌÌå·ÛÄ©ÔÚ¿ÕÆøÖгä·Ö¼ÓÈÈ£¬´ýÖÊÁ¿²»Ôٱ仯ʱ£¬³ÆÆäÖÊÁ¿Îªb g(b£¾a)£¬Ôò»ìºÏÎïÖÐCu2OµÄÖÊÁ¿·ÖÊýΪ________¡£
£¨5£©ÊµÑéС×éÓûÀûÓøúìÉ«·ÛÄ©ÖÆÈ¡½Ï´¿¾»µÄµ¨·¯(CuSO4¡¤5H2O)¡£¾­²éÔÄ×ÊÁϵÃÖª£¬ÔÚÈÜÒºÖÐͨ¹ýµ÷½ÚÈÜÒºµÄËá¼îÐÔ¶øʹCu2£«¡¢Fe2£«¡¢Fe3£«·Ö±ðÉú³É³ÁµíµÄpHÈçÏ£º

ÎïÖÊ
Cu(OH)2
Fe(OH)2
Fe(OH)3
¿ªÊ¼³ÁµípH
6.0
7.5
1.4
³ÁµíÍêÈ«pH
13
14
3.7
 
ʵÑéÊÒÓÐÏÂÁÐÊÔ¼Á¿É¹©Ñ¡Ôñ£ºA.ÂÈË®¡¡B£®H2O2¡¡C£®NaOH¡¡D£®Cu2(OH)2CO3
ʵÑéС×éÉè¼ÆÈçÏÂʵÑé·½°¸£º

ÊԻشð£º
¢ÙÊÔ¼Á1Ϊ________(Ìî×Öĸ£¬ºóͬ)£¬ÊÔ¼Á2Ϊ________¡£
¢Ú¹ÌÌåXµÄ»¯Ñ§Ê½Îª____________________________________________________¡£
¢Û²Ù×÷¢ñΪ___________________________________________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÊµÑéÌâ

ΪÁË̽¾¿SO2ÓëNa2O2µÄ·´Ó¦ÊÇ·ñÀàËÆÓÚCO2ÓëNa2O2µÄ·´Ó¦£¬¼×ͬѧÉè¼ÆÁËÈçͼËùʾµÄʵÑé×°Ö㬻شðÏÂÁÐÎÊÌ⣺

£¨1£©ÒÆ¿ªÃÞ»¨£¬½«´ø»ðÐǵÄľÌõ·ÅÔÚCÊԹܿڣ¬Î´¼ûľÌõ¸´È¼£¬¼×ͬѧÒò´ËÈÏΪSO2ÓëNa2O2µÄ·´Ó¦²»Í¬ÓÚCO2¡£Çë°´¼×ͬѧµÄ¹Ûµãд³ö·´Ó¦µÄ»¯Ñ§·½³Ìʽ                                                                       ¡£
£¨2£©ÒÒͬѧÈÏΪÎÞÂÛ·´Ó¦Ô­ÀíÈçºÎ£¬×îÖÕ¶¼ÓÐO2²úÉú£¬ÒÒͬѧµÄÀíÓÉÊÇ                               ¡£°´ÕÕÒÒͬѧµÄ¹Ûµã£¬¸Ã×°ÖÃÐè×öµÄ¸Ä½øÊÇ                                                                       
                                                                       ¡£
£¨3£©¼ÙÉèNa2O2ÍêÈ«·´Ó¦£¬·´Ó¦ºóB×°ÖÃÖйÌÌåÉú³ÉÎï¿ÉÄÜÊÇ£º¢ÙNa2SO3£»¢ÚNa2SO4£»¢ÛNa2SO3ºÍNa2SO4¡£
ÇëÉè¼ÆʵÑé·½°¸¼ìÑ飬д³öʵÑé²½ÖèÒÔ¼°Ô¤ÆÚÏÖÏóºÍ½áÂÛ£¬Íê³ÉÏÂ±í£º
ÏÞÑ¡ÊÔ¼Á£º2 mol¡¤L£­1 HClÈÜÒº£¬1 mol¡¤L£­1 HNO3ÈÜÒº£¬1 mol¡¤L£­1 BaClÈÜÒº£¬1 mol¡¤L£­1 Ba£¨NO3£©2ÈÜÒº£¬0.01 mol¡¤L£­1 KMnO4ËáÐÔÈÜÒº¡£

