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Use spherical coordinates $x=s\sin \theta \cos \varphi$, $y=s\sin \theta \sin \varphi$, $z=s\cos \theta$. The density becomes ${\rho }_{0}s\sin \theta$, and the Jacobian determinant supplies $dV=s^2\sin \theta dsd\theta d\varphi$. Therefore the mass is the volume integralUsing ${\sin }^2\theta =(1-\cos 2\theta )/2$, the three integrals are $2\pi$, $\pi /2$ and $1/4$. Hence

Expand the divergence of $\Phi \nabla \Psi -\Psi \nabla \Phi$. The cross terms cancel, leaving $\Phi \Delta \Psi -\Psi \Delta \Phi$. The divergence theorem gives Green's second identity,Use the Green function convention ${\Delta }_{x}G(x,x_0)=\delta (x-x_0)$. Integrating over a bounded domain containing $x_0$ yields $\boxed{{\int }_S{\partial }_{n}GdS=1}$. Thus zero Neumann boundary data would be incompatible with this unit Dirac delta source. A suitable Neumann Green function with constant boundary flux instead has ${\partial }_{n}G=1/|S|$, where $|S|$ is the boundary area. Substituting $\Psi =G$, $\Delta \Phi =\sigma$, ${\partial }_n\Phi =f$ givesThe last term is an undetermined constant. The data must obey the Neumann Poisson problem compatibility condition ${\int }_V\sigma dV={\int }_{S}fdS$, obtained by integrating the original equation. Under this condition an additive constant in $G$ cancels from the first two terms.

For the upper half-space, reflect the source to $x_0^{\ast }=(x_{0,1},x_{0,2},-x_{0,3})$. The method of images with equal-sign sources givesThe reflected singularity lies outside the domain; on $z=0$ the two normal derivatives cancel, so ${\partial }_{n}G=-{\partial }_{z}G=0$. Its total unit flux passes through a large hemisphere at infinity, where $G\sim -1/(2\pi R)$. This is why the bounded-domain condition on the boundary flux cannot be applied to the infinite plane alone.

With zero interior source and the given point boundary source, the boundary representation is $\Phi (x_0)=-G((0,0,0),x_0)$. The terms at infinity vanish for this resulting $1/R$ field. ThereforeTo check the singular boundary condition, $-{\partial }_z\Phi =z/[2\pi (x^2+y^2+z^2{)}^{3/2}]$ is the Poisson kernel for the upper half-space. Its integral over a horizontal plane is one and its mass concentrates at the origin as $z\downarrow 0$, giving ${\partial }_n\Phi =\delta (x)\delta (y)$ in the distributional sense. This is the Neumann point source on a half-space boundary.

Choose generalized coordinates $q_i$ measured from a static equilibrium point, and assume a conservative system with no velocity-dependent forces. Keeping terms of degree two in displacements and velocities gives the small oscillation LagrangianHere $A$ is the positive-definite symmetric mass matrix evaluated at equilibrium, and $B$ is the symmetric Hessian matrix of the potential energy there. The linear potential term is zero by equilibrium, and an additive constant has been dropped. Coordinate-dependent corrections to the kinetic matrix contribute only at higher order.

The Euler-Lagrange equations are $A\ddot{\mathbf{q}}+B\mathbf{q}=0$. For a normal mode $\mathbf{q}=\mathbf{v}{e}^{i\omega t}$, they give the generalized eigenvalue problem $(B-{\omega }^{2}A)\mathbf{v}=0$. A nonzero mode shape exists exactly whenThis is a degree-$n$ polynomial in ${\omega }^2$. At a stable equilibrium its roots are nonnegative; zero roots correspond to zero modes.

