Atom Volume Calc: Find the Volume of a Lead Atom

In summary, to find the volume of an individual atom in a tightly packed spherical arrangement, we can convert the mass of lead from g/cm^3 to g/m^3 and the mass of the atom from kg to g, and then divide the two values to get m^3. The correct answer is 3.03*10^-23 m^3.
  • #1
Kpgabriel
36
0

Homework Statement


Lead has a mass of 11.35 g per cubic centimeter of volume, and the mass of one of its atoms is 3.439×10-25 kg. If the atoms are spherical and tightly packed, what is the volume of an individual atom? ( m3)

Homework Equations


conversions

The Attempt at a Solution


For this problem I converted the 11.35 g/cm ^3 into 11350 g/m^3 and then converted the atom from 3.439*10^-25 kg into 3.439*10^-22 g. Then I divided it by the 11350 to get m^3 and the result was 3.03*10^-26 but it appears to be wrong.
 
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  • #2
Kpgabriel said:
spherical and tightly packed,
 
  • #3
Kpgabriel said:

Homework Statement


Lead has a mass of 11.35 g per cubic centimeter of volume, and the mass of one of its atoms is 3.439×10-25 kg. If the atoms are spherical and tightly packed, what is the volume of an individual atom? ( m3)

Homework Equations


conversions

The Attempt at a Solution


For this problem I converted the 11.35 g/cm ^3 into 11350 g/m^3 and then converted the atom from 3.439*10^-25 kg into 3.439*10^-22 g. Then I divided it by the 11350 to get m^3 and the result was 3.03*10^-26 but it appears to be wrong.

How many ##cm## are there in a ##m##?
 
Last edited:
  • #4
PeroK said:
How many ##cm## are there is a ##m##?
100 so it would be cubed to get m^3 and it would be 3.03*10^-29 which is the right answer
 
  • #5
If 11,35 grams fit into 1 cm3 then 3.439×10-22 g will fitt into
3.439×10-22g/ 11.35= 3.03 cm3
 
  • #6
Mathijsgri said:
If 11,35 grams fit into 1 cm3 then 3.439×10-22 g will fitt into
3.439×10-22g/ 11.35= 3.03 cm3
* 3.03*10-23 cm3
 
  • #7
Mathijsgri said:
* 3.03*10-23 cm3
But you're looking for m^3
 
  • #8
Kpgabriel said:
But you're looking for m^3
1m3= 1.000.000cm3
? = 3.03*10 - 23cm3
 
  • #9
Kpgabriel said:
But you're looking for m^3

Are you helping him now? o_O
 
  • #10
'crosswise multiplication table' i don't know if it is an english term, i translated it literally from dutch.
 
  • #11
Mathijsgri said:
1m3= 1.000.000cm3
? = 3.03*10 - 23cm3
Right. so it would be 11.35 g/cm^3 * 1000000cm^3/1 m^3 to give 11350000 g/m^3 which you divide into 3.439*10^-22 g
 
  • #12
Mathijsgri said:
'crosswise multiplication table' i don't know if it is an english term, i translated it literally from dutch.
I just call it a conversion
 
  • #13
Kpgabriel said:
Right. so it would be 11.35 g/cm^3 * 1000000cm^3/1 m^3 to give 11350000 g/m^3 which you divide into 3.439*10^-22 g
keep in mind that 1 cm3 is not equalle to 1.000.000 m3
 
  • #14
Mathijsgri said:
keep in mind that 1 cm3 is not equalle to 1.000.000 m3

I don't think you're helping here. The OP has got the correct answer. Leaving aside the question of the basic volume of a sphere as opposed to the volume it occupies in a 3D array.
 
  • #15
Kpgabriel said:
100 so it would be cubed to get m^3 and it would be 3.03*10^-29 which is the right answer

I assume that's the answer on your answer sheet?
 
  • #16
you are right i made a typo in my calculator, sorry for that.
 
  • #17
PeroK said:
I assume that's the answer on your answer sheet?
Yes it was a test question I got wrong so I was trying to find the correct method to the answer
 

FAQ: Atom Volume Calc: Find the Volume of a Lead Atom

1. What is "Atom Volume Calc"?

"Atom Volume Calc" is a scientific tool used to calculate the volume of a lead atom. It takes into account the atomic weight and density of lead to determine the volume of a single lead atom.

2. How is the volume of an atom calculated?

The volume of an atom can be calculated by using the formula V = (4/3)πr^3, where V is the volume and r is the radius of the atom. In the case of "Atom Volume Calc", the atomic weight and density of lead are also factored in to provide a more accurate calculation.

3. Why is it important to know the volume of a lead atom?

Knowing the volume of a lead atom is important for various scientific and industrial purposes. It can help in understanding the structure and properties of lead, as well as in determining its potential uses in different fields such as medicine, electronics, and construction.

4. How accurate is the calculation of "Atom Volume Calc"?

The accuracy of the calculation depends on the accuracy of the input data, such as the atomic weight and density of lead. If these values are known accurately, then the calculated volume of the lead atom should also be accurate.

5. Can "Atom Volume Calc" be used for other elements besides lead?

Yes, "Atom Volume Calc" can be used for other elements as well. However, the input data would need to be adjusted accordingly to match the atomic weight and density of the specific element being calculated.

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