ÌâÄ¿ÄÚÈÝ

A¡¢B¡¢C¡¢D¡¢E¡¢FÖܱíÇ°ËÄÖÜÆÚÖеij£¼ûÔªËØ£¬ÆäÏà¹ØÐÅÏ¢ÈçÏÂ±í£º

ÔªËØ
Ïà¹ØÐÅÏ¢
A
AÊÇÖÜÆÚ±íÖÐÔ­×Ӱ뾶×îСµÄÔªËØ
B
BÔªËصÄÔ­×Ó¼Ûµç×ÓÅŲ¼Îªns11np14
C
MµÄ»ù̬ԭ×ÓL²ãµç×ÓÊýÊÇK²ãµç×ÓÊýµÄ3±¶
D
DÊǵÚÈýÖÜÆÚÖеÚÒ»µçÀëÄÜ×îСµÄÔªËØ
E
EÊǵؿÇÖк¬Á¿×î¶àµÄ½ðÊôÔªËØ
F
ÓжàÖÖ»¯ºÏ¼Û£¬ÆäijÖָ߼ÛÑôÀë×ӵļ۵ç×Ó¾ßÓнÏÎȶ¨µÄ°ë³äÂú½á¹¹
 
£¨1£©FλÓÚÔªËØÖÜÆÚ±íÖÐλÖà            £¬Æä»ù̬ԭ×ÓºËÍâ¼Ûµç×ÓÅŲ¼Ê½Îª          £»
£¨2£©BµÄµç¸ºÐÔ±ÈMµÄ           £¨Ìî¡°´ó¡±»ò¡°Ð¡¡±£©£»B2A3·Ö×ÓÖмüÓë¼üÓë¸öÊýÖ®±ÈΪ             £»
£¨3£©Ð´³öEµÄµ¥ÖÊÓëDµÄ×î¸ß¼ÛÑõ»¯ÎïµÄË®»¯ÎïÈÜÒº·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º          £»
£¨4£©ÒÑ֪ÿ5.4gE¿ÉÓë×îµÍ¼ÛFµÄÑõ»¯Îï·´Ó¦£¬·Å³ö346.2kJµÄÈÈÁ¿¡£ÔòÇëд³ö¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ£º                   ¡£

(12·Ö£¬Ã¿¿Õ2·Ö)
£¨1£©µÚËÄÖÜÆÚ¢ø×å3d64s2
£¨2£©Ð¡  5£º1
£¨3£©2A1+2NaOH+6H2O=2Na[Al(OH)4]+3H2¡ü
£¨4£©2Al(s)+3FeO(s)=2A12O3(s)+3Fe(s)  ¡÷H=-3462kJ/mol

½âÎöÊÔÌâ·ÖÎö£ºÓÉÌâÖÐËù¸øÌõ¼þ¿ÉÖªAΪÇâÔªËØ£¬BΪ̼ԪËØ£¬CΪÑõÔªËØ£¬DΪÄÆÔªËØ£¬EΪÂÁÔªËØ£¬FΪÌúÔªËØ¡££¨2£©ÔªËصķǽðÊôÐÔԽǿ£¬Ôòµç¸ºÐÔÔ½´ó£»C2H4·Ö×ÓÖÐÓÐ1¸öË«¼üºÍ4¸öµ¥¼ü£¬¹ÊÓÐ1¸ö¦Ð¼ü£¬5¸ö¦Ò¼ü¡££¨4£©Ð´³ö»¯Ñ§·½³Ìʽ¼´¿É¼ÆËã³ö¡÷H£¬ÊéдÈÈ»¯Ñ§·½³ÌʽʱӦעÒâдÃ÷¸÷ÎïÖʵÄ״̬¡£
¿¼µã£º¿¼²éÎïÖʽṹºÍÔªËØÖÜÆÚÂÉ£¬¿¼²é¿¼Éú¶ÔÔªËØÍƶϺÍÎïÖʽṹ¼°ÐÔÖʵÄÕÆÎÕ¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

