ANWSER
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Question 1(a)(ii):
The partial pressures of the gases are calculated using their mole fractions and the total pressure (760 mmHg):
– CO₂: \(0.050 \times 760 = 38.0 \text{ mmHg}\)
– H₂O: \(0.250 \times 760 = 190.0 \text{ mmHg}\)
– HCl: \(0.051 \times 760 = 38.76 \text{ mmHg}\)
– HF: \(0.028 \times 760 = 21.28 \text{ mmHg}\)
– SO₂: \(0.017 \times 760 = 12.92 \text{ mmHg}\)
– H₂: \(0.001 \times 760 = 0.76 \text{ mmHg}\)
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Question 1(a)(iii):
Using the balanced equation \(4NH_3 + 5O_2 \rightarrow 4NO + 6H_2O\):
1. Moles of NH₃: \(n = \frac{PV}{RT} = \frac{1.30 \times 3.00}{0.0821 \times 802} = 0.059 \text{ mol}\)
2. Moles of H₂O produced: \(0.059 \times \frac{6}{4} = 0.0885 \text{ mol}\)
3. Volume of steam at 125°C and 1.00 atm: \(V = \frac{nRT}{P} = \frac{0.0885 \times 0.0821 \times 398}{1.00} = 2.89 \text{ L}\)
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Question 1(b)(i):
– Diffusion: Spreading of gas molecules through a space or another gas.
– Effusion: Escape of gas molecules through a tiny hole into a vacuum.
– Effusion rates: \( \text{Rate}_{\text{He}} / \text{Rate}_{\text{O}_2} = \sqrt{M_{\text{O}_2} / M_{\text{He}}} = \sqrt{32 / 4} = 2.83\). Helium effuses 2.83 times faster than oxygen.
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Question 1(b)(ii):
Using Graham’s Law:
\( \frac{\text{Time}_{\text{unknown}}}{\text{Time}_{\text{Br}_2}}} = \sqrt{\frac{M_{\text{unknown}}}{M_{\text{Br}_2}}} \)
\( \frac{1.50}{4.73} = \sqrt{\frac{M}{159.8}} \)
Solving: \( M = 16.0 \text{ g/mol} \). The gas is likely methane (CH₄).
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Question 1(b)(iii):
– Reaction 1: \( Mg + H_2 \rightarrow MgH_2 \) (A = MgH₂)
– Reaction 2: \( Mg + 2HCl \rightarrow MgCl_2 + H_2 \) (B = H₂, C = MgCl₂)
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Question 1(c)(i):
D₂O is toxic because it disrupts biochemical reactions due to kinetic isotope effects, altering reaction rates in metabolic pathways.
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Question 1(c)(ii):
– Rb₂O: Basic (Group 1 metal oxide).
– BeO: Amphoteric (forms Be²⁺ and O²⁻, reacts with acids/bases).
– As₂O₅: Acidic (non-metal oxide, forms acids in water).
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Question 1(c)(iii):
Hydrogen resembles:
– Group 1A: Can lose 1 electron to form H⁺ (e.g., in acids).
– Group VIIA: Can gain 1 electron to form H⁻ (e.g., in hydrides).
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Question 1(c)(iv):
Na⁺ and Mg²⁺ are stable due to their noble gas configurations. Na²⁺ and Mg³⁺ are unstable because removing additional electrons requires excessive energy.
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Question 12:
Graham’s Law of Diffusion: The rate of effusion/diffusion of a gas is inversely proportional to the square root of its molar mass: \( \text{Rate} \propto \frac{1}{\sqrt{M}} \).
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Question 14:
| Hybrid | Geometry | Angles |
|———-|—————|————–|
| sp | Linear | 180° |
| sp³d | Trigonal bipyramidal | 90°, 120° |
| sp² | Trigonal planar | 120° |
| sp³d² | Octahedral | 90° |
| sp³ | Tetrahedral | 109.5° |
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Question 3 (SO₄²⁻ Lewis structure):
– Central S atom with 4 O atoms (2 single bonds, 2 double bonds), and 2 extra electrons for the -2 charge.
Question 5 (HCO₂⁻ Lewis structure):
– Central C atom bonded to H, 2 O atoms (1 double bond, 1 single bond with negative charge).
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Question 6:
(a) HgCl₂: sp hybridization (linear).
(c) PF₃: sp³ hybridization (trigonal pyramidal).
