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A single gold atom of a particular isotope has a mass of 197.0 amu (a rarely used unit of mass which you are not required to know for this class, but which gives you a chance to deal with unfamiliar units; 1amu=1.66×10 −24

g ) (a) How many gold atoms are in a cube of this gold isotope, which is 10.0 cm on each side, if the density of gold is 19.3 g/cm 3
? Report your answer in scientific notation - and remember, you won't be reminded in the future that all answers are to be reported with the correct significant digits. (b) How many electrons are in the cube?

User Peacedog
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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.

Learn more about Gold_atom_calculation

User Luca Masera
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