ÒÑÖªCuSO4ÈÜÒº·Ö±ðÓëNa2CO3ÈÜÒº¡¢Na2SÈÜÒºµÄ·´Ó¦Çé¿öÈçÏ£º

£¨1£©CuSO4 +Na2CO3  Ö÷Òª£ºCu2++ CO32©¤ + H2O == Cu(OH)2¡ý+ CO2¡ü

                   ´ÎÒª£ºCu2+ + CO32©¤== CuCO3¡ý

£¨2£©CuSO4 + Na2S    Ö÷Òª£ºCu2+ + S2©¤== CuS¡ý

                   ´ÎÒª£ºCu2+ + S2©¤+ 2H2O == Cu(OH)2¡ý+H2S¡ü

ÏÂÁм¸ÖÖÎïÖʵÄÈܽâ¶È´óСµÄ±È½ÏÖУ¬ÕýÈ·µÄÊÇ

A£®CuS <Cu(OH)2<CuCO3                B£®CuS >Cu(OH)2>CuCO3

C£®Cu(OH)2>CuCO3>CuS                 D£®Cu(OH)2<CuCO3<CuS

 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨1£©³£ÎÂÏ£¬Èç¹ûÈ¡0.1mol?L-1HAÈÜÒºÓë0.1mol?L-1NaOHÈÜÒºµÈÌå»ý»ìºÏ£¨»ìºÏºóÈÜÒºÌå»ýµÄ±ä»¯ºöÂÔ²»¼Æ£©£¬²âµÃ»ìºÏÒºµÄpH=8£¬Çë»Ø´ðÏÂÁÐÎÊÌ⣺
¢Ù»ìºÏºóÈÜÒºµÄpH=8µÄÔ­Òò£¨ÓÃÀë×Ó·½³Ìʽ±íʾ£©£º
 
£®
¢ÚË®µçÀë³öµÄc£¨H+£©£º»ìºÏÈÜÒº
 
0.1mol?L-1NaOHÈÜÒº£¨Ì¡¢£¾¡¢=£©£®
¢ÛÒÑÖªNH4AÈÜҺΪÖÐÐÔ£¬ÓÖÖª½«HAÈÜÒº¼Óµ½Na2CO3ÈÜÒºÖÐÓÐÆøÌå·Å³ö£¬ÊÔÍƶϣ¨NH4£©2CO3ÈÜÒºµÄpH
 
7£¨Ì¡¢£¾¡¢=£©£»ÏàͬζÈÏ£¬µÈÎïÖʵÄÁ¿Å¨¶ÈµÄÏÂÁÐËÄÖÖÑÎÈÜÒº°´pHÓÉ´óµ½Ð¡µÄÅÅÁÐ˳ÐòΪ£º
 
£¨ÌîÐòºÅ£©
a£®NH4HCO3    b£®NH4A    c£®£¨NH4£©2CO3 d£®NH4Cl
£¨2£©ÄÑÈܵç½âÖÊÔÚË®ÈÜÒºÖдæÔÚ×ŵçÀëƽºâ£®ÔÚ³£ÎÂÏ£¬ÈÜÒºÀï¸÷ÖÖÀë×ÓµÄŨ¶ÈÒÔËüÃÇ»¯Ñ§¼ÆÁ¿ÊýΪ´Î·½µÄ³Ë»ý½ÐÈܶȻý³£Êý£¨Ksp£©£® ÀýÈ磺Cu£¨OH£©2?Cu2++2OH-£¬Ksp=c£¨Cu2+£©?c£¨OH-£©2=2¡Á10-20£®µ±ÈÜÒºÖи÷ÖÖÀë×ÓµÄŨ¶È´Î·½µÄ³Ë»ý´óÓÚÈܶȻýʱ£¬Ôò²úÉú³Áµí£¬·´Ö®¹ÌÌåÈܽ⣮
¢ÙijCuSO4ÈÜÒºÀïc£¨Cu2+£©=0.02mol/L£¬ÈçÒªÉú³ÉCu£¨OH£©2 ³Áµí£¬Ó¦µ÷ÕûÈÜÒºµÄpH£¾5£¬Ô­ÒòÊÇ£¨ÓÃÊý¾Ý˵Ã÷£©
 
£»
¢ÚҪʹ0.2mol/LCuSO4ÈÜÒºÖÐCu2+ ³Áµí½ÏΪÍêÈ«£¨¼´Ê¹Cu2+ Å¨¶È½µÖÁÔ­À´µÄǧ·ÖÖ®Ò»£©£¬ÔòÓ¦ÏòÈÜÒºÀï¼ÓÈëNaOHÈÜÒº£¬Ê¹ÈÜÒºµÄpH¡Ý¶àÉÙ£¿Ð´³ö¼ÆËã¹ý³Ì£º
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

(16·Ö)²¨¶û¶àÒºÊÇÒ»ÖÖ±£»¤ÐÔɱ¾ú¼Á£¬¹ã·ºÓ¦ÓÃÓÚÊ÷ľ¡¢¹ûÊ÷ºÍ»¨»ÜÉÏ£¬ÏÊÀ¶É«µÄµ¨·¯¾§ÌåÊÇÅäÖƲ¨¶û¶àÒºµÄÖ÷ÒªÔ­ÁÏ¡£ÒÑÖªCuSO4¡¤5H2OµÄ²¿·Ö½á¹¹¿É±íʾÈçÏ£º

£¨1£©Ð´³öÍ­Ô­×Ó¼Ûµç×Ó²ãµÄµç×ÓÅŲ¼Ê½____________£¬ÓëͭͬÖÜÆÚµÄËùÓÐÔªËصĻù̬ԭ×ÓÖÐ×îÍâ²ãµç×ÓÊýÓëÍ­Ô­×ÓÏàͬµÄÔªËØÓÐ__________(ÌîÔªËØ·ûºÅ)¡£

£¨2£©ÇëÔÚÉÏͼÖаÑCuSO4¡¤5H2O½á¹¹ÖеĻ¯Ñ§¼üÓöÌÏß¡°¡ª¡ª¡±±íʾ³öÀ´¡£

£¨3£©ÍùŨCuSO4ÈÜÒºÖмÓÈë¹ýÁ¿½ÏŨµÄNH3¡¤H2OÖ±µ½Ô­ÏÈÉú³ÉµÄ³ÁµíÇ¡ºÃÈܽâΪֹ£¬µÃµ½ÉîÀ¶É«ÈÜÒº¡£Ð¡ÐļÓÈëÔ¼ºÍÈÜÒºµÈÌå»ýµÄC2H5OH²¢Ê¹Ö®·Ö³ÉÁ½²ã£¬¾²Ö᣾­¹ýÒ»¶Îʱ¼äºó¿É¹Û²ìµ½ÔÚÁ½²ã¡°½»½ç´¦¡±Ï²¿Îö³öÉîÀ¶É«Cu(NH3)4SO4¡¤H2O¾§Ì塣ʵÑéÖÐËù¼ÓC2H5OHµÄ×÷ÓÃÊÇ______________________________________________¡£

£¨4£©Cu(NH3)4SO4¡¤H2O¾§ÌåÖгÊÕýËÄÃæÌåµÄÔ­×ÓÍÅÊÇ______________£¬ÔÓ»¯¹ìµÀÀàÐÍÊÇsp3µÄÔ­×ÓÊÇ____________________________¡£

£¨5£©È罫ÉîÀ¶É«ÈÜÒº¼ÓÈÈ£¬¿ÉÄܵõ½Ê²Ã´½á¹û£¿________________________________¡£

£¨6£©°ÑCoCl2ÈܽâÓÚË®ºó¼Ó°±Ë®Ö±µ½ÏÈÉú³ÉµÄCo(OH)2³ÁµíÓÖÈܽâºó£¬ÔÙ¼Ó°±Ë®£¬Ê¹Éú³É[Co(NH3)6]2+¡£´ËʱÏòÈÜÒºÖÐͨÈë¿ÕÆø£¬µÃµ½µÄ²úÎïÖÐÓÐÒ»ÖÖÆä×é³É¿ÉÓÃCoCl3¡¤5NH3±íʾ¡£°Ñ·ÖÀë³öµÄCoCl3¡¤5NH3ÈÜÓÚË®ºóÁ¢¼´¼ÓÏõËáÒøÈÜÒº£¬ÔòÎö³öAgCl³Áµí¡£¾­²â¶¨£¬Ã¿1 mol CoCl3¡¤5NH3Ö»Éú³É2 mol AgCl¡£Çëд³ö±íʾ´ËÅäºÏÎï½á¹¹µÄ»¯Ñ§Ê½£¨îܵÄÅäλÊýΪ6£©___________£¬´ËÅäºÏÎïÖеÄCo»¯ºÏ¼ÛΪ__  ____¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2010ÄêÖ£ÖÝÊÐÖÇÁÖѧУ¸ß¶þÏÂѧÆÚÆÚÄ©¿¼ÊÔ»¯Ñ§ÊÔÌâ ÌâÐÍ£ºÌî¿ÕÌâ

(16·Ö)²¨¶û¶àÒºÊÇÒ»ÖÖ±£»¤ÐÔɱ¾ú¼Á£¬¹ã·ºÓ¦ÓÃÓÚÊ÷ľ¡¢¹ûÊ÷ºÍ»¨»ÜÉÏ£¬ÏÊÀ¶É«µÄµ¨·¯¾§ÌåÊÇÅäÖƲ¨¶û¶àÒºµÄÖ÷ÒªÔ­ÁÏ¡£ÒÑÖªCuSO4¡¤5H2OµÄ²¿·Ö½á¹¹¿É±íʾÈçÏ£º

£¨1£©Ð´³öÍ­Ô­×Ó¼Ûµç×Ó²ãµÄµç×ÓÅŲ¼Ê½____________£¬ÓëͭͬÖÜÆÚµÄËùÓÐÔªËصĻù̬ԭ×ÓÖÐ×îÍâ²ãµç×ÓÊýÓëÍ­Ô­×ÓÏàͬµÄÔªËØÓÐ__________(ÌîÔªËØ·ûºÅ)¡£
£¨2£©ÇëÔÚÉÏͼÖаÑCuSO4¡¤5H2O½á¹¹ÖеĻ¯Ñ§¼üÓöÌÏß¡°¡ª¡ª¡±±íʾ³öÀ´¡£
£¨3£©ÍùŨCuSO4ÈÜÒºÖмÓÈë¹ýÁ¿½ÏŨµÄNH3¡¤H2OÖ±µ½Ô­ÏÈÉú³ÉµÄ³ÁµíÇ¡ºÃÈܽâΪֹ£¬µÃµ½ÉîÀ¶É«ÈÜÒº¡£Ð¡ÐļÓÈëÔ¼ºÍÈÜÒºµÈÌå»ýµÄC2H5OH²¢Ê¹Ö®·Ö³ÉÁ½²ã£¬¾²Ö᣾­¹ýÒ»¶Îʱ¼äºó¿É¹Û²ìµ½ÔÚÁ½²ã¡°½»½ç´¦¡±Ï²¿Îö³öÉîÀ¶É«Cu(NH3)4SO4¡¤H2O¾§Ì塣ʵÑéÖÐËù¼ÓC2H5OHµÄ×÷ÓÃÊÇ______________________________________________¡£
£¨4£©Cu(NH3)4SO4¡¤H2O¾§ÌåÖгÊÕýËÄÃæÌåµÄÔ­×ÓÍÅÊÇ______________£¬ÔÓ»¯¹ìµÀÀàÐÍÊÇsp3µÄÔ­×ÓÊÇ____________________________¡£
£¨5£©È罫ÉîÀ¶É«ÈÜÒº¼ÓÈÈ£¬¿ÉÄܵõ½Ê²Ã´½á¹û£¿________________________________¡£
£¨6£©°ÑCoCl2ÈܽâÓÚË®ºó¼Ó°±Ë®Ö±µ½ÏÈÉú³ÉµÄCo(OH)2³ÁµíÓÖÈܽâºó£¬ÔÙ¼Ó°±Ë®£¬Ê¹Éú³É[Co(NH3)6]2+¡£´ËʱÏòÈÜÒºÖÐͨÈë¿ÕÆø£¬µÃµ½µÄ²úÎïÖÐÓÐÒ»ÖÖÆä×é³É¿ÉÓÃCoCl3¡¤5NH3±íʾ¡£°Ñ·ÖÀë³öµÄCoCl3¡¤5NH3ÈÜÓÚË®ºóÁ¢¼´¼ÓÏõËáÒøÈÜÒº£¬ÔòÎö³öAgCl³Áµí¡£¾­²â¶¨£¬Ã¿1 mol CoCl3¡¤5NH3Ö»Éú³É2 mol AgCl¡£Çëд³ö±íʾ´ËÅäºÏÎï½á¹¹µÄ»¯Ñ§Ê½£¨îܵÄÅäλÊýΪ6£©___________£¬´ËÅäºÏÎïÖеÄCo»¯ºÏ¼ÛΪ__  ____¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2011-2012ѧÄê¹ã¶«Ê¡ÂÞ¶¨ÊиßÒ»ÏÂѧÆÚÆÚÖÐÖʼìÀí×Û»¯Ñ§ÊÔ¾í£¨´ø½âÎö£© ÌâÐÍ£ºÊµÑéÌâ

