Ëæ×ÅÊÀ½ç¹¤Òµ¾­¼ÃµÄ·¢Õ¹¡¢È˿ڵľçÔö£¬È«ÇòÄÜÔ´½ôÕż°ÊÀ½çÆøºòÃæÁÙÔ½À´Ô½ÑÏÖصÄÎÊÌ⣬ÈçºÎ½µµÍ´óÆøÖÐCO2µÄº¬Á¿¼°ÓÐЧµØ¿ª·¢ÀûÓÃCO2ÒýÆðÁËÈ«ÊÀ½çµÄÆÕ±éÖØÊÓ¡£

£¨1£©ÈçͼΪC¼°ÆäÑõ»¯ÎïµÄ±ä»¯¹Øϵͼ£¬Èô¢Ù±ä»¯ÊÇÖû»·´Ó¦£¬ÔòÆ仯ѧ·½³Ìʽ¿ÉΪ______________________________________£»

ͼÖб仯¹ý³ÌÄÄЩÊÇÎüÈÈ·´Ó¦________£¨ÌîÐòºÅ£©¡£

£¨2£©¼×´¼ÊÇÒ»ÖÖ¿ÉÔÙÉúÄÜÔ´£¬¾ßÓпª·¢ºÍÓ¦ÓõĹãÀ«Ç°¾°£¬¹¤ÒµÉÏ¿ÉÓÃÈçÏ·½·¨ºÏ³É¼×´¼£º

·½·¨Ò»¡¡CO£¨g£©£«2H2£¨g£©??CH3OH£¨g£©

·½·¨¶þ¡¡CO2£¨g£©£«3H2£¨g£©??CH3OH£¨g£©£«H2O£¨g£©

ÔÚ25¡æ¡¢101 kPaÏ£¬1¿Ë¼×´¼ÍêȫȼÉÕ·ÅÈÈ22.68 kJ£¬Ð´³ö¼×´¼È¼ÉÕµÄÈÈ»¯Ñ§·½³Ìʽ£º_____________________________________________£»

ij»ðÁ¦·¢µç³§CO2µÄÄê¶ÈÅÅ·ÅÁ¿ÊÇ2 200Íò¶Ö£¬Èô½«´ËCO2Íêȫת»¯Îª¼×´¼£¬ÔòÀíÂÛÉÏÓÉ´Ë»ñµÃµÄ¼×´¼ÍêȫȼÉÕ·ÅÈÈÔ¼ÊÇ________kJ£¨±£ÁôÈýλÓÐЧÊý×Ö£©¡£

£¨3£©½ðÊôîÑÒ±Á¶¹ý³ÌÖÐÆäÖÐÒ»²½·´Ó¦Êǽ«Ô­ÁϽðºìʯת»¯£ºTiO2£¨½ðºìʯ£©£«2C£«2Cl2¸ßÎÂ,TiCl4£«2CO¡¡ÒÑÖª£ºC£¨s£©£«O2£¨g£©=CO2£¨g£©¡¡¦¤H£½£­393.5 kJ¡¤mol£­1

2CO£¨g£©£«O2£¨g£©=2CO2£¨g£©¡¡¦¤H£½£­566 kJ¡¤mol£­1

TiO2£¨s£©£«2Cl2£¨g£©=TiCl4£¨s£©£«O2£¨g£©¡¡¦¤H£½£«141 kJ¡¤mol£­1

ÔòTiO2£¨s£©£«2Cl2£¨g£©£«2C£¨s£©=TiCl4£¨s£©£«2CO£¨g£©µÄ¦¤H£½________¡£

£¨4£©³ôÑõ¿ÉÓÃÓÚ¾»»¯¿ÕÆø£¬ÒûÓÃË®Ïû¶¾£¬´¦Àí¹¤Òµ·ÏÎïºÍ×÷ΪƯ°×¼Á¡£³ôÑõ¼¸ºõ¿ÉÓë³ý²¬¡¢½ð¡¢Ò¿¡¢·úÒÔÍâµÄËùÓе¥ÖÊ·´Ó¦¡£È磺

6Ag£¨s£©£«O3£¨g£©=3Ag2O£¨s£©¡¡¦¤H£½£­235.8 kJ¡¤mol£­1£¬

ÒÑÖª£º2Ag2O£¨s£©=4Ag£¨s£©£«O2£¨g£©¦¤H£½£«62.2 kJ¡¤mol£­1£¬

ÔòO3ת»¯ÎªO2µÄÈÈ»¯Ñ§·½³ÌʽΪ_________________________¡£

 

£¨1£©C£«H2O£¨g£©¸ßÎÂ,H2£«CO£¨ºÏÀí¼´¿É£¬ÈçÓëCuO¡¢FeO¡¢SiO2µÈ·´Ó¦£©¡¡¢Ù¢Û

£¨2£©CH3OH£¨l£©£«O2£¨g£©=CO2£¨g£©£«2H2O£¨l£©¡¡¦¤H£½£­725.76 kJ¡¤mol£­1¡¡3.63¡Á1014

£¨3£©£­80 kJ¡¤mol£­1

£¨4£©2O3£¨g£©=3O2£¨g£©¡¡¦¤H£½£­285 kJ¡¤mol£­1

¡¾½âÎö¡¿£¨1£©ÓÉÐÅÏ¢¿ÉÖª·´Ó¦¢ÙÊÇÖû»·´Ó¦£¬ËùÒÔӦΪÎüÈÈ·´Ó¦£»£¨2£©1 mol¼×´¼È¼ÉշųöÈÈÁ¿22.68 kJ¡Á32£½725.76 kJ£¬ÒÀ¾ÝÐÅÏ¢Éú³ÉÎïˮΪҺ̬£»ÒÀ¾Ý̼ԭ×ÓÊغãn£¨CH3OH£©£½n£¨CO2£©£½£½5¡Á1011 mol£¬·Å³öÈÈÁ¿725.76 kJ¡¤mol£­1¡Á5¡Á1011 mol£½3.63¡Á1014 kJ£»£¨3£©¶ÔÒÑÖªÈÈ»¯Ñ§·½³Ìʽ±àºÅ´ÓÇ°µ½ºó·Ö±ðΪ¢Ù¢Ú¢Û£¬¢Û£­¢Ú£«¢Ù¡Á2¼´¿ÉµÃËùÇó·½³Ìʽ£¬ËùÒÔ¦¤H£½£«141 kJ¡¤mol£­1£­£¨£­566 kJ¡¤mol£­1£©£«£¨£­393.5 kJ¡¤mol£­1£©¡Á2£»£¨4£©Ê×ÏÈд³öËùÇó·´Ó¦µÄ»¯Ñ§·½³Ìʽ2O3£¨g£©=3O2£¨g£©£¬Ç°Ê½¡Á2£«ºóʽ¡Á3¼´¿ÉµÃËùÇóÈÈ»¯Ñ§·½³Ìʽ¡£

 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2014Äê¸ß¿¼»¯Ñ§¶þÂÖ¸´Ï°Ö¸µ¼Ô¤²âѺÌâÁ·Ï°¾í£¨¶þ£© £¨½âÎö°æ£© ÌâÐÍ£ºÑ¡ÔñÌâ

ÏÂÁÐÊôÓÚÈ¡´ú·´Ó¦µÄÊÇ£¨¡¡¡¡£©

¢ÙCH3CH=CH2£«Br2CH3CHBrCH2Br

¢ÚCH3CH2OH£«3O22CO2£«3H2O

¢ÛCH3COOH£«CH3CH2OHCH3COOCH2CH3£«H2O

¢ÜC6H6£«HNO3C6H5NO2£«H2O

A£®¢Ù¢Ú B£®¢Û¢Ü C£®¢Ù¢Û D£®¢Ú¢Ü

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2014Äê¸ß¿¼»¯Ñ§¶þÂÖ¸´Ï°×¨Ìâ¹ö¶¯Á·3 ÔªËؼ°Æ仯ºÏÎïÁ·Ï°¾í£¨½âÎö°æ£© ÌâÐÍ£ºÊµÑéÌâ

ijѧϰС×éÉè¼ÆÓÃÈçͼװÖÃÑéÖ¤¶þÑõ»¯ÁòµÄ»¯Ñ§ÐÔÖÊ¡£

£¨1£©ÄÜ˵Ã÷¶þÑõ»¯Áò¾ßÓÐÑõ»¯ÐÔµÄʵÑéÏÖÏóΪ_________________________¡£

£¨2£©ÎªÑéÖ¤¶þÑõ»¯ÁòµÄ»¹Ô­ÐÔ£¬³ä·Ö·´Ó¦ºó£¬È¡ÊÔ¹ÜbÖеÄÈÜÒº·Ö³ÉÈý·Ý£¬·Ö±ð½øÐÐÈçÏÂʵÑé¡£

·½°¸¢ñ£ºÏòµÚÒ»·ÝÈÜÒºÖмÓÈëAgNO3ÈÜÒº£¬Óа×É«³ÁµíÉú³É

·½°¸¢ò£ºÏòµÚ¶þ·ÝÈÜÒºÖмÓÈëÆ·ºìÈÜÒº£¬ºìÉ«ÍÊÈ¥

·½°¸¢ó£ºÏòµÚÈý·ÝÈÜÒºÖмÓÈëBaCl2ÈÜÒº£¬²úÉú°×É«³Áµí

ÉÏÊö·½°¸ÖкÏÀíµÄÊÇ________£¨Ìî¡°¢ñ¡±¡¢¡°¢ò¡±»ò¡°¢ó¡±£©£»ÊÔ¹ÜbÖз¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ____________________________________________¡£

£¨3£©µ±Í¨Èë¶þÑõ»¯ÁòÖÁÊÔ¹ÜcÖÐÈÜÒºÏÔÖÐÐÔʱ£¬¸ÃÈÜÒºÖÐc£¨Na£«£©£½________£¨Óú¬Áò΢Á£Å¨¶ÈµÄ´úÊýʽ±íʾ£©¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2014Äê¸ß¿¼»¯Ñ§¶þÂÖ¸´Ï°×¨Ìâ¹ö¶¯Á·2 »¯Ñ§»ù±¾ÀíÂÛÁ·Ï°¾í£¨½âÎö°æ£© ÌâÐÍ£ºÌî¿ÕÌâ

ÄÜÔ´µÄ¿ª·¢ÀûÓÃÓëÈËÀàÉç»áµÄ¿É³ÖÐø·¢Õ¹Ï¢Ï¢Ïà¹Ø¡£

¢ñ.ÒÑÖª£ºFe2O3£¨s£©£«3C£¨s£©=2Fe£¨s£©£«3CO£¨g£©

¦¤H1£½a kJ¡¤mol£­1

CO£¨g£©£«O2£¨g£©=CO2£¨g£©¡¡¦¤H2£½b kJ¡¤mol£­1

4Fe£¨s£©£«3O2£¨g£©=2Fe2O3£¨s£©¡¡¦¤H3£½c kJ¡¤mol£­1

ÔòCµÄȼÉÕÈȦ¤H£½________kJ¡¤mol£­1¡£

¢ò.£¨1£©ÒÀ¾ÝÔ­µç³ØµÄ¹¹³ÉÔ­Àí£¬ÏÂÁл¯Ñ§·´Ó¦ÔÚÀíÂÛÉÏ¿ÉÒÔÉè¼Æ³ÉÔ­µç³ØµÄÊÇ________£¨ÌîÐòºÅ£©¡£

A£®C£¨s£©£«CO2£¨g£©=2CO£¨g£©

B£®NaOH£¨aq£©£«HCl£¨aq£©=NaCl£¨aq£©£«H2O£¨l£©

C£®2H2O£¨l£©=2H2£¨g£©£«O2£¨g£©

D£®2CO£¨g£©£«O2£¨g£©=2CO2£¨g£©

ÈôÒÔÈÛÈÚµÄK2CO3ÓëCO2Ϊ·´Ó¦µÄ»·¾³£¬ÒÀ¾ÝËùÑ¡·´Ó¦Éè¼Æ³ÉÒ»¸öÔ­µç³Ø£¬Çëд³ö¸ÃÔ­µç³ØµÄ¸º¼«·´Ó¦£º_____________________________________¡£

£¨2£©Ä³ÊµÑéС×éÄ£Ä⹤ҵºÏ³É°±·´Ó¦N2£¨g£©£«3H2£¨g£©2NH3£¨g£©¡¡¦¤H£½£­92.4 kJ¡¤mol£­1£¬¿ªÊ¼ËûÃǽ«N2ºÍH2»ìºÏÆøÌå20 mol£¨Ìå»ý±È1¡Ã1£©³äÈë5 LºÏ³ÉËþÖУ¬·´Ó¦Ç°Ñ¹Ç¿ÎªP0£¬·´Ó¦¹ý³ÌÖÐѹǿÓÃP±íʾ£¬·´Ó¦¹ý³ÌÖÐÓëʱ¼ätµÄ¹ØϵÈçͼËùʾ¡£

Çë»Ø´ðÏÂÁÐÎÊÌ⣺

¢Ù·´Ó¦´ïƽºâµÄ±êÖ¾ÊÇ__________________________£¨Ìî×Öĸ´úºÅ£¬ÏÂͬ£©¡£

A£®Ñ¹Ç¿±£³Ö²»±ä

B£®ÆøÌåÃܶȱ£³Ö²»±ä

C£®NH3µÄÉú³ÉËÙÂÊÊÇN2µÄÉú³ÉËÙÂʵÄ2±¶

¢Ú0¡«2 minÄÚ£¬ÒÔc£¨N2£©±ä»¯±íʾµÄƽ¾ù·´Ó¦ËÙÂÊΪ________________¡£

¢ÛÓûÌá¸ßN2µÄת»¯ÂÊ£¬¿É²ÉÈ¡µÄ´ëÊ©ÓÐ_____________________________¡£

A£®ÏòÌåϵÖа´Ìå»ý±È1¡Ã1ÔÙ³äÈëN2ºÍH2

B£®·ÖÀë³öNH3

C£®Éý¸ßζÈ

D£®³äÈ뺤ÆøʹѹǿÔö´ó

E£®¼ÓÈëÒ»¶¨Á¿µÄN2

£¨3£©25¡æʱ£¬BaCO3ºÍBaSO4µÄÈܶȻý³£Êý·Ö±ðÊÇ8¡Á10£­9ºÍ1¡Á10£­10£¬Ä³º¬ÓÐBaCO3³ÁµíµÄÐü×ÇÒºÖУ¬c£¨CO32-£©£½0.2 mol¡¤L£­1£¬Èç¹û¼ÓÈëµÈÌå»ýµÄNa2SO4ÈÜÒº£¬ÈôÒª²úÉúBaSO4³Áµí£¬¼ÓÈëNa2SO4ÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶È×îСÊÇ________mol¡¤L£­1¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2014Äê¸ß¿¼»¯Ñ§¶þÂÖ¸´Ï°×¨Ìâ¹ö¶¯Á·2 »¯Ñ§»ù±¾ÀíÂÛÁ·Ï°¾í£¨½âÎö°æ£© ÌâÐÍ£ºÑ¡ÔñÌâ

ÏÂÁÐÓйػ¯Ñ§ÊµÑé²Ù×÷ºÏÀíµÄÊÇ£¨¡¡¡¡£©

¢ÙÔÚÅäÖÆFeSO4ÈÜҺʱ³£ÏòÆäÖмÓÈëÒ»¶¨Á¿Ìú·ÛºÍÏ¡ÁòËá

¢ÚÅäÖÆ100 mL 1.00 mol/LµÄNaClÈÜҺʱ£¬¿ÉÓÃÍÐÅÌÌìƽ³ÆÈ¡5.85 g NaCl¹ÌÌå

¢ÛÎïÖÊÈÜÓÚË®µÄ¹ý³ÌÖУ¬Ò»°ãÓзÅÈÈ»òÎüÈÈÏÖÏó£¬Òò´ËÈܽâµÄ²Ù×÷²»ÄÜÔÚÁ¿Í²ÖнøÐÐ

¢ÜÓÃÊԹܼдÓÊԹܵ×ÓÉÏÂÍùÉϼÐס¾àÊԹܿÚÔ¼1/3´¦£¬ÊÖ³ÖÊԹܼ㤱úÄ©¶Ë£¬½øÐмÓÈÈ

¢Ý²»É÷½«Å¨ÁòËáÕ´ÔÚƤ·ôÉÏ£¬Á¢¼´ÓÃNaHCO3ÈÜÒº³åÏ´

¢ÞÓù㷺pHÊÔÖ½²âµÃijÈÜÒºµÄpH£½12.3

¢ßÓüîʽµÎ¶¨¹ÜÁ¿È¡20.00 mL 0.100 0 mol/L KMnO4ÈÜÒº

¢à½«Í­ÏÈÑõ»¯³ÉÑõ»¯Í­£¬ÔÙÓëÁòËá·´Ó¦À´ÖÆÈ¡ÁòËáÍ­

A£®¢Ü¢Ý¢Þ¢ß B£®¢Ù¢Ü¢Ý¢ß C£®¢Ú¢Û¢Þ¢à D£®¢Ù¢Û¢Ü¢à

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2014Äê¸ß¿¼»¯Ñ§¶þÂÖ¸´Ï°×¨Ìâ¹ö¶¯Á·1 »¯Ñ§»ù±¾¸ÅÄîÁ·Ï°¾í£¨½âÎö°æ£© ÌâÐÍ£ºÑ¡ÔñÌâ

ʵÑéÊÒÖÐÖƱ¸ÒÒÏ©µÄ·½·¨ÊÇʹÒÒ´¼ÍÑË®£¬·´Ó¦¿ÉÒÔ¼òµ¥µØ±íʾΪCH3CH2OH¨D¡úCH2=CH2¡ü£«H2O¡£ÒÑÖªCH2=CH2£¨g£©ºÍC2H5OH£¨l£©µÄȼÉÕÈÈ·Ö±ðÊÇ1 411.0 kJ¡¤mol£­1ºÍ1 366.8 kJ¡¤mol£­1¡£ÔòʵÑéÊÒÓÃC2H5OH£¨l£©ÖƱ¸CH2=CH2£¨g£©Éú³É1 molҺ̬ˮµÄ¦¤HΪ£¨¡¡¡¡£©

A£®£­44.2 kJ¡¤mol£­1

B£®£«44.2 kJ¡¤mol£­1

C£®£­2 777.8 kJ¡¤mol£­1

D£®£«2 777.8 kJ¡¤mol£­1

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2014Äê¸ß¿¼»¯Ñ§¶þÂÖ¸´Ï°×¨Ìâ¹ö¶¯Á·1 »¯Ñ§»ù±¾¸ÅÄîÁ·Ï°¾í£¨½âÎö°æ£© ÌâÐÍ£ºÑ¡ÔñÌâ

ÏÂÁл¯Ñ§ÓÃÓï±íʾÕýÈ·µÄÊÇ£¨¡¡¡¡£©

A£®Cl£­µÄ½á¹¹Ê¾Òâͼ£º

B£®¼×Íé·Ö×ÓµÄÇò¹÷Ä£ÐÍ£º

C£®ÇâÑõ»¯ÄƵĵç×Óʽ£º

D£®´ÎÂÈËáµÄ½á¹¹Ê½£ºH¡ªO¡ªCl

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2014Äê¸ß¿¼»¯Ñ§Ò»ÂÖ¸´Ï°¿ÎºóѵÁ·×¨ÌâÁ·Ï°¾í¶þ£¨½âÎö°æ£© ÌâÐÍ£ºÑ¡ÔñÌâ

ÏÂÁÐÐðÊöÖУ¬ÕýÈ·µÄÊÇ£¨ £©

A£®ÒÒ´¼¡¢ÒÒËáºÍÒÒËáÒÒõ¥ÄÜÓñ¥ºÍNa2CO3ÈÜÒº¼ø±ð

B£®¾ÛÒÒÏ©¿É·¢Éú¼Ó³É·´Ó¦

C£®ËùÓеÄÌÇÀà¡¢ÓÍÖ¬ºÍµ°°×Öʶ¼ÄÜ·¢ÉúË®½â·´Ó¦

D£®·ûºÏ·Ö×ÓʽC5H12µÄÎïÖÊÓÐ4ÖÖ

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2014Äê¸ß¿¼»¯Ñ§Ò»ÂÖ¸´Ï°¿Îºó¹æ·¶ÑµÁ·9-2Á·Ï°¾í£¨½âÎö°æ£© ÌâÐÍ£ºÑ¡ÔñÌâ

ÒÔ¸õËá¼ØΪԭÁÏ£¬µç»¯Ñ§·¨ÖƱ¸ÖظõËá¼ØµÄʵÑé×°ÖÃʾÒâͼÈçÏ£º

ÏÂÁÐ˵·¨²»ÕýÈ·µÄÊÇ£¨ £©

A£®ÔÚÒõ¼«ÊÒ£¬·¢ÉúµÄµç¼«·´Ó¦Îª£º2H2O£«2e£­=2OH£­£«H2¡ü

B£®ÔÚÑô¼«ÊÒ£¬Í¨µçºóÈÜÒºÖð½¥ÓÉ»ÆÉ«±äΪ³ÈÉ«£¬ÊÇÒòΪÑô¼«ÇøH£«Å¨¶ÈÔö´ó£¬Ê¹Æ½ºâ2CrO42-£«2H£«Cr2O72-£«H2OÏòÓÒÒƶ¯

C£®¸ÃÖƱ¸¹ý³Ì×Ü·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º4K2CrO4£«4H2O2K2Cr2O7£«4KOH£«2H2¡ü£«O2¡ü

D£®²â¶¨Ñô¼«ÒºÖÐKºÍCrµÄº¬Á¿£¬ÈôKÓëCrµÄÎïÖʵÄÁ¿Ö®±È(nk/nCr)Ϊd£¬Ôò´Ëʱ¸õËá¼ØµÄת»¯ÂÊΪ1£­

 

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