1

(2013/12/09 )

(ex: 1.60 g)

+ 1

21.5 oC 0.205 g 0.50 mol 0.02 L 1000 mL 2

21.5 oC 0.205 g 0.50 mol 0.02 L 1000 mL

3

1 1031.0 1031.00 1031.000 103

1234

33211

228.19 g5

+ 1

3.1 mL10.05 3.15 mL3

4

2

3

4

5

1.75 9.1

10.85: 10.93

172.63 1.3

171.33

: 171.34

6

22

731 0.26514 273.15760 0.0 298

.256052

11211

log (2.00 104) = log (104) + log (2.00)

= 4 + 0.301= 4.301

3

[H+] = 2.0 10-12 M

= - log (2.0 10-12)= 12 - log (2.0)= 12 - 0.30= 11.70

2

pH = - log [H+]

E1

7

STP

11

OHatm

11

NSTP

Tn

V)P(P

TnV)(VP

(K) 273.15(mol) 1(L)V(atm) 1 22

E1

8VSTP = 20.9742664

= 20.974

1.0725

11052

n = 97.0347

= 0.0 741105 = 3 0.0 (mol)

: E3

9

42.49 g43.53 g43.33 g

43.53 - 42.49 = 1.04 (g) 43.33 - 42.49 = 0.84 (g) 1.04 - 0.84 = 0.20 (g)

E3

10

42.49 g443.53 g443.33 g4 43.53 - 42.49 = 1.04 g23 43.33 - 42.49 = 0.84 g22 1.04 - 0.84 = 0.20 g22

CuO = 0.0130.013 = 1.01.0 CuO

Cu0. 0.01321793863.

84

13 255

0.0

O0. 0.0120

13 2

2516.00

0.0

11

=+ C T50.0 1.0 (45.0 36.9) = 50.0 1.0 (36.9 29.8) + C (36.9 29.8) 50.0 1.0 8.1 = 50.0 1.0 7.1 + C 7.1405 = 355 + C 7.17.1 C = 50C = 50 / 7.1 = 7.04225 = 7.0cal/oC2

E6 ()

50.0 mL 50.0 mL

29.8 oC 45.0 oC 36.9 oC 7.0 cal/oC

() 1oC

T =

12

E6 ()() HCl-NaOH H < 01.0 M HCl 50.0 mL 1.0 M NaOH 50.0 mL

1.0 M HCl 28.9 oC 1.0 M NaOH 28.5 oC

34.8 oC , H 6.5102 cal

-13 kcal/mol

13

14

15

No

16

17

1. 2.

18

HCl-NaOH

HCl(aq) + NaOH(aq) H2O(l) + NaCl(aq)

-13 kcal/mol-13.60 kcal/mol

| -13 (-13.60) | 13.60 100 % = 4 %

o o o o of f 2 f fH = [1xH (NaCl(aq) + 1xH (H O(l)] - [(1xH (HCl(aq) + 1xH (NaOH(aq) ]

= [(-407.27) + (-285.83)] - [(-167.15) + (-469.15)]= (-693.10) - (-636.30)= - 56.80 (kJ/mol)= - 13.60 (kcal/mol) Langes Handbook of Chemistry

o o of fH = mH ( ) - nH ( )

19

(3)

20

E1, E3, E10, E12, E19, E20-1

()

(4)

21

E6, E13, E16, E17

()

22

50(1. 2. 3. )(1 +510)

() 110 110

(XY)