ÌâÄ¿ÄÚÈÝ

ÀàÍÆ·¨ÊÇ¿ÆÑ§Ñ§Ï°µÄÖØÒª·½·¨Ö®Ò»¡£ÔÚѧϰÁ˱×åÔªËØµÄ¸÷ÖÖÐÔÖʺó£¬ÈçϱíËùʾ£¬±í¸ñÌṩµÄÊÇÑõ×åÔªËØµÄ²¿·ÖÐÔÖÊ¡£Çë½áºÏÔªËØÖÜÆÚÂÉÍê³ÉÏÂÁÐÎÊÌ⣺

ÔªËØ8O16S34Se52Te
µ¥ÖÊÈÛµã(¡æ)-218.4113 450
µ¥Öʷеã(¡æ)-183444.66851390
Ö÷Òª»¯ºÏ¼Û-2-2£¬+4£¬+6-2£¬+4£¬+6 
Ô­×Ó°ë¾¶Öð½¥Ôö´ó
µ¥ÖÊÓëH2·´Ó¦Çé¿öµãȼʱÒ×»¯ºÏ¼ÓÈÈ»¯ºÏ¼ÓÈÈÄÑ»¯ºÏ²»ÄÜÖ±½Ó»¯ºÏ

(1)ÎøµÄÈ۵㷶Χ¿ÉÄÜÊÇ________¡£
(2)íڵϝºÏ¼Û¿ÉÄÜÓÐ________¡£
(3)ÇâÎøËáÓнÏÇ¿µÄ»¹Ô­ÐÔ¡£Òò´Ë·ÅÔÚ¿ÕÆøÖг¤ÆÚ±£´æÒ×±äÖÊ£¬Æä¿ÉÄÜ·¢ÉúµÄ»¯Ñ§·½³ÌʽΪ£º________¡£
(4)¹¤ÒµÉÏAl2Te3¿ÉÓÃÀ´ÖƱ¸H2Te£¬Íê³ÉÏÂÁл¯Ñ§·½³Ìʽ£ºAl2Te3+________¡úAl(OH)3¡ý+  H2Te¡ü

113¡«450¡æ    £­2¡¢£«4¡¢£«6    2H2Se£«O2===2H2O£«2Se    Al2Te3£«6H2O===2Al(OH)3¡ý£«3H2Te¡ü
ѧ»á·ÖÎöÊý¾Ý£»
£¨1£©¹Û²ì¡°µ¥ÖÊÈ۵㡱һÐÐÊý¾Ý£ºÓÉ×óµ½ÓÒ£¬Öð½¥Éý¸ß£¬¹ÊÎøµÄÈ۵㷶Χ¿ÉÄÜÊÇ113¡«450¡æ£»
£¨2£©¹Û²ì¡°Ö÷Òª»¯ºÏ¼Û¡±Ò»ÐÐÊý¾Ý£¬ÇÒͬÖ÷×åÔªËØ£¬ÓÉÉϵ½Ï£¬Ê§µç×ÓÄÜÁ¦Öð½¥ÔöÇ¿£¬¹Ê»áÖð½¥±íÏÖ³öÕý¼Û£»
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø