£¨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¬ÌúÔªËصÄÀë×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
Ä꼶 | ¸ßÖÐ¿Î³Ì | Ä꼶 | ³õÖÐ¿Î³Ì |
¸ßÒ» | ¸ßÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ | ³õÒ» | ³õÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ |
¸ß¶þ | ¸ß¶þÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õ¶þ | ³õ¶þÃâ·Ñ¿Î³ÌÍƼö£¡ |
¸ßÈý | ¸ßÈýÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õÈý | ³õÈýÃâ·Ñ¿Î³ÌÍƼö£¡ |
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º¼ªÁÖÊ¡³¤´ºÍâ¹úÓïѧУ2011£2012ѧÄê¸ß¶þÏÂѧÆÚÆÚÄ©¿¼ÊÔ»¯Ñ§ÊÔÌâ ÌâÐÍ£º022
(1)ÈÛµã·ÐµãHF________HI£»ÔÒò£º________£®
(2)¸÷ÖÖº¬ÑõËáHClO¡¢HClO3¡¢H2SO3¡¢HClO4µÄËáÐÔÓÉÇ¿µ½ÈõÅÅÁÐΪ________£®
(3)ÏÂÁÐ4ÖÖÎïÖÊÈÛµã·ÐµãÓɸߵ½µÍÅÅÁÐΪ________(ÌîÐòºÅ)£®
¢Ù½ð¸Õʯ(C£C)¢ÚÕà(Ge£Ge)
¢Û¾§Ìå¹è(Si£Si)¢Ü½ð¸ÕÉ°(Si£C)
(4)¾§Ìå¾ßÓйæÔòµÄ¼¸ºÎÍâÐΣ¬¾§ÌåÖÐ×î»ù±¾µÄÖظ´µ¥Ôª³Æ֮Ϊ¾§°û£®NaCl¾§Ìå½á¹¹ÈçÏÂͼËùʾ£®¾§ÌåÖÐÿ¸öNa£«Í¬Ê±ÎüÒý×Å________¸öCl££¬ÓÉÕâЩCl£¹¹³ÉµÄ¼¸ºÎ¹¹ÐÍΪ________¾§ÌåÖÐÔÚÿ¸öCl£ÖÜΧÓëËü×î½Ó½üÇÒ¾àÀëÏàµÈµÄCl£¹²ÓÐ________¸ö£®
(5)ijԪËصļ¤·¢Ì¬Ô×ӵĵç×ÓÅŲ¼Ê½Îª1s22s22p63s23p34s1£¬Ôò¸ÃÔªËØ»ù̬Ô×ӵĵç×ÓÅŲ¼Ê½Îª________£»Æä×î¸ß¼ÛÑõ»¯Îï¶ÔÓ¦µÄË®»¯ÎïµÄ»¯Ñ§Ê½ÊÇ________£®
(6)ijԪËØÔ×ӵļ۵ç×Ó¹¹ÐÍΪ3d104S2£¬ËüÊôÓÚµÚ________ÖÜÆÚ£¬ÊÇ________×壬________ÇøÔªËØ£¬ÔªËØ·ûºÅÊÇ________£®¸ÃÔ×ÓºËÍâÓÐ________¸öÄܼ¶£¬µç×ÓÔƵÄÐÎ×´ÓÐ________ÖÖ£®
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2011-2012ѧÄ꼪ÁÖÊ¡³¤´ºÍâ¹úÓïѧУ¸ß¶þÏÂѧÆÚÆÚÄ©¿¼ÊÔ»¯Ñ§ÊÔ¾í£¨´ø½âÎö£© ÌâÐÍ£ºÌî¿ÕÌâ
£¨1£©¡¢ÈÛµã·ÐµãHF HI£»ÔÒò£º
£¨2£©¡¢¸÷ÖÖº¬ÑõËáHClO¡¢HClO3¡¢H2SO3¡¢HClO4µÄËáÐÔÓÉÇ¿µ½ÈõÅÅÁÐΪ
£¨3£©¡¢ÏÂÁÐ4ÖÖÎïÖÊÈÛµã·ÐµãÓɸߵ½µÍÅÅÁÐΪ________ £¨ÌîÐòºÅ£©
