ÏÂÁÐÓйØÎïÖÊÐÔÖʵıȽÏÕýÈ·µÄΪ (¡¡¡¡)

¢ÙͬÖ÷×åÔªËصĵ¥ÖÊ´ÓÉϵ½Ï£¬·Ç½ðÊôÐÔ¼õÈõ£¬ÈÛµãÔö¸ß

¢ÚÔªËصÄ×î¸ßÕý»¯ºÏ¼ÛÔÚÊýÖµÉϵÈÓÚËüËùÔÚµÄ×åÐòÊý

¢ÛͬÖÜÆÚÖ÷×åÔªËصÄÔ­×Ӱ뾶ԽС£¬Ô½ÄÑʧȥµç×Ó

¢ÜÔªËصķǽðÊôÐÔԽǿ£¬ËüµÄÆø̬Ç⻯ÎïË®ÈÜÒºµÄËáÐÔԽǿ

¢Ý»¹Ô­ÐÔ£ºS2£­>Cl£­>Br£­

¢ÞËáÐÔ£ºHClO4>H2SO4>H3PO4>H2SiO3

A£®¢Ù¢Û                                   B£®¢Ú¢Ü

C£®¢Û¢Þ                                   D£®¢Ý¢Þ

 

¡¾´ð°¸¡¿

C

¡¾½âÎö¡¿¿ÉÒÔÓ÷´ÀýÅųý£»¢ÙµÚIVA×åÖÐCµÄÈÛµã×î¸ß£»¢ÚÖÐÑõÔªËØÎÞ×î¸ßÕý¼Û+6£»¢ÜFÔªËطǽðÊôÐÔ×îÇ¿£¬µ«ÆäÆø̬Ç⻯ÎïË®ÈÜÒºÊÇÈõË᣻¢Ý»¹Ô­ÐÔ£ºS2£­>Br£­>Cl£­£»¹Ê´ð°¸ÎªC

 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÏÂÁÐÓйØÎïÖÊÐÔÖʵÄÃèÊöºÍ¸ÃÐÔÖʵÄÓ¦ÓþùÕýÈ·µÄÊÇ£¨¡¡¡¡£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

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»¯ºÏÎÊԻشðÏÂÁÐÎÊÌ⣺
£¨1£©Ð´³öA¡¢EÁ½ÔªËØÐγɵÄÔ­×Ó¸öÊý±ÈΪ1£º1µÄ»¯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£ºÏòËáÐÔPdCl2ÈÜÒºÖÐͨAÆû³µÎ²Æø£¬ÈôÉú³ÉºÚÉ«³Áµí£¨Pd£©£¬Ö¤Ã÷Æû³µÎ²ÆøÖк¬ÓÐBD£®Ð´³ö·´Ó¦µÄÀë×Ó·½³Ìʽ
 
£®
£¨5£©ÏÂÁÐÓйØÎïÖÊÐÔÖʵıȽÏÖУ®²»ÕýÈ·µÄÊÇ
 
£®
a£®ÈÈÎȶ¨ÐÔ£ºH2S£¾SiH4    b£®Àë×Ӱ뾶£ºNa+£¾S2-
c£®µÚÒ»µçÀëÄÜN£¾O    d£®ÔªËص縺ÐÔ£ºC£¾H
£¨6£©ÒÑÖª£º¢ÙCH3OH£¨g£©+H2O£¨g£©=CO2£¨g£©+3H2£¨g£©¡÷H=+49.0kJ/mol
¢ÚCH3OH£¨g£©+
32
O2£¨g£©=CO2£¨g£©+2H2O£¨g£©¡÷H=-192.9kJ/mol
ÓÉÉÏÊö·½³Ìʽ¿ÉÖª£®CH3OHµÄȼÉÕÈÈ
 