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Question 7:
Units of \( R \): \( \text{L·atm·K}^{-1}\text{·mol}^{-1} \).
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Question 8:
Density of butane:
\( \rho = \frac{PM}{RT} = \frac{177.4 \times 58.12}{8.314 \times 125} = 0.991 \text{ g/L} \).
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Question 9:
Charles’ Law Graph: \( V \) vs. \( T \) (K) is a straight line passing through the origin.
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Question 10:
Graph of P vs. PV: A horizontal line (since PV is constant for an ideal gas at fixed \( n \) and \( T \)).
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Question 11:
(a) Ideal Gas Law: \( P = \frac{nRT}{V} = \frac{(10/16) \times 0.0821 \times 296}{1.00} = 15.2 \text{ atm} \).
(b) Van der Waals: \( \left(P + \frac{a n^2}{V^2}\right)(V – nb) = nRT \). Solving gives \( P \approx 14.8 \text{ atm} \).
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Question 12:
– Empirical Formula: C₆H₁₀S₂O (from mass percentages).
– Molecular Formula: C₁₂H₂₀S₄O₂ (molar mass = 162 g/mol).
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Question 13:
Balmer series wavelengths (\( n_f = 2 \)):
(a) \( n = 3 \): \( \lambda = 656 \text{ nm} \).
(b) \( n = 4 \): \( \lambda = 486 \text{ nm} \).
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Question 20(a):
Partial pressure of H₂: \( 750 \text{ mmHg} – 24 \text{ mmHg} = 726 \text{ mmHg} \).
Question 20(b):
Moles of H₂: \( n = \frac{PV}{RT} = \frac{(726/760) \times 0.335}{0.0821 \times 298} = 0.013 \text{ mol} \).
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Question 21:
Total pressure:
– Moles of H₂ = \( \frac{1.00}{2.02} = 0.495 \text{ mol} \).
– Moles of O₂ = \( \frac{1.00}{32.0} = 0.031 \text{ mol} \).
– Moles of N₂ = \( \frac{1.00}{28.0} = 0.036 \text{ mol} \).
– \( P_{\text{total}} = \frac{(0.495 + 0.031 + 0.036) \times 0.0821 \times 398}{10.0} = 1.83 \text{ atm} \).
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Question 22:
Applications of Radioactivity:
(a) Medical imaging (e.g., PET scans).
(b) Cancer treatment (radiotherapy).
(c) Carbon dating.
(d) Industrial radiography.
(e) Nuclear power generation.
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Question 23:
Binding energy of Fe-56:
(a) \( \text{BE} = \Delta m \times 931.5 \text{ MeV} = 492 \text{ MeV} \).
(b) \( \text{BE/nucleon} = \frac{492}{56} = 8.79 \text{ MeV/nucleon} \).
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Question 24:
Law of Radioactive Decay: The rate of decay is proportional to the number of undecayed nuclei: \( N = N_0 e^{-\lambda t} \).
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Question 25:
(a) Half-life: \( t_{1/2} = \frac{\ln 2}{1.36 \times 10^{-11}} = 1.62 \times 10^{10} \text{ years} \).
(b) Decay rate: \( \lambda N = 1.36 \times 10^{-11} \times \frac{1}{226} \times 6.022 \times 10^{23} = 3.62 \times 10^{10} \text{ atoms/s} \).
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Question 26:
Rate constant: \( \lambda = \frac{\ln 2}{27.72 \times 86400} = 2.89 \times 10^{-7} \text{ s}^{-1} \).
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Question 27:
Pressure of O₂: \( P = \frac{nRT}{V} = \frac{(5.00/32.0) \times 0.0821 \times 308}{6.0} = 0.658 \text{ atm} = 500 \text{ mmHg} \).
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Question 28:
Moles of CO₂: \( n = \frac{PV}{RT} = \frac{(328/760) \times 0.168}{0.0821 \times 262} = 0.00338 \text{ mol} \).
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Question 29:
(a) After 7.0 days: \( m = 4.00 \times e^{-\ln 2 \times 7.0/2.62} = 0.70 \text{ mg} \).
(b) After 30 days: \( m = 4.00 \times e^{-\ln 2 \times 30/2.62} = 0.0002 \text{ mg} \).
(c) Time for 0.40 mg: \( t = \frac{2.62 \times \ln(4.00/0.40)}{\ln 2} = 8.7 \text{ days} \).
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