(13·Ö) ÒÑÖªA¡¢B¡¢C¡¢DºÍE¶¼ÊÇÔªËØÖÜÆÚ±íÖÐǰ36ºÅµÄÔªËØ£¬ËüÃǵÄÔ­×ÓÐòÊýÒÀ´ÎÔö´ó¡£AÓëÆäËû4ÖÖÔªËØ¼È²»ÔÚͬһÖÜÆÚÓÖ²»ÔÚͬһÖ÷×å¡£BºÍCÊôͬһÖ÷×壬DºÍEÊôͬһÖÜÆÚ£¬ÓÖÖªEÊÇÖÜÆÚ±íÖÐ1£­18ÁÐÖеĵÚ7ÁÐÔªËØ¡£DµÄÔ­×ÓÐòÊý±ÈEС5£¬D¸úBÐγɵľ§ÌåÆä¾§°û½á¹¹Èçͼ£¬Í¼ÖÐСÇò´ú±íD£¬´óÇò´ú±íB¡£

Çë»Ø´ð£º
(1)AÔªËØµÄÃû³ÆÊÇ________________£»
(2)BÔªËØµÄ¹ìµÀ±íʾʽÊÇ________________£¬CµÄÔ­×ӽṹʾÒâͼÊÇ________________£¬BÓëAÐγɵϝºÏÎï±ÈCÓëAÐγɵϝºÏÎï·Ðµã¸ß£¬ÆäÔ­ÒòÊÇ__________________________________________________________£»
(3)EÊôÓÚÔªËØÖÜÆÚ±íÖеÚ________ÖÜÆÚ£¬µÚ________×åµÄÔªËØ£¬ÆäÔªËØÃû³ÆÊÇ________£»ÊôÓÚÔªËØÖÜÆÚ±íÖеĠ        Çø£¨ÌîÔªËØ·ÖÇø£©£¬ËüµÄ£«2¼ÛÀë×ӵĵç×ÓÅŲ¼Ê½Îª________________£»
(4)´ÓͼÖпÉÒÔ¿´³ö£¬D¸úBÐγɵÄÀë×Ó»¯ºÏÎïµÄ»¯Ñ§Ê½Îª________________£»¸ÃÀë×Ó»¯ºÏÎï¾§ÌåµÄÃܶÈΪa g?cm£­3£¬Ôò¾§°ûµÄÌå»ýÊÇ________________(Ö»ÒªÇóÁгöËãʽ)¡£

(1)Çâ¡¡(2)ÂÔ¡¡ ·ú»¯Çâ·Ö×Ӽ䴿ÔÚÇâ¼ü£¬ÂÈ»¯Çâ·Ö×Ó¼äûÓÐÇâ¼ü¡¡
(3)ËÄ¡¡¢÷B¡¡ÃÌ   dÇø  1s22s22p63s23p63d5(4) CaF24¡Á78 g?mol£­1a g?cm£­3¡Á6.02¡Á1023 mol£­1

½âÎö

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

(13·Ö) ÒÑÖªA¡¢B¡¢C¡¢DºÍE¶¼ÊÇÔªËØÖÜÆÚ±íÖÐǰ36ºÅµÄÔªËØ£¬ËüÃǵÄÔ­×ÓÐòÊýÒÀ´ÎÔö´ó¡£AÓëÆäËû4ÖÖÔªËØ¼È²»ÔÚͬһÖÜÆÚÓÖ²»ÔÚͬһÖ÷×å¡£BºÍCÊôͬһÖ÷×壬DºÍEÊôͬһÖÜÆÚ£¬ÓÖÖªEÊÇÖÜÆÚ±íÖÐ1£­18ÁÐÖеĵÚ7ÁÐÔªËØ¡£DµÄÔ­×ÓÐòÊý±ÈEС5£¬D¸úBÐγɵľ§ÌåÆä¾§°û½á¹¹Èçͼ£¬Í¼ÖÐСÇò´ú±íD£¬´óÇò´ú±íB¡£

Çë»Ø´ð£º

 (1)AÔªËØµÄÃû³ÆÊÇ________________£»

 (2)BÔªËØµÄ¹ìµÀ±íʾʽÊÇ________________£¬CµÄÔ­×ӽṹʾÒâͼÊÇ________________£¬BÓëAÐγɵϝºÏÎï±ÈCÓëAÐγɵϝºÏÎï·Ðµã¸ß£¬ÆäÔ­ÒòÊÇ__________________________________________________________£»

 (3)EÊôÓÚÔªËØÖÜÆÚ±íÖеÚ________ÖÜÆÚ£¬µÚ________×åµÄÔªËØ£¬ÆäÔªËØÃû³ÆÊÇ________£»ÊôÓÚÔªËØÖÜÆÚ±íÖеĠ        Çø£¨ÌîÔªËØ·ÖÇø£©£¬ËüµÄ£«2¼ÛÀë×ӵĵç×ÓÅŲ¼Ê½Îª________________£»