(12·Ö)¢ñÊÔд³öÖÐѧ½×¶Î³£¼ûµÄÁ½ÖÖÖÆÈ¡ÑõÆøµÄ»¯Ñ§·½³Ìʽ£º
                                 £»                                       ¡£
¢ò.ʵÑéÊÒÖÐͨ³£ÓÃMnO2×÷´ß»¯¼Á·Ö½â¹ýÑõ»¯Ç⣬ÒÑÖªCuSO4ÈÜÒº¶Ô¹ýÑõ»¯ÇâµÄ·Ö½âÒ²¾ßÓд߻¯×÷Óã¬Ä³ÊµÑéÐËȤС×éͬѧ²ÂÏëÆäËûÑÎÈÜÒºÒ²¿ÉÄÜÔÚÕâ¸ö·´Ó¦ÖÐÆðͬÑùµÄ×÷Óã¬ÓÚÊÇËûÃÇ×öÁËÒÔÏÂ̽¾¿¡£ÇëÄã°ïÖúËûÃÇÍê³ÉʵÑ鱨¸æ£º
(1)ʵÑé¹ý³Ì£ºÔÚÒ»Ö§ÊÔ¹ÜÖмÓÈë5 mL 5%µÄH2O2ÈÜÒº£¬È»ºóµÎÈëÊÊÁ¿µÄFeCl3ÈÜÒº£¬°Ñ´ø»ðÐǵÄľÌõÉìÈëÊԹܡ£
ʵÑéÏÖÏ󣺠                                                 ¡£
ʵÑé½áÂÛ£ºFeCl3ÈÜÒº¿ÉÒÔ´ß»¯·Ö½âH2O2¡£
(2)ÒÑÖªFeCl3ÔÚË®ÖпɽâÀë³öFe3£«ºÍCl£­£¬Í¬Ñ§ÃÇÌá³öÒÔϲÂÏ룺
¼×ͬѧµÄ²ÂÏ룺ÕæÕý´ß»¯·Ö½âH2O2µÄÊÇFeCl3ÈÜÒºÖеÄH2O£»
ÒÒͬѧµÄ²ÂÏ룺ÕæÕý´ß»¯·Ö½âH2O2µÄÊÇFeCl3ÈÜÒºÖеÄFe3£«£»
±ûͬѧµÄ²ÂÏ룺ÕæÕý´ß»¯·Ö½âH2O2µÄÊÇFeCl3ÈÜÒºÖеÄCl£­¡£
ÄãÈÏΪ×î²»¿ÉÄܵÄÊÇ       Í¬Ñ§µÄ²ÂÏ룬ÀíÓÉÊÇ                            ¡£
(3)ͬѧÃǶÔÓàϵÄÁ½¸ö²ÂÏ룬ÓÃʵÑé½øÐÐÁË̽¾¿£¬²¢¼Ç¼ÈçÏÂ,ÇëÄã×Ðϸ·ÖÎöºóÌî±í£º
ʵÑé¹ý³Ì ÊµÑéÏÖÏó ½áÂÛ

ʵÑé¹ý³Ì
ʵÑéÏÖÏó
½áÂÛ
ÏòÊ¢ÓÐ5 mL 5%µÄH2O2ÈÜÒºµÄÊÔ¹ÜÖеÎÈëÉÙÁ¿µÄ HCl£¬°Ñ´ø»ðÐǵÄľÌõÉìÈëÊÔ¹Ü.
ÎÞÃ÷ÏÔÏÖÏó
              
              
ÏòÊ¢ÓÐ5 mL 5%µÄH2O2ÈÜÒºµÄÊÔ¹ÜÖеÎÈëÉÙÁ¿µÄFe2(SO4)3£¬°Ñ´ø»ðÐǵÄľÌõÉìÈëÊÔ¹Ü.
ÊÔ¹ÜÖÐÓдóÁ¿ÆøÅݲúÉú£¬´ø»ðÐǵÄľÌõ¸´È¼ 
              
