Question

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?

Answer 1

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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