B.0 Arithmetic and numerical computation

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.