ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿Ìú¼°Æ仯ºÏÎïÔÚÉú²úºÍÉú»îÖÐÓÐ׏㷺µÄÓ¦Óá£

(1)ijÑо¿ÐÔѧϰС×éÉè¼ÆÁËÈçÏÂͼËùʾװÖÃ̽¾¿¸ÖÌúµÄ¸¯Ê´Óë·À»¤£¬ÉÕ±­ÄÚÒºÌå¾ùΪ±¥ºÍʳÑÎË®¡£

¢ÙÔÚÏàͬÌõ¼þÏ£¬Èý×é×°ÖÃÖÐÌúµç¼«¸¯Ê´×î¿ìµÄÊÇ__(Ìî×°ÖÃÐòºÅ)£¬¸Ã×°ÖÃÖÐÕý¼«·´Ó¦Ê½Îª__¡£

¢ÚΪ·ÀÖ¹½ðÊôFe±»¸¯Ê´£¬¿ÉÒÔ²ÉÓÃÉÏÊö___(Ìî×°ÖÃÐòºÅ)×°ÖÃÔ­Àí½øÐзÀ»¤¡£

(2)ͨÐÅÓÃÁ×ËáÌú﮵ç³ØÓÐÌå»ýС¡¢ÖØÁ¿Çá¡¢¸ßÎÂÐÔÄÜÍ»³ö¡¢¿É¸ß±¶Âʳä·Åµç¡¢ÂÌÉ«»·±£µÈÖÚ¶àÓŵ㡣Á×ËáÌú﮵ç³ØÊÇÒÔÁ×ËáÌúï®ÎªÕý¼«²ÄÁϵÄÒ»ÖÖï®Àë×Ó¶þ´Îµç³Ø£¬·Åµçʱ£¬Õý¼«·´Ó¦Ê½ÎªM1-xFexPO4+e-+Li+=LiM1-xFexPO4£¬ÆäÔ­ÀíÈçͼËùʾ¡£

¢Ù·Åµçʱ£¬µçÁ÷ÓÉ___µç¼«¾­¸ºÔØÁ÷Ïò___µç¼«£»¸º¼«·´Ó¦Ê½Îª____¡£

¢Ú¸Ãµç³Ø¹¤×÷ʱLi+ÒÆÏò___µç¼«£»³äµçʱʯīµç¼«½ÓµçÔ´µÄ____¼«¡£

¢Û¸Ãµç³ØµÄ×Ü·´Ó¦·½³ÌʽΪ____¡£

¡¾´ð°¸¡¿¢Ù O2+2H2O+4e£­ = 4OH£­ ¢Ú¢Û Á×ËáÌúï® Ê¯Ä« LiC6£­e£­ = Li++6C Á×ËáÌúï® ¸º M1-xFexPO4+LiC6XLiM1-xFexPO4+6C

¡¾½âÎö¡¿

(1)¢ÙÖÐÌú×÷¸º¼«£¬Í­ÎªÕý¼«£¬¢ÚÖÐпΪ¸º¼«£¬ÌúΪÕý¼«£¬ÊÇÎþÉüÑô¼«µÄÒõ¼«±£»¤·¨£»¢ÛÖÐÌúΪÒõ¼«£¬Ê¯Ä«ÎªÑô¼«£¬ÊÇÍâ¼ÓµçÔ´µÄÒõ¼«±£»¤·¨¡£

(2)¢Ù·Åµçʱ£¬Ê¯Ä«µç¼«LiC6ʧȥµç×Ó±äΪï®Àë×Ó£¬Îª¸º¼«£¬×ó±ßΪÕý¼«£»¢ÚÔ­µç³ØÖÐÑôÀë×ÓÏòÕý¼«Òƶ¯£¬ÒõÀë×ÓÏò¸º¼«Òƶ¯£»¢Û½«Õý¸º¼«µç¼«·´Ó¦Ê½Ïà¼ÓµÃµ½×Ü·´Ó¦·½³Ìʽ¡£

(1)¢ÙÔÚÏàͬÌõ¼þÏ£¬¢ÙÖÐÌú×÷¸º¼«£¬Í­×÷Õý¼«£¬Ìú±»¸¯Ê´£¬¸¯Ê´½Ï¿ì£¬¢ÚÖÐпΪ¸º¼«£¬ÌúΪÕý¼«£¬ÌúÊܵ½Ò»¶¨±£»¤£¬ÊÇÎþÉüÑô¼«µÄÒõ¼«±£»¤·¨£¬¢ÛÖÐÌúΪÒõ¼«£¬Ê¯Ä«ÎªÑô¼«£¬ÌúÊܵ½±£»¤£¬ÊÇÍâ¼ÓµçÔ´µÄÒõ¼«±£»¤·¨£¬Òò´ËÈý×é×°ÖÃÖÐÌúµç¼«¸¯Ê´×î¿ìµÄÊÇ¢Ù£¬¸Ã×°ÖÃÊÇÎüÑõ¸¯Ê´£¬Òò´ËÕý¼«·´Ó¦Ê½ÎªO2+2H2O+4e£­ = 4OH£­£»¹Ê´ð°¸Îª£º¢Ù£»O2+2H2O+4e£­ = 4OH£­¡£

¢Ú¸ù¾ÝÇ°Ãæ·ÖÎöµÃµ½Îª·ÀÖ¹½ðÊôFe±»¸¯Ê´£¬¿ÉÒÔ²ÉÓÃÉÏÊö¢Ú¢Û×°ÖÃÔ­Àí½øÐзÀ»¤£»¹Ê´ð°¸Îª£º¢Ú¢Û¡£

(2)¢Ù·Åµçʱ£¬Ê¯Ä«µç¼«LiC6ʧȥµç×Ó±äΪï®Àë×Ó£¬×÷¸º¼«£¬Òò´ËµçÁ÷ÓÉÁ×ËáÌúï®(Õý¼«)µç¼«¾­¸ºÔØÁ÷Ïòʯī(¸º¼«)µç¼«£»¸º¼«·´Ó¦Ê½ÎªLiC6£­e£­ = Li++6C£»¹Ê´ð°¸Îª£ºÁ×ËáÌúﮣ»Ê¯Ä«£»LiC6£­e£­ = Li++6C¡£

¢ÚÔ­µç³ØÖÐÑôÀë×ÓÏòÕý¼«Òƶ¯£¬ÒõÀë×ÓÏò¸º¼«Òƶ¯£»Òò´Ë¸Ãµç³Ø¹¤×÷ʱLi+ÒÆÏòÕý¼«¼´Á×ËáÌú﮵缫£»·Åµçʱ£¬Ê¯Ä«µç¼«Îª¸º¼«£¬Ôò³äµçʱʯīµç¼«½ÓµçÔ´µÄ¸º¼«£»¹Ê´ð°¸Îª£ºÁ×ËáÌúﮣ»¸º¡£

¢Û¸Ãµç³ØµÄ¸º¼«·´Ó¦ÊÇLiC6£­e£­ = Li++6C£¬Õý¼«·´Ó¦ÊÇM1-xFexPO4+e£­+Li+=LiM1-xFexPO4£¬Òò´Ë×Ü·´Ó¦·½³ÌʽΪM1-xFexPO4+LiC6XLiM1-xFexPO4+6C£»¹Ê´ð°¸Îª£ºM1-xFexPO4+LiC6XLiM1-xFexPO4+6C¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿¼×´¼ÊÇÖØÒªµÄ»¯¹¤Ô­ÁÏ£¬ÔÚÓлúºÏ³ÉÖоßÓй㷺ӦÓá£

