Final answer:
The cube of gold contains 5.89 x 10^25 gold atoms, and 4.66 x10^27 electrons.
Step-by-step explanation:
The first step to find the number of gold atoms in the cube is to calculate the mass of the cube. Given that the density of gold is 19.3 g/cm^3 and the size of the cube is 10 cm on each side, the cube's volume is 1000 cm^3 . Therefore, the mass of the gold cube is 19.3 g/cm^3 x 1000 cm^3 = 19300 g . Next, convert this mass into atomic mass units (amu) as 1 amu = 1.66x10^-24 g. Hence, the mass of the cube in amu is 19300 g x (1/ 1.66x10^-24 g) = 1.16x10^28 amu. Since each gold atom has a mass of 197.0 amu, the number of gold atoms in the cube is 1.16x10^28 amu/ 197.0 amu = 5.89 x 10^25 atoms.
Part b asks for the number of electrons in the cube. Each gold atom has 79 electrons (since it's atomic number is 79). The total number of electrons is thus 79 x 5.89 x 10^25 = 4.66 x10^27 electrons.
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