To solve this problem, we need to determine the number of moles of \(H^+\) ions required for the reaction where \(MnO_4^-\) oxidizes oxalate ions \((C_2O_4^{2-})\) to carbon dioxide \((CO_2)\). The balanced redox reaction in an acidic medium is:
\(2MnO_4^- + 5C_2O_4^{2-} + 16H^+ \rightarrow 2Mn^{2+} + 10CO_2 + 8H_2O\)
Let's break down the stoichiometric coefficients:
Hence, the number of moles of \(H^+\) ions required by 1 mole of \(MnO_4^-\) is 8.
This result is verified as falling within the provided range of 8 to 8.
The balanced reaction is:
\[ 2 \text{MnO}_4^- + 5 \text{C}_2\text{O}_4^{2-} + 16 \text{H}^+ \rightarrow 2 \text{Mn}^{2+} + 10 \text{CO}_2 + 8 \text{H}_2\text{O} \]From the stoichiometry, we see that 16 moles of \(\text{H}^+\) are required for 2 moles of \(\text{MnO}_4^-\). Therefore, 8 moles of \(\text{H}^+\) are needed for each mole of \(\text{MnO}_4^-\).
200 cc of $x \times 10^{-3}$ M potassium dichromate is required to oxidise 750 cc of 0.6 M Mohr's salt solution in acidic medium. Here x = ______ .

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: $g = 10\,\text{m s}^{-2}$) 
The equivalent resistance between the points \(A\) and \(B\) in the given circuit is \[ \frac{x}{5}\,\Omega. \] Find the value of \(x\). 
Method used for separation of mixture of products (B and C) obtained in the following reaction is: 