ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿ÏÂÁйØϵʽ´íÎóµÄÊÇ( ¡¡ )

A. CO2µÄË®ÈÜÒº£ºc(H+)£¾c(HCO3-)£¾2c(CO32-)

B. µÈŨ¶ÈµÄHCNÈÜÒºÓëNaOHÈÜÒºµÈÌå»ý»ìºÏ£¬ËùµÃÈÜÒºpH>7£¬ÔòÈÜÒºÖÐÀë×ÓŨ¶È£ºc(Na+)£¾c(CN-) £¾c(OH-)£¾c(H+)

C. NaHCO3ÈÜÒºÖдæÔÚË®½âƽºâ£ºHCO3-+H2OH2CO3+OH-

D. Á½ÖÖÈõËáHXºÍHY»ìºÏºó£¬ÈÜÒºÖеÄc(H+)Ϊ(KaΪµçÀëƽºâ³£Êý) ++ c(OH-)

¡¾´ð°¸¡¿D

¡¾½âÎö¡¿

A£®CO2µÄË®ÈÜÒºÖУ¬Ì¼ËᲿ·ÖµçÀë³öÇâÀë×Ó£¬ÈÜÒº³ÊËáÐÔ£¬ÓÉÓÚÇâÀë×Ó»¹À´×ÔË®µÄµçÀë¡¢HCO3-µÄµçÀ룬Ôòc(H+)£¾c(HCO3-)£¬ÓÉÓÚµÚ¶þ²½µçÀ뼫Èõ£¬Ôòc(HCO3-)£¾£¾c(CO32-)£¬ËùÒÔ¸ÃÈÜÒºÖÐÀë×ÓŨ¶È´óСΪ£ºc(H+)£¾c(HCO3-)£¾2c(CO32-)£¬¹ÊAÕýÈ·£»B£®µÈŨ¶ÈµÄHCNÈÜÒºÓëNaOHÈÜÒºµÈÌå»ý»ìºÏ£¬ËùµÃÈÜÒºpH£¾7£¬Ôòc(OH-)£¾c(H+)£¬¸ù¾ÝµçºÉÊغã¿ÉµÃ£ºc(Na+)£¾c(CN-)£¬ÔòÈÜÒºÖÐÀë×ÓŨ¶È´óСΪ£ºc(Na+)£¾c(CN-)£¾c(OH-)£¾c(H+)£¬¹ÊBÕýÈ·£»C£®HCO3-ÄÜË®½âÉú³ÉH2CO3¡¢OH-£¬Ë®½â·½³ÌʽΪHCO3-+H2OH2CO3+OH-£¬¹ÊCÕýÈ·£»D£®Á½ÖÖÈõËáHXºÍHY»ìºÏºó£¬¸ù¾Ý¶þÕߵĵçÀëƽºâ³£Êý¿ÉÖªÈÜÒºÖеÄc(H+)Ϊ(KaΪµçÀëƽºâ³£Êý)£ºc(H+)==£¬¹ÊD´íÎó£»¹ÊÑ¡D¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿CO2µÈÎÂÊÒÆøÌåµÄÅÅ·ÅËù´øÀ´µÄÎÂÊÒЧӦÒѾ­¶ÔÈËÀàµÄÉú´æ»·¾³²úÉúºÜ´óÓ°Ïì¡£CO2µÄÀûÓÃÒ²³ÉΪÈËÃÇÑо¿µÄÈȵ㡣ÒÔCO2ºÍH2ΪԭÁϺϳɼ״¼¼¼Êõ»ñµÃÓ¦Óá£

£¨1£©ÒÑÖªCH3OH (g)+O2(g)===CO2(g)+2H2O(l ) ¡÷H1=£­363 kJ/mol

2H2(g)+O2(g)===2H2O(1) ¡÷H2=£­571.6kJ/mol

H2O(1)====H2O(g) ¡÷H3=+44 kJ/ mol

ÔòCO2(g)+3H2(g)CH3OH(g)+H2O(g)µÄ·´Ó¦ÈÈ¡÷H=___________¡£

£¨2£©¸Ã·´Ó¦³£ÔÚ230~280¡æ¡¢1.5MPaÌõ¼þϽøÐС£²ÉÓô߻¯¼ÁÖ÷Òª×é·ÖΪCuO-ZnO-Al2O3¡£´ß»¯¼Á»îÐÔ×é·ÖΪµ¥ÖÊÍ­£¬Òò´Ë·´Ó¦Ç°ÒªÍ¨ÇâÆø»¹Ô­¡£Ð´³öµÃµ½»îÐÔ×é·ÖµÄ·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º__________________________________________¡£Ê¹Óò»Í¬´ß»¯¼Áʱ£¬¸Ã·´Ó¦·´Ó¦ÈÈ¡÷H__________(Ìî¡°Ïàͬ¡±»ò¡°²»Í¬¡±)

£¨3£©¸Ã·´Ó¦¿ÉÒÔ¿´×÷ÒÔÏÂÁ½¸ö·´Ó¦µÄµþ¼Ó£º

CO2(g)+H2(g)CO(g)+H2O(g)£¬Æ½ºâ³£ÊýK1£»

CO(g)+2H2(g)CH3OH(g)£¬Æ½ºâ³£ÊýK2£»

ÔòCO2(g)+3H2(g)CH3OH(g)+H2O(g)µÄƽºâ³£ÊýK=___________(Óú¬K1¡¢K2µÄ´úÊýʽ±íʾ)

£¨4£©·´Ó¦¹ý³ÌÖУ¬·¢ÏÖβÆøÖÐ×ܻẬÓÐÒ»¶¨Å¨¶ÈµÄCO£¬ÎªÁ˼õÉÙÆäŨ¶È£¬¿ÉÒÔ²ÉÈ¡µÄ´ëʩΪ_________________________________(дһÌõ¼´¿É)

£¨5£©ÎªÁËÌá¸ß·´Ó¦ËÙÂÊ£¬²ÉÈ¡µÄ´ëÊ©¿ÉÒÔÓÐ___________¡£

A.ʹÓøßЧ´ß»¯¼Á B.ÔڽϸßѹǿϽøÐÐ C.½µµÍѹǿ D.³äÈë¸ßŨ¶ÈCO2

£¨6£©ÓÐÈËÑо¿ÁËÓõ绯ѧ·½·¨°ÑCO2ת»¯ÎªCH3OH£¬ÆäÔ­ÀíÈçͼËùʾ£º

ÔòͼÖÐAµç¼«½ÓµçÔ´___________¼«¡£ÒÑÖªBµç¼«Îª¶èÐԵ缫£¬ÔòÔÚË®ÈÜÒºÖУ¬¸Ã¼«µÄµç¼«·´Ó¦Îª______________________¡£

¡¾ÌâÄ¿¡¿£¨1£©·´Ó¦3Fe£¨s£©+4H2O£¨g£©Fe3O4£¨s£©+4H2£¨g£©ÔÚÒ»ÈÝ»ý¿É±äµÄÃܱÕÈÝÆ÷ÖнøÐУ¬ÊԻشð£º

¢ÙÔö¼ÓFeµÄÁ¿£¬Æä·´Ó¦ËÙÂÊ____£¨Ìî¡°Ôö´ó¡±¡¢¡°²»±ä¡±»ò¡°¼õС¡±£¬ÏÂͬ£©¡£

¢Ú½«ÈÝÆ÷µÄÌå»ýËõСһ°ë£¬Æä·´Ó¦ËÙÂÊ____¡£

¢Û±£³ÖÌå»ý²»±ä£¬³äÈëHe£¬Æä·´Ó¦ËÙÂÊ____¡£

¢Ü±£³Öѹǿ²»±ä£¬³äÈëHe£¬Æä·´Ó¦ËÙÂÊ_____¡£

£¨2£©°±Æø¿É×÷ΪÍÑÏõ¼Á£¬ÔÚºãκãÈÝÃܱÕÈÝÆ÷ÖгäÈëÒ»¶¨Á¿µÄNOºÍNH3£¬ÔÚÒ»¶¨Ìõ¼þÏ·¢Éú·´Ó¦£º6NO£¨g£©+4NH3£¨g£©5N2£¨g£©+6H2O£¨g£©¡£

¢ÙÄÜ˵Ã÷¸Ã·´Ó¦ÒѴﵽƽºâ״̬µÄ±êÖ¾ÊÇ____£¨Ìî×ÖĸÐòºÅ£©

a.·´Ó¦ËÙÂÊ5v£¨NH3£©=4v£¨N2£©

b.µ¥Î»Ê±¼äÀïÿÉú³É5mol N2£¬Í¬Ê±Éú³É4mol NH3

c.ÈÝÆ÷ÄÚN2µÄÎïÖʵÄÁ¿·ÖÊý²»ÔÙËæʱ¼ä¶ø·¢Éú±ä»¯

d.ÈÝÆ÷ÄÚn(NO£©£ºn£¨NH3£©£ºn£¨N2£©£ºn£¨H2O£©=6£º4£º5£º6

¢Úij´ÎʵÑéÖвâµÃÈÝÆ÷ÄÚNO¼°N2µÄÎïÖʵÄÁ¿Ëæʱ¼ä±ä»¯ÈçͼËùʾ£¬Í¼ÖÐv£¨Õý£©Óëv£¨Ä棩ÏàµÈµÄµãΪ_____£¨Ñ¡Ìî×Öĸ£©¡£

£¨3£©298Kʱ£¬ÈôÒÑÖªÉú³É±ê×¼×´¿öÏÂ2.24LNH3ʱ·Å³öÈÈÁ¿Îª4.62kJ¡£Ð´³öºÏ³É°±·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ____¡£

£¨4£©Ò»¶¨Ìõ¼þÏ£¬ÔÚ2LÃܱÕÈÝÆ÷ÄÚ£¬·´Ó¦2NO2£¨g£©=N2O4£¨g£©¡÷H=-180kJ¡¤mol-1£¬n£¨NO2£©Ëæʱ¼ä±ä»¯ÈçÏÂ±í£º

ʱ¼ä/s

0

1

2

3

4

5

n(NO2)/mol

0.040

0.020

0.010

0.005

0.005

0.005

ÓÃNO2±íʾ0¡«2sÄڸ÷´Ó¦µÄƽ¾ùËÙ¶È____¡£ÔÚµÚ5sʱ£¬NO2µÄת»¯ÂÊΪ____¡£¸ù¾ÝÉϱí¿ÉÒÔ¿´³ö£¬Ëæ×Å·´Ó¦½øÐУ¬·´Ó¦ËÙÂÊÖð½¥¼õС£¬ÆäÔ­ÒòÊÇ____¡£

¡¾ÌâÄ¿¡¿ÂÈ»¯ÑÇÍ­(CuCl)¿ÉÓÃ×÷´ß»¯¼Á¡¢É±¾ú¼Á¡¢Ã½È¾¼Á¡¢ÍÑÉ«¼Á¡£CuClÊÇÒ»ÖÖ°×É«·ÛÄ©£¬Î¢ÈÜÓÚË®¡¢²»ÈÜÓÚÒÒ´¼£¬ÔÚ³±ÊªµÄ¿ÕÆøÖÐÒ×±»Ñõ»¯¡£ÊµÑéÊÒ²ÉÓÃÈçͼװÖúÍÏÂÁв½ÖèÖƱ¸ÂÈ»¯ÑÇÍ­¡£

²½Öè1£ºÔÚÈý¾±ÉÕÆ¿ÖмÓÈë20%ÑÎËᡢʳÑΡ¢Í­Ð¼£¬¼ÓÈÈÖÁ60¡«70¡æ£¬¿ª¶¯½Á°èÆ÷£¬Í¬Ê±´Óc¿Ú»ºÂýͨÈëÑõÆø£¬ÖƵÃNa[CuCl2]ÈÜÒº¡£

²½Öè2£º·´Ó¦ÍêÈ«ºó£¬ÀäÈ´£¬¹ýÂË£¬ÂËÒºÓÃÊÊÁ¿µÄˮϡÊÍ£¬Îö³öCuCl¡£

²½Öè3£º¹ýÂË£¬·Ö±ðÓÃÑÎËá¡¢ÒÒ´¼Ï´µÓÂ˳öµÄ¹ÌÌå¡£

²½Öè4£ºÔÚÕæ¿Õ¸ÉÔïÆ÷ÖÐ60¡«70 ¡æ¸ÉÔï2h£¬ÀäÈ´ºóµÃµ½²úÆ·¡£

»Ø´ðÏÂÁÐÎÊÌ⣺

(1)ÖÊÁ¿·ÖÊýΪ20%µÄÑÎËáÃܶÈΪ1.1g/cm3£¬ÎïÖʵÄÁ¿Å¨¶ÈΪ___________£»ÅäÖÆ20%ÑÎËáÐèÒªµÄ²£Á§ÒÇÆ÷ÓУº__________¡¢ÉÕ±­¡¢²£Á§°ô¡¢½ºÍ·µÎ¹Ü¡£

(2)²½Öè1ÖÐÖƵÃNa[CuCl2]µÄ»¯Ñ§·½³ÌʽΪ________________________________¡£

(3)¸ù¾Ý¡°²½Öè2ÖÐÓÃˮϡÊÍÂËÒºÄܵõ½CuCl¡±ÍƲ⣬ÂËÒºÖдæÔÚµÄƽºâÊÇ____________¡£

(4)Îö³öµÄCuCl¾§Ìå²»ÓÃË®¶øÓÃÑÎËá¡¢ÒÒ´¼·Ö±ðÏ´µÓµÄÄ¿µÄÊÇ__________________¡£

(5)²â¶¨²úÆ·ÖÐÂÈ»¯ÑÇÍ­µÄÖÊÁ¿·ÖÊý£¬ÊµÑé¹ý³ÌÈçÏ£º

׼ȷ³ÆÈ¡ÖƱ¸µÄÂÈ»¯ÑÇÍ­²úÆ·0.25 g£¬½«ÆäÖÃÓÚ×ãÁ¿µÄFeCl3ÈÜÒºÖУ¬´ýÑùÆ·È«²¿Èܽâºó£¬¼ÓÈëÊÊÁ¿Ï¡ÁòËᣬÓÃ0.10 mol/LµÄÁòËáîæ[Ce(SO4)2]±ê×¼ÈÜÒºµÎ¶¨µ½Öյ㣬ÏûºÄÁòËáîæÈÜÒº24.50 mL£¬·´Ó¦ÖÐCe4+±»»¹Ô­ÎªCe3+¡£(ÒÑÖª£ºCuCl+FeCl3=CuCl2+FeCl2)

¢ÙÁòËáîæ±ê×¼ÈÜҺӦʢ·ÅÔÚ________(Ìî¡°Ëᡱ»ò¡°¼î¡±)ʽµÎ¶¨¹ÜÖС£

¢Ú²úÆ·ÖÐÂÈ»¯ÑÇÍ­µÄÖÊÁ¿·ÖÊýΪ________¡£

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

¾«Ó¢¼Ò½ÌÍø