ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿ÂÈ»ÇËá(HSO3Cl)ÔÚÌǾ«¡¢»Ç°·Ò©µÄÉú²úÖÐÓÐÖØÒªµÄÓ¦Óᣳ£Î³£Ñ¹ÏÂÂÈ»ÇËáΪÎÞÉ«ÓÍ×´ÒºÌ壬·ÐµãԼΪ152¡æ£¬ÎüʪÐԺ͸¯Ê´ÐÔ¼«Ç¿£¬ÔÚ¿ÕÆøÖз¢ÑÌ¡£Ñ§Ï°Ð¡×éÔÚʵÑéÊÒÓÃSO3ºÍHClÀ´ÖƱ¸HSO3Cl²¢²â¶¨²úÆ·´¿¶È¡£Éè¼ÆÈçÏÂʵÑé(¼Ð³Ö×°ÖÃÂÔÈ¥)¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©ÒÇÆ÷mµÄÃû³ÆΪ___¡£

£¨2£©ÒÑÖª£ºHSO3ClÖÐÁòÔªËØΪ£«6¼Û£¬OÔ­×ÓºÍClÔ­×ÓµÄ×îÍâ²ã¾ùÂú×ã8µç×ÓÎȶ¨½á¹¹£¬ÔòHSO3ClÖеĻ¯Ñ§¼üΪ___ (Ìî¡°Àë×Ó¼ü¡±¡¢¡°¼«ÐÔ¼ü¡±»ò¡°·Ç¼«ÐÔ¼ü¡±)¡£

£¨3£©SO3¿ÉÓÉÎåÑõ»¯¶þÁ×ÓëŨÁòËá¹²ÈÈÖƱ¸£¬·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ___¡£

£¨4£©×°ÖÃBµÄ×÷ÓÃΪ___¡£ÓÉÒÇÆ÷n¿ÉÖªÖƱ¸HSO3ClµÄ·´Ó¦Îª___ (Ìî¡°·ÅÈÈ·´Ó¦¡±»ò¡°ÎüÈÈ·´Ó¦¡±)¡£

£¨5£©HSO3Cl´¿¶ÈµÄ²â¶¨(ÒÇÆ÷mÖеõ½µÄHSO3ClÖг£ÈÜÓÐÉÙÁ¿µÄSO3)£º

i.È¡25.0g²úÆ·ÈÜÓÚË®ÖУ¬¼ÓÈë¹ýÁ¿µÄBa(NO3)2ÈÜÒº³ä·Ö·´Ó¦ºó£¬¹ýÂË¡£

ii.ÏòÂËÒºÖмÓÈë¹ýÁ¿µÄAgNO3ÈÜÒº£¬¹ýÂË¡¢Ï´µÓ¡¢¸ÉÔï¡¢³ÆÁ¿£¬²âÁ¿ËùµÃ³ÁµíAgClµÄÖÊÁ¿Îª28.7g¡£

¢ÙHSO3ClÓöË®·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ___¡£

¢Ú²úÆ·HSO3ClµÄ´¿¶ÈΪ___¡£

¡¾´ð°¸¡¿Èý¾±ÉÕÆ¿ ¼«ÐÔ¼ü P2O5£«3H2SO4£¨Å¨£©3SO3¡ü£«2H3PO4 ·ÀÖ¹¿ÕÆøÖеÄË®ÕôÆø½ø È뷴ӦװÖÃAÖУ¬²¢ÎüÊÕ»Ó·¢³öÀ´µÄHClÆøÌå ·ÅÈÈ·´Ó¦ HSO3Cl+H2O=HCl+H2SO4 93.2%

¡¾½âÎö¡¿

£¨1£©ÒÇÆ÷mµÄÃû³ÆΪÈý¾±ÉÕÆ¿¡£¹Ê´ð°¸Îª£ºÈý¾±ÉÕÆ¿£»

£¨2£©HSO3ClÖÐÁòÔªËØΪ£«6¼Û£¬OÔ­×ÓºÍClÔ­×ÓµÄ×îÍâ²ã¾ùÂú×ã8µç×ÓÎȶ¨½á¹¹£¬½á¹¹Ê½Îª£¬ÔòHSO3ClÖеĻ¯Ñ§¼üΪ¼«ÐÔ¼ü ¡£¹Ê´ð°¸Îª£º¼«ÐÔ¼ü£»

£¨3£©SO3¿ÉÓÉÎåÑõ»¯¶þÁ×ÓëŨÁòËá¹²ÈÈÖƱ¸£¬·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪP2O5£«3H2SO4£¨Å¨£©3SO3¡ü£«2H3PO4¡£¹Ê´ð°¸Îª£ºP2O5£«3H2SO4£¨Å¨£©3SO3¡ü£«2H3PO4£»

£¨4£©HSO3ClÓëË®»á¾çÁÒ·´Ó¦£¬ÓÉÒÇÆ÷mÖÐÒݳöµÄHClÎÛȾ¿ÕÆø£¬×°ÖÃBµÄ×÷ÓÃΪ·ÀÖ¹¿ÕÆøÖеÄË®ÕôÆø½øÈ뷴ӦװÖÃAÖУ¬²¢ÎüÊÕ»Ó·¢³öÀ´µÄHClÆøÌå¡£ÒÇÆ÷nµÄ×÷ÓÃÊÇÀäÄý»ØÁ÷HSO3Cl£¬¿ÉÖªÖƱ¸HSO3ClµÄ·´Ó¦Îª·ÅÈÈ·´Ó¦¡£¹Ê´ð°¸Îª£º·ÀÖ¹¿ÕÆøÖеÄË®ÕôÆø½øÈ뷴ӦװÖÃAÖУ¬²¢ÎüÊÕ»Ó·¢³öÀ´µÄHClÆøÌ壻·ÅÈÈ·´Ó¦£»

£¨5£©¢ÙHSO3ClÓöË®·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪHSO3Cl+H2O=HCl+H2SO4¡£

¢Ú¸ù¾ÝÌâÖÐÐÅÏ¢£ºHSO3Cl~AgCl£¬Éú³ÉHSO3ClµÄ´¿¶ÈΪ=93.2%¡£

¹Ê´ð°¸Îª£ºHSO3Cl+H2O=HCl+H2SO4£»93.2%¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿SO2¹ã·ºÓÃÓÚÒ½Ò©¡¢ÁòËṤҵµÈÁìÓò£¬»ØÊÕ·ÏÆøÖеÄSO2¿ÉÓÃÈçÏ·½·¨¡£

·½·¨¢ñ

ÓüîʽÁòËáÂÁAl2(SO4)x(OH)yÈÜÒºÎüÊÕ¸»¼¯SO2

·½·¨¢ò

ÔÚFe2+»òFe3+´ß»¯Ï£¬ÓÿÕÆø(O2)½«SO2Ñõ»¯ÎªH2SO4

(1)·½·¨¢ñµÄ¹ý³ÌÈçÏ¡£

¢Ù ÖƱ¸Al2(SO4)x(OH)y¡£ÏòAl2(SO4)3ÈÜÒºÖмÓÈëCaO·ÛÄ©£¬µ÷pHÖÁ3.6¡£ CaOµÄ×÷ÓÃÊÇ______

¢Ú ÎüÊÕ£ºAl2(SO4)x(OH)yÎüÊÕSO2ºóµÄ²úÎïÊÇ______(д»¯Ñ§Ê½)¡£

¢Û ½âÎü£º¼ÓÈÈ¢ÚÖвúÎ²úÉúSO2£¬Al2(SO4)x(OH)yÔÙÉú¡£

(2)·½·¨¢òÖУ¬ÔÚFe2+´ß»¯Ï£¬SO2¡¢O2ºÍH2OÉú³ÉH2SO4µÄ»¯Ñ§·½³ÌʽÊÇ______¡£

(3)·½·¨¢òÖУ¬Fe2+µÄ´ß»¯¹ý³Ì¿É±íʾÈçÏ£º

¢¡£º2Fe2++O2+SO2=2Fe3++SO42-

¢¢£º¡­¡­

¢Ù д³ö¢¢µÄÀë×Ó·½³Ìʽ£º______¡£

¢Ú ÏÂÁÐʵÑé·½°¸¿É֤ʵÉÏÊö´ß»¯¹ý³Ì¡£½«ÊµÑé·½°¸²¹³äÍêÕû¡£

a.ÏòFeCl2ÈÜÒºµÎÈëKSCN£¬Îޱ仯

b.ÏòFeCl2ÈÜҺͨÈëÉÙÁ¿SO2£¬µÎÈëKSCN£¬ÑÕÉ«±äºì¡£

c.È¡bÖÐÈÜÒº£¬______¡£

(4)·½·¨¢òÖУ¬´ß»¯Ñõ»¯ºó£¬²ÉÓõζ¨·¨²â¶¨·ÏÆøÖвÐÁôSO2µÄº¬Á¿¡£½«V L(ÒÑ»»ËãΪ±ê×¼×´¿ö)·ÏÆøÖеÄSO2ÓÃ1%µÄH2O2ÍêÈ«ÎüÊÕ£¬ÎüÊÕÒºÓÃÈçͼËùʾװÖõ樣¬¹²ÏûºÄa mL c mol/L NaOH±ê×¼Òº¡£

¢ÙH2O2Ñõ»¯SO2µÄ»¯Ñ§·½³Ìʽ______¡£

¢Ú ·ÏÆøÖвÐÁôSO2µÄÌå»ý·ÖÊýΪ______¡£

¡¾ÌâÄ¿¡¿(1)ÒÒ»ùÊ嶡»ùÃÑ(ÒÔETBE±íʾ)ÊÇÒ»ÖÖÐÔÄÜÓÅÁ¼µÄ¸ßÐÁÍéÖµÆûÓ͵÷ºÍ¼Á¡£ÓÃÒÒ´¼ÓëÒ춡ϩ(ÒÔIB±íʾ)ÔÚ´ß»¯¼ÁHZSM£­5´ß»¯ÏºϳÉETBE£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£ºC2H5OH(g)+IB(g)=ETBE(g) ¡÷H¡£»Ø´ðÏÂÁÐÎÊÌ⣺

·´Ó¦Îï±»´ß»¯¼ÁHZSM£­5Îü¸½µÄ˳ÐòÓë·´Ó¦Àú³ÌµÄ¹ØϵÈçͼËùʾ£¬¸Ã·´Ó¦µÄ¡÷H=__________ kJ¡¤mol-1¡£·´Ó¦Àú³ÌµÄ×îÓÅ;¾¶ÊÇ________(ÌîC1¡¢C2»òC3)¡£

(2)¿ª·¢Çå½àÄÜÔ´Êǵ±½ñ»¯¹¤Ñо¿µÄÒ»¸öÈȵãÎÊÌâ¡£¶þ¼×ÃÑ(CH3OCH3)ÔÚδÀ´¿ÉÄÜÌæ´ú²ñÓͺÍÒº»¯Æø×÷Ϊ½à¾»ÒºÌåȼÁÏʹÓ㬹¤ÒµÉÏÒÔCOºÍH2ΪԭÁÏÉú²úCH3OCH3¡£¹¤ÒµÖƱ¸¶þ¼×ÃÑÔÚ´ß»¯·´Ó¦ÊÒÖÐ(ѹÁ¦2.0¡«10.0Mpa£¬Î¶È230¡«280¡æ)½øÐÐÏÂÁз´Ó¦£º

