Calculate the average density of a single A1-27 atom by assuming that it is a sphere with a radius of 0.143 nmThe masses of a proton, electron, and neutron are 1.6726 x 10^-24 g, 9.1094 x 10^-28 g, and 1.6749 x 10^-24 g, where r is its radiusExpress the answer in grams per cubic centimeterThe density of aluminum is found experimentally to be 2.70 g/cm^3What does that suggest about the packing of aluminum atoms in the metal?
I think you mean tin coated MS sheet or GI sheetIn both cases there are chances of corrosion in riveting area
You wrote: Cemistry help? to much math? Too much math, and not enough spelling and grammarPerhaps you meant: Chemistry help? Too much math? Nonetheless, aluminum is Al, A1 is steak sauceAll the business with the masses of protons, neutrons and electrons is just silliness, because the mass of an aluminum atom is not the mass of its parts There is something called the mass defect which represents the missing mass', the mass that corresponds to the binding energyA better solution would be to take the atomic mass of aluminum as given on the periodic table, 26.982 amu, and convert it to grams26.982 amu x (1 g / 6.022x10^23 amu) 4.481x10^-23 g Al Density mass / volume so compute the volume of the hypothetical Al atomV 4/3(pi)(radius^3) where 0.143 nm 1.43 × 10^-8 cm thanks, Google V 1.225 x 10^-23 cm^3 D m/V 4.481x10^-23 g / 1.225 x 10^-23 cm^3 3.66 g/cm^3 Since the actual density is less, then that suggests that there are spaces around the spheres where there is no aluminum, which is the problem with packing spheres close together.
I think you mean tin coated MS sheet or GI sheetIn both cases there are chances of corrosion in riveting area
You wrote: Cemistry help? to much math? Too much math, and not enough spelling and grammarPerhaps you meant: Chemistry help? Too much math? Nonetheless, aluminum is Al, A1 is steak sauceAll the business with the masses of protons, neutrons and electrons is just silliness, because the mass of an aluminum atom is not the mass of its parts There is something called the mass defect which represents the missing mass', the mass that corresponds to the binding energyA better solution would be to take the atomic mass of aluminum as given on the periodic table, 26.982 amu, and convert it to grams26.982 amu x (1 g / 6.022x10^23 amu) 4.481x10^-23 g Al Density mass / volume so compute the volume of the hypothetical Al atomV 4/3(pi)(radius^3) where 0.143 nm 1.43 × 10^-8 cm thanks, Google V 1.225 x 10^-23 cm^3 D m/V 4.481x10^-23 g / 1.225 x 10^-23 cm^3 3.66 g/cm^3 Since the actual density is less, then that suggests that there are spaces around the spheres where there is no aluminum, which is the problem with packing spheres close together.
