1. Balance the following redox reaction in acidic conditions and identify the oxidizing and reducing agents:

\[ \text{S(s)} + \text{NO}_3^- (\text{aq}) \rightarrow \text{SO}_3^{2-} (\text{aq}) + \text{NO(g)} \]

2. Determine the final value of $ n $ in a hydrogen atom transition if the electron starts in $ n=4 $ and the atom emits a photon of light with a wavelength of 97.3 nm.

The balanced redox reaction is 3S(s) + 4NO3^-(aq) -> 3SO3 2^(aq) + 2NO(g) + 2NO2(g) + H2O(l), with S(s) as the reducing agent and NO3^- as the oxidizing agent. When applying the Rydberg formula to the calculation of the final quantum number in a hydrogen atom transition, we find that n = 2.

Explanation:

The first part of the question involves a redox reaction taking place in acidic conditions. The balanced equation is 3S(s) + 4NO3^-(aq) -> 3SO3 2^(aq) + 2NO(g) + 2NO2(g) + H2O(l). Here, S(s) is the reducing agent as it loses electrons and NO3^- is the oxidizing agent as it gains electrons.

Regarding the second part, it involves the calculation of the final value of n in the transition of an electron in a hydrogen atom. We would use the Rydberg formula, which is 1/λ = R(1/n1^2 - 1/n2^2). By rearranging and inputting the given data, we find that n2 = 2. Therefore, the final value of n in the transition is 2.

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