Chemistry HSC Reference Sheet — Explained
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The NESA Chemistry exam provides a data sheet listing physical constants and a formula sheet — but having numbers printed in front of you is not the same as knowing when to reach for them. Below, every constant and formula is explained: what each symbol means, when to use it, a worked example, and a practice question to try yourself.
Constants & Data
7The number of particles (atoms, molecules, ions) in exactly one mole of a substance.
Used in: Converting between moles and number of particles: $n = N / N_A$.
Proportionality constant relating energy to amount of gas and temperature in the ideal gas law.
Used in: Ideal gas law $PV = nRT$; any calculation mixing pressure, volume, moles and temperature.
Volume occupied by exactly one mole of ideal gas at 0 °C and 100 kPa.
Used in: Quick gas-volume calculations when temperature is 0 °C and pressure is 100 kPa.
Volume occupied by exactly one mole of ideal gas at 25 °C (298 K) and 100 kPa.
Used in: Quick gas-volume calculations at room temperature and 100 kPa (standard laboratory conditions).
The equilibrium constant for the autoionisation of water: $\text{H}_2\text{O} \rightleftharpoons \text{H}^+ + \text{OH}^-$. Valid at 25 °C.
Used in: Calculating $[\text{OH}^-]$ from $[\text{H}^+]$ (or vice versa); finding pOH; pH of strong base solutions.
Energy required to raise 1 g of liquid water by 1 K (or 1 °C).
Used in: Calorimetry: $q = mc\Delta T$ to find heat transferred to or from aqueous solutions.
Charge carried by one mole of electrons (1 mole × elementary charge).
Used in: Electrochemistry calculations: charge transferred $Q = nF$; amount of substance deposited/dissolved at an electrode.
Mole & Stoichiometry
3What each symbol means
$n$ — amount of substance (mol); $m$ — mass (g); $M$ — molar mass (g mol⁻¹).
When to use it
Any time you are given a mass and need moles, or given moles and need to find the mass of a substance.
Units:
mol (when $m$ in g and $M$ in g mol⁻¹).
Worked sample
How many moles are in 36 g of water ($M = 18.02$ g mol⁻¹)?
$n = 36 / 18.02 = 2.00$ mol.
Your turn:
What mass of NaOH ($M = 40.00$ g mol⁻¹) is needed to make 0.500 mol?
$m = nM = 0.500 \times 40.00 = 20.0$ g.
Always check units: molar mass must be in g mol⁻¹ to give moles, not kg mol⁻¹.
What each symbol means
$n$ — amount of substance (mol); $N$ — number of particles; $N_A = 6.022 \times 10^{23}$ mol⁻¹ (Avogadro's constant).
When to use it
Converting between a raw count of atoms, molecules or ions and the amount in moles.
Units:
mol
Worked sample
How many moles is $1.806 \times 10^{24}$ molecules?
$n = 1.806 \times 10^{24} / 6.022 \times 10^{23} = 3.00$ mol.
Your turn:
How many atoms are in 2.00 mol of carbon?
$N = nN_A = 2.00 \times 6.022 \times 10^{23} = 1.20 \times 10^{24}$ atoms.
Particles can be atoms, molecules, formula units or ions — re-read the question carefully.
What each symbol means
$n$ — amount of substance (mol); $c$ — molar concentration (mol L⁻¹); $V$ — volume (L).
When to use it
Finding moles of solute from a solution's concentration and volume, or working out what volume contains a required number of moles.
Units:
mol (requires $V$ in litres).
Worked sample
How many moles of HCl are in 250 mL of 0.100 mol L⁻¹ solution?
$n = 0.100 \times 0.250 = 0.0250$ mol.
Your turn:
What volume of 2.00 mol L⁻¹ NaOH contains 0.150 mol?
$V = n/c = 0.150/2.00 = 0.0750$ L = 75.0 mL.
Convert mL to L before substituting: divide by 1000.
Gases
2What each symbol means
$P$ — pressure (Pa); $V$ — volume (m³); $n$ — amount of substance (mol); $R = 8.314$ J mol⁻¹ K⁻¹; $T$ — absolute temperature (K).
When to use it
Any gas calculation where pressure, volume, moles and temperature are mixed — including finding an unknown when the other three are known, or finding molar mass from gas density data.
Units:
SI units throughout: Pa for pressure, m³ for volume, K for temperature. Alternatively, use kPa and L (since 1 kPa × 1 L = 1 J) — the same $R$ works.
Worked sample
Find the volume of 0.500 mol of ideal gas at 200 kPa and 27 °C.
