A¡¢B¡¢C¡¢DÊÇËÄÖÖ³£¼ûµÄÓлúÎÆäÖУ¬AÊÇÒ»ÖÖÆø̬Ìþ£¬ÔÚ±ê×¼×´¿öϵÄÃܶÈÊÇ1.25g/L£¬Æä²úÁ¿ÊǺâÁ¿Ò»¸ö¹ú¼ÒʯÓÍ»¯¹¤·¢Õ¹Ë®Æ½µÄ±êÖ¾Ö®Ò»£»DµÄ·Ö×ÓʽΪC2H4O2£»BºÍDÔÚŨÁòËáºÍ¼ÓÈȵÄÌõ¼þÏ·¢Éú·´Ó¦£¬Éú³ÉµÄÓлúÎïÓÐÌØÊâµÄÏãζ£»A¡¢B¡¢C¡¢DÔÚÒ»¶¨Ìõ¼þϵÄת»¯¹ØϵÈçͼËùʾ£¨·´Ó¦Ìõ¼þÒÑÊ¡ÂÔ£©£º
»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Ö¸³öÏÂÁз´Ó¦µÄÀàÐÍ£º
¢ÙÊôÓÚ
¼Ó³É·´Ó¦
¼Ó³É·´Ó¦
¡¢¢ÚÊôÓÚ
Ñõ»¯·´Ó¦
Ñõ»¯·´Ó¦
£®
£¨2£©A¡¢B¡¢DÖУ¬¼È²»ÄÜʹËáÐÔ¸ßÃÌËá¼ØÈÜÒºÍÊÉ«£¬ÓÖ²»ÄÜʹäåË®ÍÊÉ«µÄÊÇ
CH3COOH
CH3COOH
£¬·Ö×ÓÖÐËùÓÐÔ­×Ó¾ùÔÚͬһ¸öƽÃæÉϵÄÊÇ
CH2=CH2
CH2=CH2
£®
£¨3£©Ð´³öÏÂÁз´Ó¦µÄ»¯Ñ§·½³Ìʽ£º
¢Ú£º
2CH3CH2OH+O2
Cu
¡÷
2CH3CHO+2H2O
2CH3CH2OH+O2
Cu
¡÷
2CH3CHO+2H2O
¡¢
B+D£º
CH3CH2OH+CH3COOH
ŨÁòËá
¡÷
CH3COOCH2CH3+H2O
CH3CH2OH+CH3COOH
ŨÁòËá
¡÷
CH3COOCH2CH3+H2O
£®
·ÖÎö£ºA¡¢B¡¢C¡¢DÊÇËÄÖÖ³£¼ûµÄÓлúÎÆäÖУ¬AÊÇÒ»ÖÖÆø̬Ìþ£¬ÔÚ±ê×¼×´¿öϵÄÃܶÈÊÇ1.25g/L£¬Ä¦¶ûÖÊÁ¿=1.25g/L¡Á22.4L/mol=28g/mol£¬Æä²úÁ¿ÊǺâÁ¿Ò»¸ö¹ú¼ÒʯÓÍ»¯¹¤·¢Õ¹Ë®Æ½µÄ±êÖ¾Ö®Ò»£¬ÔòAΪCH2=CH2£»AÓëË®·¢Éú¼Ó³É·´Ó¦Éú³ÉB£¬ÔòBΪCH3CH2OH£¬ÒÒ´¼ÔÚCu»òAg×÷´ß»¯¼ÁÌõ¼þÏ·¢ÉúÑõ»¯·´Ó¦CH3CHO£¬CΪCH3CHO£¬CH3CHO¿É½øÒ»²½Ñõ»¯ÎïCH3COOH£¬DΪCH3COOH£¬CH3CH2OHºÍCH3COOHÔÚŨÁòËá×÷ÓÃÏ·´Ó¦Éú³ÉÒÒËáÒÒõ¥£¨CH3COOCH2CH3£©£¬¾Ý´Ë½â´ð£®
