To understand the preparation of potassium permanganate (\( KMnO_4 \)) from manganese dioxide (\( MnO_2 \)), we need to look at the multi-step reaction process involved. This involves the following steps:
\(MnO_2 + 2KOH + \frac{1}{2}O_2 \rightarrow K_2MnO_4 + H_2O\)
As per the question, the product of this first step is potassium manganate (\( K_2MnO_4 \)). This is an intermediate in the preparation of potassium permanganate.
\(3K_2MnO_4 + 2CO_2 \rightarrow 2KMnO_4 + 2K_2CO_3 + MnO_2\)
With this understanding, we can conclude that the correct product formed in the first step is potassium manganate (\( K_2MnO_4 \)).
Let's rule out the incorrect options:
Therefore, the correct answer is \( K_2{MnO}_4 \).
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 \( f : \mathbb{R} \to \mathbb{R} \) be a twice differentiable function such that \[ (\sin x \cos y)(f(2x + 2y) - f(2x - 2y)) = (\cos x \sin y)(f(2x + 2y) + f(2x - 2y)), \] for all \( x, y \in \mathbb{R}. \)
If \( f'(0) = \frac{1}{2} \), then the value of \( 24f''\left( \frac{5\pi}{3} \right) \) is: