Sciences · Topic 2.2
Chemistry — Stoichiometry and Bonding
Stoichiometry connects macroscopic quantities (grams, litres) to atomic-scale particle counts via the mole. Chemical bonding explains why substances have the properties they do. At Year 4 Advanced, you must justify calculations, evaluate experimental designs, and link bonding type to physical properties.
What You'll Learn
- Define the mole and use n = m/M to convert between mass and moles
- Balance chemical equations and use mole ratios for stoichiometric calculations
- Distinguish ionic bonding (metal + non-metal) from covalent bonding (non-metal + non-metal)
- Link bonding type and structure to physical properties (melting point, conductivity, solubility)
- Explain how five factors affect reaction rate using collision theory
- Evaluate the role of catalysts and their industrial significance
IB Assessment Focus
Criterion A: Apply stoichiometry to multi-step calculations; predict products of reactions; apply collision theory to novel contexts.
Criterion B: Design fair-test experiments to investigate rate of reaction; identify variables and justify control of each.
Criterion C: Use correct chemical notation; present results in appropriate data tables and graphs.
Criterion D: Evaluate experimental method for reliability and validity; discuss industrial and environmental implications of reaction conditions.
Key Vocabulary
| Term | Definition |
|---|---|
| Mole (mol) | SI unit for amount of substance; 1 mol = 6.02 × 10²³ particles (Avogadro's number) |
| Molar mass (M) | Mass of one mole of a substance in g/mol; numerically equals relative atomic/molecular mass |
| Stoichiometry | Quantitative study of reactants and products in chemical reactions using mole ratios |
| Ionic bond | Electrostatic attraction between oppositely charged ions formed by electron transfer (metal + non-metal) |
| Covalent bond | Shared pair of electrons between two non-metal atoms |
| Activation energy (E⊂a;) | The minimum energy collisions must possess for a reaction to occur |
| Catalyst | A substance that increases reaction rate by providing an alternative pathway with lower activation energy; is not consumed |
Sciences · Topic 2.2
The Mole Concept
The mole bridges the atomic world and the laboratory. Just as a "dozen" means 12, a "mole" means 6.02 × 10²³ — Avogadro's number. This allows chemists to count atoms by weighing them.
Core Mole Equations
n = m / M m = n × M M = m / nn = number of moles (mol) | m = mass (g) | M = molar mass (g/mol)
Volume of Gas at STP
V = n × 22.4 L (at Standard Temperature and Pressure: 0°C, 1 atm)
Calculating Molar Mass
- Find the chemical formula of the substance.
- Look up the relative atomic masses (Ar) of each element from the periodic table.
- Multiply each Ar by the number of that atom in the formula.
- Sum all values to get the molar mass M in g/mol.
Example: Calculate the molar mass of H&sub2;SO&sub4;.
- H = 1 g/mol; S = 32 g/mol; O = 16 g/mol.
- M(H&sub2;SO&sub4;) = 2(1) + 32 + 4(16) = 2 + 32 + 64 = 98 g/mol.
Critical Rule: The molar mass in g/mol is numerically the same as the relative molecular mass (Mr) but has units. Always include units in your calculations. Avogadro's number (6.02 × 10²³) applies to any particle: atoms, molecules, ions, or formula units.
Sciences · Topic 2.2
Stoichiometry — Mole Ratios
A balanced chemical equation gives the exact mole ratios in which reactants combine and products form. These ratios are the foundation of all stoichiometric calculations.
Balancing Equations
- Write the unbalanced equation with correct chemical formulas.
- Count atoms of each element on both sides.
- Add coefficients (not subscripts) to balance. Start with the most complex molecule.
- Check: all atom counts must match on both sides; all coefficients should be whole numbers in lowest ratio.
Example: Balance: Fe + O&sub2; → Fe&sub2;O&sub3;
Unbalanced: Fe (1) + O&sub2; (2 O) → Fe&sub2;O&sub3; (2 Fe, 3 O).
Balance Fe: 4Fe + O&sub2; → 2Fe&sub2;O&sub3; (4 Fe each side). Balance O: 4Fe + 3O&sub2; → 2Fe&sub2;O&sub3; (6 O each side).
Balanced: 4Fe + 3O&sub2; → 2Fe&sub2;O&sub3;
Unbalanced: Fe (1) + O&sub2; (2 O) → Fe&sub2;O&sub3; (2 Fe, 3 O).
Balance Fe: 4Fe + O&sub2; → 2Fe&sub2;O&sub3; (4 Fe each side). Balance O: 4Fe + 3O&sub2; → 2Fe&sub2;O&sub3; (6 O each side).
Balanced: 4Fe + 3O&sub2; → 2Fe&sub2;O&sub3;
Stoichiometric Calculation Method
- Write the balanced equation and identify the mole ratio between given and required substances.
- Convert given quantity to moles: n = m / M.
- Apply the mole ratio to find moles of required substance.
- Convert moles to required unit (mass: m = n × M; volume: V = n × 22.4).
Example: How many grams of H&sub2;O are produced when 4 g of H&sub2; reacts completely with excess O&sub2;?
Equation: 2H&sub2; + O&sub2; → 2H&sub2;O. Mole ratio H&sub2; : H&sub2;O = 2 : 2 = 1 : 1.
n(H&sub2;) = 4 / 2 = 2 mol. n(H&sub2;O) = 2 mol (1:1 ratio).
m(H&sub2;O) = 2 × 18 = 36 g.
Equation: 2H&sub2; + O&sub2; → 2H&sub2;O. Mole ratio H&sub2; : H&sub2;O = 2 : 2 = 1 : 1.
n(H&sub2;) = 4 / 2 = 2 mol. n(H&sub2;O) = 2 mol (1:1 ratio).
m(H&sub2;O) = 2 × 18 = 36 g.
Sciences · Topic 2.2
Ionic and Covalent Bonding
Chemical bonding determines the structure of substances, which in turn determines all their physical properties. At Year 4, you must link bonding type to specific properties with justification.
Ionic Bonding
- Occurs between metals and non-metals.
- Metal atoms lose electrons to form positive ions (cations); non-metals gain electrons to form negative ions (anions).
- Ions arrange into a regular lattice structure held by strong electrostatic forces of attraction.
- Example: NaCl — Na loses 1e¹¯ to become Na♠, Cl gains 1e¹¯ to become Cl¹¯.
Covalent Bonding
- Occurs between non-metals.
- Atoms share pairs of electrons to achieve a full outer shell.
- Single bond = 1 shared pair; double bond = 2 shared pairs; triple bond = 3 shared pairs.
- Example: H&sub2;O — oxygen shares one electron pair with each hydrogen.
Bonding and Properties
| Property | Ionic compounds | Simple covalent molecules | Giant covalent structures |
|---|---|---|---|
| Melting point | High (strong lattice forces) | Low (weak intermolecular forces) | Very high (e.g., diamond: strong C-C bonds throughout) |
| Electrical conductivity | Conducts when dissolved or molten (ions free to move); not as solid | Does not conduct (no charged particles) | Does not conduct (except graphite: delocalised electrons) |
| Solubility in water | Usually soluble (ions attracted to polar water) | Variable (polar dissolves, non-polar does not) | Insoluble (too strong to break apart) |
Critical Rule: Ionic compounds only conduct electricity when dissolved in water or melted — not as a solid. In a solid ionic lattice, ions are fixed in position and cannot carry charge. This must be explicitly stated in any answer about ionic conductivity.
Sciences · Topic 2.2
Rates of Reaction and Collision Theory
Collision theory explains why reactions have different rates and how factors change them. For a reaction to occur, particles must collide with sufficient energy (at least the activation energy) and the correct orientation.