A¡¢B¡¢C¡¢D¡¢E¡¢FÁùÖÖ¶ÌÖÜÆÚÔªËØ£¬ÆäÔ­×ÓÐòÊýÒÀ´ÎÔö´ó£¬ÆäÖÐBÓëCͬÖÜÆÚ£¬DÓëEºÍFͬÖÜÆÚ£¬AÓëDͬÖ÷×壬CÓëFͬÖ÷×壬CÔªËصÄÔ­×Ó×îÍâ²ãµç×ÓÊýÊÇ´ÎÍâ²ãµç×ÓÊýµÄÈý±¶£¬DÊÇËùÔÚÖÜÆÚÖÐÔ­×Ӱ뾶×î´óµÄÖ÷×åÔªËØ¡£ÓÖÖªÁùÖÖÔªËØËùÐγɵij£¼ûµ¥ÖÊÔÚ³£Î³£Ñ¹ÏÂÓÐÈýÖÖÊÇÆøÌ壬ÈýÖÖÊǹÌÌå¡£
Çë»Ø´ðÏÂÁÐÎÊÌâ¡£
£¨1£©ÔªËØDÔÚÖÜÆÚ±íÖеÄλÖÃ________¡£
£¨2£©C¡¢D¡¢FÈýÖÖÔªËØÐγɵļòµ¥Àë×ӵİ뾶ÓÉ´óµ½Ð¡µÄ˳ÐòÊÇ(ÓÃÀë×Ó·ûºÅ±íʾ)________________¡£
£¨3£©ÓÉA¡¢B¡¢CÈýÖÖÔªËØÒÔÔ­×Ó¸öÊý±È4¡Ã2¡Ã3Ðγɻ¯ºÏÎïX£¬XÖÐËùº¬»¯Ñ§¼üÀàÐÍÓÐ________¡£
£¨4£©¢ÙÈôEÊǽðÊôÔªËØ£¬Æäµ¥ÖÊÓëÑõ»¯Ìú·´Ó¦³£ÓÃÓÚº¸½Ó¸Ö¹ì£¬Çëд³ö·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º________________________________________________________________________¡£
¢ÚÈôEÊǷǽðÊôÔªËØ£¬Æäµ¥ÖÊÔÚµç×Ó¹¤ÒµÖÐÓÐÖØÒªÓ¦Óã¬Çëд³öÆäÑõ»¯ÎïÈÜÓÚÇ¿¼îÈÜÒºµÄÀë×Ó·½³Ìʽ£º________________________________________________________________¡£
¢ÛFC2ÆøÌåÓж¾£¬Åŷŵ½´óÆøÖÐÒ×ÐγÉËáÓ꣬д³öFC2ÓëÑõÆøºÍË®ÕôÆø·´Ó¦µÄ»¯Ñ§·½³Ìʽ________________________________________________________________________¡£

ÓйضÌÖÜÆÚÔªËØA¡¢B¡¢C¡¢D¡¢E¡¢FµÄÐÅÏ¢ÈçÏ£º

ÔªËØ
ÓйØÐÅÏ¢
A
×î¸ß¼ÛÑõ»¯Îï¶ÔÓ¦µÄË®»¯Î¼×£©ÄÜÓëÆäÆø̬Ç⻯ÎÒÒ£©·´Ó¦Éú³ÉÑÎ
B
×îÍâ²ãµç×ÓÊýÊÇ´ÎÍâ²ãµç×ÓÊýµÄ2±¶
C
M²ãÉÏÓÐ3¸öµç×Ó
D
¶ÌÖÜÆÚÔ­×Ӱ뾶×î´óµÄÖ÷×åÔªËØ
E
Æäµ¥ÖÊÊǵ­»ÆÉ«¹ÌÌå
F
×î¸ßÕý¼ÛÓë×îµÍ¸º¼Û´úÊýºÍΪ6
 
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Ð´³öʵÑéÊÒÖÆÈ¡ÒҵĻ¯Ñ§·½³Ìʽ                                                   ¡£
£¨2£©ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ                (ÌîÐòºÅ)¡£
¢ÙʵÑéÊÒ¿ÉÓÃÈçͼËùʾװÖÃÖÆÈ¡BµÄ×î¸ß¼ÛÑõ»¯Îï

¢ÚÓÃCµ¥ÖÊ×ö³ÉµÄ²Û³µ£¬¶¼¿ÉÓÃÀ´ÔËÊä¼×µÄŨÈÜÒº
¢Û CºÍÍ­ÓëÏ¡ÁòËá×é³ÉµÄÔ­µç³Ø£¬Cµç¼«±»»¹Ô­
¢Ü Dµ¥ÖÊÔÚÑõÆøÖÐȼÉÕºóµÄ²úÎï¿ÉÓÃÔÚ·À¶¾Ãæ¾ßÖÐ×÷¹©Ñõ¼Á
¢Ý¹ÄÀø³Ë×ø¹«½»³µ³öÐУ¬³«µ¼µÍ̼Éú»î£¬ÊÇ¿ØÖƺÍÖÎÀíBO2½â¾ö¡°ÎÂÊÒЧӦ¡±µÄÓÐЧ;¾¶Ö®Ò»
¢Þ DFµÄµç×ÓʽΪH¡ÃCl¡Ã
£¨3£©½«EµÄ³£¼ûÑõ»¯Î¸ÃÑõ»¯ÎïÄÜʹƷºìÈÜÒºÍÊÉ«£©Í¨ÈëÓÉCuSO4ºÍNaCl»ìºÏµÄŨÈÜÒºÖУ¬ÈÜÒºÑÕÉ«±ädz£¬Îö³ö°×É«³Áµí£¬È¡¸Ã³Áµí½øÐÐÔªËØÖÊÁ¿·ÖÊý·ÖÎö£¬¿ÉÖªÆäÖк¬Cl£º35.7%£¬Cu£º64.3%£¬Ôò¸ÃÑõ»¯ÎïÔÚÉÏÊö·´Ó¦ÖеÄ×÷ÓÃÊÇ             ¡£
A£®Æ¯°×¼Á                 B£®Ñõ»¯¼Á             C£®»¹Ô­¼Á
£¨4£©ÇëÓû¯Ñ§·½·¨¼ÓÒÔÑéÖ¤£¨3£©ÖеÄÑõ»¯Î¼òҪд³öʵÑé·½·¨¡¢ÊÔ¼Á¼°Ô¤Æڿɹ۲쵽µÄÏÖÏó     ¡£

