ÌâÄ¿ÄÚÈÝ
¡¾ÌâÄ¿¡¿X¡¢Y¡¢Z¡¢U¡¢WÊÇÔ×ÓÐòÊýÒÀ´ÎÔö´óµÄÇ°ËÄÖÜÆÚÔªËØ¡£ÆäÖÐYµÄÔ×ÓºËÍâÓÐ7ÖÖÔ˶¯×´Ì¬²»Í¬µÄµç×Ó£»X¡¢ZÖÐδ³É¶Ôµç×ÓÊý¾ùΪ2£»UÊǵÚÈýÖÜÆÚÔªË÷Ðγɵļòµ¥Àë×ÓÖа뾶×îСµÄÔªËØ£»WµÄÄÚ²ãµç×ÓÈ«³äÂú£¬×îÍâ²ãÖ»ÓÐ1¸öµç×Ó¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺
(1)X¡¢Y¡¢ZµÄµÚÒ»µçÀëÄÜ´Ó´óµ½Ð¡µÄ˳ÐòÊÇ_______ (ÓÃÔªËØ·ûºÅ±íʾ£¬ÏÂͬ)¡£
(2)д³öWµÄ¼Ûµç×ÓÅŲ¼Ê½________ £¬WͬÖÜÆÚµÄÔªËØÖУ¬ÓëWÔ×Ó×îÍâ²ãµç×ÓÊýÏàµÈµÄÔªËØ»¹ÓÐ_______¡£
(3)¸ù¾ÝµÈµç×ÓÌåÔÀí£¬¿ÉÖª»¯ºÏÎïXZµÄ½á¹¹Ê½ÊÇ______£¬YZ2-µÄVSEPRÄ£ÐÍÊÇ________¡£
(4)X¡¢Y¡¢ZµÄ¼òµ¥Ç⻯ÎïµÄ¼ü½Ç´Ó´óµ½Ð¡µÄ˳ÐòÊÇ________(Óû¯Ñ§Ê½±íʾ)£¬ÔÒòÊÇ____________________¡£
(5)ÓÉÔªËØYÓëUÔªËØ×é³ÉµÄ»¯ºÏÎïA£¬¾§°û½á¹¹ÈçͼËùʾ£¨ºÚÇò±íʾYÔ×Ó£¬°×Çò±íʾUÔ×Ó£©£¬Çëд³ö»¯ºÏÎïAµÄ»¯Ñ§Ê½_______£¬¸ÃÎïÖÊÓ²¶È´ó£¬ÍƲâ¸ÃÎïÖÊΪ______¾§Ìå¡£ÆäÖÐYÔªËصÄÔÓ»¯·½Ê½ÊÇ____________¡£
(6)UµÄ¾§ÌåÊôÁ¢·½¾§Ïµ£¬Æ侧°û±ß³¤Îª405Pm£¬ÃܶÈÊÇ2.70g¡¤cm-3£¬Í¨¹ý¼ÆËãÈ·¶¨Æ侧°ûµÄÀàÐÍ________________£¨¶Ñ¼òµ¥Á¢·½¶Ñ»ý¡¢ÌåÐÄÁ¢·½¶Ñ»ý»òÃæÐÄÁ¢·½×îÃܶѻý£©£¨¼ºÖª£º4053=6.64¡Á107)¡£
¡¾´ð°¸¡¿ N>O>C 3d104s1 K¡¢Cr CO ƽÃæÈý½ÇÐÎ CH4>NH3>H2O 3ÖÖÇ⻯ÎïµÄÖÐÐÄÔ×Ó¼Û²ãµç×Ó¶àÉÙ¾ùΪ4£¬VSEPRÄ£Ð;ùΪÕýËÄÃæÌå½á¹¹£¬µ«ÖÐÐÄÔ×ӵŵç×Ó¶ÔÊýÒÀ´ÎÔö¼Ó£¬µ¼Ö¼ü½Ç±äС AlN Ô×Ó sp3 ÃæÐÄÁ¢·½×îÃܶѻý
¡¾½âÎö¡¿X¡¢Y¡¢Z¡¢U¡¢WÊÇÔ×ÓÐòÊýÒÀ´ÎÔö´óµÄÇ°ËÄÖÜÆÚÔªËØ¡£ÆäÖÐYµÄÔ×ÓºËÍâÓÐ7ÖÖÔ˶¯×´Ì¬²»Í¬µÄµç×Ó£¬ÔòYΪµªÔªËØ£»UÊǵÚÈýÖÜÆÚÔªË÷Ðγɵļòµ¥Àë×ÓÖа뾶×îСµÄÔªËØ£¬ÔòUΪÂÁÔªËØ£»X¡¢ZÖÐδ³É¶Ôµç×ÓÊý¾ùΪ2£¬XµÄÔ×ÓÐòÊýСÓÚN£¬¶øZµÄÔ×ÓÐòÊý½éÓÚNµª¡¢ÂÁÖ®¼ä£» X¡¢ZµÄºËÍâµç×ÓÅŲ¼·Ö±ðΪ1s22s22p2£¬1s22s22p4£¬¹ÊXΪ̼ԪËØ¡¢ZΪÑõÔªËØ£»WµÄÄÚ²ãµç×ÓÈ«³äÂú£¬×îÍâ²ãÖ»ÓÐ1¸öµç×Ó£¬Ô×ÓÐòÊý´óÓÚÂÁ£¬Ö»ÄÜ´¦ÓÚµÚËÄÖÜÆÚ£¬ºËÍâµç×ÓÊýΪ2+8+18+1=29£¬¹ÊWΪÍÔªËØ¡££¨1£©Í¬ÖÜÆÚÔªËØËæÔ×ÓÐòÊýÔö´ó£¬µÚÒ»µçÀëÄܳÊÔö´óÇ÷ÊÆ£¬µ«NÔªËØ2pÄܼ¶Îª°ëÂúÎȶ¨×´Ì¬£¬ÄÜÁ¿½ÏµÍ£¬µÚÒ»µçÀëÄܸßÓÚͬÖÜÆÚÏàÁÚÔªËØ£¬¹ÊµÚÒ»µçÀëÄÜN>O>C£»£¨2£©WΪÍÔªËØ£¬¼Ûµç×ÓÅŲ¼Ê½Îª3d104s1£¬WͬÖÜÆÚµÄÔªËØÖУ¬ÓëWÔ×Ó×îÍâ²ãµç×ÓÊýÏàµÈµÄÔªËØ»¹ÓÐ3d64s1£¬ÎªKÔªËØ¡¢3d54s1ΪCrÔªËØ£»£¨3£©¸ù¾ÝµÈµç×ÓÌåÔÀí£¬¿ÉÖª»¯ºÏÎïCOÓëN2»¥ÎªµÈµç×ÓÌ壬COµÄ½á¹¹Ê½ÊÇCO£¬YZ2¡ªÎªNO2¡ª£¬ÖÐÐÄÔ×Ó¼Ûµç×Ó¶ÔÊý=2+¡Á£¨5+1-2¡Á2£©=3£¬VSEPRÄ£ÐÍÊÇƽÃæÈý½ÇÐΣ»£¨4£©X¡¢Y¡¢ZµÄ×î¼òµ¥Ç⻯Îï·Ö±ðΪCH4¡¢NH3¡¢H2O£¬ÈýÖÖÇ⻯ÎïµÄÖÐÐÄÔ×Ó¼Û²ãµç×Ó¶ÔÊý¾ùΪ4£¬VSEPRÄ£Ð;ùΪÕýËÄÃæÌåÐΣ¬µ«ÖÐÐÄÔ×ӵŵç×Ó¶ÔÊýÒÀ´ÎÔö´ó£¬µ¼Ö¼ü½Ç±äС£¬¹Ê¼ü½Ç£ºCH4>NH3>H2O£»£¨5£©¾§°ûÖÐNÔ×ÓÊýĿΪ4£¬ÂÁÔ×ÓÊýĿΪ8¡Á+6¡Á=4£¬¹Ê¸Ã»¯ºÏÎﻯѧʽΪAlN£¬¸ÃÎïÖÊÓ²¶È´ó£¬ÊôÓÚÔ×Ó¾§Ì壬YÔ×ÓÐγÉ4¸ö¹²¼Û¼ü£¬¹ÊYÔ×ÓÔÓ»¯·½Ê½Îªsp3£»£¨6£©É辧°ûÖÐÂÁÔ×ÓÊýĿΪN(Al)£¬ÔòN(Al)=g=(405¡Á10-10)3¡Á2.70g/cm3£¬½âµÃN(Al)=4.00£¬ËùÒÔÆ侧°ûµÄÀàÐÍΪÃæÐÄÁ¢·½Ãܶѻý¡£
¡¾ÌâÄ¿¡¿¹¤ÒµÈ¼ÉÕú¡¢Ê¯Ó͵Ȼ¯Ê¯È¼ÁÏÊͷųö´óÁ¿µªÑõ»¯Îï(NOx)¡¢CO2¡¢SO2µÈÆøÌ壬ÑÏÖØÎÛȾ¿ÕÆø¡£¶Ô·ÏÆø½øÐÐÍÑÏõ¡¢ÍÑ̼ºÍÍÑÁò´¦Àí¿ÉʵÏÖÂÌÉ«»·±£¡¢·ÏÎïÀûÓá£
¢ñ.ÍÑÏõ£º
ÒÑÖª£ºH2µÄȼÉÕÈÈΪ285.8kJ¡¤mol-1
N2(g)+2O2(g)£½2NO2(g)¦¤H£½+133kJ¡¤mol-1
H2O(g)£½H2O(l) ¦¤H£½-44kJ¡¤mol-1
´ß»¯¼Á´æÔÚÏ£¬H2»¹ÔNO2Éú³ÉË®ÕôÆøºÍÆäËûÎÞ¶¾ÎïÖʵÄÈÈ»¯Ñ§·½³ÌʽΪ£º____________¡£
¢ò.ÍÑ̼£º
(1)Ïò2LÃܱÕÈÝÆ÷ÖмÓÈë2molCO2ºÍ6molH2£¬ÔÚÊʵ±µÄ´ß»¯¼Á×÷ÓÃÏ£¬·¢Éú·´Ó¦£º
