£¨14·Ö£©

A¡¢B¡¢C¡¢DÊÇËÄÖÖ³£¼ûµ¥ÖÊ£¬Æä¶ÔÓ¦ÔªËØµÄÔ­×ÓÐòÊýÒÀ´ÎÔö´ó£¬ÆäÖÐB¡¢DÊôÓÚ³£¼û½ðÊô£¬ÆäÓà¾ùΪ³£¼û»¯ºÏÎJÊÇÒ»ÖÖºÚÉ«¹ÌÌ壬IµÄŨÈÜÒº¾ßÓл¹Ô­ÐÔ£¬A¡ªJµÄËùÓÐÎïÖÊÖ®¼äÓÐÈçϵÄת»¯¹ØÏµ£¨²¿·Ö·´Ó¦²úÎïÊ¡ÂÔ£©£º

£¨1£©BÔªËØºÍCÔªËØµÄ¼òµ¥Àë×Ó°ë¾¶´óС¹ØÏµÊÇ £¨ÓÃÀë×Ó·ûºÅ±íʾ£©£º              _£»

£¨2£©½«µªÔªËØÓëCÐγɵϝºÏÎïNC3¼ÓÈëË®ÖвúÉúʹºìɫʯÈïÊÔÖ½±äÀ¶µÄÆøÌ壬д³ö¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ                                        £»

£¨3£©ÓÉEµÄ±¥ºÍÈÜÒºµÎÈë·ÐË®ÖÐÐγÉ͸Ã÷ÒºÌ壬ÔÙ½«´ËÒºÌå×°ÈëUÐ͹ܣ¬²¢ÔÚUÐ͹ܵÄÁ½¶Ë²åÈëµç¼«£¬½ÓֱͨÁ÷µç£¬ÔÚÑô¼«¶Ë¿É¹Û²ìµ½µÄÏÖÏóÊÇ______________  ___£»

£¨4£©½«ÊÊÁ¿J¼ÓÈëËữµÄH2O2µÄÈÜÒºÖУ¬JÈܽâÉú³ÉËüµÄ+2¼ÛÀë×Ó£¬¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ__________________________________________________            _£»

£¨5£©Ïòº¬ÓÐ0£®1 mol GµÄÈÜÒºÖеμÓ5 mol/L µÄÑÎËáÈÜÒº£¬µÃµ½³Áµí3£®9 g £¬Ôò¼ÓÈëÑÎËáµÄÌå»ý¿ÉÄÜΪ______________________________£¨mL£©¡£

 

¡¾´ð°¸¡¿

 

£¨1£©r£¨Al3+£© < r£¨Cl£©£¨»òAl3+ < Cl£© £¨2·Ö£©

£¨2£©NCl3+ 3H2O = NH3¡ü + 3HClO  £¨3·Ö£©

£¨3£©ºìºÖÉ«±ädz£¨´ðÓÐCl2Éú³É¿É¸ø1·Ö£©£¨2·Ö£©

£¨4£©MnO2 + H2O2 + 2H+ = Mn2+ + O2¡ü + 2H2O  £¨3·Ö£©

£¨5£©10ºÍ50£¨¸÷2·Ö£©

¡¾½âÎö¡¿

 

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´³öÏÂÁÐÔªËØµÄ·ûºÅ£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£º

