Assessed mathematical skills and measurement conventions

For a sample of mass $4.75\,\mathrm{g}$ and volume $5.00\,\mathrm{cm^3}$, $ho=4.75/5.00=0.950\,\mathrm{g\,cm^{-3}}$. Converting the result gives $950\,\mathrm{g\,dm^{-3}}$: the numerical value grows because one cubic decimetre contains 1000 cubic centimetres.

At A2, derive the unit of an equilibrium or rate constant by substituting concentration units into its defining expression. Align energy units before combining quantities: convert entropy from $\mathrm{J\,mol^{-1}\,K^{-1}}$ to $\mathrm{kJ\,mol^{-1}\,K^{-1}}$ by dividing by 1000 when enthalpy is in $\mathrm{kJ\,mol^{-1}}$.

$0.0050\,\mathrm{mol\,dm^{-3}}=5.0 imes10^{-3}\,\mathrm{mol\,dm^{-3}}$. Both values have two significant figures: leading zeros locate the decimal point, while the final zero records precision. Conversion of form must not invent or discard significant figures.

Using $N=nN_A$, $0.250\,\mathrm{mol} imes6.02 imes10^{23}\,\mathrm{mol^{-1}}=1.51 imes10^{23}$ particles to three significant figures. For reciprocal data, calculate first and then report consistently: $1/48\,\mathrm{s}=0.0208\,\mathrm{s^{-1}}$, or $0.021\,\mathrm{s^{-1}}$ to two significant figures.

For a compound or polymer repeat unit, write the complete formula first, include every atom, and calculate its total $M_r$. The carbon contribution is then $12.0 imes$ the number of carbon atoms. In a hydrate, the water contribution includes both its coefficient and $M_r(\ce{H2O})$.

To find an empirical or alloy ratio, convert each mass or percentage to moles, divide every amount by the smallest, then scale to the simplest credible whole-number ratio. Equation coefficients are mole ratios, not mass ratios; balance atoms and charge before using them.

For a mixture, a weighted value is the sum of each fraction times its component value. Convert percentages to fractions or divide the summed percentage products by 100, and check that the result lies between the component values.

Estimate $(2.8 imes10^9 imes0.86)/300$ as (3 imes10^9 imes0.9)/(3 imes10^2)pprox9 imes10^6. A detailed answer near $8 imes10^6$ is therefore credible; $8 imes10^3$ signals a power-of-ten or unit error. Estimation is a check, not the final reported calculation.

For a changed experimental parameter, predict the direction before calculating. Ask which measured terms change and which definition links them. For $K_c$, temperature can change the equilibrium constant; concentration, pressure or a catalyst may shift composition or rate but do not by themselves change $K_c$ at fixed temperature.

If $[\ce{H+}]=2.5 imes10^{-3}\,\mathrm{mol\,dm^{-3}}$, then $\mathrm{pH}=2.60$. If $\mathrm{pH}=4.20$, then $[\ce{H+}]=6.3 imes10^{-5}\,\mathrm{mol\,dm^{-3}}$. Substituting back into the inverse relation is a quick calculator-entry check.

At A2, a buffer approximation can be written as $\mathrm{pH}=\mathrm{p}K_a+\log_{10}([\ce{A-}]/[\ce{HA}])$ when its assumptions apply. Use equilibrium concentrations—or amounts only when both species share the same solution volume—and preserve brackets around the complete ratio.

Replace the prefix by its factor, then convert to the target prefix. For example, $25.0\,\mathrm{cm^3}=25.0 imes10^{-3}\,\mathrm{dm^3}=0.0250\,\mathrm{dm^3}$, while $150\,\mathrm{mg}=150 imes10^{-3}\,\mathrm{g}=0.150\,\mathrm{g}$. State the unit at every stage so the direction is visible.

Moving to a smaller unit makes the numerical value larger: $1\,\mathrm{g}=1000\,\mathrm{mg}=10^6\,\mu\mathrm{g}$. Moving to a larger unit makes it smaller. Use this as a reasonableness check before accepting the exponent.

Apply the scale to the whole unit. A volume conversion is cubic: $1\,\mathrm{cm}=10^{-2}\,\mathrm{m}$, so $1\,\mathrm{cm^3}=10^{-6}\,\mathrm{m^3}$. In a denominator, the numerical effect reverses; write the conversion factor rather than relying on a memorised decimal shift.

