To solve the problem, we need to determine what product is formed when acidified KMnO4 oxidizes sulphite.
1. Understanding the Reaction:
In an acidic medium, potassium permanganate (KMnO4) is a strong oxidizing agent. It oxidizes sulphite ions (SO32-) to sulphate ions (SO42-).
2. Oxidation of Sulphite:
The sulphite ion (SO32-) gets oxidized by the KMnO4 in the presence of an acidic medium. The manganese in KMnO4 undergoes reduction from +7 oxidation state to +2 oxidation state, forming Mn2+.
3. Balanced Chemical Equation:
The balanced redox reaction in acidic solution is:
2 KMnO4 + 5 SO32- + 4 H2O → 2 Mn2+ + 5 SO42- + 4 OH-
Final Answer:
Acidified KMnO4 oxidizes sulphite (SO32-) to sulphate (SO42-).
Convert Propanoic acid to Ethane
Acidified \(KMnO_4\) oxidizes sulphite to:
A racing track is built around an elliptical ground whose equation is given by \[ 9x^2 + 16y^2 = 144 \] The width of the track is \(3\) m as shown. Based on the given information answer the following: 
(i) Express \(y\) as a function of \(x\) from the given equation of ellipse.
(ii) Integrate the function obtained in (i) with respect to \(x\).
(iii)(a) Find the area of the region enclosed within the elliptical ground excluding the track using integration.
OR
(iii)(b) Write the coordinates of the points \(P\) and \(Q\) where the outer edge of the track cuts \(x\)-axis and \(y\)-axis in first quadrant and find the area of triangle formed by points \(P,O,Q\).