½â´ð£º½â£ºA¡¢B¡¢C¡¢DÊÇËÄÖÖ³£¼ûµÄÓлúÎÆäÖУ¬AÊÇÒ»ÖÖÆø̬Ìþ£¬ÔÚ±ê×¼×´¿öϵÄÃܶÈÊÇ1.25g/L£¬Ä¦¶ûÖÊÁ¿=1.25g/L¡Á22.4L/mol=28g/mol£¬Æä²úÁ¿ÊǺâÁ¿Ò»¸ö¹ú¼ÒʯÓÍ»¯¹¤·¢Õ¹Ë®Æ½µÄ±êÖ¾Ö®Ò»£¬ÔòAΪCH2=CH2£»AÓëË®·¢Éú¼Ó³É·´Ó¦Éú³ÉB£¬ÔòBΪCH3CH2OH£¬ÒÒ´¼ÔÚCu»òAg×÷´ß»¯¼ÁÌõ¼þÏ·¢ÉúÑõ»¯·´Ó¦CH3CHO£¬CΪCH3CHO£¬CH3CHO¿É½øÒ»²½Ñõ»¯ÎïCH3COOH£¬DΪCH3COOH£¬CH3CH2OHºÍCH3COOHÔÚŨÁòËá×÷ÓÃÏ·´Ó¦Éú³ÉÒÒËáÒÒõ¥£¨CH3COOCH2CH3£©£¬
£¨1£©·´Ó¦¢ÙÊÇÒÒÏ©ÓëË®·¢Éú¼Ó³É·´Ó¦Éú³ÉÒÒ´¼£»·´Ó¦¢ÚÊÇÒÒ´¼·¢Éú´ß»¯Ñõ»¯Éú³ÉÒÒÈ©£¬
¹Ê´ð°¸Îª£º¼Ó³É·´Ó¦£»Ñõ»¯·´Ó¦£»
£¨2£©A¡¢B¡¢DÖУ¬ÒÒÏ©º¬ÓÐC=CË«¼ü£¬Äܱ»ËáÐÔ¸ßÃÌËá¼ØÑõ»¯£¬ÄÜÓëäå·¢Éú¼Ó³É·´Ó¦£¬ÄÜʹËáÐÔ¸ßÃÌËá¼ØÈÜÒºÍÊÉ«£¬ÓÖÄÜʹäåË®ÍÊÉ«£¬ÒÒ´¼Äܱ»Äܱ»ËáÐÔ¸ßÃÌËá¼ØÑõ»¯£¬Ê¹ËáÐÔ¸ßÃÌËá¼ØÈÜÒºÍÊÉ«£¬ÒÒËá¼È²»ÄÜʹËáÐÔ¸ßÃÌËá¼ØÈÜÒºÍÊÉ«£¬ÓÖ²»ÄÜʹäåË®ÍÊÉ«£»
ÒÒ´¼¡¢ÒÒËáÖк¬Óм׻ù£¬¾ßÓм×ÍéËÄÃæÌå½á¹¹£¬ËùÓÐÔ­×Ó²»¿ÉÄܹ²Ã棬ÒÒϩΪƽÃæ½á¹¹£¬ËùÓÐÔ­×Ó´¦ÓÚͬһƽÃ棬
¹Ê´ð°¸Îª£ºCH3COOH£»CH2=CH2£»
£¨3£©·´Ó¦¢ÚÊÇÒÒ´¼ÔÚCu»òAg×÷´ß»¯¼ÁÌõ¼þÏ·¢ÉúÑõ»¯·´Ó¦CH3CHO£¬·´Ó¦·½³ÌʽΪ£º2CH3CH2OH+O2
Cu
¡÷
2CH3CHO+2H2O£¬
B+D·´Ó¦ÊÇÒÒËáÓëÒÒ´¼·¢Éúõ¥»¯·´Ó¦Éú³ÉÒÒËáÒÒõ¥£¬·´Ó¦·½³ÌʽΪ£ºCH3CH2OH+CH3COOH
ŨÁòËá
¡÷
 CH3COOCH2CH3+H2O£¬
¹Ê´ð°¸Îª£º2CH3CH2OH+O2
Cu
¡÷
2CH3CHO+2H2O£»CH3CH2OH+CH3COOH
ŨÁòËá
¡÷
 CH3COOCH2CH3+H2O£®
µãÆÀ£º±¾Ì⿼²éÓлúÎïÍƶϡ¢Ï©Óë´¼¡¢È©¡¢ôÈËáÖ®¼äµÄת»¯¹ØϵµÈ£¬ÄѶȲ»´ó£¬×¢Òâ»ù´¡ÖªÊ¶µÄÀí½âÕÆÎÕ£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