For $80.0-15.846=64.154\,\mathrm{kg}$, the subtraction is limited by $80.0\,\mathrm{kg}$ to the tenths place, so $64.2\,\mathrm{kg}$ is appropriate. Writing $64.154\,\mathrm{kg}$ claims precision that the first measurement did not provide.

Precision must also fit the quantity. A calculated number of protons, neutrons, atoms or molecules represents a count, so a physically interpreted answer may need to be a whole number even if earlier data contain several significant figures.

For isotopes of masses 35 and 37 with abundances 75% and 25%, $A_r=(35 imes75+37 imes25)/100=35.5$. The result lies between the isotope masses and closer to the more abundant isotope, providing a useful check.

Calculate each titre as final minus initial burette reading. Exclude the rough value, then identify concordant accurate titres: in this Edexcel chemistry context, accepted titres agree within $0.20\,\mathrm{cm^3}$. For $24.30$, $23.80$ and $24.20\,\mathrm{cm^3}$, use $24.30$ and $24.20$ only, giving a mean of $24.25\,\mathrm{cm^3}$.

A titre is final burette reading minus initial reading. If each reading is $\pm0.05\,\mathrm{cm^3}$, the titre uncertainty is $\pm0.10\,\mathrm{cm^3}$. For an $18.95\,\mathrm{cm^3}$ titre, the percentage uncertainty is $(0.10/18.95) imes100=0.53\%$.

If a reported mass of $9.53\,\mathrm{g}$ came from two balance readings, each with $\pm0.01\,\mathrm{g}$ uncertainty, the mass by difference is $9.53\pm0.02\,\mathrm{g}$, giving the possible range $9.51$ to $9.55\,\mathrm{g}$.

With the same apparatus, reduce percentage uncertainty by measuring a larger change while keeping the chemistry valid—for example, a larger temperature rise makes a fixed thermometer uncertainty a smaller fraction of the result.

If rate $\propto[\ce{A}]$, then rate $=k[\ce{A}]$ for a fixed set of conditions. The proportionality sign does not mean the numerical values are equal: the constant $k$ supplies the scale and units.

At dynamic equilibrium, the forward and reverse rates are equal, so macroscopic concentrations remain constant. The equilibrium sign does not mean equal concentrations, complete reaction or that particles have stopped reacting.

If rate is $0.040\,\mathrm{mol\,dm^{-3}\,s^{-1}}$, $[\ce{A}]=0.010\,\mathrm{mol\,dm^{-3}}$ and $[\ce{B}]=0.050\,\mathrm{mol\,dm^{-3}}$, then $k=8.0 imes10^3\,\mathrm{dm^6\,mol^{-2}\,s^{-1}}$. The unit follows by dividing the rate unit by three concentration factors.

Check the rearrangement symbolically before substitution: multiply the final expression back by the removed factor and confirm that the original equation returns. This separates an algebra error from a calculator-entry error.

For $c=0.200\,\mathrm{mol\,dm^{-3}}$ and $V=25.0\,\mathrm{cm^3}=0.0250\,\mathrm{dm^3}$, $n=0.200 imes0.0250=5.00 imes10^{-3}\,\mathrm{mol}$. The conversion is part of the substitution, not an optional correction after calculating.

At A2, substitute equilibrium concentrations into the stated $K_c$ expression and preserve every stoichiometric power; for rates, distinguish rate from rate constant and derive the unit of $k$ from the rate equation. A multi-stage calculation should label intermediate quantities so each value can be traced to its source.

If $\Delta H_{A o C}=-120\,\mathrm{kJ\,mol^{-1}}$ and $\Delta H_{A o B}=-50\,\mathrm{kJ\,mol^{-1}}$, then $-120=-50+x$, so $x=-70\,\mathrm{kJ\,mol^{-1}}$. Substitution back gives $-50+(-70)=-120$, confirming both magnitude and sign.

The same discipline applies to an unknown in a rate equation: construct the correct relationship, isolate the unknown, solve, then test the result in the original equation with units.

If $[\ce{H+}]=3.2 imes10^{-4}\,\mathrm{mol\,dm^{-3}}$, then $\mathrm{pH}=3.49$. If $K_a=1.8 imes10^{-5}$, then $\mathrm{p}K_a=4.74$. Substituting each result into its inverse power relation checks the calculator entry.

A lower pH corresponds to a higher hydrogen-ion concentration: a decrease of one pH unit means a tenfold increase in $[\ce{H+}]$. Likewise, a lower pKa corresponds to a larger $K_a$.