ÏÖÓÐÎåÖÖ¶ÌÖÜÆÚÖ÷×åÔªËØA¡¢B¡¢C¡¢   D¡¢E£¬ÆäÔ­×ÓÐòÊýÒÀ´ÎÔö´ó¡£AÔ­×ÓÔ¼Õ¼ÓîÖæÖÐÔ­×Ó×ÜÊýµÄ88.6%£¬A+ÓÖ³ÆΪÖÊ×Ó£ºBÊÇÐγɻ¯ºÏÎïÖÖÀà×î¶àµÄÔªËØ£¬CÔªËصÄ×î¼òµ¥µÄÇ⻯ÎïYµÄË®ÈÜÒºÏÔ¼îÐÔ£®EÊǶÌÖÜÆÚÔªËØÖе縺ÐÔ×îСµÄÔªËØ¡£A¡¢B¡¢C¡¢EËÄÖÖÔªËض¼ÄÜÓëDÔªËØÐγÉÔ­×Ó¸öÊý±È²»ÏàͬµÄ³£¼û»¯ºÏÎï¡£ÊԻشðÏÂÁÐÎÊÌ⣺
£¨1£©Ð´³öA¡¢EÁ½ÔªËØÐγɵÄÔ­×Ó¸öÊý±ÈΪ1:1µÄ»¯ºÏÎïµÄµç×Óʽ____¡£
£¨2£©ÏòÂÈ»¯ÑÇÌúÈÜÒºµÎ¼Ó¹ýÁ¿µÄEµÄ×î¸ß¼ÛÑõ»¯Îï¶ÔӦˮ»¯ÎïµÄÈÜÒº£¬ÏÖÏóÊÇ____¡£
£¨3£©YÈÜÒºÏÔ¼îÐÔµÄÔ­ÒòÊÇ(ÓÃÒ»¸öÀë×Ó·½³Ìʽ±íʾ£©____¡£
£¨4£©¼ìÑéÆû³µÎ²ÆøÖк¬ÓеĻ¯ºÏÎïBDµÄ·½·¨ÊÇ£ºÏòËáÐÔPdC12ÈÜÒºÖÐͨAÆû³µÎ²Æø£¬ÈôÉú³ÉºÚÉ«³Áµí£¨Pd£©£¬Ö¤Ã÷Æû³µÎ²ÆøÖк¬ÓÐBD¡£Ð´³ö·´Ó¦µÄÀë×Ó·½³Ìʽ____¡£
£¨5£©ÏÂÁÐÓйØÎïÖÊÐÔÖʵıȽÏÖУ®²»ÕýÈ·µÄÊÇ               ¡£
a£®ÈÈÎȶ¨ÐÔ£ºH2S>SiH4  b£®Àë×Ӱ뾶£ºNa+>S2£­
c£®µÚÒ»µçÀëÄÜN>O   d£®ÔªËص縺ÐÔ£ºC>H
£¨6£©ÒÑÖª£º¢ÙCH3OH(g)+H2O(g)=CO2(g)+3H2(g)  ¡÷H=+49.0kJ/mol
¢ÚCH3OH(g)+3/2O2(g)=CO2(g)+2H2O(g)  ¡÷H=£­192£®9kJ/mol
ÓÉÉÏÊö·½³Ìʽ¿ÉÖª£®CH3OHµÄȼÉÕÈÈ____£¨Ìî¡°´óÓÚ¡±¡¢¡°µÈÓÚ¡±»òСÓÚ¡±£©192.9kJ/mol¡£ÒÑ֪ˮµÄÆø»¯ÈÈΪ44 kJ/mol£®Ôò±íʾÇâÆøȼÉÕÈȵÄÈÈ»¯Ñ§·½³ÌʽΪ____¡£

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

¾«Ó¢¼Ò½ÌÍø