ÌâÄ¿ÄÚÈÝ

ÏÖÓÐÏÂÁжÌÖÜÆÚÔªËØÐÔÖʵÄÊý¾Ý
»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÔªËØ¢ÛÔÚÖÜÆÚ±íÖÐλÖÃÊÇ__________________£»ÔªËØ¢ÜÓëÔªËØ¢ßÏà±È½Ï£¬ÆøÌ¬Ç⻯Îï½ÏÎȶ¨µÄÊÇ
____________ £¨Ìѧʽ£©£»
£¨2£©ÔªËØ¢ÙÊÇ____________£¨Ð´ÔªËØ·ûºÅ£©£¬ÔªËØ¢ÞÊÇ____________£¨Ð´ÔªËØ·ûºÅ£©£¬¶þÕß°´ÕÕÔ­×Ó¸öÊý±ÈΪ1¡Ã1ÐγɵϝºÏÎïÓëË®·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ_____________________________£»
£¨3£©ÔªËØ¢ÝÓëÔªËØ¢ÚµÄ·Ç½ðÊôÐÔÇ¿Èõ˳ÐòΪ____________£¨Ð´ÔªËØ·ûºÅ£©£¬ÔªËآݵĵ¥ÖʼÓÈëµ½ÔªËØ¢ÚµÄÇ⻯ÎïµÄË®ÈÜÒºÖУ¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ________________________________£»
£¨4£©Óõç×Óʽ±íÊ¾ÔªËØ¢ÙÇ⻯ÎïµÄÐγɹý³Ì________________________£»Ð´³öÔªËØ¢ßÇ⻯ÎïµÄµç×Óʽ
____________£»Ð´³öʵÑéÊÒÖÆÔªËØ¢Ýµ¥ÖʵĻ¯Ñ§·´Ó¦·½³Ìʽ_____________________________¡£
£¨1£©µÚ¶þÖÜÆÚIA×壻NH3
£¨2£©O£»Na£»2Na2O2+2H2O==4NaOH+O2¡ü
£¨3£©Cl£¾S£»Cl2+H2S=2HCl+S¡ý
£¨4£©¡°ÂÔ¡±
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
ÏÖÓÐÏÂÁжÌÖÜÆÚÔªËØÐÔÖʵÄÏà¹ØÊý¾Ý£º
¢Ù ¢Ú ¢Û ¢Ü ¢Ý ¢Þ
Ô­×Ó°ë¾¶£¨10-10m£© 1.30 0.82 0.99 1.11 0.90 1.18
×î¸ß»ò×îµÍ»¯ºÏ¼Û +2 +3 +7 +4 +2 +3
-1 -4
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©¢ÙºÍ¢ÛÐγɵϝºÏÎïÀàÐÍΪ
Àë×Ó
Àë×Ó
£¨Ìî¡°Àë×Ó¡±»ò¡°¹²¼Û¡±£©»¯ºÏÎ
£¨2£©ÔªËآٵĽðÊôÐÔ±ÈÔªËØ¢ÞÒª
Ç¿
Ç¿
£¨Ìî¡°Ç¿¡±»ò¡°Èõ¡±£©£¬ÊÔ´ÓÔ­×ӽṹ·½Ãæ½âÊÍÆäÔ­Òò£º
þµÄÔ­×Ó°ë¾¶´óÓÚÂÁ£¬Ô­×Ӻ˶Ô×îÍâ²ãµç×ÓµÄÎüÒýÁ¦±ÈÂÁÈõ£¬±ÈÂÁÔ­×Ó¸üÈÝÒ×ʧȥµç×Ó£¬ËùÒÔÃ¾ÔªËØµÄ½ðÊôÐÔ±ÈÂÁҪǿ
þµÄÔ­×Ó°ë¾¶´óÓÚÂÁ£¬Ô­×Ӻ˶Ô×îÍâ²ãµç×ÓµÄÎüÒýÁ¦±ÈÂÁÈõ£¬±ÈÂÁÔ­×Ó¸üÈÝÒ×ʧȥµç×Ó£¬ËùÒÔÃ¾ÔªËØµÄ½ðÊôÐÔ±ÈÂÁҪǿ
£»
£¨3£©ÔÚ×ÔÈ»½çÖУ¬ÔªËآܵĴæÔÚÐÎ̬Ϊ
»¯ºÏ̬
»¯ºÏ̬
£¬¹¤ÒµÉÏ´ÖÖÆ¸ÃÔªËØµ¥ÖʵĻ¯Ñ§·½³ÌʽΪ
SiO2+2C
 ¸ßΠ
.
 
Si+2CO¡ü
SiO2+2C
 ¸ßΠ
.
 
Si+2CO¡ü
£»
£¨4£©¢ÞµÄµ¥ÖÊÓëŨNaOHÈÜÒº·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ
2Al+2OH-+2H2O=2AlO2-+3H2¡ü
2Al+2OH-+2H2O=2AlO2-+3H2¡ü
£»
£¨5£©ÊµÑéÊÒÏÖÓÐÔªËØ¢ÙµÄµ¥ÖʺÍijδ֪½ðÊôµ¥ÖÊM£¬Çë¼òҪд³ö±È½ÏÁ½Õß½ðÊôÐÔÇ¿ÈõµÄÒ»ÖÖʵÑé·½°¸
ÔÚÁ½¸öСÉÕ±­Öзֱð¼ÓÈëÊÊÁ¿ÕôÁóË®£¨»òµÈŨ¶ÈµÄÑÎËᣩ£¬È»ºóͶÈëÐÎ×´´óСÏàͬµÄ½ðÊôƬ£¬ÈôM·´Ó¦±Èþ¾çÁÒ£¬ÔòMµÄ½ðÊôÐÔ±ÈMgÇ¿
ÔÚÁ½¸öСÉÕ±­Öзֱð¼ÓÈëÊÊÁ¿ÕôÁóË®£¨»òµÈŨ¶ÈµÄÑÎËᣩ£¬È»ºóͶÈëÐÎ×´´óСÏàͬµÄ½ðÊôƬ£¬ÈôM·´Ó¦±Èþ¾çÁÒ£¬ÔòMµÄ½ðÊôÐÔ±ÈMgÇ¿
£®
ÏÖÓÐÏÂÁжÌÖÜÆÚÔªËØÐÔÖʵÄÊý¾Ý£º
          ÔªËرàºÅ
ÔªËØÐÔÖÊ
¢Ù ¢Ú ¢Û ¢Ü ¢Ý ¢Þ ¢ß ¢à
Ô­×Ó°ë¾¶£¨10-10m£© 0.74 1.60 1.52 1.10 0.99 1.86 0.75 1.43
×î¸ß»ò×îµÍ»¯ºÏ¼Û +2 +1 +5 +7 +1 +5 +3
-2 -3 -1 -3
ÊԻشðÏÂÁÐÎÊÌ⣺
£¨1£©ÉÏÊöÔªËØÖд¦ÓÚµÚ VA×åµÄÓÐ
 
£¬´¦ÓÚµÚ¶þÖÜÆÚµÄÓÐ
 
£¨ÒÔÉϾùÓñàºÅ±íʾ£©£®
£¨2£©ÉÏÊöÔªËØÖнðÊôÐÔ×îÇ¿µÄÊÇ
 
£¨ÓñàºÅ±íʾ£©£®
£¨3£©Ð´³öÔªËØ¢Ù¡¢¢àÐγɵϝºÏÎïÓëKOHÈÜÒº·´Ó¦µÄ»¯Ñ§·½³Ìʽ
 
£®
£¨4£©¾­Ñé±íÃ÷£¬³ýÇâºÍº¤Í⣬µ±ÔªËØÔ­×ӵĵç×Ó²ãÊý£¨n£©¶àÓÚ»òµÈÓÚÔ­×Ó×îÍâ²ãµç×ÓÊý£¨m£©¼´£¨n-m£©¡Ý0ʱ£¬¸ÃÔªËØÊôÓÚ½ðÊôÔªËØ£»µ±ÔªËØÔ­×ÓµÄ×îÍâ²ãµç×ÓÊý£¨m£©¶àÓÚÔ­×ӵĵç×Ó²ãÊý£¨n£©¼´£¨n-m£©£¼0ʱ£¬¸ÃÔªËØÊôÓڷǽðÊôÔªËØ£®ÊԻشð£º
a£®µÚn£¨n¡Ý2£©ÖÜÆÚÓÐ
 
ÖַǽðÊôÔªËØ£¨Óú¬nµÄ´úÊýʽ±íʾ£¬º¬Áã×åÔªËØ£©£®
b£®¸ù¾ÝÖÜÆÚ±íÖÐÿ¸öÖÜÆÚ·Ç½ðÊôÔªËØµÄÖÖÊý£¬Ô¤²âÖÜÆÚ±íÖÐÓ¦ÓÐ
 
ÖַǽðÊôÔªËØ£¨º¬Áã×åÔªËØ£©£¬»¹ÓÐ
 
ÖÖδ·¢ÏÖ£¬Î´·¢ÏֵķǽðÊô´¦ÔÚµÚ
 
ÖÜÆÚµÚ
 
×壨ע£º½«Áã×å¿´×÷ VIIIA×壩£®

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

¾«Ó¢¼Ò½ÌÍø