SnCl4ÔÚ´ß»¯ÔÓDiels£­aider·´Ó¦¡¢·Ó¼×õ£»¯·´Ö±ÖÐÓÐÖØÒªÓ¦Óã¬Ä³»¯Ñ§Ñо¿Ð¡×éÓû̽¾¿SnCl4µÄÖƱ¸£®

[²ÎÔÄ×ÊÁÏ]

SnCl4ÔÚ³£ÎÂÏÂÊÇÎÞÉ«ÒºÌ壬·ÐµãΪ114¡æ£¬Óö³±Êª¿ÕÆø±ã·¢ÉúË®½â·´Ó¦£¬SnµÄÈÛµãΪ231¡æ£®

[ʵÑé²½Öè]

µÚÒ»²½£º½ðÊôSnµÄÖƱ¸¡ª¡ªÒÔÎýʯSnO2ΪԭÁÏ£¬ÓùýÁ¿µÄ½¹Ì¿×÷»¹Ô­¼Á£¬ÔÚ¸ßÎÂÏ¿ÉÖƵôÖÎý£®

µÚ¶þ²½£ºSnCl4µÄÖƱ¸¡ª¡ªÓøÉÔï¡¢´¿¾»µÄC12ÓëÈÛÈÚµÄSn·´Ó¦ÖƱ¸SnCl4ͬʱ·Å³ö´óÁ¿µÄÈÈ£®

ÏÂͼÊÇʵÑéÊÒÖƱ¸SnCl4µÄʵÑé×°Öãº

[ÎÊÌâ˼¿¼]

(1)µÚÒ»²½·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ________£®

(2)×°ÖÃB¡¢CÖеÄÒ©Æ··Ö±ðΪ________£®

(3)µ±·´Ó¦Éú³ÉSnCl4ʱ£¬Ó¦Ï¨Ãð________´¦µÄ¾Æ¾«µÆ£¬ÀíÓÉÊÇ________£®

[ʵÑé¸Ä½ø]

ÀÏʦ˵װÖÃEÉè¼Æ²»ÍêÕû£¬ÇëÄãÌá³öÐÞ¸ÄÒâ¼û£¬²¢ËµÃ÷Ð޸ĵÄÄ¿µÄ£º________£®

´ð°¸£º
½âÎö£º

¡¡¡¡(1)SnO2£«2CSn£«2CO¡ü

¡¡¡¡(2)±¥ºÍʳÑÎË®¡¢Å¨HSO4

¡¡¡¡(3)D¡¡ÒÀ¿¿·´Ó¦·Å³öµÄÈÈÁ¿À´Î¬³ÖSnµÄÈÛ»¯£®

¡¡¡¡[ʵÑé¸Ä½ø]

¡¡¡¡×¶ÐÎÆ¿¼ÓÉÏË«¿×Èû£¬²¢Á¬½Ó×°Óмîʯ»ÒµÄ¸ÉÔï¹Ü£¬ÒÔ·ÀÖ¹¿ÕÆøÖеÄË®ÕôÆø½øÈë¼°ÎüÊÕ¹ýÁ¿Cl2


Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¶þ¼×ÃÑ£¨CH3OCH3£©ÊÇÒ»ÖÖ¸ßЧ¡¢Çå½à£¬¾ßÓÐÓÅÁ¼»·±£ÐÔÄܵÄÐÂÐÍȼÁÏ£®¹¤ÒµÉÏÖƱ¸¶þ¼×ÃÑʱÔÚ´ß»¯·´Ó¦ÊÒÖУ¨Ñ¹Ç¿2.0¡«10.0MPa£¬Î¶È230¡«280¡æ£©½øÐеķ´Ó¦Îª£º
¢ÙCO£¨g£©+2H2£¨g£©?CH3OH£¨g£©¡÷H1=-90.7kJ?mol-1£¬
¢Ú2CH3OH£¨g£©?CH3OCH3£¨g£©+H2O£¨g£©¡÷H2=-23.5kJ?mol-1£¬
¢ÛCO£¨g£©+H2O£¨g£©?CO2£¨g£©+H2£¨g£©¡÷H3=-41.2kJ?mol-1£¬
£¨1£©´ß»¯·´Ó¦ÊÒÖÐ×Ü·´Ó¦3CO£¨g£©+3H2£¨g£©?CH3OCH3£¨g£©+CO2£¨g£©µÄ¡÷H=
-246.1KJ/mol
-246.1KJ/mol
£®
£¨2£©ÔÚζȺÍÈÝ»ý²»±äµÄÌõ¼þÏ·¢Éú·´Ó¦¢Ù£¬ÄÜ˵Ã÷¸Ã·´Ó¦´ïµ½»¯Ñ§Æ½ºâ״̬µÄÒÀ¾ÝÊÇ
ab
ab
£¨¶àÑ¡¿Û·Ö£©£®
a£®ÈÝÆ÷ÖÐѹǿ±£³Ö²»±ä£»b£®»ìºÏÆøÌåÖÐc£¨CO£©²»±ä£»c£®vÕý£¨CO£©=vÄ棨H2£©£»d£®c£¨CH3OH£©=c£¨CO£©
£¨3£©ÔÚ2LµÄÈÝÆ÷ÖмÓÈëamolCH3OH£¨g£©·¢Éú·´Ó¦¢Ú£¬´ïµ½Æ½ºâºóÈôÔÙ¼ÓÈëamolCH3OH£¨g£©ÖØдﵽƽºâʱ£¬CH3OHµÄת»¯ÂÊ
²»±ä
²»±ä
£¨Ìî¡°Ôö´ó¡±¡¢¡°¼õС¡±»ò¡°²»±ä¡±£©£®
£¨4£©850¡æʱÔÚÒ»Ìå»ýΪ10LµÄÈÝÆ÷ÖÐͨÈëÒ»¶¨Á¿µÄCOºÍH2O£¨g£©·¢Éú·´Ó¦¢Û£¬COºÍH2O£¨g£©Å¨¶È±ä»¯ÈçͼËùʾ£®
¢Ù0¡«4minµÄƽ¾ù·´Ó¦ËÙÂʦԣ¨CO£©=
0.03mol/L?min
0.03mol/L?min
£®
¢ÚÈôζȲ»±ä£¬Ïò¸ÃÈÝÆ÷ÖмÓÈë4molCO¡¢2molH2O¡¢3molCO2£¨g£©ºÍ3molH2£¨g£©£¬ÆðʼʱDÕý
£¼
£¼
VÄ棨Ìî¡°£¼¡±¡¢¡°£¾¡±»ò¡°=¡±£©£¬Çë½áºÏ±ØÒªµÄ¼ÆËã˵Ã÷ÀíÓÉ
Qc=
0.3£¬mol/L¡Á0£¬3mol/L
0.4mol/L¡Á0.2mol/L
=
9
8
£¬K=
0.12mol/L¡Á0.12mol/L
0.08mol/L¡Á0.18mol/L
=1£¬ÒòΪQc£¾K£¬Æ½ºâ×óÒÆ£¬VÕý£¼VÄæ
Qc=
0.3£¬mol/L¡Á0£¬3mol/L
0.4mol/L¡Á0.2mol/L
=
9
8
£¬K=
0.12mol/L¡Á0.12mol/L
0.08mol/L¡Á0.18mol/L
=1£¬ÒòΪQc£¾K£¬Æ½ºâ×óÒÆ£¬VÕý£¼VÄæ
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡°±­·¼Ìþ¡±ÔÚ´ß»¯¡¢ÝÍÈ¡¡¢·Ö×Ó¿ª¹ØµÈ·½ÃæÓй㷺µÄÓÃ;£®¶ÔÊ嶡»ù±­[4]·¼Ìþ½á¹¹Èçͼ1£®
¾«Ó¢¼Ò½ÌÍø
£¨1£©¶ÔÊ嶡»ù±­[4]·¼ÌþÖÐ4¸öôÇ»ù¹¹³É±­µ×£¬4¸öôÇ»ù¼äµÄ½ÏÇ¿µÄ×÷ÓÃÁ¦ÊÇ
 
