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Height above the surface is $h=r-R$. Integrate the density over all exterior spherical shells:Repeated integration by parts, or the substitution $h=h_{0}s$, givesExpand $(R+h{)}^2$ and use these moments to obtain the mass of an exponential spherical atmosphere:For $h_0\ll R$, the leading term is surface area $4\pi R^2$ times column mass ${\rho }_0{h}_0$; the other terms retain the finite-curvature corrections.

Using the same cylindrical-shell volume,For unit density, this is the axial angular momentum. To bound it, apply the integral Cauchy-Schwarz inequality to $f(r)=\sqrt{r}v(r)$ and $g(r)=r^{3/2}$:Multiplication by $(2\pi h{)}^2$ gives $L^2\le \pi h{a}^{4}K$. Since the chamber's volume is $V=\pi a^{2}h$, this is precisely

The volume form in cylindrical coordinates is $dV=rdrd\varphi dz$. Integrating around the axis and over the height gives cylindrical-shell volume $2\pi hrdr$. Hence the volume integral isFor unit density, this is the kinetic energy of the rotational motion.

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

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.

The molecular formula is $\boxed{{\mathrm{C}}_4{\mathrm{H}}_8{\mathrm{O}}_2}$, interpreting the vapour density as relative to air. The organic elemental analysis first determines a simpler empirical formula; the vapor-density molar mass measurement then fixes its multiple.

Take 100 mass units of the chemical substance and divide each elemental contribution by its rounded relative atomic mass:The empirical formula is therefore ${\mathrm{C}}_2{\mathrm{H}}_4\mathrm{O}$, with empirical formula mass approximately 44. Its calculated mass percentages are 54.55% carbon, 9.09% hydrogen and 36.36% oxygen. The small discrepancies from the given organic elemental analysis are consistent with rounded experimental data; they should not be converted into large, meaningless integer subscripts.

For equal volumes of ideal gases at equal temperature and pressure, the density ratio equals the molar mass ratio. On the air convention,Doubling ${\mathrm{C}}_2{\mathrm{H}}_4\mathrm{O}$ gives ${\mathrm{C}}_4{\mathrm{H}}_8{\mathrm{O}}_2$; its rounded relative molecular mass of 88 predicts an air-relative vapor density of approximately $88/28.97=3.04$, close to 3.05. If the stated vapor density were instead relative to hydrogen, it would imply a relative molecular mass near 6.1, smaller than the empirical formula mass and incompatible with the analysis. That interpretation cannot describe a pure, undecomposed molecular vapor with the stated composition.

Under the older equivalent weight convention $\mathrm{C}=6$, $\mathrm{O}=8$, $\mathrm{H}=1$, the same composition and formula weight could be written ${\mathrm{C}}_8{\mathrm{H}}_8{\mathrm{O}}_4$. This is a change of notation, not eight modern carbon chemical atoms in the molecule. The result specifies composition and molar mass; it does not distinguish among structural isomers, such as ethyl acetate and butyric acid.

Observe each gas volume, temperature, pressure and moisture condition, and reduce both volumes to the same temperature, pressure and dry-gas basis. Equal scale readings alone do not establish equal amounts of gas.

Read the calibrated volume of each gas in a graduated tube or eudiometer, with the eye at the meniscus to avoid parallax. Allow the gas and the surrounding bath to reach thermal equilibrium, then read a thermometer close to the gas. Heating expands a fixed amount of gas, so a larger reading could reflect a higher temperature rather than more gas. The temperature used in a gas law must be an thermodynamic temperature.

Read the atmospheric pressure with a barometer. If a gas is confined over a liquid, also observe the difference between the inside and outside liquid levels. With $h$ the internal surface height above the external surface, the total internal pressure isHere ${\rho }_{\ell }$ is the confining density of the liquid and $g$ the gravitational acceleration; the sign reverses automatically when $h$ is negative. Equalizing the levels before taking the volume reading removes this hydrostatic pressure correction. A mercury barometer can itself require temperature and calibration corrections for accurate work.

If the gas is saturated with water vapor, subtract the vapor pressure of water at its measured temperature:The observed pressure includes both the test gas and water vapor; counting it all as test gas overestimates the amount. The saturation assumption needs time for equilibration. For a dry gas, no water vapor subtraction is made; another collecting liquid may require its own vapor pressure correction. A collecting liquid must also be chosen so that it neither appreciably dissolves nor reacts with the gas: water, for instance, is unsuitable for preserving the whole amount of freely soluble Ammonia.

For dilute gases obeying the ideal gas law, the volume $V_0$ at a common chosen $T_0,p_0$ isConsequently the corrected comparison isThis is also the ratio of amounts of substance in the ideal gas approximation. Check for leaks, trapped air, changing composition and incomplete transfer. At high pressure or near condensation, real gas behaviour may require compressibility factors; with $pV=ZnRT$, the inferred amount ratio contains the further factor $Z_2/Z_1$.

