ÌâÄ¿ÄÚÈÝ


ÒÑÖª£ºÁ½¸öôÇ»ùͬʱÁ¬ÔÚͬһ̼ԭ×ÓÉϵĽṹÊDz»Îȶ¨µÄ£¬Ëü½«·¢ÉúÍÑË®·´Ó¦£º

ÏÖÓзÖ×ÓʽΪC9H8O2Br2µÄÎïÖÊM£¬ÔÚÒ»¶¨Ìõ¼þÏ¿ɷ¢ÉúÏÂÊöһϵÁз´Ó¦£º

ÒÑÖª£ºÓлúÎïAµÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª44£¬ÓлúÎïIÖ»ÓÐÒ»ÖֽṹÇÒÄÜʹäåµÄCCl4ÈÜÒºÍÊÉ«¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©G¡úHµÄ·´Ó¦ÀàÐÍÊÇ                ¡£

£¨2£©HÖеĹÙÄÜÍŵÄÃû³ÆÎª                    £»DµÄ½á¹¹¼òʽΪ               ¡£

£¨3£©Ð´³öÏÂÁз´Ó¦µÄ»¯Ñ§·½³Ìʽ£º

¢ÙA¡ú B£º                               £»

¢ÚH¡ú I£º                                ¡£

£¨4£©ÓëG»¥ÎªÍ¬·ÖÒì¹¹Ì壬±½»·ÉÏÓÐÁ½¸öÈ¡´ú»ù£¬ÇÒÓöFeCl3ÈÜÒºÏÔÉ«µÄÎïÖÊÓР     ÖÖ


£¨1£©¼Ó³É·´Ó¦ £¨»ò»¹Ô­·´Ó¦£©£¨2·Ö£©£»

£¨2£©ôÇ»ù¡¢ôÈ»ù£»   £¨4·Ö£¬¸÷2·Ö£©£»

£¨3£© ¢ÙCH3CHO£«2Ag(NH3)2OH 2Ag¡ý£«CH3COONH4£«3NH3£«H2O£»

¢Ú

£¨4·Ö£¬¸÷2·Ö£©

£¨4£©5£¨2·Ö£©


Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

A¡¢B¡¢C¡¢D¡¢E¡¢F¡¢G¶¼ÊǶÌÖÜÆÚÖ÷×åÔªËØ£¬ËüÃǵĺ˵çºÉÊýÒÀ´ÎÔö´ó¡£ÔªËØAµÄÔ­×ÓºËÄÚ½öÓÐÒ»¸öÖÊ×Ó£¬A¡¢DͬÖ÷×壬B¡¢CΪͬÖÜÆÚÔªËØ£¬ÇÒÓëAÄÜÐγÉÏàͬµç×ÓÊýµÄ»¯ºÏÎCÓëFͬÖ÷×壬FµÄÖÊ×ÓÊýΪCµÄ2±¶£¬ÔªËØEµÄ×îÍâ²ãµç×ÓÊý±ÈK²ãµç×ÓÊý¶à1£¬B¡¢C¡¢FµÄÖÊ×ÓÊýÖ®ºÍµÈÓÚE¡¢GµÄÖÊ×ÓÊýÖ®ºÍ¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺

(1)д³öÓÉÉÏÊöÖÁÉÙÁ½ÖÖÔªËØ×é³ÉµÄ¾ßÓÐÆ¯°××÷ÓõÄÎïÖʵĻ¯Ñ§Ê½________(ÖÁÉÙд³öÁ½ÖÖ)¡£

(2)A·Ö±ðÓëB¡¢C¡¢GÄÜÐγÉÏàÓ¦µÄ×î³£¼ûÈýÖÖ»¯ºÏÎÕâÈýÖÖ»¯ºÏÎïµÄ·ÐµãÓɸߵ½µÍµÄ˳ÐòΪ__________(Óû¯Ñ§Ê½±íʾ)¡£

(3)A¡¢C¡¢GÄÜÐγÉÒ»ÖÖ3Ô­×Ó·Ö×Ó£¬ÊÔд³öËüµÄµç×Óʽ                 £¬CÓëDÄÜÐγÉÒ»ÖÖÏà¶Ô·Ö×ÓÖÊÁ¿Îª78µÄ»¯ºÏÎËüµÄµç×ÓʽΪ                £»¢ÚÖÐËùº¬µÄ»¯Ñ§¼üÓУº                          ¡£

