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Oduvanchick [21]
3 years ago
14

A package of aluminum foil contains 50. ft2 of foil, which weighs approximately 8.5 oz . aluminum has a density of 2.70 g/cm3. w

hat is the approximate thickness of the foil in millimeters?
Chemistry
1 answer:
Serhud [2]3 years ago
8 0
Hello!

We have the following data:

Area (A) = 50 square feet
Mass (m) = 8.5 ounces
Density (d) = 2.70 g/cm³
Volume (V) = ?
Thickness (T) =? (in mm)

To move on, we must transform the area of 50 ft² in cm², let's see:

1 ft² ------- 929,0304 cm²
50 ft² ----- A

A = 50*929,0304

\boxed{A = 46451,52\:cm^2}\Longleftarrow(Area)

In the same way, we will convert the mass of 8.5 oz in grams, see:

1 oz -------- 28,3495 g
8,5 oz ------- m

m = 8,5*28,3495

\boxed{m = 240,97075\:g}\Longleftarrow(mass)

Knowing that the density is 2.70 g/cm³ and the mass is 240.97075 g, we will find the volume, applying the data in the density formula we have:

d =  \dfrac{m}{V} \to V =  \dfrac{m}{d}

V =  \dfrac{240,97075\:\diagup\!\!\!\!\!g}{2,70\:\diagup\!\!\!\!\!g/cm^3}

V = 89,24842593... \to \boxed{V \approx 89,25\:cm^3}

The statement wants to find the thickness of the packaging, for this we have some important data, such as: V (volume) = 89,25 cm³ and Area (A) = 46451,52 cm² and T (thickness) =? (in mm)

In the calculations of Costs in Surface Treatment of a part within the flat geometry, we will use the following formula:

V (volume) = A (Area) * T (Thickness)

89,25\:cm^3 = 46451,52\:cm^2\:*\:T

46451,52\:cm^2*T = 89,25\:cm^3

T =  \dfrac{89,25\:cm^3}{46451,52\:cm^2}

T = 0,001921358009...\:cm

We will convert to millimeters, going through a decimal place on the right

T = 0,01921358009..\:mm

\boxed{\boxed{T \approx 0,0192\:mm\:or\:T\approx \:1,92*10^{-2}\:mm}}\end{array}}\Longleftarrow(thickness)\qquad\checkmark

Hope this helps! :))





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