£®
£¨2£©¶ÔÊ嶡»ù±­[4]·¼ÌþÖУ¬Ì¼Ô­×ÓµÄÔÓ»¯·½Ê½ÓÐ
 

£¨3£©²»Í¬´óСµÄ¡°±­·¼Ìþ¡±ÄÜʶ±ðÖÐÐÔ·Ö×Ó»òÀë×Ó£¬Èç CO32-¡¢N3-£®
¢ÙCO32-µÄ¿Õ¼ä¹¹ÐÍÊÇ
 
£®
¢ÚÓëN3-»¥ÎªµÈµç×ÓÌåµÄÖÐÐÔ·Ö×ÓÓÐ
 
£¨¾ÙÒ»Àý£©£®
£¨4£©±­·¼Ìþ¿É¶ÔÏ¡ÍÁLa3+¼°Fe3+½øÐÐÝÍÈ¡£®Fe3+»ù̬ʱ£¬ºËÍâµç×ÓÅŲ¼Ê½Îª
 
£®
£¨5£©¶ÔÊ嶡»ù±­[4]·¼ÌþÓëîÑÀë×ÓÐγɵÄÅäºÏÎï½á¹¹Èçͼ2£¬ÆäÖÐÅäλԭ×ÓÊÇ
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¾«Ó¢¼Ò½ÌÍøÈçͼΪijÓлúÎïµÄ½á¹¹¼òʽ£®ÓÖÖªõ¥ÀàÔÚÍ­¸õÑõ»¯ÎCuO?CuCrO4£©´ß»¯Ï£¬ÓëÇâÆø·´Ó¦µÃµ½´¼£¬ôÊ»ùË«¼ü¿Éͬʱ±»»¹Ô­£¬µ«±½»·ÔÚ´ß»¯Ç⻯¹ý³ÌÖв»±ä£¬Æä·´Ó¦Ô­ÀíÈçÏ£ºRCOOR¡ä+2H2
Í­¸õÑõ»¯Îï
RCH2OH+HO-R¡ä¹ØÓڸû¯ºÏÎïµÄÏÂÁÐ˵·¨ÖУ¬ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A¡¢¸ÃÓлúÎïµÄ»¯Ñ§Ê½ÎªC20H14O5
B¡¢1mol¸ÃÓлúÎïÓëŨäåË®·´Ó¦£¬ÏûºÄ5molBr2
C¡¢Óë×ãÁ¿ÇâÑõ»¯ÄÆÈÜÒº³ä·Ö·´Ó¦£¬ÏûºÄ4molÇâÑõ»¯ÄÆ
D¡¢1mol¸ÃÓлúÎïÔÚÍ­¸õÑõ»¯Îï´ß»¯ÏÂÄÜÓë3molÇâÆø·¢Éú·´Ó¦

