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).
In the given figure, the blocks $A$, $B$ and $C$ weigh $4\,\text{kg}$, $6\,\text{kg}$ and $8\,\text{kg}$ respectively. The coefficient of sliding friction between any two surfaces is $0.5$. The force $\vec{F}$ required to slide the block $C$ with constant speed is ___ N.
(Given: $g = 10\,\text{m s}^{-2}$) 
Two circular discs of radius \(10\) cm each are joined at their centres by a rod, as shown in the figure. The length of the rod is \(30\) cm and its mass is \(600\) g. The mass of each disc is also \(600\) g. If the applied torque between the two discs is \(43\times10^{-7}\) dyne·cm, then the angular acceleration of the system about the given axis \(AB\) is ________ rad s\(^{-2}\).
