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 that of potassium due to the following reasons:

1. In Ca²⁺, the outer electron is being removed from a lower energy level than in K, reflecting the overall more effective nuclear charge experienced by K due to its larger outer shell.

2. Additionally, there is more electron shielding in potassium, whereby the inner electrons repel the outer electrons, making them easier to remove.

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

The formula of the hydroxide that is least soluble in water among Group 2 elements is:

$\text{Mg(OH)}_2$.

Step 1: Calculate moles of BaCl₂ and Na₂SO₄:

• Moles of BaCl₂ = Volume × Concentration = $6 \text{cm}^3 \times \frac{0.25 \text{mol}}{1000 \text{cm}^3} = 0.0015 \text{mol}$

• Moles of Na₂SO₄ = $8 \text{cm}^3 \times \frac{0.15 \text{mol}}{1000 \text{cm}^3} = 0.0012 \text{mol}$

Step 2: Write the balanced equation:

$\text{Ba}^{2+} + \text{Na}_2\text{SO}_4 \rightarrow \text{BaSO}_4(s) + 2\text{Na}^{+}$

From the equation, 1 mole of Ba²⁺ reacts with 1 mole of Na₂SO₄:

• Excess reagent: Barium Chloride ( ext{BaCl₂})

Ionic equation: $\text{Ba}^{2+} + \text{SO}_{4}^{2-} \rightarrow \text{BaSO}_4(s)$

Reagent in excess: Barium Chloride. Total volume of other reagent:

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The amount of Na₂SO₄ needed to completely react with 0.0015 mol BaCl₂:

• Moles required = 0.0015 mol (which means 0.0015 L) Therefore, volume of Na₂SO₄ = $0.0012 \text{mol} / 0.15 = 8 \text{cm}^{3}$

Isotopes have identical chemical properties because they have the same number of electrons and hence the same electron configuration.

To find the percentage abundance of the ⁸⁴Sr isotope:

1. Assume the abundance of ⁸⁴Sr is $x$.

2. As per the average atomic mass, we have: $x(84) + x(86) + (100-2x)(88) = 87.7$

3. This simplifies to:

   $84x + 86x + 8800 - 176x = 87.7$

4. Solving this gives:

   • First combine like terms: $-6x = 87.7 - 8800$

5. And solving for x gives the exact abundance.

The ion with the longest time of flight will be the heaviest ion, usually, since time of flight inversely relates to mass. Thus, the ion with the longest time of flight is ¹³⁰Ba.

1. Using the formula for kinetic energy: $KE = \frac{1}{2} mv^2$
2. Rearranging for v, we have: $v = \sqrt{\frac{2 \cdot KE}{m}}$
3. The mass of ¹³⁷Ba = $\frac{137}{6.022 \times 10^{23}}$ kg
4. Substituting into the kinetic energy equation to find the speed and subsequently the length of the tube: $L = vt = \sqrt{\frac{2 \cdot 7.35 \times 10^{-19}}{\text{mass}}} \times 2.71 \times 10^{6}$
5. This calculation gives the length of the flight tube in metres.