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Lead (II) bromide dissociates according to the following equation PbBr2 (s) ↔ Pb2+ + 2 Br-...

Lead (II) bromide dissociates according to the following equation PbBr2 (s) ↔ Pb2+ + 2 Br- Determine the value of Q, the reaction quotient based (also known as the ion product) for a solution with the concentrations below. (Pb2+) = 0.0018 M, (Br-) = 0.0044 M Report your answer in scientific notation with two significant figures. For example 1.2x10-5 would be entered as 1.2e-5

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