 (4)´ÓͼÖпÉÒÔ¿´³ö£¬D¸úBÐγɵÄÀë×Ó»¯ºÏÎïµÄ»¯Ñ§Ê½Îª________________£»¸ÃÀë×Ó»¯ºÏÎï¾§ÌåµÄÃܶÈΪa g•cm£­3£¬Ôò¾§°ûµÄÌå»ýÊÇ________________(Ö»ÒªÇóÁгöËãʽ)¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2009¡ª2010ѧÄê¹ãÖÝÊÐÆßÇøÁª¿¼¸ß¶þ»¯Ñ§ÏÂѧÆÚÆÚÄ©¼à²â ÌâÐÍ£º¼ÆËãÌâ

(12·Ö)
¢ñ£®ÒÑÖª³£ÎÂÏ£¬AgClµÄKsp£½1.8¡Á10-10£¬AgBrµÄKsp£½4.9¡Á10-13¡£
£¨1£©ÏÖÏòAgClµÄÐü×ÇÒºÖУº
¢Ù¼ÓÈëAgNO3¹ÌÌ壬Ôòc(Cl-)         £¨Ìî¡°±ä´ó¡±¡¢¡°±äС¡±»ò¡°²»±ä¡±£¬ÏÂͬ£©£»
¢ÚÈô¸Ä¼Ó¸ü¶àµÄAgCl¹ÌÌ壬Ôòc(Ag+)         £»
¢ÛÈô¸Ä¼Ó¸ü¶àµÄKBr¹ÌÌ壬Ôòc(Ag+)         £¬c(Cl-)         £»
£¨2£©ÓйØÄÑÈÜÑεÄÈܶȻý¼°Èܽâ¶ÈÓÐÒÔÏÂÐðÊö£¬ÆäÖÐÕýÈ·µÄÊÇ                   £»

A£®½«ÄÑÈܵç½âÖÊ·ÅÈë´¿Ë®ÖУ¬Èܽâ´ïµ½Æ½ºâʱ£¬Éý¸ßζȣ¬ Ksp Ò»¶¨Ôö´ó
B£®Á½ÖÖÄÑÈÜÑεç½âÖÊ£¬ÆäÖÐKspСµÄÈܽâ¶ÈÒ²Ò»¶¨Ð¡
C£®ÄÑÈÜÑεç½âÖʵÄKspÓëζÈÓйØ
D£®ÏòAgClµÄÐü×ÇÒºÖмÓÈëÊÊÁ¿µÄË®£¬Ê¹AgClÔٴδﵽÈÜ½âÆ½ºâ£¬AgClµÄKsp²»±ä£¬ÆäÈܽâ¶ÈÒ²²»±ä
¢ò£®×î½üÓÐÈËÖÆÔìÁËÒ»ÖÖȼÁÏµç³ØÊ¹ÆûÓÍÑõ»¯Ö±½Ó²úÉúµçÁ÷£¬ÆäÖÐÒ»¸öµç¼«Í¨Èë¿ÕÆø£¬ÁíÒ»¸öµç¼«Í¨ÈëÆûÓÍÕôÆø£¬µç³ØµÄµç½âÖÊÊDzôÔÓÁËY2O3µÄZrO2¾§Ì壬ËüÔÚ¸ßÎÂÏÂÄÜ´«µ¼O2-¡£»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÒÔÐÁÍéΪÆûÓ͵Ĵú±íÎÔòÕâ¸öµç³Ø·Åµçʱ±Ø·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ               ¡£
£¨2£©Õâ¸öµç³Ø¸º¼«µÄµç¼«·´Ó¦Ê½ÎªC8H18 + 25O2- -50e-==8CO2 + 9H2O,£¬Õý¼«µÄµç¼«·´Ó¦Ê½Îª      ¡£¹ÌÌåµç½âÖÊÀïO2-µÄÒÆ¶¯·½ÏòÊÇ    £¬ÏòÍâµç·Êͷŵç×ӵĵ缫ÊÇ       ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2013½ì¸£½¨ÈªÖݽú½­Êи߶þÏÂѧÆÚÆÚÖл¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÌî¿ÕÌâ