(1)Óü״¼ÖÆÈ¡¼×°·µÄ·´Ó¦Îª£ºCH3OH(g)+NH3(g)CH3NH2(g)+H2O(g)¡÷H

ÒÑÖª¸Ã·´Ó¦ÖÐÏà¹Ø»¯Ñ§¼üµÄ¼üÄÜÊý¾ÝÈçÏ£º

¹²¼Û¼ü

C¨DO

H¨DO

N¨DH

C¨DN

¼üÄÜ£¯kJ¡¤mol-1

351

463

393

293

Ôò¸Ã·´Ó¦µÄ¡÷H=______kJ¡¤mol-1

(2)Ò»¶¨Ìõ¼þÏ£¬½«2mol COºÍ6mol H2ͨÈë2LÃܱÕÈÝÆ÷Öз¢ÉúÈçÏ·´Ó¦£º

Ö÷·´Ó¦£ºCO(g)+2H2(g)CH3OH(g)¡÷H<0 ¢ñ

¸±·´Ó¦£º2CH3OH(g)CH3OCH3(g)+H2O(g)¡÷H<0 ¢ò

·´Ó¦µ½t minʱ£¬´ïµ½Æ½ºâ״̬¡£Æ½ºâʱCH3OHµÄÌå»ý·ÖÊý¦Õ(CH3OH)Ëæζȡ¢Ñ¹Ç¿µÄ±ä»¯ÈçͼËùʾ£º

¢ÙͼÖÐa___b(Ìî¡°´óÓÚ¡±»ò¡°Ð¡ÓÚ¡±)¡£Í¼ÖÐYÖá±íʾζȣ¬ÆäÀíÓÉÊÇ_________£»

¢ÚÈô·´Ó¦IIµÄƽºâ³£ÊýKÖµ±äС£¬ÔòÏÂÁÐ˵·¨ÖÐÕýÈ·µÄÊÇ___________(ÌîÐòºÅ)¡£

A£®Æ½ºâ¾ùÏòÕý·´Ó¦·½ÏòÒƶ¯ B£®Æ½ºâÒƶ¯µÄÔ­ÒòÊÇÉý¸ßÁËζÈ

C£®´ïµ½ÐÂƽºâºó£¬¦Õ(CH3OH)¼õС D£®ÈÝÆ÷ÖЦÕ(CH3OCH3)Ôö´ó

¢Ûƽºâʱ£¬MµãCH3OHµÄÌå»ý·ÖÊýΪ12.5£¥£¬c(CH3OCH3)=0.1mol¡¤L-1£¬Ôò´ËʱCOµÄת»¯ÂÊΪ_____£»ÓÃH2±íʾIµÄ·´Ó¦ËÙÂÊΪ_____mol¡¤L-1¡¤min-1¡£

(3)ÓÃNaOHÈÜÒº×öCO2̼²¶×½¼Á£¬ÔÚ½µµÍ̼ÅŷŵÄͬʱҲ»ñµÃÁËÖØÒªµÄ»¯¹¤²úÆ·Na2CO3¡£³£ÎÂÏ£¬Èôij´Î²¶×½ºóµÃµ½pH=11µÄÈÜÒº£¬ÔòÈÜÒºÖÐc()¡Ãc()=___________[ÒÑÖªH2CO3µÄµçÀëƽºâ³£ÊýΪ£ºK1=4.4¡Á107¡¢K2=5¡Á1011]£¬ÈÜÒºÖÐc(Na+)_______ c()+2c()(Ìî¡°>¡±¡°<¡±»ò¡°=¡±)¡£

¡¾ÌâÄ¿¡¿ÒÒõ£±ûͪͭÊǽðÊôÓлú»¯ºÏÎïÖÐÒ»ÖÖÖØÒªµÄ»¯ºÏÎ¹ã·ºÓ¦ÓÃÓÚ»¯¹¤¡¢Ê¯ÓÍ¡¢ÖÆÒ©¡¢µç×Ó¡¢²ÄÁÏ¡¢»úеµÈÁìÓò¡£ÊµÑéÊÒÖƱ¸Ô­ÀíÈçÏ£º

ʵÑé²½Ö裺

(1)ÖÆÈ¡ÇâÑõ»¯Í­(II)

·Ö±ð³ÆÈ¡4.000g(0.1mol)ÇâÑõ»¯ÄÆ¡¢8.000g(0.05mol)ÎÞË®ÁòËáÍ­ÓÚ250mL¡¢100mLÉÕ±­ÖУ¬¼ÓÈëÊÊÁ¿µÄÕôÁóˮʹÆäÈܽ⣬Ȼºó½«ÁòËáÍ­ÈÜÒºµ¹ÈëÇâÑõ»¯ÄÆÈÜÒºÖУ¬Ò¡ÔÈ£¬Ê¹·´Ó¦ÍêÈ«£¬ÔÙ½«³Áµí½øÐгéÂË¡£

(2)ÖƱ¸ÒÒõ£±ûͪͭ(II)

³ÆÈ¡0.1960g(2mmol)ÐÂÖƵÄÇâÑõ»¯Í­ÓÚ100mLÒÇÆ÷aÖУ¬ÔÚµªÆø±£»¤Ï¼ÓÈëÉÙÁ¿µÄËÄÇâ߻ૣ¬²¢½øÐнÁ°è£¬Ô¼Îå·ÖÖÓÖ®ºó£¬¼ÓÈë0.4000g(4mmol)ÒÒõ£±ûͪ£¬²¹³äËÄÇâß»à«Ô¼30mL£¬ÔÚ50¡æϼÓÈÈ»ØÁ÷Ô¼2h£¬È»ºóÀäÈ´ÖÁÊÒΣ¬×ªÒƵ½×¶ÐÎÆ¿ÖУ¬Óñ¡Ä¤·â¿Ú£¬·ÅÖÃ4¡«5Ì죬µÃµ½À¶É«Õë×´¾§Ìå¡£

ÒÑÖª£º

I£®ÇâÑõ»¯Í­(II)·Ö½âζÈΪ60¡æ¡£

II£®ËÄÇâß»à«Ò×»Ó·¢£¬·Ðµã66¡æ£¬´¢´æʱӦ¸ô¾ø¿ÕÆø£¬·ñÔòÒ×±»Ñõ»¯³É¹ýÑõ»¯Îï¡£

III£®ÒÒõ£±ûͪͭ(II)ÊÇÒ»ÖÖÀ¶É«Õë×´¾§Ì壬ÄÑÈÜÓÚË®£¬Î¢ÈÜÓÚÒÒ´¼£¬Ò×ÈÜÓÚ±½¡¢Âȷ¡¢ËÄÂÈ»¯Ì¼¡£66.66kPaѹÁ¦Ï£¬78¡æÉý»ª¡£

»Ø´ðÏÂÁÐÎÊÌ⣺

(1)ÖÆÈ¡ÇâÑõ»¯Í­(II)ʱ²ÉÓóéÂ˵ķ½Ê½½øÐУ¬ÒÑÖª³éÂË×°ÖÃÈçͼËùʾ£¬³éÂËÓë³£¹æ¹ýÂËÏà±ÈÆäÓŵãÊÇ____________________¡£

(2)ÖƱ¸ÒÒõ£±ûͪͭ(II)ʱ²ÉÓõªÆø±£»¤µÄÄ¿µÄ________________£¬¼ÓÈëËÄÇâ߻૵Ä×÷ÓÃÊÇ__________________¡£

(3)¼ÓÈÈ»ØÁ÷¼òÒ××°ÖÃÈçͼ(¼ÓÈȲ¿·ÖÊ¡ÂÔ)Ëùʾ£¬ÒÇÆ÷aµÄÃû³ÆÊÇ_______________£¬¼ÓÈÈ»ØÁ÷ʱ²ÉÓÃÇòÐÎÀäÄý¹Ü¶ø²»Ñ¡ÓÃÖ±ÐÎÀäÄý¹ÜµÄÔ­ÒòÊÇ________________¡£¼ÓÈÈ·½Ê½Ò˲ÉÓÃ_______________¡£ÔÚ50¡æϼÓÈÈ»ØÁ÷µÄÔ­Òò³ý·ÀÖ¹ËÄÇâ߻ૻӷ¢Í⣬»¹ÓпÉÄܵÄÔ­ÒòÊÇ_____________(Óû¯Ñ§·½³Ìʽ±íʾ)¡£

(4)ÈôÒª´¿»¯ÒÒõ£±ûͪͭ¾§Ì壬¿ÉÒÔ²ÉÓõÄʵÑé·½·¨ÊÇ___________¡£Èô´¿»¯ºóÀ¶É«Õë×´¾§ÌåµÄÖÊÁ¿Îª0.3630g(ÒÒõ£±ûͪͭµÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª262)£¬Ôò²úÂÊΪ________%¡£

