ÌâÄ¿ÄÚÈÝ
9£®Åð¼°Æ仯ºÏÎïÔÚÏÖ´ú¹¤Òµ¡¢Éú»îºÍ¹ú·ÀÖÐÓÐÖØÒªÓ¦ÓüÛÖµ£®£¨1£©ÅðÔ×ӵĵç×ÓÅŲ¼Ê½ÊÇ1s22s22p1£®
£¨2£©×î¼òµ¥µÄÅðÍéÊÇB2H6£¨ÒÒÅðÍ飩£¬½á¹¹¼ûͼ£¬ÆäÖÐBÔ×ÓµÄÔÓ»¯·½Ê½Îªsp3ÔÓ»¯£®
£¨3£©BF3ºÍBCl3¶¼ÓÐÇ¿ÁÒ½ÓÊܹµç×Ó¶ÔµÄÇãÏò£¬ÈçÈý·ú»¯ÅðÆøÌåÓë°±ÆøÏàÓö£¬Á¢¼´Éú³É°×É«¹ÌÌ壬д³ö¸Ã°×É«¹ÌÌå½á¹¹Ê½£¬²¢±ê×¢³öÆäÖеÄÅäλ¼ü£®
£¨4£©½üÄêÀ´£¬ÈËÃǿ϶¨ÅðÊÇÈ˺Ͷ¯Îï·úÖж¾µÄÖØÒª½â¶¾¼Á£®ÅðÔÚÌåÄÚ¿ÉÓë·úÐγÉÎȶ¨µÄÅäºÏÎïBF4-
¡¢£¬²¢ÒԺͷúÏàͬµÄ;¾¶²Î¼ÓÌåÄÚ´úл£¬µ«¶¾ÐԱȷúС£¬ÇÒÒ×ËæÄòÅųö£¬¹ÊÈÏΪÅð¶Ô·ú»¯Îï¾ßÓнⶾ×÷Óã®
£¨5£©¾½á¹¹Ñо¿Ö¤Ã÷£¬ÅðËᾧÌåÖÐB£¨OH£©3µ¥Ôª½á¹¹Èçͼ£¨1£©Ëùʾ£®¸÷µ¥ÔªÖеÄÑõÔ×Óͨ¹ýO-H¡OÇâ¼üÁ¬½á³É²ã×´½á¹¹Èçͼ£¨2£©Ëùʾ£®²ãÓë²ãÖ®¼äÒÔ΢ÈõµÄ·Ö×Ó¼äÁ¦Ïà½áºÏ¹¹³ÉÕû¸öÅðËᾧÌ壮
¢ÙH3BO3ÊÇÒ»ÔªÈõËᣬд³öËüÓëË®·´Ó¦µÄ»¯Ñ§·½³ÌʽH3BO3+H2O?[B£¨OH£©4]-+H+£¬
¢Ú¸ù¾Ý½á¹¹ÅжÏÏÂÁÐ˵·¨ÕýÈ·µÄÊÇbcdg£®
a£®ÅðËᾧÌåÊôÓÚÔ×Ó¾§Ìå b£®ÅðËᾧÌåÓÐÁÛƬ״µÄÍâ²ã
c£®ÅðËᾧÌåÊÇ·Ö×Ó¾§Ìå d£®ÅðËᾧÌåÓл¬Äå¸Ð£¬¿É×÷È󻬼Á
e£®ÔÚB£¨OH£©3µ¥ÔªÖУ¬BÔ×ÓÒÔsp3ÔÓ»¯¹ìµÀºÍÑõÔ×Ó½áºÏ¶ø³É
f£®H3BO3·Ö×ÓµÄÎȶ¨ÐÔÓëÇâ¼üÓйØ
g£®º¬1mol H3BO3µÄ¾§ÌåÖÐÓÐ3molÇâ¼ü
h£®·Ö×ÓÖÐÅðÔ×Ó×îÍâ²ãΪ8e-Îȶ¨½á¹¹
¢ÛÅðËá³£ÎÂÏÂΪ°×ɫƬ״¾§Ì壬ÈÜÓÚË®£¨273KʱÈܽâ¶ÈΪ6.35£©£¬ÔÚÈÈË®ÖÐÈܽâ¶ÈÃ÷ÏÔÔö´ó£¨373KʱΪ27.6£©£®Çë·ÖÎöÆäÖÐÔÒòÊÜÈÈʱÅðËᾧÌåÖеĴóÁ¿Çâ¼üÓв¿·Ö¶ÏÁÑËùÖ£¬¾§ÌåÄÚÇâ¼ü²»ÀûÓÚÎïÖÊÈܽ⣮
·ÖÎö £¨1£©¸ù¾ÝÔªËØÃû³Æ£¬ÅжÏÔªËØÔ×ӵĺËÍâµç×ÓÊý£¬ÔÙ¸ù¾ÝºËÍâµç×ÓÅŲ¼¹æÂÉÀ´Ð´£¬ÅðÔªËØΪ5ºÅÔªËØ£»
