Chemistry 3 - Quantitative Chemistry Flashcards
(581 cards)
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Relative Formula Mass (Mᵣ) definition The sum of the relative atomic masses (Aᵣ) of all atoms in a compound’s formula.
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Calculate Mᵣ of MgCl₂ (Aᵣ: Mg=24
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Cl=35.5) 24 + (2 × 35.5) = 95.
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% mass formula for an element in a compound (Aᵣ × number of atoms × 100) ÷ Mᵣ of compound.
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% mass of sodium in Na₂CO₃ (Aᵣ: Na=23
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C=12
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Practice: Mass of FeCl₂ to provide 10g iron in a 20% iron mixture (Aᵣ: Fe=56
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Cl=35.5) Step 1: 10g iron = 20% of 50g mixture. Step 2: % Fe in FeCl₂ = (56/127)×100 ≈ 44.1%. Step 3: Mass FeCl₂ = 10g ÷ 0.441 ≈ 22.7g.
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Why mass INCREASES in an unsealed reaction A reactant gas (e.g.
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oxygen) from air is added to the vessel
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Why mass DECREASES in an unsealed reaction A product gas (e.g.
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CO₂) escapes the vessel
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Definition of a mole An amount of substance containing 6.02 × 10²³ particles (Avogadro’s number).
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Mass of 1 mole of a substance Equal to its Mᵣ in grams (e.g.
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1 mole CO₂ = 44g).
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Moles formula Number of moles = mass (g) ÷ Mᵣ.
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Moles in 66g of CO₂ (Mᵣ=44) 66 ÷ 44 = 1.5 moles.
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Mass of 4 moles of carbon (Aᵣ=12) 4 × 12 = 48g.
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Law of conservation of mass No atoms are created/destroyed in reactions; total mass of reactants = total mass of products.
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Prove mass is conserved in 2Li + F₂ → 2LiF Reactants: (2×7) + 38 = 52; Products: 2×26 = 52.
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How to calculate reacting masses using moles Use moles = mass ÷ Mᵣ and the balanced equation's mole ratios.
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Mole ratio in Mg + 2HCl → MgCl₂ + H₂ 1 mole Mg : 2 moles HCl : 1 mole MgCl₂ : 1 mole H₂.
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Steps to balance equations using masses 1) Find moles (mass ÷ Mᵣ). 2) Divide by smallest moles. 3) Convert to whole numbers. 4) Write equation.
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Balance: 8.1g ZnO + 0.60g C → 2.2g CO₂ + 6.5g Zn 1) Moles: ZnO=0.10
C=0.050
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Limiting reactant definition The reactant completely used up first
stopping the reaction and limiting product amount.
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Excess reactant definition Reactant not fully used up; added to ensure limiting reactant is consumed.
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Why product mass depends on limiting reactant More limiting reactant = more product particles (directly proportional).
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Steps to calculate product mass from limiting reactant 1) Write equation. 2) Find Mᵣ. 3) Calculate moles of known. 4) Use mole ratio. 5) Convert to mass.
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Mass of Al₂O₃ from 135g Al (4Al + 3O₂ → 2Al₂O₃) 1) Moles Al=5. 2) Mole ratio (4:2) → 2.5 moles Al₂O₃. 3) Mass=2.5×102=255g.
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Concentration definition A measure of how much solute is dissolved in a given volume of solution (units: g/dm³ or mol/dm³).
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Formula for concentration (g/dm³) Concentration = mass of solute (g) ÷ volume of solvent (dm³).
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Calculate concentration: 30g NaCl in 0.2dm³ water 30 ÷ 0.2 = 150 g/dm³.
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Convert 500cm³ to dm³ 500 ÷ 1000 = 0.5 dm³.
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Mass of solute needed for 24 g/dm³ in 0.40dm³ 24 × 0.40 = 9.6 g.
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pH scale range 0 (acidic) to 14 (alkaline)
with 7 as neutral (e.g.
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Acid definition Substance with pH <7; forms H⁺ ions in water (e.g.
lemon juice
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Base vs. Alkali Base: pH >7; Alkali: base dissolved in water (forms OH⁻ ions
e.g.
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Neutralisation reaction Acid + Base → Salt + Water or H⁺(aq) + OH⁻(aq) → H₂O(l).
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How to measure pH 1) Indicator (color change
e.g.
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What ions do acids produce in water? Hydrogen ions (
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H
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+
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H
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+
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)
e.g.
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H
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C
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l
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→
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H
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+
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+
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C
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l
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−
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HCl→H
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+
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+Cl
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−
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.
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Strong acid vs. weak acid Strong: Fully ionizes (e.g.
HCl). Weak: Partially ionizes (e.g.
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pH and
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H
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+
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H
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+
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ion relationship Each pH decrease by 1 =
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10
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×
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10× increase in
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H
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+
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H
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+
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concentration.
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pH 4 vs. pH 5 acid pH 4 has 10× more
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H
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+
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H
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+
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than pH 5.
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Concentration vs. strength Concentration: Amount of acid per volume. Strength: % ionized (strong/weak).
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Dilute strong acid example 0.1M HCl (fully ionized but low concentration).
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Neutralization reaction formula
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A
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c
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i
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d
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+
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B
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a
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s
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e
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→
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S
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a
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l
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t
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+
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W
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a
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t
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e
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r
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Acid+Base→Salt+Water.
