ÒÑÖª£º¢Ù1 gÇâÆøÔÚÑõÆøÖÐÍêȫȼÉÕÉú³ÉÆø̬ˮ£¬·Å³öÈÈÁ¿120.9 kJ£¬¢ÚÖкÍÈÈΪ57.3 kJ¡¤mol£­1£¬¢ÛC(ʯīs)=C(½ð¸Õʯs)¡¡¦¤H£½£«1.90 kJ¡¤mol£­1£¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ(¡¡¡¡)¡£

A£®ÇâÆøµÄȼÉÕÈÈΪ241.8 kJ¡¤mol£­1

B£®ÇâÆøȼÉÕµÄÈÈ»¯Ñ§·½³Ìʽ£º2H2£«O2=2H2O¡¡¦¤H£½£­483.6 kJ¡¤mol£­1

C£®ÑÎËáºÍ°±Ë®»ìºÏµÄÈÈ»¯Ñ§·½³Ìʽ£ºH£«(aq)£«OH£­(aq)=H2O(l)¡¡¦¤H£½£­57.3 kJ¡¤mol£­1

D£®ÓÉ¢Û¿ÉÖª½ð¸Õʯ²»¼°Ê¯Ä«Îȶ¨

 

¡¡D

¡¾½âÎö¡¿¡¡1 g H2Ϊ0.5 mol,1 mol H2ÓëO2·´Ó¦Éú³É1 molÆø̬ˮ£¬·Å³ö241.8 kJµÄÈÈÁ¿£¬µ«È¼ÉÕÈÈÊÇÖ¸Éú³ÉÎȶ¨µÄÑõ»¯ÎïËù·Å³öµÄÈÈÁ¿£¬H2¶ÔÓ¦µÄ²úÎïÊÇҺ̬ˮ£¬A´íÎó£»ÈÈ»¯Ñ§·½³ÌʽÖбØÐë±êÃ÷ÎïÖʵľۼ¯×´Ì¬£¬B´íÎó£»ÖкÍÈÈÊÇÖ¸ÔÚÏ¡ÈÜÒºÖУ¬Ç¿Ëá¸úÇ¿¼î·¢ÉúÖкͷ´Ó¦Éú³É 1 mol H2OʱµÄ·´Ó¦ÈÈ£¬µ«Ò»Ë®ºÏ°±ÊÇÈõ¼î£¬C´íÎó£»Ê¯Ä«µÄÄÜÁ¿±È½ð¸ÕʯµÄÄÜÁ¿µÍ£¬¸üÎȶ¨£¬DÕýÈ·¡£

 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2013-2014ѧÄê¸ß¿¼»¯Ñ§¶þÂÖ¸´Ï°½­ËÕרÓà ¸ß¿¼Ä£ÄâÑÝÁ·1Á·Ï°¾í£¨½âÎö°æ£© ÌâÐÍ£ºÌî¿ÕÌâ

ÒÔÖؾ§Ê¯£¨Ö÷Òª³É·ÖΪBaSO4£©ÎªÖ÷ÒªÔ­ÁÏÖƱ¸Á¢µÂ·Û£¨ZnSºÍBaSO4µÄ»ìºÏÎµÄ¹¤ÒÕÁ÷³ÌÈçÏ£º

£¨1£©±ºÉÕʱ£¬½¹Ì¿Ðè¹ýÁ¿£¬ÆäÄ¿µÄÊÇ£º¢Ù____________________________£»¢Ú____________________________¡£

£¨2£©±ºÉÕ¹ý³ÌÖз¢ÉúµÄ·´Ó¦Ö®Ò»ÊÇ4CO£¨g£©£«BaSO4£¨s£©BaS£¨s£©£«4CO2£¨g£©£¬¸Ã·´Ó¦µÄƽºâ³£ÊýµÄ±í´ïʽΪ__________________________________________________¡£

·´Ó¦Æ÷Öз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ____________________________________¡£

£¨4£©Óù¤Òµ¼¶Ñõ»¯Ð¿£¨º¬ÉÙÁ¿FeOºÍFe2O3ÔÓÖÊ£©ºÍÁòËáÖÆÈ¡ÁòËáпÈÜҺʱ£¬ÐèÏȺó¼ÓÈëH2O2ÈÜÒººÍ̼Ëáп¾«ÖÆ£¬¼ÓÈëH2O2ÈÜҺʱ·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ

________________________________£¬¼ÓÈë̼ËáпµÄÄ¿µÄÊǽ«ÈÜÒºÖеÄFe3£«×ª»¯ÎªºìºÖÉ«³Áµí£¬Ð´³ö¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º__________________________________________¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2013-2014ѧÄê¸ß¿¼»¯Ñ§¶þÂÖ¸´Ï°½­ËÕרÓà µÚ7½²Ë®ÈÜÒºÖеÄÀë×ÓƽºâÁ·Ï°¾í£¨½âÎö°æ£© ÌâÐÍ£ºÌî¿ÕÌâ

ijζÈ(t ¡æ)ʱ£¬Ë®µÄÀë×Ó»ýΪKW£½1.0¡Á10£­13mol2¡¤L£­2£¬Ôò¸ÃζÈ(Ìî¡°´óÓÚ¡±¡¢¡°Ð¡ÓÚ¡±»ò¡°µÈÓÚ¡±)________25 ¡æ£¬ÆäÀíÓÉÊÇ_______________________________________¡£

Èô½«´ËζÈÏÂpH£½11µÄ¿ÁÐÔÄÆÈÜÒºa LÓëpH£½1µÄÏ¡ÁòËáb L»ìºÏ(Éè»ìºÏºóÈÜÒºÌå»ýµÄ΢С±ä»¯ºöÂÔ²»¼Æ)£¬ÊÔͨ¹ý¼ÆËãÌîдÒÔϲ»Í¬Çé¿öʱÁ½ÖÖÈÜÒºµÄÌå»ý±È£º

(1)ÈôËùµÃ»ìºÏҺΪÖÐÐÔ£¬Ôòa¡Ãb£½________£»´ËÈÜÒºÖи÷ÖÖÀë×ÓµÄŨ¶ÈÓÉ´óµ½Ð¡ÅÅÁÐ˳ÐòÊÇ___________________________________¡£

(2)ÈôËùµÃ»ìºÏÒºµÄpH£½2£¬Ôòa¡Ãb£½________¡£´ËÈÜÒºÖи÷ÖÖÀë×ÓµÄŨ¶ÈÓÉ´óµ½Ð¡ÅÅÁÐ˳ÐòÊÇ__________________________________________¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2013-2014ѧÄê¸ß¿¼»¯Ñ§¶þÂÖ¸´Ï°½­ËÕרÓà µÚ6½²»¯Ñ§·´Ó¦ËÙÂʺͻ¯Ñ§Æ½ºâÁ·Ï°¾í£¨½âÎö°æ£© ÌâÐÍ£ºÑ¡ÔñÌâ

ÔÚÎÒ¹úÆû³µÎ²ÆøÒѳÉΪÖ÷ÒªµÄ´óÆøÎÛȾÎʹÓÃÏ¡ÍÁµÈ´ß»¯¼ÁÄܽ«Æû³µÎ²ÆøÖеÄCO¡¢NOxºÍ̼Ç⻯ºÏÎïת»¯³ÉÎÞ¶¾ÎïÖÊ£¬´Ó¶ø¼õÉÙÎÛȾ¡£ÏòÈÝ»ýÏàͬµÄÁ½¸öÃܱÕÈÝÆ÷ÄÚ(×°ÓеÈÁ¿µÄijÖÖ´ß»¯¼Á)£¬·Ö±ð³äÈëµÈÁ¿µÄNOx¼°C3H6£¬ÔÚ²»Í¬Î¶ÈÏ£¬·¢ÉúÈçÏ·´Ó¦£º

 

¢Ù18 NO(g)£«2C3H6(g)9N2(g)£«6CO2(g)£«6H2O(g)£»

¢Ú18 NO2(g)£«4C3H6(g)9N2(g)£«12 CO2(g)£«12 H2O(g)

·Ö±ð²â¶¨²»Í¬Î¶ÈʱNOxµÄת»¯ÂÊ£¬ËùµÃµÄÇúÏßÈçͼËùʾ¡£ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ(¡¡¡¡)¡£

A£®¸Ã·´Ó¦µÄ¦¤H£¾0

B£®¼ÓÈë´ß»¯¼Á£¬NO(g)µÄת»¯ÂʱÈNO2(g)µÄµÍ

C£®·ÖÀë³öH2O(g)¿ÉÌá¸ß̼Ç⻯ºÏÎïµÄת»¯ÂÊ

D£®Ôö´óC3H6(g)µÄŨ¶È¿ÉÌá¸ßNOxµÄת»¯ÂÊ£¬¾ßÓÐʵ¼ÊÒâÒå

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2013-2014ѧÄê¸ß¿¼»¯Ñ§¶þÂÖ¸´Ï°½­ËÕרÓà µÚ5½²»¯Ñ§·´Ó¦ÓëÄÜÁ¿Á·Ï°¾í£¨½âÎö°æ£© ÌâÐÍ£ºÌî¿ÕÌâ

¼×´¼ÊÇÒ»ÖÖ¿ÉÔÙÉúÄÜÔ´£¬¾ßÓй㷺µÄ¿ª·¢ºÍÓ¦ÓÃÇ°¾°¡£

(1)¹¤ÒµÉÏÒ»°ã²ÉÓÃÏÂÁÐÁ½ÖÖ·´Ó¦ºÏ³É¼×´¼£º

·´Ó¦¢ñ£ºCO(g)£«2H2(g)CH3OH(g)¡¡¦¤H1

·´Ó¦¢ò£ºCO2(g)£«3H2(g)CH3OH(g)£«H2O(g)¡¡¦¤H2

¢ÙÉÏÊö·´Ó¦·ûºÏ¡°Ô­×Ó¾­¼Ã¡±Ô­ÔòµÄÊÇ________(Ìî¡°¢ñ¡±»ò¡°¢ò¡±)¡£

¢ÚϱíËùÁÐÊý¾ÝÊÇ·´Ó¦¢ñÔÚ²»Í¬Î¶ÈϵĻ¯Ñ§Æ½ºâ³£Êý(K)¡£

ζÈ

250 ¡æ

300 ¡æ

350 ¡æ

K

2.041

0.270

0.012

 

ÓɱíÖÐÊý¾ÝÅжϣ¬¦¤H1______0(Ìî¡°£¾¡±¡¢¡°£½¡±»ò¡°£¼¡±)¡£

¢ÛijζÈÏ£¬½«2 mol COºÍ6 mol H2³äÈë2 LµÄÃܱÕÈÝÆ÷ÖУ¬³ä·Ö·´Ó¦£¬´ïµ½Æ½ºâºó£¬²âµÃc(CO)£½0.2 mol¡¤L£­1£¬ÔòCOµÄת»¯ÂÊΪ________£¬´ËʱµÄζÈΪ________(´ÓÉϱíÖÐÑ¡Ôñ)¡£

(2)ÒÑÖªÔÚ³£Î³£Ñ¹Ï£º

¢Ù2CH3OH(l)£«3O2(g)=2CO2(g)£«4H2O(g) ¦¤H1£½£­1 275.6 kJ¡¤mol£­1

¢Ú2CO(g)£«O2(g)=2CO2(g) ¦¤H2£½£­566.0 kJ¡¤mol£­1

¢ÛH2O(g)=H2O(l)¡¡¦¤H3£½£­44.0 kJ¡¤mol£­1

д³ö¼×´¼²»ÍêȫȼÉÕÉú³ÉÒ»Ñõ»¯Ì¼ºÍҺ̬ˮµÄÈÈ»¯Ñ§·½³Ìʽ£º______________________________________¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2013-2014ѧÄê¸ß¿¼»¯Ñ§¶þÂÖ¸´Ï°½­ËÕרÓà µÚ5½²»¯Ñ§·´Ó¦ÓëÄÜÁ¿Á·Ï°¾í£¨½âÎö°æ£© ÌâÐÍ£ºÑ¡ÔñÌâ

ÏÂÁйØÓÚ»¯Ñ§·´Ó¦µÄÃèÊöÖÐÕýÈ·µÄÊÇ(¡¡¡¡)¡£

A£®ÐèÒª¼ÓÈȲÅÄÜ·¢ÉúµÄ·´Ó¦Ò»¶¨ÊÇÎüÈÈ·´Ó¦

B£®ÒÑÖªNaOH(aq)£«HCl(aq)=NaCl(aq)£«H2O(l)

¦¤H£½£­57.3 kJ¡¤mol£­1£¬Ôòº¬40.0 g NaOHµÄÏ¡ÈÜÒºÓëÏ¡´×ËáÍêÈ«Öкͣ¬Ò²·Å³ö57.3 kJµÄÈÈÁ¿

C£®CO(g)µÄȼÉÕÈÈÊÇ283.0 kJ¡¤mol£­1£¬Ôò±íʾCO(g)µÄȼÉÕÈȵÄÈÈ»¯Ñ§·½³ÌʽΪ2CO(g)£«O2(g)=2CO2(g)¡¡¦¤H£½£­283.0 kJ¡¤mol£­1

D£®ÒÑÖª2C(s)£«2O2(g)=2CO2(g)¡¡¦¤H£½a,2C(s)£«O2(g)=2CO(g)¡¡¦¤H£½b£¬Ôòb£¾a

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2013-2014ѧÄê¸ß¿¼»¯Ñ§¶þÂÖ¸´Ï°½­ËÕרÓà µÚ4½²ÎïÖʽṹԪËØÖÜÆÚÂÉÁ·Ï°¾í£¨½âÎö°æ£© ÌâÐÍ£ºÑ¡ÔñÌâ

ÔªËØR¡¢X¡¢T¡¢Z¡¢QÔÚÔªËØÖÜÆÚ±íÖеÄÏà¶ÔλÖÃÈç±íËùʾ£¬ÆäÖÐRµ¥ÖÊÔÚ°µ´¦ÓëH2¾çÁÒ»¯ºÏ²¢·¢Éú±¬Õ¨£¬ÔòÏÂÁÐÅжÏÕýÈ·µÄÊÇ(¡¡¡¡)¡£

 

A£®·Ç½ðÊôÐÔ£ºZ£¼T£¼X

B£®RÓëQµÄµç×ÓÊýÏà²î26

C£®Æø̬Ç⻯ÎïÎȶ¨ÐÔ£ºR£¼T£¼Q

D£®×î¸ß¼ÛÑõ»¯ÎïµÄË®»¯ÎïµÄËáÐÔ£ºT£¾Q

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2013-2014ѧÄê¸ß¿¼»¯Ñ§¶þÂÖ¸´Ï°½­ËÕרÓà µÚ2½²»¯Ñ§³£ÓüÆÁ¿Á·Ï°¾í£¨½âÎö°æ£© ÌâÐÍ£ºÌî¿ÕÌâ

¹¤ÒµÉÏÓú¬Ð¿ÎïÁÏ(º¬FeO¡¢CuOµÈÔÓÖÊ)¿ÉÖƵûîÐÔZnO£¬Á÷³ÌÈçÏ£º

 

(1)ÉÏÊöÁ÷³ÌÖУ¬½þ³öÓõÄÊÇ60%H2SO4(1.5 g¡¤cm£­3)£¬ÅäÖÆÕâÖÖH2SO4 100 mLÐèÒª18.4 mol¡¤L£­1µÄŨH2SO4________ mL(±£ÁôһλСÊý)¡£

(2)¼ÓÈëÑõ»¯¼ÁH2O2ºó£¬ÓÐFe(OH)3³Áµí³öÏÖ£¬Ã»ÓÐCu(OH)2³Áµí³öÏÖ£¬ÈôÈÜÒºÖÐc(Fe3£«)£½2.6¡Á10£­18 mol¡¤L£­1£¬ÔòÈÜÒºÖÐc(Cu2£«)µÄÈ¡Öµ·¶Î§ÊÇ________mol¡¤L£­1¡£(ÒÑÖªKsp[Fe(OH)3]£½2.6¡Á10£­39£¬

Ksp[Cu(OH)2]£½2.2¡Á10£­20)

(3)¼ÓÈëNH4HCO3ºóÉú³ÉµÄ³ÁµíÊÇÐÎ̬¾ùΪZna(OH)b(CO3)c(a¡¢b¡¢cΪÕýÕûÊý)µÄÁ½ÖÖ¼îʽ̼ËáпAºÍBµÄ»ìºÏÎAÖÐa£½5¡¢b£½6£¬ÔòÉú³É¼îʽ̼ËáпAµÄ»¯Ñ§·½³ÌʽΪ_______________________________________________¡£

(4)È¡Ï´µÓ¡¢ºæ¸ÉºóµÄ¼îʽ̼ËáпAºÍBµÄ»ìºÏÎï49.70 g£¬ÆäÎïÖʵÄÁ¿Îª0.10 mol£¬¸ßαºÉÕÍêÈ«·Ö½âµÃµ½37.26 g ZnO¡¢3.584 L CO2(±ê×¼×´¿öÏÂ)ºÍË®£¬Í¨¹ý¼ÆËãÇó³ö¼îʽ̼ËáпBµÄ»¯Ñ§Ê½¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2013-2014ѧÄê¸ß¿¼»¯Ñ§¶þÂÖ¸´Ï°½­ËÕרÓà µÚ13½²ÊµÑé·½°¸Éè¼ÆÓëÆÀ¼ÛÁ·Ï°¾í£¨½âÎö°æ£© ÌâÐÍ£ºÊµÑéÌâ

¹ÌÌåÏõËáÑμÓÈÈÒ×·Ö½âÇÒ²úÎï½Ï¸´ÔÓ¡£Ä³Ñ§Ï°Ð¡×éÒÔMg(NO3)2ΪÑо¿¶ÔÏó£¬Äâͨ¹ýʵÑé̽¾¿ÆäÈÈ·Ö½âµÄ²úÎÌá³öÈçÏÂ4ÖÖ²ÂÏ룺

¼×£ºMg(NO2)2¡¢NO2¡¢O2

ÒÒ£ºMgO¡¢NO2¡¢O2

±û£ºMg3N2¡¢O2

¶¡£ºMgO¡¢NO2¡¢N2

(1)ʵÑéС×é³ÉÔ±¾­ÌÖÂÛÈ϶¨²ÂÏ붡²»³ÉÁ¢£¬ÀíÓÉÊÇ_______________________¡£

²éÔÄ×ÊÁϵÃÖª£º2NO2£«2NaOH=NaNO3£«NaNO2£«H2O

Õë¶Ô¼×¡¢ÒÒ¡¢±û²ÂÏ룬Éè¼ÆÈçÏÂͼËùʾµÄʵÑé×°ÖÃ(ͼÖмÓÈÈ¡¢¼Ð³ÖÒÇÆ÷µÈ¾ùÊ¡ÂÔ)£º

 

(2)ʵÑé¹ý³Ì

¢ÙÒÇÆ÷Á¬½Óºó£¬·ÅÈë¹ÌÌåÊÔ¼Á֮ǰ£¬¹Ø±Õk£¬Î¢ÈÈÓ²Öʲ£Á§¹Ü(A)£¬¹Û²ìµ½EÖÐÓÐÆøÅÝÁ¬Ðø·Å³ö£¬±íÃ÷__________¡£

¢Ú³ÆÈ¡Mg(NO3)2¹ÌÌå3.7 gÖÃÓÚAÖУ¬¼ÓÈÈǰͨÈëN2ÒÔÇý¾¡×°ÖÃÄڵĿÕÆø£¬ÆäÄ¿µÄÊÇ________£»¹Ø±Õk£¬Óþƾ«µÆ¼ÓÈÈʱ£¬ÕýÈ·²Ù×÷ÊÇÏÈ________£¬È»ºó¹Ì¶¨ÔÚ¹ÜÖйÌÌ岿λϼÓÈÈ¡£

¢Û¹Û²ìµ½AÖÐÓкì×ØÉ«ÆøÌå³öÏÖ£¬C¡¢DÖÐδ¼ûÃ÷ÏԱ仯¡£

¢Ü´ýÑùÆ·ÍêÈ«·Ö½â£¬A×°ÖÃÀäÈ´ÖÁÊÒΡ¢³ÆÁ¿£¬²âµÃÊ£Óà¹ÌÌåµÄÖÊÁ¿Îª1.0 g¡£

¢ÝÈ¡ÉÙÁ¿Ê£Óà¹ÌÌåÓÚÊÔ¹ÜÖУ¬¼ÓÈëÊÊÁ¿Ë®£¬Î´¼ûÃ÷ÏÔÏÖÏó¡£

(3)ʵÑé½á¹û·ÖÎöÌÖÂÛ

¢Ù¸ù¾ÝʵÑéÏÖÏóºÍÊ£Óà¹ÌÌåµÄÖÊÁ¿¾­·ÖÎö¿É³õ²½È·ÈϲÂÏë______ÊÇÕýÈ·µÄ¡£

¢Ú¸ù¾ÝDÖÐÎÞÃ÷ÏÔÏÖÏó£¬Ò»Î»Í¬Ñ§ÈÏΪ²»ÄÜÈ·ÈÏ·Ö½â²úÎïÖÐÓÐO2£¬ÒòΪÈôÓÐO2£¬DÖн«·¢ÉúÑõ»¯»¹Ô­·´Ó¦£º______________(Ìîд»¯Ñ§·½³Ìʽ)£¬ÈÜÒºÑÕÉ«»áÍÊÈ¥£»Ð¡×éÌÖÂÛÈ϶¨·Ö½â²úÎïÖÐÓÐO2´æÔÚ£¬Î´¼ì²âµ½µÄÔ­ÒòÊÇ__________

____________________________________________________¡£

¢ÛС×éÌÖÂÛºó´ï³ÉµÄ¹²Ê¶ÊÇÉÏÊöʵÑéÉè¼ÆÈÔ²»ÍêÉÆ£¬Ðè¸Ä½ø×°ÖýøÒ»²½Ì½¾¿¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