Some Basic Concepts of Chemistry

Given:
- Mass of NaOH = 4 g
- Volume of solution = 250 mL = 0.250 L
- Molar mass of NaOH = 23 + 16 + 1 = 40 g mol⁻¹

Formula:
$$M = \frac{\text{No. of moles of solute}}{\text{Volume of solution in litres}}$$

Step 1: Calculate moles of NaOH.
$$n_{\text{NaOH}} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{4\text{ g}}{40\text{ g mol}^{-1}} = 0.1\text{ mol}$$

Step 2: Calculate molarity.
$$M = \frac{0.1\text{ mol}}{0.250\text{ L}} = 0.4\text{ mol L}^{-1}$$

$$\boxed{M = 0.4\text{ M}}$$

Note: Molarity depends on temperature because volume changes with temperature.

Given:
- Molarity (M) = 3 mol L⁻¹
- Density of solution = 1.25 g mL⁻¹
- Molar mass of NaCl = 23 + 35.5 = 58.5 g mol⁻¹

Step 1: Find mass of NaCl in 1 L of solution.
$$\text{Mass of NaCl} = 3\text{ mol} \times 58.5\text{ g mol}^{-1} = 175.5\text{ g}$$

Step 2: Find mass of 1 L solution.
$$\text{Mass of solution} = 1000\text{ mL} \times 1.25\text{ g mL}^{-1} = 1250\text{ g}$$

Step 3: Find mass of water (solvent).
$$\text{Mass of water} = 1250 - 175.5 = 1074.5\text{ g} = 1.0745\text{ kg}$$

Step 4: Calculate molality.
$$m = \frac{\text{No. of moles of solute}}{\text{Mass of solvent in kg}} = \frac{3\text{ mol}}{1.0745\text{ kg}}$$

$$\boxed{m = 2.79\text{ m}}$$

Note: Molality does not change with temperature since mass is unaffected by temperature.

Concept: Molar mass = sum of atomic masses of all atoms in the molecule.

Atomic masses used: H = 1, O = 16, C = 12

(i) H₂O:
$$M = 2(1) + 1(16) = 2 + 16 = \boxed{18\text{ g mol}^{-1}}$$

(ii) CO₂:
$$M = 1(12) + 2(16) = 12 + 32 = \boxed{44\text{ g mol}^{-1}}$$

(iii) CH₄:
$$M = 1(12) + 4(1) = 12 + 4 = \boxed{16\text{ g mol}^{-1}}$$

Given: Sodium sulphate = Na₂SO₄

Molar mass of Na₂SO₄:
$$M = 2(23) + 1(32) + 4(16) = 46 + 32 + 64 = 142\text{ g mol}^{-1}$$

Formula:
$$\text{Mass per cent of element} = \frac{\text{Mass of element in 1 mol of compound}}{\text{Molar mass of compound}} \times 100$$

Mass % of Na:
$$= \frac{46}{142} \times 100 = \boxed{32.39\%}$$

Mass % of S:
$$= \frac{32}{142} \times 100 = \boxed{22.54\%}$$

Mass % of O:
$$= \frac{64}{142} \times 100 = \boxed{45.07\%}$$

Verification: $32.39 + 22.54 + 45.07 = 100\%$ ✓

Given: Fe = 69.9%, O = 30.1%

Step 1: Convert mass % to moles (assume 100 g sample).
$$\text{Moles of Fe} = \frac{69.9}{55.85} = 1.251\text{ mol}$$
$$\text{Moles of O} = \frac{30.1}{16} = 1.881\text{ mol}$$

Step 2: Divide by the smallest number of moles.
$$\text{Fe}: \frac{1.251}{1.251} = 1.00$$
$$\text{O}: \frac{1.881}{1.251} = 1.504 \approx 1.5$$

Step 3: Multiply by 2 to get whole numbers.
$$\text{Fe}: 1.00 \times 2 = 2$$
$$\text{O}: 1.5 \times 2 = 3$$

Empirical formula: $\boxed{\text{Fe}_2\text{O}_3}$

Balanced equation:
$$\text{C} + \text{O}_2 \rightarrow \text{CO}_2$$
1 mol C reacts with 1 mol O₂ (32 g) to give 1 mol CO₂ (44 g).

(i) 1 mole of carbon burnt in air (excess O₂):
O₂ is in excess in air, so 1 mol C reacts completely.
$$\text{CO}_2 \text{ produced} = 1\text{ mol} = \boxed{44\text{ g}}$$

(ii) 1 mole of carbon burnt in 16 g of O₂:
$$\text{Moles of O}_2 = \frac{16}{32} = 0.5\text{ mol}$$
O₂ is the limiting reagent (only 0.5 mol available, but 1 mol needed).
$$\text{CO}_2 \text{ produced} = 0.5\text{ mol} = \boxed{0.5 \times 44 = 22\text{ g}}$$

(iii) 2 moles of carbon burnt in 16 g of O₂:
$$\text{Moles of O}_2 = \frac{16}{32} = 0.5\text{ mol}$$
2 mol C requires 2 mol O₂, but only 0.5 mol O₂ is available. O₂ is the limiting reagent.
$$\text{CO}_2 \text{ produced} = 0.5\text{ mol} = \boxed{0.5 \times 44 = 22\text{ g}}$$

Given:
- Volume = 500 mL = 0.500 L
- Molarity = 0.375 M
- Molar mass of CH₃COONa = 82.0245 g mol⁻¹

Step 1: Find moles of sodium acetate required.
$$n = M \times V = 0.375\text{ mol L}^{-1} \times 0.500\text{ L} = 0.1875\text{ mol}$$

Step 2: Find mass.
$$\text{Mass} = n \times M_{\text{molar}} = 0.1875\text{ mol} \times 82.0245\text{ g mol}^{-1}$$
$$= \boxed{15.38\text{ g}}$$

Given:
- Density = 1.41 g mL⁻¹
- Mass per cent of HNO₃ = 69%
- Molar mass of HNO₃ = 1 + 14 + 48 = 63 g mol⁻¹

Step 1: Consider 1 L (1000 mL) of solution.
$$\text{Mass of solution} = 1000 \times 1.41 = 1410\text{ g}$$

Step 2: Mass of HNO₃ in 1 L.
$$\text{Mass of HNO}_3 = \frac{69}{100} \times 1410 = 972.9\text{ g}$$

Step 3: Moles of HNO₃.
$$n = \frac{972.9}{63} = 15.44\text{ mol}$$

Step 4: Molarity.
$$M = \frac{15.44\text{ mol}}{1\text{ L}} = \boxed{15.44\text{ mol L}^{-1}}$$

Given: Mass of CuSO₄ = 100 g

Molar mass of CuSO₄:
$$M = 63.5 + 32 + 4(16) = 63.5 + 32 + 64 = 159.5\text{ g mol}^{-1}$$

Step 1: Moles of CuSO₄.
$$n = \frac{100}{159.5} = 0.6270\text{ mol}$$

Step 2: Each mole of CuSO₄ contains 1 mole of Cu.
$$\text{Moles of Cu} = 0.6270\text{ mol}$$

Step 3: Mass of Cu.
$$\text{Mass of Cu} = 0.6270 \times 63.5 = \boxed{39.81\text{ g}}$$

Step 1: Find empirical formula (same calculation as Exercise 1.3).
$$\text{Moles of Fe} = \frac{69.9}{55.85} = 1.251$$
$$\text{Moles of O} = \frac{30.1}{16} = 1.881$$

Ratio Fe : O = $\frac{1.251}{1.251} : \frac{1.881}{1.251} = 1 : 1.504 \approx 2 : 3$

Empirical formula: Fe₂O₃

Step 2: Calculate empirical formula mass.
$$M_{\text{EF}} = 2(55.85) + 3(16) = 111.7 + 48 = 159.7\text{ g mol}^{-1}$$

Step 3: Determine n.
For iron oxides, the molar mass of Fe₂O₃ ≈ 159.7 g mol⁻¹.
$$n = \frac{\text{Molar mass}}{\text{EF mass}} = \frac{159.7}{159.7} = 1$$

Molecular formula: $\boxed{\text{Fe}_2\text{O}_3}$

Given: 3 moles of C₂H₆

Each molecule of C₂H₆ contains 2 C atoms and 6 H atoms.

(i) Moles of carbon atoms:
$$= 3\text{ mol C}_2\text{H}_6 \times 2 = \boxed{6\text{ mol}}$$

(ii) Moles of hydrogen atoms:
$$= 3\text{ mol C}_2\text{H}_6 \times 6 = \boxed{18\text{ mol}}$$

(iii) Number of molecules of ethane:
$$= 3\text{ mol} \times 6.022 \times 10^{23}\text{ mol}^{-1}$$
$$= \boxed{18.066 \times 10^{23} = 1.8066 \times 10^{24}\text{ molecules}}$$

Given:
- Mass of sugar = 20 g
- Volume = 2 L
- Molar mass of C₁₂H₂₂O₁₁ = 12(12) + 22(1) + 11(16) = 144 + 22 + 176 = 342 g mol⁻¹

Step 1: Moles of sugar.
$$n = \frac{20}{342} = 0.05848\text{ mol}$$

Step 2: Molarity.
$$M = \frac{0.05848\text{ mol}}{2\text{ L}} = \boxed{0.0293\text{ mol L}^{-1}}$$

Given:
- Density of methanol = 0.793 kg L⁻¹ = 793 g L⁻¹
- Volume of solution to be prepared = 2.5 L
- Molarity = 0.25 M
- Molar mass of methanol (CH₃OH) = 12 + 4 + 16 = 32 g mol⁻¹

Step 1: Find moles of methanol needed.
$$n = M \times V = 0.25\text{ mol L}^{-1} \times 2.5\text{ L} = 0.625\text{ mol}$$

Step 2: Find mass of methanol needed.
$$\text{Mass} = 0.625\text{ mol} \times 32\text{ g mol}^{-1} = 20\text{ g}$$

Step 3: Find volume of methanol.
$$V = \frac{\text{Mass}}{\text{Density}} = \frac{20\text{ g}}{793\text{ g L}^{-1}} = \boxed{0.02522\text{ L} = 25.22\text{ mL}}$$

Given: Mass of air = 1034 g cm⁻²

Concept: Pressure = Force/Area = (mass × g)/Area

Step 1: Convert mass per unit area to SI units.
$$1034\text{ g cm}^{-2} = 1034 \times 10^{-3}\text{ kg} \times (10^{-2}\text{ m})^{-2}$$
$$= 1034 \times 10^{-3}\text{ kg} \times 10^{4}\text{ m}^{-2}$$
$$= 1034 \times 10\text{ kg m}^{-2} = 10340\text{ kg m}^{-2}$$

Step 2: Calculate pressure (using g = 9.8 m s⁻²).
$$P = \frac{F}{A} = \frac{m \times g}{A} = 10340\text{ kg m}^{-2} \times 9.8\text{ m s}^{-2}$$
$$= \boxed{1.013 \times 10^{5}\text{ Pa}}$$

SI unit of mass: Kilogram (kg)

Definition: The kilogram is the SI unit of mass. It is defined as the mass equal to the international prototype of the kilogram (a platinum-iridium alloy cylinder kept at the International Bureau of Weights and Measures, Sèvres, France).

In the revised SI (2019), the kilogram is defined by fixing the numerical value of the Planck constant $h = 6.62607015 \times 10^{-34}$ J s (i.e., kg m² s⁻¹), thereby defining the kilogram in terms of fundamental constants.

Matching:

| Prefix | Multiple |
|--------|----------|
| (i) micro | $10^{-6}$ |
| (ii) deca | $10$ |
| (iii) mega | $10^{6}$ |
| (iv) giga | $10^{9}$ |
| (v) femto | $10^{-15}$ |

Significant figures are the meaningful digits in a measured or calculated quantity that are known with certainty plus one uncertain (estimated) digit.

Rules for counting significant figures:
1. All non-zero digits are significant. (e.g., 285 has 3 significant figures)
2. Zeros between non-zero digits are significant. (e.g., 2005 has 4 significant figures)
3. Leading zeros (zeros to the left of the first non-zero digit) are NOT significant. (e.g., 0.0025 has 2 significant figures)
4. Trailing zeros in a number with a decimal point ARE significant. (e.g., 4.500 has 4 significant figures)
5. Trailing zeros in a whole number without a decimal point may or may not be significant (ambiguous).

Significance: Significant figures indicate the precision of a measurement.

Given: Contamination = 15 ppm by mass

Concept: 1 ppm = 1 part per million by mass = $\frac{1\text{ g}}{10^6\text{ g}}$ of solution

(i) Per cent by mass:
$$15\text{ ppm} = \frac{15}{10^6} \times 100\% = \frac{15 \times 10^{-4}}{100} \times 100\%$$
$$= \boxed{1.5 \times 10^{-3}\%}$$

(ii) Molality:

Molar mass of CHCl₃ = 12 + 1 + 3(35.5) = 12 + 1 + 106.5 = 119.5 g mol⁻¹

In 10⁶ g of water sample:
- Mass of CHCl₃ = 15 g
- Mass of water (solvent) ≈ (10⁶ − 15) g ≈ 10⁶ g (since contamination is very small)

$$\text{Moles of CHCl}_3 = \frac{15}{119.5} = 0.1255\text{ mol}$$

$$\text{Mass of water} = 10^6\text{ g} = 10^3\text{ kg}$$

$$m = \frac{0.1255\text{ mol}}{10^3\text{ kg}} = \boxed{1.255 \times 10^{-4}\text{ mol kg}^{-1}}$$

Scientific notation: $N \times 10^n$ where 1 \leq N < 10

(i) $0.0048 = \boxed{4.8 \times 10^{-3}}$

(ii) $234000 = \boxed{2.34 \times 10^{5}}$

(iii) $8008 = \boxed{8.008 \times 10^{3}}$

(iv) $500.0 = \boxed{5.000 \times 10^{2}}$

(v) $6.0012 = \boxed{6.0012 \times 10^{0}}$ (already in scientific notation)

(i) 0.0025:
Leading zeros are not significant. Only 2 and 5 are significant.
$$\boxed{2\text{ significant figures}}$$

(ii) 208:
All digits including the zero between 2 and 8 are significant.
$$\boxed{3\text{ significant figures}}$$

(iii) 5005:
Zeros between non-zero digits are significant.
$$\boxed{4\text{ significant figures}}$$

(iv) 126,000:
Trailing zeros in a whole number without decimal point are ambiguous; conventionally taken as not significant.
$$\boxed{3\text{ significant figures}}$$

(v) 500.0:
The decimal point makes all trailing zeros significant.
$$\boxed{4\text{ significant figures}}$$

(vi) 2.0034:
All digits are significant (zeros between non-zero digits count).
$$\boxed{5\text{ significant figures}}$$

(i) 34.216 → 3 significant figures:
The 4th digit is 1 (< 5), so drop it.
$$\boxed{34.2}$$

(ii) 10.4107 → 3 significant figures:
The 4th digit is 1 (< 5), so drop.
$$\boxed{10.4}$$

(iii) 0.04597 → 3 significant figures:
Leading zeros are not significant. The significant digits are 4, 5, 9, 7.
Rounding to 3 sig figs: 4th significant digit is 7 (≥ 5), so round up 9 → 10, carry over.
$$0.04597 \rightarrow \boxed{0.0460}$$

(iv) 2808 → 3 significant figures:
The 4th digit is 8 (≥ 5), so round up.
$$2808 \rightarrow \boxed{2810}$$

(a) Law obeyed:

For a fixed mass of dinitrogen (14 g), the masses of dioxygen that combine are 16 g, 32 g (cases i and ii).
Ratio of O₂ = 16 : 32 = 1 : 2 (simple whole number ratio)

For 28 g of dinitrogen, masses of O₂ are 32 g and 80 g (cases iii and iv).
Ratio of O₂ = 32 : 80 = 2 : 5 (simple whole number ratio)

This obeys the Law of Multiple Proportions.

Statement: When two elements combine to form two or more different compounds, the masses of one element that combine with a fixed mass of the other element bear a simple whole number ratio to one another.

---

(b) Unit conversions:

(i) 1 km = ? mm = ? pm
$$1\text{ km} = 10^3\text{ m} = 10^3 \times 10^3\text{ mm} = \boxed{10^6\text{ mm}}$$
$$1\text{ km} = 10^3\text{ m} = 10^3 \times 10^{12}\text{ pm} = \boxed{10^{15}\text{ pm}}$$

(ii) 1 mg = ? kg = ? ng
$$1\text{ mg} = 10^{-3}\text{ g} = 10^{-3} \times 10^{-3}\text{ kg} = \boxed{10^{-6}\text{ kg}}$$
$$1\text{ mg} = 10^{-3}\text{ g} = 10^{-3} \times 10^{9}\text{ ng} = \boxed{10^{6}\text{ ng}}$$

(iii) 1 mL = ? L = ? dm³
$$1\text{ mL} = \boxed{10^{-3}\text{ L}}$$
$$1\text{ mL} = 1\text{ cm}^3 = (10^{-1}\text{ dm})^3 = 10^{-3}\text{ dm}^3 = \boxed{10^{-3}\text{ dm}^3}$$

Given:
- Speed of light, $c = 3.0 \times 10^8\text{ m s}^{-1}$
- Time, $t = 2.00\text{ ns} = 2.00 \times 10^{-9}\text{ s}$

Formula: Distance = Speed × Time

$$d = c \times t = 3.0 \times 10^8\text{ m s}^{-1} \times 2.00 \times 10^{-9}\text{ s}$$

$$= 6.0 \times 10^{-1}\text{ m}$$

$$= \boxed{0.60\text{ m}}$$

Reaction: A + B₂ → AB₂

1 atom/mol of A reacts with 1 molecule/mol of B₂ in a 1:1 ratio.

(i) 300 atoms of A + 200 molecules of B₂:
A needed for 200 B₂ = 200 atoms. A available = 300 (excess).
Limiting reagent: B₂

(ii) 2 mol A + 3 mol B₂:
B₂ needed for 2 mol A = 2 mol. B₂ available = 3 mol (excess).
Limiting reagent: A

(iii) 100 atoms of A + 100 molecules of B₂:
Ratio is exactly 1:1. Both are consumed completely.
No limiting reagent (both are in stoichiometric amounts)

(iv) 5 mol A + 2.5 mol B₂:
A needed for 2.5 mol B₂ = 2.5 mol. A available = 5 mol (excess).
Limiting reagent: B₂

(v) 2.5 mol A + 5 mol B₂:
B₂ needed for 2.5 mol A = 2.5 mol. B₂ available = 5 mol (excess).
Limiting reagent: A

Balanced equation: $\text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g)$

Given:
- Mass of N₂ = $2.00 \times 10^3$ g
- Mass of H₂ = $1.00 \times 10^3$ g

Step 1: Calculate moles of reactants.
$$n_{\text{N}_2} = \frac{2000}{28} = 71.43\text{ mol}$$
$$n_{\text{H}_2} = \frac{1000}{2} = 500\text{ mol}$$

Step 2: Identify limiting reagent.
From stoichiometry, 1 mol N₂ requires 3 mol H₂.
71.43 mol N₂ requires $71.43 \times 3 = 214.3$ mol H₂.
Available H₂ = 500 mol > 214.3 mol.

∴ N₂ is the limiting reagent.

(i) Mass of NH₃ produced:
1 mol N₂ gives 2 mol NH₃.
$$n_{\text{NH}_3} = 2 \times 71.43 = 142.86\text{ mol}$$
$$\text{Mass of NH}_3 = 142.86 \times 17 = \boxed{2428.6\text{ g} \approx 2.43 \times 10^3\text{ g}}$$

(ii) Yes, H₂ will remain unreacted.

(iii) Mass of unreacted H₂:
$$\text{H}_2\text{ consumed} = 71.43 \times 3 = 214.3\text{ mol}$$
$$\text{H}_2\text{ remaining} = 500 - 214.3 = 285.7\text{ mol}$$
$$\text{Mass of H}_2\text{ remaining} = 285.7 \times 2 = \boxed{571.4\text{ g}}$$

0.50 mol Na₂CO₃:
This refers to a definite amount (number of moles) of Na₂CO₃.
$$\text{Mass} = 0.50\text{ mol} \times 106\text{ g mol}^{-1} = 53\text{ g of Na}_2\text{CO}_3$$
No mention of volume or solvent — it is simply a quantity of substance.

0.50 M Na₂CO₃:
This refers to a concentration — a solution containing 0.50 moles of Na₂CO₃ dissolved in enough water to make 1 litre of solution.
$$\text{It contains } 0.50\text{ mol Na}_2\text{CO}_3\text{ per litre of solution}$$

Key difference: 0.50 mol is an amount of substance, while 0.50 M is a concentration (moles per litre of solution).

Balanced equation:
$$2\text{H}_2(g) + \text{O}_2(g) \rightarrow 2\text{H}_2\text{O}(g)$$

By Gay-Lussac's Law of Gaseous Volumes, volumes react in the same ratio as molar coefficients.

From the equation: 2 volumes H₂ + 1 volume O₂ → 2 volumes H₂O

Given: 10 volumes H₂ and 5 volumes O₂.

Check stoichiometry: 10 volumes H₂ requires $\frac{10}{2} = 5$ volumes O₂. Exactly 5 volumes O₂ are available — both are completely consumed.

$$\text{Volumes of H}_2\text{O produced} = 10\text{ volumes}$$

$$\boxed{10\text{ volumes of water vapour are produced.}}$$

The basic SI unit of length is metre (m) and of mass is kilogram (kg).

(i) 28.7 pm:
$$1\text{ pm} = 10^{-12}\text{ m}$$
$$28.7\text{ pm} = 28.7 \times 10^{-12}\text{ m} = \boxed{2.87 \times 10^{-11}\text{ m}}$$

(ii) 15.15 pm:
$$15.15\text{ pm} = 15.15 \times 10^{-12}\text{ m} = \boxed{1.515 \times 10^{-11}\text{ m}}$$

(iii) 25365 mg:
$$1\text{ mg} = 10^{-6}\text{ kg}$$
$$25365\text{ mg} = 25365 \times 10^{-6}\text{ kg} = \boxed{2.5365 \times 10^{-2}\text{ kg}}$$

Concept: Number of atoms = $\frac{\text{mass}}{\text{molar mass}} \times N_A$ (for monoatomic) or $\times 2N_A$ (for diatomic)

For 1 g of each substance:

(i) Au: Molar mass = 197 g mol⁻¹
$$n = \frac{1}{197} = 5.08 \times 10^{-3}\text{ mol atoms}$$

(ii) Na: Molar mass = 23 g mol⁻¹
$$n = \frac{1}{23} = 4.35 \times 10^{-2}\text{ mol atoms}$$

(iii) Li: Molar mass = 6.941 g mol⁻¹
$$n = \frac{1}{6.941} = 1.44 \times 10^{-1}\text{ mol atoms}$$

(iv) Cl₂: Molar mass = 71 g mol⁻¹; each molecule has 2 atoms
$$n = \frac{1}{71} \times 2 = 2.82 \times 10^{-2}\text{ mol atoms}$$

Comparison: Li (0.144) > Na (0.0435) > Cl₂ (0.0282) > Au (0.00508)

$$\boxed{\text{(iii) 1 g of Li has the largest number of atoms.}}$$

Given:
- Mole fraction of ethanol, $x_{\text{ethanol}} = 0.040$
- Density of water = 1 g mL⁻¹ = 1 kg L⁻¹
- Molar mass of ethanol (C₂H₅OH) = 46 g mol⁻¹
- Molar mass of water = 18 g mol⁻¹

Step 1: Find mole fraction of water.
$$x_{\text{water}} = 1 - 0.040 = 0.960$$

Step 2: Assume 1 mol total. Then moles of ethanol = 0.040 mol, moles of water = 0.960 mol.

Step 3: Find mass of water (solvent).
$$\text{Mass of water} = 0.960 \times 18 = 17.28\text{ g} = 0.01728\text{ L}$$
(using density = 1 g mL⁻¹)

Step 4: Calculate molarity.
$$M = \frac{\text{moles of ethanol}}{\text{volume of solution in L}}$$

Volume of solution ≈ volume of water (dilute solution) = $\frac{17.28\text{ g}}{1\text{ g mL}^{-1}} = 17.28\text{ mL} = 0.01728\text{ L}$

$$M = \frac{0.040\text{ mol}}{0.01728\text{ L}} = \boxed{2.31\text{ mol L}^{-1}}$$

Concept: 1 mole of ¹²C = 12 g and contains $6.022 \times 10^{23}$ atoms.

$$\text{Mass of one }^{12}\text{C atom} = \frac{12\text{ g}}{6.022 \times 10^{23}}$$

$$= \frac{12}{6.022 \times 10^{23}}\text{ g}$$

$$= \boxed{1.993 \times 10^{-23}\text{ g}}$$

(i) $\dfrac{0.02856 \times 298.15 \times 0.112}{0.5785}$:

For multiplication/division, the answer should have the same number of significant figures as the number with the least significant figures.

- 0.02856 → 4 sig figs
- 298.15 → 5 sig figs
- 0.112 → 3 sig figs
- 0.5785 → 4 sig figs

Least = 3 significant figures.
$$\boxed{3\text{ significant figures}}$$

(ii) $5 \times 5.364$:

- 5 is an exact number (or 1 sig fig)
- 5.364 → 4 sig figs

If 5 is exact (counting number), the answer has 4 significant figures.
$$\boxed{4\text{ significant figures}}$$

(iii) $0.0125 + 0.7864 + 0.0215$:

For addition/subtraction, the answer should have the same number of decimal places as the number with the fewest decimal places.