$T = 27 + 273 = 300$ K. Using kPa and L: $V = nRT/P = 0.500 \times 8.314 \times 300 / 200 = 6.24$ L.
Your turn:
A 2.00 L flask at 25 °C contains gas at 150 kPa. How many moles?
$n = PV/RT = 150 \times 2.00 / (8.314 \times 298) = 0.121$ mol.
Temperature must be in Kelvin: $T(\text{K}) = T(°\text{C}) + 273$. Forgetting this is the single most common error in gas calculations.
What each symbol means
$V$ — volume of gas (L); $n$ — amount of substance (mol); $V_m$ — molar volume at the given conditions (22.71 L mol⁻¹ at 0 °C / 100 kPa; 24.79 L mol⁻¹ at 25 °C / 100 kPa).
When to use it
Quick calculations at standard temperature and pressure — avoids the full ideal gas law when $T$ and $P$ are fixed at the standard conditions.
Units:
L
Worked sample
What volume does 3.00 mol of H₂ occupy at 25 °C and 100 kPa?
$V = 3.00 \times 24.79 = 74.4$ L.
Your turn:
How many moles of O₂ are in 11.36 L at STP (0 °C, 100 kPa)?
$n = 11.36 / 22.71 = 0.500$ mol.
Use $V_m = 24.79$ L mol⁻¹ for 25 °C (room temperature) and $V_m = 22.71$ L mol⁻¹ for 0 °C — both at 100 kPa. Check the question's stated conditions.
Solutions & Concentration
3What each symbol means
$c$ — molar concentration (mol L⁻¹); $n$ — amount of solute (mol); $V$ — volume of solution (L).
When to use it
Making up a standard solution, calculating concentration from titration data, or determining how many moles are in a given volume.
Units:
mol L⁻¹ (also written M)
Worked sample
4.00 g of NaOH ($M = 40.00$ g mol⁻¹) is dissolved to make 500 mL of solution. Find $c$.
$n = 4.00/40.00 = 0.100$ mol; $c = 0.100/0.500 = 0.200$ mol L⁻¹.
Your turn:
What is the concentration of a solution containing 0.360 mol of glucose in 1.50 L?
$c = 0.360/1.50 = 0.240$ mol L⁻¹.
Titration endpoint: at equivalence point $n_{\text{acid}} \times$ stoich ratio $= n_{\text{base}}$. Write $c_1V_1 \times \text{ratio} = c_2V_2$ for polyprotic acids/bases.
What each symbol means
$c_1$ — initial concentration (mol L⁻¹); $V_1$ — initial volume (L); $c_2$ — final concentration (mol L⁻¹); $V_2$ — final volume (L).
When to use it
Calculating the concentration after diluting a solution, or determining what volume of stock solution to dilute.
Units:
Volumes in same unit on both sides (L or mL — must match).
Worked sample
25.0 mL of 2.00 mol L⁻¹ HCl is diluted to 500 mL. Find $c_2$.
$c_2 = c_1V_1/V_2 = 2.00 \times 25.0/500 = 0.100$ mol L⁻¹.
Your turn:
What volume of 6.00 mol L⁻¹ H₂SO₄ is needed to make 250 mL of 0.600 mol L⁻¹ solution?
$V_1 = c_2V_2/c_1 = 0.600 \times 250/6.00 = 25.0$ mL.
The moles of solute are conserved on dilution — that is the physical basis of this formula.
What each symbol means
$m_{\text{solute}}$ — mass of solute (g); $m_{\text{solution}}$ — mass of solution (g). 1 ppm = 1 mg kg⁻¹ = 1 mg L⁻¹ (for dilute aqueous solutions where $\rho \approx 1$ g mL⁻¹).
When to use it
Expressing very low concentrations of trace substances (pollutants, minerals in water, blood alcohol).
Units:
Dimensionless (mg kg⁻¹ for solids; mg L⁻¹ for dilute aqueous solutions).
Worked sample
A water sample contains 0.003 g of lead in 1.000 kg (1000 g). Express in ppm.
$\text{ppm} = 0.003/1000 \times 10^6 = 3$ ppm.
Your turn:
Convert 8.5 mg L⁻¹ of fluoride in water to ppm (dilute solution).
8.5 ppm (1 mg L⁻¹ ≈ 1 ppm for dilute aqueous solutions).
For dilute aqueous solutions, mg L⁻¹ and ppm are numerically equal — you can use either form.
Energetics & Calorimetry
3What each symbol means
$q$ — heat transferred (J); $m$ — mass of solution or substance (g); $c$ — specific heat capacity (J g⁻¹ K⁻¹; for water $c = 4.18$ J g⁻¹ K⁻¹); $\Delta T$ — temperature change (K or °C, same magnitude).
When to use it
Calorimetry experiments: calculating heat released or absorbed by a reaction from the temperature change of the surrounding water/solution.
Units:
Joules (J); convert to kJ by dividing by 1000.
Worked sample
100 g of solution rises from 21.0 °C to 27.5 °C on dissolving a salt. Find $q$.
$q = 100 \times 4.18 \times (27.5 - 21.0) = 100 \times 4.18 \times 6.5 = 2717$ J $= 2.72$ kJ.
Your turn:
200 g of water cools from 85 °C to 20 °C. How much heat did the water lose?
$q = 200 \times 4.18 \times 65 = 54\,340$ J $= 54.3$ kJ.
Sign convention: if temperature rises, the surroundings gained heat ($q > 0$), so the reaction released heat ($\Delta H < 0$, exothermic). Flip the sign for the enthalpy of reaction.
What each symbol means
$\Delta H$ — molar enthalpy change (kJ mol⁻¹); $q$ — heat transferred to surroundings (kJ, from $q = mc\Delta T$); $n$ — moles of the substance specified in the equation.
When to use it
Converting a calorimetry result ($q$) into a standard molar enthalpy ($\Delta H$) for reporting, or for comparing with literature values.
Units:
kJ mol⁻¹
Worked sample
Burning 0.500 g of ethanol ($M = 46.07$ g mol⁻¹) heated 200 g of water by 13.2 °C. Find $\Delta H_{\text{comb}}$.
$q = 200 \times 4.18 \times 13.2 = 11035$ J $= 11.04$ kJ; $n = 0.500/46.07 = 0.01085$ mol; $\Delta H = -11.04/0.01085 = -1017$ kJ mol⁻¹.
Your turn:
2.72 kJ of heat was released when 0.050 mol of NaOH dissolved. Find $\Delta H_{\text{soln}}$.
$\Delta H = -2.72/0.050 = -54.4$ kJ mol⁻¹.
The negative sign makes $\Delta H$ negative (exothermic) when heat is released to the surroundings — a common source of sign errors.
What each symbol means
$\Delta H_{\text{rxn}}$ — enthalpy change for the target reaction; $\sum \Delta H_{\text{products}}$ — sum of standard enthalpies of formation of products; $\sum \Delta H_{\text{reactants}}$ — sum of standard enthalpies of formation of reactants.
When to use it
Calculating $\Delta H$ of a reaction from standard enthalpies of formation, or combining thermochemical equations (multiply, reverse, add) when a direct measurement is not available.
Units:
kJ mol⁻¹
Worked sample
Given $\Delta H_f^\circ$: CO₂(g) = −393.5, H₂O(l) = −285.8, C₂H₅OH(l) = −277.7 kJ mol⁻¹. Find $\Delta H_{\text{comb}}$ of ethanol.
C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O. $\Delta H = [2(-393.5) + 3(-285.8)] - [(-277.7) + 0] = [-787 - 857.4] + 277.7 = -1366.7$ kJ mol⁻¹.
Your turn:
Using Hess's Law, what sign of $\Delta H$ indicates an endothermic reaction?
Positive $\Delta H$ (products have higher enthalpy than reactants).
When reversing a thermochemical equation, change the sign of $\Delta H$. When multiplying coefficients, multiply $\Delta H$ by the same factor.
Equilibrium & Acids/Bases
7What each symbol means
$[\text{X}]$ — equilibrium molar concentration of species X (mol L⁻¹); $a, b, c, d$ — stoichiometric coefficients from the balanced equation $a\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D}$.
When to use it
Writing an equilibrium expression for any reversible reaction; comparing $Q$ vs $K$ to predict direction of shift; calculating equilibrium concentrations.
Units:
$K_{eq}$ is dimensionless in modern convention (activities); in HSC problems it carries units of (mol L⁻¹)^{Δn} — omit units unless asked.
Worked sample
Write $K_{eq}$ for N₂(g) + 3H₂(g) ⇌ 2NH₃(g).
$K_{eq} = \dfrac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}$.
Your turn:
Write $K_{eq}$ for 2SO₂(g) + O₂(g) ⇌ 2SO₃(g). If $K_{eq} = 280$ at a certain temperature, what does a large $K$ indicate?
$K_{eq} = [\text{SO}_3]^2/([\text{SO}_2]^2[\text{O}_2])$. Large $K$ indicates products are heavily favoured at equilibrium.
Pure solids and liquids (including water as solvent) are not included in the $K$ expression — their concentrations are constant.
