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A solution of permanganate is standardized by titration with oxalic acid $\left( \mathrm { H } _ { 2 } \mathrm { C } _ { 2 } \mathrm { O } _ { 4 } \right).$ It required 28.97 $\ \mathrm { mL}$ of the permanganate solution to react completely with 0.1058 g of oxalic acid. The unbalanced equation for the reaction is

\mathrm { MnO } _ { 4 } ^ { - } ( a q ) + \mathrm { H } _ { 2 } \mathrm { C } _ { 2 } \mathrm { O } _ { 4 } ( a q ) \stackrel { \mathrm {Acidic} } { \longrightarrow } \mathrm { Mn } ^ { 2 + } ( a q ) + \mathrm { CO } _ { 2 } ( \mathrm { g } )

What is the molarity of the permanganate solution?

A system is tested for faults once per hour. If there is no fault, none will be detected. If there is a fault, the probability is 0.8 that it will be detected. The tests are independent of one another. a. If there is a fault, what is the probability that it will be detected in 3 hours or less? b. Given that a fault has gone undetected for 2 hours, what is the probability that it will be detected in the next hour? c. What is the mean number of tests that must be conducted in order to detect a fault?

On August 10, a credit card account had a balance of $345. A purchase of$56 was made on August 15, and $157 was charged on August 27. A payment of$75 was made on August 15. The interest on the average daily balance is 1.25% per month. Find the finance charge on the September 10 bill.

Question

A solution of permanganate is standardized by titration with oxalic acid $\left(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\right).$ It required 28.97 mL of the permanganate solution to react completely with 0.1058 g of oxalic acid. The unbalanced equation for the reaction is

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Answered 1 year ago

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First we need to determine oxidation numbers.

$\text{MnO}_4^{-} (\text{aq})+ \text{H}_2\text{C}_2\text{O}_4 (\text{aq}) \rightarrow \text{Mn}^{+2} (\text{aq}) + \text{CO}_2 (\text{g})$

$\textbf{Oxygen}$ is always -2 except in peroxides. Compounds must always have neutral charge. Knowing that we can calculate oxidation number for any compound. For $MnO_4^{-}$ we see that we have 4 molecules of oxygen so its oxidation number $-2\times{4}=-8$ but we see a negative charge so we add $+1$ and now we have $-8+1=-7$ .

As we mentioned, compounds always need to have neutral charge, so we need to cancel that negative charge, so that means that oxidation number for $\textbf{Manganese}$ is +7.

Using that logic we calculate oxidation numbers for the rest of the elements:

$\textbf{Hydrogen}$ is always +1 and we know that Oxygen is -2, so we calculate that the oxidation number for $\textbf{Carbon}$ is +3.

For our products, we have already written charge for $\textbf{Manganese}$ +2, and we calculate that the oxidation number for $\textbf{Carbon}$ is +4.