Explain the type of reaction represented by the following equation:
$(i) \ CuSO_{4} + 2NaOH \longrightarrow Cu(OH)_{2} + Na_{2}SO_{4}$
$(i) \ CuSO_{4} + 2NaOH \longrightarrow Cu(OH)_{2} + Na_{2}SO_{4}$
(N/A) The given reaction is a $Double \ Displacement \ Reaction$.
In a double displacement reaction, two reactant compounds exchange their ions to form two new compounds.
In this specific reaction, the $Cu^{2+}$ ion from copper sulfate exchanges with the $Na^{+}$ ion from sodium hydroxide to form copper hydroxide $(Cu(OH)_{2})$ and sodium sulfate $(Na_{2}SO_{4})$.
In a double displacement reaction, two reactant compounds exchange their ions to form two new compounds.
In this specific reaction, the $Cu^{2+}$ ion from copper sulfate exchanges with the $Na^{+}$ ion from sodium hydroxide to form copper hydroxide $(Cu(OH)_{2})$ and sodium sulfate $(Na_{2}SO_{4})$.
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