(4) D¡¢EÁ½ÖÖÔªËØµÄÔ­×Ó¶¼ÄÜÐγÉÏàÓ¦µÄ¼òµ¥Àë×Ó£¬ÔòÁ½Àë×Ó°ë¾¶´óС¹ØÏµÎª________(ÓÃÀë×Ó·ûºÅ±íʾ)£»½«D¡¢EÁ½ÖÖÔªËØµÄµ¥ÖÊͬʱͶÈëË®ÖУ¬³ä·Ö·´Ó¦ºó£¬²âµÃÈÜÒºÖÐÖ»ÓÐÒ»ÖÖÈÜÖÊ£¬ÇÒÎÞ¹ÌÌåÎïÖÊÊ£Ó࣬ÔòËùͶÈ뵽ˮÖеÄDµÄµ¥ÖʺÍEµÄµ¥ÖʵÄÖÊÁ¿Ö®±ÈΪ________£¬ÉÏÊö·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ________________¡£


   ÖйúÆøÏó¾ÖµÄÊý¾ÝÏÔʾ£¬2013ÄêÈ«¹úƽ¾ùÎíö²ÌìÊýΪ52ÄêÀ´Ö®×î¡£ÐγÉÎíö²µÄÖ÷Òª³É·ÝΪ£ºÉú²úÉú»îÖÐÅÅ·ÅµÄ·ÏÆø¡¢Æû³µÎ²Æø¼°Ñï³¾µÈ¡£

£¨1£©ÓÃCH4¿ÉÒÔÏû³ýÆû³µÎ²ÆøÖеªÑõ»¯ÎïµÄÎÛȾ¡£

ÒÑÖª£ºCH4(g)+2NO2(g)£½N2(g)£«CO2(g)+2H2O(l)  ¡÷H£½£­955 kJ¡¤mol£­1

2NO2(g)£½N2O4(g)   ¡÷H£½£­56.9 kJ¡¤mol£­1

д³öCH4´ß»¯»¹Ô­N2O4(g)Éú³ÉN2ºÍH2O(l)µÄÈÈ»¯Ñ§·½³Ìʽ                   ¡£

£¨2£©ÒÑÖª£ºCO(g)£«H2O(g)CO2(g)£«H2(g)   ¡÷H£½£­41kJ¡¤mol£­1£¬Ä³Î¶ÈÏ£¬ÏòÈÝ»ýΪ2LµÄÃܱÕÈÝÆ÷ÖгäÈë2.0molCO(g)ºÍ2.0molH2O(g)£¬ÔÚtminʱ´ïµ½Æ½ºâ£¬²âµÃ·Å³öÁË32.8kJÈÈÁ¿£¬Ôòt minÄÚÓÃH2±íʾµÄƽ¾ù·´Ó¦ËÙÂÊΪ         £¬ÓÉ´Ë¿ÉÖªÔÚ¸ÃζÈÏ·´Ó¦CO2(g)£«H2(g)CO(g)£«H2O(g)µÄ»¯Ñ§Æ½ºâ³£ÊýΪ         ¡£ÏàͬÌõ¼þÏ£¬ÏòͬһÃܱÕÈÝÆ÷ÖгäÈë1.0molCO2ºÍ1.0molH2·´Ó¦´ïµ½Æ½ºâºó£¬ÎüÊÕµÄÈÈÁ¿Îª       kJ¡£

£¨3£©¼îʽÁòËáÂÁ·¨ÑÌÆøÍÑÁò¹¤ÒÕÖ÷ÒªÓÐÒÔÏÂÈý²½

¢ÙÏòAl2(SO4)3ÈÜÒºÖÐͶÈë·Ûĩ״ʯ»Òʯ£¬Éú³É¼îʽÁòËáÂÁ[Al2(SO4)3¡¤Al2O3]ÈÜÒº¡£

¢Ú¼îʽÁòËáÂÁÎüÊÕSO2£¬Al2(SO4)3¡¤Al2O3+3SO2£½Al2(SO4)3¡¤Al2(SO3)3£¬Çëд³öAl2(SO4)3¡¤Al2O3Óë¹ýÁ¿ÉÕ¼îÈÜÒº·´Ó¦µÄ»¯Ñ§·½³Ìʽ                     ¡£

¢Û½«Al2(SO4)3¡¤Al2 (SO3)3Ñõ»¯³ÉAl2(SO4)3£¬¿ÉÑ¡ÓÃÑõ»¯¼ÁΪ          £¨Ìî´úºÅ£©

a. Å¨ÁòËá       b. KMnO4ÈÜÒº           c. 5%µÄH2O2ÈÜÒº     d. ¿ÕÆø

¸Ã²½·´Ó¦µÄÄ¿µÄÊÇ                               ¡£

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

¾«Ó¢¼Ò½ÌÍø