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Acid + metal oxide example
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2
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H
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C
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l
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+
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C
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u
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O
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→
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C
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u
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C
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l
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2
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+
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H
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2
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O
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2HCl+CuO→CuCl
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2
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+H
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2
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O.
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Acid + metal hydroxide example
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H
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2
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S
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O
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4
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+
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2
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K
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O
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H
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→
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K
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2
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S
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O
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4
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+
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2
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H
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2
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O
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H
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2
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SO
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4
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+2KOH→K
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2
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SO
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4
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+2H
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2
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O.
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Acid + carbonate products Salt + Water + Carbon dioxide (
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C
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O
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2
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CO
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2
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).
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Making copper chloride
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C
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u
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O
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+
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2
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H
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C
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l
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→
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C
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u
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C
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l
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2
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+
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H
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2
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O
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CuO+2HCl→CuCl
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2
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+H
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2
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O; filter excess CuO
crystallize.
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Testing for
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C
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O
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2
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CO
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2
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Bubble through limewater → turns cloudy.
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Key Notes:
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Reactivity series order (high to low) Potassium (K)
Sodium (Na)
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How reactivity is determined By how easily metals lose electrons to form positive ions.
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Metal + acid reaction products Salt + hydrogen gas (e.g.
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F
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e
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+
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2
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H
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C
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l
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→
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F
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e
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C
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l
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2
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+
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H
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2
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Fe+2HCl→FeCl
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2
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+H
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2
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).
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Observations of reactivity More reactive metals (e.g.
Mg) react vigorously; less reactive (e.g.
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Metal + water reaction products Metal hydroxide + hydrogen (e.g.
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C
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a
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+
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2
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H
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2
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O
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→
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C
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a
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(
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O
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H
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)
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2
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+
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H
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2
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Ca+2H
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2
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O→Ca(OH)
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2
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+H
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2
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).
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Metals reacting with water K
Na
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Oxidation definition Gain of oxygen (e.g.
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2
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M
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g
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+
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O
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2
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→
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M
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g
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O
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2Mg+O
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2
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→2MgO).
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Reduction definition Loss of oxygen (e.g.
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2
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C
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u
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O
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+
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C
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→
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2
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C
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u
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+
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C
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O
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2
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2CuO+C→2Cu+CO
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2
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).
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Extraction method depends on reactivity
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Above carbon (e.g.
Al): Electrolysis (expensive).
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Below carbon (e.g.
Fe): Reduction using carbon (e.g.
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Why carbon reduces metal oxides Carbon can only remove oxygen from metals less reactive than itself.
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Example of carbon reduction
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2
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F
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e
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2
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O
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3
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+
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3
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C
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→
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F
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e
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+
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3
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C
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O
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2
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2Fe
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2
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O
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3
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+3C→4Fe+3CO
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2
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.
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Unreactive metals in nature Gold (Au) found as pure metal; no extraction needed.
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OIL RIG mnemonic Oxidation Is Loss (of e⁻)
Reduction Is Gain (of e⁻).
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Redox example (Fe + HCl) Fe oxidized:
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F
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e
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→
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F
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e
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2
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+
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+
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2
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e
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−
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Fe→Fe
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2+
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+2e
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−
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H⁺ reduced:
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2
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H
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+
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+
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2
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e
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−
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→
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H
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2
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2H
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+
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+2e
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−
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→H
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2
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.
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Displacement reaction rule More reactive metal displaces less reactive metal from its compound (see reactivity series).
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Fe + CuSO₄ reaction
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F
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e
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+
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C
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u
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S
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O
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4
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→
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F
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e
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S
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O
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4
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+
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C
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u
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Fe+CuSO
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4
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→FeSO
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4
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+Cu
Fe oxidized, Cu²⁺ reduced.
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Electrolysis definition Splitting ionic compounds using electricity
requires molten/aqueous electrolyte.
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Electrode reactions Cathode (-): Reduction (gain e⁻). Anode (+): Oxidation (lose e⁻).
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Molten PbBr₂ electrolysis Cathode:
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P
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b
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2
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+
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+
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2
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e
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−
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→
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P
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b
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Pb
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2+
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+2e
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−
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→Pb. Anode:
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2
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B
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r
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−
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→
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B
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r
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2
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+
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2
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e
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−
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2Br
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−
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→Br
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2
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+2e
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−
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.
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Aluminium extraction Electrolysis of molten Al₂O₃ (mixed with cryolite).
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Cathode:
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A
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l
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3
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+
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+
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3
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e
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−
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→
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A
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l
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Al
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3+
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+3e
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−
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→Al.
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Anode:
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2
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O
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2
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−
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→
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O
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2
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+
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4
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e
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−
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2O
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2−
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→O
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2
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+4e
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−
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.
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Aqueous electrolysis rules
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Cathode: H⁺ discharged if metal > H (e.g., Na⁺ → H₂)
else metal forms (e.g., Cu²⁺ → Cu).
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Anode: Halides form X₂ (e.g., Cl₂)
else OH⁻ → O₂.
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NaCl(aq) electrolysis products Cathode: H₂ gas. Anode: Cl₂ gas.
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Testing electrolysis gases
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Cl₂: Bleaches litmus.
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H₂: "Squeaky pop" with splint.
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O₂: Relights glowing splint.
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Balancing half-equations Ensure e⁻ balance (e.g.,
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4
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O
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H
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−
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→
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O
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2
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+
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2
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H
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2
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O
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+
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4
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e
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−
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4OH
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−
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→O
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2
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+2H
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2
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579
O+4e
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−
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).