(13·Ö)£¨1£©A¡¢B¡¢DΪ¶ÌÖÜÆÚÔªËØ£¬Çë¸ù¾ÝÐÅÏ¢»Ø´ðÎÊÌ⣺

ÔªËØ

A

B

D

ÐÔÖÊ»ò½á¹¹ÐÅÏ¢

¹¤ÒµÉÏͨ¹ý·ÖÀëҺ̬¿ÕÆø»ñµÃÆäµ¥ÖÊ£¬µ¥ÖÊÄÜÖúȼ

ÆøÌ¬Ç⻯ÎïµÄË®ÈÜÒºÏÔ¼îÐÔ

Ô­×ÓÓÐÈý¸öµç×Ӳ㣬¼òµ¥Àë×ÓÔÚ±¾ÖÜÆÚÖа뾶×îС

¢ÙµÚÒ»µçÀëÄÜ£ºA      B£¨Ìî¡°>¡±¡¢¡°£½¡±¡¢¡°<¡±£©£¬»ù̬DÔ­×ӵĵç×ÓÅŲ¼Ê½Îª          ¡£

¢ÚBºÍDÓɹ²¼Û¼üÐγɵÄij»¯ºÏÎïBDÔÚ2200¡æ¿ªÊ¼·Ö½â£¬BDµÄ¾§ÌåÀàÐÍΪ                 ¡£

£¨2£©·¢Õ¹ÃºµÄÒº»¯¼¼Êõ±»ÄÉÈë¡°Ê®¶þÎ广»®¡±£¬ÖпÆÔºÉ½Î÷ú»¯Ëù¹ØÓÚúҺ»¯¼¼ÊõµÄ¸ßЧ´ß»¯¼ÁÑз¢ÏîĿȡµÃ»ý¼«½øÕ¹¡£ÒÑÖª£ºÃº¿ÉÒÔÏÈת»¯ÎªÒ»Ñõ»¯Ì¼ºÍÇâÆø£¬ÔÙÔÚ´ß»¯¼Á×÷ÓÃϺϳɼ״¼£¨CH3OH£©,´Ó¶øÊµÏÖÒº»¯¡£

   ¢Ùijº¬Í­Àë×ÓµÄÀë×ӽṹÈçÓÒͼËùʾ£º

ÔÚ¸ÃÀë×ÓÄÚ²¿Î¢Á£¼ä×÷ÓÃÁ¦µÄÀàÐÍÓУº          £¨Ìî×Öĸ£©¡£

 A£®Àë×Ó¼ü  B.¼«ÐÔ¼ü C.·Ç¼«ÐÔ¼ü  D.Åäλ¼ü  E.·¶µÂ»ªÁ¦  F.Çâ¼ü

¢ÚúҺ»¯»ñµÃ¼×´¼,ÔÙ¾­´ß»¯µÃµ½ÖØÒª¹¤ÒµÔ­Áϼ×È©(HCHO)£¬¼×´¼µÄ·ÐµãΪ65¡æ£¬¼×È©µÄ·ÐµãΪ£­21¡æ£¬Á½Õß¾ùÒ×ÈÜÓÚË®¡£¼×´¼µÄ·Ðµã±È¼×È©¸ßÊÇÒòΪ¼×´¼·Ö×Ӽ䴿ÔÚ×ÅÇâ¼ü£¬¶ø¼×È©·Ö×Ó¼äûÓÐÇâ¼ü¡£¼×´¼ºÍ¼×È©¾ùÈÜÓÚË®£¬ÊÇÒòΪËüÃǾù¿ÉÒÔºÍË®ÐγɷÖ×Ó¼äÇâ¼ü¡£ÇëÄã˵Ã÷¼×È©·Ö×Ó¼äûÓÐÇâ¼üµÄÔ­ÒòÊÇ                                                  ¡£

¢Û¼×´¼·Ö×ÓÖУ¬½øÐÐsp3ÔÓ»¯µÄÔ­×ÓÓР        £¬¼×È©ÓëH2·¢Éú¼Ó³É·´Ó¦£¬µ±Éú³É1mol¼×´¼Ê±£¬¶ÏÁѵġǼüµÄÊýĿΪ            

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2011-2012ѧÄê½­ËÕÊ¡ÉϸԸ߼¶ÖÐѧ¸ß¶þµÚһѧÆÚÆÚÖп¼ÊÔ»¯Ñ§ÊÔ¾í£¨Ñ¡ÐÞ£© ÌâÐÍ£º¼ÆËãÌâ

(9·Ö)ÏÖÓÐA¡¢BÁ½ÖÖÁ´Ìþ±¥ºÍÒ»Ôª´¼µÄ»ìºÏÎï0.3mol£¬ÆäÖÊÁ¿Îª13.8g¡£ÒÑÖªAºÍB̼ԭ×ÓÊý¾ù²»´óÓÚ4£¬ÇÒA<B¡£
¢Å»ìºÏÎïÖÐAµÄ·Ö×ÓʽΪ_______________¡£
¢ÆÈôn(A)¡Ãn(B)=1¡Ã1ʱ£¬BµÄ½á¹¹¼òʽΪ____________________________¡£
¢ÇÈôn(A)¡Ãn(B)¡Ù1¡Ã1ʱ£¬BµÄ·Ö×ÓʽΪ_________________£¬n(A)¡Ãn(B)=___________¡£

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