How Can You Calculate the Area of a Gold Leaf and the Length of a Gold Fiber?

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Homework Help Overview

The discussion revolves around calculating the area of a gold leaf and the length of a gold fiber, given specific mass and density values. The problem involves understanding the relationships between mass, volume, and dimensions in the context of gold's properties.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the calculation of volume using mass and density, questioning the correct application of units. There is discussion on how to relate the thickness of the gold leaf to its area and how to derive the length of the cylindrical fiber from its radius.

Discussion Status

Some participants have provided guidance on unit conversions and the relationships between volume, area, and dimensions. There is an ongoing exploration of the correct formulas needed for both parts of the problem, with no explicit consensus reached yet.

Contextual Notes

Participants emphasize the importance of unit consistency and the need to maintain the volume of gold throughout the calculations. There is a mention of potential confusion regarding the interpretation of density units.

blanny
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Gold, which has a mass of 19.32g for each cubic centimeter of volume, is the most ductile metal and can be pressed into a thin leaf or drawn out into a long fiber.

a) If the sample of Gold, with a mass of 27.63 g is pressed into a leaf of 1.000 micrometer thickness, what is the area of the leaf?

Would it be correct to use 27.63g / 19.32g = cubic centimeters? I don't think this would be right because it doesn't take into account the 1.000 micrometer...

b) If, instead the Gold is drawn into a cylindrical fiber with radius 2.500 micrometers, what is the length of the fiber?

What equations would I need to solve this problem? Thanks, this problem has been really bugging me.

blanny
 
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Avoid being sloppy with units: grams divided grams is dimensionless.

Given the density and the mass you can determine the volume of the gold (again, watch those units!). The principle you seem to be missing is that the volume of the gold remains unchanged. In (a) the product of the thickness times the area is the volume and in (b) the cross-sectional area of the cylinder times its length is also the volume.
 
blanny said:
Would it be correct to use 27.63g / 19.32g = cubic centimeters?
Yes and no. Look at your units.

27.63g / 19.32 g, the grams cancel and you're left with a unitless number of 1.430

But 19.32 is not grams. It is "grams for each cubic centimeter" or g/cc aka g/cm^3

Use the correct units and look at the units in your answer:

27.63g / (19.32 g/cc) . The grams cancel. The cc is the denominator of a denominator, which puts it in the numerator, and you get 1.430 cc.

So you were correct in your assumption that the answer would yield cubic centimeters.

Additional formulas needed:

Area: length x width
Volume: length x width x height
or since lenth x width = area
Volume = area x height

for part b you will need the formula for volume of a cylinder:
area of a circle x height
area of a circle = pi x radius^2
 
Last edited:
Tide beat me to it in pointing out your units :rolleyes:
You got to love this forum! Welcome.
 

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