Ò»¡¢Ñ¡ÔñÌâ
µÚËÄÕ »¯Ñ§Æ½ºâìØºÍGibbsº¯Êý
£¨ £©01.ijζÈÏ·´Ó¦SO2(g)+1/2O2(g) SO3(g)µÄK¦È=50£»Í¬Î¶ÈÏ·´Ó¦2SO3 2SO2+O2µÄK¦ÈÊÇ
4 2
a 2500 b 100 c 4¡Á10-d 2¡Á10-¦È1
£¨ £©02.¿ÉÄæ·´Ó¦2NO(g) N2(g)+O2(g),¡÷rH m =-180kJ?mol -,¶ÔÆäÄæ·´Ó¦À´Ëµ£¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ
¦È¦È
a ÉýÎÂKÔö´ó b ÉýÎÂK¼õС c ¼ÓѹƽºâÔòÒÆ¶¯ d Ôö¼ÓN2Ũ¶È£¬NO½âÀë¶ÈÔö¼Ó
£¨ £©03.ÈÝÆ÷ÖмÓÈëÏàͬÎïÖʵÄÁ¿µÄNOºÍCl2£¬ÔÚÒ»¶¨Î¶ÈÏ·¢Éú·´Ó¦NO(g)+1/2Cl2(g) NOCl(g),
´ïµ½Æ½ºâʱ£¬¶Ô¸÷ÓйØÎïÖʵķÖѹÅжÏÕýÈ·µÄÊÇ a p(NO)=p(Cl2) b p(NO)=p(NOCl) c p(NO)£¼p(Cl2) d p(NO)£¾p(Cl2)
£¨ £©04.ijζÈÏÂÆ½ºâA+B G+FµÄ¡÷rHm£¼0£¬Éý¸ßÎÂ¶ÈÆ½ºâÄæÏòÒÆ¶¯µÄÔÒòÊÇ
a r£¨Õý£©¼õС£¬r(Äæ)Ôö´ó b k£¨Õý£©¼õС£¬k£¨Ä棩Ôö´ó c r£¨Õý£©ºÍr£¨Ä棩¶¼Ôö´ó d r£¨Õý£©Ôö¼ÓµÄ±¶ÊýСÓÚr£¨Ä棩Ôö¼ÓµÄ±¶Êý £¨ £©05.¿ÉÄæ·´Ó¦A+2B 2C£¬ÒÑ֪ijζÈÏ£¬Õý·´Ó¦ËÙÂʳ£Êýk£¨Õý£©=1£¬Äæ·´Ó¦ËÙÂʳ£Êýk£¨Ä棩
=0.5£¬Ôò´¦ÓÚÆ½ºâ״̬µÄÌåϵÊÇ
33 33
a [A]=1mol?dm-£¬[B]=[C]=2mol?dm-b [A]=2mol?dm-£¬[B]=[C]=1mol?dm-
-3-3 -3-3
c [A]=[C]=2mol?dm£¬[B]=1mol?dmd [A]=[C]=1mol?dm£¬[B]=2mol?dm
£¨ £©06.·´Ó¦2SO2(g)+O2(g) 2SO3(g),ƽºâºó±£³ÖÌå»ý²»±ä£¬¼ÓÈë¶èÐÔÆøÌåHeʹ×ÜѹÔö¼Ó1 ±¶£¬Ôò
a ƽºâÏòÓÒÒÆ¶¯ b ƽºâÏò×óÒÆ¶¯ c ƽºâ²»·¢ÉúÒÆ¶¯ d ÎÞ·¨ÅжÏ
£¨ £©07.ºÏ³É°±·´Ó¦3H2(g)+N2(g) 2NH3(g)£¬ÔÚºãѹϽøÐÐʱ,ÈôÏòÌåϵÖÐÒýÈëë²Æø,Ôò°±µÄ²úÂÊ
a ¼õС b Ôö´ó c ²»±ä d ÎÞ·¨ÅжÏ
£¨ £©08.298K·´Ó¦BaCl2?H2O(s) BaCl2(s)+H2O(g)´ïƽºâʱ£¬p(H2O)=330Pa,Ôò·´Ó¦µÄ¡÷rG¦ÈΪm
-1 -1 -1 -1
a -14.2kJ?molb 14.2kJ?molc 142kJ?mold -142kJ?mol
¦È
£¨ £©09.ÏÂÁз´Ó¦ÖУ¬KֵСÓÚKpÖµµÄÊÇ
a H2(g)+Cl2(g) 2HCl(g) b H2(g)+S(g) H2S(g) c CaCO3(s) CaO(s)+CO2(g) d C(s)+O2(g) CO2(g)
¦È1
£¨ £©10.ÒÑÖª·´Ó¦NO£¨g£©+CO(g) 1/2N2(g)+CO2(g)µÄ¡÷rH 373.2kJ?mol -£¬ÈôÒªÌá¸ßÓж¾ÆøÌåNO m =-
ºÍCOµÄת»¯ÂÊ£¬¿É²ÉÈ¡µÄ´ëÊ©ÊÇ
a µÍεÍѹ b µÍθßѹ c ¸ßθßѹ d ¸ßεÍѹ
£¨ £©11.»¯ºÏÎïAµÄË®ºÏ¾§ÌåA?3H2OÍÑË®·´Ó¦¹ý³ÌÈçÏ£º¢ÙA?3H2O(s) A?2H2O(s)+H2O(g) K1 ¦È¢ÚA?2H2O(s) A?H2O(s)+H2O(g) K¦È £»¢ÛA?H2O(s) A£¨s£©+H2O(g) K¦È£¬ÎªÊ¹A?2HO(s) 2 3 2
¦È
¾§Ìå±£³ÖÎȶ¨£¨²»·¢Éú³±½â»ò·ç»¯£©£¬ÔòÈÝÆ÷ÖÐË®ÕôÆøÑ¹p(H2O)ÓëÆ½ºâ³£ÊýKµÄ¹ØÏµÓ¦Âú×ã
¦È ¦È ¦È ¦È
a K2£¾p(H2O)/p£¾K3 ¦Èb p(H2O)/p£¾K2 ¦È c p(H2O)/p¦È£¾K1 ¦È d K1¦È£¾p(H2O)/p£¾K2 ¦È
£¨ £©12.ij·ÅÈÈ·´Ó¦£¬Î¶ÈÉý¸ß10¡æ£¬Ôò
a ¶Ô·´Ó¦ÎÞÓ°Ïì b ƽºâ³£ÊýÔö´ó1 ±¶ c ƽºâ³£Êý¼õС d ²»¸Ä±ä·´Ó¦ËÙÂÊ £¨ £©13.Ò»ÃܱÕÈÝÆ÷ÖУ¬ÓÐA¡¢B¡¢CÈýÖÖÆøÌ彨Á¢ÁË»¯Ñ§Æ½ºâ£¬ËüÃǵķ´Ó¦ÊÇA(g)+B(g) C(g),Ïàͬ
ζÈÏ£¬Ìå»ýËõС2/3£¬Ôòƽºâ³£ÊýKΪÔÀ´µÄ a 3 ±¶ b 2 ±¶ c 9 ±¶ d ²»±ä
¦È 1£¨ £©14.·´Ó¦NO2£¨g£©+NO(g) N2O3(g)µÄ¡÷rHm=-40.5kJ?mol -£¬·´Ó¦´ïµ½Æ½ºâʱ£¬ÏÂÁÐÒòËØÖпÉʹ
ƽºâÄæÏòÒÆ¶¯µÄÊÇ
a T¡¢VÒ»¶¨,ѹÈëÄÊÆø b TÒ»¶¨,V±äС c V¡¢PÒ»¶¨,TÉý¸ß d P¡¢TÒ»¶¨,ѹÈëNO
£¨ £©15.ΪÁËÌá¸ß·´Ó¦CO(g)+H2O(g) CO2(g)+H2(g)ÖÐCOµÄת»¯ÂÊ£¬¿ÉÒÔ²ÉÈ¡
a Ôö¼ÓCOµÄŨ¶È b Ôö¼ÓË®ÕôÆøµÄŨ¶È c °´±ÈÀýÔö¼ÓË®ÕôÆøºÍCOµÄŨ¶È d ÈýÖÖ°ì·¨¶¼ÐÐ
£¨ £©16.ÔÚºãÎÂÏ£¬½«2NO2£¨g£© N2O4(g)ÆøÌ寽ºâϵͳµÄ×ÜѹÁ¦Ôö¼ÓÒ»±¶£¬ÔòÆäÌå»ý
a µÈÓÚÔÌå»ýµÄ1/2 b СÓÚÔÌå»ýµÄ1/2 c µÈÓÚÔÌå»ýµÄ2/3 d ´óÓÚÔÌå»ýµÄ1/2 £¨ £©17.½«¹ÌÌåNH4NO3£¨s£©ÈÜÓÚË®ÖУ¬ÈÜÒº±äÀ䣬Ôò¸Ã¹ý³ÌµÄ¡÷G£¬¡÷H£¬¡÷SµÄ·ûºÅÒÀ´ÎΪ
a +£¬-£¬- b +£¬+£¬- c -£¬+£¬- d -£¬+£¬+
-1
£¨ £©18.±ùµÄÈÛ»¯ÈÈΪ330.5J?g£¬0¡æÊ±½«1.0gË®Äý½áΪ±ùʱµÄ¡÷SΪ
1 1 1
a -330.5J?K-b -1.21J?K-c 0 d 1.21J?K- ¦È
£¨ £©19.±ê׼״̬Ï£¬ÏÂÁÐÀë×ÓÖÐSm Ϊ0 µÄÊÇ
a Na+ b Cu2+ c H+ d Cl-
£¨ £©20.ÏÂÁÐÎïÖÊÖУ¬Ä¦¶ûìØ×î´óµÄÊÇ
a MgF2 b MgO c MgSO4 d MgCO3
5
146 J?mol ?K £¬·´Ó¦´ïƽºâʱ¸÷ÎïÖÊ £¨ £©21.·´Ó¦2NO(g)+O2(g)=2NO2(g)£¬¡÷rHm= -114 KJ?mol £¬¡÷rSm =-¦È
·Öѹ¾ùΪP£¬Ôò·´Ó¦µÄζÈÊÇ a 780¡æ b 508¡æ c 482¡æ d 1053¡æ
¦È
£¨ £©22.ÏÂÁз´Ó¦ÖУ¬¡÷rS×î´óµÄÊÇ m b 2SO2(g)+O2(g)¡ú2SO3(g) a C(s)+O2(g)¡úCO2£¨g£© c 3H2(g)+N2(g)¡ú2NH3(g) d CuSO4(s)+5H2O(l)¡úCuSO4?5H2O(s) £¨ £©23.ÒºÌå·ÐÌÚ¹ý³ÌÖУ¬ÏÂÁм¸ÖÖÎïÀíÁ¿ÖÐÊýÖµÔö¼ÓµÄÊÇ
a ÕôÆøÑ¹ b Ħ¶û×ÔÓÉÄÜ c Ħ¶ûìØ d ÒºÌåÖÊÁ¿ £¨ £©24.ÏÂÁйý³ÌÖУ¬¡÷G=0 µÄÊÇ
a °±ÔÚË®ÖнâÀë´ïƽºâ b ÀíÏëÆøÌåÏòÕæ¿ÕÅòÕÍ c ÒÒ´¼ÈÜÓÚË® d Õ¨Ò©±¬Õ¨ £¨ £©25.ÓÉË®¡¢Ë®Æø¡¢ÃºÓͼ°±ù×é³ÉµÄϵͳÊÇ
a ËÄÏàϵͳ b Á½Ïàϵͳ c ÈýÏàϵͳ d µ¥Ïàϵͳ £¨ £©26.Ìõ¼þÏàͬµÄͬһ·´Ó¦ÓÐÁ½ÖÖ²»Í¬Ð´·¨
£¨1£©N2£¨g£©+3H2£¨g£©¡ú2NH3£¨g£©£¬¡÷G1£»£¨2£©1/2N2£¨g£©+3/2H2£¨g£©¡úNH3£¨g£©£¬¡÷G2¡£ÔòÓÐ a ¡÷G1=¡÷G2 b ¡÷G1=£¨¡÷G2£©2 c ¡÷G1=1/2¡÷G2 d ¡÷G1=2¡÷G2
£¨ £©27.×Ô·¢½øÐеĺãκãѹ»¯Ñ§·´Ó¦£¬Æä±ØÒªÌõ¼þÊÇ
a ¡÷rS£¼0 b ¡÷rH£¼0 c ¡÷rH£¼T¡÷rS d ¡÷fG£¾0 £¨ £©28.ÒÑÖª298KʱÏÂÁз´Ó¦£º4NH3£¨g£©+5O2£¨g£©¡ú4NO£¨g£©+6H2O£¨l£©
1
¡÷fG ¦Èm /KJ?mol- -16.5 +86 -237 £»¿ÉÒÔÅжϴ˷´Ó¦ÔÚ±ê׼״̬Ï a ²»×Ô·¢½øÐÐ b ×Ô·¢½øÐÐ c ´¦ÓÚÆ½ºâ״̬ d ÄÑÒÔÅжÏ
¦È -1 ¦È
-1 -1
¶þ¡¢Ìî¿ÕÌâ
01.ºÏ³É°±·´Ó¦£ºN2(g)+3H2(g) 873Kʱ£¬K¦È= 3
¡£
¦È 4
2NH3(g),ÔÚ673Kʱ£¬K2=5.7¡Á10 -£¬ÔÚ473Kʱ£¬K1¦È=0.61£¬Ôò
1
02.ÔÚ20¡æÊ±£¬¼×´¼µÄÕôÆøÑ¹Îª11.83kPa£¬Ôò¼×´¼Æø»¯¹ý³ÌµÄ¡÷rG ¦Èm Ϊ kJ?mol -¡£¼×´¼ÔÚÕý
1³£·ÐµãʱµÄ¡÷rG ¦Èm Ϊ kJ?mol -¡£
-1 ¦È-1 -1¦È03.ÒÑÖª»·ÎìÍéµÄÆø»¯¹ý³Ì¡÷rHm=28.7kJ?mol £¬¡÷rSm=88J?mol ?K ¡£»·ÎìÍéµÄÕý³£·ÐµãΪ ¡æ£¬
ÔÚ25¡æÊ±±¥ºÍÕôÆøÑ¹Îª
kPa.
¦È
SO2(g)+1/2O2(g),¡÷rHm=98.7kJ?mol
-104.ÒÑÖª·´Ó¦SO3(g) .Ìîдϱí
K¦È ƽºâÒÆ¶¯·½Ïò
Ôö¼Ó×Üѹ Éý¸ßÎÂ¶È ¼Ó´ß»¯¼Á
k£¨Õý£© k£¨Ä棩 r£¨Õý£© r£¨Ä棩 £»
£»
1
05.ÔÚ100¡æ¡¢ºãѹÌõ¼þÏÂ,Ë®µÄÆø»¯ÈÈΪ2.26kJ?g-,1molË®ÔÚ100¡æÊ±Æø»¯£¬Ôò¸Ã¹ý³ÌQ=
¡÷H= £»¡÷S= £»¡÷G= ¡£
Ò»
06.2molHg(l)ÔڷеãζÈ(630K)ʱÕô·¢ÎüÊÕµÄÈÈÁ¿Îª109.12kJ¡£Ôò¹¯µÄ±ê׼Ħ¶ûÕô·¢ìÊ¡÷vapHm¡ð=
¸Ã¹ý³Ì¶Ô»·¾³×ö¹¦W= £»¡÷U= £»¡÷S= £»¡÷G= ¡£
£»
07.³£Î³£Ñ¹ÏÂ,ZnºÍCuSO4ÈÜÒºÔÚ¿ÉÄæµç³ØÖз´Ó¦£¬·ÅÈÈ6.0kJ,×öµç¹¦200kJ,Ôò´Ë¹ý³ÌµÄ¡÷rS= ¡÷rG=
¡£
08.ÔÚ25¡æÊ±£¬ÈôÁ½¸ö·´Ó¦µÄƽºâ³£ÊýÖ®±ÈΪ10£¬ÔòÁ½¸ö·´Ó¦µÄ¡÷rG ¦Èm Ïà²î ¦È
09.ÒÑÖª·´Ó¦CaCO3(s)=CaO(s)+CO2(g),ÔÚ298Kʱ£¬¡÷rGm=130kJ?mol
-1
-1
kJ?mol ¡£
¦È -1
,1200Kʱ¡÷rGm=-15.3kJ?mol £¬
Ôò¸Ã·´Ó¦µÄ¡÷rHm=
¦È
£»¡÷rSm=
¦È
¡£
¡£
10.½«ÏÂÁÐÎïÖʰ´Ä¦¶ûìØÖµÓÉСµ½´óÅÅÁУ¬Æä˳ÐòΪ LiCl(s),Li(s),Cl2(g),I2(g),Ne(g).