CO2(g)+3H2(g)CH3OH(l)+H2O(l)
¢Ù¸Ã·´Ó¦×Ô·¢½øÐеÄÌõ¼þÊÇ_____________(Ìî¡°µÍΡ±¡¢¡°¸ßΡ±»ò¡°ÈÎÒâζȡ±)
¢ÚÏÂÁÐÐðÊöÄÜ˵Ã÷´Ë·´Ó¦´ïµ½Æ½ºâ״̬µÄÊÇ____________¡£(Ìî×Öĸ)
a.»ìºÏÆøÌåµÄƽ¾ùʽÁ¿±£³Ö²»±ä b.CO2ºÍH2µÄÌå»ý·ÖÊý±£³Ö²»±ä
c.CO2ºÍH2µÄת»¯ÂÊÏàµÈ d.»ìºÏÆøÌåµÄÃܶȱ£³Ö²»±ä
e.1molCO2Éú³ÉµÄͬʱÓÐ3mol H¡ªH¼ü¶ÏÁÑ
¢ÛCO2µÄŨ¶ÈËæʱ¼ä£¨0¡«t2£©±ä»¯ÈçÏÂͼËùʾ£¬ÔÚt2ʱ½«ÈÝÆ÷ÈÝ»ýËõСһ±¶£¬t3ʱ´ïµ½Æ½ºâ£¬t4ʱ½µµÍζȣ¬t5ʱ´ïµ½Æ½ºâ£¬Ç뻳öt2¡«t6 CO2Ũ¶ÈËæʱ¼äµÄ±ä»¯¡£_____________
¢Æ¸Ä±äζȣ¬Ê¹·´Ó¦CO2(g)+3H2(g)CH3OH(g)+H2O(g) ¦¤H©‚0ÖеÄËùÓÐÎïÖʶ¼ÎªÆø̬¡£Æðʼζȡ¢Ìå»ýÏàͬ£¨T1¡æ¡¢2LÃܱÕÈÝÆ÷£©¡£·´Ó¦¹ý³ÌÖв¿·ÖÊý¾Ý¼ûÏÂ±í£º
·´Ó¦Ê±¼ä | CO2(mol) | H2(mol) | CH3OH(mol) | H2O(mol) | |
·´Ó¦¢ñ£ººãκãÈÝ | 0min | 2 | 6 | 0 | 0 |
10min | 4.5 | ||||
20min | 1 | ||||
30min | 1 | ||||
·´Ó¦¢ò£º¾øÈȺãÈÝ | 0min | 0 | 0 | 2 | 2 |
¢Ù´ïµ½Æ½ºâʱ£¬·´Ó¦¢ñ¡¢¢ò¶Ô±È£ºÆ½ºâ³£ÊýK(I)______K(II)£¨Ìî¡°©ƒ¡±¡°©‚¡±»ò¡°£½¡±ÏÂͬ£©£»Æ½ºâʱCH3OHµÄŨ¶Èc(I)____ c(II)¡£
¢Ú¶Ô·´Ó¦¢ñ£¬Ç°10minÄÚµÄƽ¾ù·´Ó¦ËÙÂÊv(CH3OH)£½_______¡£ÔÚÆäËûÌõ¼þ²»±äµÄÇé¿öÏ£¬Èô30minʱֻ¸Ä±äζÈT2¡æ£¬´ËʱH2µÄÎïÖʵÄÁ¿Îª3.2mol£¬ÔòT1___T2(Ìî¡°>¡±¡¢¡°<¡±»ò¡°=¡±)¡£Èô30minʱֻÏòÈÝÆ÷ÖÐÔÙ³äÈë1molCO2(g)ºÍ1molH2O(g)£¬Ôòƽºâ_____Òƶ¯(Ìî¡°ÕýÏò¡±¡°ÄæÏò¡±»ò¡°²»¡±)¡£
¢ÇÀûÓÃÈ˹¤¹âºÏ×÷Óÿɽ«CO2ת»¯Îª¼×Ëᣬ·´Ó¦ÔÀíΪ2CO2+2H2O=2HCOOH+O2,
×°ÖÃÈçͼËùʾ£º
¢Ùµç¼«2µÄµç¼«·´Ó¦Ê½ÊÇ____________£»
¢ÚÔÚ±ê×¼×´¿öÏ£¬µ±µç¼«2ÊÒÓÐ11.2L CO2·´Ó¦¡£ ÀíÂÛÉϵ缫1ÊÒÒºÌåÖÊÁ¿_____(Ìî¡°Ôö¼Ó¡±»ò¡°¼õÉÙ¡±______g¡£