A single gold atom has a mass of 197.0 amu. How many gold atoms are in a cube of gold that is 10.0 mm on each side (about the size of a sugar cube) if the density of gold is 19.3 g/cm3

5.90x10²² atoms of gold are present in the cube

Explanation:

First, we must find the volume of the cube in cm³. With density we can find the mass of the gold and the moles using its molar mass. As 1 mol = 6.022x10²³ atoms we can find the number of atoms:

Volume in cm³:

(10.0mm)³ = 1000mm³

1mm³ = 0.001cm³

1000mm³ * (0.001cm³ / 1mm³) = 1cm³

Mass gold:

1cm³ * (19.3g/cm³) = 19.3g Gold

Moles Gold:

19.3g * (1mol / 197.0g) = 0.0980 moles Gold

Atoms gold:

0.0980 moles Gold * (6.022x10²³ atoms / mol) =

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

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