·´Ó¦¢¡£ºCO(g)+2H2(g)CH3OH(g) ¦¤H1=-99kJ¡¤mol1

·´Ó¦¢¢£º2CH3OH(g)CH3OCH3(g)+H2O(g) ¦¤H2=-23.5kJ¡¤mol1

·´Ó¦¢££ºCO(g)+H2O(g)CO2(g)+H2(g) ¦¤H3=-41.2kJ¡¤mol1

¢ÙÔÚ¸ÃÌõ¼þÏ£¬Èô·´Ó¦1µÄÆðʼŨ¶È·Ö±ðΪ£ºc(CO)=0.6mol¡¤L1£¬c(H2)=1.4mol¡¤L1£¬8minºó´ïµ½Æ½ºâ£¬COµÄת»¯ÂÊΪ50%£¬Ôò8minÄÚH2µÄƽ¾ù·´Ó¦ËÙÂÊΪ__________¡£

¢ÚÔÚt¡æʱ£¬·´Ó¦2µÄƽºâ³£ÊýΪ400£¬´ËζÈÏ£¬ÔÚ1LµÄÃܱÕÈÝÆ÷ÖмÓÈëÒ»¶¨µÄ¼×´¼£¬·´Ó¦µ½Ä³Ê±¿Ì²âµÃ¸÷×é·ÖµÄÎïÖʵÄÁ¿Å¨¶ÈÈçÏ£º

ÎïÖÊ

CH3OH

CH3OCH3

H2O

c(mol¡¤L1)

0.46

1.0

1.0

´Ëʱ¿ÌvÕý___vÄæ(Ìî¡°£¾¡±¡°£¼¡±»ò¡°£½¡±)£¬Æ½ºâʱc(CH3OCH3)µÄÎïÖʵÄÁ¿Å¨¶ÈÊÇ___¡£

¢Û´ß»¯·´Ó¦ÊÒµÄ×Ü·´Ó¦3CO(g)+3H2(g)CH3OCH3(g)+CO2(g)£¬COµÄƽºâת»¯ÂʦÁ(CO)Óëζȡ¢Ñ¹Ç¿µÄ¹ØϵÈçͼËùʾ£¬Í¼ÖÐX´ú±í___(Ìζȡ±»ò¡°Ñ¹Ç¿¡±)£¬ÇÒL1___L2(Ìî¡°£¾¡±¡°£¼¡±»ò¡°£½¡±)¡£

¢ÜÔÚ´ß»¯¼ÁµÄ×÷ÓÃÏÂͬʱ½øÐÐÈý¸ö·´Ó¦£¬·¢ÏÖËæ×ÅÆðʼͶÁϱȵĸı䣬¶þ¼×ÃѺͼ״¼µÄ²úÂÊ(²úÎïÖеÄ̼ԭ×ÓÕ¼ÆðʼCOÖÐ̼ԭ×ӵİٷÖÂÊ)³ÊÏÖÈçͼµÄ±ä»¯Ç÷ÊÆ¡£ÊÔ½âÊÍͶÁϱȴóÓÚ1.0Ö®ºó¶þ¼×ÃѲúÂʺͼ״¼²úÂʱ仯µÄÔ­Òò£º______________¡£

¡¾ÌâÄ¿¡¿Í­µÄ¶àÖÖ»¯ºÏÎïÔÚÉú²úÉú»îÖж¼Óй㷺ÓÃ;¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©Cu2OºÍCuOÊÇÍ­µÄÁ½ÖÖÑõ»¯Î¿É»¥Ïàת»¯¡£ÒÑÖª£º

