ÌâÄ¿ÄÚÈÝ

¾§Ìå¾ßÓйæÔòµÄ¼¸ºÎÍâÐΣ¬¾§ÌåÖÐ×î»ù±¾µÄÖØ¸´µ¥Ôª³ÆÎª¾§°û¡£NaCl¾§Ìå½á¹¹ÈçÏÂͼËùʾ¡£ÒÑÖªFexO¾§Ìå¾§°û½á¹¹ÎªNaClÐÍ£¬ÓÉÓÚ¾§ÌåȱÏÝ£¬xֵСÓÚ1¡£²âÖªFexO¾§ÌåÃܶÈΪ¦Ñ=5.71 g¡¤cm-3£¬¾§°û±ß³¤Îª4.28¡Á10-10 m¡£

£¨1£©FexOÖÐxÖµ(¾«È·ÖÁ0.01)Ϊ____________¡£

£¨2£©¾§ÌåÖеÄFen+·Ö±ðΪFe2+¡¢Fe3+,ÔÚFe2+ºÍFe3+µÄ×ÜÊýÖУ¬Fe2+ËùÕ¼·ÖÊý£¨ÓÃСÊý±íʾ£¬¾«È·ÖÁ0.001£©Îª____________________¡£

£¨3£©´Ë¾§Ì廯ѧʽΪ______________¡£

£¨4£©Óëij¸öFe2+(»òFe3+)¾àÀë×î½üÇҵȾàÀëµÄO2-Χ³ÉµÄ¿Õ¼ä¼¸ºÎÐÎ×´ÊÇ___________¡£

£¨5£©ÔÚ¾§ÌåÖУ¬ÌúÔªËØµÄÀë×Ó¼ä×î¶Ì¾àÀëΪ____________m¡£

½âÎö£º¸ù¾ÝNaCl¾§Ìå½á¹¹£¬1¸öNaCl¾§°ûÊÇÓÉ8¸öСÁ¢·½Ì塲ÈçÏÂͼ(1)¡³¹¹³ÉµÄ¡£Ã¿¸öСÁ¢·½Ì庬12¸öNaCl£¬Í¬ÀíÔÚFexO¾§ÌåÖУ¬Ã¿¸öСÁ¢·½Ì庬12¸öFexO£¬ÍêÕû¾§ÌåÖС²ÈçÏÂͼ(2)¡³£¬Fe¡¢O½»Ìæ³öÏÖ£¬x=1¡£

£¨1£©Ã¿¸öСÁ¢·½ÌåµÄÖÊÁ¿Îª£ºm=¦ÑV

=5.71 g¡¤cm-3¡Á()3

=5.60¡Á10-23 g¡£

M£¨FexO£©=2m¡¤NA=2¡Á5.60¡Á10-23 g¡Á6.02¡Á1023mol-1=67.4 g¡¤mol-1¡£

56.0 g¡¤mol-1¡Áx+16.0 g¡¤mol-1¡Á1=67.4 g¡¤mol-1

x=0.92¡£

£¨2£©Éè1 mol Fe0.92OÖУ¬N(Fe2+)=y mol,

Ôò£ºN(Fe3+)=(0.92-y)mol¡£¸ù¾Ý»¯ºÏÎïÖи÷ÔªËØÕý¸º»¯ºÏ¼Û´úÊýºÍΪÁãµÄÔ­Ôò£º2¡Áy mol+3¡Á(0.92-y mol)+(-2)¡Á1 mol=0,y=0.76¡£

¹ÊFe2+ËùÕ¼·ÖÊýΪ0.76/0.92=0.826¡£

£¨3£©ÓÉÓÚFe2+Ϊ0.76,ÔòFe3+Ϊ0.92-0.76=0.16£¬¹Ê»¯Ñ§Ê½Îª£º¡£

£¨4£©FeÔÚ¾§ÌåÖÐËùÕ¼¿Õ϶µÄ¼¸ºÎÐÎ״ΪÕý°ËÃæÌå¡£Èçͼ(3)Ëùʾ¡£

ͼ(3)

£¨5£©Èçͼ(1)£¬ÌúÔªËØÀë×Ó¼äµÄ¾àÀë

r==1.41¡Á4.28¡Á10-10 m¡Á=3.02¡Á10-10 m¡£

´ð°¸£º£¨1£©0.92

£¨2£©0.826¡¡

£¨3£©

£¨4£©Õý°ËÃæÌå

£¨5£©3.02¡Á10-10


Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø