Rhenium has an atomic number of 75 - AQA - A-Level Chemistry - Question 2 - 2022 - Paper 1

To find the relative isotopic mass of the second isotope, we can set up the equation using the relative abundances:

Let the relative isotopic mass of ¹⁸⁷Re be x.

total relative abundance = 10 + 17 = 27

Using the formula for relative atomic mass:

o Relative atomic mass = rac{(Isotopic mass_1 imes Abundance_1 + Isotopic mass_2 imes Abundance_2)}{Total abundance}

So,

186.3 = \frac{(185 \times 10 + x \times 17)}{27}

Multiply through by 27:

186.3 \times 27 = 1850 + 17x

5030.1 = 1850 + 17x

So, solving for x:

17x = 5030.1 - 1850

17x = 3180.1

x = \frac{3180.1}{17}

x \approx 186.0

Thus, the relative isotopic mass of the other rhenium isotope is approximately 186.0.

The isotopes of rhenium have the same chemical properties because they have the same number of electrons and the same electron configuration, which determines chemical behavior.

First, calculate the mass of the ¹⁸⁷Re ion using the equation:

a mass_ = \frac{Relative\ isotopic\ mass}{Avogadro\ constant} = \frac{187}{6.022 \times 10^{23}} \approx 3.1 \times 10^{-25} kg.

Now, using the kinetic energy equation, rearrange to find v:

KE = ½ mv² \Rightarrow v = \sqrt{\frac{2 \times KE}{m}}

Substituting the values:

v = \sqrt{\frac{2 \times 1.153 \times 10^{-13} J}{3.1 \times 10^{-25} kg}} \approx 6.67 \times 10^{6} m/s.

Finally, calculate the time:

time = \frac{Distance}{Speed} = \frac{1.450 m}{6.67 \times 10^{6} m/s} \approx 2.17 \times 10^{-7} s.

The relative abundance of ¹⁸⁷Re is determined by the time taken for the ion to travel through the flight tube, as lighter ions arrive at the detector faster than heavier ions, which leads to a measurement of the number of ions detected and consequently their abundance.