ijÑо¿Ð¡×éÓÃÈçͼËùʾװÖýøÐÐÍ­ÓëŨÁòËá·´Ó¦µÄʵÑéÑо¿¡£
£¨1£©Ð´³öÊÔ¹ÜBÖеÄʵÑéÏÖÏó______________________¡£
£¨2£©Ð´³öAÖз´Ó¦µÄ»¯Ñ§·½³Ìʽ______________________¡£
£¨3£©¼ÌÐøÏòAÊÔ¹ÜÖмÓÈëH2O2£¬·¢ÏÖͭƬÈܽ⣬·´Ó¦µÄÀë×Ó·½³ÌʽΪ£º______________________¡£ÈôÈÔ²»²¹³äŨÁòËᣬֻҪÇóʹͭƬÈܽ⣬Ҳ¿ÉÒÔ¼ÓÈ루ÌîдÁ½ÖÖÊôÓÚ²»Í¬Àà±ðÎïÖʵĻ¯Ñ§Ê½£©___________¡¢
___________¡£
£¨4£©BÊԹܿڵÄÃÞ»¨Ó¦Õ´ÓеÄÊÔ¼ÁÊÇ___________¡£
£¨5£©Ð¡×é³ÉÔ±Ïò·´Ó¦ºóµÄÈÜÒºÖмÓÈë×ãÁ¿µÄÑõ»¯Í­£¬Ê¹Ê£ÓàµÄÁòËáÈ«²¿×ª»¯ÎªÁòËáÍ­£¬¹ýÂ˺󣬽«ÂËÒº¼ÓÈÈŨËõ£¬ÀäÈ´½á¾§ÖƵÃÁòËáÍ­¾§Ì壨CuSO4¡¤xH2O£©¡£Ð¡×é³ÉÔ±²ÉÓüÓÈÈ·¨²â¶¨¸Ã¾§ÌåÀï½á¾§Ë®xµÄÖµ¡£
¢ÙÔÚËûÃǵÄʵÑé²Ù×÷ÖУ¬ÖÁÉÙ³ÆÁ¿___________´Î£»
¢ÚÏÂÃæÊÇÆäÖÐÒ»´ÎʵÑéµÄÊý¾Ý£º
¸ù¾ÝÉϱíÊý¾Ý¼ÆËãÅжÏxµÄʵ²àÖµ±ÈÀíÂÛÖµ£¨x=5£©________£¨Ìî¡°Æ«´ó¡±¡¢¡°Æ«Ð¡¡±£©£¬Õâ´ÎʵÑéÖвúÉúÎó²îµÄÔ­Òò¿ÉÄÜÊÇ___________£¨ÌîÐòºÅ£©
A£®ÁòËáÍ­¾§ÌåÖк¬Óв»»Ó·¢µÄÔÓÖÊ
B£®ÊµÑéÇ°¾§Ìå±íÃæÓÐʪ´æË®
C£®¼ÓÈÈʱÓо§Ìå·É½¦³öÈ¥
D£®¼ÓÈÈʧˮºó¶ÖÃÔÚ¿ÕÆøÖÐÀäÈ´
£¨1£©Æ·ºìÈÜÒºÍÊÉ«
£¨2£©Cu+2H2SO4(Ũ)CuSO4+SO2¡ü+2H2O
£¨3£©Cu+H2O2+2H+==Cu2++2H2O£»Fe2O3£»NaNO3
£¨4£©NaOH£¨ÈÜÒº£©»òKMnO4£¨ÈÜÒº£©£¨ÆäËüºÏÀí´ð°¸Ò²¿ÉÒÔ£©
£¨5£©¢Ù4£»¢ÚÆ«´ó£»BC
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ijÑо¿Ð¡×éÓÃÈçͼËùʾװÖÃ̽¾¿SO2ºÍFe£¨NO3£©3ÈÜÒº·´Ó¦µÄÔ­Àí£®
ÒÑÖª£º1.0mol/LFe£¨NO3£©3ÈÜÒºµÄpH=1£®Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©×°ÖÃAÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ
Na2SO3+H2S04£¨Å¨£©=Na2S04+SO2¡ü+H2O
Na2SO3+H2S04£¨Å¨£©=Na2S04+SO2¡ü+H2O
£®
£¨2£©ÎªÅųý¿ÕÆø¶ÔʵÑéµÄ¸ÉÈÅ£¬µÎ¼ÓŨÁòËá֮ǰӦ½øÐеIJÙ×÷ÊÇ
´ò¿ªµ¯»É¼Ð£¬Ïò×°ÖÃÖÐͨÈëÒ»¶Îʱ¼äµÄN2£¬¹Ø±Õµ¯»É¼Ð
´ò¿ªµ¯»É¼Ð£¬Ïò×°ÖÃÖÐͨÈëÒ»¶Îʱ¼äµÄN2£¬¹Ø±Õµ¯»É¼Ð
£®
£¨3£©×°ÖÃBÖвúÉúÁË°×É«³Áµí£¬Æä³É·ÖÊÇ
BaS04
BaS04
£¬ËµÃ÷SO2¾ßÓÐÐÔ
»¹Ô­ÐÔ
»¹Ô­ÐÔ
£®
£¨4£©·ÖÎöBÖвúÉú°×É«³ÁµíµÄÔ­Òò£º
¼ÙÉè1£º
SO2ºÍFe3+¡¢ËáÐÔÌõ¼þÏÂNO3-¶¼·´Ó¦
SO2ºÍFe3+¡¢ËáÐÔÌõ¼þÏÂNO3-¶¼·´Ó¦
£»
¼ÙÉè2£ºSO2ÓëFe3+·´Ó¦£»
¼ÙÉè3£ºSO2ÓëËáÐÔÌõ¼þϵÄN
O
-
3
·´Ó¦£®
¢Ù°´¼ÙÉè2£¬×°ÖÃBÖз´Ó¦µÄÀë×Ó·½³ÌʽÊÇ
SO2+2Fe3++Ba2++2H2O=BaSO4¡ý+2Fe2++4H+
SO2+2Fe3++Ba2++2H2O=BaSO4¡ý+2Fe2++4H+
£®
¢Ú°´¼ÙÉè3£¬Ö»Ð轫װÖÃBÖеÄFe£¨NO3£©3ÈÜÒºÌæ»»³ÉµÈÌå»ýµÄÏÂÁÐÈÜÒº£¬ÔÚÏàͬÌõ¼þϽøÐÐʵÑ飮ӦѡÔñµÄÊÔ¼ÁÊÇ
c
c
£¨ÌîÐòºÅ£©£®
a.0.1mol/LÏ¡ÏõËá
b.1.5mol/L Fe£¨NO3£©2ÈÜÒº
c.6.0mol/L NaNO3ÈÜÒººÍ0.2mol/LÑÎËáµÈÌå»ý»ìºÏµÄÈÜÒº£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â

