Finding number of atoms and volume

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Homework Help Overview

The problem involves calculating the number of atoms in a cubic centimeter of copper, as well as visualizing the arrangement of these atoms in smaller cubes. The subject area pertains to concepts in density, mass, and atomic structure.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to use the density and mass of a copper atom to find the number of atoms in a given volume, expressing frustration with the textbook's clarity. Some participants suggest focusing on the first part of the problem as a unit conversion issue. Others question the setup of the calculations and encourage inventorying unit conversion factors.

Discussion Status

The discussion is ongoing, with participants exploring different approaches to the problem. Some guidance has been offered regarding unit conversions, but there is no explicit consensus on the correct method yet.

Contextual Notes

The original poster expresses difficulty with the textbook and the problem's requirements, indicating a potential lack of clarity in the provided material. There is an emphasis on needing the answer from part (a) to proceed with parts (b) and (c).

fernanhen
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Homework Statement



The mass of a copper atom is 5.30 10-25 kg, and the density of copper is 8 920 kg/m3 .
(a) Determine the number of atoms in 1 cm3 of copper.

_______Cu—atom/cm3



(b) Visualize the one cubic centimeter as formed by stacking up identical cubes, with one copper atom at the center of each. Determine the volume of each cube.


_______cm3/Cu—atom




(c) Find the edge dimension of each cube, which represents an estimate for the spacing between atoms.

_________cm



Homework Equations



Density=mass/Volume



The Attempt at a Solution




Well, the book the book we are using for this class "Physics for Scientists and Engineers" by Serway has no right being used in an Introductory Physics class.

I re-read the relevant chapter THREE times and still haven't figured out how to solve this.

Therefore, I used my intuition alone.

Using the info I have:

3.06*10^(-25)kg= (8,920kg/m^3)*1cm^3 = M=D*V

I canceled the volumes but that won't left over any missing variables.

I googled for alternatives and my textbook absolutely does not point us in any direction regarding this question. Please help.

Thanks in advance.
 
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You need the answer from a) to do b) and c). So focus on that first. It is essentially a unit conversion problem.
 
lewando said:
You need the answer from a) to do b) and c). So focus on that first. It is essentially a unit conversion problem.

Yes, I just figured that this morning. Thanks.

But now I am having trouble with that.

I equated:

(8920kg/m^3)*(m/cm)*(cm-atom)/kg=Cu-atom/cm^3

Then 8920/m^3*1m/100*5.24*10^(-27)=Cu-atom/cm^3

But now what? Does my setup even look right?
 
fernanhen said:
I equated:

(8920kg/m^3)*(m/cm)*(cm-atom)/kg=Cu-atom/cm^3

Then 8920/m^3*1m/100*5.24*10^(-27)=Cu-atom/cm^3

But now what? Does my setup even look right?

Not really. You need to inventory all of your unitary conversion factors and apply them correctly.
1 = 1 Cu atom / 5.30 10-25 kg
1 = 1m3 / 1x106 cm3
1 = 8920 kg / 1m3

You are trying to convert 1 cm3 [of Cu] to Cu atoms. If you had a unitary conversion factor
in the form of:
1 = X Cu atoms / Y cm3 you would be close to being finished with part a). Can you try this using the above unitary factors?
 
Last edited:

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