Solve Gold Density Problem: Area & Length of Gold Leaf & Fiber

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The discussion revolves around calculating the area of a gold leaf and the length of a cylindrical fiber made from gold, given its density of 19.32 g/cm³. For a mass of 2.274 g and a leaf thickness of 8.678 μm, the area calculated is 0.01356 m². The challenge lies in determining the length of the fiber with a radius of 2.200 μm, where the relationship between area and length is incorrectly stated in the initial attempts. The correct approach requires acknowledging the two-sided nature of the leaf's surface area.

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1. Problem: Gold, which has a density of 19.32 g/cm3, is the most ductile metal and can be pressed into a thin leaf or drawn out into a long fiber. (a) If a sample of gold with a mass of 2.274 g, is pressed into a leaf of 8.678 μm thickness, what is the area (in m2) of the leaf? (b) If, instead, the gold is drawn out into a cylindrical fiber of radius 2.200 μm, what is the length (in m) of the fiber?


2. Homework Equations : a=pi x r^2 and (a)(l)= part a



3. The Attempt at a Solution : part a= 0.01356, and I tried plugging it into the equation (1.521x10^-11 m^2)(l)=0.01356

If anyone could find the error in part B! Please Help
 
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JudyyNunez said:
2. Homework Equations : a=pi x r^2 and (a)(l)= part a

The "(a)(l)= part a" looks wrong. The answer to part a is an area. You can't say "area times length = area".

3. The Attempt at a Solution : part a= 0.01356

That looks right (except you forgot the units).

I don't understand what you did for part (b).
 
what is the area (in m2) of the leaf?
Don't forget that the leaf has two sides, so there are two surface areas. This could be a trick question! :smile:

So state your answer as ... m2 per side for each of 2 sides.

If you knew the shape, you could also work out the area of its other very narrow side/s. :wink:
 

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