Chemistry Problem density of aluminum?

Question

A solid aluminum sphere has a mass of 89g

Use the density of aluminum to find the radius of the sphere in inches.

Answer 1

Density = mass /  volume

Density of Aluminum = 2.7 g/mL

Volume = mass / density

Volume = 89g / 2.7g/mL

Volume = 32.96mL

Converting to cubic meters,

32.96mL x (1L / 1000mL) x (1 x 10^-3 m^3 / 1 L) = 3.296 x 10^-5 m^3

Volume of sphere = 4/3 π r^3

r^3 = 3V / 4π

r = cube root of (3V / 4π)

r = cube root of [(3 x 3.296 x 10^-5 m^3) / 4π]

r = cube root of 7.87 x 10^-6 cubic meters

r = 0.0199 meters

Converting to inches

1 m = 39.37 inches

0.0199 meters x (39.37 inches / 1 meter) = 0.78 inches

Answer: 0.78 inches

Answer 2

density=mass/volume

Answer 3

density

D = m / V

V = m / D

volume of a sphere

V = 4/3 pi r^2

m / D = 4/3 pi r^3

r = (3m/4piD)^(1/3)

r = (3 x 89g / 4/ pi / 2.70 g/cm^3)^(1/3)

r = 1.99 cm

(technically 2.0 cm to 2 sig figs)

Inches???  Surely you jest.  This is chemistry, after all.  

r = 1.99 cm x (1 in / 2.54 cm) = 0.783 inches

(technically 0.78 inches to 2 sig figs)

Answer 4

Density of Aluminium = 2.7g/cc.

89g / 2.7 g/cc = 33cc.

Volume of sphere = 1.333 x 3.142 x r³ = 33cc.

4.2 x r³ = 33cc

r³ = 33cc / 4.2 =  ÃƒÂ‚³âˆš8 = 2cm / 2.54cm/" = 0.79" = 0.8".

This is no jest, what's wrong with inches...??

Answer 5

ρ(solid Al) = 2700 kg/m³

V(sphere) = (4/3)πR³

ρ = m/V ──► V = m/ρ

(4/3)πR³ = m/ρ

R³ = 3m / (4πρ)

R = [3m / (4πρ)]^(1/3) = [3(0.089) / (4π(2700))]^(1/3) = 0.0199 m

= 19.9 mm