Mass of an atomic layer of a baloon

  • Thread starter Thread starter Myrddin
  • Start date Start date
  • Tags Tags
    Atomic Mass
AI Thread Summary
To estimate the mass of aluminum required for a single atomic layer on a toy balloon, the balloon is modeled as a sphere with a radius of 10 cm. The surface area is calculated, and the volume of the atomic layer is determined by multiplying the surface area by the approximate radius of an aluminum atom. The mass is then found by using the volume and the density of aluminum, which is 2.7 x 10^3 kg/m^3. The discussion highlights a potential oversight in not incorporating the molar mass of aluminum, which is 27 grams per mole, into the calculations. The method proposed effectively uses density and volume to find mass, but further clarification on the role of molar mass is sought.
Myrddin
Messages
24
Reaction score
0
An inflated smooth toy balloon is coated with a single atomic layer of aluminium. Esitimate the mass of Al that would be required. The density of Al is 2.7 x10^3kg/m^3 and its molar mass if 27 grams for mol.


I assusmed the balloon was a sphere of radius 10cm, found surface area then mutliplied by approximate radius of an atom to find volume of the layer. Then used volume x density to find mass, but this method doesn't make use of the molar mass in the question? Any other methods you can see?
 
Physics news on Phys.org
Myrddin said:
I assumed the balloon was a sphere of radius 10cm, found surface area then multliplied by approximate radius of an atom to find volume of the layer. Then used volume x density to find mass, but this method doesn't make use of the molar mass in the question? Any other methods you can see?

The approximate radius of the aluminium atom follows from the molar mass and density.

ehild
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top