ʵÑé²½Öè
Ô¤ÆÚÏÖÏóºÍ½áÂÛ
²½Öè1£ºÈ¡BÖеÄÉÙÁ¿¹ÌÌåÑùÆ·ÓÚÊÔ¹ÜÖУ¬µÎ¼Ó×ãÁ¿ÕôÁóË®£¬Èܽ⣬ȻºóÈ¡ÉÙÁ¿´ý²âÒº·Ö±ðÖÃÓÚ¢ñ¡¢¢òÊÔ¹ÜÖÐ
¹ÌÌåÍêÈ«Èܽâ
²½Öè2£ºÍù¢ñÊÔ¹ÜÖмÓÈë                           £¬ÔٵμӠ                    
                                                                       £¬
ÔòÖ¤Ã÷Éú³ÉÎïÖк¬Na2SO4
 
²½Öè3£ºÍù¢òÊÔ¹ÜÖР                                                                      
 
                                                                       
Èô                                                                       £¬
ÔòÖ¤Ã÷Éú³ÉÎïÖÐÓÐNa2SO3£»Èô
 
                                                                       
 
Ôò˵Ã÷Éú³ÉÎïÖÐûÓÐNa2SO3¡£
 
 
£¨4£©Éú³ÉÎïÖÐÑÇÁòËáÄƺ¬Á¿µÄ²â¶¨£º
¢ÙÈ¡a gÉú³ÉÎïÅäÖƳÉ100 mLÈÜÒº£¬È¡10.00 mL¸ÃÈÜÒºÓÚ׶ÐÎÆ¿ÖУ¬¼ÓÈ뼸µÎµí·ÛÈÜÒº×÷ָʾ¼Á£¬ÓÃ0.010 0 mol¡¤L£­1µâË®½øÐе樣¬µÎ¶¨ÖÕµãÏÖÏóΪ                                 ¡£¼Ç¼Êý¾Ý£¬Öظ´µÎ¶¨2´Î£¬Æ½¾ùÏûºÄµâË®20.00 mL¡£
¢Ú¼ÆË㣺Éú³ÉÎïÖÐÑÇÁòËáÄƵÄÖÊÁ¿·ÖÊýΪ           ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÊµÑéÌâ

Ò»°ã²â¶¨ÑùÆ·Öгɷֺ¬Á¿µÄʵÑéÓ¦Öظ´2¡«3´Î¡£ÎªÁ˲ⶨijÇâÑõ»¯ÄƹÌÌåÖлìÓеÄ̼ËáÄƵÄÖÊÁ¿·ÖÊý£¬¼×¡¢ÒÒ¡¢±ûÈýλͬѧ·Ö±ðÉè¼ÆÁËÈçÏÂʵÑé·½°¸£º

¢ñ£®¼×ͬѧµÄ·½°¸ÈçͼËùʾ£º
£¨1£©ÈçºÎ¼ìÑéA×°ÖõÄÆøÃÜÐÔ£¿_____________________________________________¡£
£¨2£©¼×ͬѧÖظ´½øÐÐÁËÈý´ÎʵÑ飬µÃµ½Ì¼ËáÄƵÄÖÊÁ¿·ÖÊýµÄÊý¾Ý´æÔڽϴóµÄÆ«²î£¬ÄãÈÏΪ¿ÉÄÜÒýÆð²âÁ¿½á¹ûÆ«µÍµÄÔ­ÒòÊÇ_______£¨ÌîÐòºÅ£©¡£
A£®×°ÖÃÄÚÔ­ÓпÕÆøÖеĶþÑõ»¯Ì¼ÆøÌåÒ²±»¼îʯ»ÒÎüÊÕ
B£®×°ÖÃÍâ¿ÕÆøÖеÄË®ÕôÆøºÍ¶þÑõ»¯Ì¼±»¼îʯ»ÒÎüÊÕ
C£®·´Ó¦Íê³Éºó£¬×°ÖÃÖеĶþÑõ»¯Ì¼Ã»ÓÐÈ«²¿±»¼îʯ»ÒÎüÊÕ
D£®¼ÓÈëÏ¡ÁòËáµÄÁ¿²»×ã¡¢·´Ó¦²»³ä·Ö
£¨3£©ÎªÁËÈü׵ÄʵÑé²âÁ¿½á¹û¸ü׼ȷ£¬ÔÚÆäËûʵÑé²½Ö趼ÕýÈ·µÄÌõ¼þÏ£¬ÄãÈÏΪͼÖеÄʵÑé×°ÖÃÓ¦¸ÃÈçºÎ¸Ä½ø£º______________¡£
¢ò£®ÒÒͬѧµÄ·½°¸ÊÇ£º´ÓͼÖÐËùÌṩµÄ×°ÖÃÖÐÑ¡ÔñʵÑé×°Ö㬴úÌæ¼×ͬѧʵÑé×°ÖÃÖеÄB¡¢C£¬Í¨¹ý²â¶¨·Å³öµÄ¶þÑõ»¯Ì¼µÄÌå»ý£¨²»¿¼ÂǶþÑõ»¯Ì¼ÈÜÓÚË®£©À´¼ÆËã¡£

