ÈçͼËùʾ£¬C³£ÎÂÏÂÊÇÎÞÉ«ÒºÌ壬E¡¢XÊǵ¥ÖÊ£¬E¡¢F¡¢H¾ùΪÎÞÉ«ÆøÌ壮AºÍBÓÉÏàͬµÄÔªËØ×é³É£®
¸ù¾ÝͼÖÐËùʾµÄת»¯¹Øϵ£¬ÊԻشðÏÂÁÐÎÊÌ⣺


£¨1£©CµÄ½á¹¹Ê½ÊÇ
 
£¬FµÄµç×ÓʽÊÇ
 
£¬·´Ó¦¢ÛµÄ·´Ó¦ÀàÐÍÊÇ
 
£®
£¨2£©Ð´³ö¸÷ÎïÖʵĻ¯Ñ§Ê½£ºH
 
£¬I
 
£¬D
 
£¬J
 
£®
£¨3£©ÊµÑéÊÒÅäÖÆKÈÜҺʱҪ¼ÓÈëÉÙÁ¿µÄ
 
£¬ÅäÖÆJÈÜҺʱҪ¼ÓÈëÉÙÁ¿µÄ
 
£®
£¨4£©Ð´³ö·´Ó¦¢Ú¢ÜµÄÀë×Ó·½³Ìʽ£º
¢Ú
 
£»
¢Ü
 
£®
¿¼µã£ºÎÞ»úÎïµÄÍƶÏ
רÌ⣺ÍƶÏÌâ
·ÖÎö£ºÈçͼËùʾ£¬C³£ÎÂÏÂÊÇÎÞÉ«ÒºÌ壬E¡¢XÊǵ¥ÖÊ£¬E¡¢F¡¢H¾ùΪÎÞÉ«ÆøÌ壮BºÍÇâÑõ»¯±µÈÜÒº·´Ó¦Éú³ÉÆøÌåFΪ°±Æø£¬ËµÃ÷BΪï§ÑΣ¬CΪˮ£¬FΪ°±ÆøºÍE·´Ó¦Éú³ÉC£¨ÎªË®£©ºÍH£¬·ÖÎöÅжÏEΪO2£¬HΪNO£¬DΪNO2£¬ËùÒÔA·Ö½âÉú³ÉÑõÆø¡¢¶þÑõ»¯µª¡¢Ë®µÄÎïÖÊÊÇÏõËᣬAºÍBÓÉÏàͬµÄÔªËØ×é³É£¬ËùÒÔBΪNH4NO3£¬GΪBa£¨NO3£©2£¬XΪÌúµ¥ÖÊ£¬ÉÙÁ¿Ìú·´Ó¦Éú³ÉKΪÏõËáÌú£»¹ýÁ¿Ìú·´Ó¦Éú³ÉJΪÏõËáÌú£»ÒÀ¾Ýת»¯¹Øϵ·ÖÎö»Ø´ðÑ¡ÏîÎÊÌ⣮
½â´ð£º ½â£ºÈçͼËùʾ£¬C³£ÎÂÏÂÊÇÎÞÉ«ÒºÌ壬E¡¢XÊǵ¥ÖÊ£¬E¡¢F¡¢H¾ùΪÎÞÉ«ÆøÌ壮BºÍÇâÑõ»¯±µÈÜÒº·´Ó¦Éú³ÉÆøÌåFΪ°±Æø£¬ËµÃ÷BΪï§ÑΣ¬CΪˮ£¬FΪ°±ÆøºÍE·´Ó¦Éú³ÉC£¨ÎªË®£©ºÍH£¬·ÖÎöÅжÏEΪO2£¬HΪNO£¬DΪNO2£¬ËùÒÔA·Ö½âÉú³ÉÑõÆø¡¢¶þÑõ»¯µª¡¢Ë®µÄÎïÖÊÊÇÏõËᣬAºÍBÓÉÏàͬµÄÔªËØ×é³É£¬ËùÒÔBΪNH4NO3£¬GΪBa£¨NO3£©2£¬XΪÌúµ¥ÖÊ£¬ÉÙÁ¿Ìú·´Ó¦Éú³ÉKΪÏõËáÌú£»¹ýÁ¿Ìú·´Ó¦Éú³ÉJΪÏõËáÌú£»
£¨1£©CΪˮ£¬Ë®·Ö×ӵĽṹʽΪ£º£»FΪNH3£¬°±ÆøµÄµç×ÓʽΪ£º£»·´Ó¦¢ÛÊÇBa£¨NO3£©2+H2SO4£¨Å¨£©=BaSO4¡ý+2HNO3£¬ÊôÓÚ¸´·Ö½â·´Ó¦£»
¹Ê´ð°¸Îª£º£»£»¸´·Ö½â·´Ó¦£»

