ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿Á×ÄÜÐγÉÖڶ൥ÖÊÓ뻯ºÏÎï¡£»Ø´ðÏÂÁÐÎÊÌ⣺

(1)Á×Ôڳɼüʱ£¬Äܽ«Ò»¸ö3sµç×Ó¼¤·¢½øÈë3dÄܼ¶¶ø²Î¼Ó³É¼ü£¬Ð´³ö¸Ã¼¤·¢Ì¬Ô­×ӵĺËÍâµç×ÓÅŲ¼Ê½__ ¡£

(2)ºÚÁ×ÊÇÒ»ÖÖ¶þά²ÄÁÏ£¬ÆäÖÐÒ»²ãµÄ½á¹¹Èçͼ1Ëùʾ¡£

¢ÙºÚÁ×ÖÐPÔ­×ÓµÄÔÓ»¯·½Ê½Îª _________ ¡£Ã¿Ò»²ãÄÚPÐγÉÁùÔª»·±Ë´ËÏà½Ó£¬Æ½¾ùÿ¸ö¿Õ¼äÁùÔª»·Öк¬ÓеÄÁ×Ô­×ÓÊÇ ____¸ö¡£

¢ÚÓÃ4-¼×Ñõ»ùÖصª±½ËÄ·úÅðËáÑΣ¨Èçͼ2£©´¦ÀíºÚÁ×ÄÉÃײÄÁÏ£¬¿ÉÒÔ±£»¤ºÍ¿ØÖÆÆäÐÔÖÊ¡£

¸ÃÑεĹ¹³ÉÔªËØÖÐC¡¢N¡¢O¡¢FµÄµç¸ºÐÔÓÉ´óµ½Ð¡Ë³ÐòΪ__£¬1mol¸ÃÑÎÑôÀë×Óº¬ÓеĦҼüµÄÊýĿΪ______ £¬¸ÃÑÎÒõÀë×ӵļ¸ºÎ¹¹ÐÍÊÇ__¡£

(3)Á×îÆ¿ó¿ÉÌáÈ¡Ï¡ÍÁÔªËØîÆ(Y)£¬Ä³Á×îÆ¿óµÄ½á¹¹ÈçÏ£º

¸ÃÁ×îÆ¿óµÄ»¯Ñ§Ê½Îª__£¬ÓëPO43¡ª»¥ÎªµÈµç×ÓÌåµÄÒõÀë×ÓÓÐ__ £¨Ð´³öÁ½ÖÖÀë×ӵĻ¯Ñ§Ê½£©¡£ÒÑÖª¾§°û²ÎÊýa= 0.69 nm£¬c=0.60 nm£¬°¢·ü¼ÓµÂÂÞ³£ÊýΪNA£¬Ôò¸ÃÁ×îÆ¿óµÄÃܶÈΪ__g.cm¡ª3£¨Áгö¼ÆËãʽ£©¡£

¡¾´ð°¸¡¿1s22s22p63s13p33d1 sp3 2 F>O>N>C 17NA ÕýËÄÃæÌå YPO4 SO42-¡¢ClO4-¡¢BrO4-¡¢IO4-¡¢SiO44-

¡¾½âÎö¡¿

(1)Á×Ϊ15ºÅÔªËØ£¬»ù̬PÔ­×ӵĺËÍâµç×ÓÅŲ¼Ê½Îª1s22s22p63s23p3£»Á×Ôڳɼüʱ£¬Äܽ«Ò»¸ö3sµç×Ó¼¤·¢½øÈë3dÄܼ¶¶ø²Î¼Ó³É¼ü£¬¸Ã¼¤·¢Ì¬Ô­×ӵĺËÍâµç×ÓÅŲ¼Ê½Îª1s22s22p63s13p33d1£»

(2)¢ÙºÚÁ×ÖÐÿ¸öPÔ­×ÓÓëÁíÍâÈý¸öPÔ­×ÓÐγɹ²¼Û¼üÇÒÓÐÒ»¶Ô¹Âµç×Ó¶Ô£¬PÔ­×ÓµÄÔÓ»¯·½Ê½Îªsp3£»Ã¿Ò»²ãÄÚPÐγÉÁùÔª»·±Ë´ËÏà½Ó£¬Æ½¾ùÿ¸ö¿Õ¼äÁùÔª»·Öк¬ÓеÄÁ×Ô­×ÓÊǸö£»