£¨2£©¸ù¾Ý¼Û²ãµç×Ó¶Ô»¥³âÀíÂÛÈ·¶¨ÆäÔÓ»¯ÀàÐÍ£»
£¨3£©ÅðÔªËؾßÓÐȱµç×ÓÐÔ£¬Æ仯ºÏÎï¿ÉÓë¾ßÓйµç×ӶԵķÖ×Ó»òÀë×ÓÐγÉÅäºÏÎBF3ÄÜÓëNH3·´Ó¦Éú³ÉBF3•NH3£¬NÔ×Óº¬Óйµç×Ó¶Ô£»
£¨4£©ÅäºÏÎïÖÐÅäλ¼üԽǿ£¬ÅäºÏÎïÔ½Îȶ¨£¬·úΪ×î»îÆõķǽðÊô£¬ÅðÔÚÌåÄÚ¿ÉÓë·úÐγÉÎȶ¨µÄÅäºÏÎï·úÅðËá¸ùÀë×Ó£»
£¨5£©¢ÙÅðËáΪһԪÈõËᣬÔÚË®ÈÜÒºÀïºÍË®½áºÏÐγÉÅäλ¼ü£¬²¿·ÖµçÀë³öÒõÑôÀë×Ó£»
¢Úa£®ÅðËᾧÌåÖдæÔÚH3BO3·Ö×Ó£¬¸ù¾Ý¾§ÌåÖдæÔÚµÄ΢Á£È·¶¨¾§ÌåÀàÐÍ£»
b£®ÅðËᣨH3BO3£©ÊÇÒ»ÖÖƬ²ã×´½á¹¹°×É«¾§Ì壬ÓëʯīÏàËƵIJã×´½á¹¹£»
c£®ÅðËᾧÌåÖдæÔÚH3BO3·Ö×Ó£¬ÅðËᾧÌåÊÇ·Ö×Ó¾§Ì壻
d£®¸ù¾ÝÅðËᾧÌåΪƬ²ã×´½á¹¹·ÖÎö£»
e£®¸ù¾ÝBÔ×ӵļ۲ãµç×Ó¶ÔÅжÏÆäÔÓ»¯ÀàÐÍ£»
f£®·Ö×ÓµÄÎȶ¨ÐÔÓ뻯ѧ¼üÓйأ»
g£®ÀûÓþù̯·¨¼ÆË㺬1molH3BO3µÄ¾§ÌåÖеÄÇâ¼ü£¬º¬1mol H3BO3µÄ¾§ÌåÖÐÓÐ3molÇâ¼ü£»
h£®Óɽṹ¿ÉÖª£¬ÅðÔ×Ó×îÍâ²ãÖ»ÓÐ3¸öµç×Ó£¬ÓëÑõÔ×ÓÐγÉ3¶Ô¹²Óõç×Ó¶Ô£»
¢ÛÅðËᣨH3BO3£©ÊÇÒ»ÖÖƬ²ã×´½á¹¹£¬¼ÓÈÈÆÆ»µÁËÅðËᾧÌåÄÚ·Ö×ÓÖ®¼äµÄÇâ¼ü£»
½â´ð ½â£º£¨1£©BÔªËØΪ5ºÅÔªËØ£¬Ô×ÓºËÍâÓÐ5¸öµç×Ó£¬·ÖÁ½²ãÅŲ¼£¬µÚÒ»²ã2¸ö£¬µÚ¶þ²ã3¸ö£¬ËùÒÔºËÍâµç×ÓÅŲ¼Ê½Îª£º1s22s22p1£¬
¹Ê´ð°¸Îª£º1s22s22p1£»
£¨2£©ÒÒÅðÍé·Ö×ÓÖÐÿ¸öÅðÔ×Óº¬ÓÐ4¸ö¹²¼Û¼ü£¬ËùÒÔBÔ×Ó²ÉÓÃsp3ÔÓ»¯£¬
¹Ê´ð°¸Îª£ºsp3ÔÓ»¯£»
£¨3£©ÅðÔªËؾßÓÐȱµç×ÓÐÔ£¬Æ仯ºÏÎï¿ÉÓë¾ßÓйµç×ӶԵķÖ×Ó»òÀë×ÓÐγÉÅäºÏÎBF3ÄÜÓëNH3·´Ó¦Éú³ÉBF3•NH3£¬BÓëNÖ®¼äÐγÉÅäλ¼ü£¬NÔ×Óº¬Óйµç×Ó¶Ô£¬ËùÒÔµªÔ×ÓÌṩ¹Âµç×Ó¶Ô£¬BF3•NH3½á¹¹Ê½Îª£º£¬
¹Ê´ð°¸Îª£ºBF3•NH3£»£»
£¨4£©È±µç×Ó»¯ºÏÎï¾ßÓкÜÇ¿µÄ½ÓÊܵç×ÓµÄÄÜÁ¦£¬ÅðÓë·úÔªËØÐγɵÄBF3Ϊȱµç×Ó»¯ºÏÎïÓë·úÀë×ÓÐγÉBF4-£¬ÅðÔªËؾßÓÐȱµç×ÓÐÔ£¬ËùÒÔÅðÔÚÌåÄÚ¿ÉÓë·úÐγÉÎȶ¨µÄÅäºÏÎïBF4-£¬
¹Ê´ð°¸Îª£ºBF4-£»
£¨5£©¢ÙÅðËáΪһԪÈõËᣬÔÚË®ÈÜÒºÀºÍË®µçÀë³öµÄÇâÑõ¸ùÀë×ÓÐγÉÅäλ¼ü£¬ÆäµçÀë·½³ÌʽΪ£ºH3BO3+H2O?[B£¨OH£©4]-+H+£¬
¹Ê´ð°¸Îª£ºH3BO3+H2O?[B£¨OH£©4]-+H+£»
¢Úa£®ÅðËᾧÌåÖдæÔÚH3BO3·Ö×Ó£¬ÇҸþ§ÌåÖдæÔÚÇâ¼ü£¬ËµÃ÷ÅðËáÓÉ·Ö×Ó¹¹³É£¬ÊÇ·Ö×Ó¾§Ì壬Ô×Ó¾§ÌåÄÚÖ»Óй²¼Û¼ü£¬¹Êa´íÎó£»
b£®ÅðËᣨH3BO3£©ÊÇÒ»ÖÖƬ²ã×´½á¹¹°×É«¾§Ì壬²ãÄڵġ°H3BO3¡±Î¢Á£Ö®¼äͨ¹ýÇâ¼üÏàÁ¬£¬¹ÊbÕýÈ·£»
c£®ÅðËᾧÌåÖдæÔÚH3BO3·Ö×Ó£¬ÅðËᾧÌåÊÇ·Ö×Ó¾§Ì壬¹ÊcÕýÈ·£»
d£®ÅðËᣨH3BO3£©ÊÇÒ»ÖÖƬ²ã×´½á¹¹°×É«¾§Ì壬Ƭ²ã×´½á¹¹¾§ÌåÓл¬Äå¸Ð£¬¿É×÷È󻬼Á£¬¹ÊdÕýÈ·£»
e£®ÔÚB£¨OH£©3µ¥ÔªÖУ¬BÖ»ÐγÉÁË3¸öµ¥¼ü£¬Ã»Óйµç×Ó¶Ô£¬ËùÒÔ²ÉÈ¡sp2ÔÓ»¯£¬¹Êe´íÎó£»
f£®·Ö×ÓµÄÎȶ¨ÐÔÓë·Ö×ÓÄÚµÄB-O¡¢H-O¹²¼Û¼üÓйأ¬È۷еãÓëÇâ¼üÓйأ¬¹Êf´íÎó£»
g£®1¸öÅðËá·Ö×ÓÐγÉÁË6¸öÇâ¼ü£¬µ«Ã¿¸öÇâ¼üÊÇ2¸öÅðËá·Ö×Ó¹²Óõģ¬ËùÒÔƽ¾ùº¬3¸öÇâ¼ü£¬Ôòº¬ÓÐ1molH3BO3µÄ¾§ÌåÖÐÓÐ3molÇâ¼ü£¬¹ÊgÕýÈ·£»
h£®ÅðÔ×Ó×îÍâ²ãÖ»ÓÐ3¸öµç×Ó£¬ÓëÑõÔ×ÓÐγÉ3¶Ô¹²Óõç×Ó¶Ô£¬Òò´ËBÔ×Ó²»ÊÇ8e-Îȶ¨½á¹¹£¬¹Êh´íÎó£»
¹Ê´ð°¸Îª£ºbcdg£»
¢Û¾§ÌåÄÚÇâ¼ü²»ÀûÓÚÎïÖÊÈܽ⣬¼ÓÈÈÆÆ»µÁËÅðËᾧÌåÄÚ·Ö×ÓÖ®¼äµÄÇâ¼ü£¬ËùÒÔ¼ÓÈÈʱ£¬ÅðËáµÄÈܽâ¶ÈÔö´ó£¬
¹Ê´ð°¸Îª£ºÊÜÈÈʱÅðËᾧÌåÖеĴóÁ¿Çâ¼üÓв¿·Ö¶ÏÁÑËùÖ£¬¾§ÌåÄÚÇâ¼ü²»ÀûÓÚÎïÖÊÈܽ⣮
