Observing Aluminum Foil Lab
Observing Aluminum Foil Lab
Alexandra Bellissimo
Mrs. Taylor D1P4
Date preformed: September 11th 2014
Date of Submission: September 19th 2014
The purpose of this experiment is to calculate the thickness of aluminum foil in centimeters.
In this lab we measured the thickness by plugging in the quantitative properties of our piece of aluminum foil into the equation for volume and then isolated “h” to solve for height.
The thinking behind this method is that by using accurate measurements and proper significant digits we can calculate and accurate thickness.
Materials:
Electronic Scale Ruler
Scissors Aluminum Foil
Procedure:
1. Materials were gathered
2. Aluminum foil was cut into a 10x10 square
3. Foil was weighed at 3.4 grams
4. Equation for density was rearranged to solve for volume: D=m/v. We then substituted the mass and density of the foil (D=2.70g/mL)
5. Volume formula was then rearranged to solve for height: V= lxwxh/lxw. We then substituted volume, length, density and width.
Observations:
Qualitative: Quantitative:
Metallic/Silver in colour Weight – 3.4 grams
Solid Surface area: 10x10cm
Odorless
Sample calculations:
V=0.43/2.70
V=0.16
h=10x10/0.16 h=0.16/100 h=0.0016cm
My conclusion is that the density of my piece of aluminum foil is 0.0016 cm thick.
Alexandra Bellissimo
Mrs. Taylor D1P4
Date preformed: September 11th 2014
Date of Submission: September 19th 2014
The purpose of this experiment is to calculate the thickness of aluminum foil in centimeters.
In this lab we measured the thickness by plugging in the quantitative properties of our piece of aluminum foil into the equation for volume and then isolated “h” to solve for height.
The thinking behind this method is that by using accurate measurements and proper significant digits we can calculate and accurate thickness.
Materials:
Electronic Scale Ruler
Scissors Aluminum Foil
Procedure:
1. Materials were gathered
2. Aluminum foil was cut into a 10x10 square
3. Foil was weighed at 3.4 grams
4. Equation for density was rearranged to solve for volume: D=m/v. We then substituted the mass and density of the foil (D=2.70g/mL)
5. Volume formula was then rearranged to solve for height: V= lxwxh/lxw. We then substituted volume, length, density and width.
Observations:
Qualitative: Quantitative:
Metallic/Silver in colour Weight – 3.4 grams
Solid Surface area: 10x10cm
Odorless
Sample calculations:
V=0.43/2.70
V=0.16
h=10x10/0.16 h=0.16/100 h=0.0016cm
My conclusion is that the density of my piece of aluminum foil is 0.0016 cm thick.