¢ÚÔªËØÔ­×ӵĵõç×ÓÄÜÁ¦Ô½Ç¿£¬Ôòµç¸ºÐÔÔ½´ó£¬Ô­×ӵõç×ÓÄÜÁ¦´óСΪ£ºF£¾O£¾N£¾C£¬Ôòµç¸ºÐÔ´óСΪ£ºF£¾O£¾N£¾C£»4-¼×Ñõ»ùÖصª±½ËÄ·úÅðËáÑÎÑôÀë×ÓÖк¬ÓÐÒ»¸öµªµªÈý¼ü¡¢Ò»¸öµªÌ¼¼ü¡¢±½»·ÉÏÓÐËĸö̼Çâ¼ü¡¢±½»·ÉÏÁù¸ö̼̼¼ü¡¢Á½¸ö̼Ñõ¼ü¡¢Èý¸ö̼Çâ¼ü£¬¹Ê1mol¸ÃÑÎÑôÀë×Óº¬ÓеĦҼüµÄÊýĿΪ17NA£»¸ÃÑÎÒõÀë×ÓBF4-£¬BÉϵŵç×Ó¶ÔÊýΪ=0£¬¼Û²ãµç×Ó¶ÔÊýΪ0+4=4£¬BΪsp3ÔÓ»¯£¬¹ÊBF4-µÄ¿Õ¼ä¹¹ÐÍΪÕýËÄÃæÌ壻

(3)¸ù¾Ý¾ù̯·¨ÇóË㣬¸ÃÁ×îÆ¿óÒ»¸ö¾§°ûº¬ÓиöY£¬¸öPO43-£¬¹Ê¸ÃÁ×îÆ¿óµÄ»¯Ñ§Ê½ÎªYPO4£¬µÈµç×ÓÌåÊÇÖ¸¼Ûµç×Ó×ÜÊýºÍÔ­×Ó×ÜÊý£¨ÇâµÈÇáÔ­×Ó²»¼ÆÔÚÄÚ£©ÏàͬµÄ·Ö×Ó¡¢Àë×Ó»òÔ­×ÓÍÅ£¬ÓëPO43¡ª»¥ÎªµÈµç×ÓÌåµÄÒõÀë×ÓÓÐSO42-¡¢ClO4-¡¢BrO4-¡¢IO4-¡¢SiO44-£»ÒÑÖª¾§°û²ÎÊýa= 0.69 nm£¬c=0.60 nm£¬°¢·ü¼ÓµÂÂÞ³£ÊýΪNA£¬Ôò¸ÃÁ×îÆ¿óµÄÃܶÈΪg.cm¡ª3¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿ÊµÑéÊÒÀï¿ÉÓÃÈçÏÂͼËùʾµÄ×°ÖÃÖÆÈ¡ÂÈËá¼Ø¡¢´ÎÂÈËáÄÆ£¬²¢ÑéÖ¤ÂÈË®µÄÐÔÖÊ¡£

ͼÖТÙΪÂÈÆø·¢Éú×°Ö㬢ڵÄÊÔ¹ÜÖÐÊ¢ÓеÄÈÜÒº£¬²¢ÖÃÓÚÈÈˮԡÖУ¬¢ÛµÄÊÔ¹ÜÖÐÊ¢ÓеÄÈÜÒº£¬²¢ÖÃÓÚ±ùˮԡÖУ¬¢ÜµÄÊÔ¹ÜÖмÓÓÐ×ÏɫʯÈïÊÔÒº£¬¢ÝΪβÆøÎüÊÕ×°Öá£

£¨1£©ÖÆÈ¡ÂÈÆøʱ£¬ÔÚÉÕÆ¿ÖÐÏȼÓÈëÒ»¶¨Á¿µÄ¶þÑõ»¯Ã̹ÌÌ壬ÔÙͨ¹ý_________£¨ÌîдÒÇÆ÷Ãû³Æ£©ÏòÉÕÆ¿ÖмÓÈëÊÊÁ¿µÄ____________£¨ÌîдÊÔ¼ÁÃû³Æ£©¡£

£¨2£©Îª³ýÈ¥ÂÈÆøÖлìÓеÄÂÈ»¯ÇâÆøÌ壬¿ÉÔڢٺ͢ÚÖ®¼ä°²×°Ê¢ÓÐ__________£¨Ñ¡Ìî×Öĸ±àºÅ£©µÄ¾»»¯×°Öá£

a.¼îʯ»Ò b.±¥ºÍʳÑÎË® c.ŨÁòËá d.±¥ºÍ̼ËáÇâÄÆÈÜÒº

£¨3£©±¾ÊµÑéÖÆÈ¡´ÎÂÈËáÄƵÄÀë×Ó·½³ÌʽÊÇ____________________¡£

£¨4£©±È½ÏÖÆÈ¡ÂÈËá¼ØºÍ´ÎÂÈËáÄƵÄÌõ¼þ£¬¿ÉÒÔ³õ²½µÃµ½µÄ½áÂÛÊÇ__¡£

£¨5£©·´Ó¦Íê±Ï¾­ÀäÈ´ºó£¬¢ÚµÄÊÔ¹ÜÖÐÓдóÁ¿¾§ÌåÎö³ö£¬Í¼ÖзûºÏ¸Ã¾§ÌåÈܽâ¶ÈËæζȱ仯¹æÂɵÄÇúÏßÊÇ___________£¨Ñ¡Ìî×Öĸ£©£»´Ó¢ÚµÄÊÔ¹ÜÖзÖÀë¸Ã¾§ÌåµÄ²Ù×÷ÊÇ___________£¨ÌîдʵÑé²Ù×÷Ãû³Æ£©¡£