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.

Use a documented examination of the whole body and lungs, with a lung flotation test as corroboration rather than an infallible verdict. Previous breathing, live birth and death by drowning are three different propositions. The question asks how to investigate; it does not supply a body on which to report an actual positive or negative result.

First establish the apparent age and development of the child, the condition of the remains and any evidence about birth, recovery from the water or attempted resuscitation. If the child is evidently older, independent evidence of survival after birth may settle much of the issue. In a newly born child, the forensic problem is harder. Record decomposition, injuries and any manipulation before opening the chest. Avoid inflating or unnecessarily compressing the lungs before their original state has been documented.

On opening the chest, observe the position, extent, colour and consistency of the lungs. Unexpanded newborn lungs are generally compact and do not fill the chest as expanded lungs do. Aerated regions may be more expanded, spongy and crepitant. Cut surfaces may show small air bubbles. These observations support a judgement about aeration, but disease, collapse and decomposition complicate their interpretation. Examine both sides and different regions, because a few breaths may expand only part of the lungs.

For the historical lung flotation test, place the removed thoracic organs in water, then examine the lungs separately, their lobes and representative small pieces. Record what floats and what sinks at each stage, using a vessel deep enough for free flotation. The physical principle is reduction of average density by retained gas. Testing only a whole organ can miss a small aerated region. The test belongs to the pre-1865 forensic toolkit: the contemporary encyclopedia discussed lung flotation and explicitly recognized both artificial inflation and putrefaction as alternative explanations for floating.^[ref]

There are two main classes of error. Floating can result from gas introduced by attempted artificial respiration or generated after death, so it does not by itself prove spontaneous breathing. Look for decomposition elsewhere and compare the distribution of gas in the lungs and other organs. A floating piece of liver can support the suspicion of generalized gas formation, but a sinking comparison organ does not guarantee that all pulmonary gas arose during life. Historical claims that squeezing a fragment always distinguishes breathed air from artificially introduced air were too confident; compression is not a decisive separator of the causes of aeration.^[ref]

Sinking does not establish that the child never breathed. A few ineffective breaths, limited expansion, or heavy, fluid-filled or diseased lung tissue can leave the material tested denser than water. A later prospective autopsy study reported four false-negative tests in which lungs sank despite prior life reported by medical staff. That observation demonstrates why a negative result cannot stand alone; the study's overall performance in its own material is not a universal accuracy guarantee for decomposed bodies recovered from water.^[ref]

Use the whole examination and independent evidence to reach a qualified conclusion. Feeding, healing changes or reliable observations of the living child may support survival, but they are different evidence from lung aeration. A child can show signs of life without establishing effective breathing, and an inflated lung does not establish a particular cause of death. Immersion may have occurred before or after death; neither flotation nor the body's location alone proves drowning. Where decomposition or attempted resuscitation prevents a confident distinction, report the limit rather than manufacture certainty.

At the heated bottom, thermal conduction first transfers energy into the adjacent gas. A warmer parcel ordinarily has lower density than its cooler surroundings at approximately the same local pressure, so buoyancy drives it upward. Cooler gas replaces it below, creating natural convection. Motion and mixing transport warm gas through the vessel, with conduction continuing across smaller temperature differences. In a sealed rigid vessel the mean pressure can rise as heating continues; “less dense at the same local pressure” is the buoyancy comparison, not a claim that the vessel's overall pressure stays fixed.

A useful control heats an otherwise similar vessel at the top: the resulting warm-over-cold arrangement is stably stratified, suppressing the overturning seen with bottom heating. Heat still spreads by thermal conduction; absence of bulk overturning does not mean absence of heat transfer.

For unwarmed cold outside air, use distributed high-level inlets or upper window openings that let the incoming stream mix with warmer room air before descending into the occupied region. A concentrated low-level cold stream can pass directly across people's feet without first mixing. Provide a separate route for outgoing air and avoid an inlet-to-outlet shortcut that leaves the occupied region poorly ventilated. The UK Health Security Agency's guidance specifically notes that a top window opening allows incoming air to warm as it mixes, reducing cold draughts.^[ref]

This is not a universal rule that every inlet belongs at the ceiling. A low inlet beside a heat source can warm the air before it rises, and deliberately designed displacement systems use other arrangements. Wind, opening heights, available pressure differences and the exhaust path determine the actual flow. The relevant principles are controlled air replacement, buoyancy and mixing, not height in isolation.

