ÌâÄ¿ÄÚÈÝ

(10·Ö)ÖÐѧ»¯Ñ§Öм¸ÖÖ³£¼ûÎïÖʵÄת»¯¹ØϵÈçÏ£º(Ìáʾ£ºAµ¥ÖÊÓпɱä¼Û̬ÇÒÓëÁò·´Ó¦Éú³ÉµÍ¼Û̬)

½«DÈÜÒºµÎÈë·ÐË®Öпɵõ½ÒÔFΪ·ÖÉ¢ÖʵĺìºÖÉ«½ºÌå¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺
(1)ºìºÖÉ«½ºÌåFÁ£×ÓÖ±¾¶´óСµÄ·¶Î§£º________¡£
(2)BµÄ»¯Ñ§Ê½£º________¡£
(3)д³ö£ºDµÄÈÜÒºÓ백ˮ·´Ó¦µÄÀë×Ó·½³Ìʽ£º
________________________________________________________________________£»
CµÄÈÜÒºÓëË«ÑõË®·´Ó¦µÄÀë×Ó·½³Ìʽ£º
________________________________________________________________________¡£
(4)д³ö¼ø¶¨EÖÐÑôÀë×ÓµÄʵÑé·½·¨ºÍÏÖÏó£º
________________________________________________________________________¡£
(1)1¡«100 nm
(2)FeS
(3)Fe3£«£«3NH3¡¤H2O===Fe(OH)3¡ý£«3NH¡¡2Fe2£«£«H2O2£«2H£«===2Fe3£«£«2H2O
(4)È¡ÉÙÁ¿EÈÜÒºÓÚÒ»ÊÔ¹ÜÖУ¬¼ÓÈëÊÊÁ¿Å¨NaOHÈÜÒº£¬¼ÓÈȲúÉúµÄÆøÌåÄÜʹʪÈóµÄºìɫʯÈïÊÔÖ½±äÀ¶É«£¬ÔòÖ¤Ã÷ÈÜÒºÖк¬NH?
ÓÉFΪFe(OH)3ÖªAΪFe£¬BΪFeS£¬CΪFeSO4£¬DΪFe2(SO4)3£¬EΪ(NH4)2SO4¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
(12·Ö)ÏÂÁÐÊÇÖÎÁÆ¿ÚÇ»Ñ×Ö¢µÄÒ©ÎﻪËØƬ(Î÷µØµâƬ)µÄ²¿·ÖʹÓÃ˵Ã÷£º
ʹÓÃ˵Ã÷Êé
Ö÷Òª³É·Ö
»îÐÔ·Ö×Óµâ(I2)£¬º¬Á¿1.5 mg/Ƭ
Öü²Ø
Õڹ⡢Ãܱա¢ÔÚÒõÁ¹´¦±£´æ
ÓÐЧÆÚ
¶þÄê
Çë¸ù¾ÝÉÏÊö˵Ã÷»Ø´ð£º
(1)ÍƶϻªËØƬ________(Ìî¡°ÊÇ¡±»ò¡°²»ÊÇ¡±)°×É«¡£
(2)ijͬѧÓû֤ʵ¸ÃҩƬÖÐȷʵº¬ÓзÖ×ӵ⣬Éè¼Æ²¢Íê³ÉÈçÏÂʵÑ飺
¢Ù½«Ò©Æ¬ÑÐËé¡¢Èܽ⡢¹ýÂË£¬ËùµÃÂËÒº·Ö×°Óڼס¢ÒÒÁ½ÊÔ¹ÜÖб¸Óá£
¢ÚÔÚ¼×ÊÔ¹ÜÖмÓÈëÏÂÁÐÒ»ÖÖÊÔ¼Á£¬Õñµ´¡¢¾²Ö㬹۲쵽ҺÌå·Ö²ã£¬ÈôÉϲãÒºÌåÑÕɫΪ________É«£¬ÔòËù¼ÓÊÔ¼ÁΪÏÂÁÐÖеÄ________£¬ÓÉ´Ë˵Ã÷´æÔÚ·Ö×ӵ⡣
A£®±½ 
B£®¾Æ¾«
C£®ËÄÂÈ»¯Ì¼ 
D£®ÑÎËá
¢ÛÔÚÒÒÊÔ¹ÜÖеμÓ________ÈÜÒº£¬ÈÜÒº±äÀ¶É«£¬Ò²ÄÜ˵Ã÷´æÔÚ·Ö×ӵ⡣
(3)µâÔڵؿÇÖÐÖ÷ÒªÒÔNaIO3µÄÐÎʽ´æÔÚ£¬ÔÚº£Ë®ÖÐÖ÷ÒªÒÔI£­µÄÐÎʽ´æÔÚ¡£¼¸ÖÖÁ£×ÓÖ®¼äÓÐÈçÏÂת»¯¹Øϵ£º

¢ÙÈçÒÔ;¾¶¢ñÖÆÈ¡I2£¬¼ÓµÄCl2ÒªÊÊÁ¿£¬Èç¹ýÁ¿¾Í»á·¢Éú;¾¶¢óµÄ¸±·´Ó¦£¬Ôڸø±·´Ó¦µÄ²úÎïÖУ¬IOÓëCl£­ÎïÖʵÄÁ¿Ö®±ÈΪ1¡Ã6£¬ÔòÑõ»¯¼ÁºÍ»¹Ô­¼ÁÎïÖʵÄÁ¿Ö®±ÈΪ________¡£
¢ÚÒÑÖªIOÔÚͨ³£Çé¿öÏÂÊDZȽÏÎȶ¨µÄ£¬¶øÔÚËáÐÔÈÜÒºÖÐÔò¾ßÓнÏÇ¿µÄÑõ»¯ÐÔ¡£ÏÂÁÐÀë×ÓÔÚËáÐÔÌõ¼þÏ¿ɱ»ÆäÑõ»¯µÄÓÐ________(¶àÑ¡µ¹¿Û·Ö)¡£
A£®Fe3£«  B£®F£­  C£®ClO  D£®I£­

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

¾«Ó¢¼Ò½ÌÍø