µÚ6Õ - ÏàƽºâÏ°Ìâ¼°½â´ð - ͼÎÄ ÏÂÔر¾ÎÄ

µÚÎåÕ Ïàƽºâ

Ò»¡¢Ìî¿ÕÌâ

1¡¢Ò»¶¨Î¶ÈÏ£¬ÕáÌÇË®ÈÜÒºÓë´¿Ë®´ïµ½Éø͸ƽºâʱµÄ×ÔÓɶÈÊýµÈÓڣߣߣߣߣߣߣߣߣߣߡ£ 2¡¢´¿ÎïÖÊÔÚÒ»¶¨Î¶ÈÏÂÁ½Ï๲´æʱµÄ×ÔÓɶÈÊýµÈÓڣߣߣߣߣߣߣߣߣߣߡ£

3¡¢NaCl(S)ºÍº¬ÓÐÏ¡ÑÎËáµÄNaCl±¥ºÍË®ÈÜÒºµÄƽºâϵͳ£¬Æä¶ÀÁ¢×é·ÖÊýÊǣߣߣߣߣߣߣߡ£ 4¡¢ÉèÏÂÁл¯Ñ§·´Ó¦Í¬Ê±¹²´æʱ²¢´ïµ½Æ½ºâ£¨900-1200K£©£º

CaCO3?s??CaO?s??CO2?g? CO2?g??H2?g??CO?g??H2O?g?

H2O?g??CO?g??CaO(s)?CaCO3(s)?H2?g?

Ôò¸ÃϵͳµÄ×ÔÓɶÈÊýΪ£ß£ß£ß£ß£ß£ß¡£

5¡¢º¬KNO3ºÍNaClµÄË®ÈÜÒºÓë´¿Ë®´ïµ½Éø͸ƽºâʱ£¬Æä×é·ÖÊýΪ£ß£ß£ß£ß£¬ÏàÊýΪ£ß£ß£ß£¬ ×ÔÓɶÈÊýΪ£ß£ß£ß£ß¡£

6¡¢ÔÚÇâºÍʯīµÄϵͳÖУ¬¼ÓÒ»´ß»¯¼Á£¬H2ºÍʯī·´Ó¦Éú³ÉnÖÖ̼Ç⻯ºÏÎ´ËϵͳµÄ¶ÀÁ¢ ×é·ÖÊýΪ£ß£ß£ß£ß£ß£ß¡£

7¡¢ÍêÈ«»¥ÈܵÄ˫ҺϵÖУ¬ÔÚxB=0.6´¦£¬Æ½ºâÕôÆøѹÓÐ×î¸ßÖµ£¬ÄÇô×é³ÉΪxB=0.4µÄÈÜÒºÔÚ ÆøҺƽºâʱ£¬xB£¨g£©¡¢xB£¨l£©¡¢xB(×Ü)µÄ´óС˳ÐòΪ£ß£ß£ß£ß£ß£ß¡£½«xB=0.4µÄÈÜÒº½øÐÐ ¾«Áóʱ£¬Ëþ¶¥½«µÃµ½£ß£ß£ß£ß£ß£ß¡£

8¡¢¶ÔÓÚÉø͸ƽºâϵͳ£¬ÏàÂɵÄÐÎʽӦд³É£ß£ß£ß£ß£ß£ß¡£

9¡¢NH4Cl¹ÌÌå·Ö½â´ïµ½Æ½ºâʱ£¬NH4Cl?s??HCl(g)?NH3(g)£¬ÏµÍ³µÄ¶ÀÁ¢×é·ÖÊýΪ£ß

£ß£ß£¬×ÔÓɶÈΪ£ß£ß£ß¡£

10¡¢½«AlCl3ÈÜÓÚË®ÖУ¬È«²¿Ë®½â£¬Éú³ÉAl(OH)3³Áµí£¬´Ëϵͳ×ÔÓɶÈÊýf=£ß£ß£ß£ß¡£ 11¡¢ÒÑÖª100oCʱˮµÄ±¥ºÍÕôÆøѹΪ101.325KPa,Óù«Ê½£ß£ß£ß£ß£ß£ß£ß£ß£ß¿ÉÇó³ö25oCʱ Ë®µÄ±¥ºÍÕôÆøѹ¡£

´ð°¸

1¡¢2 2¡¢0 3¡¢2

1

4¡¢3 5¡¢3;2:4 6¡¢2

7¡¢xB£¨g£©>xB(×Ü)>xB£¨l£© xB=0.6ºã·Ð»ìºÏÎï 8¡¢f?k???3 9¡¢1;1 10¡¢2 11¡¢lnp2?rHm?p1R?11???T?T??

2??1¶þ¡¢µ¥Ñ¡Ìâ

1¡¢ÓÒͼΪH2OA.-(NH4)2SO4B.µÄ·Ðµã-×é³Éͼ¡£ÈçºÎ´ÓwB=0.4 µÄÈÜÒºÖÐÌáÈ¡½Ï¶àµÄ¾«ÖÆ(NH4)2SO4¹ÌÌå?( ) A.½µÎÂÖÁ-18.3¡æÒÔÏ B.ÔÚÃܱÕÈÝÆ÷ÖÐƽºâÕô·¢ C.ÔÚ³¨¿ªÈÝÆ÷Öж¨ÎÂÕô·¢ ÔÙ½µÎÂÖÁ-18.3¡æÒÔÉÏ D.ÏÈÕô·¢Ò»²¿·ÖË®·Ý£¬ÔÙ ½µÎÂÖÁ-18.3¡æÒÔÏÂ

2¡¢ÈçÓÒͼËùʾ£¬µ±Ë®´¦ÔÚÈýÏàµãƽºâʱ£¬Èôϵͳ·¢Éú¾øÈÈÅòÕÍ£¬Ë®µÄÏà̬½«ÈçºÎ±ä»¯£¿ ( )

A.ÆøÏà¡¢¹ÌÏàÏûʧ£¬È«²¿±ä³ÉҺ̬£» B.ÆøÏà¡¢ÒºÏàÏûʧ£¬È«²¿±ä³É¹Ì̬£» C.ÒºÏàÏûʧ£¬¹ÌÏà¡¢ÆøÏ๲´æ£» D.¹ÌÏàÏûʧ£¬ÒºÏà¡¢ÆøÏ๲´æ

3¡¢¶Ô¼òµ¥µÍ¹²ÈÛÌåϵ£¬ÔÚ×îµÍ¹²È۵㣬µ±Î¶ȼÌÐøÏ ½µÊ±£¬Ìåϵ´æÔÚ( )

A.Ò»Ïà B.¶þÏà C.Ò»Ïà»ò¶þÏà D.ÈýÏà

4¡¢ÒÑÖª´¿AºÍ´¿BµÄ±¥ºÍÕôÆøѹpA*

ÖÐ

2

²»¶Ï¼ÓÈëB£¬ÔòÈÜÒºµÄÕôÆøѹ( )

A.²»¶ÏÔö´ó B.²»¶Ï¼õС C.ÏÈÔö´óºó¼õС D.ÏȼõСºóÔö´ó 5¡¢ÔÚÃèÊöºã·Ð»ìºÏÎïʱ£¬ÏÂÁи÷µãºÎÕß²»¶Ô?( )

A. ºã·Ðµã f*=0£¬¶¨Ñ¹ÏÂÊÇÒ»¸öÎÞ±äÁ¿µã B. ²»¾ßÓÐÃ÷È·µÄ×é³É C. ƽºâʱÆøÏàºÍÒºÏà×é³ÉÏàͬ D. ºã·ÐµãËæѹÁ¦¸Ä±ä¶ø¸Ä±ä

6¡¢ÒÒ֪ij»ìºÏÎïµÄ×îµÍ¹²ÈÛµãµÄ×é³ÉΪº¬B40%¡£º¬BΪ20%µÄA¡¢B»ìºÏÎï½µÎÂÖÁ×îµÍ

¹²ÈÛµãʱ£¬´ËʱÎö³öµÄ¹ÌÌåΪ( ) A.¹ÌÌåA B.¹ÌÌåB

C.×îµÍ¹²ÈÛ»ìºÏÎï D.×é³ÉСÓÚ40%µÄA¡¢B¹ÌÌå»ìºÏÎï 7¡¢²¿·Ö»¥ÈÜ˫Һϵ£¬Ò»¶¨Î¶ÈÏÂÈô³öÏÖÁ½Ïàƽºâ£¬Ôò( ) A.ÌåϵµÄ×é³ÉÒ»¶¨

B.Á½ÏàµÄ×é³ÉÓëÌåϵµÄ×Ü×é³ÉÎ޹ء£ÇÒÁ½ÏàµÄÁ¿Ö®±ÈΪ³£Êý C.Á½ÏàµÄ×é³ÉÓëÌåϵµÄ×Ü×é³É¹ØÓС£ÇÒÁ½ÏàÖÊÁ¿·ÖÊýÖ®±ÈΪ³£Êý D.Á½ÏàµÄ×é³É²»¶¨ 8¡¢ºã·Ð»ìºÏÎï( ) A.ÆøÒºÁ½ÏàµÄÁ¿Ïàͬ

B.ÆøÒºÁ½ÏàÖÐijÖÖ×é·ÖBµÄÎïÖʵÄÁ¿Ïàͬ C.ÆøÒºÁ½ÏàµÄ×é³ÉÏàͬ

D.ÆøÒºÁ½ÏàµÄ×é³ÉÏàͬ£¬ÔÚP-XͼºÍÔÚT-XͼÉϾùΪͬһֵ 9¡¢ÓÒͼÖÐPQÏßÉÏ( ) A.f*=0 B.f*=1

C.AºÍB²»»¥ÈÜ D.AºÍBÍêÈ«»¥ÈÜ

10¡¢ÔÚË«×é·ÖÌåϵP-XͼÉÏ£¬ÈôÕôÆøѹ-×é³ÉÇúÏ߶ÔÀ­ÎÚ¶û¶¨

ÂɲúÉúºÜ´óµÄ¸ºÆ«²î£¬ÓÐÒ»¼«Ð¡µã£¬Ôò( ) A.½Ð×îµÍºã·Ð»ìºÏÎï B.½Ð×î¸ßºã·Ð»ìºÏÎï

C.Ëù¶ÔÓ¦µÄ×é³ÉÔÚÈκÎÇé¿ö϶¼²»·¢Éú±ä»¯ D.ÔڸõãÆøÒºÁ½ÏàµÄÁ¿Ïàͬ

3

11¡¢ÔÚÓе͹²ÈÛµã´æÔÚµÄÌåϵ£¬Èô²½ÀäÇúÏßÉϳöÏÖƽ̨£¬´Ëʱ£¬Ìåϵ´æÔÚµÄÏàÊý( ) A.1 B.2 C.3 D.2»ò3 12¡¢ÄÜ·ñ´Óº¬80%µÄBi-Cd»ìºÏÎïÖзÖÀë³ö´¿Cd?( ) A.ÄÜÍêÈ«·ÖÀ룬µÃµ½´¿Cd B.Äܲ¿·Ö·ÖÀ룬µÃµ½´¿Cd

C.²»ÄÜ·ÖÀ룬µÃ²»µ½´¿Cd D.½öͨ¹ý¼ÓÈȵķ½·¨±ã¿ÉµÃµ½Ò»²¿·Ö´¿Cd 13¡¢ÔÚË«×é·ÖÌåϵT-XͼÉÏ£¬ÈôÓÐÒ»¼«Ð¡µã£¬Ôò¸Ãµã( ) A.½Ð×îµÍºã·Ð»ìºÏÎï B.½Ð×î¸ßºã·Ð»ìºÏÎï

C.Ëù¶ÔÓ¦µÄ×é³ÉÔÚÈκÎÇé¿ö϶¼²»·¢Éú±ä»¯ D.ÔڸõãÆøÒºÁ½ÏàµÄÁ¿Ïàͬ

14¡¢¶¨Ñ¹Ï£¬ÔÚ×îµÍ¹²È۵㣬ϵͳ´æÔÚ( )

A.Ò»Ïà B.¶þÏà C.ÈýÏà D.ËÄÏà 15¡¢ÏÂͼΪÆøҺƽºâÏàͼ£¬Í¼ÖÐMµã( ) A.´ú±íѹÁ¦Îªp1ʱÆøÏàµÄ×é³É B.´ú±íѹÁ¦Îªp1ʱҺÏàµÄ×é³É

C.ÊÇѹÁ¦Îªp1ʱÆøÏàÖÐÎïÖÊBµÄÎïÖʵÄÁ¿ D.ÊÇѹÁ¦Îªp1ʱҺÏàÖÐÎïÖÊBµÄÁ¿

16¡¢ÒÑÖª±½µÄlgp~1/TµÄͼÖУ¬ÆäÖ±ÏßбÂÊΪ-1.767¡Á103£¬

Ôò±½µÄĦ¶ûÕô·¢ÈÈΪ( )

A.33.83J¡¤mol-1 B.33.83kJ¡¤mol-1 C.14.29 J¡¤mol-1 D.14.29kJ¡¤mol-1 17¡¢ÏÂͼΪÆøҺƽºâÏàͼ£¬Í¼ÖÐMµã( ) A.´ú±íζÈΪT1ʱÆøÏàµÄ×é³É B.´ú±íζÈΪT1ʱҺÏàµÄ×é³É

C.ÊÇζÈΪT1ʱÆøÏàÖÐÎïÖÊBµÄÎïÖʵÄÁ¿ D.ÊÇζÈΪT1ʱҺÏàÖÐÎïÖÊBµÄÎïÖʵÄÁ¿

18¡¢ÈôA¡¢B¶þ×é·ÖÐγÉÈýÖÖÎȶ¨»¯ºÏÎÔòµ±A-BµÄÈÜÒºÀäȴʱ£¬ ×î¶à¿ÉͬʱÎö³ö ( )

A.Ò»¸ö¹ÌÏà B.¶þ¸ö¹ÌÏà C.Èý¸ö¹ÌÏà D.Ëĸö¹ÌÏà 19¡¢µ±Ë®´¦ÔÚÈýÏàµã¶øƽºâʱ£¬Í»È»Ôö´óѹÁ¦£¬Ë®µÄÏà̬½«ÈçºÎ±ä»¯?( )

4

A.ÆøÏà¡¢¹ÌÏàÏûʧ£¬È«²¿±ä³ÉҺ̬ B.ÆøÏà¡¢ÒºÏàÏûʧ£¬È«²¿±ä³É¹ÌÏà C.ÒºÏà¡¢¹ÌÏàÏûʧ£¬È«²¿±ä³ÉÆø̬ D.¹ÌÏàÏûʧ£¬ÆøÒºÁ½Ï๲´æ 20¡¢ÈçÏÂͼËùʾ£¬ÔÚÏàºÏÈ۵㣬ÐγɵĻ¯ºÏÎïµÄ²½ÀäÇúÏß ÉϳöÏÖ( ) A.¹Õµã B.ƽ̨ C.ÎÞ¹ÕµãºÍƽ̨ D.Ò»¸ö¹ÕµãÁ½¸öƽ̨

21¡¢ÈçÓÒͼËùʾ£¬ÔÚABCÏàÇøÄÚ£¬µ±Î¶ÈÒ»¶¨Ê±£¬

Á½ÏàµÄ×é³É( )

A.Ò»¶¨ B.²»Ò»¶¨ C.¾ö¶¨ÓÚϵͳµÄ×Ü×é³É D.Ïàͬ¡£ 22¡¢ÓÒͼÖУ¬ÆøÒºÁ½Ïà×é³ÉÏàͬµÄµãΪ( ) A.A¡¢Bµã B.Cµã C.A¡¢B¡¢Cµã

D.ÆøÏàÏߺÍÒºÏàÏßÉϸ÷µã

23¡¢Èô²½ÀäÇúÏß³öÏÖƽ̨£¬ ´ËʱÌåϵµÄÌõ¼þ×ÔÓɶÈÊýΪ

( )

A.0 B.1 C.2 D.3

24¡¢ÔÚ101325PaѹÁ¦Ï£¬ ±ùÓë±»¿ÕÆø±¥ºÍÁ˵ÄË®³ÉƽºâµÄζȽÐ( ) A.Äý¹Ìµã B.±ùµã C.ÈýÏàµã D.¶µã 25¡¢ÓÒͼÖУ¬×ÔÓɶȶ¼ÎªÁãµÄÈ«²¿µÄµãΪ ( ) A.J¡¢E1¡¢C¡¢E2¡¢K B.M¡¢E1¡¢N¡¢O¡¢E2¡¢P

C.MNÏß¡¢OPÏß(¶Ëµã³ýÍâ)Éϸ÷µã¼°J¡¢C¡¢K D.MNÏß¡¢OPÏß(°üº¬¶Ëµã)Éϸ÷µã¼°J¡¢C¡¢K

5

26¡¢ÓÉÏÂͼ¿É¿´³ö£¬½«xB=0.8µÄCH3COOC2H5A.ºÍC2H5OHB.

×é³ÉµÄÈÜÒº½øÐзÖÁó£¬ÔòÄܵõ½( ) A.´¿CH3COOC2H5

B.×îµÍºã·Ð»ìºÏÎïºÍ´¿CH3COOC2H5 C.×îµÍºã·Ð»ìºÏÎïºÍ´¿C2H5OH D.´¿C2H5OHºÍCH3COOC2H5

27¡¢ÔÚÏÂͼÖУ¬ ½«×é³ÉΪxµÄÈÜÒº½øÐÐÕôÁóʱ£¬×î³õÁó

³öÒºµÄ×é³ÉΪ( )

A.a B.b C.c D.d

28¡¢µ±ÏÂͼÖУ¬wB=0.2ʱ£¬ÏÂÁÐÐðÊöÖв»ÕýÈ·µÄÊÇ( ) A.ÈÜÒº½µÎÂÖÁ0¡æÒÔÏ£¬¿ÉµÃµ½´¿¹ÌÌåH2O B.ÈÜÒº½µÎÂÖÁ-18.3¡æʱ£¬ÈÜÒºµÄ×é³ÉΪ¶¨Öµ

C.¹ÌÌå»ìºÏÎïÉýÎÂÖÁ-18.3¡æʱ£¬¹ÌÌåH2OÏÈÈÛ »¯£¬BºóÈÛ»¯

D.¹ÌÌå»ìºÏÎïÉýÎÂÖÁ-18.3¡æʱ£¬¹ÌÌåH2OºÍBͬ ʱ¿ªÊ¼ÈÛ»¯£¬ÖÁÈ«²¿ÈÛ»¯ºó£¬Î¶ÈÓÖ¼ÌÐøÉÏÉý 29¡¢¶ÔÓÒͼÃèÊöÕýÈ·µÄÊÇ( )

A.MNÏßÉϸ÷µã£¬ÏµÍ³´æÔÚÈýÏàƽºâ B.M¡¢Nµã£¬Ìåϵ´æÔÚÁ½Ïàƽºâ C. E1¡¢E2µãÌåϵ´æÔÚÁ½Ïàƽºâ D.J¡¢C¡¢KµãÌåϵ´æÔÚÁ½Ïàƽºâ 30¡¢Á½ÏàƽºâµÄ±êÖ¾ÊÇ( )

A.p(¦Á)=p(¦Â) B.T(¦Á)=T(¦Â) C.¦Ìi(¦Á)=¦Ìi(¦Â) D.xi(¦Á)+ xi(¦Â)=1

31¡¢ÔÚË®µÄp-TÏàͼÖУ¬H2O(l)µÄÕôÆøѹÇúÏß´ú±íµÄÊÇ( ) A.¦µ=1£¬f=2 B.¦µ=2£¬f=1 C.¦µ=3£»f=0 D.¦µ=2£¬f=2

32¡¢ÔÚ²»¿¼ÂÇÖØÁ¦³¡¡¢µç´Å³¡µÈÍâ½çÒòËØʱ£¬¶¨Î¶¨Ñ¹Ï£¬ÏàÂɵıí´ïʽΪ( ) A.f=K-¦µ B.f=K-¦µ+1 C.f=K-¦µ+2 D.f=K-¦µ-2 33¡¢¿Ë-¿Ë·½³ÌµÄ׼ȷ±íʾʽΪ( )

A.dp¡¤dT=¦¤Sm/¦¤Vm B.dp/dT=¦¤Hm/(T¦¤Vm)

6

C.dlnp/dT=p¦¤Hm/(RT2) D.dlnp/dT=¦¤Hm/(RT2) 34¡¢¿ËÀ­ÅåÁú·½³Ì±íÃ÷( )

A.Á½ÏàƽºâʱµÄƽºâѹÁ¦Ëæζȶø±ä»¯µÄ±ä»¯ÂÊ B.ÈÎÒâ״̬ÏÂѹÁ¦Ëæζȵı仯ÂÊ C.ËüÊÊÓÃÓÚÈκÎÎïÖʵÄÁ½Ïàƽºâ D.ÒÔÉÏ˵·¨¶¼²»¶Ô

35¡¢ÏÂÁÐÌåϵÖÐÄÄÒ»¸öÊǾùÏàÌåϵ( ) A.Ë®ÎíºÍË®ÕôÆø»ìºÏÔÚÒ»Æ𣻠B.Èé×´Òº

C.Ë®ºÍ±»Ë®±¥ºÍµÄ¿ÕÆø D.Á½ÖÖ²»Í¬µÄ½ðÊôÐγɵĹÌÈÛÌå

´ð°¸

1 C 12 B 23 A 34 A

Èý¡¢¶àÑ¡Ìâ

1¡¢¾ßÓÐ×î¸ßºã·ÐµãµÄ¶þ×é·ÖÍêȫ˫Һϵͳ( ) A.ÈÜÒºµÄÕôÆøѹ¶ÔÀ­ÎÚ¶û¶¨ÂɲúÉúÕýÆ«²î B.ÈÜÒºµÄÕôÆøѹ¶ÔÀ­ÎÚ¶û¶¨ÂɲúÉú¸ºÆ«²î C.ÈÜÒºÖеĸ÷×é·Ö¶ÔÀ­ÎÚ¶û¶¨ÂɲúÉú½Ï´óµÄÆ«²î D.P-XͼÉϾßÓм«´óµã E.P-XͼÉϾßÓм«Ð¡

2¡¢ÔÚË«×é·ÖÌåϵp-XͼÉÏ£¬ÈôÕôÆøѹ-×é³ÉÇúÏ߶ÔÀ­ÎÚ¶û¶¨ÂɲúÉúºÜ´óµÄÕýÆ«²î£¬ÓÐÒ»¼«´óµã£¬Ôò¸ÃÌåϵ( )

A.½Ð×îµÍºã·Ð»ìºÏÎï B.½Ð×î¸ßºã·Ð»ìºÏÎï

2 C 13 A 24 B 35 D 3 B 14 C 25 C 4 D 15 B 26 C 5 B 16 B 27 D 6 C 17 A 28 D 7 C 18 B 29 D 8 C 19 A 30 C 9 A 20 B 31 B 10 B 21 A 32 C 11 D 22 C 33 D 7

C.Ëù¶ÔÓ¦µÄ×é³ÉÊÇΨһ²»±äµÄ D.ÔڸõãÆøÒºÁ½ÏàµÄ×é³ÉÏàͬ E.ÔڸõãÆøÒºÁ½ÏàµÄÁ¿Ïàͬ

3¡¢ÓÒͼΪH2OA.-(NH4)2SO4B.µÄ·Ðµã-×é³Éͼ¡£ÈçºÎ´ÓwB=0.2 µÄÈÜÒºÖÐÌâÈ¡½Ï¶àµÄ(NH4)2SO4

¹ÌÌå?( ) A.½µÎÂÖÁ-18.3¡æÒÔÏ B.ÔÚÃܱÕÈÝÆ÷ÖÐƽºâÕô·¢ C.ÔÚ³¨¿ªÈÝÆ÷Öж¨ÎÂÕô·¢

D.Õô·¢µôÒ»²¿·ÖË®·Ý£¬´ýÈÜÒº³öÏÖ

½á¾§Ê±£¬ÔÙ½ÂÎÂÖÁ-18.3¡æÒÔÉÏ

E.ÏÈÕô·¢µôÒ»²¿·ÝË®·Ý£¬´ýÈÜÒººÜ

Ũʱ£¬ÔÙ½µÎÂÖÁ0¡æÒÔÏÂ

4¡¢0¡æʱ±ùµÄÈÛ»¯ÈÈΪ6008J¡¤mol-1¡£ÔÚ¸ÃζÈϱùµÄĦ¶ûÌå»ýΪ19.625cm3¡¤mol-1£¬ÒºÌ¬Ë®µÄĦ¶ûÌå»ýΪ18.018cm3¡¤mol-1£¬ÔòѹÁ¦¶ÔÓÚÈÛµãµÄ¹ØϵÊÇ( ) A.ѹÁ¦Ôö´ó£¬ÈÛµãÉý¸ß

B.ѹÁ¦Ôö´ó101325PaʱµÄ±ùµÄÈÛµãÉý¸ß0.00753K C.ѹÁ¦Ôö´ó£¬±ùµÄÈ۵㽵µÍ

D.ѹÁ¦Ôö´ó101325Paʱ±ùµÄÈ۵㽵µÍ0.00753K E.ѹÁ¦Ôö´ó101325Pa£¬±ùµÄÈ۵㲻±ä 5¡¢ÔÚÏÂͼÖУ¬Qµã( ) A.ΪÏàµã B.ΪÎïϵµã

C.Ëù´ú±íµÄÌåϵÊÇÓÉaµãµÄÒºÏàºÍbµãµÄÆøÏàËù×é³É D.Ëù±íʾµÄÌåϵµÄ×é³ÉΪÁ½ÏàµÄ×é³ÉaºÍbµÄ×ÜºÍ E.È·¶¨ºó£¬Ò»¶¨Î¶ÈÏ£¬Á½ÏàµÄÖÊÁ¿Ö®±ÈΪ³£Êý

6¡¢ÏÂͼΪCO2µÄÏàͼ£¬ °Ñ¸ÖÆ¿ÖеÄÒºÌåCO2ÔÚ¿ÕÆøÖÐÅç³öµÄ˲¼ä( ) A.´ó²¿·Ö³ÉΪÆøÌå B.È«²¿³ÉΪÆøÌå C.È«²¿³ÉΪ¹ÌÌå D.ÓÐÉÙ²¿·ÖÒºÌå´æÔÚ

8

E.ÎÞÒºÌå´æÔÚ

7¡¢È·ÇеØ˵£¬ Îȶ¨»¯ºÏÎïÊÇÖ¸¸Ã»¯ºÏÎï( ) A. ÔÚ¼ÓÈȹý³ÌÖв»Ò×·Ö½â B. ÈÛµã¸ß

C. ÈÛµã½Ï·Ö½âÎÂ¶È¸ß D. ÈÛ»¯Ê±ÒºÏàÓë¹ÌÏàÓÐÏàͬµÄ×é³É E. ¾ßÓÐÏàºÏÈÛµã

8¡¢µ±Ë®´¦ÔÚÈýÏàµãƽºâʱ£¬ÏµÍ³·¢Éú¾øÈÈÅòÕÍ£¬Î¶ȽµµÍ0.1K£¬Ë®µÄÏà̬½«ÈçºÎ±ä»¯? ( ) A.ÏàÊý¼õÉÙ B.ÏàÊýÔö¼Ó C.×ÔÓɶÈÊý¼õÉÙ D.×ÔÓɶÈÊýÔö¼Ó

E.ÆøÏàÏûʧ£¬ÒºÏà¡¢¹ÌÏ๲´æ

9¡¢ÏÂͼÊÇÒ»¸ö²»ÍêÈ«µÄÏàͼ£¬ ¹À¼ÆÒ»ÏÂAºÍBÒªÐÎ³É ( ) A.Ò»ÖÖÎȶ¨»¯ºÍÎï B.¶þÖÖÎȶ¨»¯ºÏÎï C.Ò»ÖÖ×îµÍ¹²ÈÛ»ìºÏÎï D.ÈýÖÖ×îµÍ¹²ÈÛ»ìºÏÎï E.Ò»ÖÖ×îµÍºã·Ð»ìºÏÎï

10¡¢µ±Ë®´¦ÔÚÈýÏàµã´¦£¬ÏµÍ³·¢Éú¾øÈÈѹËõ£¬Î¶ȸıä0.1K£¬Ë®µÄÏà̬½«ÈçºÎ±ä»¯( ) A.±äΪһÏà B.±äΪÁ½Ï๲´æ C.ÒºÏàÏûʧ D.ѹÁ¦»áÉý¸ß»ò½µµÍ E.¹ÌÏàÏûʧ

11¡¢¹ØÓÚ²½ÀäÇúÏßµÄÏÂÊö˵·¨ÖУ¬ºÎÕß²»Í×( ) A.²½ÀäÇúÏß³öÏÖתÕÛ£¬Òâζ×ÅÓÐÏà±ä

B.Ö»Óв½ÀäÇúÏßÉϳöÏÖƽ̨ʱ£¬²ÅÒâζ×ÅϵͳÖÐÓÐÏà±ä C.²½ÀäÇúÏßÉϳöÏÖתÕÛʱ£¬ÏµÍ³µÄÌõ¼þ×ÔÓɶÈf*=0 D.²½ÀäÇúÏßÉϳöÏÖƽ̨ʱ£¬Ìõ¼þ×ÔÓɶÈf*½«¼õС E.²½ÀäÇúÏßÉϳöÏÖ¹Õµãʱ£¬Ìõ¼þ×ÔÓɶÈf*½«¼õС

12¡¢ÔÚ¶þ×é·ÖË«ÒºT-x(»ò p-x)ÏàͼÖУ¬×´Ì¬µãÓɵ¥ÏàÇø½øÈëÁ½ÏàÇøʱ£¬ÌåϵµÄ×ÔÓɶÈÊý

9

( )

A. Ôö¼Ó B. ¼õÉÙ C. f* =0 D. f* =1 E. f* =2 13¡¢ÈçÓÒͼËùʾ£¬ Cµã½ÐÁÙ½çÈܽâζȡ£ÔÚCµãÒÔÉÏ( ) A.´æÔÚÒ»Ïà B.´æÔÚÁ½Ïà C.f*=0 D.f*=1 E.f*=2

14¡¢ÔÚÓÒͼÖдæÔÚ( ) A.6¸öÁ½ÏàÇø B.6¸ö¹ÌÒºÁ½ÏàÇø C.9¸öÁ½ÏàÇø D.ˮƽÏßÉÏf*=1 E.´¹Ö±ÏßÉÏf*=0

15¡¢ÓÉÓÒͼ¿É¿´³ö£¬A¡¢B¼ä¿ÉÐγÉÈýÖÖ»¯ºÏÎËùÒÔ½«Ò»¶¨Á¿µÄAºÍB»ìºÏºó£¬ÌåϵÖлá

´æÔÚ( ) A.3ÖÖ»¯ºÏÎï B.3-5ÖÖ»¯ºÏÎï C.2ÖÖ»¯ºÏÎï D.1-2ÖÖ»¯ºÏÎï E.ÈýÏàƽºâµÄ״̬

16¡¢ÓÒͼΪH2OA.-(NH4)2SO4B.µÄ·Ðµã-×é³Éͼ£¬wB=0.2µÄÈÜÒº½µÎ¹ý³ÌÖÐ( ) A.Ö»»á³öÏÖH2O(s)£¬²»»á³öÏÖ(NH4)2SO4(s) B.²»µ«»á³öÏÖH2O(s)£¬»¹»á³öÏÖ(NH4)2SO4(s) C.Ê×ÏȳöÏÖH2O(s)£¬È»ºó³öÏÖ(NH4)2SO4(s) D.Èô³öÏÖ¹ÌÌ壬Ôò±ØΪһÏà E.Èô³öÏÖ¹ÌÌ壬Ôò±ØΪ¶þÏà

17¡¢ÓÒͼΪH2OA.-(NH4)2SO4 B.µÄ·Ðµã-×é³Éͼ£¬wB =0.5

µÄÈÜÒº£¬½µÎ¹ý³ÌÖÐ( )

10

A.Ö»»á³öÏÖ(NH4)2SO4(s)£¬²»»á³öÏÖH2O(l) B.²»µ«»á³öÏÖ(NH4)2SO4 (s)£¬»¹»á³öÏÖH2O(s) C.Èô³öÏÖ¹ÌÌ壬ÔòÌåϵ±ØΪһÏà D.Èô³öÏÖ¹ÌÌ壬ÔòÌåϵ±ØΪ¶þÏà

E.Èô³öÏÖ¹ÌÌ壬ÌåϵÖпÉÄÜ»á³öÏÖ¶þÏà»òÈýÏà 18¡¢¼ÙÉèһƽºâÌåϵÖÐÓÐK¸ö×é·Ö£¬¦µ¸öÏ࣬ÈçK¸ö×é·Ö

ÔÚÿÏàÖоù´æÔÚ£¬ÔòÒªÃèÊö´ËÌåϵµÄ״̬£¬Ðè×ÔÓɶÈÊýӦΪ( ) A.f=¦µ(K-1)+2 B.f=K (¦µ-1)+2 C.f=K-¦µ+1 D.f=K-¦µ+2 E.f=K-¦µ+n (nΪ×ÔÈ»Êý)

19¡¢Ó÷ÖÁóµÄ·½·¨½«50%µÄÒÒ´¼Ë®ÈÜÒº½øÐÐÌá´¿£¬¿É( ) A.ÔÚ²ÐÒºÖеõ½´¿ÒÒ´¼ B.ÔÚÁó³öÒºÖеõ½´¿ÒÒ´¼ C.ͬʱµÃµ½´¿Ë®ºÍ´¿ÒÒ´¼ D.ͬʱµÃµ½Ë®ºÍºã·Ð»ìºÏÎï E.ÔÚÁó³öÒºÖеõ½ºã·Ð»ìºÏÎï ´ð°¸ 1 BE 11 BC

ËÄ¡¢¼ò´ðÌâ

1¡¢¶ÔÓÚ´¿Ë®£¬µ±Ë®Æø¡¢Ë® ¡¢±ùÈýÏ๲´æʱ£¬×ÔÓɶÈÓɶÈΪ¶àÉÙ£¿ÄãÔõÑùÀí½â£¿ 2¡¢ÓÐË®Æø±ä³ÉҺ̬ˮ£¬ÊÇ·ñÒ»¶¨Òª¾­¹ýÁ½Ïàƽºâ̬£¬ÊÇ·ñ»¹ÓÐÆäËü;¾¶¡£

´ð£º´ÓË®µÄÏàͼ¿ÉÖª£¬½«Ë®ÆûÉýγ¬¹ý647.2Kºó£¬ÔÚ¼Óѹ³¬¹ý2.2¡Á107Pa£¬È»ºóÔÙ½µÎ ½µÑ¹£¬¿É²»¾­¹ýÁ½Ïàƽºâ̬½«Ë®Æûת±ä³ÉҺ̬ˮ.

3¡¢ÎªÊ²Ã´¾ßÓÐ40£¥CdµÄBi-CdÌåϵ,Æä²½ÀäÇúÏßµÄÐÎ×´Óë´¿Bi¼°´¿CdµÄÏàͬ? 4¡¢ÔõÑù¿É´ÓCd 80£¥µÄBi- Cd»ìºÏÎïÖзÖÀë³öCdÀ´£¿ÄÜ·ñÈ«²¿·ÖÀë³öÀ´? 5¡¢ÏÂÁÐÁ½ÖÖÌåϵ¸÷Óм¸ÖÖ×é·Ö¼°¼¸¸ö×ÔÓɶȣ¿ A. NaH2PO4ÈÜÓÚË®ÖгÉΪÓëË®ÆøƽºâµÄ²»±¥ºÍÈÜÒº B. AlCl3ÈÜÓÚË®Öв¢·¢ÉúË®½â³Áµí³öAl(OH)3¹ÌÌå

11

2 AD 12 BD 3 CD 13 AE 4 CD 14 BC 5 BE 15 DE 6 AE 16 BC 7 DE 17 BE 8 AD 18 AD 9 BD 19 DE 10 BC ´ð£ºA£¨²»¿¼ÂÇË®½â£©

C = 2£¨NaH2PO4¡¢H2O£© P= 2

f = 2-2+2 = 2 (T, p , x ÈÎÁ½¸ö) B AlCl3 + 3H2O £½ Al(OH)3 + 3HCl

C = 4 R = 1 R¡ä= 0 P = 2

f =£¨4-1-0£©-2+2 = 3 £¨ T , pºÍ x£©

6¡¢270KµÄ¹ýÀäË®½Ï±ùÎȶ¨»¹ÊDz»Îȶ¨£¿ºÎÕߵĻ¯Ñ§ÊƸߣ¿¸ß¶àÉÙ£¿ ´ð£º270KµÄ¹ýÀäË®½Ï±ù²»Îȶ¨£¬¹ýÀäË®µÄ»¯Ñ§ÊƸߡ£ ¡ß270K¹ýÀäË®µÄ±¥ºÍÕôÆûѹ p1> ±ùµÄ±¥ºÍÕôÆûѹ p2

??G??2??1??Vdp??p1p2p2p1pRTdp?2.70?8.314ln2?0 pp1¹ýÀäË®»á×Ô·¢µØ½á±ù

7¡¢ËµÃ÷Ë®µÄÈýÏàµãÓëËüµÄ±ùµãµÄÇø±ð¡£ 8¡¢Ö¸³öÏÂÁÐÌåϵµÄ×ÔÓɶȣ¬²¢ËµÃ÷±äÁ¿ÊÇʲô£¿ £¨1£© ÔÚ101325PaµÄѹÁ¦Ï£¬Ë®ÓëË®ÕôÆø´ïƽºâ£» £¨2£© Һ̬ˮºÍË®ÕôÆø´ïƽºâ£»

£¨3£© ÔÚ101325PaµÄѹÁ¦Ï£¬I2ÔÚҺ̬ˮºÍCCl4ÖеķÖÅäÒÑ´ïƽºâ£¨ÎÞ¹ÌÌåµâ´æÔÚ£©£» £¨4£© NH3£¨g£©¡¢H2£¨g£©¡¢N2£¨g£©ÒÑ´ïƽºâ¡£

9¡¢¾ßÓÐ×î¸ß·ÐµãµÄAºÍB¶þ×é·ÖÌåϵ£¬×î¸ßºã·ÐÎïΪC£¬×îºóµÄ²ÐÁôÎïÊÇʲô£¿ÎªÊ²Ã´£¿ ´ð£º¸ù¾Ý¾ßÓÐ×î¸ßºã·ÐµãµÄÍêÈ«»¥ÈÜÁ½ÔªÏµµÄÏàͼ¿ÉÖª£¬²ÐÁôҺΪºã·ÐÎïC£¬²»¹ÜAºÍBÁ½ÎïµÄº¬Á¿¸÷Ϊ¶àÉÙ£¬µ«CÊÇ×î¸ß·Ðµã£¬×îºóÁôÏÂCÎï¡£ 10¡¢Ë®µÄÏàͼÈçÓÒ¡£

ÐðÊöµãK´ú±íµÄÎïϵÔÚµÈνµÑ¹¹ý³ÌÖеÄ״̬¡¢ ÏàÊý¼°×ÔÓɶÈÊýµÄ±ä»¯¡£

12

11¡¢ÓÒͼΪ¶þ×é·ÖϵͳÆø¡¢ÒºÆ½ºâµÄѹÁ¦-×é³Éͼ¡£½«Æ䶨ÐÔ ×ª»¯ÎªÎ¶È-×é³Éͼ¡£¸Ãϵͳ¾­¾«Áóºó£¬Ëþ¶¥½«µÃºÎ×é·Ö£¿ 12¡¢ÓÐÏÂÁл¯Ñ§·´Ó¦´æÔÚ£º

N2?g??3H2?g??2NH3?g? NH4HS(s)?NH3?g??H2S(g) NH4Cl(s)?NH3(g)?HCl(g)

ÔÚÒ»¶¨Î¶ÈÏ£¬Ò»¿ªÊ¼Ïò·´Ó¦Æ÷ÖзÅÈëNH4HS£¬NH4ClÁ½ÖÖ¹ÌÌåÒÔ¼°ÎïÖʵÄÁ¿Ö®±ÈΪ3£º 1µÄÇâÆøºÍµªÆø¡£Îʴﵽƽºâʱ£¬×é·ÖÊýΪ¶àÉÙ£¿×ÔÓɶÈÊýΪ¶àÉÙ£¿

13¡¢ÒÑÖªCO2µÄÁÙ½çζÈΪ31.1oC,ÁÙ½çѹÁ¦Îª7.4¡Á106Pa£¬ÈýÏàµãΪ-56.6 oC¡¢5.18¡Á105P£¬ ÊÔ»­³öCO2ÏàͼµÄʾÒâͼ£¬²¢ËµÃ÷£º£¨1£©ÔÚÊÒμ°³£ÎÂÏ£¬ÈôѸËٵؽ«ÖüÓÐÆøÒº¹²´æµÄCO2 ¸ÖÆ¿·§ÃÅ´ò¿ª£¬·Å³öÀ´µÄCO2¿ÉÄÜ´¦ÓÚʲô״̬£¿£¨2£©Èô»ºÂýµØ°Ñ·§ÃÅ´ò¿ª£¬·Å³öÀ´µÄ CO2´¦ÓÚʲô״̬£¿£¨3£©ÊÔ¹À¼ÆÔÚʲôζȡ¢Ñ¹Á¦·¶Î§ÄÚ£¬CO2ÄÜÒÔҺ̬´æÔÚ£¿ 14¡¢¶ÔÓÚÏÂÁи÷Ìåϵ£¬ÇóÆä×é·ÖÊý¼°×ÔÓɶÈÊý£º

(1) NH4Cl(s)¡¢NH3(aq)¡¢Cl2(aq)¡¢H2O(l)¡¢H2O(g)¡¢H2O(aq)¡¢NH3(g)¡¢OH-(aq)¡¢NH4OH(aq)¡£ (2)NaCl(s)¡¢KBr(s)¡¢K+(aq)¡¢Na+(aq)¡¢Cl-(aq)¡¢Br-(aq)¡¢H2O(l)¡¢O2(g)¡£

15¡¢FeCl3ºÍH2OÄÜÐγÉËÄÖÖ¾ßÓÐÏàºÏÈÛµãµÄË®ºÏÎFeCl3¡¤6H2O(s)¡¢2FeCl3¡¤7H2O(s)¡¢

2FeCl3¡¤5H2O(s)ºÍFeCl3¡¤2H2O(s)£¬ÎʸÃÌåϵµÄ×é·ÖÊýÊǶàÉÙ?¸ÃÌåϵºãѹÌõ¼þÏÂ×î¶àÄÜÓм¸Ï๲´æ?Óм¸¸öµÍ¹²ÈÛµã?

´ð°¸

1¡¢¸ù¾ÝÏàÂÉÆä×ÔÓÉ¶È f=1-3+2=0£¬¼´ËµÃ÷¸ÃÌåϵµÄ×é³É£¬Î¶ÈÓëѹÁ¦¾ù²»ÈÎÒâ¸Ä±ä£¬¼´È· ¶¨Îªc£¨H2O£©=1£¬T=273.16K£¬p=611Pa

3¡¢½«ÌåϵÀäÈ´µ½413Kʱ,±ãÓйÌÏà³öÏÖ,µ«Îö³öµÄ¹ÌÌå³É·Ö ÓëÒºÏà³É·ÖÏàͬ,ÒºÏà×é³É²» ±ä,f=0,ζȲ»±ä,²½ÀäÇúÏß³öÏÖƽ̨,Ö±ÖÁÈ«²¿Äý¹ÌζÈϽµ.ËùÒÔ²½ÀäÐÎ×´Óë´¿Bi¼° ´¿CdµÄÏàͬ.

4¡¢½«Ìåϵ¼ÓÈÈÈÛ»¯£¨©ƒ563K£©,ÔÙʹÆ仺ÂýÀäÈ´, £¨½Á°èʹҺÏà×é³É¾ùÔÈ£©,µ±ÌåϵÀäµ½Îï ϵµãÓëBCÏßÏཻ´¦Óд¿CdÎö³ö,µ½©ƒ413Kʱ,¿ÉÒÔ·ÖÀë³ö´¿¹Ì̬Cd.µ«²»ÄÜ°Ñ CdÈ«²¿ ·ÖÀë³öÀ´,ÒòΪµ½413KʱCdÓëBi³É×îµÍ¹²ÈÛÎº¬Cd 40 £¥£©Îö³ö.¾­¼ÆËã¿ÉÖª×î¶à ÄÜÓÐ66.7 £¥µÄCd¿ÉÒÔ·ÖÀë³öÀ´.

13

7¡¢Ë®µÄÏàͼÖУ¬±ù¡¢Ë®¡¢Ë®ÕôÆøÈýÏ๲´æµÄÏàƽºâµã³ÆΪˮµÄÈýÏàµã¡£±ùµãÔòÊÇÔÚ101.325KPaѹÁ¦Ï±»¿ÕÆø±¥ºÍÁ˵ÄË®µÄÄý¹Ìµã£¬Ö÷ÒªÇø±ðÈçÏ£º ÈýÏàµã ϵͳÎÂ¶È ÏµÍ³Ñ¹Á¦ ƽºâÏà ? f 0.01oC 0.610KPa s=1=g 3 0 (273.16k) ±ùµã ~0.6 oC 101.325KPa s=1 2 1 (273.15) (Íâѹ) 8¡¢£¨1£©C=1£¨H2O£©

R=0 R`=0 K=C-R-R'=1

?=2£¨ÆøÒº£©

f=K-?+1=1-2+1=0 ÎÞ±äÁ¿ £¨2£©C=1£¨H2O£©

R=0 R`=0 K=C-R-R`=1 ?=2 £¨Æø¡¢Òº£©

f=K-?+2=1-2+2=1 ±äÁ¿ÎªÎ¶ȻòѹÁ¦ £¨3£©C=3 £¨I2¡¢H2O¡¢CCl4£©

R=0 R`=0 K=C-R-R`=3

?=2 £¨Òº¡¢Òº£©

f=3-2+1=2 ±äÁ¿ÎªÎ¶ÈT»òI2ÔÚÆäÖÐÒ»¸öÒºÏàÖеÄŨ¶È¡£ £¨4£©C=3 £¨H2¡¢N2¡¢NH3£© R=1 R`=0

K=C-R-R`=2

?=1

f=K-?+2=2-1+2=3 ±äÁ¿ÎªÎ¶ÈT£¬Ñ¹Á¦PºÍÈÎÒâÒ»ÖÖÎïÖʵÄŨ¶È¡£

14

10¡¢ÈçͼËùʾ×÷Ò»Ìõ¸¨ÖúÏß Kµã£ºÒºÏ࣬?=1£¬f=C-?+1=2 aµã£º¹Ì¡¢ÒºÁ½Ïàƽºâ£¬?=2£¬f=1 bµã£º¹ÌÏ࣬?=1£¬f=2

cµã£º¹Ì¡¢ÆøÁ½Ïàƽºâ£¬?=2£¬f=1 dµã£ºÆøÏ࣬?=1£¬f=2

11¡¢ÓÉѹÁ¦-×é³Éͼ£¬BÎïÖʵı¥ºÍÕôÆøѹ¸ßÓÚAÎïÖÊ£¬¹ÊÔÚ

ζÈ×é³ÉͼÖУ¬AÎïÖʵķеã¸ßÓÚBÎïÖÊ£¬²¢ÇÒͼÖгöÏÖÒ»×î¸ßµã£¨¶ÔÓ¦ÓÚѹÁ¦-×é³ÉͼÖеÄ×îµÍµã£©£¬¾ßÓиõã×é³ÉCµÄ»ìºÏÎï³Æºã·Ð»ìºÏÎ¶ÔÓ¦µÄζȳÆ×î¸ßºã·Ðµã¡£ÈôҺ̬»ìºÏÎï×é³ÉÂäÔÚB£¬CÖ®¼ä£¬Ôò¾­Áóºó£¬ÔÚËþ¶¥Äܵõ½´¿ÎïÖÊB£»ÈôҺ̬»ìºÏÎï×é³ÉÂäÔÚA£¬CÖ®¼ä£¬Ôò¾­¾«Áóºó£¬ÔÚËþ¶¥Äܵõ½´¿ÎïÖÊA£¬¶¼²»ÄÜͬʱµÃµ½Á½¸ö´¿ ×é·Ö¡£

12¡¢ÎïÖÖÊýS=4£¨NH4HS£¬NH4Cl£¬H2£¬N2£©

R`=1£¨n(H2):n(N2)=3:1£©,R=0, ?=3(Á½¹ÌÏ࣬һÆøÏà) ËùÒÔ£ºK=S-R-R`=3£¬f=K-?+1=1 13¡¢ÏàͼÈçÏ£¬OΪÈýÏàµã¡£

£¨1£©¸ÖÆ¿ÖеÄCO2£¨298K£¬6.4¡Á106Pa£©£¬ÎªOCÏßÉϵÄDµã£¬¼´ÆøÒº¹²´æ¡£Ñ¸ËÙ´ò¿ª

·§ÃÅ£¬Ñ¹Á¦Öè¼õÖÁ1¡Á105Pa,ϵͳ¾øÈÈÅòÕͽµÎ£¬Óв¿·ÖCO2£¨g£©Ö±½Óת»¯ÎªCO2£¨s£©£¬ÔÚÏàͼÉϼ´Bµã£¨105Pa£¬-78 oC£©¡£ÊµÑéÊÒÖÐÖƱ¸¸É±ù¾ÍÊǸù¾ÝÕâÒ»Ô­Àí¡£ £¨2£©Èô»ºÂý´ò¿ª·§ÃÅ£¬ÏµÍ³ºãÎÂÕô·¢£¬µ±CO2£¨g£©µÄÁ÷Á¿²»´óʱ£¬³öÀ´µÄӦΪCO2

£¨g£©£¬ÏµÍ³ÓÉDµãÖÁFµã¡£

£¨3£©CO2£¨l£©´æÔÚµÄÌõ¼þ£ºt£º-56.6 oC¡«31.3 oC £»p£º5.18¡Á105Pa¡«7.4¡Á106Pa¡£

15

13Ìâͼ 14¡¢(1) K=S-R-R£§=8-4-1=3 f=K-¦µ+2=3-3+2=1

(2) K=S-R-R£§=7-2-1=4 f=K-¦µ+2=4-4+2=2 15¡¢K=2

¡à¦µ(×î´ó)=3

µ±Ñ¹Á¦Ò»¶¨Ê± f=K-¦µ+1¾ÝÌâÒâ f=2-¦µ+1=0 ¹²ÓÐ5¸ö¹²È۵㡣 Îå¡¢ÅжÏÌâ

1¡¢Ò»¸öº¬ÓÐK+¡¢Na+¡¢NO3-¼°SO42-Àë×ӵIJ»±¥ºÍË®ÈÜÒº£¬Æä×é·ÖÊýKΪ4¡£( ) 2¡¢ÔÚ101325KPaµÄѹÁ¦Ï£¬I2ÔÚҺ̬ˮºÍCCl4Öдﵽ·ÖÅäƽºâ£¨ÎÞ¹Ì̬µâ´æÔÚ£©£¬Ôò¸ÃϵͳµÄ×ÔÓɶÈÊýΪf=1¡£( )

3¡¢ºã·Ð»ìºÏÎïÓ뻯ºÏÎïÒ»Ñù£¬¾ßÓÐÈ·¶¨µÄ×é³É¡£( )

4¡¢ÓÉCaCO3(s)¡¢CaO(s)¡¢BaCO3(s)¡¢¼°CO2£¨g£©¹¹³ÉµÄƽºâÎïϵµÄ×ÔÓɶÈΪ0¡£( ) 5¡¢FeCl3ºÍH2OÄÜÐγÉFeCl3¡¤6H2O¡¢2FeCl3¡¤7H2O¡¢2FeCl3¡¤5H2O¡¢FeCl3¡¤2H2OËÄÖÖË®ºÏÎÔÚºãѹÏ£¬×î¶à¿ÉÄÜƽºâ¹²´æµÄÏàÊýΪ4¡£( )

6¡¢¹²·ÐÎïÊÇ»ìºÏÎ¶ø²»ÊÇ»¯ºÏÎÆä×é³ÉËæѹÁ¦¸Ä±ä¶ø¸Ä±ä¡£( ) 7¡¢¿ËÀ­±´ÁúÒé³Ì

dP?HmÊÊÓÃÌõ¼þÊǶà×é·ÖµÄ¶àÏàƽºâϵͳ¡£( ) ?dTT?Vm16

8¡¢¿ËÀ­±´ÁúÒé³Ì

dP?Hm£¬ÆäѹÁ¦Ëæζȵı仯ÂÊÒ»¶¨´óÓÚ0¡£( ) ?dTT?Vm9¡¢ÏàÂÉÊÇÈÈÁ¦Ñ§Öлù±¾¶¨ÂÉÖ®Ò»£¬ËüÊÊÓÃÓÚÈκκê¹Ûϵͳ¡£( )

10¡¢¾ßÓÐ×îµÍ·ÐµãµÄAºÍB¶þ×é·Öϵͳ£¬ºã·ÐÎïΪC£¬¾«ÁóºóµÄ²ÐÒºÊÇC¡£( )

´ð°¸

Îå¡¢ÅжÏÌâ 1 ¡Á

Áù¡¢Ö¤Ã÷Ìâ 1¡¢ Ö¤Ã÷£º

ÉèÒºÌåÔÚζÈT¡¢ÍâѹÁ¦PÏÂÓëÆøÏà³Êƽºâ£¬Æä±¥ºÍÕôÆøѹÁ¦Îªpsat£¬µ±ÍâѹÁ¦ÓÉP±äµ½p+dpʱ£¬Á½ÏàÓÖ´ïеÄƽºâ£¬ÕôÆøѹ±äΪpsat+dpsat,¼´£º

G?0??? Æø£¨T£¬psat£© Òº£¨T£¬P£© ??2 ¡Á 3 ¡Á 4 ¡Ì 5 ¡Á 6 ¡Ì 7 ¡Á 8 ¡Á 9 ¡Á 10 ¡Á

dG(l) dG(g) G?0???rÆø£¨T£¬psat+d psat£© Òº£¨T£¬p+dp£©???dG?l??T??dG?g??T

¼´Vm?l?dp?Vm?g?dpsat

dpsatV?l??m

dpVm?g?ÈôÕôÆø¿´×÷ÀíÏëÆøÌ壬Vm?g??dln?psat/Pa?Vm?l?? dpRTRT£¬´úÈëÉÏʽµÃ£º psatÓÉÓÚVm(l)<< Vm(g),ËùÒÔdpsat/dp<<1,¼´ÍâѹÁ¦¶ÔÕôÆøѹµÄÓ°ÏìºÜС£¬Í¨³£¿ÉºöÂÔ²»¼Æ¡£ 2¡¢ Ö¤Ã÷£º ¸ù¾Ýd(?H)??????T??????H??dT????dP£¬Á½±ßͬ³ýÒÔdTµÃ£º

??T?p??P?Td??H?????H??????H???dP????? £¨1£© ????dT??T?P??P?T?dT?ÓÉdH?TdS?Vdp£¬µÃ£º

17

??H???S???V??????T?V?V?T?? ??p???p??T??P??T??T¶ÔÓÚµ¥×é·ÖÁ½Ïàƽºâϵͳ£¬ÉÏʽ±äΪ£º

????H??????V????V?T??P???T? £¨2£©

??T??P¸ù¾ÝCP¶¨Òå¿ÉµÃ£º

????H????T???CP £¨3£© ??P¶Ôµ¥×é·ÖÁ½Ïàƽºâ£º

dp?H £¨4£© ?dTT?V½«£¨2£©£¬£¨3£©£¬£¨4£©Èýʽ´úÈ루1£©Ê½µÃ£º

??ln?V/m3?d??H??H??CP???H?? dTT?T??P??3¡¢ Ö¤Ã÷£º

ÉèÎïÖÊÔÚT£¬PÏÂʱ£¬Á½Ïàƽºâ£¬ÔÚT+dT,p+dpʱ£¬Á½ÏàÓÖ´ïеÄƽºâ¡£

G?0??? T£¬P ¦Á Ïà ??¦ÂÏà

dG(¦Á) G?0??? T+dT,P+Dp ??¦ÁÏà dG(¦Â) ¦ÂÏà

ÒòΪ dG??SdT?VdPËùÒÔ?S???dT?V???dP??S???dT?V???dP ËùÒÔ ?V????V????dP??S????S????dT ¼´

dPS????S????Sm?? dTV????V????Vm¶Ô¿ÉÄæ¹ý³Ì£¬?S?ËùÒÔ

Æß¡¢¼ÆËãÌâ 1¡¢ ½â£º

?Hmdp? dTT?Vm?Hm Tlog101.325/3.17=¡÷Hm/2.303¡Á8.314£¨1/298-1/373.2£©

18

¡÷Hm=42.7kJ¡¤mol-1 2¡¢½â£º

Vm(¹Ì)=128/1.145; ¡÷Vm(Òº)=128/0.981

¡÷Vm= Vm (Òº) - Vm (¹Ì) =0.0187L¡¤mol-1=0.0187¡Á103m3¡¤mol-1 ¡÷Hm=128 ¡Á150=19.2 ¡Á103J¡¤mol-1 ¡÷T/ ¡÷p=T ¡÷Vm/ ¡÷Hm

=353 ¡Á0.0187/£¨19.2 ¡Á103 ¡Á103£©=3.44¡Á10-7K¡¤Pa-1 3¡¢½â£º

ÈýÏ๲´æʱ-1871.2/T+7.7096=-1425.7/T+5.4366 ½âÖ® T=196K£»

©S(p/p¦È)=(-1871.2/196)+7.7096= -1.8373 P=1473.6Pa

¡÷Hm.ÈÛÈÚ=¡÷Hm.Éý»ª-¡÷Hm.Æû»¯ =1871.2-1425.7¡Á8.314¡Á2.303

-1

=8530J¡¤mol

4¡¢ ½â£º

¸ÃÌåϵµÄ×Ü×é³ÉΪº¬·Ó: 60/(90+60)¡Á100%=40%

¸ù¾ÝÌâÒâ¼°¸Ü¸Ë¹æÔò,Á½ÕßÖÊÁ¿Ö®±ÈΪ[(1-0.449)-0.4]/½â£º(0.4-0.168)=0.65 5¡¢ ½â£º

µ±³öÏÖÁ½Ïàʱ£¬ÈÜÒº¿ªÊ¼±ä»ì×Ç£¬¼´£º0.832>H2O%>0.449 ÉèÔÙ¼ÓÈëˮΪxg£¬Ôò [100(1-0.8)+x]/[100(1-0.8)+x+100¡Á0.8]>0.449 ½âÖ®£¬µÃ x>45.2g 6¡¢½â£º

Éè¼ÓÈë±½°·xg,ÔòÌåϵµÄ×Ü×é³Éº¬±½°·x/(50+x)£¬¸ù¾Ý¸Ü¸Ë¹æÔò: 49.3/[(50+x)-49.3]=[0.916-x/(50+x)]/[x/(50+x)-0.072] ½âÖ®£¬µÃ£ºx=51g 7¡¢ ½â£º

19

¾ÝÌâÒâÇó³öB¡£B=10.704 µ±T=343.2Kʱ£¬Çó³öp¡£p=1456Pa

½«ÕôÆøѹ·½³ÌÓë¿Ë-¿Ë·½³Ì£ºlnp=¦¤vapHm?/2.303RT+B±È½Ï, µÃ ¦¤vapHm=49553J¡¤mol-1 ¦¤vapSm?=109.1 J¡¤K-1¡¤mol-1 8¡¢ ½â£º

?0ln(p2/p1)= ¦¤vapHm(T2-T1)/(2.303RT1T2) ?0¦¤vapHm=30796J

¡ß¦¤vapHm?/Tb=88 J¡¤K-1¡¤mol-1 ¡àT2=30796/88=350K 9¡¢ ½â£º

ln(p2/p1)=¦¤vapHm?(T2-T1)/(2.303RT1T2) ¡à Tb=307.1K

ln(p2/101.3)=360.2¡Á74(309.8-307.9)/(2.303¡Á8.314¡Á307.9¡Á309.8) p2=108¡Á103Pa=108 kPa 10¡¢½â£º

(1) ÔÚÈýÏàµã£¬¹ÌÒºÁ½ÏàµÄÕôÆøѹÏàµÈ£¬¹Ê½«ÉÏÊö¶þ·½³ÌÁªÁ¢£¬Çó½âµÃ:

T=542.0K p=18865Pa

(2) ½«ÉÏÊöÁ½·½³ÌÓë¿Ë-¿Ë·½³Ì±È½Ï,µÃ£º¦¤sgHm=108181 J¡¤mol-1 ¦¤vapHm?=62515 J¡¤mol-1 ¦¤fusHm?=45666 J¡¤mol-1 11¡¢½â£º

ln(p2/p1)=¦¤vapHm?(T2-T1)/(2.303RT1T2)

ln(1520/101.3)=2255¡Á18(T2-273)/(2.303¡Á8.314¡Á373T2) T2=471K 12¡¢½â£º

dp/dT=¦¤vapHm?p/RT2

dp/dT=2255¡Á18¡Á101325/(8.314¡Á373.22)Pa¡¤K-1

20

dp/dT=3552Pa¡¤K-1 13¡¢½â£º

ln(p2/p1)=¦¤vapHm?(T2-T1)/(2.303RT1T2)

ln(100/101.325)=591.2¡Á74(T2-391)/(2.303¡Á8.314¡Á391T2) T2=390K 14¡¢½â£º

ÈýÏàµãµÄѹÁ¦£ºÈýÏàµãζÈʱNH3(l)ºÍNH3(s)µÄÕôÆøѹ¡£ËùÒÔÈýÏàµãµÄζȺÍѹÁ¦¼´ÎªÉÏÁ½·½³ÌÁªÁ¢Ö®ºóµÄ½â¡£ ½âÖ®, µÃ£ºT=195.2K p=5.93¡Á103Pa 15¡¢½â£º

ÈÜÒºµÄ×é³ÉΪ£º100¡Á94%/(100+100)=0.47

¸ù¾Ý¸Ü¸Ë¹æÔò£ºml(0.47-0.43)=ms(1-0.47)¼´(200-ms)¡Á0.04=ms¡Á0.53 ¡à ms=14.0g£¬¹ÌÎö³ö¾§Ìå14.0g¡£ 16¡¢½â£º

£¨1£©ÈýÏàµãʱ£¬p(s)=p(l)

¼´11.971-2310K/T=10.097-1784K/T

½âµÃT=279.2K

½«T´úÈëÉÏÊöÈÎÒ»ÕôÆøѹÓëζȵĹØϵʽÖУ¬µÃP=4977Pa ?Hmp?£¨2£©ÓÉ¿Ë-¿Ë·½³Ì lg??C ?Pa?????2.303RT²¢¶Ô´ËÌâ¸ø¹Øϵʽ£¬ÓÐ

?subHm?2.303?8.314?2310kJ?mol-1?44.23kJ?mol?1

?vapHm?2.303?8.314?1784kJ¡¤mol?1?34.16J¡¤mol?1

¹Ê ?H??H??H?10.07kJ¡¤mol?1 fusmsubmvapm?fusSm??fusHm/T?36.07J¡¤K?1¡¤mol?1

21

°Ë¡¢×ÛºÏÌâ 1¡¢½â£º

40% Sn 2¡¢½â£º

£¨a£©ÔÚ1173Kʱ£¬AgÔÚҺ̬SnÖÐÈܽâ¶ÈÔ¼ 90%£¬¹Ì̬ÈÜÒºÖÐÔ¼º¬Sn3%¡£ £¨b£©773Kʱ£¬²»ÄÜ´æÔÚAg3Sn £¨c£©673kʱ£¬±¥ºÍÈÜÒºÖк¬ÒøÔ¼ 30%£¬»»Ëã³ÉAg3SnµÄ°Ù·ÖÂÊΪ [(3¡Á108+118)/3¡Á108]¡Á30%=40.9%

0 Sn 22

Sn + Ag3 Sn Sn 12

33 f=2 117T 3 / K 973 l M l +? F ¹Ì Agf=1 f=1 773 Sn573 +lE f=0 f=1 20 l + Ag3 Sn Sn + Ag3Sn 3 G 0 H Sn ÈÜ ÌåK f=0 ?(Ag)©‡

40 60 Ag3

100 8? 0 Ag

123

1173 40% 3

973 l 773 l Sn+l + Ag3 ¹Ì ÈÜ Ìå ? 573 Ag3 S8246100

?(Ag)©‡ Ag3

Ag

3¡¢½â£º (1)ÔÚÏàͼÉϱêÃ÷

(2)0.580 l (3)Sb2Cd3 l+Sb 2Cd3 4¡¢½â£º

(1)ÕýÈ·±êÃ÷¸÷ÇøÏàºÍ×ÔÓɶÈÊý¡£ ¢ñÇø£ºÈÜÒº£¬f*=2£» ¢òÇø£ºÈÜÒº+A(s)£¬f*=1£» ¢óÇø£ºÈÜÒº+B(s)£¬ f*=1£» ¢ôÇø£ºA(s)+ B(s)£¬ f*=1¡£ (2)ÏÈÎö³öB£¨¶ÔÏõ»ù±½£© 5¡¢½â£º

£¨1£©AºÍB»áÐγɲ»Îȶ¨»¯ºÏÎïT¡£N£¬P£¬Q¸÷µãµÄÏà̬£¬ÏàÊý¼°ÕâЩµãËù´ú±íµÄÒâÒåÁÐ

ÓÚ ÏÂ±í£º µã N Ïàµã LN+A£¨S£©+T£¨S£© ÏàÊý 3 µãËù´ú±íµÄÒâÒå AÓëTµÄ×îµÍ¹²È۵㠡¤x Sb+Sb2Cd3 Sb2Cd3+Cd l+Sb l+Cd 23

Q P LP+ T£¨S£©+B£¨S£© LP 3 1 TµÄ²»ÏàºÏÈÛµã תÈÛ·´Ó¦µÄÒºÏàµã ÆäÖÐLNºÍLP·Ö±ð±íʾ×é³ÉΪN£¬PµÄÈÛÒº¡£

£¨2£©ÓÉdµã´ú±íµÄÈÛÒº½µÎµ½Zµãºó£¬¿ªÊ¼Îö³ö¹ÌÌåB£¬¼ÌÐø½µÎ£¬B²»¶ÏÎö³ö¶øÈÛÒº×é

³ÉÑØZPÏ߱仯£¬½µÎµ½Xµãºó²»Îȶ¨»¯ºÏÎïT¿ªÊ¼Îö³ö£¬´ËʱÏÂÁзÅÈȵÄתÈÛ·´Ó¦½øÐУº

B?s??Lp?T?s?

Õâʱ£¬ÈýÏ๲´æ£¬Î¶Èά³Ö²»±ä£¬LPÈÛÒºÏàµÄ×é³ÉΪP¡£Ö±µ½B£¨s£©Ïûʧºó£¬Î¶ÈÔÙ¼ÌÐøϽµ£¬T²»¶ÏÎö³ö£¬¶øÓë֮ƽºâµÄÒºÏà×é³ÉÑØPNÏ߱仯£¬Î¶Ƚµµ½yµãºó£¬¿ªÊ¼Îö³ö¹ÌÌåA£¬´ËʱLN+A£¨S£©+T£¨S£©ÈýÏ๲´æ£¬Î¶Èά³Ö²»±ä£¬´ý×é³ÉΪNµÄLNÈÛÒºÏàÏûʧºóζȲżÌÐøϽµ¡£ 6¡¢½â£º

¸÷ÏàÇøÏà̬ÁÐÓÚÏÂ±í£º ÏàÇø 1 2 3 4 5 Ïà̬ ÈÛÒºL L+±ù ±ù+MX¡¤6H2O£¨s£© L+MX¡¤6H2O£¨s£© L+MX¡¤6H2O£¨s£© ÏàÇø 6 7 8 9 10 Ïà̬ MX¡¤6H2O£¨s£©+MX¡¤2H2O£¨s£© L+MX¡¤2H2O£¨s£© MX¡¤2H2O£¨s£©+MX¡¤H2O£¨s£© L+MX¡¤H2O£¨s£© MX¡¤H2O£¨s£©+ MX£¨s£© 7¡¢½â£º

C ÈçͼËùʾ×öÒ»Ìõ¸¨ÖúÏß p/pa . K A µã£ºÒºÏ࣬¦µ=1£¬f=C-¦µ+1=2 KKµã£ºÒºÏ࣬¦µ=1£¬f=C-¦µ+1=2 £¬f=1 a µã£º¹Ì¡¢ÒºÁ½Ïàƽºâ£¬ ¦µ =2 a bµã£º¹ÌÏ࣬£¬£¬f=1 a µã£º¹Ì¡¢ÒºÁ½Ïàƽºâ£¬ ¦µ =1 f=2 ¦µ =2 b . £¬f=1 c µã£º¹Ì¡¢ÆøÁ½Ïàƽºâ£¬ ¦µ =1 f=2 ¦µ =2 c O bµã£º¹ÌÏ࣬£¬£¬ d µã£ºÆøÏ࣬ ¦µ =1 f=2 cµã£º¹Ì¡¢ÆøÁ½Ïàƽºâ£¬¦µ=2£¬f=1 B d . dµã£ºÆøÏ࣬¦µ=1£¬f=2 T/K 8¡¢½â£º

£¨1£©¸÷ÇøÎȶ¨Ïà̬ÈçͼËùʾ¡£

24

£¨2£©ÈýÏàÏßde:1???s??c?s?

ÈýÏàÏßfg:1?B?s??c?s?

£¨3£©a,bÁ½µãµÄÀäÈ´ÇúÏߣ¬Ïà±ä»¯¼°×ÔÓɶȱ仯ÈçÏÂͼËùʾ¡£

l T/K

l+a(s) l+C(s) d e l+C(s) l+B(s) a(s) f g

B(s)+C(s)

A C B a WB b l,f=2

l , f=2

l+a(s),f=1 l+c(s),f=1

a(s)Îö³ö C(s)Îö³ö lÏûʧ

lÏûʧ

f=0 B(s) Îö³ö f=0 B(s)+C(s),f=1 C(s)Îö³ö C(s)+a(s),f=1

9¡¢½â£º

£¨1£©ÈÜҺŨ¶ÈΪ10%£¬¼´º¬75gNH4Cl¡£×î¶àÎö³ö±ù250g¡£ £¨2£©Ã»ÓбùÎö³ö¡£

£¨3£©NH4Cl¹ÌÌåÓëÈÜÒº¶þÏàƽºâ¡£¹ÌÏàΪ´¿NH4Cl¹ÌÌ壬ҺÏàŨ¶ÈΪ20%¡£¹ÌÏàÖØÁ¿Îª6.25g£¬

ÒºÏàÖØÁ¿Îª93.75g¡£

ÎïÀí»¯Ñ§Ï°Ì⣨Ïàƽºâ£©

Ò»£®Ñ¡Ôñ

1.ÔÚ¦Á¡¢¦ÂÁ½ÏàÖоùÓÐAºÍBÁ½ÖÖÎïÖÊ£¬´ïµ½Ïàƽºâʱ£¬ÏÂÁи÷ʽÕýÈ·µÄÊÇ £¨1£© ¡£ £¨1£©¦Ì¦ÁB=¦Ì¦ÂB£¨2£©¦Ì¦ÁA=¦Ì¦ÂA £¨3£©¦Ì¦ÁB=¦Ì¦ÁA£¨4£©¦Ì¦ÁB=¦Ì¦ÂA

2£®×é·ÖB´Ó¦ÁÏàÀ©É¢µ½¦ÂÏàÖУ¬ÔòÏÂÁÐ˵·¨ÖÐÕýÈ·µÄÊÇ £¨4£© ¡£

25

£¨1£©×ÜÊÇ´ÓŨ¶È¸ßµÄÏàÀ©É¢µ½Å¨¶ÈµÍµÄÏࣨ2£©Æ½ºâʱÁ½ÏàµÄŨ¶ÈÏàµÈ £¨3£©×ÜÊÇ´ÓŨ¶ÈµÍµÄÏàÀ©É¢µ½Å¨¶È¸ßµÄÏࣨ4£©×ÜÊǴӸ߻¯Ñ§ÊÆÒÆÏòµÍ»¯Ñ§ÊÆ 3. ÊÒÎÂÏ°±»ù¼×Ëá立ֽⷴӦΪ NH2CO2NH4(s)====2NH3(g)+CO2 (g)

ÈôÔÚ300KʱÏòϵͳÖмÓÈëÒ»¶¨Á¿µÄ°±»ù¼×Ëá粒ÌÌ壬Ôò´ËϵͳµÄÎïÖÖÊýSºÍ×é·ÖÊýCӦΪ £¨3£© ¡£

£¨1£©1£¬1£¨2£©3£¬2£¨3£©3£¬1£¨4£©3£¬3

4£®½«¿ËÀ­±´Áú·½³ÌÓ¦ÓÃÓÚË®µÄÒº¹ÌÁ½Ï࣬ËæѹÁ¦µÄÔö³¤£¬Ë®µÄÄý¹Ìµã½« £¨2£© ¡£ £¨1£©ÉÏÉý£¨2£©Ï½µ£¨3£©²»±ä£¨4£©ÎÞ·¨ÅжÏ

5.ÔÚÒ»¶¨Î¶ÈÏ£¬ÔÚË®ºÍCCl4×é³ÉµÄ»¥²»ÏàÈܵÄϵͳÖУ¬ÏòË®²ãÖмÓÈë1:1µÄKIºÍI2£¬´ËϵͳµÄ×ÔÓɶÈÊÇ £¨2£© ¡£ £¨1£©1£¨2£©2£¨3£©3£¨4£©4

6.¶ÔÓÚ¶þ×é·Öϵͳ£¬ÄÜƽºâ¹²´æµÄ×î¶àÏàÊýΪ £¨4£© ¡£ £¨1£©1£¨2£©2£¨3£©3£¨4£©4

7.¶ÔÓÚºã·Ð»ìºÏÎÏÂÁÐ˵·¨ÖдíÎóµÄÊÇ £¨4£© ¡£ £¨1£©²»¾ßÓÐÈ·¶¨µÄ×é³É £¨2£©Æ½ºâʱÆøÏàºÍÒºÏà×é³ÉÏàͬ £¨3£©Æä·ÐµãËæÍâѹµÄ¸Ä±ä¶ø¸Ä±ä £¨4£©Ó뻯ºÏÎïÒ»Ñù¾ßÓÐÈ·¶¨µÄ×é³É

¶þ£®Ìî¿Õ

1. ÔÚË®µÄƽºâÏàͼÖС£ ÏßÊÇË®µÄÕô·¢ÇúÏߣ¬ ÏßÊDZùµÄÉý»ªÇúÏߣ¬

ÏßÊDZùµÄÈÚ»¯ÇúÏߣ¬ µãÊÇË®µÄÈýÏàµã¡£

26

.

2.½«¹ýÁ¿NH4HCO3(s)·ÅÈëÃܱÕÕæ¿ÕÈÝÆ÷ÄÚ£¬50¡æʱ£¬°´NH4HCO3(s)°´ÏÂʽ½øÐзֽ⣺NH4HCO3(s)=NH3(g)+CO2(g)+H2O(g)¡£ ´ïƽºâºó Ôò¸ÃÌåϵµÄÏàÊýP= £¬ ¶ÀÁ¢×é·ÖÊýC= £¬ ×ÔÓɶÈÊýF= ¡££¨2£¬1£¬0£©

3.(2)×îºÏÊʵÄ×é³ÉÓ¦Åä³Éw£¨Áª±½ÃÑ£©=0.78£¬ÕâÒ»×é³É¾ßÓÐ×îµÍ¹²ÈÛµã12¡æ£¬ÆäÄý¹Ìµã×î

µÍ

£¬

Ëù

ÒÔ

²»

ÖÁ

ÓÚ

Òò

Ϊ

Äý

¹Ì

¶ø

¶Â

Èû

¹Ü

µÀ

¡£

4. ÇóÏÂÁÐϵͳ¢Û´ÓXB=0.5¿ªÊ¼ÕôÁó£¬Áó³öÎï»ò²ÐÁôÎï·Ö±ðÊÇʲô£¿

27

£¨1£©TM£¬TN£¬£¨2£©Ëù´¦µÄT£¬p¼°ÆøÒºÏà×é³ÉyB¡¢xB£¬´ïÁ½ÏàƽºâÆøÒºÏàµÄÏà¶ÔÁ¿²»Í¬£¬£¨3£©0¡£

6.Ë®ºÍÕáÌÇϵͳÏàͼÈçÏ£º

28

(2)²»¿ÉÄÜ£¨3£©¶¨ÎÂÕô·¢£¬0.8£¬½µÎ£¬=13.9¡æ

29

Èý£®¼ò´ð

1.´¿ÌúµÄÏàͼ¼ûÏÂͼ5-3¡£·ÖÎöͼÖи÷µã¡¢Ïß¡¢ÃæµÄÏàƽºâ¹Øϵ¡£

2.

30

2.ÓÃÏàÂÉÅжÏÏÂÁÐÏàͼÊÇ·ñÕýÈ·£¿

£¨1£©´íÎó¡£ÈýÏàÏß²»Ë®Æ½£¬ÕâµÈÓÚ˵ÔÚÈýÏàƽºâʱζȿɱ䡣¾ÝÏàÂÉF=2-3+1=0 £¨2£©´íÎó¡£´¿×é·ÖÓй̶¨È۵㡣

3.ÊÔ˵Ã÷ÔÚ¹Ì-ҺƽºâϵͳÖУ¬Îȶ¨»¯ºÏÎï¡¢²»Îȶ¨»¯ºÏÎïÓë¹ÌÈÜÌåÈýÕßÖ®¼äµÄÇø±ð¡£

31

£¨1£©Îȶ¨»¯ºÏÎïÓй̶¨µÄ×é³ÉºÍÈ۵㣬ÔÚ¹ÌÏàºÍÒºÏ඼´æÔÚ¡£µ±Îȶ¨µÍÓÚÆäÈÛµãʱ²»»á·¢Éú·Ö½â£¬ÔÚÏàͼÉÏÓÐ×î¸ßÈ۵㣬³ÆΪÏàºÏÈ۵㣬ÔÚ¸ÃζÈÈÛ»¯Ê±ÐγɵÄÒºÏà×é³ÉºÍ¹ÌÏà×é³ÉÏàͬ

£¨2£©²»Îȶ¨»¯ºÏÎïÖ»ÔÚ¹ÌÏà´æÔÚ£¬ÔÚÒºÏ಻´æÔÚ£¬¼ÓÈÈʱÔÚÈÛµãÒÔÇ°¾Í·Ö½âΪҺÏàºÍÁíÒ»¸ö¹ÌÌ壬´ËÒºÏà×é³ÉºÍÔ­À´¹Ì̬»¯ºÏÎïµÄ×é³É²»Ïàͬ£¬ÔÚÏàͼÉÏûÓÐ×î¸ßÈ۵㣬¶ø³ÊT×ÖÐΣ¬ÆäˮƽÏßËù´ú±íµÄζȼ´Îª×ªÈÛζȣ¬³ÆΪ²»ÏàºÏÈÛµã

£¨3£©¹ÌÈÜÌå¼´¹Ì̬ÈÜÒº£¬ÊÇ»ìºÏÎ²»ÊÇ»¯ºÏÎ×é³É²»¹Ì¶¨£¬¿ÉÔÚÒ»¶¨·¶Î§Äڱ仯¡£

4. ÊÔ˵Ã÷µÍ¹²ÈÛ¹ý³ÌÓëתÈÛ¹ý³ÌÓкÎÒìͬ£¿µÍ¹²ÈÛÎïÓë¹ÌÈÜÌåÖ®¼äÓкÎÇø±ð£¿ £¨1£©µÍ¹²ÈÛ¹ý³ÌÓëתÈÛ¹ý³ÌÏà֮ͬ´¦£º¶¼ÊÇÁ½¸ö¹ÌÏàÓëÒ»¸öÒºÏàƽºâ¹²´æ¡£

£¨2£©²»Í¬Ö®´¦£ºµÍ¹²ÈÛ¹ý³ÌÒºÏà×é³É½éÓÚÁ½¹ÌÌå×é³ÉÖ®¼ä£¬×ªÈÛ¹ý³ÌÒºÏà×é³É½éÓÚÒºÏàºÍÁíÒ»ÏàÖ®¼ä£»µÍ¹²ÈÛµãÊÇÒºÏàÄܹ»´æÔÚµÄ×îµÍζȣ¬×ªÈÛζÈÔòΪ²»Îȶ¨»¯ºÏÎïÄܹ»´æÔÚµÄ×î¸ßζȡ£

£¨3£©¹ÌÈÜÌåÊÇÒ»ÖÖ×é·ÖÈÜÓÚÁíÒ»ÖÖ×é·Ö¶øÐγɵĹÌ̬ÈÜÒº£¬Îªµ¥Ï࣬¶øµÍ¹²ÈÛÎïÊÇÁ½ÖÖ¹ÌÌåµÄ»ìºÏÎΪÁ½Ïà¡£

5.ͨ³£ÖÊÁ¿·ÖÊý0.98µÄŨÁòËá¹ã·ºÓ¦ÓÃÓÚ¹¤Òµ¡¢Ò½Ò©¡£ÓÉÏÂͼ¿ÉÒÔ¿´³ö98%µÄÁòËá½á¾§Î¶ÈԼΪ273K£¬×÷Ϊ²úÆ·ÈÝÒ×ÔÚ¶¬½á£¬¶ÂÈû¹ÜµÀ£¬ÔõÑù×ö²ÅÄܱÜÃâ¡£

¼¾¶³

32

£¨92.5%£¬238K£©

½â£ºÔËÓÃÄý¹ÌµãϽµµÄÔ­Àí£¬¶¬¼¾³£²ÉÓÃ0.925µÄÁòËá×÷Ϊ²úÆ·£¬ÕâÖÖÁòËáµÄÄý¹Ìµã´ó¸ÅÔÚ238k£¬ÔÚÒ»°ãµÄµØ·½´æ·ÅºÍÔËÊ䶼²»ÖÁÓÚ¶³½á£¬¿ÉÒÔ¼õÉÙËðʧ£¬´´Ôì¸ü¶àµÄЧÒæ¡£

µÚÎåÕ ÏàƽºâÁ·Ï°Ìâ

Ò»¡¢ÅжÏÌ⣺

1£®ÔÚÒ»¸ö¸ø¶¨µÄϵͳÖУ¬ÎïÖÖÊý¿ÉÒÔÒò·ÖÎöÎÊÌâµÄ½Ç¶ÈµÄ²»Í¬¶ø²»Í¬£¬µ«¶ÀÁ¢×é·ÖÊýÊÇ Ò»¸öÈ·¶¨µÄÊý¡£

2£®µ¥×é·ÖϵͳµÄÎïÖÖÊýÒ»¶¨µÈÓÚ1¡£ 3£®×ÔÓɶȾÍÊÇ¿ÉÒÔ¶ÀÁ¢±ä»¯µÄ±äÁ¿¡£ 4£®ÏàͼÖеĵ㶼ÊÇ´ú±íϵͳ״̬µÄµã¡£

5£®ºã¶¨Ñ¹Á¦Ï£¬¸ù¾ÝÏàÂɵóöijһϵͳµÄf = l£¬Ôò¸ÃϵͳµÄζȾÍÓÐÒ»¸öΨһȷ¶¨µÄÖµ¡£ 6£®µ¥×é·ÖϵͳµÄÏàͼÖÐÁ½ÏàƽºâÏ߶¼¿ÉÒÔÓÿËÀ­±´Áú·½³Ì¶¨Á¿ÃèÊö¡£

7£®¸ù¾Ý¶þԪҺϵµÄp¡«xͼ¿ÉÒÔ׼ȷµØÅжϸÃϵͳµÄÒºÏàÊÇ·ñÊÇÀíÏëÒºÌå»ìºÏÎï¡£ 8£®ÔÚÏàͼÖÐ×Ü¿ÉÒÔÀûÓøܸ˹æÔò¼ÆËãÁ½Ïàƽ³©Ê±Á½ÏàµÄÏà¶ÔµÄÁ¿¡£ 9£®¸Ü¸Ë¹æÔòÖ»ÊÊÓÃÓÚT¡«xͼµÄÁ½ÏàƽºâÇø¡£¡£

