ÌâÄ¿ÄÚÈÝ

A¡¢B¡¢C¡¢DÊÇËÄÖÖ¶ÌÖÜÆÚÔªËØ£¬ËüÃǵÄÔ­×ÓÐòÊýÒÀ´ÎÔö´ó£¬ÆäÖÐA¡¢C¼°B¡¢D·Ö±ðÊÇͬһÖ÷×åÔªËØ£¬B¡¢DÁ½ÔªËصÄÔ­×ÓºËÖÐÖÊ×ÓÊýÖ®ºÍÊÇA¡¢CÁ½ÔªËØÔ­×ÓºËÖÐÖÊ×ÓÊýÖ®ºÍµÄÁ½±¶£¬ÓÖÖªËÄÖÖÔªËصĵ¥ÖÊÖÐÓÐÁ½ÖÖÆøÌå¡¢Á½ÖÖ¹ÌÌå¡£

Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©B¡¢DÔÚÖÜÆÚ±íÖÐͬ´¦ÔÚ_____×壬DµÄÔ­×ӽṹʾÒâͼÊÇ        £¬ÏÂÁпÉÒÔÑéÖ¤BÓëDÁ½ÔªËØÔ­×ӵõç×ÓÄÜÁ¦Ç¿ÈõµÄʵÑéÊÂʵÊÇ              £¨Ìîд±àºÅ£©£»

A£®±È½ÏÕâÁ½ÖÖÔªËصÄÆø̬Ç⻯ÎïµÄ·Ðµã

B£®±È½ÏÕâÁ½ÖÖÔªËصÄÔ­×ӵĵç×Ó²ãÊý

C£®±È½ÏÕâÁ½ÖÖÔªËصÄÆø̬Ç⻯ÎïµÄÎȶ¨ÐÔ

D£®±È½ÏÕâÁ½ÖÖÔªËصĵ¥ÖÊÓëÇ⻯ºÏµÄÄÑÒ×

£¨2£©Ð´³öÁ½ÖÖ¾ùº¬A¡¢B¡¢C¡¢DËÄÖÖÔªËصĻ¯ºÏÎïÏ໥¼ä·¢Éú·´Ó¦£¬ÇÒÉú³ÉÆøÌåµÄÀë×Ó·½³Ìʽ                                                       £»

£¨3£©A¡¢B¡¢D¼ä¿ÉÐγɼס¢ÒÒÁ½ÖÖ΢Á££¬ËüÃǾùΪ¸ºÒ»¼ÛË«Ô­×ÓÒõÀë×Ó£¬ÇÒ¼×ÓÐ18¸öµç×Ó£¬ÒÒÓÐl0¸öµç×Ó£¬Ôò¼×ÓëÒÒ·´Ó¦µÄÀë×Ó·½³ÌʽΪ                        £»

£¨4£©DÔªËصÄÆø̬Ç⻯ÎïºÍÆäµÍ¼ÛÑõ»¯ÎïËù·¢ÉúµÄ·´Ó¦ÖÐÑõ»¯²úÎïÓ뻹ԭ²úÎïµÄÎïÖʵÄÁ¿Ö®±ÈΪ                  ¡£

£¨1£©¢öA £¨2·Ö£©     £¨2·Ö£©  BCD£¨3·Ö£¬Ö»ÒªÓдíÎ󲻵÷֣©

£¨2£©HSO£«H£«=H2O£«SO2¡ü £¨3·Ö£©

£¨3£©HS£­£«OH£­=S2£­£«H2O £¨3·Ö£©

£¨4£©2£º1  £¨2·Ö£©


½âÎö:
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
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£®
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
£®

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø