Postsecondary • 8mo.
Hi there!
I need explanation for question 5 and 6 as soon as possible because tomorrow is my exam and I’m not able to understand how to answer them. If someone could kindly explain me in details in steps.
The answers are below the questions but I don’t get how to solve them?
thank u so much!
Explanation from Alloprof
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a)
Density of gold d=19.3 g cm-³
Area of the sheet S=124.5 ft²
h thickness of the sheet to be determined
m mass of gold used m=28.35 g
we have the formula
mass= density x Volume
Volume = Area x thickness
m=dxSxh h=m/d x S
units given
m : g
d : g cm-³
S: ft²
we have to change S in cm²
1 ft = 12 inch
1 inch =2.54 cm
we obtain
1 ft = 12 x 25.4 cm 1ft=30.48 cm
1ft²= 30.48² cm²
1ft²= 929.0304 cm²
S=(124.5 x 929.0305) cm²
h= 28.35 g/(19.3 g cm-³ (124.5 x 929.0305) cm²)
h=28.35 g/(19.3 g cm-³ x 124.5 x 929.0305 cm²)
h=(28.35 /(19.3 x 124.5 x 929.0305 )) (g/g cm-³ cm²)
h=(0.000012699) cm
1 cm =0.01 m
h=(0.000012699) x 0.01 m
h=0.00000012699 m
h=126.99 nm =127 nm
b)
atom gold diameter =288 pm
thickness of the sheet=(number of atoms) x (atom gold diameter)
other wise
h=n*Diameter
n=h/Diameter
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c)
m = V x d
mass = Volume x density
28.35 g = V x 19.3 g cm-³
V=28.35 g/19.3 g cm-³
V=(28.35/19.3) g/g cm-³
V=1.469 cm³
we consider this volume as a cube
V=a³
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