A¡¢B¡¢C¡¢DÊÇËÄÖÖ¶ÌÖÜÆÚÔªËØ£¬ËüÃǵÄÔ­×ÓÐòÊýÒÀ´ÎÔö´ó£¬ÆäÖУ¬A¡¢C¼°B¡¢D·Ö±ðÊÇͬÖ÷×åÔªËØ£»AÔªËØµÄÔ­×Ó°ë¾¶ÊÇËùÓÐÖ÷×åÔªËØÖÐÔ­×Ó°ë¾¶×îСµÄ£»B¡¢DÁ½ÔªËصÄÔ­×ÓºËÖÐÖÊ×ÓÊýÖ®ºÍÊÇA¡¢CÁ½ÔªËØÔ­×ÓºËÖÐÖÊ×ÓÊýÖ®ºÍµÄ2±¶£»ËÄÖÖÔªËØËùÐγɵĵ¥ÖÊÖÐA¡¢Bµ¥ÖÊÊÇÆøÌ壬C¡¢Dµ¥ÖÊÊǹÌÌ壮
£¨1£©B¡¢CÁ½ÔªËصÄÃû³ÆÎª£º
Ñõ¡¢ÄÆ
Ñõ¡¢ÄÆ
£»
£¨2£©Óõç×Óʽ±íʾC2DµÄÐγɹý³Ì£º
£®
£¨3£©ÓÉB¡¢CÁ½ÔªËØËùÐγɵÄÔ­×Ó¸öÊý±ÈΪ1£º1µÄ»¯ºÏÎïÊôÓÚ
Àë×Ó
Àë×Ó
£¨Ìî¡°Àë×Ó¡±»ò¡°¹²¼Û¡±£©»¯ºÏÎ´æÔڵĻ¯Ñ§¼üµÄÖÖÀàÓÐ
Àë×Ó¼ü¡¢·Ç¼«ÐÔ¼ü
Àë×Ó¼ü¡¢·Ç¼«ÐÔ¼ü
£¬Ð´³öËüÓëË®·´Ó¦µÄÀë×Ó·½³Ìʽ
2Na2O2+2H2O=4Na++4OH-+O2¡ü
2Na2O2+2H2O=4Na++4OH-+O2¡ü
£®
£¨4£©ÊµÑéÊÒ³£ÓÃA¡¢BÁ½ÔªËØËùÐγɵÄÔ­×Ó¸öÊýΪ1£º1µÄ»¯ºÏÎïÀ´ÖƱ¸Ò»ÖÖ³£¼ûÆøÌ壬Æä»¯Ñ§·½³ÌʽΪ
2H2O2
 MnO2 
.
 
2H2O+O2¡ü
2H2O2
 MnO2 
.
 
2H2O+O2¡ü
£®´ËʵÑéÖг£¼ÓÈëÒ»ÖÖºÚÉ«·ÛδÎïÖÊ£¬Æä×÷ÓÃÊÇ
´ß»¯¼Á
´ß»¯¼Á
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

A¡¢B¡¢C¡¢DÊÇËÄÖÖ¶ÌÖÜÆÚÔªËØ£¬E¡¢FÊǹý¶ÉÔªËØ£®A¡¢B¡¢CͬÖÜÆÚ£¬C¡¢DͬÖ÷×壬AµÄÔ­×ӽṹʾÒâͼΪ£º£¬BÊÇͬÖÜÆÚ³ýÏ¡ÓÐÆøÌåÍâ°ë¾¶×î´óµÄÔªËØ£¬CµÄ×îÍâ²ãÓРÈý¸ö³Éµ¥µç×Ó£¬E¡¢FµÄÍâΧµç×ÓÅŲ¼Ê½·Ö±ðΪ3d54s1£¬3d64s2£¬»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©AµÄ»ù̬ԭ×ӵĵç×ÓÅŲ¼Ê½ÊÇ
1s22s22p63s23p2
1s22s22p63s23p2
£»
£¨2£©BµÄ×î¸ß¼ÛÑõ»¯ÎïµÄ»¯Ñ§Ê½Îª
Na2O
Na2O
£¬CµÄ×îµÍ¸º»¯ºÏ¼ÛΪ
-3
-3
£®
£¨3£©ÓùìµÀ±íʾʽ±íʾDÔ­×ÓºËÍâµç×ÓÅŲ¼
£»
£¨4£©½âÊÍΪʲôEµÄÍâΧµç×ÓÅŲ¼Ê½Îª3d¡ä54s1£¬¶ø²»Îª3d44s2£¿
EµÄÍâΧµç×Ó³Ê3d¡ä54s1ʱ£¬3d¡¢4s¹ìµÀÉϵĵç×Ó´¦ÓÚ°ëÂú״̬£¬Õû¸öÌåϵµÄÄÜÁ¿×îµÍ
EµÄÍâΧµç×Ó³Ê3d¡ä54s1ʱ£¬3d¡¢4s¹ìµÀÉϵĵç×Ó´¦ÓÚ°ëÂú״̬£¬Õû¸öÌåϵµÄÄÜÁ¿×îµÍ
£»
£¨5£©FÔ­×ӽṹʾÒâͼÊÇ
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¾«Ó¢¼Ò½ÌÍøa¡¢b¡¢c¡¢dÊÇËÄÖÖ¶ÌÖÜÆÚÔªËØ£¬a¡¢b¡¢dͬÖÜÆÚ£¬c¡¢dͬÖ÷×壮aµÄÔ­×ӽṹʾÒâͼΪ£¬bÓëcÐγɵϝºÏÎïΪ b3c£¬ÆäÖÐbµÄ»¯ºÏ¼ÛΪ+1£®ÏÂÁбȽÏÖÐÕýÈ·µÄÊÇ£¨¡¡¡¡£©

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×Ó°ë¾¶£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°¸