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17.12. Read the percentage associated with the range of each mineral Limestone
g
_

from Figure 17.13, and calculate the difference to determine the m = 2.7 (40 cm × 23 cm × 20 cm)(0.90)
percentage of each mineral. cm 3 _
g
= 2.7(40 × 23 × 20) (cm × cm × cm)

Granite cm 3
(
3 )
1 kg
Biotite: 100%−83% = 17% = 4.5 × 10 g _
4

Sodium-rich plagioclase: 83%−63% = 20% 1 × 10 g
Potassium feldspar: 63%−25% = 38% = 45 kg
Quartz: 25%−0% = 17% Marble
g
_

m = 2.7 (40 cm × 23 cm × 20 cm)(1)
17.13. Determine the volume of the countertop by the formula for a cm 3
g
_
rectangular solid, then use the density equation to solve for the = 2.7 (40 × 23 × 20)(1) (cm × cm × cm)

3
mass of the countertop. Th e specific gravity has the same 1 kg cm

3 )
(
4

numerical value as the density. = 5.0 × 10 g _
1 × 10 g
L = 4.57 m = 50 kg
W = 0.75 m
17.15. Determine the volume from the density.
H = 2.3 cm
m _
tonne
SG granite = 2.75 ρ = 2.75 _ ρ =

V
m = ? m 3 m _
m _ m = 1.50 tonne ∴ V =
ρ = and V = L × W × H ρ
V V = ?
1.50 tonne
∴ m = ρ(L × W × H) = _

tonne

2.75 _

3
Convert all dimensions from m to cm. m
_ _
1.50 tonne
_ =
100 cm
_

L = 4.57 m ( ) 2.75 tonne
1 m
m 3
= 457 cm −1 3
= 5.45 × 10 m
_
100 cm
W = 0.75 m ( )

1 m Determine the diameter from the formula for the volume of a
= 75 cm sphere.
m = ρ(L × W × H) 4 _ 3 D _
g
2
3
_ V = πr and r =
= 2.75 (457 cm × 75 cm × 2.3 cm) 4 _ 3
D _
cm 3 ∴ V = π ( )

g
_ 3 2

= 2.75 (457 × 75 × 2.3) (cm × cm × cm)
3
4 _ _
cm 3 V = π
D

3 )
(
1 kg
5

= 2.2 × 10 g _ 3 D _ 3 8
1 × 10 g
V = π
6
2
= 2.2 × 10 kg 6 _ D _ 6 _
3

V = π
π
6 π
17.14. Determine the volume of the stone by the formula for a
6V _
3
rectangular solid, and multiply by the percentage that is calcite. In D =
π
the case of the marble, assume it is 100 percent calcite. Th en use
3 6V _

the density equation to solve for the mass of the building stone. D = √
π
Th e specific gravity has the same numerical value as the density.

3
(Th e specific gravity of calcite was provided in exercise 17.10.) Convert volume of stone from m to cm . 3

3 )
(
6
3 1 × 10 cm
L = 40 cm 5.45 × 10 m _ 3
−1

W = 23 cm 1 m
5
H = 20 cm 5.45 × 10 cm 3
SG calcite = 2.7 3 )
5
3
6(5.45 × 10 cm

D = √ __
π

Limestone:
3 6 3
%calcite = 90% = √ 3.27 × 10 cm
m = ? 3 6 3
= √ 1.04 × 10 cm
Marble:
2
= 1.01 × 10 cm
%calcite = 100%
m = ?
m _
ρ = and V = L × W × H(%calcite)
V
∴ m = ρ(L × W × H)(%calcite)
E-37 APPENDIX E Solutions for Group A Parallel Exercises 679