Assign E or Z to the following molecule:

Step 1: Identify the groups attached to each carbon atom of the double bond. The molecule is: $Cl \quad \quad Br$ $\quad \backslash \quad /$ $\quad \quad C = C$ $\quad / \quad \backslash$ $\quad H \quad \quad Cl$

Step 2: Assign priorities to the groups on each carbon atom using the Cahn-Ingold-Prelog (CIP) rules. Priority is determined by the atomic number of the atom directly attached to the double bond carbon. Higher atomic number means higher priority.

For the left carbon atom: • Group 1: Chlorine (Cl), atomic number = 17. • Group 2: Hydrogen (H), atomic number = 1. Priority: Cl (high) > H (low).

For the right carbon atom: • Group 1: Bromine (Br), atomic number = 35. • Group 2: Chlorine (Cl), atomic number = 17. Priority: Br (high) > Cl (low).

Step 3: Compare the positions of the higher priority groups on each side of the double bond. • On the left carbon, the high priority group (Cl) is on the top. • On the right carbon, the high priority group (Br) is on the top.

Since both higher priority groups (Cl and Br) are on the same side of the double bond, the configuration is Z (from German "zusammen," meaning "together").

The final answer is $\boxed{Z}$.