Gold, which has a density of 19.32gcm3, is the most ductile metal and can be...

Victor Wall

Victor Wall

Answered

2022-01-07

Gold, which has a density of 19.32gcm3, 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 27.63 g, is pressed into a leaf of 1.000μm thickness, what is the area of the leaf?
b) If, instead, the gold is drawn out into a cylindrical fiber of radius 2.500μm, what is the length of the fiber?

Answer & Explanation

Matthew Rodriguez

Matthew Rodriguez

Expert

2022-01-08Added 32 answers

a) Using the provided information about the density of gold, the sample size, thickness, and the following
equations and conversion factors, find the area of the gold leaf:
V=ellwh=Ah
m=pV
1μ=106m
Goldp=19.32gcm3
First, find the volume of the sample and then find the area of the sample.
V=mp=27.6g19.32g/cm3(0.01m1cm)3
1.429×106m3
V=AhA=Vh=1.429×106m31x106m1.429m2
b) Using the provided information from part a), the radius of the cylinder, and the following equation for the
volume of a cylinder, find the length of the fiber
V=πr2hh=Vπr2
h=1.429×106m3π(2.5×106m)272778m

Ethan Sanders

Ethan Sanders

Expert

2022-01-09Added 35 answers

From the equation(1 - 8), the density (p) of the gold is defined as the mass of the gold (m) per its volume (V):
p=mV=19.32gcm3(1)
(a) The volume of the thin leaf is found to be;
Veaf=(area)(thickness)=Aeaf×teaf(2)
From equation (1), the volume of the thin leaf will be;
Veaf=meafpeaf=27.63g19.32g/cm3=1.43cm3
Now, by converting the volume to SI units, we get
VLeaf=(1.43cm3)(1m100cm)3=1.43×106m3
The thickness of the thin leaf:  telleaf=1.00μm=1×106m Therefore, from the equation (2), we can find the area of the thin leaf as following;
Aelleaf=VLeaftLeaf=1.43×106m31×106m1.43m2
(b) The volume of cylindrical fiber is equal to the volume of the thin leaf (same gold)
VFiber=VLeaf
As known, the volume of the cylindrical fiber is;
VFiber=(crosssectionalarea)(length)=AFiber(3)
This cross-section area is circular and given by: A=πr2,wherer=2.5μm=2.5×106m Thus from equation (3), we have
Fiber=VFiberπr2=1.43×106m3(3.14)(2.5×106m)2=7.286×104m
ellFiber=7.286×104mor72.866km

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