¢Ù½ð¸Õʯ£¨C¡ªC£©¢ÚÕࣨGe¡ªGe£©
¢Û¾§Ìå¹è£¨Si¡ªSi£©¢Ü½ð¸ÕÉ°£¨Si¡ªC£©
£¨4£©¾§Ìå¾ßÓйæÔòµÄ¼¸ºÎÍâÐΣ¬¾§ÌåÖÐ×î»ù±¾µÄÖظ´µ¥Ôª³Æ֮Ϊ¾§°û¡£NaCl¾§Ìå½á¹¹ÈçͼËùʾ¡£
¾§ÌåÖÐÿ¸öNa+ͬʱÎüÒý×Å______¸öCl-£¬ÓÉÕâЩ Cl-¹¹³ÉµÄ¼¸ºÎ¹¹ÐÍΪ ¾§ÌåÖÐÔÚÿ¸öCl-ÖÜΧÓëËü×î½Ó½üÇÒ¾àÀëÏàµÈµÄCl-¹²ÓÐ________¸ö¡£
£¨5£©Ä³ÔªËصļ¤·¢Ì¬Ô×ӵĵç×ÓÅŲ¼Ê½Îª1s2s2p3s3p4s£¬Ôò¸ÃÔªËØ»ù̬Ô×ӵĵç×ÓÅŲ¼Ê½Îª £»Æä×î¸ß¼ÛÑõ»¯Îï¶ÔÓ¦µÄË®»¯ÎïµÄ»¯Ñ§Ê½ÊÇ
£¨6£©Ä³ÔªËØÔ×ӵļ۵ç×Ó¹¹ÐÍΪ3d104S2£¬ËüÊôÓÚµÚ ÖÜÆÚ£¬ÊÇ ×壬 ÇøÔªËØ£¬ÔªËØ·ûºÅÊÇ ¡£¸ÃÔ×ÓºËÍâÓÐ ¸öÄܼ¶£¬µç×ÓÔƵÄÐÎ×´ÓÐ ÖÖ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2013½ì¼ªÁÖÊ¡¸ß¶þÏÂѧÆÚÆÚÄ©¿¼ÊÔ»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÌî¿ÕÌâ
£¨1£©¡¢ÈÛµã·ÐµãHF HI£»ÔÒò£º
£¨2£©¡¢¸÷ÖÖº¬ÑõËáHClO¡¢HClO3¡¢H2SO3¡¢HClO4µÄËáÐÔÓÉÇ¿µ½ÈõÅÅÁÐΪ
£¨3£©¡¢ÏÂÁÐ4ÖÖÎïÖÊÈÛµã·ÐµãÓɸߵ½µÍÅÅÁÐΪ________ £¨ÌîÐòºÅ£©
¢Ù½ð¸Õʯ£¨C¡ªC£©¢ÚÕࣨGe¡ªGe£©
¢Û¾§Ìå¹è£¨Si¡ªSi£©¢Ü½ð¸ÕÉ°£¨Si¡ªC£©
£¨4£©¾§Ìå¾ßÓйæÔòµÄ¼¸ºÎÍâÐΣ¬¾§ÌåÖÐ×î»ù±¾µÄÖظ´µ¥Ôª³Æ֮Ϊ¾§°û¡£NaCl¾§Ìå½á¹¹ÈçͼËùʾ¡£
¾§ÌåÖÐÿ¸öNa+ͬʱÎüÒý×Å______¸öCl-£¬ÓÉÕâЩ Cl-¹¹³ÉµÄ¼¸ºÎ¹¹ÐÍΪ ¾§ÌåÖÐÔÚÿ¸öCl-ÖÜΧÓëËü×î½Ó½üÇÒ¾àÀëÏà µÈµÄCl-¹²ÓÐ________¸ö¡£
£¨5£©Ä³ÔªËصļ¤·¢Ì¬Ô×ӵĵç×ÓÅŲ¼Ê½Îª1s2s2p3s3p4s£¬Ôò¸ÃÔªËØ»ù̬Ô×ӵĵç×ÓÅŲ¼Ê½Îª £»Æä×î¸ß¼ÛÑõ»¯Îï¶ÔÓ¦µÄË®»¯ÎïµÄ»¯Ñ§Ê½ÊÇ
£¨6£©Ä³ÔªËØÔ×ӵļ۵ç×Ó¹¹ÐÍΪ3d104S2£¬ËüÊôÓÚµÚ ÖÜÆÚ£¬ÊÇ ×壬 ÇøÔªËØ£¬ÔªËØ·ûºÅÊÇ ¡£¸ÃÔ×ÓºËÍâÓÐ ¸öÄܼ¶£¬µç×ÓÔƵÄÐÎ×´ÓÐ ÖÖ
²é¿´´ð°¸ºÍ½âÎö>>
°Ù¶ÈÖÂÐÅ - Á·Ï°²áÁбí - ÊÔÌâÁбí
ºþ±±Ê¡»¥ÁªÍøÎ¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨Æ½Ì¨ | ÍøÉÏÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | µçÐÅթƾٱ¨×¨Çø | ÉæÀúÊ·ÐéÎÞÖ÷ÒåÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | ÉæÆóÇÖȨ¾Ù±¨×¨Çø
Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com