Chapter 7 Review TEST
B
KEY
A. Matching
1. f.
2. i
3. b
4. e
5. h
6. c
7. j
8. a
9. g
10. d
B. Multiple Choice
11. d
12. b
13. a
14. c
15. b
16. b
17. a
18. d
19. a
20. c
21. c
22. b
23. d
24. a
25. c
26. b
27. d
28. c
C. Problems
29.
0.25
mole x 6.02 x 1023
molecules = 1.50 x 1023
molecules
1 mole
30.
6.25 mole x 98.1 g
= 613 g
1 mole
31.
15.0 kg x
1000 g
x 1 mole
x 22.4 L
= 7636 L
1
kg
44.0
g
1 mole
= 7640 L (3 SF)
32.
0.650 g x
22.4 L
= 14.6 g/mole
1
L
1 mole
33.
(3.75 x 1015
atoms)
x 1
mole
x 197.0 g =
6.02 x 1023
atoms 1 mole
1.23 x 10-6 g
34.
24.3 g + 28.0 g + 96.0 g =
148.3 g/mole for Mg(NO3)2
Mg: 24.3 g / 148.3 g x 100 = 16.4%
N: 28.0 g/ 148.3 g x 100 = 18.9%
O: 96.0 g / 148.3 g x 100 = 64.7%
35.
27.3 g C
x 1 mole C
= 2.28 mole C
12.0 g C
72.7 g O
x 1 mole O
= 4.54 mole O
16.0 g O
Formula:
C2.28 O4.54
2.28/2.28 = 1
and
4.54/2.28 = 2
So........ empirical formula is CO2
36.
56.38 g P x 1
mole P = 1.82 mole P
31.0 g P
43.62 g O x 1 mole
O = 2.73 mole O
16.0 g O
Empirical formula:
P1.82O2.73 or PO1.5 so........ P2O3
Gram molecular mass = 110.0 g so that...........
molecular formula = 219.9 g
110.0 g
= 2 x P2O3 which is P4O6
D. Essay
37. The mass of a single atom of an element is the atomic mass given on the periodic table, expressed in atomic mass units. The mass of 6.02 x 1023, or Avogadro’s number of atoms of that element has the same numerical value as the gram atomic mass, but expressed in grams.