Model Answer: The law of conservation of mass states that mass is neither created nor destroyed in a chemical reaction; the total mass of reactants equals the total mass of products. For example, when hydrogen reacts with oxygen to form water: 2 H₂(g) + O₂(g) → 2 H₂O(l). In mole terms, 2 moles of H₂ (2 × 2.016 ≈ 4.032 g) plus 1 mole of O₂ (32.00 g) give 2 moles of H₂O (2 × 18.016 ≈ 36.032 g). Total mass before = 36.032 g; total mass after = 36.032 g, which confirms conservation of mass. In laboratory experiments, this law is validated by closed-system reactions where no mass is lost to the surroundings.

Model Answer: Balancing a chemical equation requires ensuring that atoms of each element are equal on both sides using stoichiometric coefficients while keeping formulas unchanged. Steps: (1) Write the unbalanced (skeleton) equation, (2) Count atoms of each element on reactant and product sides, (3) Start balancing with elements that appear in only one reactant and one product (often C, H), (4) Balance oxygen last and (5) Check coefficients for simplest whole-number ratios. Example: Combustion of ethanol skeleton: C₂H₅OH + O₂ → CO₂ + H₂O. Balance C: 2 CO₂. Balance H: C₂H₅OH has 6 H → 3 H₂O. Now O count: reactants' O = x (from O₂) + 1 (in ethanol); products O = 2×2 (from CO₂) + 3×1 (from H₂O) = 7. So we need (7−1)=6 O atoms from O₂ → 3 O₂. Balanced equation: C₂H₅OH + 3 O₂ → 2 CO₂ + 3 H₂O.

Model Answer: Chemical reactions are classified by the pattern of change:
Combination (synthesis): two or more reactants form one product. Example: 2 Mg(s) + O₂(g) → 2 MgO(s).
Decomposition: one compound breaks into simpler substances. Example: 2 KClO₃(s) → 2 KCl(s) + 3 O₂(g) (on heating).
Displacement (single): a more reactive element displaces a less reactive element from its compound. Example: Zn(s) + CuSO₄(aq) → ZnSO₄(aq) + Cu(s).
Double displacement (metathesis): exchange of ions between two salts, often forming a precipitate. Example: AgNO₃(aq) + NaCl(aq) → AgCl(s)↓ + NaNO₃(aq). Each equation should be balanced and include state symbols where appropriate.

Model Answer: A redox (oxidation–reduction) reaction involves transfer of electrons between species. Oxidation is loss of electrons (or gain of oxygen / loss of hydrogen) and reduction is gain of electrons (or gain of hydrogen / loss of oxygen). Consider: CuO + H₂ → Cu + H₂O. Here H₂ is oxidised to H₂O (it gains oxygen — effectively loses electrons) and CuO is reduced to Cu (it loses oxygen — effectively gains electrons). In ionic terms, Cu²⁺ gains electrons to form Cu(0) (reduction) while H₂ loses electrons to form 2 H⁺ which combine with O²⁻ to form water (oxidation). This conceptual framework helps identify oxidants and reductants in reactions.

Model Answer: A catalyst increases the rate of a chemical reaction without being consumed; it provides an alternative reaction pathway with lower activation energy, enabling more reactant molecules to overcome the energy barrier at a given temperature. Catalysts do not change the position of equilibrium but help the system reach equilibrium faster. Example: Decomposition of hydrogen peroxide is slow but is markedly accelerated by manganese dioxide (MnO₂) or by the enzyme catalase in biological systems: 2 H₂O₂(aq) → 2 H₂O(l) + O₂(g) (MnO₂ as catalyst). The catalyst provides surface or intermediate complexes that break O–O bonds more readily and regenerate at the end of reaction.

Model Answer: Stoichiometry relates quantities of reactants and products via mole ratios from a balanced equation. Reaction: CaCO₃(s) + 2 HCl(aq) → CaCl₂(aq) + CO₂(g) + H₂O(l). Mole ratio CaCO₃ : CO₂ is 1 : 1, so 1 mole of CaCO₃ yields 1 mole of CO₂. Molar mass CaCO₃ ≈ 100.09 g mol⁻¹, CO₂ ≈ 44.01 g mol⁻¹. From 100 g CaCO₃ ≈ 100 / 100.09 ≈ 0.999 mol → CO₂ produced ≈ 0.999 × 44.01 ≈ 43.97 g (≈ 44.0 g). This demonstrates mole ratio application to mass calculations in reactions.

Model Answer: Laboratory tests employ characteristic reactions: Hydrogen (H₂) — collects over water or by displacement and produces a 'pop' with a burning splint (small explosion) indicating presence of H₂. Oxygen (O₂) — supports combustion; a glowing splint relights in oxygen-rich gas. Carbon dioxide (CO₂) — turns limewater (Ca(OH)₂) milky due to CaCO₃ formation, and can extinguish a burning candle. Each test is qualitative and widely used in Class 10 practicals to confirm gas identity.

Model Answer: Precipitation reactions occur when two soluble ionic solutions are mixed and an insoluble product (precipitate) forms. Solubility rules (general) help predict precipitates: most nitrates (NO₃⁻) are soluble; most chlorides are soluble except AgCl, PbCl₂. Examples: (1) AgNO₃(aq) + NaCl(aq) → AgCl(s)↓ + NaNO₃(aq) — AgCl is white precipitate. (2) BaCl₂(aq) + Na₂SO₄(aq) → BaSO₄(s)↓ + 2 NaCl(aq) — BaSO₄ is white insoluble precipitate. Observing cloudiness or solid settling confirms precipitation.

Model Answer: Corrosion is the gradual chemical or electrochemical deterioration of metals due to reaction with their environment; for iron, it commonly forms hydrated iron(III) oxide (rust) in presence of moisture and oxygen: 4 Fe + 3 O₂ + x H₂O → 2 Fe₂O₃·x H₂O. Causes include moisture, dissolved salts (which increase electrical conductivity), acidic conditions and galvanic coupling with other metals. Preventive methods: (1) Coating/painting to exclude air and moisture, (2) Galvanisation — coating with a more reactive metal like zinc which provides sacrificial protection (zinc oxidises preferentially), and (3) Cathodic protection using sacrificial anodes. Regular maintenance and protective design reduce corrosion risk.

Model Answer: Decomposition reactions split a compound into simpler substances. Thermal decomposition occurs on heating (e.g., CaCO₃ → CaO + CO₂), often requiring heat input. Electrolytic decomposition uses electric current to decompose compounds (e.g., electrolysis of molten NaCl → Na + Cl₂), used industrially. Photochemical decomposition is driven by light (e.g., AgCl(s) decomposes under sunlight to Ag(s) + ½ Cl₂ in photographic processes). Each type involves different energy sources causing bond breakage and formation of products relevant to laboratory and industrial contexts.

Model Answer: The activity series ranks metals by their tendency to lose electrons (be oxidised). A metal higher in the series can displace a metal ion lower in the series from its salt solution. For instance zinc is above copper; therefore Zn(s) + CuSO₄(aq) → ZnSO₄(aq) + Cu(s) occurs because Zn more readily loses electrons (is oxidised) and reduces Cu²⁺ to Cu. Conversely, Cu cannot displace Zn²⁺ from ZnSO₄. The series helps predict which single displacement reactions are spontaneous under given conditions.

Model Answer: Rancidity is the oxidative deterioration of fats and oils leading to unpleasant odour and taste. Unsaturated fatty acids react with oxygen (autoxidation) producing peroxides and aldehydes responsible for rancid smell. Prevention methods: (1) Add antioxidants (e.g., BHA, BHT, vitamin E) which scavenge radicals and slow oxidation, (2) Reduce exposure to air, light and heat by storing oils in airtight, opaque containers in cool places, and (3) Hydrogenation (industrial) to reduce unsaturation, though this changes nutritional properties. Proper handling and preservatives extend shelf life.

Model Answer: Exothermic reactions release heat to the surroundings (temperature of surroundings rises); example: combustion of carbon (C + O₂ → CO₂ + heat). Endothermic reactions absorb heat (temperature falls); example: thermal decomposition of CaCO₃. Experimentally, measure temperature change of the reaction mixture or surrounding vessel using a thermometer: if temperature increases, reaction is exothermic; if it decreases, reaction is endothermic. Calorimetry provides quantitative heat measurement. Observing condensation (in exothermic) or required continuous heating (in endothermic) are practical indicators in labs.

Model Answer: A skeleton equation lists reactants and products with their formulas but without stoichiometric coefficients. Balancing converts it into a stoichiometrically correct equation. Example: Skeleton: Fe + O₂ → Fe₂O₃. Steps: count atoms: Fe:1 (LHS), 2 (RHS); O:2 (LHS), 3 (RHS). Balance Fe by placing coefficient 2 before Fe: 2 Fe + O₂ → Fe₂O₃. Now Fe:2 vs 2 OK. O:2 vs 3 not equal — find common multiple: 6 O atoms. Multiply RHS compound by 2 → Fe₂O₃ ×2 → 2 Fe₂O₃. Then LHS Fe must be 4: 4 Fe + 3 O₂ → 2 Fe₂O₃. Balanced and coefficients are lowest whole numbers. This procedure systematically balances multi-element reactions.

