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A sequence of long waves commonly generated by rapid displacement of water, including by earthquakes and landslides.

Turgor pressure is the internal hydrostatic pressure above the external pressure in a biological cell. When water enters a walled biological cell by osmosis, expansion loads the cell wall. Rising pressure can oppose further net entry while the cell membrane retains its selective permeability. Mechanical support therefore does not require the cell wall to block water.

As an original mechanical check, idealize a small flat patch of cell wall of surface area $A$ under uniform turgor pressure $\Delta P=P_{\mathrm{in}}-P_{\mathrm{out}}$. Its outward force is $F=\Delta PA$. Doubling $A$ at fixed pressure doubles the load; equal internal and external pressures give zero net pressure load. This is a force balance, not a measured wall strength or a complete model of a curved, growing biological cell. It says nothing about which solutes cross the cell membrane.

Increasing cloud droplet number at fixed liquid-water content and cloud thickness reduces mean droplet size and increases total extinction area, raising shortwave optical depth and cloud albedo. This idealized response produces negative shortwave radiative forcing. Changes in cloud amount, water content and dynamics are separate adjustments.^[ref]

Consider an original worked thermometer calibration, using the height $X$ of a liquid column responding through thermal expansion. Stipulate two equilibrated reference baths whose corrected thermometer readings are $0{}^{\circ }\mathrm{C}$ and $40{}^{\circ }\mathrm{C}$, in degree Celsius units. Suppose the corresponding column heights are $12\mathrm{mm}$ and $52\mathrm{mm}$. These are constructed inputs, not observed measurements. Assume the response is linear over this interval, with the same immersion and reading procedure throughout. Applying thermometer calibration givesFor $X=37\mathrm{mm}$, first subtract the zero-reference height, then divide by the full reference span: $(37-12)/(52-12)=25/40$. Multiplying by the $40{}^{\circ }\mathrm{C}$ span gives $t_C=25{}^{\circ }\mathrm{C}$, or $T=298.15\mathrm{K}$ in kelvin. Thus equal $37\mathrm{mm}$ readings in separately equilibrated samples test equal temperatures, regardless of how much internal energy those samples contain. The Celsius offset and modern absolute-scale definition are described in degree Celsius and kelvin.

The two endpoints determine the assumed line, but do not establish linearity. An independent middle reference and a return to the initial bath can test curvature and drift. If the height has an error of $0.2\mathrm{mm}$ while reference values and endpoints are held exact in this calculation, the inferred temperature error is $0.2{}^{\circ }\mathrm{C}$; that is only the height contribution, not a full measurement error budget. A boiling-water reference needs a measured atmospheric pressure correction, rather than an automatic $100{}^{\circ }\mathrm{C}$ assignment. The modern kelvin is not defined by either bath.

A U-tube differential manometer contains a connected column of liquid with one gas connection above each liquid meniscus. Write $p_L,p_R$ for the left and right pressures, and $z_L,z_R$ for the meniscus heights on an upward vertical coordinate. Assume constant mass density $\rho$ and downward gravitational acceleration of magnitude $g$; neglect the gas column's mass density and differences in capillary pressure. The signed vertical length difference $\Delta h=z_R-z_L$ then givesLarger left pressure means a lower left liquid meniscus: $p_L>p_R$ implies $z_L<z_R$. Equal pressures give equal meniscus heights.

The balance follows directly from hydrostatic pressure. At a common height $z_0$ below both surfaces, the two points in the connected homogeneous liquid have equal pressure:Subtracting cancels the common bottom depth and leaves $p_L-p_R=\rho g(z_R-z_L)$. The same-height rule follows from hydrostatic equilibrium, $\nabla p=-\rho g{\mathbf{e}}_z$: a horizontal displacement changes no pressure, and for constant $\rho$ the quantity $p+\rho gz$ is constant throughout the connected liquid. Equal height alone is insufficient for unrelated vessels or for fluid flow with significant acceleration.

For an original constructed example, choose water of stipulated mass density $\rho =1000\mathrm{kg}{\mathrm{m}}^{-3}$, gravitational acceleration $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$, and vertical length difference $\Delta h=0.060\mathrm{m}$. ThenThe SI pressure unit is the pascal: the product has units $\mathrm{kg}{\mathrm{m}}^{-3}\times \mathrm{m}{\mathrm{s}}^{-2}\times \mathrm{m}=\mathrm{kg}{\mathrm{m}}^{-1}{\mathrm{s}}^{-2}=\mathrm{Pa}$. Swapping the two connections gives $-588.6\mathrm{Pa}$ and reverses which liquid meniscus is lower. In equal-bore arms, volume conservation makes each surface move $30\mathrm{mm}$ from its initial equal-height position; using only one surface's displacement would halve the inferred pressure. With unequal arm areas $A_L,A_R$, a left fall $a$ and right rise $b$ instead obey $A_{L}a=A_{R}b$, while the vertical length difference is still $a+b$.