£¨6£©ÊµÑéÖпɹ۲쵽¢ÜµÄÊÔ¹ÜÖÐÈÜÒºÑÕÉ«»á·¢Éú±ä»¯£º×î³õÈÜÒºÓÉ×ÏÉ«±äΪ_________£¬ËæºóÈÜÒºÖð½¥±äΪÎÞÉ«£¬ÊÇÒòΪ___________________¡£

¡¾ÌâÄ¿¡¿25 ¡æʱ£¬²¿·ÖÎïÖʵĵçÀë³£ÊýÈç±íËùʾ£º

»¯Ñ§Ê½

CH3COOH

H2CO3

µçÀë³£Êý

1.7¡Á10£­5

K1£½4.3¡Á10£­7

K2£½5.6¡Á10£­11

Çë»Ø´ðÏÂÁÐÎÊÌ⣺

(1)ÏàͬpHµÄCH3COONa¡¢NaHCO3¡¢Na2CO3Ũ¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòΪ__________

(2)³£ÎÂÏÂ0.1mol¡¤L£­1µÄCH3COOHÈÜÒºÔÚ¼ÓˮϡÊ͹ý³ÌÖУ¬ÏÂÁбí´ïʽµÄÊý¾ÝÒ»¶¨±äСµÄÊÇ________(Ìî×Öĸ£¬ÏÂͬ)¡£

A£®c(H£«) B£®c(H£«)/c(CH3COOH) C£®c(H£«)¡¤c(OH£­) D£®c(H£«)/c(OH£­)

(3)ÏÖÓÐ10mLpH=2µÄÑÎËáÈÜÒº£¬½øÐÐÒÔϲÙ×÷£º

a.ÓëpH=5µÄÑÎËáµÈÌå»ý»ìºÏ£¬ÔòpH=_____________£»

b.Ïò¸ÃÑÎËáÈÜÒºÖмÓÈë10mL0.02mol/LCH3COONaÈÜÒº£¬Ôò¸Ã»ìºÏÈÜÒºÖдæÔÚµÄÎïÁÏÊغãʽΪ__________________£»

c.Ïò¸ÃÑÎËáÈÜÒºÖмÓÈëµÈÌå»ýµÈŨ¶ÈµÄNa2CO3ÈÜÒº£¬Ôò»ìºÏÈÜÒºÖдæÔڵĵçºÉÊغãʽΪ_______________________£»

d. ÏÂÁйØÓÚÌå»ý¶¼Îª10mL£¬pH=2µÄA(ÑÎËá)ºÍB(CH3COOH)ÈÜҺ˵·¨ÕýÈ·µÄÊÇ_____ (ÌîдÐòºÅ)¡£

¢ÙÓëµÈÁ¿µÄп·´Ó¦¿ªÊ¼·´Ó¦Ê±µÄËÙÂÊA=B

¢ÚÓëµÈÁ¿µÄп·´Ó¦£¨Ð¿ÍêÈ«Èܽ⣬ûÓÐÊ£ÓࣩËùÐèÒªµÄʱ¼äA>B

¢Û¼ÓˮϡÊÍ100±¶£¬pH´óС±È½Ï£º4=A>B>2

¢ÜÎïÖʵÄÁ¿Å¨¶È´óС±È½Ï£ºA>B

¢Ý·Ö±ðÓë10mLpH=12µÄNaOHÈÜÒº³ä·Ö·´Ó¦ºóµÄÈÜÒºpH´óС±È½Ï£ºA<B

e. ÏÖÓñê×¼HClÈÜÒºµÎ¶¨°±Ë®£¬Ó¦Ñ¡ÓÃ________ָʾ¼Á£¬ÏÂÁвÙ×÷»áµ¼Ö²ⶨ½á¹ûÆ«¸ßµÄÊÇ___¡£

A£®Î´ÓÃHCl±ê×¼ÈÜÒºÈóÏ´µÎ¶¨¹Ü

B£®µÎ¶¨Ç°×¶ÐÎÆ¿ÄÚÓÐÉÙÁ¿Ë®

C£®µÎ¶¨Ç°µÎ¶¨¹Ü¼â×첿·ÖÓÐÆøÅÝ£¬µÎ¶¨ºóÆøÅÝÏûʧ

D£®¹Û²ì¶ÁÊýʱ£¬µÎ¶¨Ç°ÑöÊÓ£¬µÎ¶¨ºó¸©ÊÓ

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

¾«Ó¢¼Ò½ÌÍø