µãÆÀ ±¾Ì⿼²éÁËÓйØÅðµÄ֪ʶ£¬²àÖØ¿¼²éÔ×ÓºËÍâµç×ÓÅŲ¼¡¢ÔÓ»¯ÀíÂÛµÄÓ¦Óá¢ÅðËᾧÌ徧ÌåÀàÐ͵ÄÅжϡ¢Ó°Ïì·Ö×ÓÎȶ¨ÐÔµÄÒòËصÈ֪ʶµã£¬ÌâÄ¿ÄѶÈÖеȣ¬×¢Òâ·Ö×ÓµÄÎȶ¨ÐÔÓ뻯ѧ¼üÓйأ¬ÎïÖʵÄÈ۷еãÓëÇâ¼üÓйأ¬×¢ÒâÅðËᣨH3BO3£©ÊÇÒ»ÖÖƬ²ã×´½á¹¹°×É«¾§Ì壮
A£® | pHÏàͬµÄ¢ÙCH3COONa¡¢¢ÚNaHCO3¡¢¢ÛC6H5ONaÈý·ÝÈÜÒºÖеÄc£¨Na+£©£º¢Û£¾¢Ú£¾¢Ù | |
B£® | ½«ÎïÖʵÄÁ¿Å¨¶È¾ùΪ0.1mol•L-1µÄNa2CO3ÈÜÒº¡¢NaHCO3ÈÜÒºµÈÌå»ý»ìºÏËùµÃÈÜÒºÖУº2c£¨OH-£©-2c£¨H+£©=3c£¨H2CO3£©+c£¨HCO3-£©-c£¨CO32-£© | |
C£® | µÈŨ¶È¡¢µÈÌå»ýµÄNa2CO3ºÍNaHCO3»ìºÏ£º$\frac{c£¨HC{O}_{3}^{-}£©}{c£¨{H}_{2}C{O}_{3}£©}$£¼$\frac{c£¨C{O}_{3}^{2-}£©}{c£¨HC{O}_{3}^{-}£©}$ | |
D£® | ½«×ãÁ¿AgCl·Ö±ð·ÅÈ룺¢Ù5mLË® ¢Ú10mL 0.2mol/L MgCl2 ¢Û20mL 0.3mol/LÑÎËá ÖÐÈܽâÖÁ±¥ºÍc£¨Ag+£©£º¢Ù£¾¢Ú£¾¢Û |
A£® | ÍÑÁò·´Ó¦¡÷H£¼0 | |
B£® | n£¨NH3£©/n£¨SO2£©£ºa£¼b£¼c | |
C£® | ÏàͬÌõ¼þÏ£¬´ß»¯¼Á»îÐÔÔ½´ó£¬SO2µÄƽºâת»¯ÂÊÔ½¸ß | |
D£® | ¼°Ê±´ÓÌåϵÖгýȥˮ£¬Æ½ºâ³£ÊýÔö´ó |
A£® | AlCl3ÈÜÒºÓëÉÕ¼îÈÜÒº·´Ó¦£¬µ±n£¨OH-£©£ºn£¨Al3+£©=7£º2ʱ£¬2Al3++7OH-¨TAl£¨OH£©3¡ý+AlO2-+2H2O | |
B£® | Cl2ÓëFeBr2ÈÜÒº·´Ó¦£¬µ±n£¨Cl2£©£ºn£¨FeBr2£©=1£º1ʱ£¬2Fe2++4Br-+3Cl2¨T2Fe3++2Br2+6Cl- | |
C£® | CuCl2ÈÜÒºÓëNaHSÈÜÒº·´Ó¦£¬µ±n£¨CuCl2£©£ºn£¨NaHS£©=1£º2ʱ£¬Cu2++2HS-¨TCuS¡ý+H2S¡ü | |
D£® | FeÓëÏ¡ÏõËá·´Ó¦£¬µ±n£¨Fe£©£ºn£¨HNO3£©=1£º2ʱ£¬3Fe+2NO3-+8H+¨T3Fe2++2NO¡ü+4H2O |
A£® | ŨÁòËá | B£® | Ï¡ÏõËá | C£® | ŨÏõËá | D£® | Ï¡ÁòËá |