£¨2£©ÎïÖʵĻ¯Ñ§Ê½HΪNO£¬IΪBaSO4£¬DΪNO2£¬JΪ·¢Fe£¨NO3£©2£¬¹Ê´ð°¸Îª£ºNO£»BaSO4£»NO2£»Fe£¨NO3£©2£»
£¨3£©ÊµÑéÊÒÅäÖÆKΪFe£¨NO3£©3ÈÜҺʱҪ¼ÓÈëÉÙÁ¿µÄÏ¡ÏõËáÒÖÖÆÌúÀë×ÓµÄË®½â£¬ÅäÖÆJΪFe£¨NO3£©2ÈÜҺʱҪ¼ÓÈëÉÙÁ¿µÄÌú·Û·ÀÖ¹ÑÇÌúÀë×Ó±»Ñõ»¯£¬¼ÓÈëÏ¡ÏõËáÒÖÖÆ ÑÇÌúÀë×ÓµÄË®½â£¬
¹Ê´ð°¸Îª£ºÏ¡ÏõË᣻Ìú·ÛºÍÏ¡ÏõË᣻
£¨4£©·´Ó¦¢ÚÊÇÌúºÍŨÏõËá¼ÓÈÈ·¢ÉúµÄ·´Ó¦£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪ£ºFe+6H++3NO3-
  ¡÷  
.
 
Fe3++3NO2¡ü+3H2O£»·´Ó¦¢ÜÊÇÏõËáÌúºÍÌú·´Ó¦Éú³ÉÏõËáÑÇÌúµÄ·´Ó¦£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪ£º2Fe3++Fe=3Fe2+£»
¹Ê´ð°¸Îª£ºFe+6H++3NO3-
  ¡÷  
.
 
Fe3++3NO2¡ü+3H2O£»2Fe3++Fe=3Fe2+£®
µãÆÀ£º±¾Ì⿼²éÁËÎïÖÊת»¯¹ØϵµÄ·ÖÎöÍƶϣ¬ÎïÖÊÐÔÖʵÄ×ÛºÏÓ¦Óã¬Á÷³ÌͼµÄ·ÖÎöÍƶÏÖÐC³£ÎÂÏÂÊÇÎÞÉ«ÒºÌ壬E¡¢XÊǵ¥ÖÊ£¬E¡¢F¡¢H¾ùΪÎÞÉ«ÆøÌ壮AºÍBÓÉÏàͬµÄÔªËØ×é³ÉÊǽâÌâ¹Ø¼ü£¬ÌâÑÛÔںͼӦÉú³ÉµÄÆøÌåÊÇ°±Æø£¬ÌâÄ¿ÄѶȽϴó£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÏÂÁз´Ó¦ÖУ¬ÊôÓÚÑõ»¯»¹Ô­·´Ó¦£¬µ«Ë®¼È²»ÊÇÑõ»¯¼Á£¬ÓÖ²»ÊÇ»¹Ô­¼ÁµÄÊÇ£¨¡¡¡¡£©
A¡¢SO3+H2O=H2SO4
B¡¢2Na2O2+2H2O=4NaOH+O2¡ü
C¡¢2F2+2H2O=4HF+O2
D¡¢Mg+2H2O=Mg£¨OH£©2+H2¡ü

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÏÂÁз´Ó¦µÄÀë×Ó·½³ÌʽÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A¡¢CuSO4ÈÜÒºÎüÊÕH2SÆøÌ壺Cu2++H2S¡úCuS¡ý+2H+
B¡¢Ca£¨HCO3£©2ÈÜÒººÍÉÙÁ¿µÄNaOHÈÜÒº»ìºÏ£ºCa2++2HC
O
-
3
+2OH-
¡úCaCO3¡ý+CO32-+2H2O
C¡¢ÁòËáÑÇÌúÈÜÒºÖмÓÈëÓÃÁòËáËữµÄ¹ýÑõ»¯ÇâÈÜÒº£ºFe2++2H++H2O2¡úFe2++2H2O
D¡¢µí·Ûµâ»¯¼ØÈÜÒºÔÚ¿ÕÆøÖбäÀ¶£º4I-+O2+2H2O¡ú2I2+4OH-

