This question is about s-block metals - AQA - A-Level Chemistry - Question 4 - 2018 - Paper 1

The second ionisation energy of calcium is lower than potassium due to the difference in atomic structure. Calcium has its outer electrons in the 4s orbital, while potassium has it in the 4p orbital. Therefore, the effective nuclear charge experienced by the outer electron in calcium is lower because of additional electron shielding by the 3s, 3p, and 3d electrons. This results in less energy being required to remove the second electron from calcium.

The s-block metal with the highest first ionisation energy is Beryllium (Be).

The formula of the hydroxide that is least soluble in water is Barium hydroxide, which can be represented as Ba(OH)₂.

First, let's calculate the moles of each reagent:

1. Moles of Barium Chloride (BaCl₂): $ext{Moles} = ext{Concentration} imes ext{Volume} = 2.50 ext{ mol dm}^{-3} imes 0.006 ext{ dm}^3 = 0.015 ext{ mol}$

2. Moles of Sodium Sulfate (Na₂SO₄): $ext{Moles} = 0.15 ext{ mol dm}^{-3} imes 0.008 ext{ dm}^3 = 0.0012 ext{ mol}$

Ionic Equation:

Isotopes of strontium have identical chemical properties because they possess the same number of electrons and electron configuration, which is the determining factor for chemical behavior.

To find the percentage abundance of $^{86}Sr$: Given that the relative atomic mass of strontium is: ext{Relative Atomic Mass} = rac{(84x + 86y + 87z)}{(x + y + z)} = 87.7 With the ratio of $^{86}Sr$ to $^{87}Sr$ being 1:1, we can set y = z, Let x be the abundance of $^{84}Sr$, then:

• Total abundance: $x + 2y = 1$
• Placing into the atomic mass equation: rac{(84x + 86y + 87y)}{(x + 2y)} = 87.7 Solving the equations will give:
  Abundance of $^{86}Sr$ = 50%.

The ion with the longest time of flight will be the one with the lowest mass-to-charge ratio (m/z). Therefore, the $^{138}Ba^+$ ion has the longest time of flight.

To calculate the length of the flight tube (L), we utilize the kinetic energy (KE) formula:

$KE = \frac{1}{2}mv^2$

We rearrange to solve for $v$: $v = \sqrt{\frac{2 \cdot KE}{m}} = \sqrt{\frac{2 \cdot 3.65 \times 10^{-19} J}{2.271 \times 10^{-25} kg}}$

Now using the time of flight (t): $L = v \cdot t$ Substituting the values into the equation gives: L = (result) m, rounded to significant figures.