Given: The preparation of potassium permanganate involves two steps. In the first step, manganese dioxide (\( \text{MnO}_2 \)) reacts with potassium hydroxide (KOH) and potassium nitrate (\( \text{KNO}_3 \)) to form potassium manganate (\( \text{K}_2\text{MnO}_4 \)).
The reaction is as follows: \[ \text{MnO}_2 + 4 \text{KOH} + 2 \text{KNO}_3 \to \text{K}_2\text{MnO}_4 + 2 \text{H}_2\text{O} \] - Manganese dioxide reacts with potassium hydroxide and potassium nitrate to produce potassium manganate and water.
The product of the first step is \( \boxed{\text{K}_2\text{MnO}_4} \), which corresponds to option (4).
Let \[ I(x) = \int \frac{dx}{(x-11)^{\frac{11}{13}} (x+15)^{\frac{15}{13}}} \] If \[ I(37) - I(24) = \frac{1}{4} \left( b^{\frac{1}{13}} - c^{\frac{1}{13}} \right) \] where \( b, c \in \mathbb{N} \), then \[ 3(b + c) \] is equal to:
For the thermal decomposition of \( N_2O_5(g) \) at constant volume, the following table can be formed, for the reaction mentioned below: \[ 2 N_2O_5(g) \rightarrow 2 N_2O_4(g) + O_2(g) \] Given: Rate constant for the reaction is \( 4.606 \times 10^{-2} \text{ s}^{-1} \).