A hydrometer is a sealed floating instrument with a weighted bulb and a narrow graduated stem. The ballast keeps the stem upright. By Archimedes' principle, its gravitational weight equals the buoyancy:It therefore sinks farther into a less dense liquid, displacing more volume, and floats higher in a denser one. The stem makes a small volume change readily visible as a change in immersion. If its cross-sectional area is $A$, then $V_{\mathrm{displaced}}=V_0+Az$ over a uniform part of the stem. A density scale is consequently not generally uniform in $z$.

Float the clean instrument freely in a sufficiently deep sample, without its touching the wall or bottom. Remove adhering bubbles, allow the temperature and motion to settle, and read the liquid surface at eye level using the meniscus convention specified for that instrument. A calibration for a particular temperature or reference density cannot be transferred without correction: both the sample and the instrument expand. An ordinary hydrometer measures density or specific gravity; a special scale for a particular chemical solution does not make it chemically selective. In particular, dissolved substances other than glucose can affect the density of urine.

Take dry air as unity, and compare each gas at the same temperature and pressure. In the ideal gas approximation, $\rho =pM/(RT)$, so the relative density isHere $M$ is molar mass. The following values use modern approximate molar masses; rounding and historical atomic weight conventions can shift the last figures.

“Olefiant gas” is confirmed as a historical synonym of Ethylene in the NIST Chemistry WebBook, SRD 69. In this gas list, “carbonic acid” means carbon dioxide, not an isolated vapour of aqueous ${\mathrm{H}}_2{\mathrm{CO}}_3$. “Alcohol” and “ether” refer to ethanol and diethyl ether, rather than their entire modern chemical classes.^[ref]

The order is therefore Ammonia, then carbon monoxide and Ethylene nearly tied, then air, then carbon dioxide and nitrous oxide nearly tied, then ethanol vapour, then diethyl ether vapour. Nearly equal molar masses do not imply equal chemical behaviour.

If instead molecular hydrogen is chosen as unity, divide each molar mass by about $2.016\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$ and state that convention. In the older nominal ${\mathrm{H}}_2=2$ approximation, the corresponding values are about $22,14,22,14.5,8.5,14,23,37$ in the source order. These are a different reference scale, not different substances.

The vapour comparisons are not comparisons of the liquid densities. Pure ethanol and diethyl ether must be considered at conditions where they remain gaseous, or by a low-pressure gas approximation. Their liquids can be lighter than water while their vapours are heavier than air.

Examples heavier than water are Mercury, glycerol and concentrated sulfuric acid; examples lighter than water are ethanol and diethyl ether. “Heavier” here means greater mass for the same volume, rather than greater total mass in arbitrarily sized samples.

Representative room-temperature relative densities, with water as unity, are

These are illustrative values, not exact constants independent of temperature or composition. Dilution changes the density of sulfuric acid, glycerol and ethanol solutions.^[ref][ref]

A distinction from simply observing a floating layer is essential: ethanol is miscible with water, so it mixes rather than remaining as a permanent upper layer. Diethyl ether has limited water solubility and can form an upper layer. More broadly, gases are also fluids, but the natural comparison intended here is between liquids; air at ordinary conditions is much less dense than liquid water.

A weak solution is a function $u\in H_0^1(\Omega )$ satisfying the weak formulationMultiplication by a test function and integration by parts give this identity from the classical equation. Conversely it is the equation in the sense of distributions, with the zero boundary condition encoded by the zero-boundary Sobolev space. Testing only against $C_0^{\infty }(\Omega )$ is equivalent, by its density in $H_0^1$ and continuity of both sides.

One suitable platform is a monolayer molybdenum disulfide semiconductor. An electrostatic accumulation gate supplies conduction electrons whose motion is confined to the atomic sheet; source and drain contacts act as reservoirs. A clean insulating substrate and dielectric or encapsulation reduce disorder and prevent gate leakage. Two split gates deplete the adjacent sheet and leave a narrow conducting gap, thereby confining the in-plane transverse direction while retaining transport along the gap.

Apply a small source-drain voltage, measure the current and determine linear-response conductance while tuning the split-gate voltage and carrier density. A graphene sheet is another two-dimensional carrier system, but its lack of an ordinary gap complicates purely electrostatic confinement; etched boundaries or a suitable gapped architecture may be needed. A two-dimensional material does not by itself guarantee ballistic transport or resolved subbands.^[ref]

Along the in-plane hard axis, the ideal uniform-magnetization model gives reversible rotation from the easy direction toward the applied field. With uniaxial anisotropy energy density $K_u{\sin }^2\theta$, minimization givesfollowed by saturation at $\pm M_s$. The right panel of the preceding figure therefore has a nearly linear central section, essentially zero remanence and ideally no hysteresis. Actual domains, misalignment and defects can produce rounding or a small loop. This is an in-plane hard axis; the much larger out-of-plane demagnetizing field is a different effect.