10£®¶ÔÓÚ¶þÔª»¥ÈÜҺϵ£¬Í¨¹ý¾«Áó·½·¨×Ü¿ÉÒԵõ½Á½¸ö´¿×é·Ö¡£

11£®¶þԪҺϵÖУ¬ÈôA×é·Ö¶ÔÀ­ÎÚ¶û¶¨ÂɲúÉúÕýÆ«²î£¬ÄÇôB×é·Ö±Ø¶¨¶ÔÀ­ÎÚ¶û¶¨Âɲú Éú¸ºÆ«²î¡£

12£®ºã·ÐÎïµÄ×é³É²»±ä¡£

13£®ÈôA¡¢BÁ½ÒºÌåÍêÈ«²»»¥ÈÜ£¬ÄÇôµ±ÓÐB´æÔÚʱ£¬AµÄÕôÆøѹÓëϵͳÖÐAµÄĦ¶û·Ö Êý³ÉÕý±È¡£

14£®ÔÚ¼òµ¥µÍ¹²ÈÛÎïµÄÏàͼÖУ¬ÈýÏàÏßÉϵÄÈκÎÒ»¸öϵͳµãµÄÒºÏà×é³É¶¼Ïàͬ¡£ 15£®Èý×é·Öϵͳ×î¶àͬʱ´æÔÚ5¸öÏà¡£

¶þ¡¢µ¥Ñ¡Ì⣺

- -

1£®H2O¡¢K+¡¢Na+¡¢Cl¡¢IÌåϵµÄ×é·ÖÊýÊÇ£º (A) K = 3 £» (B) K = 5 £» (C) K = 4 £» (D) K = 2 ¡£

2£®¿ËÀÍÐÞ˹£­¿ËÀ­²®Áú·½³Ìµ¼³öÖУ¬ºöÂÔÁËҺ̬Ìå»ý¡£´Ë·½³ÌʹÓÃʱ£¬¶ÔÌåϵËù´¦µÄΠ¶ÈÒªÇó£º

(A) ´óÓÚÁÙ½çÎÂ¶È £» (B) ÔÚÈýÏàµãÓë·ÐµãÖ®¼ä £» (C) ÔÚÈýÏàµãÓëÁÙ½çζÈÖ®¼ä £» (D) СÓڷеãÎÂ¶È ¡£

33

3£®µ¥×é·Ö¹Ì£­ÒºÁ½ÏàƽºâµÄp¡«TÇúÏßÈçͼËùʾ£¬Ôò£º (A) Vm(l) = Vm(s) £» (B) Vm(l)£¾Vm(s) £» (C) Vm(l)£¼Vm(s) £» (D) ÎÞ·¨È·¶¨ ¡£

4£®ÕôÆûÀäÄýΪҺÌåʱËù·Å³öµÄDZÈÈ£¬¿ÉÓÃÀ´£º (A) ¿ÉʹÌåϵ¶Ô»·¾³×öÓÐÓù¦ £» (B) ¿Éʹ»·¾³¶ÔÌåϵ×öÓÐÓù¦ £» (C) ²»ÄÜ×öÓÐÓù¦ £» (D) ²»ÄÜÅж¨ ¡£

5£®Ñ¹Á¦Éý¸ßʱ£¬µ¥×é·ÖÌåϵµÄÈ۵㽫ÈçºÎ±ä»¯£º (A) Éý¸ß £» (B) ½µµÍ £» (C) ²»±ä £» (D) ²»Ò»¶¨ ¡£

6£®ÁòËáÓëË®¿É×é³ÉÈýÖÖ»¯ºÏÎH2SO4¡¤H2O(s)¡¢H2SO4¡¤2H2O(s)¡¢H2SO4¡¤4H2O(s)£¬ÔÚp Ï£¬ÄÜÓëÁòËáË®ÈÜÒº¹²´æµÄ»¯ºÏÎï×î¶àÓм¸ÖÖ£º (A) 1 ÖÖ £» (B) 2 ÖÖ £» (C) 3 ÖÖ £» (D) 0 ÖÖ ¡£

7£®ÔÚ101325PaµÄѹÁ¦Ï£¬I2ÔÚҺ̬ˮÓëCCl4ÖеÄÈܽâÒѴﵽƽºâ(ÎÞ¹ÌÌåI2´æÔÚ)£¬´ËÌå ϵµÄ×ÔÓɶÈΪ£º (A) 1 £» (B) 2 £» (C) 3 £» (D) 0 ¡£

8£®NaClË®ÈÜÒººÍ´¿Ë®£¬¾­°ë͸Ĥ´ïµ½Éø͸ƽºâ£¬¸ÃÌåϵµÄ×ÔÓɶÈÊýÊÇ£º (A) f = 1 £» (B) f = 2 £» (C) f = 3 £» (D) f = 4 ¡£

9£®¶ÔÓÚÏÂÁÐƽºâϵͳ£º¢Ù¸ßÎÂÏÂË®±»·Ö½â£»¢Úͬ¢Ù£¬Í¬Ê±Í¨ÈëһЩH2(g) ºÍO2(g)£»¢ÛH2

ºÍO2ͬʱÈÜÓÚË®ÖУ¬Æä×éÔªÊýKºÍ×ÔÓɶÈÊýfµÄÖµÍêÈ«ÕýÈ·µÄÊÇ£º (A) ¢ÙK = 1£¬f = 1 ¢ÚK = 2£¬f = 2 ¢ÛK = 3£¬f = 3 £» (B) ¢ÙK = 2£¬f = 2 ¢ÚK = 3£¬f = 3 ¢ÛK = 1£¬f = 1 £» (C) ¢ÙK = 3£¬f = 3 ¢ÚK = 1£¬f = 1 ¢ÛK = 2£¬f = 2 £» (D) ¢ÙK = 1£¬f = 2 ¢ÚK = 2£¬f = 3 ¢ÛK = 3£¬f = 3 ¡£

10£®ÔÚÏÂÁÐÌåϵÖÐ×ÔÓɶÈf = 2µÄÌåϵÊÇ£º (A) 298Kʱ£¬H2O(l)H2O(g) £» (B) S(s)S(l)S(g) £» (C) C2H5OH(l) ÓëH2O(l) µÄ»ìºÏÎï £»

(D) Ò»¶¨Á¿µÄPCl5(g) ·Ö½âƽºâʱ£ºPCl5(g) = PCl3(g) + Cl2(g) ¡£

11£®Ä³ÌåϵÖÐÓÐNa2CO3Ë®ÈÜÒº¼°Na2CO3¡¤H2O(s)¡¢Na2CO3¡¤7H2O(s)¡¢Na2CO3¡¤10H2O(s)Èý ÖֽᾧˮºÏÎï¡£ÔÚpÏ£¬f = K - ¦µ + 1 = 2 - 4 + 1 = -1£¬ÕâÖÖ½á¹û±íÃ÷£º (A) Ìåϵ²»ÊÇ´¦ÓÚƽºâ̬ £» (B) Na2CO3¡¤10 H2O(s) ²»¿ÉÄÜ´æÔÚ £» (C) ÕâÖÖÇé¿öÊDz»´æÔÚµÄ £» (D) Na2CO3¡¤7H2O(s) ²»¿ÉÄÜ´æÔÚ ¡£

12£®ÏàͼÓëÏàÂÉÖ®¼äÊÇ£º

(A) ÏàͼÓÉʵÑé½á¹û»æÖƵóö£¬Ïàͼ²»ÄÜÎ¥±³ÏàÂÉ £» (B) ÏàͼÓÉÏàÂÉÍƵ¼µÃ³ö£»

(C) ÏàͼÓÉʵÑé½á¹û»æÖƵóö£¬ÓëÏàÂÉÎÞ¹Ø £» (D) Ïàͼ¾ö¶¨ÏàÂÉ ¡£

13£®ÏÂÁÐÐðÊöÖдíÎóµÄÊÇ£º

(A) Ë®µÄÈýÏàµãµÄζÈÊÇ273.15K£¬Ñ¹Á¦ÊÇ610.62 Pa £» (B) ÈýÏàµãµÄζȺÍѹÁ¦½öÓÉϵͳ¾ö¶¨£¬²»ÄÜÈÎÒâ¸Ä±ä £» (C) Ë®µÄ±ùµãζÈÊÇ0¡æ(273.15K)£¬Ñ¹Á¦ÊÇ101325 Pa £» (D) Ë®µÄÈýÏàµãf = 0£¬¶ø±ùµãf = 1 ¡£

14£®Na 2CO3¿ÉÐγÉÈýÖÖË®ºÏÑΣºNa2CO3¡¤H2O¡¢Na2CO3¡¤7H2O¡¢NaCO3¡¤10H2O£¬ÔÚ³£Ñ¹Ï£¬

34

½«Na2CO3ͶÈë±ù£­Ë®»ìºÏÎïÖдïÈýÏàƽºâʱ£¬ÈôÒ»ÏàÊDZù£¬Ò»ÏàÊÇNa2CO3Ë®ÈÜÒº£¬ ÔòÁíÒ»ÏàÊÇ£º

(A) Na2CO3 £» (B) Na2CO3¡¤H2O £» (C) Na2CO3¡¤7H2O£» (D) Na2CO3¡¤10H2O¡£

15£®Èçͼ£¬¶ÔÓÚÓұߵIJ½ÀäÇúÏ߶ÔÓ¦ÊÇÄĸöÎïϵµãµÄ ÀäÈ´¹ý³Ì£º

(A) aµãÎïϵ £» (B) bµãÎïϵ £» (C) cµãÎïϵ £» (D) dµãÎïϵ ¡£

16£®Èçͼ£¬¶ÔÓÚÐγɼòµ¥µÍ¹²ÈÛ»ìºÏÎïµÄ¶þÔªÏàͼ£¬ µ±ÎïϵµÄ×é³ÉΪx£¬ÀäÈ´µ½t¡æʱ£¬¹ÌÒº¶þÏàµÄÖØ Á¿Ö®±ÈÊÇ£º

(A) w(s)¡Ãw(l) = ac¡Ãab £» (B) w(s)¡Ãw(l) = bc¡Ãab £» (C) w(s)¡Ãw(l) = ac¡Ãbc £» (D) w(s)¡Ãw(l) = bc¡Ãac ¡£

17£®Èçͼ£¬¶ÔÓÚÐγɼòµ¥µÍ¹²ÈÛ»ìºÏÎïµÄ¶þÔªÏàͼ£¬µ±Îï ϵµã·Ö±ð´¦ÓÚC¡¢E¡¢Gµãʱ£¬¶ÔÓ¦µÄƽºâ¹²´æµÄÏàÊýΪ£º (A) Cµã1£¬Eµã1£¬Gµã1 £» (B) Cµã2£¬Eµã3£¬Gµã1 £» (C) Cµã1£¬Eµã3£¬Gµã3 £» (D) Cµã2£¬Eµã3£¬Gµã3 ¡£

18£®ÔÚÏàͼÉÏ£¬µ±Îïϵ´¦ÓÚÄÄÒ»¸öµãʱֻÓÐÒ»¸öÏࣺ (A) ºã·Ðµã £» (B) È۵㠣» (C) ÁÙ½çµã £» (D) µÍ¹²È۵㠡£

19£®¼×¡¢ÒÒ¡¢±ûÈý¸öСº¢¹²³ÔÒ»Ö§±ù¹÷£¬ÈýÈËÔ¼¶¨£º¢Å¸÷³ÔÖÊÁ¿µÄÈý·ÖÖ®Ò»£»¢ÆÖ»×¼Îü£¬ ²»×¼Ò§£»¢Ç°´ÄêÁäÓÉСµ½´ó˳ÐòÏȺó³Ô¡£½á¹û£¬ÒÒÈÏΪÕâÖ»±ù¹÷ûÓзÅÌÇ£¬¼×ÔòÈÏΪ Õâ±ù¹÷·Ç³£Ì𣬱ûÈÏΪËûÁ©¿´·¨Ì«¾ø¶Ô»¯¡£ÔòÈýÈËÄêÁ䣺 (A) ¼××î´ó£¬ÒÒ×îС £» (B) ¼××îС£¬ÒÒ×î´ó £» (C) ±û×î´ó£¬¼××îС £» (D) ±û×îС£¬ÒÒ×î´ó ¡£

20£®ÈçͼAÓëBÊÇÁ½×é·ÖºãѹϹÌÏಿ·Ö»¥ ÈÜÄý¾ÛÌåϵÏàͼ£¬Í¼ÖÐÓм¸¸öµ¥ÏàÇø£º (A) (A) 1¸ö £» (B) (B) (B) 2¸ö £» (C) (C) (C) 3¸ö £» (D) (D) (D) 4¸ö ¡£

21£®ÓÐÒ»Ðγɲ»Îȶ¨»¯ºÏÎïµÄË«×é·ÖAÓëBÄý¾ÛÌåϵ£¬ÏµÍ³µÄ×é³É¸ÕÇÉÓë²»Îȶ¨»¯ºÏÎï µÄ×é³ÉÏàͬ£¬µ±Æä´ÓҺ̬ÀäÈ´µ½²»Ïà³ÆÈ۵㣬ϵͳÄÚ½¨Á¢ÈçÏÂƽºâ£º ÒºÏà + A(s) = AxBy(²»Îȶ¨»¯ºÏÎï)£¬Èç¹ûÔÚ´ËʱϵͳÓÉÍâ½çÎüÈ¡ÈÈʱ£¬ÔòÉÏÊöµÄƽºâ½«£º (A) Ïò×óÒƶ¯ £» (B) ÏòÓÒÒƶ¯ £» (C) ²»Òƶ¯ £» (D) ÎÞ·¨Åж¨ ¡£

22£®AÓëB¿ÉÒÔ¹¹³É2ÖÖÎȶ¨»¯ºÏÎïÓë1ÖÖ²»Îȶ¨»¯ºÏÎÄÇôAÓëBµÄÌåϵ¿ÉÒÔÐγÉ

¼¸Öֵ͹²ÈÛ»ìºÏÎ

(A) 2ÖÖ £» (B) 3ÖÖ £» (C) 4ÖÖ £» (D) 5ÖÖ ¡£

23£®ÈçͼAÓëBÊÇÁ½×é·ÖºãѹϹÌÏಿ·Ö»¥ÈÜÄý¾ÛÌåϵ Ïàͼ£¬Óм¸¸öÁ½¹ÌÏàƽºâÇø£º (A) (A) 1¸ö £» (B) 2¸ö £»

35

(B) (B) 3¸ö £» (D) 4¸ö ¡£

24£®ÔÚµÚÒ»ÖÖÎïÖÊÖмÓÈëµÚ¶þÖÖÎïÖʺ󣬶þÕßµÄÈ۵㷢Éúʲô±ä»¯? (A) ×ÜÊÇϽµ £» (B) ×ÜÊÇÉÏÉý £»

(C) ¿ÉÄÜÉÏÉýÒ²¿ÉÄÜϽµ £» (D) ·þ´ÓÀ­ÎÚ¶û¶¨ÂÉ ¡£

25£®ÈçͼÊÇFeOÓëSiO2µÄºãѹÏàͼ£¬ÄÇô´æÔÚ¼¸¸öÎÈ ¶¨»¯ºÏÎ

(A) (A) 1¸ö £» (B) (B) (B) 2¸ö £» (C) (C) (C) 3¸ö £» (D) (D) (D) 4¸ö ¡£

26£®A¼°B¶þ×é·Ö×é³ÉµÄÄý¾ÛÌåϵÄÜÉú³ÉÈýÖÖÎȶ¨µÄ»¯ºÏ

ÎÔòÓÚ³£Ñ¹ÏÂÔÚÒºÏ࿪ʼÀäÈ´µÄ¹ý³ÌÖУ¬×î¶àÓм¸ÖÖ¹ÌÏàͬʱÎö³ö£¿ (A) 4ÖÖ £» (B) 5ÖÖ £» (C) 2ÖÖ £» (D) 3ÖÖ ¡£

27£®ÔÚζÈΪTʱ£¬A(l) ÓëB(l) µÄ±¥ºÍÕôÆøѹ·Ö±ðΪ30.0kPaºÍ35.0kPa£¬AÓëBÍêÈ« »¥ÈÜ£¬µ±xA = 0.5ʱ£¬pA = 10.0kPa£¬pB = 15.0kPa£¬Ôò´Ë¶þԪҺϵ³£Ñ¹ÏµÄT¡«xÏàͼΪ£º

28£®Á½×é·ÖÀíÏëÈÜÒº£¬ÔÚÈκÎŨ¶ÈÏ£¬ÆäÕôÆøѹ£º

(A) ºã´óÓÚÈÎÒ»´¿×é·ÖµÄÕôÆøѹ £» (B) ºãСÓÚÈÎÒ»´¿×é·ÖµÄÕôÆøѹ £» (C) ½éÓÚÁ½¸ö´¿×é·ÖµÄÕôÆøѹ֮¼ä £» (D) ÓëÈÜÒº×é³ÉÎÞ¹Ø ¡£

29£®ÉèAºÍB¿ÉÎö³öÎȶ¨»¯ºÏÎïAxByºÍ²»Îȶ¨»¯ºÏ ÎïAmBn£¬ÆäT¡«xͼÈçͼËùʾ£¬ÆäÖа¢À­²®Êý×Ö´ú ±íÏàÇø£¬¸ù¾ÝÏàͼÅжϣ¬Òª·ÖÀë³ö

´¿¾»µÄ»¯ºÏÎïAmBn£¬ÎïϵµãËù´¦µÄÏàÇøÊÇ£º (A) 9 £» (B) 7 £» (C) 8 £» (D) 10 ¡£

30£®ÒºÌåAÓëBÐγÉÕôÆøѹÕýÆ«²îºÜ´óµÄÈÜÒº£¬ÔÚ¾«ÁóËþÖо«Áóʱ£¬Ëþ¸ªµÃµ½µÄÊÇ£º (A) ºã·Ð»ìºÏÎï £» (B) ´¿A £» (C) ´¿B £» (D) ´¿A»ò´¿B ¡£

31£®ÈçͼAÓëBÊÇÁ½×é·ÖºãѹϹÌÏಿ·Ö»¥ÈÜ

Äý¾ÛÌåϵÏàͼ£¬Í¼ÖÐÓм¸¸öÁ½ÏàÇø£º (A) 1¸ö £» (B) 2¸ö £» (C) 3¸ö £» (D) 4¸ö ¡£

32£®Ë®ÕôÆøÕôÁóͨ³£ÊÊÓÃÓÚijÓлúÎïÓëË®×é³ÉµÄ£º (A) ÍêÈ«»¥ÈÜ˫Һϵ £» (B) »¥²»ÏàÈÜ˫Һϵ £» (C) ²¿·Ö»¥ÈÜ˫Һϵ £» (D) ËùÓÐ˫Һϵ ¡£

33£®ÏÂͼÊǶþÔªÄý¾ÛÌåϵÏàͼ£¬ÆäÖÐÎïϵµãÓëÏàµãºÏÒ»µÄÊÇ£º (A) Fµã£¬Gµã £» (B) Iµã£¬Dµã £» (C) Hµã£¬Dµã £» (D) Hµã£¬Gµã ¡£

36

34£®AÓëBÊÇÁ½ÖÖ»¥²»ÏàÈܵÄÁ½ÖÖÒºÌ壬AµÄÕý³£·Ðµã80¡æ£¬

BµÄÕý³£·Ðµã120¡æ¡£°ÑA¡¢B»ìºÏ×é³ÉÒ»¸öÌåϵ£¬ÄÇôÕâ¸ö»ìºÏÎïµÄÕý³£·ÐµãΪ£º (A) СÓÚ80¡æ£» (B) ´óÓÚ120¡æ£» (C) ½éÓÚ80¡æÓë120¡æÖ®¼ä £» (D) ÎÞ·¨È·¶¨·¶Î§ ¡£

35£®ÓÒͼÊÇÈýҺ̬ºãκãѹÏàͼ£¬ac¡¢be°ÑÏàͼ·Ö³ÉÈý¸öÏàÇø ¢Ù¡¢¢Ú¡¢¢Û£¬Ã¿¸öÏàÇø´æÔÚµÄÏàÊýÊÇ£º (A) (A) ¢ÙÇø1¡¢¢ÚÇø1¡¢¢ÛÇø1 £» (B) ¢ÙÇø1¡¢¢ÚÇø3¡¢¢ÛÇø2 £»

(B) (B) ¢ÙÇø1¡¢¢ÚÇø2¡¢¢ÛÇø2 £» (D) ¢ÙÇø1¡¢¢ÚÇø2¡¢¢ÛÇø1 ¡£

36£®½ð(ÈÛµã1063¡æ)ÓëÍ­(ÈÛµã1083¡æ)ÐγɺϽð£»È¡º¬½ðÁ¿ 50£¥µÄÈÚÈÛÌåÀäÈ´£¬Ê×ÏÈÎö³ö¹ÌÈÜÌåµÄº¬½ðÁ¿ÊÇ£º (A) ´óÓÚ50£¥ £» (B) СÓÚ50 % £» (C) µÈÓÚ50£¥ £» (D) ²»Ò»¶¨ ¡£

37£®H2O£­NaCl£­Na2SO4µÄÎïϵÖÐNa2SO4ÓëH2OÄÜÐγÉË®ºÏ ÎïNa2SO4¡¤10 H2O(D)¡£ÏàͼÖУ¬ÔÚDBCÇøÖдæÔÚµÄÊÇ£º (A) Ë®ºÏÎïDÓëÈÜÒº £»

(B) Ë®ºÏÎïD¡¢NaClºÍ×é³ÉΪFµÄÈÜÒº £» (C) Ë®ºÏÎïD¡¢NaClºÍNa2SO4ÈýÏ๲´æ £» (D) ´¿NaCl¡¢´¿Na2SO4ºÍË®ÈÜÒº ¡£

38£®H2O£­KNO3£­NaNO3ÎïϵµÄÏàͼÈçÏ¡£ÄÇô

ÔÚBECÇøÄÚƽºâµÄÏàÊÇ£º (A) ´¿NaNO3ÓëÆä±¥ºÍÈÜÒº £» (B) ´¿KaNO3ÓëÆä±¥ºÍÈÜÒº £»

(C) º¬KNO3¡¢NaNO3Óë²»±¥ºÍÈÜÒº £» (D) º¬KNO3¡¢NaNO3ÓëË«±¥ºÍÈÜÒº(E) ¡£

39£®·ÖÅ䶨Âɲ»ÊÊÓÃÏÂÁÐÄĸöÌåϵ£º (A) I2ÈܽâÔÚË®ÓëCCl4ÖÐ £»

(B) Na2CO3ÈܽâÔÚÕý¸ýÍéºÍ¶þÒÒ¶þ´¼ÃÑÖÐ £» (C) NH4ClÈܽâÔÚË®Óë±½ÖÐ £» (D) Br2ÈܽâÔÚCS2ÓëË®ÖÐ ¡£

40£®ÈçͼÊǺãκãѹϵÄÈý×é·ÖÑÎË®ÌåϵÏàͼ£¬¸´ÑοÉÐÎ ³ÉË®ºÏÎ´æÔÚ¼¸¸öÈýÏàƽºâÇø£º

(A) 2¸ö £» (B) 3¸ö £» (C) 4¸ö £» (D) 5¸ö ¡£

Èý¡¢¶àÑ¡Ì⣺

1£®Í¨³£Ëù˵µÄË®µÄ±ùµãΪ0¡æµÄº¬ÒåÊÇʲô£¿

(A) ÔÚ101.33kPaѹÁ¦Ï£¬±ùºÍ´¿Ë®Æ½ºâʱµÄÎÂ¶È £» (B) ±ù¡¢Ë®¡¢Ë®ÕôÆøÈýÏàƽºâʱµÄÎÂ¶È £» (C) ±ùµÄÕôÆøѹºÍË®µÄÕôÆøѹÏàµÈµÄÎÂ¶È £»

(D) ѹÁ¦Îª101.3kPaϱ»¿ÕÆø±¥ºÍÁ˵ÄË®ºÍ±ùƽºâµÄÎÂ¶È £» (E) ѹÁ¦ÔÚ101.3kPaʱ£¬±ù¡¢Ë®¡¢Ë®ÕôÆøÈýÏ๲´æʱµÄÎÂ¶È ¡£

2£®ÏÂÁÐϵͳ¦µ = 1µÄÊÇ£º

(A) Pb£­Sn£­SbÐγɵĹÌÈÜÌå £» (B) ¦Á-SiO2ºÍ¦Â-SiO2 £»

(C) ×óÐýÆÏÌÑÌǺÍÓÒÐýÆÏÌÑÌÇ £»(D) ÂÈ»¯ÄƺÍÕáÌÇÈÜÓÚË® £»(E) °×ÌÇÓëÃæ·Û»ìºÏÎï ¡£

37

3£®¿Ë£­¿Ë·½³Ì¿ÉÊÊÓÃÓÚÏÂÁÐÄÄЩÌåϵ£¿ (A) I2(s) I2(g) £» (B) C(ʯī) C(½ð¸Õʯ) £» (C) I2(s) I2(l) £» (D) I2(l) I2(g) £» (E) I2(g)(n,T1,p1) I2(g) (n,T2,p2) £» (F) I2(l) + H2 2HI(g) ¡£

4£®¶ÔÓÚÁ½×é·ÖÄý¾ÛÌåϵµÄ²½ÀäÇúÏßÐÎ×´£¬ÏÂÁÐÅжÏÕýÈ·µÄÊÇ£º (A) ²½ÀäÇúÏßÉÏ£¬Ò»¶¨´æÔÚ¡°¹Õµã¡± £»

(B) Ò»Ìõ²½ÀäÇúÏßÉÏÓÐƽ̨£¬¾ÍÒ»¶¨ÓС°¹Õµã¡± £» (C) Ò»Ìõ²½ÀäÇúÏßÉÏÓС°¹Õµã¡±£¬¾ÍÒ»¶¨ÓС°Æ½Ì¨¡± £» (D) Ò»Ìõ²½ÀäÇúÏßÉÏ¿ÉÒÔÓС°¹Õµã¡±¶øÎÞ¡°Æ½Ì¨¡± £» (E) Ò»Ìõ²½ÀäÇúÏßÉÏ¿ÉÒÔÓС°Æ½Ì¨¡±¶øÎÞ¡°¹Õµã¡± ¡£ 5£®ÔÚCaF2£­CaCl2µÄÄý¾ÛϵͳÏàͼÖУ¬ÎïϵµãΪa£¬µ± ½µµÍϵͳζÈʱ£¬²»ÕýÈ·µÄÊÇ£º (A) (A) aµã¦µ = 1£¬f* = 2 £» (B) bµã¦µ = 1£¬f* = 1 £»

(B) (B) cµã¦µ = 2£¬f* = 1 £» (C) (C) dµã¦µ = 2£¬f* = 1 £» (E) cµã¦µ = 3£¬f* = 0 ¡£

6£®¹ØÓڸܸ˹æÔòµÄÊÊÓÃÐÔ£¬ÏÂÁÐ˵·¨´íÎóµÄÊÇ£º

(A) ÊÊÓÃÓÚµ¥×é·ÖÌåϵµÄÁ½¸öƽºâÏà £» (B) ÊÊÓÃÓÚ¶þ×é·ÖÌåϵµÄÁ½ÏàƽºâÇø £» (C) ÊÊÓÃÓÚ¶þ×é·ÖÌåϵÏàͼÖеÄÈκÎÇø £»(D) ÊÊÓÃÓÚÈý×é·ÖÌåϵÖеÄÁ½¸öƽºâÏà £» (E) ÊÊÓÃÓÚÈýÏàƽºâÏßÉϵÄÈÎÒâÁ½¸öƽºâÏà ¡£

7£®¼ÓѹÓÚ±ù£¬±ù¿É²¿·ÖÈÛ»¯³ÉË®£¬ÕâÊÇʲôԭÒò£¿ (A) ¼ÓѹÉúÈÈ £» (B) ±ùµÄ¾§¸ñÊÜѹ±ÀÀ£ £» (C) ±ùÈÛ»¯Ê±ÎüÈÈ £» (D) ±ùµÄÃܶÈСÓÚË® £» (E) ±ùµÄÃܶȴóÓÚË® ¡£

8£®AºÍB¶þÔªÄý¾ÛϵÏàͼÈçͼËùʾ£¬ÔÚÏÂÁÐÐðÊöÖдíÎóµÄÊÇ£º (A) 1ΪҺÏ࣬¦µ = 1£¬f = 2 £»

(B) Òª·ÖÀë³ö´¿A£¬Îïϵµã±ØÐëÔÚ3ÇøÄÚ £» (C) Òª·ÖÀë³ö´¿AmBn£¬Îïϵµã±ØÐëÔÚ6ÇøÄÚ £» (D) J¡¢F¡¢E¡¢IºÍSÖîµãf = 0 £»

(E) GCÖ±Ïß¡¢DIÖ±ÏßÉϵĵ㣬f = 0 ¡£

9£®20¡æʱ±½ºÍ¼×±½µÄÕôÆøѹ·Ö±ðΪ9959.2PaºÍ2973.1Pa£¬±½(A)ºÍ¼×±½(B)ÐγɵÄÈÜÒº ½üËÆÓÚÀíÏëÈÜÒº£¬µ±x(±½) = 0.5ʱ£¬ÆøÏàÖб½µÄ·Öѹp(±½) = 4979.6Pa£¬¼×±½µÄ·Öѹ

p(¼×±½) = 1486.6Pa£¬ÔÚͼÖУ¬p¡«xͼºÍT¡«xͼÕýÈ·µÄÊÇ£º

10£®ÏÂÊö˵·¨ÖУ¬ÄÄÒ»¸öÊÇ´íÎóµÄ£¿

(A) ͨ¹ýÏàͼ¿ÉÒÔÈ·¶¨Ò»¶¨Ìõ¼þÏÂÌåϵÓɼ¸Ï๹³É £» (B) Ïàͼ¿É±íʾ³öƽºâʱÿһÏàµÄ×é³ÉÈçºÎ £» (C) Ïàͼ¿É±íʾ³ö´ïµ½ÏàƽºâËùÐèµÄʱ¼ä³¤¶Ì £»

(D) ͨ¹ý¸Ü¸Ë¹æÔò¿ÉÔÚÏàͼÉϼÆËã¸÷ÏàµÄÏà¶ÔÊýÁ¿¶àÉÙ £» (E) ÏàͼÉÏÎïϵµãÓëÏàµã¶¼ÊÇ·Ö¿ªµÄ ¡£

38

ËÄ¡¢Ö÷¹ÛÌ⣺

1£®ÔÚ25¡æʱ£¬A¡¢B¡¢CÈýÖÖÎïÖÊ(Ï໥֮¼ä²»·¢Éú·´Ó¦)ËùÐγɵÄÈÜÒºÓë¹Ì̬AºÍÓÉB¡¢C×é³ÉµÄÆøÏàͬʱ´ïµ½Æ½ºâ£¬ÎÊ£º

(1) ´ËÌåϵµÄ×ÔÓɶÈΪ¶àÉÙ£¿ (2) ´ËÌåϵÖÐÄÜƽºâ¹²´æµÄ×î¶àÓм¸Ïࣿ

2£®»¬±ùЬÏÂÃæµÄ±ùµ¶Óë±ù½Ó´¥Ã泤Ϊ7.68cm£¬¿íΪ0.00245cm¡£

(1) Èô»¬±ùÕßÌåÖØΪ60kg£¬ÊÔÇóÊ©ÓÚ±ùÃæµÄѹǿΪ¶àÉÙ£¿(Ë«½Å»¬ÐÐ)

-(2) ÔÚ¸ÃѹǿÏ£¬±ùµÄÈÛµãÊǶàÉÙ£¿ËÈÖª±ùµÄĦ¶ûÈÛ»¯ÈÈΪ6009.5 J¡¤mol1£¬±ùµÄÃܶÈΪ

--0.92g¡¤cm3£¬Ë®µÄÃܶÈΪ1.0 g¡¤cm3 ¡£

3£®ÒÒõ£ÒÒËáÒÒõ¥CH3COCH2COOC2H5ÊÇÓлúºÏ³ÉµÄÖØÒªÊÔ¼Á£¬ËüµÄÕôÆøѹ·½³ÌΪ£º lnp = -5960/T + B£¬pµÄµ¥Î»ÊÇPa£¬´ËÊÔ¼ÁÔÚÕý³£·Ðµã181¡æʱ ²¿·Ö·Ö½â£¬µ«ÔÚ70¡æÊÇÎȶ¨µÄ£¬¿ÉÔÚ70¡æʱ¼õѹÕôÁóÌá´¿£¬ ѹǿӦ½µµ½¶àÉÙ£¿¸ÃÊÔ¼ÁµÄĦ¶ûÆø»¯ÈÈÊǶàÉÙ£¿

4£®ÒºÌåAÓëBÐγɷÇÀíÏëÈÜÒº£¬ÔÚÒ»¶¨Î¶ÈÏ£¬ÕôÆø×ÜѹÓëÈÜ Òº×é³ÉµÄ¹ØϵͼÈçÓÒ£º

(1) ´ÖÂÔ»­³öpA¡«xB ¡¢pB¡«xBÏß £»

(2) ´ÖÂÔ»­³öT¡«xB ºÍT¡«yBÏß(·Ðµã¡«×é³ÉÏß) ¡£ 5£®ÔÚ16¡«78¡æÖ®¼ä£¬ÒºÌ¬¹¯µÄÕôÆøѹ¹«Ê½Îª £ºlg(p/p) = 7.648£­3328/T £­0.848lg(T/K)£¬

-¹¯µÄÈýÏàµãζÈΪ243.3K£¬Ä¦¶ûÈÛÈÚÈÈΪ2320 J¡¤mol1£¬ÊÔÇó¹Ì̬¹¯ÔÚ -78¡æʱµÄÕôÆøѹ¡£

6£®ÏÂÃæÊÇÁ½×é·ÖÄý¾ÛÌåϵÏàͼ£¬×¢Ã÷ÿ¸öÏàͼÖÐÏà̬£¬²¢Ö¸³öÈýÏàƽºâÏß¡£

7£®NiÓëMoÐγɻ¯ºÏÎïMoNi£¬ÔÚ1345¡æʱ·Ö½â³ÉMoÓ뺬53£¥ MoµÄÒºÏ࣬ÔÚ1300¡æÓÐΨһ×îµÍ¹²È۵㣬¸ÃÎÂÏÂƽºâÏàΪMoNi£¬º¬48£¥ MoµÄÒºÏàºÍº¬32£¥ MoµÄ¹ÌÈÛÏ࣬ËÈÖªNiµÄÈÛµã1452¡æ£¬MoµÄÈÛµãΪ2535¡æ£¬»­³ö¸ÃÌåϵµÄ´ÖÂÔÏàͼ[t¡æ¡«cͼ]¡£

8£®AgClÓëLiClÌåϵ²»Ðγɻ¯ºÏÎ¹Ì̬²¿·Ö»¥ÈÜ£¬ÔÚ480¡æʱ£¬ÈÛÈÚÌåÓë·Ö±ðº¬15£¥¼°30£¥ AgClµÄ¹ÌÈÛÌå³Éƽºâ£¬AgClÓëLiClµÄÈ۵㠷ֱðΪ455¡æÓë610¡æ£¬ÊÔ»æÖÆÆäÏàͼ¡£

9£®Èçͼ£¬ÊÇNaCl£­(NH4)2SO4£­H2OÔÚ298K¡¢ 101.325kPaʱµÄÏàͼ¡£ÏÖÓÐNaClÓë(NH4)2SO4 »ìºÏÑÎ100¿Ë£¬ÆäÖÐ(NH4)2SO4º¬Á¿Îª25%£¬Îï ϵµãÏ൱ͼÖÐfµã£¬ÀûÓÃÏàͼ¼ÆË㣬¿ÉÒÔ×î¶àÌá ´¿µÃµ½¶àÉÙ¿ËNaCl¾§Ì壿

10£®CaCO3ÔÚ¸ßηֽâΪCaOºÍCO2¡£

(1) ÓÉÏàÂÉÖ¤Ã÷ÎÒÃÇ¿ÉÒÔ°ÑCaCO3ÔÚ±£³Ö¹Ì¶¨Ñ¹Á¦µÄCO2ÆøÁ÷ÖмÓÈȵ½Ï൱µÄζȶø²»Ê¹CaCO3·Ö½â£»

(2) Ö¤Ã÷µ±CaCO3ÓëCaOµÄ»ìºÏÎïÓëÒ»¶¨Ñ¹Á¦µÄCO2¹²´æʱÓÐÇÒ½öÓÐÒ»¸öƽºâζȡ£

11£®ÔÚ101.325kPaÏ£¬CaCO3·Ö½â³ÉCaO(s) ºÍCO2(g)£¬ÔÚ1169Kʱ´ïµ½·Ö½âƽºâ¡£

39

(1) »­³öÁ½×é·ÖCaO£­CO2ÔÚ101.325PaʱµÄµÈѹÏàͼ£» (2) ±ê³ö¸÷¸öÏàÇøµÄÏà̬¡£

12£®NaCl£­H2O¶þ×é·ÖÌåϵµÄµÍ¹²ÈÛµãΪ -21.1¡æ£¬´Ëʱ±ù¡¢NaCl¡¤2H2O(s) ºÍŨ¶ÈΪ22.3£¥(ÖØÁ¿°Ù·ÖÊý)µÄNaClË®ÈÜҺƽºâ¹²´æ£¬ÔÚ -9¡æʱÓÐÒ»²»ÏàºÏÈ۵㣬ÔÚ¸ÃÈÛµãζÈʱ£¬²»Îȶ¨»¯ºÏÎïNaCl¡¤2H2O·Ö½â³ÉÎÞË®NaClºÍ27£¥ µÄNaClË®ÈÜÒº£¬ÒÑÖªÎÞË®NaClÔÚË®ÖеÄÈܽâ¶ÈÊÜζȵÄÓ°Ïì²»´ó(µ±Î¶ÈÉý¸ßʱ£¬Èܽâ¶ÈÂÔÓÐÔö¼Ó)¡£

(1) Çë»æÖÆÏàͼ£¬²¢Ö¸³öͼÖÐÏß¡¢ÃæµÄÒâÒ壻

(2) ÈôÔÚ±ùˮƽºâÌåϵÖмÓÈë¹ÌÌåNaCl×÷ÖÂÀä¼Á¿É»ñµÃ×îµÍζÈÊǼ¸¶È£¿

(3) ÈôÓÐ1000g 28£¥ µÄNaClÈÜÒº£¬ÓÉ160¡æÀäµ½ -10¡æ£¬Îʴ˹ý³ÌÖÐ×î¶àÄÜÎö³ö¶àÉÙ

´¿NaCl £¿

12£®ÏÖÓд¦ÓÚ263KµÄ1 mol¹ýÀäË®£¬¼Ù¶¨Ë®×Ô·¢µØ½á¾§£¬½á¾§¹ý³ÌÒ»Ö±ÑÓÐøµ½Ìåϵƽºâ¡£ (1) Ìåϵ´ïµ½×îÖÕζÈΪ¶àÉÙ£¿(2) ƽºâʱÓм¸Ï๲´æ£¿ (3) ¼ÆËãƽºâʱ¸÷ÏàµÄÁ¿£»(4) ¼ÆËã½á¾§¹ý³ÌµÄ¦¤Sm¡£

----ÒÑÖªCp,m[H2O(l)] = 75.31 J¡¤K1¡¤mol1£»Cp,m[H2O(s)] = 37.66 J¡¤K1¡¤mol1£»

-ÈÛ»¯ÈȦ¤Hm[H2O(s)] = 6.02 kJ¡¤mol1 ¡£

--13£®ÒÑÖªBiºÍCdµÄÈÛµãºÍÈÛÈÚìÊ·Ö±ðÊÇ544.2K¡¢594.1KÒÔ¼°11.00 kJ¡¤mol1¡¢5.98 kJ¡¤mol1£¬Ô¤ÑÔÓÉÕâÁ½¸ö½ðÊôËùÐγɵÄ×îµÍ¹²ÈÛ»ìºÏÎïµÄζȺÍ×é³É(ÉèÐγÉÀíÏë»ìºÏÎï)¡£ÊµÑéֵΪ140¡æºÍCdµÄº¬Á¿(ÖÊÁ¿°Ù·ÖÊý)Ϊ40£¥ ¡£

14£®Ö¸³öÏÂÁи÷ͼÖÐËùÐγɵĻ¯ºÏÎïµÄ¾­Ñéʽ£¬²¢ËµÃ÷¸÷ÏàÇøÊÇÓÉÄÄЩÏà×é³ÉµÄ£¿ (1) ͼ¢ñÖÐÉèÉú³ÉµÄ»¯ºÏÎïΪX £»(2) ûÓл¯ºÏÎ

(3) ͼ¢óÖÐÉèÉú³ÉX¡¢YÁ½ÖÖ»¯ºÏÎ

15£®·ÓË®ÌåϵÔÚ60¡æ·Ö³ÉÁ½ÒºÏ࣬µÚÒ»Ïຬ16.8£¥ (ÖÊÁ¿°Ù·ÖÊý) µÄ·Ó£¬µÚ¶þÏຬ44.9£¥ µÄË®¡£

(1) Èç¹ûÌåϵÖк¬90gË®ºÍ60g·Ó£¬ÄÇôÿÏàÖØÁ¿Îª¶àÉÙ£¿

(2) Èç¹ûҪʹº¬80£¥ ·ÓµÄ100gÈÜÒº±ä³É»ë×Ç£¬±ØÐë¼ÓË®¶àÉÙ¿Ë£¿

16£®80¡æʱäå±½ºÍË®µÄÕôÆøѹ·Ö±ðΪ8.825kPaºÍ47.335kPa£¬äå±½µÄÕý³£·ÐµãÊÇ156¡æ¡£¼ÆË㣺

(1) ä屽ˮÕôÆøÕôÁóµÄζȣ¬ÒÑ֪ʵÑéÊҵĴóÆøѹΪ101.325kPa£»

-(2) ÔÚÕâÖÖË®ÕôÆøÕôÁóµÄÕôÆøÖÐäå±½µÄÖÊÁ¿·ÖÊý¡£ÒÑÖªäå±½µÄĦ¶ûÖÊÁ¿Îª156.9 g¡¤mol1£» (3) Õô³ö10kgäå±½ÐèÏûºÄ¶àÉÙǧ¿ËË®ÕôÆø£¿

µÚÎåÕ ÏàƽºâÁ·Ï°Ìâ´ð°¸

Ò»¡¢ÅжÏÌâ´ð°¸£º

40

1£®¶Ô¡£ 2£®´í¡£

3£®´í£¬Ó¦ÊÇÔÚÒ»¶¨·¶Î§ÄÚ¿ÉÒÔ¶ÀÁ¢±ä»¯µÄÇ¿¶È±äÊý¡£ 4£®´í£¬Ò²Óдú±íÏàµÄµã¡£

5£®´í£¬f = l±íʾÓÐÒ»¸öÇ¿¶È±äÊý¿É¶ÀÁ¢±ä»¯¡£ 6£®¶Ô¡£

7£®¶Ô£¬ÔÚp¡«xͼÖÐÀíÏëҺ̬»ìºÏÎïµÄÒºÏàÏßÊÇÖ±Ïß¡£ 8£®¶Ô¡£

9£®´í£¬ÆäËûÏàͼµÄÁ½ÏàƽºâÇøÒ²¿ÉÒÔʹÓøܸ˹æÔò¡£ 10£®´í£¬¶ÔÓкã·ÐÎïµÄϵͳ£¬×î¶àÖ»Äܵõ½Ò»¸ö´¿×é·Ö¡£ 11£®´í£¬Í¨³£Á½¸ö×é·Ö¶¼Í¬Ê±²úÉúÕý(»ò¸º)Æ«²î¡£ 12£®´í£¬ºã·Ð×é³ÉÓëѹÁ¦Óйء£

13£®´í£¬Òò²»»¥ÈÜ£¬pA = pA¡¯£¬Óë×é³ÉÎ޹ء£ 14£®¶Ô¡£ 15£®¶Ô¡£

¶þ¡¢µ¥Ñ¡Ìâ´ð°¸£º

1. C£» 2. C£» 3. C£» 4. C£» 5. D£» 6. B£» 7. B£» 8. C£» 9. D£» 10.D£» 11.A£» 12.A£» 13.A£» 14.D£» 15.C£» 16.C£» 17.B£» 18.C£» 19.B£» 20.C£» 21.A£» 22.B£» 23.D£» 24.C£» 25.D£» 26.C£» 27.B£» 28.C£» 29.B£» 30.D£» 31.C£» 32.B£» 33.D£» 34.A£» 35.D£» 36.B£» 37.C£» 38.D£» 39.C£» 40.B¡£

Èý¡¢¶àÑ¡Ìâ´ð°¸£º 1. D£» 2. AD£» 3. AD£» 4. DE£» 5. BC£» 6. AC£» 7. D£» 8. BC£» 9. CD£» 10. CE¡£

ËÄ¡¢Ö÷¹ÛÌâ´ð°¸£º

1£®½â£ºN = 3 C = 3 ¦µ = 3 (s£¬l£¬£ç)

(1) f* = 3-3 + 1 = 1 £»(2) f*min = 0 £¬3-¦µmax + 1 = 0 £¬¦µmax = 4 ¡£

-2£®½â£º(1) p = W/2S = 60/(7.62 ¡Á 0.00245 ¡Á 3) = 1607 kg¡¤cm2 = 1549 ¡Á 104 Pa (2) dp/dT = ¦¤H/T¦¤V £¬dp = (¦¤H/T¦¤V) ¡¤dT »ý·Ö£ºp2-p1 = (¦¤H/¦¤V) ¡Á ln(T2/T1)£¬p2 = 15749 ¡Á 104 Pa£¬p1 = 101325 Pa T1 = 273.15 K£¬ ¦¤V = 18.02/1.00 - 18.02/0.92 = -1.567 cm3 ln(T2/T1) = (p2-p1) ¡Á ¦¤V/¦¤H

- = (15746 ¡Á 104-101325) ¡Á (-1.567 ¡Á 106)/6009.5 = -0.04106 T2 = 0.9598 T1 = 0.9598 ¡Á 273.15 = 262.16 K ¼´£º t2 = - 11¡æ

3£®½â£ºT = 273 + 181 = 454 K £¬p = 101325 Pa ln101325 = - 5960/454 + B£¬B = 24.654

lnp = - 5960/T +24.654£¬T2 = 273 + 70 = 343K

lnp2 = - 5960/343 + 24.654 = 7.278£¬p2 = 1448.1 Pa

5960?H??lnp???lnp????2????T?p?T?pRT2T??ÒòΪ £¬ÓÉ¿Ë-¿Ë·½³Ì£º

-1

¦¤Hm,vap(Æø»¯) = 5960R = 5960 ¡Á 8.314 = 49551 J¡¤mol

4£®½â£º

41

5£®½â£º¦¤Hm(Æø»¯) = a + bT + cT2

dlnpl/dT = ¦¤Hm(Æø)/RT2 = a/RT2 + b/RT + c/R (1)

2.303lgpl = 7.648¡Á2.303 £­3328¡Á2.303/T £­0.848¡Á2.303lgT lnpl = 7.648¡Á2.303 £­7664/T£­8.848lnT

dlnpl/dT = 7664/T2-0.848/T Óë(1)ʽ±È½Ï£¬

a = 7664R = 63718 b = - 0.848R = -7.05 c = 0 ¦¤Hm(Æø»¯) = 63718-7.05 T

¦¤Hm(Éý»ª) = ¦¤Hm(ÈÛÈÚ) + ¦¤Hm(Æø»¯) = 2320 + 63718-7.05 T = 66038-7.05 T dlnps/dT = ¦¤Hm(Éý»ª)/RT2 = 66038/RT2-7.05/RT = 7943/T2-0.848/T »ý·Ö £º lnps = - 7943/T-0.848lnT + C

lgps = - 3449/T-0.848lgT + C' ÈýÏàµã T = 243.3K lgps = lgpL -2339/243.3-0.848lg243.3 + C' = 7.648-3328/243.3-0.848lg243.3

¡àC' = 8.145 lgps = - 3449/T-0.848lgT + 8.145 µ± T = 273.2-78 = 196.2 K lgps = - 3449/195.2 = - 0.848lg195.2 + 8.145 = - 11.460

---ps = 3.92 ¡Á 1012 p= 3.29 ¡Á 1012 ¡Á 101325 = 3.98 ¡Á 107 Pa

6£®½â£º

7£®½â£º »¯ºÏÎïÖк¬MoΪ 62 % ¡£

8. ½â£º

9£®½â£ºÔÚͼÉÏÁª½á Cf£¬½»ADÏßÓÚgµã£¬Á¿³öCf = 3.7 cm£¬Cg = 2.6 cm£¬ Óøܸ˹æÔò¼ÆËãgµãÎïϵÖÊÁ¿£¬w(g) = 100 ¡Á 3.7/2.6 = 142.3 g £¬¼ÓË® 42.3 g Á¿³öAD = 2.6 cm £¬Dg = 0.9 cm£¬ w (NaCl) = 142.3 ¡Á 0.9/2.6 = 49.26 g

10£®½â£º(1) ¸ù¾ÝÌâÒ⣬ÌåϵÖÐÖ»´æÔÚCaCO3ºÍCO2

S = 2 £¬R = R¡¯ = 0 K = S-R-R¡¯ = 2ÒòΪѹÁ¦p¹Ì¶¨£¬

ÇÒ¦µ = 2[CaCO3(s)£¬CO2 (g)] ËùÒÔ£º f = K-¦µ + 1 = 2-2 + 1 = 1

Õâ˵Ã÷ÌåϵÉÐÓÐÒ»¸ö×ÔÓɶȣ¬´Ë¼´ÎªÎ¶ȣ¬ÔÚζȿÉ×ÔÓɱ仯µÄÇé¿öÏ£¬ ÌåϵÖÐCaCO3²»·Ö½â¡£

(2) ÌåϵÖÐÓÐCaCO3(s)£¬CaO(s)ºÍCO2 (g)

42

ͬʱ´æÔÚ»¯Ñ§Æ½ºâ CaCO3(s) CaO(s) + CO2 (g) ¹Ê S = 3 £¬R¡¯ = 0 £¬R = 1 K = S-R¡¯-R = 2

ÒòΪѹÁ¦p¹Ì¶¨£¬ÇÒ¦µ = 3[CaCO3(s)£¬CaO(s)£¬CO2 (g)] ËùÒÔ f = K-¦µ + 1 = 2-3 + 1 = 0

11£®½â£º

¢ñÇø£ºCaO(s) + CO2(g) ¢òÇø£ºCaO(s) + CaCO3(s) ¢óÇø£ºCO2(g) + CaCO3(s)

11£®½â£º(1) ͼÖÐµÄ ac ΪˮµÄ±ùµãϽµÇúÏߣ»ec Ϊˮ

»¯ÎïNaCl¡¤2H2OµÄÈܽâÇúÏߣ»eh ΪNaClµÄÈܽâ¶ÈÇúÏߣ»bd ΪÈýÏàÏߣ¬ÏßÉÏÈÎÒâÒ» µã´ú±í±ù¡¢Ë®»¯ÎïºÍ¾ßÓÐ c µã×é³ÉµÄNaClÈÜÒºÈýÏàƽºâ¹²´æ£»eg ΪÈýÏàÏߣ¬ÏßÉÏ ÈÎÒâÒ»µã´ú±íNaCl¡¢Ë®»¯ÎïºÍ¾ßÓÐeµã×é³ÉµÄNaClÈÜÒºÈýÏàƽºâ¹²´æ¡£

¢ñÊÇÒºÏàÇø£»¢òÊÇ¹Ì (NaCl) ҺƽºâÇø£»¢óÊDZùҺƽºâÇø£»¢ôÊÇ¹Ì (NaCl¡¤2H2O) Һƽ ºâÇø¡£

(2) ÓÉÏàͼ¿ÉÖª£¬ÔÚ±ùˮƽºâÌåϵÖмÓÈë NaCl£¬Ëæ׿ÓÈëNaClµÄÔö¶à£¬ÌåϵÎÂ¶È ÑØcÏßϽµ£¬ÖÁcµãζȽµÎª×îµÍ£¬Õâʱ ÌåϵµÄζÈΪ -21.1¡æ¡£

(3) ÔÚÀäÈ´µ½-9¡æʱ£¬ÒÑ´ïµ½µÍ¹²È۵㣬 ´ËʱÒÑ¿ªÊ¼ÓÐNaCl¡¤2H2OÎö³ö£¬µ½-10¡æ£¬ ´¿NaClÒÑÏûʧ£¬Òò´ËÔÚÀäÈ´µ½-10¡æ¹ý³Ì ÖУ¬×î¶àÎö³öµÄ´¿NaCl¿ÉÓɸܸ˹æÔò¼Æ

Ë㣺w (Òº) ¡Á 1 = w (NaCl) ¡Á 72 w (NaCl) / w×Ü = 1/(72 + 1) = 1/73 w (NaCl) = w×Ü/ 73 = 1000/73 = 13.7 g

¼´ÀäÈ´µ½-10¡æ¹ý³ÌÖУ¬×î¶àÄÜÎö³ö´¿NaCl 13.7 g¡£

12£®½â£º(1) ÓÉÓÚ 1.00mol ¹ýÀäË®Éý¸ßµ½ 0.01¡æ (¼´273.16 K) ʱÐèÒªµÄÈÈÁ¿Ô¼Îª£º

-75.31 ¡Á (273.16-263) = 765.15 J¡¤mol1 £¬

-СÓÚ1mol¹ýÀäË®½á¾§Îª±ù·Å³öµÄÈÈÁ¿ 6020 J¡¤mol1 £¬ËùÒÔÎÞ·¨Ê¹¹ýÀäˮȫ²¿½á³É±ù£¬ ×îÖÕζÈΪ0.01¡æ(273.16 K)µÄ±ùË®¹²´æÌåϵ¡£ (2) ƽºâʱÊÇÈýÏà (±ù¡¢Ë®¡¢ÕôÆø) ¹²´æ¡£

(3) ¼ÙÉèÓÐx mol H2O(s) Éú³É£¬ÓÉÓÚ¹ýÀäË®½á±ù¹ý³ÌºÜ¿ì£¬¹ÊÌåϵÊǾøÈȵģ¬¼´ 263K

ʱ½á³öx mol±ùµÄ·ÅÈÈ£¬Ê¹±ùË®ÌåϵÉýÎÂÖÁ 273.16K ¡£ 75.31 ¡Á 10.16 ¡Á (1-x) + 37.66 ¡Á 10.16¡¤x = x¡¤[6020 + (37.66-75.31) ¡Á 10.16] x = 0.127 ÔòƽºâʱÓÐ 0.127 mol ±ùºÍ 0.873 mol Ë®¡£

(4) ½á¾§¹ý³ÌµÄ¦¤Sm°üÀ¨Á½²¿·Ö£¬µÚÒ»²¿·ÖÊÇ 1mol¹ýÀäË®ÓÉ 263KÉýÎÂÖÁ 273.16K£º

T2Cp,m?H2O(l)?273.16dT?75.3?ln--T263 = 2.855 J¡¤¦¤S1 = T1K1¡¤mol1

--µÚ¶þ²¿·ÖÊÇ x mol Ë®½á³É±ù£º¦¤S2 = -¦¤Hm¡¤x / T = -6020x / 273.16 = -2.799 J¡¤K1¡¤mol1

--ËùÒԽᾧ¹ý³ÌµÄ ¦¤Sm = 2.855-2.799 = 0.056 J¡¤K1¡¤mol1

13£®½â£ºlnx(Bi ) = - (11.00 ¡Á 103/R) ¡Á (1/T-1/544.2)

?43

lnx(Cd) = - (5.98 ¡Á 103/R) ¡Á (1/T-1/594.1)

x(Bi) + x(Cd) = 1 ½âµÃ£ºT = 404.7 K¡Ö131.5¡æ x(Bi) = 0.4325£¬»»Ëã³ÉCdµÄÖÊÁ¿°Ù·ÖÊýΪ41.4£¥ ¡£

14£®½â£º(1)ÓÉͼ¢ñÖªXΪA3B¡£

1£ºL + A(s)£»2£ºA(s) + A3B(s)£»3£ºA3B(s) + L£»4£ºA3B(s) + B(s)£»5£ºB(s) + L (2) ÓÉͼ¢òÖª£º 1£º¦Á£»2£ºL + ¦Á£»3£ºL + ¦Â£»4£º¦Â£»5£º¦Á + ¦Â (3) ÓÉͼ¢óÖªXΪA2B£¬YΪAB ¡£

1£ºL£»2£º¦Á£»3£ºL + ¦Á£»4£ºL + A2B(s)£»5£º¦Á + A2B(s)£» 6£ºL + AB(s)£» 7£ºA2B(s) + AB(s)£» 8£ºL+ AB(s)£» 9£ºL+B(s) 10£ºAB(s) + B(s)

15£®½â£ºÔÚ¢ñÏàÖк¬16.8£¥ µÄ·Ó£»ÔÚ¢òÏàÖк¬ 44.9£¥ µÄË®£¬Ôòº¬·ÓΪ 55.1£¥ ¡£ (1) ÉèÎïϵµãΪ o £¬ÆäÖзӺ¬Á¿Îª£º 60/150 = 40£¥ ÓÚÊÇ w1 + w2 = 150 ÇÒ w1/w2 = (55.1-40) / (40-16.8) ½âµÃ£ºw1 = 59.1 g £¬w2 = 90.9 g

(2) 80 / [ 100 + w(Ë®) ] = 55.1£¥ Ðè¼ÓË® w(Ë®) = 45.2 g

16£®½â£º(1) lnp(äå±½) = -A/T + B

ÓÉ£ºln 8.825(kPa) = - A/353 + B£»ln 101.325(kPa) = -A/429 + B µÃ£ºlnp(äå±½) = - 4863.4/T + 22.86 (1)

**

¶ÔH2O¶øÑÔ£º353K£¬p = 47.335 kPa £»373K£¬p = 101.325 kPa µÃ£ºlnp*(Ë®) = -5010.6/T + 24.96 (2)

**

ÓÖÒòΪ£ºp(äå±½) + p(Ë®) = 101.325 kPa (3)

*

ÁªÁ¢·½³Ì(1)¡¢(2)¡¢(3) £¬½âµÃp(äå±½) = 15.66 kPa £¬p*(Ë®) = 85.71 kPa ´úÈëµÃ£ºT = 368.4 K = 95.2¡æ

(2) ÓÉp*(äå±½) = p×Üy(äå±½) £»p*(Ë®) = p×Üy(Ë®)

¿ÉµÃ£ºp*(äå±½) / p*(Ë®) = y(äå±½) / y(Ë®) = n(äå±½) / n(Ë®)

= [w(äå±½)/M(äå±½)] / [w(Ë®)/M(Ë®)] *

w(äå±½) / w(Ë®) = [p(äå±½)¡¤M(äå±½) / p*(Ë®)¡¤M(Ë®)]

= (15.66 ¡Á 156.9) / (85.71 ¡Á 18) = 1.593

ËùÒÔ w(äå±½) = 1.593 / 2.593 = 61.4£¥

(3) w(Ë®) = w(äå±½) / 1.593 = 10 / 1.593 = 6.28 kg

ÎïÀí»¯Ñ§²âÑ飨ËÄ£©

Ò»¡¢Ñ¡ÔñÌâ¡£ÔÚÌâºóÀ¨ºÅÄÚ£¬ÌîÉÏÕýÈ·´ð°¸´úºÅ¡£

1¡¢ ÁòËáÓëË®¿ÉÐγÉH2SO4?H2O(s)£¬H2SO4?2H2O(s)£¬H2SO4?4H2O(s)ÈýÖÖË®ºÏÎÎÊÔÚ101325 PaµÄѹÁ¦Ï£¬ÄÜÓëÁòËáË®ÈÜÒº¼°±ùƽºâ¹²´æµÄÁòËáË®ºÏÎï×î¶à¿ÉÓжàÉÙÖÖ£¿( ) (1) 3ÖÖ£» (2) 2ÖÖ£» (3) 1ÖÖ£» (4) ²»¿ÉÄÜÓÐÁòËáË®ºÏÎïÓë֮ƽºâ¹²´æ¡£

2¡¢ ×é·ÖA(¸ß·Ðµã)Óë×é·ÖB(µÍ·Ðµã)ÐγÉÍêÈ«»¥ÈܵĶþ×é·Öϵͳ£¬ÔÚÒ»¶¨Î¶ÈÏ£¬Ïò´¿BÖмÓÈëÉÙÁ¿µÄA£¬ÏµÍ³ÕôÆøѹÁ¦Ôö´ó£¬Ôò´ËϵͳΪ£º( )¡£

£¨1£© ÓÐ×î¸ßºã·ÐµãµÄϵͳ£» £¨2£© ²»¾ßÓкã·ÐµãµÄϵͳ£» £¨3£© ¾ßÓÐ×îµÍºã·ÐµãµÄϵͳ¡£

3¡¢Éè·´Ó¦ aA(g ) == yY(g) + zZ(g),ÔÚ101.325 kPa¡¢300 KÏ£¬AµÄת»¯ÂÊÊÇ600 KµÄ2±¶£¬¶øÇÒÔÚ300 KÏÂϵͳѹÁ¦Îª101 325 PaµÄת»¯ÂÊÊÇ2¡Á101 325 PaµÄ2 ±¶£¬¹Ê¿ÉÍƶϸ÷´Ó¦£¨ £©¡£

44

£¨1£©Æ½ºâ³£ÊýÓëζȣ¬Ñ¹Á¦³É·´±È£» £¨2£©ÊÇÒ»¸öÌå»ýÔö¼ÓµÄÎüÈÈ·´Ó¦ £» £¨3£©ÊÇÒ»¸öÌå»ýÔö¼ÓµÄ·ÅÈÈ·´Ó¦£»

£¨4£©Æ½ºâ³£ÊýÓëζȳÉÔÚÕý±È£¬ÓëѹÁ¦³É·´±È¡£

4¡¢Ä³·´Ó¦A(s) == Y(g) + Z(g)µÄ?rGÓëζȵĹØϵΪ?rG= (£­45 000£«110 T/K) J ¡¤mol -1

£¬ ÔÚ±ê׼ѹÁ¦ÏÂ, Òª·ÀÖ¹¸Ã·´Ó¦·¢Éú£¬Î¶ȱØÐë £º ( ) ¡£ (1) ¸ßÓÚ 136 ¡æ £» (2) µÍÓÚ184 ¡æ £» (3) ¸ßÓÚ184 ¡æ£» (4) µÍÓÚ136 ¡æ£»

(5) ¸ßÓÚ136 ¡æ¶øµÍÓÚ 184 ¡æ¡£

5¡¢ ½«¹ÌÌåNH4HCO3(s) ·ÅÈëÕæ¿ÕÈÝÆ÷ÖУ¬µÈÎÂÔÚ400 K£¬NH4HCO3 °´ÏÂʽ·Ö½â²¢´ïµ½Æ½ºâ£º NH4HCO3(s) === NH3(g) + H2O(g) + CO2(g) ϵͳµÄ×é·ÖÊýCºÍ×ÔÓɶÈÊý?Ϊ£º( )¡£ (£±)£Ã£½£²£¬? £½£±£» (£²)£Ã£½£²£¬? £½£²£» (£³)£Ã£½£±£¬? £½£°£» (£´)£Ã£½£³£¬? £½£²¡£ 6¡¢ ÒÑÖªµÈη´Ó¦

¢Ù CH4(g) == C(s) + 2H2(g)

¢Ú CO(g) + 2H2(g) == CH3OH(g)

ÈôÌá¸ßϵͳ×ÜѹÁ¦£¬ÔòƽºâÒƶ¯·½ÏòΪ£¨ £©¡£ £¨1£©¢ÙÏò×ó £¬¢ÚÏòÓÒ £» £¨2£©¢ÙÏòÓÒ£¬¢ÚÏò×ó£» £¨3£©¢ÙºÍ¢Ú¶¼ÏòÓÒ¡£

¶þ¡¢¼ÆËãÌâ¡£Çë¼ÆËãÏÂÁи÷Ìâ¡£ ( ±¾ ´ó Ìâ7·Ö )

ÔÚ323 Kʱ£¬ÏÂÁз´Ó¦ÖÐNaHCO3(s)ºÍCuSO4-5H2O(s)µÄ·Ö½âѹÁ¦·Ö±ðΪ4 000 PaºÍ6052 Pa£º

·´Ó¦¢Ù 2NaHCO3(s) ==Na2CO3(s) + H2O(g) + CO2(g) ·´Ó¦¢Ú CuSO4-5H2O(s) ==CuSO4-3H2O(s) + 2H2O(g) Çó£º

(1) ·´Ó¦¢ÙºÍ¢ÚµÄK¢ÙºÍK¢Ú£»

(2) ½«·´Ó¦¢Ù£¬¢ÚÖеÄËÄÖÖ¹ÌÌåÎïÖÊ·ÅÈëÒ»Õæ¿ÕÈÝÆ÷ÖУ¬Æ½ºâºóCO2µÄ·ÖѹÁ¦Îª¶àÉÙ£¨T=323 K£©£¿

Èý¡¢¼ÆËãÌâ¡£Çë¼ÆËãÏÂÁи÷Ìâ¡£ ( ±¾ ´ó Ìâ11·Ö )

ÒÑÖª·´Ó¦ÔÚ800 Kʱ½øÐУº A (s) + 4B (g) == 3Y (s) + 4Z (g) ÓйØÊý¾ÝÈçÏ£º

ÎïÖÊ A(s) B(s) Y(s) Z(s) O?fHm298K?? OSm298K?? Cp,m298~800KJ?mol?1?K?1193.00 28.33 30.88 36.02 ?? kJ?mol?1-1116.71 0 0 -241.84 J?mol?K151.46 130.58 27.15 188.74 ?1?145

£¨1£©¼ÆËãϱíÖеÄÊý¾Ý ÎÂ¶È O?fHmOSmO?rGm K kJ?mol?1 298 K 800 K J?mol?1?K?1 kJ?mol?1

£¨2£©800 Kʱ£¬½«A(s)ºÍY(s)ÖÃÓÚÌå»ý·ÖÊý·Ö±ðΪw(B)=0.50, w(Z)=0.40, w(¶èÐÔÆøÌå)=0.10µÄ»ìºÏÆøÌåÖУ¬ÉÏÊö·´Ó¦½«ÏòÄĸö·½Ïò½øÐУ¿£¨p=100 kPa£©

ËÄ¡¢¼ÆËãÌâ¡£Çë¼ÆËãÏÂÁи÷Ìâ¡£ ( ±¾ ´ó Ìâ13·Ö ) äå±½ÓëË®µÄ»ìºÏÎïÔÚ101.325 kPaÏ·еãΪ95.7?C,ÊÔ´ÓÏÂÁÐÊý¾Ý¼ÆËãÁó³öÎïÖÐÁ½ÖÖÎïÖʵÄÖÊÁ¿±È.(äå±½ºÍË®ÍêÈ«²»»¥ÈÜ) t / ?C p*(H2O) / kPa 92 100 75.487 101.325 ¼ÙÉèË®µÄÕô·¢ìÊ?vapHmÓëζÈÎ޹أ¬äå±½¡¢Ë®µÄĦ¶ûÖÊÁ¿·Ö±ðΪ157.0 g?mol-1£¬18.02 g?mol-1¡£

Îå¡¢¼ÆËãÌâ¡£Çë¼ÆËãÏÂÁи÷Ìâ¡£ ( ±¾ ´ó Ìâ8·Ö )

ÔÚp=101.3 kPa£¬85¡æʱ£¬Óɼױ½(A)¼°±½(B)×é³ÉµÄ¶þ×é·ÖҺ̬»ìºÏÎï¼´´ïµ½·ÐÌÚ¡£¸ÃҺ̬»ìºÏÎï¿ÉÊÓΪÀíÏëҺ̬»ìºÏÎï¡£ÊÔ¼ÆËã¸ÃÀíÏëҺ̬»ìºÏÎïÔÚ101.3 kPa¼°85¡æ·ÐÌÚʱµÄÒºÏà×é³É¼°ÆøÏà×é³É¡£ÒÑÖª85¡æʱ´¿¼×±½ºÍ´¿±½µÄ±¥ºÍÕôÆøѹ·Ö±ðΪ46.00 kPaºÍ116.9 kPa¡£

Áù¡¢(±¾Ð¡Ìâ12·Ö)

¸ù¾Ýͼ(a)£¬Í¼(b)»Ø´ðÏÂÁÐÎÊÌâ

£¨1£© Ö¸³öͼ(a)ÖУ¬KµãËù´ú±íµÄϵͳµÄ×Ü×é³É£¬Æ½ºâÏàÊý¼°Æ½ºâÏàµÄ×é³É£» £¨2£© ½«×é³Éx(¼×´¼)=0.33µÄ¼×´¼Ë®ÈÜÒº½øÐÐÒ»´Î¼òµ¥ÕôÁó¼ÓÈȵ½85 ?CÍ£Ö¹ÕôÁó£¬ÎÊÁó³öÒºµÄ×é³É¼°²ÐÒºµÄ×é³É£¬Áó³öÒºµÄ×é³ÉÓëÒºÏà±È·¢ÉúÁËʲô±ä»¯£¿Í¨¹ýÕâÑùÒ»´Î¼òµ¥ÕôÁóÊÇ·ñÄܽ«¼×´¼ÓëË®·Ö¿ª£¿

46

£¨3£© ½«(2)ËùµÃµÄÁó³öÒºÔÙÖØмÓÈȵ½78 ?C£¬ÎÊËùµÃµÄÁó³öÒºµÄ×é³ÉÈçºÎ£¿Óë(2)ÖÐËùµÃµÄÁó³öÒºÏà±È·¢ÉúÁËʲô±ä»¯£¿

£¨4£© ½«(2)ËùµÃµÄ²ÐÒºÔٴμÓÈÈ µ½91 ?C£¬ÎÊËùµÃµÄ²ÐÒºµÄ×é³ÉÓÖÈçºÎ£¿Óë(2)ÖÐËùµÃµÄ²ÐÒºÏà±È·¢ÉúÁËʲô±ä»¯£¿

Óû½«¼×´¼Ë®ÈÜÒºÍêÈ«·ÖÀ룬Ҫ²Éȡʲô²½Ö裿

Æß¡¢(±¾Ð¡Ìâ9·Ö)

AºÍBÁ½ÖÖÎïÖʵĻìºÏÎïÔÚ101 325PaÏ·еã-×é³ÉͼÈçͼ£¬Èô½«1 molAºÍ4 molB»ìºÏ£¬ÔÚ101325 PaÏÂÏȺó¼ÓÈȵ½t1=200 ?C£¬t2=400 ?C£¬t3=600 ?C£¬¸ù¾Ý·Ðµã-×é³Éͼ»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©ÉÏÊö3¸öζÈÖУ¬Ê²Ã´Î¶ÈÏÂƽºâϵͳÊÇÁ½Ïàƽºâ£¿ÄÄÁ½Ïàƽºâ£¿¸÷ƽºâÏàµÄ×é³ÉÊǶàÉÙ£¿¸÷ÏàµÄÁ¿ÊǶàÉÙ(mol)£¿

£¨2£©ÉÏÊö3¸öζÈÖУ¬Ê²Ã´Î¶ÈÏÂƽºâϵͳÊǵ¥ÏࣿÊÇʲôÏࣿ

°Ë¡¢ÎÊ´ðÌâ¡£ ( ±¾ ´ó Ìâ9·Ö )

ÒÑÖª CaF2 - CaCl2 Ïàͼ£¬Óû´ÓCaF2 - CaCl2ϵͳÖеõ½»¯ºÏÎïCaF2? CaCl2µÄ´¿´â½á¾§¡£ÊÔÊöÓ¦²Éȡʲô´ëÊ©ºÍ²½Ö裿

¾Å¡¢¼ÆËãÌâ¡£Çë¼ÆËãÏÂÁи÷Ìâ¡£ ( ±¾ ´ó Ìâ10·Ö )

ÒÑÖªÏÂÁÐÁ½·´Ó¦µÄKÖµÈçÏ£º

FeO(s) + CO(g) == Fe(s) + CO2(g) K1 Fe3O4(s) + CO(g) == FeO(s) + CO2(g) K2 TK 873 973 K1 K2 0.871 1.15 0.678 1.77

¶øÇÒ Á½·´Ó¦µÄ ??BCp,m=0 ÊÔÇó£º

(1) ÔÚʲôζÈÏÂFe(s) , FeO(s), Fe3O4(s), CO(g)¼°CO2(g)È«²¿´æÔÚÓÚƽºâϵͳÖУ»

47

p(CO2)(2) ´ËζÈÏÂ

p(CO)=£¿

ÎïÀí»¯Ñ§²âÑ飨ËÄ£©´ð°¸

Ò»¡¢Ñ¡ÔñÌâ¡£ÔÚÌâºóÀ¨ºÅÄÚ£¬ÌîÉÏÕýÈ·´ð°¸´úºÅ¡£ 1¡¢½â£º£¨3£© £¨2·Ö£© 2¡¢½â£º(3) (2·Ö) 3¡¢½â£º£¨3£© £¨2·Ö£© 4¡¢½â£º(2) £¨2·Ö£© 5¡¢½â£º(3) £¨2·Ö£© 6¡¢½â£º£¨1£© £¨2·Ö£© ¶þ¡¢¼ÆËãÌâ¡£Çë¼ÆËãÏÂÁи÷Ìâ¡£ ( ±¾ ´ó Ìâ7·Ö )

1½â£º(1)¶ÔÓÚ·´Ó¦¢Ù£¬p(H2O)= p(CO2)= 2p(×Ü£©

4000Pa???p(×Ü)?O???p(H2O)?p(CO3)?????p??2?100000Pa??O22??=4¡Á10-4 ??(p)???=Ôò K¢Ù==?¶ÔÓÚ·´Ó¦¢Ú£¬p(H2O)=p(×Ü)

226052Pa?p(H2O)???100000Pa??O2??=3.66¡Á10-3 £¨3·Ö£© (p)Ôò K¢Ú==

(2) ËÄÖÖ¹ÌÌåÎïÖÊ·ÅÈëÒ»Õæ¿ÕÈÝÆ÷ÖÐÔÚ323 KÏ´ﷴӦƽºâºó£¬ÏµÍ³ÓÐÈçÏÂÌØÐÔ£º ¢Ù ζȲ»±ä£¬·´Ó¦¢ÙºÍ¢ÚµÄK²»±ä£¨²»¿¼ÂÇÆøÌåµÄ·ÇÀíÏëÐÔ£©£» ¢Ú ¶ÔÓÚ·´Ó¦¢Ú£¬K²»±ä£¬p(H2O)Ò಻±ä£¬¹Êp(H2O)=6 052 Pa£»

¢Û ¶ÔÓÚ·´Ó¦¢ÙºÍ¢Ú£¬ÔÚͬһÈÝÆ÷ÄÚ´ïƽºâʱ£¬Ë®ÕôÆøÖ»ÄÜÓÐÒ»¸ö·ÖѹÁ¦p(H2O)£¬Ëü¼ÈÂú×ã·´Ó¦¢ÚµÄK£¬Ò²Í¬Ê±Âú×ã·´Ó¦¢ÙµÄK£¬¹Ê¶ÔÓÚ·´Ó¦¢Ù£¬p(H2O)=6 052 Pa£¬Òò´Ë£¬ÎÒÃÇÓУº

226052Pa?p?CO2? K¢Ù=?p(H2O)?p(CO2)?/(p)2=

?100000Pa?2=4¡Á10-4

4?106Pa2Ôò p(CO2)=6052Pa=661 Pa £¨7·Ö£© ¼´¼×±½º¬Á¿Îª w (¼×±½)=0.80 (ÖÊÁ¿·ÖÊý)¡£ £¨8·Ö£© Èý¡¢¼ÆËãÌâ¡£Çë¼ÆËãÏÂÁи÷Ìâ¡£ ( ±¾ ´ó Ìâ8·Ö ) ½â£º£¨1£© ??rHm??rSm??rGm-1 kJ?mol K ¶È 298K 800K -1 kJ?mol J?K-1?mol?1 149.35 114.4 162.6 93.90 100.9 39.29 2.07?10-18 2.72?10-3 £¨4·Ö£©

?0.4p???OO40.5p??=0.410 (p/p)(p/p)AB £¨2£©J==

J> K£¬A > 0£¬¹Ê·´Ó¦Ïò×ó½øÐС£ £¨8·Ö£©

48

(pZ/pO)4(pY/pO)34

ËÄ¡¢¼ÆËãÌâ¡£Çë¼ÆËãÏÂÁи÷Ìâ¡£ ( ±¾ ´ó Ìâ13·Ö )

?101.325kPa??vapHm?????R½â£ºln?75.487kPa?=

??11???373.2K365.2K????

??101.325kPa?11?vapHm???????p*?Ë®???373.2K368.9K???=?? R ln?½â³ö£º95.7?Cʱp?(Ë®)=86.5 kPa £» p?(äå±½) =(103.325-86.65) kPa=14.68 kPa £¬ £¨5·Ö£© 86.5?18.02g?mol-1p(Ë®)m(Ë®)M(Ë®)-1 m(äå±½)= p(äå±½)?M(äå±½)= 14.68?157.0g?mol

m (Ë®) £ºm (äå±½) =1.68£º1 ¡£ £¨10·Ö£©

Îå¡¢¼ÆËãÌâ¡£Çë¼ÆËãÏÂÁи÷Ìâ¡£ ( ±¾ ´ó Ìâ8·Ö )

½â£º ÓɸÃҺ̬»ìºÏÎï¿ÉÊÓΪÀíÏëҺ̬»ìºÏÎ¸÷×é·Ö¾ù·ûºÏÀ­ÎÚ¶û¶¨ÂÉ£¬¹Ê

*** p=pA?pB=pA+ (pB£­pA)xB

*p?pA** xB=pB?pA xA=1£­ xB=0.220 £¨3·Ö£©

?(101.3?46.0)kPa?0.780(116.9?46.0)kPa

*pBxBpB116.9kPa?0.780????0.900pp101.3kPaÆøÏà×é³É£¬ÓÉʽ yB

yA=1£­ yB=0.100 £¨8·Ö£©

Áù¡¢(±¾Ð¡Ìâ12·Ö) ½â£º

£¨1£©Èçͼ(a)Ëùʾ£¬Kµã´ú±íµÄ×Ü×é³Éx(CH3OH)=0.33ʱ£¬ÏµÍ³ÎªÆø¡¢ÒºÁ½Ïàƽºâ, LµãΪƽºâÒºÏà, x(CH3OH)=0.15£¬GµãΪƽºâÆøÏ࣬y(CH3OH)=0.52£» (3·Ö)

£¨2£©ÓÉͼ(b)¿ÉÖª£¬Áó³öÒº×é³ÉyB,1=0.52£¬²ÐÒº×é³ÉxB,1=0.15¡£¾­¹ý¼òµ¥ÕôÁó£¬Áó³öÒºÖм״¼º¬Á¿±ÈÔ­Òº¸ß£¬¶ø²ÐÒº±ÈÔ­ÒºµÍ£¬Í¨¹ýÒ»´Î¼òµ¥ÕôÁ󣬲»ÄÜʹ¼×´¼ÓëË®ÍêÈ«·Ö¿ª£» (6·Ö)

£¨3£©Èô½«(2)ËùµÃµÄÁó³öÒºÔÙÖØмÓÈȵ½78¡æ£¬ÔòËùµÃÁó³öÒº×é³ÉyB,2=0.67£¬Óë(2)ËùµÃÁó³öÒºÏà±È£¬¼×´¼º¬Á¿ÓÖ¸ßÁË£» (7·Ö)

£¨4£©Èô½«(2)ÖÐËùµÃ²ÐÒºÔÙ¼ÓÈȵ½91¡æ£¬ÔòËùµÃµÄ²ÐÒº×é³ÉxB,2=0.07£¬Óë(2)ÖÐËùµÃµÄ²ÐÏà±È£¬¼×´¼º¬Á¿ÓÖ¼õÉÙÁË£» (9·Ö) £¨5£©Óû½«¼×´¼Ë®ÈÜÒºÍêÈ«·ÖÀ룬¿É½«Ô­Òº½øÐжà´Î·´¸´ÕôÁó»ò¾«Áó¡£ (12·Ö)

Æß¡¢(±¾Ð¡Ìâ9·Ö)

49

½â£º(1)t2=400?Cʱ£¬Æ½ºâϵͳÊÇÁ½Ïàƽºâ¡£´ËʱÊÇÒº-ÆøÁ½Ïàƽºâ¡£¸÷ƽºâÏàµÄ×é³É£¬ÈçÏÂͼËùʾ£¬Îª xB(l) =0.88 , yB=0.50

n(l)GK0.8?0.50 n(g)=KL=0.88?0.8

n(g)+n(l)=5 mol

½âµÃ£ºn(l)= 4.0mol£»n(g)=1.0 mol £¨6·Ö£© (2) t1=200 ?Cʱ£¬´¦ÓÚÒºÏࣻ t 3 =600 ?Cʱ£¬´¦ÓÚÆøÏà¡£ £¨9·Ö£©

°Ë¡¢ÎÊ´ðÌâ¡£

( ±¾ ´ó Ìâ9·Ö )

½â£º±ØÐëÑ¡¶¨ÈÜÒºµÄ×é³ÉÔÚº¬CaCl2 ԼΪ w(CaCl2)= 0.60~0.80 Ö®¼ä¡£½ñ¼Ù¶¨Ñ¡×é³ÉΪa Ö®ÈÜÒº£¬´ÓaÀäÈ´ÏÂÀ´ÓëFD ÏßÏཻ£¬µ±Ô½¹ýFDÏߺó±ãÓйÌÏàCaF2? CaCl2Îö³ö£¬ÈÜÒº×é³ÉÑØFD Ï߸ı䣬´ýζȽµµ½GDH£¨¼´ÈýÏàµãζȣ©ÏßÒÔÉÏÒ»µãµãʱ£¬½«¹ÌÌå´ÓÈÜÒºÖзÖÀ룬¼´¿ÉµÃµ½´¿´âµÄCaF2? CaCl2 ½á¾§¡£ £¨9·Ö£©

¾Å¡¢¼ÆËãÌâ¡£Çë¼ÆËãÏÂÁи÷Ìâ¡£ ( ±¾ ´ó Ìâ10·Ö )

½â£º(1) K1=p(CO2)/p(CO)

K2=p(CO2)/p(CO)

È«²¿ÎïÖÊ´æÔÚÓÚ·´Ó¦Æ½ºâϵͳÖУ¬ÓɱØÈ»ÓÐK1= K2£¬´Ëʱ֮ζȾÍΪËùÒªÇóµÄζȡ£ ln K1=

?O?rHm,1RTO?rHm,2+C1

?RT+C2 ln K2=

½«Êý¾Ý´úÈëÉÏʽ£¬·Ö±ðÇóµÃ£º

C1=£­4.29 ? rH,1 =£­31 573 J-mol-1

C2=+4.91 ? rH,2=35 136 J-mol-1 £¨4·Ö£©

?O?rHm,1RT£«C1=

?O?rHm,2RT£«C2

OO?rHm,2??rHm,1Ôò T=

=872 K £¨6·Ö£©

p(CO)= K= K 2 (2) ÔÚ872 Kʱ p(CO)12

¹ÊËã³öK1»òK2(ÔÚ872 K)£¬µÃ

p(CO)=1.071 £¨10·Ö£© 2 p(CO)R(C2?C1)ÏàƽºâÁ·Ï°Ìâ

1. ÔÚ¶¨Ñ¹Ï£¬NaCl¾§ÌåºÍÕáÌǾ§ÌåÓëËüÃǵı¥ºÍ»ìºÏË®ÈÜҺƽºâ¹²´æʱ£¬¶ÀÁ¢×é·ÖÊýCºÍÌõ¼þ×ÔÓɶÈF?£º´ða£»

(a) C=3£¬ F?=1 (b) C=3£¬ F?=2 (c) C=4£¬ F?=2 (d) C=4£¬ F?=3 ×¢Ò⣺Èç¹ûÉÏÊöÌâÄ¿¸ÄΪ£ºÔÚ¶¨Ñ¹Ï£¬NaCl¾§ÌåºÍÕáÌǾ§ÌåÓëËüÃǵĹý±¥ºÍ»ìºÏË®ÈÜҺƽºâ¹²´æʱ£¬ÏàÂÉ»¹ÊÇ·ñÊÊÓã¿

50