Please Please help me! I don't understand these questions that are left and I need to get a good grade on it, can you please read it over and help me with it by telling me what to do and providing the answer so I can make sure it is done right?
Given this table:
A. Mass of piece of aluminum (g): 64.96g
B. Volume of water in cylinder: 70mL
C. Volume of water plus aluminum (mL): 92.5mL
D. Volume of piece of aluminum (mL0 (C-B)
E. Length (l) of aluminum foil (cm):11.9 cm
F. Width (w) of aluminum foil (cm): 9.4 cm
G. Mass of aluminum foil (g): 0.67g
H. Density of aluminum (A/D) (g/cm^3): 2.88 g/cm^3
I. Volume of foil (cm^3) (G/H): 0.23 cm^3
J. Height (thickness) of foil (cm): ?
K. Moles of aluminum in foil: ?
L. Atoms of aluminum in foil: ?
I must do these problems:
4. Find the height (thinkness) of the foil (h) in cm from the relation Volume = l x w x h and record the result.
5. One aluminum atom is 2.5 x 10^-8 cm thinck. Find the thickness of the foil in atoms, which is given by Number of atoms thick = height (J) x 1 atom/ 2.5 x 10^-8 cm
6. Compute the number of moles of aluminum and the total number of atoms of aluminum in your piece of aluminum foil.
7. If the population of the world is 5.6 x 10^9 individuals, how many atoms of aluminum could you distribute to each person from your sample of aluminum foil?
Given this table:
A. Mass of piece of aluminum (g): 64.96g
B. Volume of water in cylinder: 70mL
C. Volume of water plus aluminum (mL): 92.5mL
D. Volume of piece of aluminum (mL0 (C-B)
E. Length (l) of aluminum foil (cm):11.9 cm
F. Width (w) of aluminum foil (cm): 9.4 cm
G. Mass of aluminum foil (g): 0.67g
H. Density of aluminum (A/D) (g/cm^3): 2.88 g/cm^3
I. Volume of foil (cm^3) (G/H): 0.23 cm^3
J. Height (thickness) of foil (cm): ?
K. Moles of aluminum in foil: ?
L. Atoms of aluminum in foil: ?
I must do these problems:
4. Find the height (thinkness) of the foil (h) in cm from the relation Volume = l x w x h and record the result.
5. One aluminum atom is 2.5 x 10^-8 cm thinck. Find the thickness of the foil in atoms, which is given by Number of atoms thick = height (J) x 1 atom/ 2.5 x 10^-8 cm
6. Compute the number of moles of aluminum and the total number of atoms of aluminum in your piece of aluminum foil.
7. If the population of the world is 5.6 x 10^9 individuals, how many atoms of aluminum could you distribute to each person from your sample of aluminum foil?
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Problem 4.
Volume of aluminum is given as C - B = 92.5 mL - 70 mL = 22.5 mL = 22.5 cm^3 (1mL = 1cm^3)
E gives the length: 11.9 cm
F gives the width: 9.4 cm
V = lwh; solving for h: h = V/lw
h = 22.5 cm^3 / (11.9 cm * 9.4 cm) = 0.2011442875022349 cm (Adjust significant figures as necessary).
Problem 5.
N = h/a; N = number of atoms, a = thickness of one aluminum atom.
N = 0.2011442875022349 cm / (2.5 * 10^-8 cm) = 8045771.5 atoms thick. [8.05 * 10^6 Atoms thick]
Problem 6.
Number of mols = mass/molar mass.
Mass is given in A: 64.96g
Molar mass is found on the periodic table: 26.98 g/mol
Number of mols = 64.96g / (26.98 g/mol) = 2.4077 mol = 2.41 mol
Total number of atoms: 6.022*10^23 atoms in one mol (Avogadro's number)
2.41 mol * 6.022*10^23 atoms/mol = 1.4499 * 10^24 atoms = 1.45*10^24 atoms
Problem 7.
1.45*10^24 atoms/(5.6*10^9 people) = 2.589^10^14 atoms/person = 2.59*10^14 atoms/person
Volume of aluminum is given as C - B = 92.5 mL - 70 mL = 22.5 mL = 22.5 cm^3 (1mL = 1cm^3)
E gives the length: 11.9 cm
F gives the width: 9.4 cm
V = lwh; solving for h: h = V/lw
h = 22.5 cm^3 / (11.9 cm * 9.4 cm) = 0.2011442875022349 cm (Adjust significant figures as necessary).
Problem 5.
N = h/a; N = number of atoms, a = thickness of one aluminum atom.
N = 0.2011442875022349 cm / (2.5 * 10^-8 cm) = 8045771.5 atoms thick. [8.05 * 10^6 Atoms thick]
Problem 6.
Number of mols = mass/molar mass.
Mass is given in A: 64.96g
Molar mass is found on the periodic table: 26.98 g/mol
Number of mols = 64.96g / (26.98 g/mol) = 2.4077 mol = 2.41 mol
Total number of atoms: 6.022*10^23 atoms in one mol (Avogadro's number)
2.41 mol * 6.022*10^23 atoms/mol = 1.4499 * 10^24 atoms = 1.45*10^24 atoms
Problem 7.
1.45*10^24 atoms/(5.6*10^9 people) = 2.589^10^14 atoms/person = 2.59*10^14 atoms/person