ijÑо¿Ð¡×éÓÃÈçͼËùʾװÖÃÖÆÈ¡ÂÈÆø²¢Ì½¾¿ÂÈÆøµÄÐÔÖÊ£®

Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÖÆÈ¡ÂÈÆøʱ£¬ÔÚÉÕÆ¿Àï¼ÓÈëÒ»¶¨Á¿µÄ¶þÑõ»¯Ã̺ÍNaCl£¬Í¨¹ý
·ÖҺ©¶·
·ÖҺ©¶·
£¨ÌîдÒÇÆ÷Ãû³Æ£©ÏòÉÕÆ¿ÖмÓÈëÊÊÁ¿µÄŨÁòËᣮ
£¨2£©Ä³Í¬Ñ§²éÔÄ×ÊÁϺó·¢ÏÖŨÁòËáÓë¶þÑõ»¯ÃÌ·´Ó¦²úÉúÑõÆø£®ÏÖÐèÒªÄã°ï¸ÃͬѧÉè¼ÆÒ»¸ö·½°¸£¬¼ÈÒªÍê³ÉºóÐøʵÑ飬ÓÖҪ̽¾¿ÓÐÑõÆø²úÉúµÄʵÑ飮ÒÇÆ÷Á¬½Ó˳ÐòΪa¡ú
c
c
¡ú
b
b
¡úd¡úe¡ú
i
i
¡ú
h
h
¡ú
g
g
¡ú
f
f
¡úl¡úm£®¼ìÑéÊÇ·ñÓÐÑõÆø²úÉúµÄ²Ù×÷ÊÇ
ÔÚm´¦£¬Óôø»ðÐǵÄľÌõ¿´ÊÇ·ñ¸´È¼£¬ÈôÄÜ£¬ËµÃ÷ÓÐÑõÆøÉú³É
ÔÚm´¦£¬Óôø»ðÐǵÄľÌõ¿´ÊÇ·ñ¸´È¼£¬ÈôÄÜ£¬ËµÃ÷ÓÐÑõÆøÉú³É
£¬Ð´³öŨÁòËáÓë¶þÑõ»¯ÃÌ·´Ó¦²úÉúÑõÆøµÄ»¯Ñ§·½³Ìʽ£º
2Mn02+2H2SO4£¨Å¨£©
 ¼ÓÈÈ 
.
 