To read a calibration graph, locate the measured response on its axis, draw to the best-fit line, then project to the concentration axis. Show both construction lines. If the sample was diluted, the graph gives the diluted concentration; apply the dilution factor afterwards to recover the original concentration.

At A2, compare initial rates at controlled concentrations. If doubling $[\ce{A}]$ leaves rate unchanged, doubles it or quadruples it, the order in $\ce{A}$ is 0, 1 or 2 respectively. Translate that pattern into $ext{rate}=k[\ce{A}]^m[\ce{B}]^n$.

Interpret each spectrum using its own axes and conventions. Peak position and signal size may carry different meanings in mass, infrared or NMR spectra, so identify what the supplied spectrum measures before assigning a chemical feature.

Decide between a straight line and a smooth curve from the pattern and the stated model. A best-fit line should balance scatter rather than pass through every point. Retain a suspected anomaly unless there is evidence to exclude it; do not bend the fit solely to capture that point.

Include the origin only when it is a supplied point or the chemical relationship justifies it. An axis may use a clearly marked break, but the numerical scale must remain uniform on each section. Preserve transformed labels such as $1/t\,/\,\mathrm{s^{-1}}$ or $1/T\,/\,\mathrm{K^{-1}}$.

For an A2 zero-order concentration-time graph, $[\ce{A}]=[\ce{A}]_0-kt$. The straight-line gradient is $-k$ and the intercept is the initial concentration. If concentration falls from $0.80$ to $0.20\,\mathrm{mol\,dm^{-3}}$ over $300\,\mathrm{s}$, the gradient is $-2.0 imes10^{-3}\,\mathrm{mol\,dm^{-3}\,s^{-1}}$, so $k=2.0 imes10^{-3}\,\mathrm{mol\,dm^{-3}\,s^{-1}}$.

A straight concentration-time line has the same gradient throughout, so the concentration changes by the same amount per unit time. A falling reactant concentration gives a negative graph gradient; the rate of disappearance is reported as its positive magnitude.

A best-fit line changes from $0.600$ to $0.360\,\mathrm{mol\,dm^{-3}}$ in $120\,\mathrm{s}$. Its gradient is $(0.360-0.600)/120=-2.00 imes10^{-3}\,\mathrm{mol\,dm^{-3}\,s^{-1}}$; the disappearance rate, and zero-order $k$, is $2.00 imes10^{-3}\,\mathrm{mol\,dm^{-3}\,s^{-1}}$.

For the initial-rates method, compare experiments in which only one reactant concentration changes. If multiplying $[\ce{A}]$ by a factor $f$ multiplies the initial rate by $f^m$, then $m$ is the order in $\ce{A}$. Repeat for other reactants, then combine the orders in the rate equation.

When doubling $[\ce{A}]$ at constant $[\ce{B}]$ quadruples the initial rate, $2^m=4$, so $m=2$. If changing $[\ce{B}]$ does not change the rate, $n=0$ and the $[\ce{B}]^0$ factor may be omitted.

Lone pairs repel more strongly than bonding pairs because their electron density is concentrated near one nucleus. One lone pair changes tetrahedral electron-region geometry into a trigonal-pyramidal molecular shape; two produce a bent shape and compress the bond angle further.

Isomers share a molecular formula but differ in arrangement. First fix connectivity: different carbon skeletons, functional-group positions or substituent positions are structural isomers. With the same connectivity, restricted rotation or a chiral three-dimensional arrangement can produce stereoisomers.

For a substituted ring, fix one substituent as position 1, place the remaining substituent(s) systematically, then remove drawings related only by rotation or reflection of the ring. For two identical substituents on benzene, the distinct relative positions are 1,2-, 1,3- and 1,4-. This symmetry check prevents duplicate counting.

At A2, inspect each tetrahedral carbon and trace its four attached groups. It is a chiral centre only when all four groups are different. A valid pair of enantiomer drawings keeps every bond connection unchanged and reverses the complete three-dimensional arrangement.

A carbon-carbon double bond prevents free rotation. Geometric isomerism requires each double-bonded carbon to have two different substituents. The $E/Z$ system compares the higher-priority substituent on each carbon and remains usable where cis/trans labels are ambiguous.

For the single-centre cases required here, a tetrahedral carbon bonded to four different groups is asymmetric and gives two non-superimposable mirror arrangements. A mirror plane through the complete molecule would map one half onto the other and indicates that the structure is not chiral.

Apply the same spatial test to molecular or complex-ion representations: identify which positions are equivalent by symmetry, then decide whether two drawings superimpose or preserve a genuine geometric or mirror-image difference.