What each symbol means
Same form as $K_{eq}$ but uses current (non-equilibrium) concentrations. $Q < K$: reaction proceeds forward; $Q > K$: reaction proceeds in reverse; $Q = K$: system is at equilibrium.
When to use it
Predicting which direction a reaction will shift when conditions change (e.g., adding or removing a reactant).
Units:
Same convention as $K_{eq}$.
Worked sample
For N₂ + 3H₂ ⇌ 2NH₃, $K = 6.2 \times 10^{-2}$. At a moment, $[\text{N}_2] = 0.1$, $[\text{H}_2] = 0.1$, $[\text{NH}_3] = 0.1$ mol L⁻¹. Which way will the reaction shift?
$Q = 0.1^2/(0.1 \times 0.1^3) = 0.01/0.0001 = 100 > K$. Reaction shifts in reverse (towards reactants).
Your turn:
If $Q < K$, which direction does the reaction proceed?
Forward — towards products.
Calculate $Q$ by substituting the current concentrations (not equilibrium values) into the same expression as $K$.
What each symbol means
$[\text{H}^+]$ — molar concentration of hydrogen ions (mol L⁻¹, at 25 °C); equivalently $[\text{H}_3\text{O}^+]$.
When to use it
Converting between $[\text{H}^+]$ and pH, or calculating pH of a strong acid/base solution.
Units:
pH is dimensionless.
Worked sample
Find the pH of 0.050 mol L⁻¹ HCl (strong acid, fully dissociates).
$[\text{H}^+] = 0.050$ mol L⁻¹; $\text{pH} = -\log_{10}(0.050) = 1.30$.
Your turn:
What is $[\text{H}^+]$ when pH = 3.40?
$[\text{H}^+] = 10^{-3.40} = 3.98 \times 10^{-4}$ mol L⁻¹.
To reverse: $[\text{H}^+] = 10^{-\text{pH}}$. A one-unit decrease in pH corresponds to a 10-fold increase in $[\text{H}^+]$.
What each symbol means
$\text{pOH} = -\log_{10}[\text{OH}^-]$; the sum equals 14 because $K_w = [\text{H}^+][\text{OH}^-] = 1.0 \times 10^{-14}$ at 25 °C, so $-\log K_w = \text{pH} + \text{pOH} = 14$.
When to use it
Finding pH of a strong base solution (calculate pOH first, then subtract from 14), or finding $[\text{OH}^-]$ from pH.
Units:
Dimensionless.
Worked sample
Find the pH of 0.020 mol L⁻¹ NaOH.
$[\text{OH}^-] = 0.020$ mol L⁻¹; $\text{pOH} = -\log(0.020) = 1.70$; $\text{pH} = 14 - 1.70 = 12.30$.
Your turn:
What is $[\text{OH}^-]$ when pH = 11.00?
$\text{pOH} = 14 - 11 = 3$; $[\text{OH}^-] = 10^{-3} = 1.00 \times 10^{-3}$ mol L⁻¹.
This relationship only holds exactly at 25 °C (where $K_w = 10^{-14}$). At other temperatures $K_w$ changes.
What each symbol means
$K_w$ — ionic product of water (mol² L⁻²); $[\text{H}^+]$ and $[\text{OH}^-]$ in mol L⁻¹ at 25 °C.
When to use it
Finding $[\text{OH}^-]$ from $[\text{H}^+]$ or vice versa; understanding neutral pH = 7 at 25 °C.
Units:
mol² L⁻²
Worked sample
Find $[\text{OH}^-]$ in a solution where $[\text{H}^+] = 2.5 \times 10^{-4}$ mol L⁻¹.
$[\text{OH}^-] = 1.0 \times 10^{-14} / 2.5 \times 10^{-4} = 4.0 \times 10^{-11}$ mol L⁻¹.
Your turn:
At pH = 9.0, what is $[\text{OH}^-]$?
$[\text{H}^+] = 10^{-9}$; $[\text{OH}^-] = 10^{-14}/10^{-9} = 10^{-5} = 1.0 \times 10^{-5}$ mol L⁻¹.
In a neutral solution at 25 °C, $[\text{H}^+] = [\text{OH}^-] = 1.0 \times 10^{-7}$ mol L⁻¹, so pH = 7.
What each symbol means
$K_a$ — acid dissociation constant; $[\text{H}^+]$ — hydrogen ion concentration (mol L⁻¹); $[\text{A}^-]$ — conjugate base concentration (mol L⁻¹); $[\text{HA}]$ — equilibrium concentration of weak acid (mol L⁻¹).
When to use it
Calculating the pH of a weak acid solution, finding degree of dissociation, or comparing acid strengths (larger $K_a$ = stronger weak acid).
Units:
mol L⁻¹
Worked sample
For acetic acid, $K_a = 1.8 \times 10^{-5}$. Find $[\text{H}^+]$ in 0.100 mol L⁻¹ CH₃COOH.
$x^2 / (0.100 - x) \approx x^2/0.100 = 1.8 \times 10^{-5}$; $x = \sqrt{1.8 \times 10^{-6}} = 1.34 \times 10^{-3}$ mol L⁻¹; pH = 2.87.
Your turn:
If $K_a = 6.3 \times 10^{-4}$ and $[\text{H}^+] = 7.94 \times 10^{-3}$ mol L⁻¹ at equilibrium, what is [HA] if $[\text{A}^-] = [\text{H}^+]$?
$[\text{HA}] = [\text{H}^+][\text{A}^-]/K_a = (7.94\times10^{-3})^2/6.3\times10^{-4} = 0.100$ mol L⁻¹.
For weak acids: if degree of dissociation < 5 %, the approximation $[\text{HA}] \approx c_{\text{initial}}$ is valid and avoids solving a quadratic.
What each symbol means
$\text{p}K_a$ — negative base-10 logarithm of the acid dissociation constant. Smaller $\text{p}K_a$ = stronger acid.
When to use it
Comparing relative strengths of weak acids; calculating $K_a$ from a tabulated $\text{p}K_a$ value.
Units:
Dimensionless.
Worked sample
Acetic acid has $K_a = 1.8 \times 10^{-5}$. Find p$K_a$.
$\text{p}K_a = -\log(1.8 \times 10^{-5}) = 4.74$.
Your turn:
Convert p$K_a = 9.25$ (ammonium ion) to $K_a$.
$K_a = 10^{-9.25} = 5.6 \times 10^{-10}$ mol L⁻¹.
At the half-equivalence point in a titration, pH = p$K_a$ for a weak acid — useful for identifying the acid from a titration curve.
Electrochemistry
3What each symbol means
$Q$ — electric charge (coulombs, C); $I$ — current (amperes, A); $t$ — time (seconds, s).
When to use it
Calculating the total charge passed through an electrolytic cell from current and time, as the first step in all electrolysis yield calculations.
Units:
Coulombs (C).
Worked sample
A current of 2.50 A is passed for 30.0 minutes. Find $Q$.
$Q = It = 2.50 \times (30.0 \times 60) = 2.50 \times 1800 = 4500$ C.
Your turn:
What current is needed to pass 3600 C in 20 minutes?
$I = Q/t = 3600/1200 = 3.00$ A.
Convert minutes to seconds before substituting: $t(\text{s}) = t(\text{min}) \times 60$.
What each symbol means
$n_e$ — moles of electrons transferred; $Q$ — charge (C); $F = 96\,485$ C mol⁻¹ (Faraday constant).
When to use it
Converting total charge into moles of electrons, then using the half-equation stoichiometry to find moles of substance deposited or dissolved at an electrode.
Units:
mol
Worked sample
How many moles of electrons are transferred when 4500 C of charge is passed?
$n_e = 4500/96485 = 0.0466$ mol.
Your turn:
How many moles of Cu²⁺ are deposited (Cu²⁺ + 2e⁻ → Cu) by 0.0466 mol of electrons?
$n_{\text{Cu}} = 0.0466/2 = 0.0233$ mol.
Use the balanced half-equation to relate moles of electrons to moles of product. For Cu²⁺ it is 2 electrons per ion; for Ag⁺ it is 1.
What each symbol means
$m$ — mass of substance deposited or dissolved (g); $n$ — moles of substance (mol, from Faraday's law and half-equation stoichiometry); $M$ — molar mass (g mol⁻¹).
When to use it
Final step in electrolysis calculations — converting moles of deposited/dissolved substance to mass.
Units:
g
Worked sample
How many grams of copper ($M = 63.55$ g mol⁻¹) are deposited by 0.0233 mol?
$m = 0.0233 \times 63.55 = 1.48$ g.
Your turn:
A current of 1.00 A flows for 965 s through a silver nitrate solution. Find the mass of Ag deposited ($M_{\text{Ag}} = 107.87$ g mol⁻¹; Ag⁺ + e⁻ → Ag).
$Q = 1.00 \times 965 = 965$ C; $n_e = 965/96485 = 0.01000$ mol; $n_{\text{Ag}} = 0.01000$ mol; $m = 0.01000 \times 107.87 = 1.08$ g.
Chain the three steps: $Q = It$ → $n_e = Q/F$ → $n_{\text{substance}} = n_e / z$ (where $z$ is electrons per ion) → $m = nM$.