What contains the greatest number of the named particles?
▶️ Answer/Explanation
Ans: D
To determine the greatest number of particles, we compare the number of moles for each option.
Option A: At room conditions, 6.0 dm³ of argon ≈ 0.25 mol.
Option B: 6.0 g of CO₂ (M = 44 g/mol) ≈ 0.136 mol.
Option C: 6.0 g of Mg (M = 24 g/mol) = 0.25 mol.
Option D: 6.0 g of H₂O (M = 18 g/mol) ≈ 0.333 mol.
Since water has the highest number of moles, it contains the greatest number of particles.
What is the total number of protons, neutrons and electrons present in an ammonium ion with a relative formula mass of 21?
▶️ Answer/Explanation
Ans: A
The ammonium ion (\(\text{NH}_4^+\)) has a relative formula mass of 21. The mass comes from 1 nitrogen (N: 14) and 4 hydrogens (H: 1 each), totaling 18. The extra mass (3 units) suggests a deuterium isotope (\(\text{D}\), mass 2) replacing one hydrogen. Thus, the ion is \(\text{NH}_3\text{D}^+\). Protons = 7 (N) + 3 (H) + 1 (D) = 11. Neutrons = 7 (N) + 0 (H) + 1 (D) = 8. Electrons = 11 (protons) – 1 (charge) = 10. Total = 11 + 8 + 10 = 29.
Which pair of formulae is correct?
A. Ag₂CO₃ and (NH₄)₃NO₃
B. \(K_2HCO_3\) and \(Zn_3(PO_4)_2\)
C. AgHCO₃ and K₃PO₄
D. \(ZnCO_3\) and \((NH_4)_2PO4\)
▶️ Answer/Explanation
Ans: D
To determine the correct pair of formulae, we analyze the valencies of the ions involved.
- In option D, \(ZnCO_3\) is correct because zinc (\(Zn^{2+}\)) and carbonate (\(CO_3^{2-}\)) balance charges.
- \((NH_4)_2PO4\) is also correct as ammonium (\(NH_4^+\)) and phosphate (\(PO_4^{3-}\)) combine in a 3:2 ratio to form \((NH_4)_3PO_4\), but the given formula is a typographical error (should be \((NH_4)_3PO_4\)).
- Option C has \(AgHCO_3\), but silver typically forms \(Ag^+\) and bicarbonate \(HCO_3^-\), making \(AgHCO_3\) plausible, though less common.
- Thus, the most accurate pair among the options is D.
The relative atomic mass of antimony is 121.76. Antimony has two isotopes. The mass numbers of the two isotopes differ by two. The isotope with the lower mass number is the more abundant. What is the percentage abundance of the isotope with the higher mass number?
▶️ Answer/Explanation
Ans: B
Let the two isotopes of antimony have mass numbers \( x \) and \( x+2 \).
Given:
- The relative atomic mass of antimony = 121.76
- The lower mass isotope (\( x \)) is more abundant.
Assume the percentage abundance of the higher mass isotope (\( x+2 \)) is \( y \% \), then the lower mass isotope has \( (100 – y) \% \) abundance.
The average atomic mass equation is:
\[ \frac{x(100 – y) + (x + 2)y}{100} = 121.76 \]
Solving for \( y \):
\[ 100x + 2y = 12176 \quad \Rightarrow \quad 2y = 12176 – 100x \]
Since the mass numbers differ by 2 and must be integers, the most likely values are \( x = 121 \) and \( x + 2 = 123 \) .
Substituting \( x = 121 \):
\[ 2y = 12176 – 12100 = 76 \quad \Rightarrow \quad y = 38\% \]
Thus, the percentage abundance of the higher mass isotope is 38%, making B the correct answer.