With the stated parameters, the mass and restoring matrices areThe generalized eigenvalue problem givesTherefore the two normal angular frequencies areThe second row of the modal equation is $-{\omega }^{2}x+(1-{\omega }^2)\varphi =0$. It gives the mode shapesEach normal mode has arbitrary small amplitude $a$ and phase $\delta$. The bob's horizontal displacement is $X=x+\varphi$ to first order. Thus in the lower mode, bob and cart move in phase, with bob amplitude $1+\sqrt{2}$ times the cart amplitude. In the higher mode they move oppositely, because $X=(1-\sqrt{2})x$; the bob amplitude is $\sqrt{2}-1$ times the cart amplitude. Its vertical displacement is quadratic in the angle and is absent at this order.

Label the equal masses $1,2$ and the mass $2m$ by $3$, in cyclic order. Let their equilibrium angles be $0,2\pi /3,4\pi /3$, with angular displacements $q_1,q_2,q_3$. Because the hoop has unit radius, its three arc lengths areThe springs themselves are constrained to the hoop, so these are arc lengths. If their common natural length is ${\ell }_0$, Hooke's law gives $V=\frac{k}{2}{\sum }_i({\ell }_i-{\ell }_0{)}^2$. Since the arc lengths have fixed sum $2\pi$, all terms linear in the $q_i$ cancel. Dropping the constant,Thus the unspecified common natural length does not affect the motion. The mass and restoring matrices in the generalized eigenvalue problem areExchange of the two equal masses separates symmetric and antisymmetric normal modes. Direct multiplication gives the complete setThe first is rigid rotation of all three beads: every arc length stays fixed, giving a zero mode. In the second, the equal masses move oppositely while the heavy bead stays fixed; the net force on that bead cancels. In the third, the equal masses move together and the heavy bead moves oppositely with equal angular amplitude. Both nonzero normal modes have zero total angular momentum. These are the normal modes of three beads connected by springs on a circular hoop with $M=2m$.

Use the mode shapes from part (b), and put $\omega =\sqrt{2k/m}$. The initial velocity decomposes asThere is no antisymmetric component because the initial state treats the equal masses identically. Initial displacements are zero, so the zero mode contributes a linear drift and the nonzero normal mode contributes a sine:In terms of actual angular positions,The conserved angular momentum is initially $2m\Omega$ and the total moment of inertia is $4m$, so the common angular velocity must be $\Omega /2$, agreeing with the drift. The relative displacements remain of order $\Omega /\omega$, even when the common angle becomes large. Hence small $\Omega /\omega$ preserves the cyclic ordering and small spring extensions at arbitrarily late times. In the ideal arc-spring model the displayed energies are exactly quadratic in the relative angles, so this motion remains exact while that ordering persists.

For three unit masses, $M=I$ andThe uniform vector has $11^{T}1=31$, whereas every vector with coordinate sum zero is annihilated by $11^T$. One orthonormal choice of normal modes is thereforeIf $k=1$ is retained from (b), the angular frequencies are $1,2,2$. The equal upper frequencies follow from degenerate normal modes from permutation symmetry: the stiffness is scalar on the two-dimensional sum-zero subspace. In the uniform mode the inter-bar strings do not change length, but all three beam connections stretch, giving the nonzero lower frequency. The motion is an arbitrary sine-cosine superposition of these modes about the gravitational equilibrium.

For complete neutralization, the required masses of anhydrous potassium carbonate, anhydrous sodium carbonate and magnesium oxide are in the ratio $138:106:40=69:53:20$. Here “carbonate of potash” means potassium carbonate, “carbonate of soda” means sodium carbonate, and “caustic magnesia” means magnesium oxide, rather than magnesium hydroxide or a magnesium carbonate preparation. The hydration state of the sodium carbonate matters considerably.

Each formula unit of either carbonate consumes two protons; each formula unit of magnesium oxide also consumes two protons. Thus one mole of each base neutralizes one mole of sulfuric acid when the final salts are the normal sulfates:The carbon dioxide evolved in the first two chemical reactions escapes, whereas magnesium oxide produces no gas. The stoichiometric coefficients, rather than the visibly different behaviour of the bases, determine the required amounts.

Let $m$ be the mass of pure ${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$ in the given chemical solution. Using the rounded relative atomic masses $\mathrm{H}=1$, $\mathrm{C}=12$, $\mathrm{O}=16$, $\mathrm{Na}=23$, $\mathrm{Mg}=24$ and $\mathrm{K}=39$ givesThus 98 mass units of pure sulfuric acid require respectively 138, 106 and 40 mass units. Present atomic weights slightly change these rounded values, without changing the one-to-one mole ratios. If $m_s$ denotes the mass of the supplied acid and $w$ its mass fraction of ${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$, substitute $m=w{m}_s$. A volume alone is insufficient without its concentration or an equivalent density and composition.

If “carbonate of soda” is instead the crystalline decahydrate, sodium carbonate decahydrate, its rounded molar mass is $286\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$, so the corresponding factor is $286/98$, rather than $106/98$. Likewise an impure base requires correction for its assayed content. The unspecified given quantity can therefore be handled exactly by proportional reasoning; no missing numerical measurement is needed.

A suitable historical answer gives both the chemical transformation and the means of isolating a reasonably pure product. The following are preparative principles, with modern chemical formulas; they are not instructions for self-preparation of medicines. In particular, hydrogen cyanide, mercury compounds and arsenic compounds present serious toxicity hazards. Hydrogen cyanide is extremely volatile, can be absorbed by several routes, and its odour is an unreliable warning.^[ref]

For hydrocyanic acid, a customary early nineteenth-century route is the controlled distillation of potassium ferrocyanide with dilute sulfuric acid and water. Hydrogen cyanide passes into the distillate; nonvolatile salts remain behind. A generating flask, efficient laboratory condenser and chilled receiver are needed because hydrogen cyanide is readily lost. The contemporary account of hydrocyanic acid preparation in Pharmacopoeia Castrensis Ruthenica in The British and Foreign Medical Review (1842) specifically describes the ferrocyanide–sulfuric acid method and its losses on imperfect condensation. The concentration must be determined analytically before any defined dilution: a nominal starting mixture does not establish the final strength.^[ref] At the level of the released cyanide, the essential acid-base reaction is $\mathrm{C}{\mathrm{N}}^{-}+{\mathrm{H}}^{+}\to \mathrm{HCN}$. This equation alone is not a complete equation for decomposing the stable ferrocyanide coordination complex. Concentrated sulfuric acid can lead to different decomposition chemistry, so the specified dilution is meaningful.

For formic acid, gently heat oxalic acid with glycerol and some water, then separate the aqueous formic acid by distillation. The useful net transformation isThe glycerol participates through intermediate esters and is regenerated rather than consumed in this net equation. Marcellin Berthelot's 1856 account describes the carbon dioxide evolution, subsequent water addition and distillation, and reuse of the residual glycerol. Overheating is disadvantageous because the formic acid can decompose further to carbon monoxide and water. The method therefore separates a volatile acid from a much less volatile carrier; it is not merely the dry destructive heating of oxalic acid.^[ref]

For oxalic acid, oxidize a carbohydrate, such as sucrose or starch, with nitric acid; concentrate the resulting chemical solution appropriately and cool it to obtain crystals, followed by recrystallization. Nitrogen oxides evolve, and excessive oxidation can carry carbon onward to carbon dioxide, so an idealized single-product equation would conceal losses. The use of starch and nitric acid for oxalic acid production is explicitly documented in an 1850 account of nitric oxide recovery.^[ref] An alternative already established before 1864 is alkaline treatment of sawdust, producing an oxalate which can be purified through sparingly soluble calcium oxalate.^[ref] Decomposing the isolated calcium oxalate with sulfuric acid separates much of the calcium as calcium sulfate:Filter off the calcium sulfate and crystallize the oxalic acid from the filtrate. The usual crystals contain two water of crystallization; a calculation for the anhydrous acid must not treat their entire mass as ${\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_4$.