Ñ¡Ôñ×î¼ò×°ÖõÄÁ¬½Ó˳ÐòΪ_______¡£
¢ó£®±ûͬѧµÄ·½°¸ÊÇ£º³ÆÈ¡ÑùÆ·m g£¬²¢Èܽ⣬¼ÓÈë¹ýÁ¿ÂÈ»¯±µÈÜÒº£¬¹ýÂË¡¢Ï´µÓ¡¢ºæ¸É¡¢³ÆÁ¿£¬µÃ¹ÌÌån g¡£
£¨1£©ÅäÖÆ100 mL 0£®10 mol/L BaCl2ÈÜÒºµÄʵÑéÖÐËùÐèµÄ²£Á§ÒÇÆ÷³ýÉÕ±­¡¢²£Á§°ô¡¢½ºÍ·µÎ¹Ü¡¢Á¿Í²Í⻹ÓÐ_______£¨ÌîÒÇÆ÷Ãû³Æ£©¡£
£¨2£©»ìºÏÎïÖÐ̼ËáÄƵÄÖÊÁ¿·ÖÊýΪ£¨ÓÃm¡¢n±íʾ£©_______¡£
£¨3£©Ca2+¡¢Ba2+¶¼¿ÉÒÔʹ³ÁµíÍêÈ«£¬ÄÜ·ñʹÓÃÂÈ»¯¸ÆÈÜÒº´úÌæÂÈ»¯±µÈÜÒº£¿_______ £¨Ìî¡°ÄÜ¡±»ò¡°·ñ¡±£©£¬Ô­ÒòÊÇ£º_____________________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÊµÑéÌâ

ijͬѧÉè¼ÆÁËÒÔÏÂÁ÷³ÌÀ´¼ìÑé̼ËáÄÆ·ÛÄ©ÖпÉÄܺ¬ÓÐÉÙÁ¿ÂÈ»¯ÄƺÍÇâÑõ»¯ÄÆÖеÄÒ»ÖÖ»òÁ½ÖÖÔÓÖÊ¡£

(1)²½Öè1ËùÐè²£Á§ÒÇÆ÷ÊÇ______£»²½Öè3µÄ²Ù×÷Ãû³ÆÊÇ______¡£
(2)¶Ô¼ìÑé̼ËáÄÆ·ÛÄ©ÖпÉÄܺ¬ÓеÄÔÓÖÊÌá³öºÏÀí¼ÙÉ裺
¼ÙÉè1£ºÖ»º¬ÓÐÂÈ»¯ÄÆ£»
¼ÙÉè2£ºÖ»º¬ÓÐ___________£»
¼ÙÉè3£ºÂÈ»¯ÄƺÍÇâÑõ»¯Äƶ¼º¬ÓС£
(3)Éè¼ÆʵÑé·½°¸£¬½øÐÐʵÑé¡£
ÏÞÑ¡ÒÔÏÂÊÔ¼Á£ºÂÈ»¯±µÈÜÒº¡¢ÏõËá±µÈÜÒº¡¢·Ó̪ÊÔÒº¡¢Ï¡ÏõËᡢϡÑÎËᡢϡÁòËá¡¢ÏõËáÒøÈÜÒº¡£»Ø´ðÏÂÁÐÎÊÌ⣺
¢Ù¼ÓÈë¹ýÁ¿ÊÔ¼ÁAµÄÃû³ÆÊÇ______¡£
¢ÚÌîдϱí:

ʵÑé²½Öè
Ô¤ÆÚÏÖÏóºÍ½áÂÛ
²½Öè4:
 
²½Öè5:
 
 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÊµÑéÌâ

¸ù¾ÝMgÓëCO2µÄ·´Ó¦ÍƲ⣬NaÒ²ÄÜÔÚCO2ÖÐȼÉÕ£¬ÇÒ¹ÌÌå²úÎï¿ÉÄÜΪC¡¢Na2OºÍNa2CO3ÖеÄÁ½ÖÖ»òÈýÖÖ¡£Ä³ÐËȤС×éÓÃÈçÏÂ×°ÖÿªÕ¹Á½¸ö½×¶ÎµÄʵÑé̽¾¿¡£

¡¾ÊµÑéI¡¿Òý·¢CO2ÓëNa·´Ó¦¡£²Ù×÷²½ÖèΪ£º
¢Ù°´Í¼Á¬½Ó×°Ö㬼ìÑé×°ÖÃÆøÃÜÐÔ£¬²¢Íù¸÷×°ÖÃÖмÓÈëÊÔ¼Á£»      
¢Ú´ò¿ª×°ÖÃaÉÏ»îÈûÒ»¶Îʱ¼ä£»
¢Ûµãȼd´¦¾Æ¾«µÆ£¬Ê¹CO2ÓëNa³ä·Ö·´Ó¦¡£Í£Ö¹¼ÓÈÈ£¬¼ÌÐøͨÆøÌåʹ˫ͨ¹ÜÀäÈ´¡£
»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©×°ÖÃaÖÐÓõ½µÄ²£Á§ÒÇÆ÷ÓÐÊԹܺ͠                     £»
£¨2£©×°ÖÃbÖÐ×°ÈëµÄÊÔ¼ÁÊÇ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡          £»
£¨3£©×°ÖÃcµÄ×÷ÓÃÊÇ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡          £»
£¨4£©²½Öè¢ÚÖУ¬µ±¹Û²ìµ½¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡                Ê±£¬²ÅÄܽøÈë²½Öè¢Û¡£
¡¾ÊµÑé¢ò¡¿Ì½¾¿·´Ó¦²úÎï¼°·´Ó¦»úÀí¡£È¡·´Ó¦ºó˫ͨ¹ÜÖйÌÌåÎïÖÊ29.2 g½øÐÐÈçÏÂʵÑ飺
¢Ù×Ðϸ¹Û²ì¹ÌÌ壬·¢ÏÖÓкÚÉ«¿ÅÁ££»
¢Ú½«¹ÌÌåÈÜÓÚ×ãÁ¿µÄË®²¢¹ýÂË¡¢Ï´µÓ£¬µÃµ½1.8 gÂËÔü£¨¸ÉÔ£»
¢Û½«ÂËÒº¼ÓˮϡÊÍÅä³É250 mLµÄÈÜÒº£»
¢ÜÈ¡ÉÙÁ¿¢ÛµÄÈÜÒº£¬ÏȼÓ×ãÁ¿BaCl2ÈÜÒº£¬¹Û²ìµ½°×É«³Áµí,ËùµÃ¹ÌÌåÖÊÁ¿Îª3.94g£»ÔÙ¼Ó¼¸µÎ·Ó̪ÊÔÒº£¬ÈÜÒº±äºì£»
¢ÝÈ¡25.00 mL¢ÛµÄÈÜÒº£¬µÎ¼Ó¼×»ù³È×÷Ϊָʾ¼Á£¬ÓÃ3.0 mol¡¤L-1ÑÎËáµÎ¶¨£¬ÏûºÄÑÎËáÌå»ýΪ20.00 mL¡£
»Ø´ðÏÂÁÐÎÊÌ⣺
£¨5£©ÈçºÎÅжϢÝÖеĵζ¨Öյ㡡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡                   
£¨6£©·´Ó¦ºó×°ÖÃdÖеĹÌÌåÎïÖÊΪ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡     £»Í¨¹ý¼ÆËã·ÖÎö£º29.2 g²ÐÁô¹ÌÌåÖУ¬¸÷×é·ÖµÄÖÊÁ¿¸÷ÊǶàÉÙ?¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡                   

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