Model Answer: To prepare NaCl by neutralisation: add measured dilute HCl to a stoichiometric amount of NaOH (or vice versa) with stirring until neutralisation (use indicator or pH meter) — reaction: HCl + NaOH → NaCl + H₂O. After neutralisation, filter if insoluble impurities present, then evaporate the water by gentle heating until saturated solution remains. Allow to cool slowly or leave for crystallisation; NaCl crystals form as solution cools. Separate crystals by filtration, wash with a little cold distilled water to remove adhering mother liquor, and dry in a desiccator or oven at low temperature. Using pure reagents and careful crystallisation yields dry pure salt suitable for analysis.

Model Answer: Subscripts define the identity and composition of a chemical species; altering them changes the substance itself (e.g., H₂O to H₂O₂ changes water to hydrogen peroxide). Balancing must use coefficients to change amounts (moles) of reactants/products without changing identities. For example, balance H₂ + O₂ → H₂O: incorrect method would change product formula to H₂O₂ to balance oxygen — this is wrong because it changes the product to a different compound. Correct approach uses coefficients: 2 H₂ + O₂ → 2 H₂O, preserving chemical identities and obeying conservation of mass.

Model Answer: Balanced equations provide mole ratios essential for quantitative calculations: predicting product yields, required reagent amounts, and limiting reagent analysis. Example: Suppose we need 44.0 g CO₂ (1.0 mol) from reaction CaCO₃ → CaO + CO₂. Balanced stoichiometry is 1:1 (1 mol CaCO₃ produces 1 mol CO₂). Therefore to obtain 1.0 mol CO₂, we need 1.0 mol CaCO₃ ≈ 100.09 g. If only 150 g CaCO₃ available, maximum CO₂ = 150 / 100.09 ≈ 1.499 mol → 1.499 × 44.01 ≈ 65.95 g. Such calculations are central for laboratory and industrial stoichiometric planning and safety.

Model Answer: Electrolysis uses an external electric current to drive non-spontaneous chemical reactions. In molten ionic compounds, ions are free to move and carry charge to electrodes. For molten NaCl, electrolysis yields sodium metal at the cathode and chlorine gas at the anode: Cathode: Na⁺ + e⁻ → Na; Anode: 2 Cl⁻ → Cl₂ + 2 e⁻; Net: 2 NaCl(l) → 2 Na(l) + Cl₂(g). The molten state is required because in solid NaCl ions are fixed in lattice and cannot move; aqueous electrolysis would produce H₂ and O₂ instead due to water's easier reduction/oxidation unless special conditions are used (chlor-alkali process uses brine and membrane cells to obtain NaOH, Cl₂, H₂).

Model Answer: Understanding chemical reactions is crucial for pollution control, industry, and daily life. Example 1 — acid–base reactions and neutralisation: knowledge helps treat acidic industrial effluents by neutralisation before discharge, preventing ecological harm. Example 2 — corrosion and prevention: understanding rusting mechanisms informs infrastructure protection (bridges, pipelines) through coatings and sacrificial anodes, saving economic costs and preventing failures. Additionally, stoichiometry ensures efficient resource use and waste minimisation in chemical manufacturing. These examples show how Chapter 1 concepts translate to environmental stewardship and practical engineering solutions.

Model Answer: Experiment: Reaction of zinc with copper(II) sulfate solution to demonstrate single displacement. Procedure: Place 50 mL of 0.1 M CuSO₄ in a beaker (blue solution). Add a clean zinc strip to the solution. Observe for a change in solution colour and formation of a brown deposit on zinc. Balanced equation: Zn(s) + CuSO₄(aq) → ZnSO₄(aq) + Cu(s). Observations: blue colour fades as Cu²⁺ is reduced to brown copper metal on the zinc surface; zinc dissolves forming colourless ZnSO₄. After some time, copper flakes can be collected, washed and dried. Safety: wear goggles and gloves, handle solutions carefully, dispose waste per school guidelines (do not pour concentrated heavy-metal solutions down sink), and wash hands after experiment. This demonstration illustrates reactivity series and electron transfer in displacement.