£¨Ìî¡°´óÓÚ¡±¡¢¡°µÈÓÚ¡±»òСÓÚ¡±£©192.9kJ/mol£®ÒÑ֪ˮµÄÆø»¯ÈÈΪ44kJ/mol£®Ôò±íʾÇâÆøȼÉÕÈȵÄÈÈ»¯Ñ§·½³ÌʽΪ
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÏÂÁÐÓйØÎïÖÊÐÔÖʵÄÓ¦Óò»ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A¡¢Å¨H2SO4¾ßÓÐÎüË®ÐÔ£¬¿ÉÓÃÀ´¸ÉÔï°±ÆøB¡¢Na2CO3ÈÜÒº¾ßÓмîÐÔ£¬¿ÉÓÃÓÚ³ýÆ÷Ãó±íÃæÓÍÎÛC¡¢ï®ÖÊÁ¿Çá¡¢±ÈÄÜÁ¿´ó£¬½ðÊôï®ÓÃ×÷µç³Ø¸º¼«²ÄÁÏD¡¢ClO2¾ßÓÐÇ¿Ñõ»¯ÐÔ£¬¿ÉÓÃÓÚ×ÔÀ´Ë®µÄÏû¶¾É±¾ú

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2013-2014ѧÄêÖØÇìÊÐÈýÏ¿ÃûУÁªÃ˸ßÈý12ÔÂÁª¿¼»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÌî¿ÕÌâ

W¡¢M¡¢N¡¢X¡¢Y¡¢ZÊÇÎåÖÖ³£¼ûµÄ¶ÌÖÜÆÚÔªËØ£¬ÆäÔ­×Ӱ뾶ËæÔ­×ÓÐòÊý±ä»¯ÈçÓÒͼËùʾ¡£ÒÑÖªWµÄÒ»ÖÖºËËصÄÖÊÁ¿ÊýΪ18£¬ÖÐ×ÓÊýΪ10£»MºÍNeÔ­×ӵĺËÍâµç×ÓÊýÏà²î1£»NÓëXÏàÁÚ£»XµÄµ¥ÖÊÊÇÒ»ÖÖ³£¼ûµÄ°ëµ¼Ìå²ÄÁÏ£»YµÄ×îÍâ²ãµç×ÓÊýÊÇÆä×îÄÚ²ãµç×ÓÊýµÄ3±¶£»ZµÄ·Ç½ðÊôÐÔÔÚͬÖÜÆÚÖ÷×åÔªËØÖÐ×î´ó¡£

£¨1£©ÔªËØXÔÚÖÜÆÚ±íÖеÄλÖÃÊÇ_______________£»ÔªËØYµÄÀë×ӽṹʾÒâͼΪ____________

£¨2£©ÏÂÁÐÓйØÎïÖÊÐÔÖʵıȽÏÖУ®ÕýÈ·µÄÊÇ________________

A£®M¡¢X¡¢ZÐγɵĵ¥ÖÊÈ۷еã M£¾X£¾Z

B£®Ç⻯ÎïµÄÈÈÎȶ¨ÐÔ£ºW£¾X

C£®X·Ö±ðÓëWºÍZÐγɵĻ¯ºÏÎïÖл¯Ñ§¼üÀàÐÍ£¬¾§ÌåÀàÐ;ùÏàͬ

D£®ZÔªËغ¬ÑõËáµÄËáÐÔÒ»¶¨Ç¿ÓÚYÔªËصĺ¬ÑõËá

£¨3£©ÔªËØWÓëM°´Ô­×Ó¸öÊý±È1:1ÐγɵĻ¯ºÏÎïAµÄµç×ÓʽΪ______________¡£

£¨4£©X¡¢ZÓëÇâÈýÖÖÔªËØ°´1:2:2ÐγɵÄÎåÔ­×Ó»¯ºÏÎïÊôÓÚ_______·Ö×Ó£¨Ìî ¡°¼«ÐÔ¡± ¡°·Ç¼«ÐÔ¡±£©

£¨5£©ÓëZͬ×åµÄÔªËض¼ÄÜÐγÉÇ⻯ÎÆäÖÐË®ÈÜÐÔ×îÇ¿µÄÊÇ______£¨Ìѧʽ£©

£¨6£©º¬ÓÐNÔªËصÄÁ½ÖÖ³£¼ûÀë×ÓÄÜ·¢ÉúË«Ë®½â·´Ó¦£¬ÊÔд³ö¸ÃÀë×Ó·½³Ìʽ ________________¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2012-2013ѧÄê°²»ÕÊ¡³ØÖÝÊиßÈýÉÏѧÆÚÆÚĩͳ¿¼Àí×Û»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÌî¿ÕÌâ

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____¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