i.2Cu2O(s)£«O2(g)=4CuO(s) ¡÷H=-292.0kJ¡¤mol-1

ii.C(s)£«2CuO(s)=Cu2O(s)£«CO(g) ¡÷H=£«35.5kJ¡¤mol-1

ÈôCOµÄȼÉÕÈÈΪ283.0kJ¡¤mol-1£¬ÔòC(s)µÄȼÉÕÈÈΪ___¡£

£¨2£©Cu2OºÍCuO³£ÓÃ×÷´ß»¯¼Á¡£

¢ÙÖÊ×Ó½»»»Ä¤È¼Áϵç³Ø(PEMFC)µÄȼÁÏÆøÖгýº¬ÓÐH2Í⻹º¬ÓÐÉÙÁ¿µÄCOºÍCO2£¬ÆäÖÐCOÊÇPEMFC´ß»¯¼ÁµÄÑÏÖض¾»¯¼Á£¬¿ÉÓÃCuO/CeO2×÷´ß»¯¼ÁÓÅÏÈÑõ»¯ÍѳýCO¡£160¡æ¡¢ÓÃCuO/CeO2×÷´ß»¯¼Áʱ£¬Ñõ»¯COµÄ»¯Ñ§·½³ÌʽΪ___£»·Ö±ðÓÃHIO3ºÍH3PO4¶ÔCuO/CeO2½øÐд¦Àí£¬ÔÚÒ»¶¨Ìõ¼þÏ£¬ÀûÓò»Í¬´ß»¯¼Á½øÐÐCOÑõ»¯µÄ¶Ô±ÈʵÑ飬µÃÈçͼÇúÏߣ¬ÆäÖд߻¯¼Á___ (Ìî¡°b¡±»ò¡°c¡±)´ß»¯ÐÔÄÜ×îºÃ£»120¡æʹÓô߻¯¼Áb½øÐÐÑõ»¯£¬ÈôȼÁÏÆøÖÐCOµÄÌå»ý·ÖÊýΪ0.71%£¬ÆøÌåÁ÷ËÙΪ2000mL¡¤h-1£¬Ôò1hºó£¬COÌå»ýΪ___mL¡£

¢ÚÔÚCu2O´ß»¯×÷ÓÃϺϳÉCH3OH£¬·´Ó¦ÈçÏ£ºCO(g)£«2H2(g)CH3OH(g) ¡÷H=-90.0kJ¡¤mol-1£¬ÓÐÀûÓÚÌá¸ß¸Ã·´Ó¦COµÄƽºâת»¯ÂʵÄÌõ¼þÊÇ___(Ìî±êºÅ)¡£

A.¸ßεÍѹ B.µÍθßѹ C.¸ßθßѹ D.µÍεÍѹ

T¡æʱ£¬½«COºÍH2°´Ò»¶¨±ÈÀý»ìºÏºóͶÈëÈÝ»ýΪ2LµÄºãÈÝÃܱÕÈÝÆ÷ÖУ¬COµÄÆðʼŨ¶ÈΪ1.0mol¡¤L-1£¬Æ½ºâʱ£¬²âµÃÌåϵÖУ¬n(H2)=1.4mol£¬n(CH3OH)=1.7mol£¬·´Ó¦´ïµ½Æ½ºâʱCOµÄת»¯ÂÊΪ___£¬Èô·´Ó¦´ïµ½Æ½ºâ״̬ºó£¬±£³ÖÆäËûÌõ¼þ²»±ä£¬ÔÙ³äÈë0.2molCOºÍ0.2molCH3OH£¬Æ½ºâÏò___(Ìî¡°Õý¡±»ò¡°Ä桱)·´Ó¦·½ÏòÒƶ¯£¬ÀíÓÉÊÇ___¡£

£¨3£©CuS³ÊºÚÉ«£¬ÊÇ×îÄÑÈܵÄÎïÖÊÖ®Ò»£¬ÓÉÓÚËüµÄÄÑÈÜÐÔʹµÃһЩ¿´ËƲ»¿ÉÄܵķ´Ó¦¿ÉÒÔ·¢Éú¡£Ïò0.01mol¡¤L-1CuSO4ÈÜÒºÖУ¬³ÖÐøͨÈëH2Sά³Ö±¥ºÍ(H2S±¥ºÍŨ¶ÈΪ0.1mol¡¤L-1)£¬·¢Éú·´Ó¦£ºH2S(aq)£«Cu2£«(aq)CuS(s)£«2H£«(aq)£¬¸Ã·´Ó¦µÄ»¯Ñ§Æ½ºâ³£ÊýKΪ___(±£Áô2λÓÐЧÊý×Ö)¡£ÒÑÖª£ºKa1(H2S)=1.1¡Á10-7£¬Ka2(H2S)=1.3¡Á10-13£¬Ksp(CuS)=6.3¡Á10-36¡£

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

¾«Ó¢¼Ò½ÌÍø