- 0.0125 → 4 decimal places
- 0.7864 → 4 decimal places
- 0.0215 → 4 decimal places

All have 4 decimal places, so the answer has 4 decimal places.
$$0.0125 + 0.7864 + 0.0215 = 0.8204$$
$$\boxed{4\text{ significant figures}}$$

Avogadro's number: $N_A = 6.022 \times 10^{23}$ mol⁻¹

(i) 52 moles of Ar:
$$\text{Number of atoms} = 52\text{ mol} \times 6.022 \times 10^{23}\text{ mol}^{-1}$$
$$= \boxed{3.131 \times 10^{25}\text{ atoms}}$$

(ii) 52 u of He:
Molar mass of He = 4 u = 4 g mol⁻¹
$$\text{Number of atoms} = \frac{52\text{ u}}{4\text{ u/atom}} = 13\text{ atoms}$$
$$= \boxed{13\text{ atoms}}$$

(iii) 52 g of He:
$$n = \frac{52\text{ g}}{4\text{ g mol}^{-1}} = 13\text{ mol}$$
$$\text{Number of atoms} = 13 \times 6.022 \times 10^{23} = \boxed{7.829 \times 10^{24}\text{ atoms}}$$

Step 1: Find mass of C and H from combustion products.

Moles of CO₂ = $\frac{3.38}{44} = 0.0768$ mol → Moles of C = 0.0768 mol
Mass of C = $0.0768 \times 12 = 0.9216$ g

Moles of H₂O = $\frac{0.690}{18} = 0.0383$ mol → Moles of H = $0.0383 \times 2 = 0.0767$ mol
Mass of H = $0.0767 \times 1 = 0.0767$ g

Step 2: Find mole ratio C : H.
$$C : H = 0.0768 : 0.0767 \approx 1 : 1$$

(i) Empirical formula: $\boxed{\text{CH}}$

Step 3: Find molar mass.

At STP, 1 mole of gas occupies 22.4 L.
$$\text{Moles of gas in 10.0 L} = \frac{10.0}{22.4} = 0.4464\text{ mol}$$
$$\text{Molar mass} = \frac{11.6\text{ g}}{0.4464\text{ mol}} = \boxed{26.0\text{ g mol}^{-1}}$$

Step 4: Find molecular formula.

Empirical formula mass of CH = 12 + 1 = 13 g mol⁻¹
$$n = \frac{26.0}{13} = 2$$

(iii) Molecular formula: $\boxed{\text{C}_2\text{H}_2}$ (ethyne/acetylene)

Given:
- Volume of HCl = 25 mL = 0.025 L
- Molarity of HCl = 0.75 M
- Molar mass of CaCO₃ = 100 g mol⁻¹

Step 1: Find moles of HCl.
$$n_{\text{HCl}} = 0.75\text{ mol L}^{-1} \times 0.025\text{ L} = 0.01875\text{ mol}$$

Step 2: From stoichiometry, 2 mol HCl reacts with 1 mol CaCO₃.
$$n_{\text{CaCO}_3} = \frac{0.01875}{2} = 9.375 \times 10^{-3}\text{ mol}$$

Step 3: Find mass of CaCO₃.
$$\text{Mass} = 9.375 \times 10^{-3}\text{ mol} \times 100\text{ g mol}^{-1}$$
$$= \boxed{0.9375\text{ g}}$$

Given:
- Mass of MnO₂ = 5.0 g
- Molar mass of MnO₂ = 55 + 32 = 87 g mol⁻¹
- Molar mass of HCl = 1 + 35.5 = 36.5 g mol⁻¹

Step 1: Find moles of MnO₂.
$$n_{\text{MnO}_2} = \frac{5.0}{87} = 0.05747\text{ mol}$$

Step 2: From stoichiometry, 4 mol HCl reacts with 1 mol MnO₂.
$$n_{\text{HCl}} = 4 \times 0.05747 = 0.2299\text{ mol}$$

Step 3: Find mass of HCl.
$$\text{Mass of HCl} = 0.2299\text{ mol} \times 36.5\text{ g mol}^{-1}$$
$$= \boxed{8.39\text{ g}}$$