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7. 2 ·Ö (3881)

[´ð] (D) £¨2·Ö£© 8. 2 ·Ö (6601)

[´ð] (A)

±íÃæÕÅÁ¦? ËæÎ¶ÈÖ®Ôö¸ß¶øÏ½µ£¬ÓÉ ?p = 2? /R Öª£¬ÓÒ¶ËÀäȴʱ Æä¸½¼ÓѹÁ¦Ôö¼Ó¶ø×ó¶Ë²»±ä£¬¹ÊÏò×óÒÆ¶¯¡£ 9. 2 ·Ö (7119)

[´ð] (A) 10 2 ·Ö (7028) 7028

[´ð] (C) £¨2·Ö£© ¶þ¡¢Ìî¿ÕÌâ ( ¹²10Ìâ 20·Ö ) 11. 2 ·Ö (7079)

[´ð] »¯Ñ§Îü¸½µÄÇ¿¶È£»ÖеÈÇ¿¶ÈµÄ»¯Ñ§Îü¸½¡£ 12. 2 ·Ö (4170)

[´ð] Zn(s)|ZnCl2(m2)||AgNO3(m1)|Ag(s) £¨2·Ö£© 13. 2 ·Ö (4077)

[´ð] NaCl £¨2·Ö£© 14. 2 ·Ö (5657)

[´ð] 52.9 kJ¡¤mol-1 15. 2 ·Ö (4685)

[´ð] E= 0.323 V K= 8.46¡Á1010

$

$

16. 2 ·Ö (3906)

[´ð] Ôö¼Ó£¬¼õÉÙ £¨2·Ö£© 17. 2 ·Ö (5662)

[´ð] k(cat)/k0 =exp{[-E(cat)-E]/RT}=1013 (2·Ö) 18. 2 ·Ö (4435)

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Ag©¦AgNO3(m)©¦Ag(Au)[a(Ag)=0.120)] 19. 2 ·Ö (3907)

[´ð] 7.92¡Á10-8

20. 2 ·Ö (5310)

[´ð] 1 Èý¡¢¼ÆËãÌâ ( ¹² 8Ìâ 50·Ö ) 21. 10 ·Ö (4298)

[´ð]

(1) Zn(s) + 2AgCl(s) = 2Ag(s) + ZnCl2 (2·Ö) (2) ?rGm= -zFE = -195.9 kJ¡¤mol-1 ?rSm= zF(?E/?T)p= -94.96 J¡¤K-1¡¤mol-1 ?rHm= ?rGm+ T?rSm= -224.2 kJ¡¤mol-1 ?rUm= ?rHm- ?(pV) = ?rHm= -224.2 kJ¡¤mol-1

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?rFm= ?rGm= -195.9 kJ¡¤mol-1 Qr= T?rSm= -28.30 kJ¡¤mol-1 (6·Ö) (3) Qp= ?rUm+ W = ?rUm= -224.2 kJ¡¤mol-1 ÈÈÁ¦Ñ§±ä»¯ÖµÍ¬ÉÏ (2·Ö) 22 10 ·Ö (4698) 4698 [´ð] (1) Hg(l)|Hg22+(a2)||Hg2+(a1)|Hg(l) (3·Ö) (2) E = 0.854 V ¨C 0.798 V= 0.056 V £¨3·Ö£© (3) K=exp(zEF/RT) = 78.4 (z = 2) £¨4·Ö£© 23. 10 ·Ö (4959) [´ð] (1) ?(Zn2+/Zn) =? (Zn2+/Zn) +(RT/2F)ln? (Zn2+) =-0.793 V $$$ ?(Cd2+/Cd) =? (Cd2+/Cd) +(RT/2F)ln? (Cd2+) =-0.433 V $ ?(H+/H2) =? (H+/H2)+(RT/F)ln? (H+)-? (H2)= -1.014 V (6·Ö) $ ËùÒÔCdÊ×ÏÈÔÚÒõ¼«Îö³ö (2) µ±Zn¿ªÊ¼Îö³öʱ ?(Cd2+/Cd) =? (Cd2+/Cd) +(RT/2F)ln? (Cd2+) =-0.793 V $ ? (Cd2+) =6.0¡Á10-14, m(Cd2+)=6.0¡Á10-14 mol¡¤kg-1 (4·Ö) 24. 5 ·Ö (5487) [´ð] 2A == A2 t=0 p0 t=t p (p0-p) pT=(p0+p) p=2pt - p0 k=25. 5 ·Ö (6715) £¨5·Ö£© [´ð] ln(p/p¡ã)= 2? M/(RT?R') p = 99887 Pa (3·Ö) p < p£¬²»ÄÜÕô·¢³ö (2·Ö) $26. 5 ·Ö (6707) [´ð] 2? /R'= ?gh ? = 0.020 N¡¤m-1 (5·Ö) 27. 5 ·Ö (4644) [´ð] µç³Ø·´Ó¦ AgCl(s) £½ Ag+ + Cl- lnK= zFE/RT K= 1.8 ¡Á 10-10 (2·Ö)

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$

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I= 0.01 mol¡¤kg-1 lg?? = -0.509(1¡Á1)I1/2 ?? = 0.889 (2·Ö) AgCl µÄÈÜҺŨ¶È m/m= (K?)1/2/?? = 1.48¡Á10-5

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m = 1.48¡Á10-5 mol¡¤kg-1 (1·Ö) 28. 5 ·Ö (4018)

[´ð] lg?? = -0.509©¦z+ z-©¦I1/2 ¶ø I = 0.6 mol¡¤kg-1 (3·Ö) ¡ßlg?? = -0.509¡Á2¡Á2¡Á 0.61/2 ¡à?? = 0.06 (2·Ö) 29. 5 ·Ö (5518)

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