ÌâÄ¿ÄÚÈÝ

£¨7·Ö£©¸ù¾ÝϱíÐÅÏ¢»Ø´ðÒÔÏÂÎÊÌ⣺²¿·Ö¶ÌÖÜÆÚÔªËصÄÔ­×Ӱ뾶¼°Ö÷Òª»¯ºÏ¼Û¡£
ÔªËØ
A
B
C
D
Ô­×Ӱ뾶(nm)
0.130
0.118
0.090
0.102
Ö÷Òª»¯ºÏ¼Û
£«2
£«3
£«2
£«6£¬£­2
ÔªËØ
E
F
G
H
Ô­×Ӱ뾶(nm)
0.073
0.154
0.037
0.099
Ö÷Òª»¯ºÏ¼Û
£­2
£«1
£«1
£«7£¬£­1
(1)E¡¢F¡¢GÈýÔªËØÐγɵĻ¯ºÏÎïÖл¯Ñ§¼üÀàÐÍÊÇ______¡£
(2)B¡¢HÁ½ÔªËصÄ×î¸ß¼ÛÑõ»¯ÎïËù¶ÔÓ¦µÄË®»¯ÎïÏ໥·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ________¡£
(3)ʵÑéÊÒÖÐÖÆÈ¡Hµ¥ÖÊ·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ________¡£
(4)½«A¡¢BÁ½ÖÖÔªËصĵ¥ÖÊÓõ¼ÏßÁ¬½Ó½þÈëNaOHÈÜÒºÖУ¬·¢ÏÖµ¼ÏßÖÐÓеçÁ÷²úÉú£¬Æ为¼«·´Ó¦Ê½Îª________¡£
(1)Àë×Ó¼üºÍ¹²¼Û¼ü   (2)Al(OH)3£«3H£«===Al3£«£«3H2O
(3)MnO2£«4HCl(Ũ)MnCl2£«Cl2¡ü£«2H2O  (4)Al£­3e£­£«4OH£­===AlO2£­£«2H2O
¸ù¾ÝÔªËصĽṹ¼°ÓйØÐÔÖÊ¿ÉÖª£¬AÊÇMg£¬BÊÇAl£¬CÊÇBe£¬DÊÇS£¬EÊÇO£¬FÊÇNa£¬GÊÇH£¬HÊÇCl¡£
£¨1£©E¡¢F¡¢GÈýÔªËØÐγɵĻ¯ºÏÎïÊÇÇâÑõ»¯ÄÆ£¬º¬ÓÐÀë×Ó¼üºÍ¼«ÐÔ¼ü¡£
£¨2£©B¡¢HÁ½ÔªËصÄ×î¸ß¼ÛÑõ»¯ÎïËù¶ÔÓ¦µÄË®»¯Îï·Ö±ðÊÇÇâÑõ»¯ÂÁºÍ¸ßÂÈËᣬËùÒÔ¶þÕß·´Ó¦µÄ·½³ÌʽÊÇAl(OH)3£«3H£«===Al3£«£«3H2O¡£
£¨3£©ÊµÑéÊÒÖÆÈ¡ÂÈÆøµÄ»¯Ñ§·½³ÌʽÊÇMnO2£«4HCl(Ũ)MnCl2£«Cl2¡ü£«2H2O¡£
£¨4£©Ã¾µÄ»îÆÃÐÔÇ¿ÓÚÂÁµÄ£¬µ«Ã¾ºÍÇâÑõ»¯ÄÆÈÜÒº²»·´Ó¦£¬¶øÂÁÊÇ¿ÉÒÔ·´Ó¦µÄ£¬ËùÒÔÂÁÊǸº¼«£¬Ã¾ÊÇÕý¼«¡£Ôò¸º¼«µç¼«·´Ó¦Ê½ÊÇAl£­3e£­£«4OH£­===AlO2£­£«2H2O¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
£¨12·Ö£©ÒÑÖªÔªËصÄijÖÖÐÔÖÊ¡°X¡±ºÍÔ­×Ӱ뾶¡¢½ðÊôÐÔ¡¢·Ç½ðÊôÐÔµÈÒ»Ñù£¬Ò²ÊÇÔªËصÄÒ»ÖÖ»ù±¾ÐÔÖÊ¡£ÏÂÃæ¸ø³ö13ÖÖÔªËصÄXµÄÊýÖµ£º
ÔªËØ
Al
B
Be
C
Cl
F
Li
XµÄÊýÖµ
1.5
2.0
1.5
2.5
2.8
4.0
1.0
ÔªËØ
Mg
Na
O
P
S
Si
 