For calomel, intimately combine mercury(II) sulfate, metallic Mercury and dry sodium chloride, then collect the volatile mercury(I) chloride in a cooler receiving chamber. The net change isThe extra metallic Mercury lowers the average oxidation state from +2 to +1. The product is purified by appropriate sublimation and washing to remove soluble contamination, particularly mercury(II) chloride. Both the chemical composition and the fine subdivision of the product matter: condensation through a chamber can give a powder rather than an adherent crystalline crust. This physical part of calomel preparation is described in the 1843 primary account in Polytechnisches Journal volume 87 (1843); the account's historical therapeutic claims are not adopted here.^[ref]

For corrosive sublimate, heat dry mercury(II) sulfate with sodium chloride and collect the mercury(II) chloride in the cooler part of a suitable sublimation apparatus:Here no metallic Mercury is added. Nonvolatile sodium sulfate stays behind, while mercury(II) chloride is transferred and condensed. Further sublimation or recrystallization separates impurities. ^[ref] The crucial distinction from calomel preparation is whether Mercury remains in the +2 oxidation state or is converted to +1.

For the solid called arsenious acid, roast an arsenic-bearing ore in air and collect the volatile arsenic trioxide in cooler chambers; purify by resublimation. The essential oxidation is represented byActual ores also contain other chemical elements, so their roasting equations and gaseous by-products depend on composition. A contemporary 1860 report describes roasting arseniferous ores, collecting the volatile arsenic trioxide in chambers and refining it by another sublimation. It also demonstrates that apparently good material could contain antimony(III) oxide: physical appearance alone does not establish purity.^[ref] The product intended here is dry $\mathrm{A}{\mathrm{s}}_2{\mathrm{O}}_3$; dissolving it does not make the bottled solid a pure crystalline hydrate ${\mathrm{H}}_3\mathrm{As}{\mathrm{O}}_3$.

For tartar emetic, react antimony(III) oxide with potassium hydrogen tartrate in water. A small amount of water first gives intimate contact in a paste; subsequent heating with more water completes formation of the soluble potassium antimony tartrate. Filter away unreacted insoluble matter and cool the clear filtrate to crystallize the product. This sequence is directly described in a January 1864 discussion of the newly issued British Pharmacopoeia, rather than inferred from its later editions.^[ref] In the traditional antimonyl notation the underlying reaction, before assigning crystal water of crystallization, isThis notation expresses the overall composition; tartrate coordination complexes and water of crystallization need the qualification given in 4. The historical pharmaceutical use of tartar emetic does not make it a safe contemporary emetic.^[ref]

For potassium chlorate, pass chlorine into hot concentrated potassium hydroxide. The disproportionation gives potassium chloride together with potassium chlorate:Concentration and cooling permit fractional crystallization of the much less soluble cold-water potassium chlorate, leaving much of the potassium chloride in the mother liquor; recrystallization improves purity. Cold dilute alkali instead favours hypochlorite, so the heating condition is essential. An economical contemporary manufacturing alternative uses hot calcium hydroxide and potassium chloride; the overall accounting isThe 1850 account in Polytechnisches Journal volume 118 (1850) describes a steam-heated stirred vessel and recovery by cooling, rather than an electrical manufacturing process introduced retrospectively.^[ref]

Cow's milk contains water, dispersed fat, casein, soluble milk proteins, lactose and mineral constituents, with smaller amounts of other substances. These constituents occupy different physical states, which makes their separation possible: the fat is present as an emulsion, much of the casein as colloidal casein micelles, and the lactose and many salts in chemical solution. Vitamins, enzymes and small nitrogenous constituents accompany these main fractions. The proportions vary between samples and are not fixed by the name cow's milk.^[ref]

First let fresh unhomogenized milk stand so that the less dense fat globules rise as cream; skimming gives a fat-enriched layer and skimmed milk beneath. This separates fractions rather than pure chemical substances: cream still contains water and other constituents. Churning cream coalesces the fat globules into butter; extraction of a separately prepared dried portion with a suitable organic solvent provides another way to isolate fat. A modern centrifugal separator accelerates the same density-based separation, but it is not needed to explain an 1864 bench separation.

Next add dilute acetic acid carefully to skimmed milk. Lowering its pH toward the isoelectric point of casein, about 4.6, produces a curd that can be collected by filtration and washed. Excess acid, retained fat, and shifts of calcium phosphate between curd and liquid limit purity. Rennet offers another route to a curd, but its enzyme-catalysed action and the mineral content of the resulting curd differ from simple acid precipitation; the two should not be treated as identical mechanisms.^[ref]

The liquid after removal of the curd is whey. Heat this fraction under suitable acidity to induce protein denaturation and aggregate much of its residual whey protein; collect the coagulum by filtration. Older descriptions commonly grouped this fraction under “albumin”, but modern milk chemistry distinguishes several proteins. Heating alone does not make every minor constituent fall into a single pure fraction.^[ref]

Concentrate the deproteinized whey and cool it to crystallize lactose. Filter and wash the crystals, then purify them by recrystallization; dissolved salts and residual organic matter remain substantially in the mother liquor. This provides a separation from most protein, not an automatic complete separation from all minerals.^[ref]

Finally use separate measured portions for water and mineral determinations. Controlled drying gives an operational total solids measurement; subtraction from the initial mass estimates the water lost. Ignition of a dried portion destroys most organic matter and leaves ash for mineral analysis. Ash is a transformed residue: it does not preserve all original salts unchanged, and some constituents can volatilize. Calcium, phosphorus, sodium, potassium and chloride are among the important mineral constituents. Skimming, curd formation, heat coagulation, lactose crystallization, drying and ashing separate or determine the principal fractions; none alone constitutes a complete pure-component analysis.

A molecule is an electrically neutral bound assembly of more than one atom. It can contain identical atoms, as in ${\mathrm{O}}_2$, or different atoms, as in ${\mathrm{H}}_{2}O$. Discrete molecules are not a universal description of matter: an extended ionic crystal does not consist of separate neutral molecules corresponding to each empirical formula unit. The IUPAC definition also specifies binding sufficient to support a vibrational state.^[ref]

Elasticity is a body's tendency to recover its original configuration when the deforming force is removed. Within a linear regime, Hooke's law makes stress proportional to strain, but elasticity does not require exact linearity. The elastic limit matters: permanent plastic deformation is a different response.

Inertia is the property expressed by Newton's first law: without a resultant external force, a body's velocity remains constant in an inertial frame of reference. Mass measures translational inertia. It is not a force acting against motion, and a moving body does not require a continuing forward force merely to keep moving.

A force describes an interaction that changes momentum. In Newtonian mechanics,A force has magnitude and direction; its point of application also matters for the torque and hence the body's rotational response.

The relevant claim concerns a machine that repeats its complete operating cycle while delivering a continuing net output of work without a corresponding supply of energy. A machine cannot supply net work in a cycle without taking in energy or depleting an energy store. This follows from conservation of energy, an experimentally supported physical law rather than a theorem of pure logic.

Let $E$ include the machine's internal energy, kinetic energy and potential energy. With $Q$ the net heat received and $W$ the net work delivered, the first law of thermodynamics givesFor a complete cycle all stores return to their initial values, so $\Delta E=0$ and $W=Q$. If there is neither heat input nor another energy input, $W=0$. A descending mass, unwinding spring, consumed fuel or discharged battery supplies energy, but it cannot be omitted from the accounting while calling the result perpetual motion.

Changing an overbalanced wheel's geometry does not evade the argument. A conservative force gives zero net work around a closed configuration cycle, while friction dissipates usable mechanical energy. Resetting a mechanism or returning a raised weight costs work that must be included.

