The unit cell of copper corresponds to a face centered cube of edge length 3.596 $$\mathring{A}$$ with one copper atom at each lattice point. The calculated density of copper in kg/m$$^3$$ is ______. [Molar mass of Cu: 63.54 g; Avogadro Number $$= 6.022 \times 10^{23}$$]

For a face-centered cubic (FCC) unit cell, there are 4 atoms per unit cell (8 corner atoms contributing $$\frac{1}{8}$$ each, and 6 face atoms contributing $$\frac{1}{2}$$ each: $$8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 4$$).

The density formula for a unit cell is $$\rho = \frac{Z \times M}{N_A \times a^3}$$, where $$Z = 4$$ is the number of atoms per unit cell, $$M = 63.54$$ g/mol is the molar mass of copper, $$N_A = 6.022 \times 10^{23}$$ mol$$^{-1}$$ is Avogadro's number, and $$a = 3.596$$ Å $$= 3.596 \times 10^{-8}$$ cm is the edge length.

First, we calculate $$a^3 = (3.596 \times 10^{-8})^3 = (3.596)^3 \times 10^{-24}$$ cm$$^3$$. Computing $$(3.596)^3 = 3.596 \times 3.596 \times 3.596 = 12.931 \times 3.596 = 46.504$$. So $$a^3 = 46.504 \times 10^{-24}$$ cm$$^3$$.

Substituting into the density formula: $$\rho = \frac{4 \times 63.54}{6.022 \times 10^{23} \times 46.504 \times 10^{-24}} = \frac{254.16}{6.022 \times 46.504 \times 10^{-1}} = \frac{254.16}{28.005} = 9.076$$ g/cm$$^3$$.

Converting to kg/m$$^3$$: $$\rho = 9.076 \times 1000 = 9076$$ kg/m$$^3$$.

The calculated density of copper is $$\boxed{9076}$$ kg/m$$^3$$.