Problem: You are given a cube of silver metal that measures 1.050 cm on each edge. The density of silver is 10.5 g/cm3.How many atoms are in this cube?

FREE Expert Solution

We’re asked to determine how many atoms are present in a cube of silver metal that measures 1.050 cm on each edge given its density.

We can calculate the # of Ag (silver) atoms using the following steps:

Step 1: Determine the volume of the silver cube. Recall that the volume of a cube is given by the equation:

$\overline{){\mathbf{Volume}}{\mathbf{=}}{{\mathbf{l}}}^{{\mathbf{3}}}}$

l = length of edge of a cube

Step 2: Determine the mass of the silver cube from the given density from the equation:

$\overline{){\mathbf{density}}{\mathbf{=}}\frac{\mathbf{mass}}{\mathbf{volume}}}\phantom{\rule{0ex}{0ex}}$

This means that: Mass = density x volume

Step 3: Determine atoms Ag. The flow of the solution will be like this:

mass Ag  (molar mass Ag)  → moles Ag (Avogadro’s number)  → atoms Ag

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Problem Details

You are given a cube of silver metal that measures 1.050 cm on each edge. The density of silver is 10.5 g/cm3.

How many atoms are in this cube?