A charged colloid in a polar electrolyte is surrounded by an electric double layer. For a negatively charged surface, counterions are positive and co-ions negative; their distributions reverse for a positive surface. A compact Stern layer contains specifically adsorbed or closely approaching ions, followed by a diffuse layer where electrostatic attraction competes with thermal motion. Mobile charge integrated over the surroundings balances the particle charge in an overall neutral system.

For a mean-field dilute electrolyte, $n_i(r)=n_{i,\infty }\exp [-z_{i}e\psi (r)/(k_{B}T)]$ with Poisson's equation relates charge density and potential. At weak potential, the diffuse potential outside a sphere decays approximately as $\psi (r)\propto e^{-\kappa (r-a)}/r$. The inverse screening length obeys ${\kappa }^2=e^2{\sum }_i{z}_i^2{n}_{i,\infty }/(ϵk_{B}T)$: added salt shortens the Debye length. The zeta potential is the potential at the hydrodynamic slip plane, not necessarily the surface potential. The simple model neglects ion size, correlations and chemical charge regulation; a polar solvent with almost no dissociated ions need not provide a short screening length.

For a $(10,0)$ tube, $C=10a=2.46\mathrm{nm}$. A cut normal to its axis crosses ten axial load-bearing C-C bonds in the ideal unstrained honeycomb geometry. Assuming equal load sharing and simultaneous failure at $8\mathrm{nN}$ per bond,The graphene areal mass is ${\sigma }_m=0.75\mathrm{mg}{\mathrm{m}}^{-2}=7.5\times 10^{-7}\mathrm{kg}{\mathrm{m}}^{-2}$, giving linear mass densitysince the tex unit has $1\mathrm{tex}=10^{-6}\mathrm{kg}{\mathrm{m}}^{-1}$. Thus the ideal carbon nanotube tenacity isThis neglects vacancies, curvature-induced changes in bond strength, thermal activation, imperfect load sharing and structural relaxation at failure. An aligned macroscopic fibre is a network of finite tubes or bundles: sliding, weak intertube shear transfer, imperfect alignment, defects and unequal load participation limit its strength. The nominal $10\mu \mathrm{m}$ fibre diameter alone does not turn it into one defect-free covalent cylinder, explaining why a value near $2N/tex$ can be much lower than this ideal single-tube estimate.

For fluid density ${\rho }_f$, speed $U$ and characteristic length $d$, inertial kinetic energy scales as ${\rho }_f{d}^3{U}^2$ while viscous work over distance $d$ scales as $\eta U{d}^2$. Their ratio is the Reynolds number,Smaller particles at a fixed speed have smaller Re, so viscous stresses dominate fluid inertia when $\mathrm{Re}\ll 1$. Increasing speed increases Re and eventually makes inertia more important, rather than making viscous dominance stronger.

Particle inertia is a separate issue. Its particle velocity-relaxation time is ${\tau }_p={\rho }_p{d}^2/(18\eta )$; comparing it with a travel time $d/U$ gives a Stokes number ${\tau }_{p}U/d$. This uses particle density ${\rho }_p$, whereas the fluid Reynolds number uses fluid density ${\rho }_f$.

For radius $a=10\mathrm{nm}$ and gold density $2.0\times 10^4\mathrm{kg}{\mathrm{m}}^{-3}$, $m=8.38\times 10^{-20}\mathrm{kg}$. The supplied Stokes model gives $\zeta =3.96\times 10^{-12}\mathrm{kg}{\mathrm{s}}^{-1}$ and $D=1.05\times 10^{-9}{\mathrm{m}}^2{\mathrm{s}}^{-1}$ at $300\mathrm{K}$. Therefore the standard one-second diffusive rms distances are $45.7\mu \mathrm{m}$ in one coordinate and $79.2\mu \mathrm{m}$ in three dimensions.

Using the source's supplied shorthand literally with $m⟨v^2⟩=3k_{B}T$ instead gives $\sqrt{3k_{B}Tt/\zeta }=56.0\mu \mathrm{m}$. That shorthand lacks the factor two in the standard long-time mean-square displacement; it should not be identified with either of the preceding standard rms conventions. None of these rms values is a cumulative path length.

An ideal drag-free fall from rest gives $g{t}^2/2=4.91\mathrm{m}$. Actual gravitational settling in the assumed viscous model is much smaller: ${\tau }_p=m/\zeta =2.12\times 10^{-8}\mathrm{s}$, $u_t=mg/\zeta =0.208\mu \mathrm{m}{\mathrm{s}}^{-1}$, and displacement in one second is approximately $0.208\mu \mathrm{m}$. Thus diffusion overwhelms settling, even though ideal vacuum fall exceeds the rms diffusion distance. For rarefied-gas nanoparticle drag, at a 20-nm size in argon, continuum no-slip drag may require a rarefaction/slip correction; pressure and gas mean-free-path data are absent, so these numerical transport estimates are conditional on the supplied Stokes model.