A¡¢B¡¢C¡¢DÊÇËÄÖÖ¶ÌÖÜÆÚÔªËØ£¬EÊǹý¶ÉÔªËØ£®A¡¢B¡¢CͬÖÜÆÚ£¬C¡¢DͬÖ÷×壬AµÄÔ­×ӽṹʾÒâͼΪ£º£¬BÊÇͬÖÜÆÚµÚÒ»µçÀëÄÜ×îСµÄÔªËØ£¬CµÄ×îÍâ²ãÓÐÈý¸ö³Éµ¥µç×Ó£¬EµÄÍâΧµç×ÓÅŲ¼Ê½Îª3d64s2£®»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Ð´³öÏÂÁÐÔªËصķûºÅ£ºA
Si
Si
  B
Na
Na
  C
P
P
   D
N
N

£¨2£©Óû¯Ñ§Ê½±íʾÉÏÊöÎåÖÖÔªËØÖÐ×î¸ß¼ÛÑõ»¯Îï¶ÔӦˮ»¯ÎïËáÐÔ×îÇ¿µÄÊÇ
HNO3
HNO3
£¬¼îÐÔ×îÇ¿µÄÊÇ
NaOH
NaOH
£®
£¨3£©ÓÃÔªËØ·ûºÅ±íʾDËùÔÚÖÜÆÚ£¨³ýÏ¡ÓÐÆøÌåÔªËØÍ⣩µÚÒ»µçÀëÄÜ×î´óµÄÔªËØÊÇ
F
F
£¬µç¸ºÐÔ×î´óµÄÔªËØÊÇ
F
F
£®
£¨4£©DµÄÇ⻯Îï±ÈCµÄÇ⻯ÎïµÄ·Ðµã
¸ß
¸ß
£¨Ìî¡°¸ß¡°»ò¡°µÍ¡°£©£¬Ô­Òò
°±Æø·Ö×ÓÖ®¼äº¬ÓÐÇâ¼ü
°±Æø·Ö×ÓÖ®¼äº¬ÓÐÇâ¼ü

£¨5£©EÔªËØÔ­×ӵĺ˵çºÉÊýÊÇ
26
26
£¬EÔªËØÔÚÖÜÆÚ±íµÄµÚ
ËÄ
ËÄ
ÖÜÆÚ£¬µÚ
¢ø
¢ø
×壬ÒÑÖªÔªËØÖÜÆÚ±í¿É°´µç×ÓÅŲ¼·ÖΪsÇø¡¢pÇøµÈ£¬ÔòEÔªËØÔÚ
d
d
Çø£®
£¨6£©A¡¢B¡¢C×î¸ß¼ÛÑõ»¯ÎïµÄ¾§ÌåÀàÐÍÊÇ·Ö±ðÊÇ
Ô­×Ó
Ô­×Ó
¾§Ìå¡¢
Àë×Ó
Àë×Ó
¾§Ìå¡¢
·Ö×Ó
·Ö×Ó
¾§Ìå
£¨7£©»­³öDµÄºËÍâµç×ÓÅŲ¼Í¼
£¬ÕâÑùÅŲ¼×ñÑ­ÁË
ÅÝÀû
ÅÝÀû
Ô­ÀíºÍ
ºéÌØ
ºéÌØ
¹æÔò£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â

A¡¢B¡¢C¡¢DÊÇËÄÖÖ³£¼ûµ¥ÖÊ£¬Æä¶ÔÓ¦ÔªËصÄÔ­×ÓÐòÊýÒÀ´ÎÔö´ó£¬ÆäÖÐB¡¢DÊôÓÚ³£¼û½ðÊô£¬ÆäÓà¾ùΪ³£¼û»¯ºÏÎJÊÇÒ»ÖÖºÚÉ«¹ÌÌ壬IµÄŨÈÜÒº¾ßÓл¹Ô­ÐÔ£¬´ÓA-IµÄËùÓÐÎïÖÊÖ®¼äÓÐÈçϵÄת»¯¹Øϵ£º

»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Ð´³öÎïÖÊCµÄ¹¹³ÉÔªËØÔÚÖÜÆÚ±íÖеÄλÖÃ
µÚÈýÖÜÆÚ¡¢µÚ¢÷A×å
µÚÈýÖÜÆÚ¡¢µÚ¢÷A×å
£®
£¨2£©Ð´³öBÓëF·´Ó¦µÄ»¯Ñ§·½³Ìʽ
8Al+3Fe3O4
 ¸ßΠ
.
 
9Fe+4Al2O3
8Al+3Fe3O4
 ¸ßΠ
.
 
9Fe+4Al2O3
£®
£¨3£©ÓÉEµÄ±¥ºÍÈÜÒº¿ÉÒÔÖƵýºÌ壬¾ßÌåÖƱ¸·½·¨ÊÇ£º
½«±¥ºÍFeCl3ÈÜÒºÖðµÎ¼ÓÈë·ÐË®ÖУ¬¼ÌÐø¼ÓÈÈÖÁÒºÌå±äΪºìºÖÉ«
½«±¥ºÍFeCl3ÈÜÒºÖðµÎ¼ÓÈë·ÐË®ÖУ¬¼ÌÐø¼ÓÈÈÖÁÒºÌå±äΪºìºÖÉ«
£¬Óû¯Ñ§·½³Ìʽ±íʾ¸Ã¹ý³ÌµÄÔ­Àí£º
Fe3++3H2O
  ¡÷  
.
 
Fe£¨OH£©3£¨½ºÌ壩+3H+
Fe3++3H2O
  ¡÷  
.
 
Fe£¨OH£©3£¨½ºÌ壩+3H+
£®ÈôÒªÌá´¿¸Ã½ºÌ壬²ÉÓõIJÙ×÷·½·¨½Ð
ÉøÎö
ÉøÎö
£®´ËÒºÌå¾ßÓеÄÐÔÖÊÊÇ
abd
abd
£¨ÌîдÐòºÅ×Öĸ£©
a£®¹âÊøͨ¹ý¸ÃÒºÌåʱÐγɹâÁÁµÄ¡°Í¨Â·¡±?
b£®ÏòÒºÌåÖÐÖðµÎ¼ÓÈë×ãÁ¿ÇâµâËᣬÏÈÓгÁµí²úÉú£¬ºó³ÁµíÖð½¥Èܽ⣬ÔÙµÎÈ뼸µÎµí·ÛÈÜÒº£¬ÈÜÒº±äΪÀ¶É«
c£®Ïò¸ÃÒºÌåÖмÓÈëÏõËáÒøÈÜÒº£¬ÎÞ³Áµí²úÉú?
d£®½«¸ÃÒºÌå¼ÓÈÈ¡¢Õô¸É¡¢×ÆÉÕ£¬µÃºì×ØÉ«¹ÌÌå?
ÁíÈ¡ÉÙÁ¿ÉÏÊö½ºÌåÖÃÓÚUÐ͹ÜÖУ¬°´ÈçͼװÖÃͼÁ¬½ÓºÃ×°Öã®Í¨µçһС¶Îʱ¼äºó£¬X¼«¸½½üµÄÏÖÏóÊÇ
ºìºÖÉ«¼ÓÉî
ºìºÖÉ«¼ÓÉî
£®
£¨4£©È¡ÉÙÁ¿ÉÏÊö½ºÌåÖÃÓÚÊÔ¹ÜÖУ¬ÏòÊÔ¹ÜÖеμÓÒ»¶¨Á¿Ï¡ÑÎËᣬ±ßµÎ±ßÕñµ´£¬¿ÉÒÔ¿´µ½ÈÜÒºÑÕÉ«Öð½¥±ädz£¬×îÖյõ½»ÆÉ«µÄÈÜÒº£¬·¢Éú´Ë±ä»¯µÄÀë×Ó·½³ÌʽΪ
Fe£¨OH£©3+3H+=Fe3++3H2O
Fe£¨OH£©3+3H+=Fe3++3H2O
£®
£¨5£©GµÄË®ÈÜÒºÏÔ
¼î
¼î
ÐÔ£¨ÌîËá»ò¼î£©£»Ô­ÒòÊÇ£¨ÓÃÀë×Ó·½³Ìʽ±í Ê¾£©
AlO2-+2H2OAl£¨OH£©3+OH-
AlO2-+2H2OAl£¨OH£©3+OH-
£®
£¨6£©JÔÚH2O2·Ö½â·´Ó¦ÖÐ×÷´ß»¯¼Á£®Èô½«ÊÊÁ¿J¼ÓÈëËữµÄH2O2µÄÈÜÒºÖУ¬JÈܽâÉú³ÉËüµÄ+2¼ÛÀë×Ó£¬¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ
MnO2+H2O2+2H+¨TMn2++O2¡ü+2H2O
MnO2+H2O2+2H+¨TMn2++O2¡ü+2H2O
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

