Data and Results
Part I: Determining the density of a metal cylinder
Part II: Submarine Adventure
Mass of graduated cylinder
39.0834 g
Volume of ocean water
2.10 mL
Mass of water
2.143 g
Weight of balloon w/weights
138.590 g
Density of ocean water
1.020 g/mL
Volume of balloon needed to match
135.804 cm^3
Final volume of submarine
95.950 cm^3
Final density of submarine
1.45 g/mL
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Calculations
In order to find the volume of the metal cylinders, volume by measurement would
work with the formula: V
metal cylinders
= ╥ r
2
h
V
metal cylinder 1
= ╥ (1.24)
2
(0.575) = 2.778 cm
3
V
metal cylinder 2
= ╥ (0.622)
2
(5.10) = 6.209 cm
3
Mass of the water can be found by subtracting the mass of the 10mL graduated
cylinder from the mass of the graduated cylinder with water.
Mass
water
= 41.226g39.083g= 2.143 g
The density of the “ocean water” can be found through this formula:
Density
ocean water
= grad. cylinder with water mass(g) grad.cylinder mass(g)
Volume of “ocean water”(mL)
= 41.226g39.083g
= 1.020 g/mL
2.10 mL
The volume of the balloon needed to match the saltwater density can be found
through this equation:
Density
ocean water
= Density
submarine
= Balloon+pellets mass(g)
4/3*╥r
3
1.020 g/mL= 138.590g
r
3
= 32.437 r=3.189 d=6.378
4/3*╥r
3
mL
Plug in the diameter into the volume equation and the volume of the
balloon need to match the saltwater density would be 135.804 cm
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 Fall '08
 dillon
 Chemistry, ocean water, 12.340 g/cm3, 8.808 g/cm3

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