              

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2014½ì¹ã¶«Ê¡ÂÞ¶¨ÊиßÒ»ÏÂѧÆÚÆÚÖÐÖʼìÀí×Û»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÊµÑéÌâ

(12·Ö)¢ñÊÔд³öÖÐѧ½×¶Î³£¼ûµÄÁ½ÖÖÖÆÈ¡ÑõÆøµÄ»¯Ñ§·½³Ìʽ£º

                                  £»                                        ¡£

¢ò.ʵÑéÊÒÖÐͨ³£ÓÃMnO2×÷´ß»¯¼Á·Ö½â¹ýÑõ»¯Ç⣬ÒÑÖªCuSO4ÈÜÒº¶Ô¹ýÑõ»¯ÇâµÄ·Ö½âÒ²¾ßÓд߻¯×÷Óã¬Ä³ÊµÑéÐËȤС×éͬѧ²ÂÏëÆäËûÑÎÈÜÒºÒ²¿ÉÄÜÔÚÕâ¸ö·´Ó¦ÖÐÆðͬÑùµÄ×÷Óã¬ÓÚÊÇËûÃÇ×öÁËÒÔÏÂ̽¾¿¡£ÇëÄã°ïÖúËûÃÇÍê³ÉʵÑ鱨¸æ£º

(1)ʵÑé¹ý³Ì£ºÔÚÒ»Ö§ÊÔ¹ÜÖмÓÈë5 mL 5%µÄH2O2ÈÜÒº£¬È»ºóµÎÈëÊÊÁ¿µÄFeCl3ÈÜÒº£¬°Ñ´ø»ðÐǵÄľÌõÉìÈëÊԹܡ£

ʵÑéÏÖÏó£º                                                   ¡£

ʵÑé½áÂÛ£ºFeCl3ÈÜÒº¿ÉÒÔ´ß»¯·Ö½âH2O2¡£

(2)ÒÑÖªFeCl3ÔÚË®ÖпɽâÀë³öFe3£«ºÍCl£­£¬Í¬Ñ§ÃÇÌá³öÒÔϲÂÏ룺

¼×ͬѧµÄ²ÂÏ룺ÕæÕý´ß»¯·Ö½âH2O2µÄÊÇFeCl3ÈÜÒºÖеÄH2O£»

ÒÒͬѧµÄ²ÂÏ룺ÕæÕý´ß»¯·Ö½âH2O2µÄÊÇFeCl3ÈÜÒºÖеÄFe3£«£»

±ûͬѧµÄ²ÂÏ룺ÕæÕý´ß»¯·Ö½âH2O2µÄÊÇFeCl3ÈÜÒºÖеÄCl£­¡£

ÄãÈÏΪ×î²»¿ÉÄܵÄÊÇ        ͬѧµÄ²ÂÏ룬ÀíÓÉÊÇ                             ¡£

(3)ͬѧÃǶÔÓàϵÄÁ½¸ö²ÂÏ룬ÓÃʵÑé½øÐÐÁË̽¾¿£¬²¢¼Ç¼ÈçÏÂ,ÇëÄã×Ðϸ·ÖÎöºóÌî±í£º

ʵÑé¹ý³Ì ÊµÑéÏÖÏó ½áÂÛ

ʵÑé¹ý³Ì

ʵÑéÏÖÏó

½áÂÛ

ÏòÊ¢ÓÐ5 mL 5%µÄH2O2ÈÜÒºµÄÊÔ¹ÜÖеÎÈëÉÙÁ¿µÄ HCl£¬°Ñ´ø»ðÐǵÄľÌõÉìÈëÊÔ¹Ü.

ÎÞÃ÷ÏÔÏÖÏó

               

              

ÏòÊ¢ÓÐ5 mL 5%µÄH2O2ÈÜÒºµÄÊÔ¹ÜÖеÎÈëÉÙÁ¿µÄFe2(SO4)3£¬°Ñ´ø»ðÐǵÄľÌõÉìÈëÊÔ¹Ü.

ÊÔ¹ÜÖÐÓдóÁ¿ÆøÅݲúÉú£¬´ø»ðÐǵÄľÌõ¸´È¼ 

              

              

 

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