6
µÚÎåÕ Ëá¼îƽºâ
Ò»¡¢Ñ¡ÔñÌâ
3
( )01.½«0.1mol?dm-µÄÏÂÁÐÈÜÒº¼ÓˮϡÊÍÒ»±¶ºó£¬pH±ä»¯×îСµÄÊÇ
a HCl b H2SO4 c HNO3 d HAc ( )02.ÔÚHAc-NaAc×é³ÉµÄ»º³åÈÜÒºÖУ¬Èô[HAc]£¾[Ac-]£¬Ôò¸Ã»º³åÈÜÒºµÖ¿¹Ëá»ò¼îµÄÄÜÁ¦Îª
a ¿¹ËáÄÜÁ¦£¾¿¹¼îÄÜÁ¦ b ¿¹ËáÄÜÁ¦£¼¿¹¼îÄÜÁ¦ c ¿¹Ëá¼îÄÜÁ¦Ïàͬ d ÎÞ·¨ÅÐ¶Ï ( )03.ÏÂÁÐÀë×ÓÖмîÐÔ×îÇ¿µÄÊÇ
a CN-b Ac-c NO- d NH+ 2 4
-( )04.H2AsOµÄ¹²éî¼îÊÇ 4
a H3AsO4 b HAsO2-c AsO3-d H2AsO3- 4 4
( )05.ÔÚ»ìºÏÈÜÒºÖÐ,ijÈõËáHXÓëÆäÑÎNaXµÄŨ¶ÈÏàͬ,ÒÑÖª·´Ó¦X-+H2O HX+OH-µÄƽºâ³£ÊýΪ
10
1.0¡Á10-,Ôò´ËÈÜÒºµÄpHֵΪ a 2 b 4 c 5 d 10
( )06.½«µÈÌå»ýµÄHClÈÜÒº£¨pH=3£©ºÍNaOHÈÜÒº£¨pH=10£©»ìºÏºó£¬ÈÜÒºpH½éÓÚÄÄ×éÊý¾ÝÖ®¼ä£¿
a 1¡«2 b 3¡«4 c 6¡«7 d 11¡«12 ( )07.ÈõËáÐÔË®ÈÜÒºÖÐÇâÀë×ÓŨ¶È¿É±íʾΪ
¦È POH-1414-pOH
d 10 a 14-pOH b Kw/pOH c 10
( )08.H3PO4ÈÜÒºÖмÓÈëÒ»¶¨Á¿NaOHºó£¬ÈÜÒºµÄpH=10.00,Ôò¸ÃÈÜÒºÖÐÏÂÁÐÎïÖÖŨ¶È×î´óµÄÊÇ
-2-3- a H3PO4 b H2POc HPOd PO4 4 4
-8 ¦È-14-3 2-¦È( )09.ij¶þÔªÈõËáH2AµÄKa(1)=6¡Á10 £¬Ka(2)=8¡Á10 £¬ÈôÆäŨ¶ÈΪ0.05mol?dm ,Ôò[A ]Ϊ
-8-3 143 83 143
a 6¡Á10mol?dmb 8¡Á10-mol?dm-c 3¡Á10-mol?dm-d 4¡Á10-mol?dm--3
( )10.½«0.1mol?dmNaAcÈÜÒº¼ÓˮϡÊÍʱ£¬ÏÂÁи÷ÏîÊýÖµÖÐÔö´óµÄÊÇ
a [Ac-]/[OH-] b [OH-]/[Ac-] c [Ac-] d [OH-] -3
( )11.ÔÚ0.1mol?dm°±Ë®ÖмÓÈëµÈÌå»ý¡¢µÈŨ¶ÈµÄÏÂÁÐÈÜÒººó£¬Ê¹»ìºÏÒºpH×î´óÔòÓ¦¼ÓÈë
a HCl b H2SO4 c HNO3 d HAc
33
( )12.0.4mol?dm-HAcÈÜÒºÖÐH+Ũ¶ÈÊÇ0.1mol?dm-HAcÈÜÒºÖÐH+Ũ¶ÈµÄ
a 1 ±¶ b 2 ±¶ c 3 ±¶ d 4 ±¶ ( )13.ÔÚÏÂÁÐÈÜÒºÖÐHCN½âÀë¶È×î´óµÄÊÇ
333
a 0.1mol?dm-KCN b 0.1mol?dm-KCN ºÍ0.1mol?dm-KCl»ìºÏÈÜÒº -3-3-3
c 0.2mol?dmNaCl d 0.1mol?dmKClºÍ0.2mol?dmNaCl»ìºÏÈÜÒº
¦È 7 11
( )14.ÒÑÖªH2CO3µÄKa(1)=4.2¡Á10 -,Ka¦È(2)=5.6¡Á10 -¡£Ôò0.1mol?dm- 3NaHCO3ÈÜÒºµÄpHΪ
a 5.6 b 7.0 c 8.3 d 13.0 ( )15.ÏÂÁÐÎïÖÊÖÐÊôÓÚÈõµç½âÖʵÄÊÇ
a BaCO3 b (NH4)2S c HgCl2 d Ca(NO3)2
( )16.²»Êǹ²éîËá¼î¶ÔµÄÒ»×éÎïÖÊÊÇ
- + - 2- _
b NaOH,Na c HS,S a NH3,NH2 d H2O,OH
( )17.ÏÂÁÐÎïÖÊÖУ¬¼ÈÊÇÖÊ×ÓËᣬÓÖÊÇÖÊ×Ó¼îµÄÊÇ
2-+3- c Sa H2O b NH4 d PO 4
-5¦È¦È
( )18.ÒÑÖªKb(NH3)=1.8¡Á10 £¬ÔòÆä¹²éîËáµÄKa ֵΪ
-9 10 10 5
a 1.8¡Á10b 1.8¡Á10-c 5.6¡Á10-d 5.6¡Á10-( )19.ͬŨ¶ÈÑÎNaA£¬NaB£¬NaC£¬NaDµÄË®ÈÜÒºpHÒÀ´ÎÔö´ó£¬ÔòͬŨ¶ÈÏÂÁÐÏ¡ËáÖÐÀë½â¶È×î´óµÄÊÇ
a HD b HC c HB d HA
( )20.ÅðµÄ±»¯Îï×÷Ϊ·Ò×˹ËᣬÆäËáÐÔÇ¿Èõ¹ØÏµÕýÈ·µÄÒ»×éÊÇ
a BF3>BCl3>BBr3 b BF3 ( )21¸ù¾ÝËá¼îÖÊ×ÓÀíÂÛÔÚË®ÈÜÒºÖÐ×îÈõµÄ¼îÊÇ c PO3-d NO-a ClO4-b SO2-4 4 3 ( )22NH+µÄ¹²éî¼îΪ 4 + a NH3 b NH2 ¡ªc NH4 d NH2OH ( )23.°´ÕÕËá¼îÖÊ×ÓÀíÂÛ£¬ÏÂÁÐÎïÖÊÖпÉ×÷ΪËáµÄÊÇ a S2-b NO-c [Fe(H2O)6]2+ d PO3-3 4 2+ ( )24.°´ÕÕËá¼îÖÊ×ÓÀíÂÛ£¬[Cr(OH2)5OH]µÄ¹²éîËáÊÇ a [Cr(OH2)6]3+ b [Cr(OH2)4(OH)2]+ c [Cr(OH2)3(OH)3] d [Cr(OH)4]- 7 ¦È 5 ( )25.NH3?H2O µÄKb=1.8¡Á10 -£¬Èô½«0.09molµÄNH4ClÈÜÓÚ0.5dm3Ë®ÖУ¬´Ëʱ 5 a ÈÜÒºµÄpHֵΪ9 c NH4ClµÄË®½â³£ÊýΪ5.6¡Á10- -8 b ÈÜÒºµÄpHֵΪ5 d NH4ClµÄË®½â³£ÊýΪ5.6¡Á10 --( )26.¸ù¾Ý·´Ó¦£ºNH+ HO¡úÅжϣ¬ÕâËÄÖÖÎïÖÊÖУ¬¼îÐÔ½ÏÇ¿µÄÊÇ 2 2 NH+OH3 - -d H2O a OHb NH3 c NH2 ( )27.ÔÚÏÂÁÐÈܼÁÖУ¬´×ËáµÄµçÀë³£Êý×î´óµÄÊÇ a Òº°± b Һ̬·ú»¯Çâ c ´¿ÁòËá d Ë® ( )28.°´ÕÕÖÊ×ÓÀíÂÛ£¬ÏÂÁи÷ÎïÖÖÖÐÄÄÒ»¸öûÓй²éî¼î d SO2-a H3SO+ b H2SO4 c HSO4- 4 4 ( )29.ÏÂÁÐŨ¶ÈÏàͬµÄÑÎÈÜÒºÖУ¬pHÖµ×î¸ßµÄÊÇ a NaCl b K2CO3 c Na2SO4 d Na2S ( )30ÏÂÁÐŨ¶ÈÏàͬµÄËáʽÑÎÈÜÒºÖÐ,³ÊËáÐÔµÄÊÇ a NaHCO3 b NaH2PO4 c Na2HPO4 d NaHS ( )31.pOHΪ3.00µÄÇ¿¼îÈÜÒºÓëpOHΪ2.00 µÄÇ¿¼îÈÜÒºµÈÌå»ý»ìºÏ£¬»ìºÏÒºµÄpHֵΪ a 11.26 b 11.50 c 11.74 d 11.92 ¦È -5 -3 ( )32.ÒÑÖªHAcµÄKa=1.8¡Á10 ,Ôò0.18mol?dm NaAcË®ÈÜÒºµÄpHֵΪ a 9 b 8 c 5 d10 ( )33.°´ÕÕËá¼îÖÊ×ÓÀíÂÛ£¬ÏÂÁÐÎïÖÊÖмȿÉ×÷ΪËáÓÖ¿É×÷Ϊ¼îµÄÊÇ a [AI(H2O)6]3+ b [Cr(H2O)6]3+ c HCO3 -d PO4 3-( )34.ÏÂÁдëÊ©ÖУ¬¿ÉÒÔʹÈõËáÇ¿¼îÑκÍÈõ¼îÇ¿ËáÑεÄË®½â¶È¶¼Ôö´óµÄÊÇ a Éý¸ßÎÂ¶È b ½µµÍÎÂ¶È c Ôö¼ÓÑεÄŨ¶È d Éý¸ßÈÜÒºµÄpHÖµ ( )35.ÓöË®·´Ó¦·Å³öÆøÌå²¢Éú³É³ÁµíµÄÊÇ a SnCl2 b Bi(NO3)3 c Mg3N2 d (NH4)2SO4 ( )36.ÔÚHAcÈÜÒºÖмÓÈëÉÙÁ¿NaAc¹ÌÌ壬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ a HAcµçÀë¶ÈÔö´ó b HAcµÄK¦Èa ¼õС c ÈÜÒºµÄpHÖµ¼õС d ÈÜÒºµÄpHÖµÔö´ó ( )37.