Converting from mass/volume to length

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SUMMARY

The discussion centers on calculating the thickness of a gold leaf film using its volume and area. The density of gold is given as 19.3 g/cm³, and the mass of the gold leaf is 1.93 mg, resulting in a volume of 10^(-4) cm³. The area mentioned in the problem is incorrectly stated as 14.5 cm³ instead of cm². Correcting this allows for the calculation of thickness using the formula: thickness = volume/area, leading to a final thickness of 690 angstroms.

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  • Understanding of density and its formula (density = mass/volume)
  • Knowledge of unit conversions, particularly between centimeters and angstroms
  • Familiarity with volume and area concepts in geometry
  • Basic algebra skills for solving equations
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  • Study the properties of gold, including its density and applications in materials science
  • Explore geometric calculations involving volume and area
  • Review common errors in physics problems and how to identify them
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Homework Statement


A piece of gold leaf (density 19.3 g/cm^3) weighing 1.93 mg can be beaten into a transparent film covering an area of 14.5 cm^3. what is the volume of 1.93 mg of gold? What is the thickness of the transparent film in angstroms? I only need help with the second question. The volume of 1.93 mg of gold is 10^(-4) cm^3, but I don't know how to convert to length on the second question.


Homework Equations


Angstrom=10^(-10) meters
d=mass over volume


The Attempt at a Solution


I tried to cheat by taking the third root of 14.5 cm^3 in order to get to cm:

(14.5cm^3)^(1/3)=2.44 cm

2.44 cm (10^(-2)m/1cm)(1 Angstrom/10^(-10) meters)= 2.44 x 10^8 anstroms

The answer is 690 angstroms though and i don't know how
 
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dchau503 said:

Homework Statement


A piece of gold leaf (density 19.3 g/cm^3) weighing 1.93 mg can be beaten into a transparent film covering an area of 14.5 cm^3. what is the volume of 1.93 mg of gold? What is the thickness of the transparent film in angstroms? I only need help with the second question. The volume of 1.93 mg of gold is 10^(-4) cm^3, but I don't know how to convert to length on the second question.


Homework Equations


Angstrom=10^(-10) meters
d=mass over volume


The Attempt at a Solution


I tried to cheat by taking the third root of 14.5 cm^3 in order to get to cm:

(14.5cm^3)^(1/3)=2.44 cm

2.44 cm (10^(-2)m/1cm)(1 Angstrom/10^(-10) meters)= 2.44 x 10^8 anstroms

The answer is 690 angstroms though and i don't know how

There is at least one typo in the problem statement. What are the units of area?

Once you fix that, the volume of the leaf is just the area multiplied by the thickness, right?
 
dchau503 said:

Homework Statement


A piece of gold leaf (density 19.3 g/cm^3) weighing 1.93 mg can be beaten into a transparent film covering an area of 14.5 cm^3. what is the volume of 1.93 mg of gold? What is the thickness of the transparent film in angstroms? I only need help with the second question. The volume of 1.93 mg of gold is 10^(-4) cm^3, but I don't know how to convert to length on the second question.

Homework Equations


Angstrom=10^(-10) meters
d=mass over volume

The Attempt at a Solution


I tried to cheat by taking the third root of 14.5 cm^3 in order to get to cm:

(14.5cm^3)^(1/3)=2.44 cm

2.44 cm (10^(-2)m/1cm)(1 Angstrom/10^(-10) meters)= 2.44 x 10^8 anstroms

The answer is 690 angstroms though and i don't know how

One problem may be the area of 14.5... cm^3 is not area.
If I assume that to be cm^2 the problem works out fine.
I think of it as a very shallow box of gold with a volume of LxWxH where you know L and W and just need H so that the volume becomes what it's supposed to be. LxW is area, so AxH should = Volume.

Good?
 


berkeman said:
There is at least one typo in the problem statement. What are the units of area?

Once you fix that, the volume of the leaf is just the area multiplied by the thickness, right?

There's no area measurement listed in the original question. But that would make it a lot easier, if the problem did list it.

Edit: actually, the problem did say the "area of 14.5 cm^3"...but that's weird because a cubed unit signifies volume, not area.
 


dchau503 said:
There's no area measurement listed in the original question. But that would make it a lot easier, if the problem did list it.

Edit: actually, the problem did say the "area of 14.5 cm^3"...but that's weird because a cubed unit signifies volume, not area.

So the units must be wrong...
 


Jakeus314 said:
One problem may be the area of 14.5... cm^3 is not area.
If I assume that to be cm^2 the problem works out fine.
I think of it as a very shallow box of gold with a volume of LxWxH where you know L and W and just need H so that the volume becomes what it's supposed to be. LxW is area, so AxH should = Volume.

Good?

Your advice put me on the right track, but I still don't have the complete answer:

I did 14.5 cm^3 x (thickness) = 1X10^(-4) cm^3

From the equation, I solved that thickness is 6.9 x 10^(-6) cm^3. I got the right significant figures, but not the right units. How do I convert that answer into angstroms?
 
dchau503 said:
Your advice put me on the right track, but I still don't have the complete answer:

I did 14.5 cm^3 x (thickness) = 1X10^(-4) cm^3

From the equation, I solved that thickness is 6.9 x 10^(-6) cm^3. I got the right significant figures, but not the right units. How do I convert that answer into angstroms?

14.5 cm^3 is an error on someones part.
6.9 x 10^(-6) cm^3 is not a thickness.

Solving for thickness,
14.5 cm^2 x (thickness in cm) = 1*10^(-4) cm^3
Gives a thickness in cm.
Convert that to angstroms since those are the requested units.

cm^3 / cm^2. Yields cm
 
Jakeus314 said:
14.5 cm^3 is an error on someones part.
6.9 x 10^(-6) cm^3 is not a thickness.

Solving for thickness,
14.5 cm^2 x (thickness in cm) = 1*10^(-4) cm^3
Gives a thickness in cm.
Convert that to angstroms since those are the requested units.

cm^3 / cm^2. Yields cm

Thanks for the answer. I finally understand the problem and it's solution, even with the typo in the problem. :(

Crash-course study books frequently have mistakes in them, lol.
 

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