A¡¢B¡¢C¡¢DÊÇËÄÖÖÓлúÎËüÃǵķÖ×ÓÖоùº¬ÓÐ2¸ö̼ԭ×Ó£¬ÆäÖÐAºÍBÊÇÌþ£®ÔÚ±ê×¼×´¿öÏ£¬A¶ÔÇâÆøµÄÏà¶ÔÃܶÈÊÇ13£¬BÓëHCl·´Ó¦Éú³ÉC£¬CºÍD»ìºÏºó¼ÓÈëNaOH²¢¼ÓÈÈ¿ÉÉú³ÉB£®
£¨1£©ÅжÏA¡¢B¡¢C¡¢D¸÷ÊÇÄÄÖÖÓлúÎд³öËüÃǵĽṹ¼òʽ£»
£¨2£©Ð´³öÌâÖÐËùÉæ¼°µÄÓйط´Ó¦µÄ»¯Ñ§·½³Ìʽ£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

A¡¢B¡¢C¡¢DÊÇËÄÖÖ¶ÌÖÜÆÚÔªËØ£¬ËüÃǵÄÔ­×ÓÐòÊýÒÀ´ÎÔö´ó£¬ÆäÖÐA¡¢C¼°B¡¢D·Ö±ðÊÇͬÖ÷×åÔªËØ£®ÓÖÖªB¡¢DÁ½ÔªËصÄÔ­×ÓºËÖÐÖÊ×ÓÊýÖ®ºÍÊÇA¡¢CÁ½ÔªËصÄÔ­×ÓºËÖÐÖÊ×ÓÊýÖ®ºÍµÄ2±¶£®ÕâËÄÖÖÔªËصĵ¥ÖÊÖг£ÎÂÏÂÓÐÁ½ÖÖÊÇÆøÌ壬ÆäÖÐÒ»ÖÖÆøÌåÒ×ȼÉÕ£¬ÁíÍâÁ½ÖÖΪ¹ÌÌ壬ÆäÖÐÒ»ÖÖÊÇ»ÆÉ«¾§Ì壬Öʴ࣬Ò×ÑгɷÛÄ©£®
£¨1£©Ð´³öÔªËØ·ûºÅ£ºA
H
H
B
O
O
C
Na
Na
D
S
S
£®
£¨2£©Ð´³öÁ½ÖÖ¾ùº¬A¡¢B¡¢C¡¢DËÄÖÖÔªËصĻ¯ºÏÎïÏ໥·´Ó¦·Å³öÆøÌåµÄ»¯Ñ§·½³Ìʽ£º
NaHSO4+NaHSO3=Na2SO4+SO2¡ü+H2O
NaHSO4+NaHSO3=Na2SO4+SO2¡ü+H2O
£®
£¨3£©Óõç×Óʽ±íʾ»¯ºÏÎïC2DÐγɹý³Ì
£®
£¨4£©ÓÃAºÍBÁ½ÔªËصĵ¥ÖÊ¿ÉÒÔÖƳÉȼÁϵç³Ø£¬µç³ØÖÐ×°ÓÐŨKOHÈÜÒº£®Óöà¿×µÄ¶èÐԵ缫½þÈëŨKOHÈÜÒºÖУ¬Á½¼«¾ùÓÐÌØÖƵķÀÖ¹ÆøÌå͸¹ý¸ôĤ£¬ÔÚÒ»¼«Í¨ÈëAµÄµ¥ÖÊ£¬ÁíÒ»¼«Í¨ÈëBµÄµ¥ÖÊ£®Ð´³öͨÈëBµ¥ÖʵÄÒ»¼«µç¼«·´Ó¦Ê½ÊÇ£º
O2+4e-+2H2O=4OH-
O2+4e-+2H2O=4OH-
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

a¡¢b¡¢c¡¢dÊÇËÄÖÖ¶ÌÖÜÆÚÔªËØ£®a¡¢b¡¢dͬÖÜÆÚ£¬c¡¢dͬÖ÷×壮aµÄÔ­×ӽṹʾÒâͼΪ¾«Ó¢¼Ò½ÌÍø£¬bÓëcÐγɻ¯ºÏÎïµÄµç×ÓʽΪ¾«Ó¢¼Ò½ÌÍøÏÂÁбȽÏÖÐÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A¡¢Ô­×Ӱ뾶£ºa£¾c£¾d£¾bB¡¢×î¸ß¼Ûº¬ÑõËáµÄËáÐÔc£¾d£¾aC¡¢Ô­×ÓÐòÊý£ºa£¾d£¾b£¾cD¡¢µ¥ÖʵÄÑõ»¯ÐÔa£¾b£¾d£¾c

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