2MnSO4+O2¡ü+2H20
2Mn02+2H2SO4£¨Å¨£©
 ¼ÓÈÈ 
.
 
2MnSO4+O2¡ü+2H20
£®
£¨3£©ÊµÑéÖпɹ۲쵽¢ÛµÄ׶ÐÎÆ¿ÀïÈÜÒºµÄÑÕÉ«±ä»¯Îª
×ÏÉ«¡úºìÉ«¡úÎÞÉ«¡ú»ÆÂÌ
×ÏÉ«¡úºìÉ«¡úÎÞÉ«¡ú»ÆÂÌ
£®
£¨4£©È¡³ö¢ÜÖеļ¯ÆøÆ¿½øÐÐÂÈÆøÓëÍ­µÄ·´Ó¦ÊµÑ飬Ӧ¸Ã½øÐеIJÙ×÷ÊÇ
ÓÃÛáÛöǯ¼ÐסһÊøϸͭ˿£¬×ÆÈÈ
ÓÃÛáÛöǯ¼ÐסһÊøϸͭ˿£¬×ÆÈÈ
ºóÁ¢¿Ì·ÅÈë³äÂúÂÈÆøµÄ¼¯ÆøÆ¿ÖУ®ÊµÑéÖÐÓÐͬѧÌá³öÓ¦¸ÃÔÚ¼¯ÆøÆ¿µ×²¿ÏÈ·ÅÉÙÁ¿µÄË®»òϸɳ£¬ÄãÈÏΪÊÇ·ñÒ»¶¨Òª·Å£¿
²»ÐèÒª
²»ÐèÒª
£¨Ìî¡°ÊÇ¡±»ò¡°·ñ¡±£©£»ÀíÓÉÊÇ
ʵÑéÖÐÉú³ÉµÄÊÇÑ̶ø²»ÊÇζȽϸߵĹÌÌåÈÛÈÚÎ²»»áʹƿµ×Õ¨ÁÑ
ʵÑéÖÐÉú³ÉµÄÊÇÑ̶ø²»ÊÇζȽϸߵĹÌÌåÈÛÈÚÎ²»»áʹƿµ×Õ¨ÁÑ
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ijÑо¿Ð¡×éÓÃÈçͼËùʾװÖýøÐÐÍ­ÓëŨÁòËá·´Ó¦µÄʵÑéÑо¿£®
£¨1£©Ð´³öÊÔ¹ÜBÖеÄʵÑéÏÖÏó
Æ·ºìÈÜÒºÍÊÉ«
Æ·ºìÈÜÒºÍÊÉ«
£®
£¨2£©Ð´³öAÖз´Ó¦µÄ»¯Ñ§·½³Ìʽ
Cu+2H2SO4
  ¡÷  
.
 
CuSO4+SO2¡ü+2H2O
Cu+2H2SO4
  ¡÷  
.
 
CuSO4+SO2¡ü+2H2O

£¨3£©¼ÌÐøÏòAÊÔ¹ÜÖмÓÈëH2O2£¬·¢ÏÖͭƬÈܽ⣬·´Ó¦µÄÀë×Ó·½³ÌʽΪ£º
Cu+H2O2+2H+=Cu2++2H2O
Cu+H2O2+2H+=Cu2++2H2O
£®
ÈôÈÔ²»²¹³äŨÁòËᣬֻҪÇóʹͭƬÈܽ⣬Ҳ¿ÉÒÔ¼ÓÈ루ÌîдÁ½ÖÖÊôÓÚ²»Í¬Àà±ðÎïÖʵĻ¯Ñ§Ê½£©
Fe2O3
Fe2O3
¡¢
NaNO3
NaNO3
£®
£¨4£©BÊԹܿڵÄÃÞ»¨Ó¦Õ´ÓеÄÊÔ¼ÁÊÇ
NaOH£¨ÈÜÒº£©
NaOH£¨ÈÜÒº£©
£®
£¨5£©Ð¡×é³ÉÔ±Ïò·´Ó¦ºóµÄÈÜÒºÖмÓÈë×ãÁ¿µÄÑõ»¯Í­£¬Ê¹Ê£ÓàµÄÁòËáÈ«²¿×ª»¯ÎªÁòËáÍ­£¬¹ýÂ˺󣬽«ÂËÒº¼ÓÈÈŨËõ£¬ÀäÈ´½á¾§ÖƵÃÁòËáÍ­¾§Ì壨CuSO4?xH2O£©£®Ð¡×é³ÉÔ±²ÉÓüÓÈÈ·¨²â¶¨¸Ã¾§ÌåÀï½á¾§Ë®xµÄÖµ£®
¢ÙÔÚËûÃǵÄʵÑé²Ù×÷ÖУ¬ÖÁÉÙ³ÆÁ¿
4
4
´Î£»
¢ÚÏÂÃæÊÇÆäÖÐÒ»´ÎʵÑéµÄÊý¾Ý£º
ÛáÛöÖÊÁ¿ ÛáÛöÓ뾧ÌåµÄ×ÜÖÊÁ¿ ¼ÓÈȺóÛáÛöÓë¹ÌÌå×ÜÖÊÁ¿
11.7g 22.7g 18.9g
¸ù¾ÝÉϱíÊý¾Ý¼ÆËãÅжÏxµÄʵ²àÖµ±ÈÀíÂÛÖµ£¨x=5£©
ƫС
ƫС
£¨Ìî¡°Æ«´ó¡±¡¢¡°Æ«Ð¡¡±£©£¬Õâ´ÎʵÑéÖвúÉúÎó²îµÄÔ­Òò¿ÉÄÜÊÇ
AD
AD
£¨ÌîÐòºÅ£©
A£®ÁòËáÍ­¾§ÌåÖк¬Óв»»Ó·¢µÄÔÓÖÊ        B£®ÊµÑéÇ°¾§Ìå±íÃæÓÐʪ´æË®
C£®¼ÓÈÈʱÓо§Ìå·É½¦³öÈ¥                D£®¼ÓÈÈʧˮºó¶ÖÃÔÚ¿ÕÆøÖÐÀäÈ´£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

ijÑо¿Ð¡×éÓÃÈçͼËùʾװÖÃÖÆÈ¡ÂÈÆø²¢Ì½¾¿ÂÈÆøµÄÐÔÖÊ£®

Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÖÆÈ¡ÂÈÆøʱ£¬ÔÚÉÕÆ¿Àï¼ÓÈëÒ»¶¨Á¿µÄ¶þÑõ»¯Ã̺ÍNaCl£¬Í¨¹ý______£¨ÌîдÒÇÆ÷Ãû³Æ£©ÏòÉÕÆ¿ÖмÓÈëÊÊÁ¿µÄŨÁòËᣮ
£¨2£©Ä³Í¬Ñ§²éÔÄ×ÊÁϺó·¢ÏÖŨÁòËáÓë¶þÑõ»¯ÃÌ·´Ó¦²úÉúÑõÆø£®ÏÖÐèÒªÄã°ï¸ÃͬѧÉè¼ÆÒ»¸ö·½°¸£¬¼ÈÒªÍê³ÉºóÐøʵÑ飬ÓÖҪ̽¾¿ÓÐÑõÆø²úÉúµÄʵÑ飮ÒÇÆ÷Á¬½Ó˳ÐòΪa¡ú______¡ú______¡úd¡úe¡ú______¡ú______¡ú______¡ú______¡úl¡úm£®¼ìÑéÊÇ·ñÓÐÑõÆø²úÉúµÄ²Ù×÷ÊÇ______£¬Ð´³öŨÁòËáÓë¶þÑõ»¯ÃÌ·´Ó¦²úÉúÑõÆøµÄ»¯Ñ§·½³Ìʽ£º______£®
£¨3£©ÊµÑéÖпɹ۲쵽¢ÛµÄ׶ÐÎÆ¿ÀïÈÜÒºµÄÑÕÉ«±ä»¯Îª______£®
£¨4£©È¡³ö¢ÜÖеļ¯ÆøÆ¿½øÐÐÂÈÆøÓëÍ­µÄ·´Ó¦ÊµÑ飬Ӧ¸Ã½øÐеIJÙ×÷ÊÇ______ºóÁ¢¿Ì·ÅÈë³äÂúÂÈÆøµÄ¼¯ÆøÆ¿ÖУ®ÊµÑéÖÐÓÐͬѧÌá³öÓ¦¸ÃÔÚ¼¯ÆøÆ¿µ×²¿ÏÈ·ÅÉÙÁ¿µÄË®»òϸɳ£¬ÄãÈÏΪÊÇ·ñÒ»¶¨Òª·Å£¿______£¨Ìî¡°ÊÇ¡±»ò¡°·ñ¡±£©£»ÀíÓÉÊÇ______£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ijÑо¿Ð¡×éÓÃÈçͼËùʾװÖýøÐÐÍ­ÓëŨÁòËá·´Ó¦µÄʵÑéÑо¿¡£

£¨1£©Ð´³öÊÔ¹ÜBÖеÄʵÑéÏÖÏó                                              ¡£

£¨2£©Ð´³öAÖз´Ó¦µÄ»¯Ñ§·½³Ìʽ£º                                          ¡£

£¨3£©³ä·Ö·´Ó¦ºó£¬ÈôAÊÔ¹ÜÖÐÓÐͭƬʣÓ࣬¼ÌÐøÏòAÖмÓÈë

NaNO3£¬·¢ÏÖͭƬÈܽ⣬·´Ó¦µÄÀë×Ó·½³Ìʽ

Ϊ                                           ¡£

£¨4£©³ä·Ö·´Ó¦ºó£¬ÈôAÊÔ¹ÜÖÐÎÞͭƬʣÓ࣬

µ«³öÏÖ°×É«»ì×Ç£¬¸Ã°×É«¹ÌÌåÊÇ               £¬

³õ²½È·Èϸð×É«¹ÌÌåÊÇʲôÎïÖʵÄʵÑé²Ù×÷·½·¨

ÊÇ                                        ¡£

£¨5£©BÊԹܿڵÄÃÞ»¨Ó¦Õ´ÓеÄÊÔ¼ÁÊÇ         ¡£

£¨6£©Ð¡×é³ÉÔ±Ïò£¨4£©·´Ó¦ºóµÄÈÜÒºÖмÓÈë×ãÁ¿µÄÑõ»¯Í­¡£Ê¹Ê£ÓàµÄÁòËáÈ«²¿×ª»¯ÎªÁòËáÍ­£¬¹ýÂ˺󣬽«ÂËÒº¼ÓÈÈŨËõ£¬ÀäÈ´ºó½á¾§ÖƵÃÁòËáÍ­¾§Ì壨CuSO4¡¤xH2O£©¡£Ð¡×é³ÉÔ±²ÉÓüÓÈÈ·¨²â¶¨¸Ã¾§ÌåÀï½á¾§Ë®xµÄÖµ

¢ÙÔÚËûÃǵÄʵÑé²Ù×÷ÖУ¬ÖÁÉÙ³ÆÁ¿ËĴΣ¬×îºóÁ½´Î³ÆÁ¿µÄÄ¿µÄÊÇ                     

¢ÚÏÂÃæÊÇÆäÖÐÒ»´ÎʵÑéµÄÊý¾Ý£º

ÛáÛöÖÊÁ¿

ÛáÛöÓ뾧ÌåµÄ×ÜÖÊÁ¿

¼ÓÈȺóÛáÛöÓë¹ÌÌå×ÜÖÊÁ¿

11.0g

37.8g

27.0g

¸ù¾ÝÉϱíÊý¾Ý¼ÆËãÅжÏxµÄʵ²âÖµ±ÈÀíÂÛÖµ£¨x£½5£©          £¨Ìî¡°Æ«´ó¡±¡¢¡°Æ«Ð¡¡±£©¡£

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