Five Factors Affecting Rate of Reaction
| Factor | Effect on rate | Collision theory explanation |
|---|---|---|
| Temperature | Increase → faster | Particles have more kinetic energy: more frequent collisions AND higher proportion exceed activation energy |
| Concentration | Increase → faster | More particles per unit volume → more frequent collisions |
| Surface area | Increase → faster | More particles exposed at the surface → more collision opportunities for reacting particles |
| Catalyst | Present → faster | Provides alternative reaction pathway with lower activation energy; more collisions have sufficient energy to react |
| Pressure (gases) | Increase → faster | Same as concentration: more particles per volume → more frequent collisions |
Catalyst: Strength and Limitation
Industrial significance: Catalysts lower energy costs in industrial processes (e.g., iron catalyst in the Haber process for ammonia synthesis). They make reactions economically viable at lower temperatures and pressures, reducing energy consumption and CO&sub2; emissions.
Limitation: Catalysts can be poisoned (deactivated) by impurities; they are specific to particular reactions; some are expensive metals (e.g., platinum). Catalysts don't change the equilibrium position of reversible reactions — only how quickly equilibrium is reached.
Industrial significance: Catalysts lower energy costs in industrial processes (e.g., iron catalyst in the Haber process for ammonia synthesis). They make reactions economically viable at lower temperatures and pressures, reducing energy consumption and CO&sub2; emissions.
Limitation: Catalysts can be poisoned (deactivated) by impurities; they are specific to particular reactions; some are expensive metals (e.g., platinum). Catalysts don't change the equilibrium position of reversible reactions — only how quickly equilibrium is reached.
Critical Rule: A catalyst is NOT consumed in the reaction. It is chemically unchanged at the end and can be recovered and reused. This is what distinguishes a catalyst from a reactant. A catalyst speeds up reaction by lowering activation energy, NOT by increasing the frequency of collisions.
Sciences · Topic 2.2
Worked Examples
Detailed calculations and evaluations showing Year 4 Advanced standard.
EXAMPLE 1Calculate the number of moles in 88 g of CO&sub2;. (C = 12, O = 16)
+
Full Solution
Step 1 — Molar mass of CO&sub2;:
M(CO&sub2;) = 12 + 2(16) = 12 + 32 = 44 g/mol.
Step 2 — Number of moles:
n = m / M = 88 / 44 = 2 mol.
Interpretation: 88 g of CO&sub2; contains 2 moles, or 2 × 6.02 × 10²³ = 1.204 × 10²&sup4; molecules of CO&sub2;.
M(CO&sub2;) = 12 + 2(16) = 12 + 32 = 44 g/mol.
Step 2 — Number of moles:
n = m / M = 88 / 44 = 2 mol.
Interpretation: 88 g of CO&sub2; contains 2 moles, or 2 × 6.02 × 10²³ = 1.204 × 10²&sup4; molecules of CO&sub2;.
EXAMPLE 2What mass of iron(III) oxide (Fe&sub2;O&sub3;) is produced from 56 g of iron? (Fe = 56, O = 16)
+
Full Solution
Balanced equation: 4Fe + 3O&sub2; → 2Fe&sub2;O&sub3;
Mole ratio Fe : Fe&sub2;O&sub3; = 4 : 2 = 2 : 1.
n(Fe): n = 56 / 56 = 1 mol.
n(Fe&sub2;O&sub3;): Using ratio 2:1 → n(Fe&sub2;O&sub3;) = 1/2 = 0.5 mol.
m(Fe&sub2;O&sub3;): M(Fe&sub2;O&sub3;) = 2(56) + 3(16) = 112 + 48 = 160 g/mol.
m = 0.5 × 160 = 80 g.
Mole ratio Fe : Fe&sub2;O&sub3; = 4 : 2 = 2 : 1.
n(Fe): n = 56 / 56 = 1 mol.
n(Fe&sub2;O&sub3;): Using ratio 2:1 → n(Fe&sub2;O&sub3;) = 1/2 = 0.5 mol.
m(Fe&sub2;O&sub3;): M(Fe&sub2;O&sub3;) = 2(56) + 3(16) = 112 + 48 = 160 g/mol.
m = 0.5 × 160 = 80 g.
EXAMPLE 3Explain why NaCl has a high melting point but a low-molecular-weight covalent molecule like water has a low boiling point.
+
Full Solution
NaCl: Consists of Na♠ and Cl¹¯ ions arranged in a giant ionic lattice. The lattice is held by strong electrostatic forces of attraction between oppositely charged ions in all directions. A large amount of energy is needed to overcome these forces and separate the ions — hence NaCl has a high melting point (801°C).
Water (H&sub2;O): Consists of individual H&sub2;O molecules with covalent O-H bonds within each molecule. The forces between molecules are weak intermolecular forces (hydrogen bonds in water's case, which are still weaker than ionic bonds). Little energy is needed to separate the molecules — hence water has a relatively low boiling point (100°C at 1 atm).
Key distinction: When melting or boiling a substance, you break intermolecular forces (for covalent molecular substances) or lattice forces (for ionic compounds) — NOT the covalent bonds within molecules. This is why even though H&sub2;O has strong covalent O-H bonds (within the molecule), it has a low boiling point because the between-molecule forces are weaker than ionic lattice forces.
Water (H&sub2;O): Consists of individual H&sub2;O molecules with covalent O-H bonds within each molecule. The forces between molecules are weak intermolecular forces (hydrogen bonds in water's case, which are still weaker than ionic bonds). Little energy is needed to separate the molecules — hence water has a relatively low boiling point (100°C at 1 atm).
Key distinction: When melting or boiling a substance, you break intermolecular forces (for covalent molecular substances) or lattice forces (for ionic compounds) — NOT the covalent bonds within molecules. This is why even though H&sub2;O has strong covalent O-H bonds (within the molecule), it has a low boiling point because the between-molecule forces are weaker than ionic lattice forces.
EXAMPLE 4A student investigates how concentration affects the rate of reaction between HCl and marble chips. Evaluate the design.
+
Full Solution
Reaction: CaCO&sub3; + 2HCl → CaCl&sub2; + H&sub2;O + CO&sub2;
The student could measure the volume of CO&sub2; produced per minute, or the change in mass (as CO&sub2; escapes).
Variables: Independent = HCl concentration. Dependent = rate of CO&sub2; production (cm³/min). Controlled = temperature, marble chip size (same surface area), volume of HCl, mass of marble chips.
Strengths: Measuring CO&sub2; volume gives continuous data and allows a rate vs time graph; mass loss is easy to measure and gives clear data.
Limitations: Gas collection may lose some CO&sub2; before the tube is connected; measuring mass loss is affected by the balance sensitivity; marble chips are not perfectly uniform in size — surface area is difficult to control exactly. The student should use at least 3 trials per concentration for reliability and calculate mean rate.
Conclusion: The design is broadly valid but requires careful attention to controlling surface area (use crushed marble of known mass and similar size) and three repeats to identify and remove anomalies.
The student could measure the volume of CO&sub2; produced per minute, or the change in mass (as CO&sub2; escapes).
Variables: Independent = HCl concentration. Dependent = rate of CO&sub2; production (cm³/min). Controlled = temperature, marble chip size (same surface area), volume of HCl, mass of marble chips.
Strengths: Measuring CO&sub2; volume gives continuous data and allows a rate vs time graph; mass loss is easy to measure and gives clear data.
Limitations: Gas collection may lose some CO&sub2; before the tube is connected; measuring mass loss is affected by the balance sensitivity; marble chips are not perfectly uniform in size — surface area is difficult to control exactly. The student should use at least 3 trials per concentration for reliability and calculate mean rate.
Conclusion: The design is broadly valid but requires careful attention to controlling surface area (use crushed marble of known mass and similar size) and three repeats to identify and remove anomalies.
EXAMPLE 5Using collision theory, explain why increasing temperature has a greater effect on reaction rate than increasing concentration.
+
Full Solution
Increasing concentration increases the number of particles per unit volume, so collisions become more frequent. However, the energy of collisions remains the same — so only the collision frequency increases.
Increasing temperature has a dual effect: (1) particles move faster, so collisions are more frequent; AND (2) a greater proportion of particles now have energy exceeding the activation energy (Ea). According to the Maxwell-Boltzmann distribution, even a small temperature rise significantly increases the fraction of particles with E ≥ Ea.
Conclusion: Temperature affects both collision frequency AND the proportion of successful collisions (those with enough energy). Concentration only affects frequency. This dual mechanism makes temperature a more powerful factor in changing reaction rate, particularly for reactions with high activation energies.
Increasing temperature has a dual effect: (1) particles move faster, so collisions are more frequent; AND (2) a greater proportion of particles now have energy exceeding the activation energy (Ea). According to the Maxwell-Boltzmann distribution, even a small temperature rise significantly increases the fraction of particles with E ≥ Ea.
Conclusion: Temperature affects both collision frequency AND the proportion of successful collisions (those with enough energy). Concentration only affects frequency. This dual mechanism makes temperature a more powerful factor in changing reaction rate, particularly for reactions with high activation energies.
EXAMPLE 6Calculate the volume of O&sub2; produced at STP when 2.45 g of KClO&sub3; decomposes. (K=39, Cl=35.5, O=16)
+
Full Solution
Balanced equation: 2KClO&sub3; → 2KCl + 3O&sub2;
Mole ratio KClO&sub3; : O&sub2; = 2 : 3.
M(KClO&sub3;): 39 + 35.5 + 3(16) = 39 + 35.5 + 48 = 122.5 g/mol.
n(KClO&sub3;): 2.45 / 122.5 = 0.02 mol.
n(O&sub2;): 0.02 × (3/2) = 0.03 mol.
V(O&sub2;) at STP: 0.03 × 22.4 = 0.672 L = 672 cm³.
Mole ratio KClO&sub3; : O&sub2; = 2 : 3.
M(KClO&sub3;): 39 + 35.5 + 3(16) = 39 + 35.5 + 48 = 122.5 g/mol.
n(KClO&sub3;): 2.45 / 122.5 = 0.02 mol.
n(O&sub2;): 0.02 × (3/2) = 0.03 mol.
V(O&sub2;) at STP: 0.03 × 22.4 = 0.672 L = 672 cm³.
Sciences · Topic 2.2
Practice Q&A
Attempt each question before revealing the model answer.
CALCULATEHow many moles are in 36 g of H&sub2;O? (H=1, O=16)
+
Model Answer
M(H&sub2;O) = 2(1) + 16 = 18 g/mol. n = 36 / 18 = 2 mol.
APPLYExplain why a catalyst is not considered a reactant.
+
Model Answer
A catalyst speeds up a reaction by providing an alternative reaction pathway with lower activation energy, but it is not consumed during the reaction. It is chemically unchanged at the end and can be recovered. A reactant, by definition, is consumed and converted into products during the reaction. Because the catalyst can be reused, it is classified separately from reactants.
EXPLAINWhy does NaCl conduct electricity when molten but not when solid?
+
Model Answer
In solid NaCl, Na♠ and Cl¹¯ ions are fixed in the ionic lattice and cannot move — no mobile charge carriers, so no electrical conductivity. When melted, the lattice breaks down and ions become free to move toward oppositely charged electrodes, allowing charge to flow. The same applies when dissolved in water: ions dissociate and become mobile in solution.
CALCULATEWhat mass of CO&sub2; is produced when 2 mol of C&sub3;H&sub8; (propane) burns completely? (C=12, O=16, H=1)
+
Model Answer
Balanced equation: C&sub3;H&sub8; + 5O&sub2; → 3CO&sub2; + 4H&sub2;O. Mole ratio C&sub3;H&sub8; : CO&sub2; = 1 : 3.
n(CO&sub2;) = 2 × 3 = 6 mol. M(CO&sub2;) = 12 + 32 = 44 g/mol. m(CO&sub2;) = 6 × 44 = 264 g.
n(CO&sub2;) = 2 × 3 = 6 mol. M(CO&sub2;) = 12 + 32 = 44 g/mol. m(CO&sub2;) = 6 × 44 = 264 g.
EVALUATEEvaluate the use of surface area as a variable in a marble chip experiment.
+
Model Answer
Strength: Surface area directly affects reaction rate; using the same mass of marble in different chip sizes (whole, crushed, powdered) provides a clear, observable relationship between surface area and rate.
Limitation: Surface area is difficult to measure precisely for irregularly shaped chips. Using mass as a proxy for surface area is not accurate because two chips of the same mass may have different surface areas. This introduces uncertainty. A more controlled method would use chips of identical measured surface area (geometric shapes) or use chemical etching to estimate surface area.
Conclusion: The variable is valid but the method has limitations in controlling surface area exactly; results show the trend but quantitative comparison of different experiments is unreliable.
Limitation: Surface area is difficult to measure precisely for irregularly shaped chips. Using mass as a proxy for surface area is not accurate because two chips of the same mass may have different surface areas. This introduces uncertainty. A more controlled method would use chips of identical measured surface area (geometric shapes) or use chemical etching to estimate surface area.
Conclusion: The variable is valid but the method has limitations in controlling surface area exactly; results show the trend but quantitative comparison of different experiments is unreliable.
COMPARECompare the electrical conductivity of diamond and graphite. Both are giant covalent structures of carbon. Explain the difference.
+
Model Answer
Diamond: Each carbon atom forms 4 covalent bonds in a tetrahedral arrangement. All outer-shell electrons are used in bonding — there are no free/delocalised electrons. Diamond does not conduct electricity.
Graphite: Each carbon atom forms 3 covalent bonds in a hexagonal layer structure. The 4th outer-shell electron from each carbon atom is delocalised, free to move between layers. These mobile electrons act as charge carriers, allowing graphite to conduct electricity.
Conclusion: Both are giant covalent carbon structures, but graphite's layered structure with delocalised electrons makes it a conductor — a unique exception among covalent substances.
Graphite: Each carbon atom forms 3 covalent bonds in a hexagonal layer structure. The 4th outer-shell electron from each carbon atom is delocalised, free to move between layers. These mobile electrons act as charge carriers, allowing graphite to conduct electricity.
Conclusion: Both are giant covalent carbon structures, but graphite's layered structure with delocalised electrons makes it a conductor — a unique exception among covalent substances.
DISCUSSEvaluate the industrial significance of catalysts, with reference to a specific example.
+
Model Answer
Example — Haber process: Nitrogen and hydrogen react to form ammonia: N&sub2; + 3H&sub2; ⇌ 2NH&sub3;. Without a catalyst (iron with promoters), this reaction requires very high temperatures to proceed at a useful rate — making it economically unviable.
Industrial significance: The iron catalyst allows the reaction to proceed at 400–500°C rather than thousands of degrees, dramatically reducing fuel consumption and production costs. Ammonia from the Haber process is essential for fertiliser production, supporting global food supply for billions of people.
Environmental benefit: Lower temperatures mean less energy and lower CO&sub2; emissions per tonne of ammonia produced.
Limitation: The catalyst is sensitive to catalyst poisons (e.g., sulfur compounds) which block active sites. Impurities in feedstock gases must be carefully removed, adding processing cost. Additionally, the equilibrium yield at 400°C is only ~15% — unreacted gases must be recycled, requiring additional equipment.
Industrial significance: The iron catalyst allows the reaction to proceed at 400–500°C rather than thousands of degrees, dramatically reducing fuel consumption and production costs. Ammonia from the Haber process is essential for fertiliser production, supporting global food supply for billions of people.
Environmental benefit: Lower temperatures mean less energy and lower CO&sub2; emissions per tonne of ammonia produced.
Limitation: The catalyst is sensitive to catalyst poisons (e.g., sulfur compounds) which block active sites. Impurities in feedstock gases must be carefully removed, adding processing cost. Additionally, the equilibrium yield at 400°C is only ~15% — unreacted gases must be recycled, requiring additional equipment.
JUSTIFYWhy do ionic compounds typically have higher melting points than simple covalent compounds?
+
Model Answer
Ionic compounds form a giant ionic lattice held together by strong electrostatic forces of attraction between oppositely charged ions, acting in all three dimensions. To melt an ionic compound, enough thermal energy must be supplied to overcome the lattice energy — this is large, giving high melting points (e.g., NaCl: 801°C, MgO: 2852°C).
Simple covalent compounds (e.g., H&sub2;O, CO&sub2;) consist of discrete molecules. The forces between molecules are weak intermolecular forces (van der Waals, dipole-dipole, hydrogen bonds) — far weaker than ionic bonds. Only these intermolecular forces need to be overcome during melting, so much less energy is required — resulting in low melting points.
Exception: Giant covalent structures (e.g., diamond, SiO&sub2;) have very high melting points because covalent bonds throughout the whole structure must be broken.
Simple covalent compounds (e.g., H&sub2;O, CO&sub2;) consist of discrete molecules. The forces between molecules are weak intermolecular forces (van der Waals, dipole-dipole, hydrogen bonds) — far weaker than ionic bonds. Only these intermolecular forces need to be overcome during melting, so much less energy is required — resulting in low melting points.
Exception: Giant covalent structures (e.g., diamond, SiO&sub2;) have very high melting points because covalent bonds throughout the whole structure must be broken.
Sciences · Review
Flashcard Review
Tap each card to reveal the answer. Try to answer from memory first.
State the mole formula triangle.
n = m / M | m = n × M | M = m / n
n = moles (mol), m = mass (g), M = molar mass (g/mol)
n = moles (mol), m = mass (g), M = molar mass (g/mol)
Tap to reveal
What is Avogadro's number?
6.02 × 10²³ — the number of particles in one mole of any substance (atoms, molecules, ions, or formula units).
Tap to reveal
What is the volume of one mole of any gas at STP?
22.4 litres (22 400 cm³) at Standard Temperature and Pressure (0°C, 1 atm).
Tap to reveal
What type of bonding occurs between a metal and a non-metal?
Ionic bonding — the metal loses electrons (forms cation) and the non-metal gains electrons (forms anion). Held by electrostatic attraction.
Tap to reveal
Why does ionic NaCl conduct when dissolved but not as a solid?
In solution, ions are free to move and carry charge. In solid form, ions are locked in the lattice and cannot move — no electrical conductivity.
Tap to reveal
What is a catalyst? Key property?
A substance that increases reaction rate by providing an alternative pathway with lower activation energy. It is NOT consumed — chemically unchanged at the end of the reaction.
Tap to reveal
State five factors that increase the rate of reaction.
1. Temperature (increase) 2. Concentration (increase) 3. Surface area (increase) 4. Catalyst (add) 5. Pressure — for gases (increase)
Tap to reveal
What does a balanced equation's coefficients tell you?
The mole ratio of reactants to products. E.g., 2H&sub2; + O&sub2; → 2H&sub2;O means 2 mol H&sub2; reacts with 1 mol O&sub2; to produce 2 mol H&sub2;O.
Tap to reveal
What is activation energy?
The minimum energy that colliding particles must have for a reaction to occur. A catalyst lowers E⊂a;, increasing the proportion of successful collisions.
Tap to reveal
Why does increasing temperature increase reaction rate more effectively than increasing concentration?
Concentration only increases collision frequency. Temperature increases BOTH collision frequency AND the proportion of particles with energy ≥ activation energy — a dual effect.
Tap to reveal
Molar mass of H&sub2;SO&sub4;? (H=1, S=32, O=16)
M = 2(1) + 32 + 4(16) = 2 + 32 + 64 = 98 g/mol
Tap to reveal
Why does graphite conduct electricity but diamond does not?
Graphite: each C forms 3 bonds; the 4th electron is delocalised between layers — mobile charge carriers. Diamond: each C forms 4 bonds; all electrons used in bonding — no free electrons.
Tap to reveal
What is covalent bonding?
The sharing of electron pairs between two non-metal atoms so both achieve a full outer shell. Can be single (1 pair), double (2 pairs), or triple (3 pairs) bonds.
Tap to reveal
State the first step in any stoichiometry calculation.
Write and balance the chemical equation. Then use the coefficients to identify the mole ratio between the given and required substances.
Tap to reveal
Why do simple covalent compounds have low melting points?
Simple covalent molecules are held together by weak intermolecular forces (van der Waals, dipole-dipole). Little energy is needed to separate the molecules. Note: the covalent bonds within molecules are strong — but they are NOT broken on melting.
Tap to reveal
Sciences · Practice Test
Practice Test — 20 Questions
0Score / 20
Q 1 / 20
Correct
Wrong
Score