Indefinitely continuing motion without useful output is a different possibility: an ideal frictionless oscillator or freely moving body need not stop. Connecting a load removes energy and changes that situation. A cyclic machine supplied with heat from only one reservoir and converting all that heat into work is excluded instead by the Kelvin-Planck statement of the Second law of thermodynamics. The first-law and second-law objections address different proposed energy balances.

The 1864 solid units are the avoirdupois pound, avoirdupois ounce and grain; the liquid units are the Imperial measures. The grain is unchanged, while the old apothecaries' pound and ounce are not the units of the new formulary. ^[ref]

The relations for solids areTheir approximate French metric equivalents are:

For liquids,The approximate metric volumes are:

A millilitre is a cubic centimetre, and a litre is a cubic decimetre. These displayed equivalents are rounded; small differences from a period table's last digits reflect rounding and historical calibration conventions. A fluid ounce is a volume and an avoirdupois ounce is a mass. They cannot be interchanged for an arbitrary liquid without its density. Likewise, a drop is not a defined minim: its volume depends on the liquid and the dropper.

The 1864 preparations contain either crude opium or its aqueous Extractum Opii; those quantities must be distinguished from a quantity of isolated morphine. The following are the officinal opium preparations. Ratios for solids mean the mass fraction in the compounded preparation; liquid strengths specify mass per volume or the actual ingredients used.

The arithmetic uses the new avoirdupois ounce of 437.5 grains, not the old apothecaries' ounce of 480 grains. These are historical formulation proportions, not modern assays of morphine delivered after solvent extraction. Confectio Opii was omitted in 1864, and this edition's Vinum Opii contains no spices. The separate morphine hydrochloride preparations—Liquor Morphiae Hydrochloratis, Suppositoria Morphiae, Trochisci Morphiae, and Trochisci Morphiae et Ipecacuanhae—contain an isolated alkaloid salt rather than crude opium. ^[ref]

The seven named mixtures of the 1864 British Pharmacopoeia are Mistura Ammoniaci, Mistura Amygdalae, Mistura Creasoti, Mistura Cretae, Mistura Ferri Composita, Mistura Guaiaci and Mistura Scammonii. Their compositions are as follows; avoirdupois ounces are used for solids and fluid ounces for liquids.

The category is an official preparation name, not every possible bedside combination of medicines. Camphor water and acacia mucilage are not additional entries under Mistura: the 1864 names are Aqua Camphorae and Mucilago Acaciae. ^[ref]

Pulvis Jalapae Compositus contains powdered jalap 5 avoirdupois ounces, potassium hydrogen tartrate 9 avoirdupois ounces, and powdered ginger 1 avoirdupois ounce. Thus the mass proportions are 5:9:1, and jalap supplies one third of the powder's mass. This is a compound purgative, not pure jalap resin. ^[ref]

Pilula Scillae Composita contains powdered squill 1¼ avoirdupois ounces, powdered ginger, ammoniacum and hard soap 1 avoirdupois ounce each, with treacle 2 avoirdupois ounces, or sufficient to make a pill mass. Using the specified 2 avoirdupois ounces of treacle, the whole is 6¼ avoirdupois ounces and contains one fifth squill by mass. The allowance of sufficient treacle makes consistency a compounding requirement, not a claim that every moist pill batch has an analytically exact 20% squill content. ^[ref]

Liquor Morphiae Hydrochloratis contains morphine hydrochloride 4 grains, dilute hydrochloric acid 8 minims, rectified spirit 2 fluid drachms and distilled water 6 fluid drachms. Its conventional strength is ½ grain of morphine hydrochloride per fluid drachm, not ½ grain of free morphine. The small quantity of hydrochloric acid maintains an acid solution. ^[ref]

Liquor Strychniae contains strychnine 4 grains, dilute hydrochloric acid 6 minims, rectified spirit 2 fluid drachms and distilled water 6 fluid drachms. Dissolve the strychnine in the acid mixed with part of the water, using the prescribed assistance of heat, then add the spirit and remaining water. Its conventional strength is ½ grain of strychnine per fluid drachm. Acid promotes solubility by forming a soluble salt of the alkaloid. ^[ref]

For the last two solutions, the stated historical strengths use the nominal 8 fluid drachms of spirit and water. The extra few minims of dilute acid must still appear in the actual recipe; treating all ingredient volumes as perfectly additive would not give an exact modern analytical concentration.

