In this question Q is used to represent a halogen atom. Magnesium and calcium each form a compound with chlorine and a compound with bromine.
One of these compounds contains:
● the element in Group 2 with the higher first ionisation energy and
● the element in Group 17 with the higher Q–Q bond energy.
What is the formula of this compound?
▶️ Answer/Explanation
Ans: A
Magnesium (Mg) has a higher first ionisation energy than calcium (Ca) because ionisation energy decreases down Group 2. Chlorine (Cl) has a higher Cl–Cl bond energy than bromine (Br) because bond energy decreases down Group 17. Therefore, the compound must contain Mg and Cl, giving the formula MgCl₂ (Option A).
In this question you may assume that nitrogen behaves as an ideal gas. One atmosphere pressure = 101 kPa. Which volume does 1.0 g of nitrogen occupy at 50°C and a pressure of 2.0 atmospheres?
▶️ Answer/Explanation
Ans: C
Using the ideal gas equation \( pV = nRT \), we first find the number of moles \( n \) of nitrogen: \( n = \frac{1.0 \text{ g}}{28 \text{ g/mol}} = 0.0357 \text{ mol} \). The pressure \( p = 2.0 \text{ atm} = 202 \text{ kPa} \), and temperature \( T = 50°C = 323 \text{ K} \). Substituting into the equation, \( V = \frac{nRT}{p} = \frac{0.0357 \times 8.314 \times 323}{202} \approx 0.47 \text{ L} = 470 \text{ cm}^3 \). Thus, the correct answer is C.
A 0.216 g sample of aluminium carbide reacts with an excess of water to produce methane gas. This is the only carbon-containing product formed in the reaction. This methane gas burns completely in \(O_2\) to form \(H_2O\) and \(CO_2\) only. The volume of \(CO_2\) produced at room temperature and pressure is 108 cm³. What is the formula of aluminium carbide?
▶️ Answer/Explanation
Ans: D
1. The reaction of aluminium carbide (\(Al_xC_y\)) with water produces methane (\(CH_4\)): \[ Al_xC_y + H_2O \rightarrow Al(OH)_3 + CH_4 \]
2. The methane burns to form \(CO_2\): \[ CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O \]
3. Given \(108 \, cm^3\) of \(CO_2\) at RTP (24,000 cm³/mol), moles of \(CO_2\) = \(\frac{108}{24000} = 0.0045 \, mol\). This equals moles of \(CH_4\) and thus moles of carbon in \(Al_xC_y\).
4. Moles of aluminium carbide = \(\frac{0.216}{27x + 12y}\). Since \(y \times \text{moles of } Al_xC_y = 0.0045\), solving gives \(x:y \approx 4:3\), confirming \(Al_4C_3\).
Which of these samples of gas contains the same number of atoms as 1 g of hydrogen gas?
▶️ Answer/Explanation
Ans: C
1 g of hydrogen gas (\(H_2\), \(M_r = 2\)) contains \(\frac{1}{2} \times N_A\) molecules, where \(N_A\) is Avogadro’s number. Since each \(H_2\) molecule has 2 atoms, the total number of atoms is \(N_A\). For neon (Ne, \(M_r = 20\)), 20 g contains \(N_A\) atoms. Thus, 20 g of neon has the same number of atoms as 1 g of hydrogen gas.