¹ØÓÚËá¼îָʾ¼Á´íÎóµÄ˵·¨ÊÇ a ¿ÉÒÔָʾËá¼îµÎ¶¨ÖÕµã b ¿ÉÒÔָʾÈÜÒºµÄpH c Ò»°ãÊÇÓлúËá»òÓлú¼î d ¾ùΪÓÐÑÕÉ«µÄÈõËá»òÈõ¼î ( )38.Ҫʹ´×ËáµÄµçÀëÆ½ºâÏòÓÒÒÆ¶¯ÇÒpHÖµ¼õСµÄ·½·¨ÓÐ a ¼ÓÑÎËá b ¼ÓË® c Êʵ±ÉýΠd ÀäÈ´ ( )39.ÔÚÏ¡ÊÍijµç½âÖÊÈÜҺʱ£¬¸ÃÈÜÒºµ¼µçÐԵı仯ÊÇ a ÏÈÖð½¥ÔöÇ¿£¬ºóÖð½¥¼õÈõ b Öð½¥ÔöÇ¿ c Öð½¥¼õÈõ d ÎÞÃ÷ÏԱ仯 +-( )40.ÔÚH2CO3 H£«HCO3 µçÀëÆ½ºâÌåϵÖУ¬ÏÂÁÐÌõ¼þÄÜʹÆäµçÀëÆ½ºâÏòÄæ·½ÏòÒÆ¶¯µÄÊÇ a ¼ÓNaOH b ÓÃˮϡÊÍ c ¼ÓÑÎËá dÉý¸ßÎÂ¶È + ( )41.Ò»ÔªÈõËáµÄHŨ¶ÈÓëËáµÄŨ¶ÈCÖ®¼äµÄ¹ØÏµÊÇ a ÏàµÈ b CСÓÚH+ c C´óÓÚH+Ũ¶È d ÎÞ·¨ÅÐ¶Ï ( )42.пÓë´×Ëá·´Ó¦²úÉúÇâÆø£¬ÏòÈÜÒºÖмÓÈëÒ»¶¨Á¿µÄ´×ËáÄÆºó£¬²úÉúÇâÆøµÄËÙ¶È»á a ±ä¿ì b ±äÂý c ²»±ä d Ïȱä¿ìºó±äÂý ( )43.¸ù¾ÝËá¼îµç×ÓÀíÂÛ£¬ÏÂÁÐÎïÖÊÖв»¿É×÷ΪLewis ¼îµÄÊÇ a H2O b NH3 c Ni2+ d CN- ( )44.Ìá³öÅäλÀíÂ۵ĿÆÑ§¼ÒÊÇ a ±«ÁÖ£¨Panling£© b ά¶ûÄÉ(Werner) c·Ò×˹(Lewis) d²£¶û(Bohr) ( )45.ÔÚ[Co(NH3)5Cl](NO3)2ÀÖÐÐÄÀë×ÓµÄÑõ»¯Ì¬Îª a+1 b +2 c +3 d +4 ( )46.ÔÚÅäÀë×Ó[Co(en)(C2O4)2]-ÖУ¬ÖÐÐÄÀë×ÓµÄÅäλÊýΪ a 3 b 4 c 5 d 6 ( )47.ÔÚÅäºÏÎï[Pt(NH3)6][PtCl4]ÖУ¬Á½¸öÖÐÐÄÀë×Ó²¬µÄÑõ»¯Ì¬ÊÇ a ¶¼ÊÇ+8 b ¶¼ÊÇ+6 c ¶¼ÊÇ+4 d ¶¼ÊÇ+2 ( )48.ÏÂÁз´Ó¦£ºCuSO4+4NH3¡ú[Cu(NH3)4]SO4 ÖУ¬×÷Ϊ·Ò×˹ËáµÄÊÇ a Cu2+ b NH3 c SO2-d Cu(NH3)2+2 4 ( )49.ÏÂÁÐÎïÖÊÖв»ÄÜ×÷ΪÅäλÌåµÄÊÇ a ÁòÇèËá¼Ø b Áò´úÁòËáÄÆ c ¼×Íé d Ò»Ñõ»¯Ì¼ £¨ £©50.¸ù¾ÝÈíÓ²Ëá¼îÀíÂÛ£¬ÏÂÁÐÎïÖÊÊôÓÚÈíËáµÄÊÇ a H+ bAg+ c NH3 d AsH3 8