Solution to Set 4: Difference between revisions
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<math>{1.40\times 10^5 mol\over 1 m^3} {6.022\times 10^{23}\over 1 mol}= 8.43\times 10^{28} atoms</math> | <math>{1.40\times 10^5 mol\over 1 m^3} {6.022\times 10^{23}\over 1 mol}= 8.43\times 10^{28} atoms</math> | ||
c. Since the bonds between the atoms are small | c. Since the bonds between the atoms are small compared to the diameter of the atom, we neglect the bonds for an estimate.The structure is fcc and the length of each side of the cube is | ||
The structure is fcc and the length of each side of the cube is | <math>2r=362 pm</math>, where r is the radius of one Cu atom. | ||
<math>2r=362 pm</math>, where r is the radius of one Cu atom | |||
d. Atomic radius | |||
<math>p = {m_a\over v_s)</math>, where <math>v_s = {4\over 3} \pi r^3</math> | |||
Revision as of 17:54, 20 February 2009
Problem 1
Cu, density , atomic mass , fcc structure
a. In , the number of moles is
b. Atoms in
c. Since the bonds between the atoms are small compared to the diameter of the atom, we neglect the bonds for an estimate.The structure is fcc and the length of each side of the cube is , where r is the radius of one Cu atom.
d. Atomic radius Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p = {m_a\over v_s)} , where