¡¾ÌâÄ¿¡¿µª»¯ÂÁ(AlN)ÊÇÒ»ÖÖÐÔÄÜÓÅÒìµÄÐÂÐͲÄÁÏ£¬ÔÚÐí¶àÁìÓòÓй㷺ӦÓã¬Ç°¾°¹ãÀ«¡£Ä³»¯Ñ§Ð¡×éÄ£Ä⹤ҵÖƵª»¯ÂÁÔ­ÀíÓûÔÚʵÑéÊÒÖƱ¸µª»¯ÂÁ²¢¼ìÑéÆä´¿¶È¡£

²éÔÄ×ÊÁÏ£º¢ÙʵÑéÊÒÓñ¥ºÍNaNO2ÈÜÒºÓëNH4C1ÈÜÒº¹²ÈÈÖÆN2£ºNaNO2+NH4Cl NaCl+N2¡ü+2H2O

¢Ú¹¤ÒµÖƵª»¯ÂÁ£ºAl2O3+3C+N2 2AlN+3CO£¬µª»¯ÂÁÔÚ¸ßÎÂÏÂÄÜË®½â¡£

¢ÛAlNÓëNaOH±¥ºÍÈÜÒº·´Ó¦£ºAlN+NaOH+H2O= NaAlO2+NH3¡ü¡£

I£®µª»¯ÂÁµÄÖƱ¸

(1)ʵÑéÖÐʹÓõÄ×°ÖÃÈçÉÏͼËùʾ£¬Çë°´ÕÕµªÆøÆøÁ÷·½Ïò½«¸÷ÒÇÆ÷½Ó¿ÚÁ¬½Ó£ºe¡úc¡úd¡ú___(¸ù¾ÝʵÑéÐèÒª£¬ÉÏÊö×°ÖÿÉʹÓöà´Î)¡£

(2)B×°ÖÃÄÚµÄXÒºÌå¿ÉÄÜÊÇ____£¬E×°ÖÃÄÚÂÈ»¯îÙÈÜÒºµÄ×÷ÓÿÉÄÜÊÇ_____¡£

¢ò£®µª»¯ÂÁ´¿¶È(º¬Al2O3¡¢CÔÓÖÊ)µÄ²â¶¨

(·½°¸i)¼×ͬѧÓÃÏÂͼװÖòⶨAlNµÄ´¿¶È(²¿·Ö¼Ð³Ö×°ÖÃÒÑÂÔÈ¥)£¬ÇÒ²»¿¼ÂÇNH3ÔÚNaOHŨÈÜÒºÖеÄÈܽ⡣

(3)Ϊ׼ȷ²â¶¨Éú³ÉÆøÌåµÄÌå»ý£¬Á¿Æø×°ÖÃ(ÐéÏß¿òÄÚ)ÖеÄYÒºÌå¿ÉÒÔÊÇ___¡£

a£®CCl4 b£®H2O c£® NH4Cl±¥ºÍÈÜÒº d£® Ö²ÎïÓÍ

(4)Èô×°ÖÃÖзÖҺ©¶·Óëµ¼Æø¹ÜÖ®¼äûÓе¼¹ÜAÁ¬Í¨£¬¶ÔËù²âAlN´¿¶ÈµÄÓ°ÏìÊÇ____(Ìî¡°Æ«´ó¡±¡°Æ«Ð¡¡±»ò¡°²»±ä¡±)

(·½°¸ii)ÒÒͬѧÈÏΪÉÏÊö·½°¸²»¿ÉÐУ¬Éè¼ÆÒÔϲ½Öè²â¶¨ÑùÆ·ÖÐAlNµÄ´¿¶È¡£

(5)²½Öè¢ÚͨÈë¹ýÁ¿_____(Ìѧʽ)ÆøÌå¡£

(6)²½Öè¢Û¹ýÂËËùÐèÒªµÄÖ÷Òª²£Á§ÒÇÆ÷ÓÐ______¡£

(7)ÑùÆ·ÖÐAlNµÄ´¿¶ÈÊÇ_____(Óú¬m1¡¢m2¡¢m3µÄ±í´ïʽ±íʾ)¡£

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

¾«Ó¢¼Ò½ÌÍø