Acidified \(KMnO_4\) oxidizes sulphite to:
\(SO_3^{2-} \)
\(SO_4^{2-} \)
\(SO_2(g) \)
\(S_2O_8^{2-} \)
Solution:
The reaction between acidified potassium permanganate (KMnO4) and sulphite (SO32-) occurs as an oxidation-reduction (redox) process. In this reaction, KMnO4 acts as an oxidizing agent, converting the sulphite ion to another form.
The oxidation state of sulphur in SO32- is +4. The goal of the oxidation process is to increase the oxidation state of sulphur. Acidified KMnO4 effectively oxidizes sulphite ions to peroxodisulfate ions (S2O82-). This involves the formation of a compound where two sulphur atoms are bonded via an oxygen-oxygen bond, and each sulphur atom is in the +6 oxidation state.
The balanced redox reaction is:
2KMnO4 + 5SO32- + 6H+ → 2Mn2+ + 5SO42- + 3H2O
Correct answer:
S2O82-
Convert Propanoic acid to Ethane
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 \(P,O,Q\).