XµÄÊýÖµ
1.2
0.9
3.5
2.1
2.5
1.7
 
ÊÔ½áºÏÔªËØÖÜÆÚÂÉ֪ʶÍê³ÉÏÂÁÐÎÊÌ⣺
£¨1£©¾­Ñé¹æÂɸæËßÎÒÃÇ£ºµ±Ðγɻ¯Ñ§¼üµÄÁ½Ô­×ÓÏàÓ¦ÔªËصÄX²îÖµ´óÓÚ1.7ʱ£¬ËùÐγɵÄÒ»°ãΪÀë×Ó¼ü£»µ±Ð¡ÓÚ1.7ʱ£¬Ò»°ãΪ¹²¼Û¼ü¡£ÊÔÍƶÏBeCl2ÖеĻ¯Ñ§¼üÀàÐÍÊÇ                ¡£
£¨2£©¸ù¾ÝÉϱí¸ø³öµÄÊý¾Ý£¬¼òÊöÖ÷×åÔªËصÄXµÄÊýÖµ´óСÓëÔªËصĽðÊôÐÔÖ®¼äµÄ¹Øϵ                                        £»¼òÊöµÚ¶þÖÜÆÚÔªËØ(³ý¶èÐÔÆøÌåÍâ)µÄXµÄÊýÖµ´óСÓëÔ­×Ӱ뾶֮¼äµÄ¹Øϵ                                    ¡£
£¨3£©Ä³»¯ºÏÎï·Ö×ÓÖк¬ÓÐS¡ªN¼ü£¬ÄãÈÏΪ¸Ã¹²Óõç×Ó¶ÔÆ«ÏòÓÚ      Ô­×Ó(ÌîÔªËØ·ûºÅ)¡£
£¨4£©Ð´³öOÔ­×ӵĵç×ÓÅŲ¼Í¼                            
£¨5£©ClÔªËصÄ×î¸ßÕý¼ÛΪ     £¬Æä×î¸ß¼ÛÑõ»¯ÎïµÄË®»¯ÎïµÄ»¯Ñ§Ê½Îª         ¡£
£¨6£©ÈôÒª¼ø¶¨Ä³»¯ºÏÎïÖÐÊÇ·ñº¬ÓÐSÔªËØ£¬ÐèÓà            µÄ·½·¨£»ÈôҪȷ¶¨C2H6OµÄ½á¹¹£¬ÐèÓà            µÄ·½·¨£¬·Ö×ÓʽΪC2H6OµÄÁ½ÖÖ²»Í¬µÄ½á¹¹¼òʽ·Ö±ðΪ             £¬               £¬ÕâÁ½Öֽṹ»¥Îª              ¡£
£¨Ã¿¿Õ3·Ö£¬¹²15·Ö£©Ô­×ÓÐòÊýÒÀ´ÎÔö´óµÄËÄÖÖ³£¼ûÔªËØW¡¢X¡¢Y¡¢Z£¬¾ù¿ÉÐγÉÒ»ÖÖ»ò¶àÖÖÑõ»¯Îï¡£ÆäÖÐW¡¢YµÄÑõ»¯ÎïÊǵ¼ÖÂËáÓêµÄÖ÷ÒªÎïÖÊ£¬XµÄÑõ»¯Îï¼È¿ÉÒÔºÍÇ¿Ëá¡¢Òà¿ÉÒÔºÍÇ¿¼î·´Ó¦£¬ZÔò¾ßÓÐשºìÉ«ºÍºÚÉ«µÄÁ½ÖÖÑõ»¯Îï¡£
£¨1£©WÔªËØÔÚÔªËØÖÜÆÚ±íµÄλÖÃΪ                  £¬Óõç×Óʽ±íʾÆäÆø̬Ç⻯Îï                                                   ¡£
£¨2£©º¬ÓÐXµÄijÖÖÎïÖʳ£ÓÃ×÷¾»Ë®¼Á£¬ÓÃÀë×Ó·½³Ìʽ±íʾÆ侻ˮµÄÔ­ÀíÊÇ                       
                                                            ¡£
£¨3£©YµÄÒ»ÖÖÑõ»¯ÎÊôÓÚËáÐÔÑõ»¯ÎËü¼È¾ßÓÐÑõ»¯ÐÔ£¬Ò²¾ßÓл¹Ô­ÐÔ£¬»¹¾ßÓÐƯ°×ÐÔ¡£Éè¼ÆʵÑ飬ÑéÖ¤¸ÃÑõ»¯Îï¾ßÓл¹Ô­ÐÔ£¬Æä²Ù×÷·½·¨ÊÇ                                                     
                                                                                                                                                ¡£   
£¨4£©Ò»¶¨ÎïÖʵÄÁ¿µÄÏ¡ÏõËáÇ¡Äܽ«ZµÄשºìÉ«µÄÑõ»¯ÎïÑõ»¯£¬×ÔÉí±»»¹Ô­³ÉNO¡£Ôò·´Ó¦ÖÐÑõ»¯¼ÁÓ뻹ԭ¼ÁµÄÎïÖʵÄÁ¿µÄ±ÈΪ           ¡£

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

¾«Ó¢¼Ò½ÌÍø