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

»¯ºÏÎïHÊÇÒ»ÖÖÏãÁÏ£¬´æÔÚÓÚ½ðéÙÖУ¬¿ÉÓÃÈçÏ·Ïߺϳɣº

Íê³ÉÏÂÁи÷Ì⣮
£¨1£©C4H10µÄÒ»ÂÈ´úÍéµÄͬ·ÖÒì¹¹ÌåÓÐ
 
ÖÖ¡¢¾ÅÂÈ´úÍéµÄͬ·ÖÒì¹¹ÌåÓÐ
 
ÖÖ£®
£¨2£©FµÄ½á¹¹¼òʽÊÇ
 
£»·´Ó¦£¨1£©µÄ·´Ó¦ÀàÐÍÊÇ
 
£»
£¨3£©·´Ó¦£¨2£©µÄ»¯Ñ§·½³ÌʽΪ
 
£»
£¨4£©ÓëG¾ßÓÐÏàͬ¹ÙÄÜÍŵÄGµÄ·¼ÏãÀàͬ·ÖÒì¹¹Ìå¹²ÓÐ
 
ÖÖ£¨²»°üÀ¨G£©£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÔÚ100mL0.1mol/LµÄ´×ËáÈÜÒºÖУ¬Óûʹ´×ËáµÄµçÀë³Ì¶ÈÔö´ó£¬ÇâÀë×ÓŨ¶È¼õС£¬¿É²ÉÓõķ½·¨ÊÇ£¨¡¡¡¡£©
A¡¢¼ÓÈÈ
B¡¢¼ÓÈëlmol/LµÄ´×ËáÈÜÒºl00mL
C¡¢¼ÓÈëÉÙÁ¿µÄ0.5mol/LµÄÁòËá
D¡¢¼ÓÈëÉÙÁ¿µÄlmol/LµÄNaOHÈÜÒº

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÏÂÁÐÀë×Ó·½³ÌʽÊéдÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A¡¢Ì¼ËáÇâÄƵÄË®½â·´Ó¦£ºHCO3-+H2O¨TCO32-+H3O+
B¡¢ÏõËáÒøÈÜÒºÖеμӹýÁ¿°±Ë®£ºAg++2NH3?H2O=[Ag£¨NH3£©2]++2H2O
C¡¢ÒÔÍ­×÷µç¼«µç½âÁòËáÍ­ÈÜÒº£º2Cu2++2H2O
 µç½â 
.
 
2Cu+O2¡ü+4H+
D¡¢ÏòÃ÷·¯[KAl£¨SO4£©2]ÈÜÒºÖÐÖðµÎ¼ÓÈëBa£¨OH£©2ÈÜÒºÖÁSO42-Ç¡ºÃÍêÈ«³Áµí£º2Al3++3SO42-+3Ba2++6OH-=2Al£¨OH£©3¡ý+3BaSO4¡ý

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

[»¯Ñ§Óë¼¼Êõ]
º£ÑóÊÇÒ»×ù¾Þ´óµÄ±¦²Ø£¬ÂȼҵºÍ½ðÊôþµÄÖƱ¸Ô­Á϶¼À´×ÔÓÚº£Ë®£®

£¨1£©Àë×ÓĤµç½â¼¼ÊõÖð½¥³ÉΪÁËÂȼҵµÄÖ÷ÒªÉú²ú¼¼Êõ£¬ÔÚÏÂͼÀë×ÓĤµç½â²ÛÖУ¬¢Ü±íʾ
 
£¬¢ß·Ö±ð
 
£®
£¨2£©ÒÑÖªº£Ë®Öк¬Na+¡¢Cl-¡¢Ca2+¡¢Mg2+¡¢SO2-4µÈÀë×Ó£¬ÏÂÃæÊǺ£Ë®ÀûÓõçÉøÎö·¨»ñµÃµ­Ë®µÄÔ­Àíͼ£¬µç¼«Îª¶èÐԵ缫£®Çë·ÖÎöÏÂÁÐÎÊÌ⣺
¢ÙÑôÀë×Ó½»»»Ä¤ÊÇÖ¸
 
£¨ÌîA»òBÖ®Ò»£©£®
¢Úд³öͨµçºóÑô¼«ÇøµÄµç¼«·´Ó¦Ê½£º
 
£®
£¨3£©ÎªÁ˳ýÈ¥º£Ë®ÖеÄCa2+¡¢Mg2+£¬ÓÃÀë×Ó½»»»Ê÷Ö¬À´½øÐд¦Àí£¬Ë®ÖеÄCa2+¡¢Mg2+Óë½»»»Ê÷Ö¬µÄ
 
ÆðÀë×Ó½»»»×÷Ó㬵±ÒõÀë×Ó½»»»Ê÷֬ʧЧºó¿É·ÅÈë
 
ÈÜÒºÖÐÔÙÉú£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨1£©ÒÑÖªÏÂÁз´Ó¦ÔÚÒ»¶¨Ìõ¼þÏ¿ÉÒÔ·¢Éú£º
H2O2+2Fe2++2H+=2Fe3++2H2O     
H2O2+2Fe3+=2Fe2++O2¡ü+2H+
ÔÚÒÔÉÏ·´Ó¦ÖÐFe2+ʵ¼ÊÉÏÆð×Å
 
×÷Óã¬×Ü·´Ó¦Ê½Îª
 

£¨2£©I2ÓëFe2+Ò»Ñù·¢ÉúÉÏÊöÀàËÆ·´Ó¦£¬Àà±È£¨1£©ÔÚÏÂÃæÌîÈëÅäƽµÄºÏÊʵĻ¯Ñ§·´Ó¦·½³Ìʽ£ºH2O2+I2=2HIO£¬
 
£¬×Ü·´Ó¦Ê½Îª
 

£¨3£©ÔÚÁòËáºÍKIµÄ»ìºÏÈÜÒºÖмÓÈë×ãÁ¿µÄH2O2£¬·Å³ö´óÁ¿µÄÎÞÉ«ÆøÌ壬ÈÜÒº³Ê×ØÉ«£¬²¢¿Éʹµí·Û±äÀ¶É«£®ÓÐѧÉúÈÏΪ¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽΪ£ºH2O2+2I-=O2¡ü+2H++I2£¬Õâ¸ö·½³ÌʽÕýÈ·Âð£¿
 
£¬ÀíÓÉÊÇ
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

·°¼°»¯ºÏÎïÓÃ;¹ã·º£®¹¤ÒµÉϳ£Óú¬ÉÙÁ¿Al2O3µÄ·°Ìú¿ó£¨FeO?V2O5£©¼îÈÛ·¨ÌáÈ¡V2O5£®¼òÒªÁ÷³ÌÈçÏ£º

ÒÑÖª£º¢Ù±ºÉÕʱ¿É·¢Éú·´Ó¦£ºV2O5+Al2O3+2Na2CO3 
 ¸ßΠ
.
 
2NaVO3+2NaAlO2+2CO2¡ü
¢Ú³£ÎÂÏÂÎïÖʵÄÈܽâ¶È£ºNaVO3¡«21.2g/100gË®£»HVO3¡«0.008g/100gË®
£¨1£©¡°½þ³öÔüB¡±µÄÖ÷Òª³É·ÖÊÇ
 
£®£¨Ð´»¯Ñ§Ê½£©
£¨2£©Éú²úÖУ¬²»Ö±½ÓÓÃH2SO4½þÅÝ¡°ÉÕÔüA¡±»ñÈ¡HVO3µÄÔ­ÒòÊÇ
 
£®
£¨3£©¡°²Ù×÷¢Ù¡±°üÀ¨
 
¡¢Ï´µÓ£®Èç¹û²»Ï´µÓ£¬Ôò²úÆ·ÖпÉÄܺ¬ÓеĽðÊôÑôÀë×ÓÊÇ
 
¡¢
 
£®ÏÂÁÐ×°Ö㨲¿·Ö¼Ð³ÖÒÇÆ÷Ê¡È¥£©¿ÉÓÃÔÚʵÑéÊÒ½øÐС°²Ù×÷¢Ú¡±µÄÊÇ
 
£®£¨ÌîÐòºÅ£©

£¨4£©NaVO3ÓÃÓÚÔ­Ó͵ÄÍÑÁò¼¼Êõ£¬ÓÉV2O5ÈÜÓÚNaOHÈÜÒºÖÐÖÆÈ¡£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪ£º
 
£®
£¨5£©V2O5»¹¿ÉÓÃÓÚ½«¹èÌúÁ¶³É·°Ìú£®Éú²úÖмÓÈëCaO¿É½ÚÄܼõÅÅ£®Óйط´Ó¦ÈçÏ£º
2V2O5£¨l£©+5Si£¨s£©+5CaO£¨s£©=4V£¨s£©+5CaSiO3£¨s£©£»¡÷H1=-2096.7kJ/mol
ÒÑÖª£ºCaSiO3£¨s£©=CaO£¨s£©+SiO2£¨s£©£»¡÷H2=+92.5kJ/mol
Ôò£º2V2O5£¨l£©+5Si£¨s£©=4V£¨s£©+5SiO2£¨s£©£»¡÷H3=
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