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º0119 Ô¿¼Ìâ ÌâÐÍ£ºÊµÑéÌâ

2007Äêŵ±´¶û»¯Ñ§½±ÊÚÓèµÂ¹ú¿Æѧ¼Ò¹þµÂ¡¤°£Ìضû£¬ÒÔ±íÕÃËûÔÚ±íÃ滯ѧÑо¿ÁìÓò×ö³öµÄ¿ªÍØÐÔ¹±Ïס£±íÃ滯ѧÁìÓòµÄÑо¿¿ÉÒÔ°ïÖúÎÒÃÇÁ˽âÖÚ¶à·×ÔӵĻ¯Ñ§¹ý³Ì£¬Èç¸ÖÌúΪºÎ»áÉúÐ⣬ȼÁϵç³ØÈçºÎ×÷ÓÃÒÔ¼°Ä³Ð©·´Ó¦µÄ´ß»¯¼ÁÊÇÈçºÎ·¢»ÓÆ书Äܵģ¬ÉõÖÁ¿ÉÒÔ½âÊͳôÑõ²ãµÄÏûºÄµÈ¡£
£¨1£©Ä³Ð£»¯Ñ§Ñо¿ÐÔѧϰС×éµÄͬѧÔÚ¼¼ÊõÈËÔ±µÄÖ¸µ¼Ï£¬°´ÏÂÁÐÁ÷³Ì̽¾¿²»Í¬´ß»¯¼Á¶ÔNH3»¹Ô­NO·´Ó¦µÄ´ß»¯ÐÔÄÜ¡£
Èô¿ØÖÆÆäËûʵÑéÌõ¼þ¾ùÏàͬ£¬ÔÚ´ß»¯·´Ó¦Æ÷ÖÐ×°Óв»Í¬µÄ´ß»¯¼Á£¬½«¾­´ß»¯·´Ó¦ºóµÄ»ìºÏÆøÌåͨ¹ýµÎÓзÓ̪µÄÏ¡ÁòËáÈÜÒº£¨ÈÜÒºµÄÌå»ýºÍŨ¶È¾ùÏàͬ£©¡£Îª±È½Ï²»Í¬´ß»¯¼ÁµÄ´ß»¯ÐÔÄÜ£¬ÐèÒª²âÁ¿²¢¼Ç¼µÄÊý¾ÝÊÇ________________¡£
£¨2£©ÔÚÆû³µµÄÅÅÆø¹ÜÉÏ°²×°¡°´ß»¯×ª»»Æ÷¡±£¨Óò¬¡¢îٺϽð×÷´ß»¯¼Á£©£¬ËüµÄ×÷ÓÃÊÇʹCO¡¢NO·´Ó¦Éú³É¿ÉÄÜÓë´óÆø»·¾³ÖÐÑ­»·µÄÎÞ¶¾ÆøÌ壬²¢´ÙʹÌþÀàÎïÖʳä·ÖȼÉÕ¡£
д³öCOºÍNO·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º________________£¬¸Ã·´Ó¦ÖÐ×÷Ñõ»¯¼ÁµÄÎïÖÊÊÇ___________¡£
ÓÃCH4´ß»¯»¹Ô­NOxÒ²¿ÉÏû³ýµªÑõ»¯ÎïµÄÎÛȾ¡£ÀýÈ磺
CH4£¨g£©+4NO2£¨g£©==4NO£¨g£©+CO2£¨g£©+2H2O£¨g£©£»¡÷H1=-574kJ/mol
CH4£¨g£©+4NO£¨g£©==2N2£¨g£©+CO2£¨g£©+2H2O£¨g£©£»¡÷H2
Èô1molCH4»¹Ô­NO2ÖÁN2£¬Õû¸ö¹ý³ÌÖзųöµÄÈÈÁ¿Îª867kJ£¬Ôò¡÷H2=________________¡£
£¨3£©¢ÙÓÐÈËÈÏΪ£º¸ÃÏîÑо¿¿ÉÄÜʹ°±µÄºÏ³É·´Ó¦ÔÚÌú´ß»¯¼Á±íÃæ½øÐÐʱЧÂÊ´ó´óÌá¸ß£¬´Ó¶øʹԭÁÏÀûÓÃÂÊ´ó´óÌá¸ß¡£ÇëÄãÓ¦Óû¯Ñ§»ù±¾ÀíÂ۶Դ˹۵ã½øÐÐÆÀ¼Û£º________________________________¡£
¢ÚºÏ³É°±¹¤ÒµÉú²úÖÐËùÓõĦÁ-Fe´ß»¯¼ÁµÄÖ÷Òª³É·ÝÊÇFeO¡¢Fe2O3£¬µ±´ß»¯¼ÁÖÐFe2+ÓëFe3+µÄÎïÖʵÄÁ¿Ö®±ÈΪ1©U2ʱ£¬Æä´ß»¯»îÐÔ×î´ó£¬ÒÔFe2O3ΪԭÁÏÖƱ¸ÉÏÊö´ß»¯¼Á£¬¿ÉÏòÆäÖмÓÈëÊÊÁ¿µÄÌ¿·Û£¬·¢Éú·´Ó¦ÈçÏ£º2Fe2O3+C==4FeO+CO2ΪÖƵÃÕâÖÖ»îÐÔ×î´óµÄ´ß»¯¼Á£¬Ó¦Ïò480g Fe2O3·ÛÄ©ÖмÓÈëÌ¿·ÛµÄÖÊÁ¿ÊÇ
________________

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