inder: Q3 HO Chemistry - Law of Conservation of Matter 4. \( \mathrm{KOH}+\ldots \mathrm{HBr} \rightarrow \) KBr \( +\ldots \quad \mathrm{H}_{2} \mathrm{O} \) Interpretation:

\[ \mathrm{KOH} + \mathrm{HBr} \rightarrow \mathrm{KBr} + \mathrm{H_2O} \] 1. First, write the unbalanced reaction: \[ \mathrm{KOH} + \mathrm{HBr} \rightarrow \mathrm{KBr} + \mathrm{H_2O} \] 2. Count the atoms on the left-hand side: - Potassium (K): 1 from \(\mathrm{KOH}\) - Oxygen (O): 1 from \(\mathrm{KOH}\) - Hydrogen (H): 1 from \(\mathrm{KOH}\) \(+\) 1 from \(\mathrm{HBr}\) \(=\) 2 - Bromine (Br): 1 from \(\mathrm{HBr}\) 3. Count the atoms on the right-hand side: - Potassium (K): 1 from \(\mathrm{KBr}\) - Bromine (Br): 1 from \(\mathrm{KBr}\) - Oxygen (O): 1 from \(\mathrm{H_2O}\) - Hydrogen (H): 2 from \(\mathrm{H_2O}\) 4. Both sides have the same number of each type of atom. This confirms that the reaction is balanced in accordance with the Law of Conservation of Matter, which states that matter cannot be created or destroyed, only transformed. Thus, the balanced equation is: \[ \mathrm{KOH} + \mathrm{HBr} \rightarrow \mathrm{KBr} + \mathrm{H_2O} \]