Answer:
15.0 µm
Step-by-step explanation:
Density = mass/volume
D = m/V Multiply each side by V
DV = m Divide each side by D
V = m/D
Data:
m = 1.091 g
D = 7.28 g/cm³
l = 10.0 cm
w = 10.0 cm
Calculation:
(a) Volume of foil
V = 1.091 g × (1 cm³/7.28 g)
= 0.1499 cm³
(b) Thickness of foil
The foil is a rectangular solid.
V = lwh Divide each side by lw
h = V/(lw)
= 0.1499/(10 × 10)
= 1.50 × 10⁻³ cm Convert to millimetres
= 0.015 mm Convert to micrometres
= 15.0 µm
The foil is 15.0 µm thick.
Aluminum foil is often incorrectly termed tin foil. If the density of tin is 7.28g/cm³, the thickness of a piece of tin foil is 15.0 μm that measures 10.0 cm by 10.0 cm and has a mass of 1.091 g.
To find the volume use the expression
Volume = [tex]\frac{\text{Mass}}{\text{Density}}[/tex]
Volume = [tex]\frac{1.091\ g}{7.28\ g/cm^3}[/tex]
Volume = 0.15 cm³
The aluminum foil is rectangular solid.
Volume of rectangle = length × width × height
0.15 cm³ = 10 cm × 10 cm × height
Height = [tex]\frac{0.15\ cm^3}{10\ cm \times 10\ cm}[/tex]
Height = 1.5 × 10⁻³ cm
= 0.015mm [convert cm to mm]
= 15.0 μm [convert mm to μm]
Thus from the above conclusion we can say that Aluminum foil is often incorrectly termed tin foil. If the density of tin is 7.28g/cm³, the thickness of a piece of tin foil is 15.0 μm that measures 10.0cm by 10.0 cm and has a mass of 1.091 g.
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