Question

part a
choose the sketch of the mass spectrum of rhenium where the total intensity will then be normalized to give the data in the table

Explanation:

To solve this, we need to recall how mass spectra for isotopes work, especially normalization. The key is the relative intensities of the isotopes of rhenium (Re has isotopes \(^{185}\text{Re}\) and \(^{187}\text{Re}\) with natural abundances: \(^{185}\text{Re}\) is about 37.4% and \(^{187}\text{Re}\) is about 62.6%). When normalized, the most abundant isotope is set to 100% intensity. So \(^{187}\text{Re}\) (more abundant) should have the 100% peak, and \(^{185}\text{Re}\) should be around \(\frac{37.4}{62.6}\approx 60\%\) of that (since 37.4/62.6 ≈ 0.6, so ~60% relative to 100% for \(^{187}\text{Re}\)).

Looking at the graphs:

• The second graph (with \(^{187}\text{Re}\) at 100% and \(^{185}\text{Re}\) at ~60% intensity) matches the relative abundances. The first graph has both ~50%, third has \(^{185}\text{Re}\) lower, fourth has \(^{185}\text{Re}\) higher (100%, \(^{187}\text{Re}\) lower) which is wrong.

Rhenium has isotopes \(^{185}\text{Re}\) (≈37.4% abundance) and \(^{187}\text{Re}\) (≈62.6% abundance). In a normalized mass spectrum, the more abundant isotope (\(^{187}\text{Re}\)) is set to 100% intensity. The \(^{185}\text{Re}\) intensity relative to \(^{187}\text{Re}\) is \(\frac{37.4}{62.6}\approx 60\%\), so the graph with \(^{187}\text{Re}\) at 100% and \(^{185}\text{Re}\) at ~60% intensity (the second graph) is correct.

Answer:

The second graph (with \(^{187}\text{Re}\) at 100% intensity and \(^{185}\text{Re}\) at ~60% intensity, labeled with the second radio button)