Volumetric analysis determines an amount of substance from the measured volume of a reagent of known strength that reacts with it in a known proportion. A measured aliquot is placed in a suitable vessel, the standard solution is delivered from a burette, and the volume required to reach the chosen titration endpoint is read. A pipette measures the aliquot; a volumetric flask prepares or dilutes the standard solution.

For a reaction $aA+bB⟶\text{products}$, if the concentration of the titrant $B$ is $c_B$ and the corrected volume delivered is $V_B$, thenWith volume in litres and concentration in moles per litre, the answer is an amount in moles. The corresponding mass is $m_A=M_A{n}_A$, using the appropriate molar mass. For an unknown solution of aliquot volume $V_A$,These are modern symbols for the same proportional reasoning that a nineteenth-century analyst could express with equivalent weights and a specified standard solution.

A suitable reaction is sufficiently rapid, complete and selective, with a recognisable endpoint. In an acid-base titration, an indicator can change colour near neutralization; other reactions use precipitation, redox or a colour intrinsic to a reagent. The observed titration endpoint must represent the required equivalence point closely enough. The reagent's nominal preparation alone is insufficient if its concentration changes in storage; standardisation of a solution, a reagent blank where needed, and correct reading of the meniscus control systematic error. A titration measures the intended analyte only insofar as interfering substances have been excluded or accounted for.

Fresh rainwater is ordinarily soft because evaporation leaves most nonvolatile dissolved salts behind; “always” is an idealization, not an exceptionless law. Water hardness is principally caused by dissolved calcium and magnesium ions. Water vapour rising from a hard-water source does not carry these ions with it as vapour. Condensation consequently supplies low-mineral water before contact with dust, surfaces or soil. Mineral particles, sea spray and a collecting surface can add salts, so actual collected rainwater need not have zero water hardness. Nor does softness establish purity or freedom from microorganisms.^[ref]

The historically appropriate quantitative principle is soap titration of water hardness. Soluble soap reacts with hardness-producing ions to form poorly soluble salts, using up soap before a persistent lather can form. For a representative carboxylate soap,Magnesium behaves similarly. Once the dissolved hardness-producing ions have substantially reacted, additional soap can sustain the specified lather. The endpoint is therefore an operational observation under standardized shaking, standing time, temperature and sample volume; it is not the first transient bubble.

Use a measured volume of sample in a stoppered vessel with enough headspace for shaking. Add a soap solution of known strength incrementally from a burette, shaking in a reproducible way, until the chosen persistent-lather endpoint appears. Determine a blank with a corresponding amount of soft or distilled water, because some soap is needed for the lather even when no hardness salts are present. Standardize the soap against a reference solution containing a known calcium amount, expressed as an equivalent mass of calcium carbonate. This standardization is essential: the apparent strength cannot safely be inferred from the mass of an incompletely characterized commercial soap.

Let $V_s$ be the sample volume in millilitres, $V$ the soap titre in millilitres and $B$ the corresponding blank. If $f$ is the standardized capacity in milligrams of calcium carbonate equivalent per millilitre of soap solution, thenFor a standard containing $m_0$ milligrams of calcium carbonate equivalent and requiring blank-corrected titre $V_0-B_0$, $f=m_0/(V_0-B_0)$. This makes the measurement quantitative without assuming an unreported soap concentration or inventing a sample titre. Historical results could instead be stated as grains of calcium carbonate equivalent per gallon, provided the gallon convention was specified.

Repeating the examination after controlled boiling, cooling, removal of the precipitate and restoration of the original volume estimates the residual or permanent hardness. The reduction estimates the boiling-removable fraction, often called temporary hardness. A representative reaction isCarbonate equilibria and magnesium chemistry make this more subtle than assuming every dissolved salt is destroyed by boiling. Very hard samples may need measured dilution; magnesium, other metal ions and contamination can disturb the lather endpoint. Thus the soap result is an experimentally calibrated hardness equivalent, not an exact identification of every original salt.

Burn a known mass of pure, suitably dry material completely, collect the resulting carbon dioxide and water separately, calculate the carbon and hydrogen masses, and obtain oxygen by difference under the stated three-element assumption. This is organic elemental analysis, not an identification of molecular structure.

Weigh a representative portion of mass $m$ in a clean combustion boat or suitable carrier. In the classical apparatus, a heated combustion tube contains an oxidizing material such as copper(II) oxide, with a controlled stream of clean, dry oxygen-bearing gas where the arrangement requires it. The purpose is complete conversion of carbon into carbon dioxide and hydrogen into water, without unburnt material escaping. An equation for a model substrate iswith the oxygen amount understood as the net external requirement when positive.

Downstream, a preweighed water-absorption tube collects the water. A following preweighed alkaline absorber collects the carbon dioxide; the order prevents water from being incorrectly attributed to the carbon dioxide absorber. Guards exclude atmospheric moisture and carbon dioxide during handling, and a final clean-gas sweep transfers products still held in the apparatus. Let the blank-corrected mass increases be $a$ for the water collector and $b$ for the carbon dioxide collector.

Using the simple historical atomic-mass ratios $\mathrm{C}=12$, $\mathrm{H}=1$, $\mathrm{O}=16$,Consequently,Modern work uses the appropriate modern molar masses in the same ratios. No sample mass or absorption results are supplied, so no numerical percentages are justified.

The oxygen already present in the sample cannot be obtained merely by measuring the oxygen in the combustion products: the oxidant supplies additional oxygen. Difference works here only because the sample is assumed to contain exclusively carbon, hydrogen and oxygen and to be free of unaccounted water, solvent, ash and other elements. If the combustion leaves a mineral residue, that mass cannot silently be reported as oxygen. A separate check of purity and residue therefore tests an assumption of the calculation.

Incomplete combustion, escaped vapour, exhausted absorbers, moisture entering from the carrier gas, and contamination of the carbon dioxide absorber all bias the result. A blank and an independent repeat with a suitable reference material check recovery. Oxygen by difference accumulates the errors in both other determinations; forcing the percentages to sum to 100 is not independent proof that the analysis is correct.

The volume $V$ is the three-dimensional space occupied by the body. Its mass $m$ measures its inertia: in Newtonian mechanics the resultant force is $m\mathbf{a}$. Its gravitational weight is the force exerted on it by gravity,For a uniform body, $m=\rho V$, where $\rho$ is its mass density; more generally $m=\int \rho dV$. Volume can change with temperature or pressure while the body's mass remains fixed in the classical closed-system model. Equal volumes of different materials need not have equal masses.

The same mass need not have the same gravitational weight. The local gravitational acceleration changes between planets and with position on Earth. Outside a spherical Earth, $g(r)=G{M}_{\oplus }/r^2$, so increasing altitude reduces it. Earth's rotation and shape also affect terrestrial weighing: a scale at rest on the rotating Earth responds to the effective acceleration including the centrifugal contribution, rather than simply to a spherical gravitational field.

Distinguish gravitational weight from apparent weight. A supporting scale measures a contact force. In a vertically accelerating cabin, upward acceleration $b$ gives $N=m(g+b)$ while contact is maintained. During common free fall, $N=0$ although $mg$ remains nonzero. Immersion in a fluid can similarly reduce the support needed through buoyancy, without removing the gravitational attraction. These distinctions explain why a change in a balance reading is not automatically a change of mass.