If the right port is open to the local atmosphere, $p_L-p_R$ is the left gauge pressure. Only with that reference and known atmospheric pressure can one obtain absolute pressure fromFor an explicitly stipulated atmospheric pressure of $100000\mathrm{Pa}$, the example gives $100588.6\mathrm{Pa}$ absolute pressure. If the right port is connected to another vessel or a pressure tapping, the same reading is a differential pressure and supplies no absolute pressure without knowing that reference.^[ref]

A valence electron belongs to the electronic arrangement used to describe a chemical atom's chemical bonds and ordinary chemical reactions. For the neutral hydrogen, carbon, nitrogen, and oxygen chemical atoms used here, the valence electrons occupy the outermost occupied electron shell. Their respective electron configurations are $1s^1$, $1s^{2}2s^{2}2p^2$, $1s^{2}2s^{2}2p^3$, and $1s^{2}2s^{2}2p^4$, giving one, four, five, and six valence electrons. See Chemistry 2e, 6.4: electron configurations and valence electrons.

Lewis structures show these valence electrons while omitting the inner electrons. Consequently, eight drawn valence electrons in water do not mean eight total electrons: neutral water has ten, with two inner oxygen electrons omitted. Neutral ammonia likewise has ten total electrons but eight valence electrons; neutral carbon dioxide has twenty-two total electrons but sixteen valence electrons.

This distinction makes the small counts useful: connections and possible lone pairs can be organized without representing every electron or solving the entire many-electron wavefunction. The outermost-electron shell shortcut applies to the examples above, rather than serving as a universal definition for all chemical elements; the electronic arrangement relevant to chemical bonds can also involve other electron subshells.

The VSEPR model predicts approximate molecular geometry by arranging the electron domains around a central chemical atom to reduce their mutual repulsion. Start with a Lewis structure, count the domains, and then distinguish the resulting electron pair geometry from the positions of the chemical atoms themselves. A lone pair counts as a domain even though it is not another chemical atom.^[ref]

Three worked counts have the same total of eight valence electrons. In methane, carbon's four plus the contributions of four hydrogen chemical atoms give $4+4=8$: four bond pairs, no lone pairs, and tetrahedral molecular geometry. In ammonia, nitrogen's five plus three from hydrogen give $5+3=8$: three bond pairs and one lone pair, producing trigonal pyramidal molecular geometry. In water, oxygen's six plus two from hydrogen give $6+2=8$: two bond pairs and two lone pairs, producing bent molecular geometry. All three have four electron domains; their molecular geometries differ.

The regular tetrahedral directions can be represented by the unit vectorsEach pair has dot product $-1/3$, so the ideal bond angle is $\theta =\arccos (-1/3)\approx 109.47^{\circ }$. This is a geometric calculation, rather than a prediction that every four-domain molecule has that exact bond angle. Lone pairs and different kinds of chemical bonds distort the ideal arrangement; water's H–O–H angle is about $104.5^{\circ }$.^[ref]

The model is attractive because small-number bookkeeping produces a spatial prediction without first solving for every molecular orbital. Shape then helps explain chemical polarity and which hydrogen bond donor or hydrogen bond acceptor sites are accessible. It is a qualitative model, not a full calculation of the electronic energy or a universal solution for every chemical compound. Gillespie and Nyholm's 1957 review ^[ref] is a historical anchor for its development.

An ideal gas law measurement obtains amount of substance without knowing the chemical formula in advance:Here $p$ is vapor pressure, $V$ is its occupied volume, $T$ is absolute temperature, $R$ is the molar gas constant, and $m$ is the mass of vapor occupying that volume at those conditions.

Santa Monica College's teaching experiment uses a $125\mathrm{mL}$ Erlenmeyer flask with a pinhole foil cap, a $600\mathrm{mL}$ water bath on a hot plate, a thermometer, and a barometer. Excess volatile liquid displaces air. A dissolved dye loses its colour and deposits when the liquid evaporates; three more minutes of heating follow. Cooling condenses retained vapor. Subtracting the dry assembly mass from the cooled assembly mass supplies $m$. The flask volume comes from brim-full water mass and mass density at its measured temperature. Controls include a dry exterior, excluding bath water, and three trials.^[ref]

The inference assumes vapor pressure equals atmospheric pressure, vapor and bath reach thermal equilibrium, the retained vapor mass is recovered on cooling, and the vapor obeys the ideal gas law. Residual air, thermal expansion of the flask, and vapor temperature below bath temperature are explicit error questions in the teaching experiment. Repeating trials assesses repeatability; it does not remove a shared bias.^[ref]

A volumetric flask prepares a chemical solution at a specified total volume, using a calibration mark.

Its mark specifies contained volume under its calibration conditions. In contrast, a volumetric pipette specifies a delivered portion. The narrow neck makes a small change of volume visible as a larger change of liquid height. The final chemical solution volume includes the solute; “dilute to $100\mathrm{mL}$” does not mean “add $100\mathrm{mL}$ of water”.^[ref]

For an original dilution example, a fully transferred $10.00\mathrm{mL}$ aliquot of $0.100\mathrm{mol}{\mathrm{L}}^{-1}$ acid contains $1.000\mathrm{mmol}$. Making the final total $100.0\mathrm{mL}$ gives analytical concentration $0.0100\mathrm{mol}{\mathrm{L}}^{-1}$. A subsequent $10.00\mathrm{mL}$ aliquot contains $0.1000\mathrm{mmol}$ and needs $1.000\mathrm{mL}$ of $0.100\mathrm{mol}{\mathrm{L}}^{-1}$ base for one-to-one neutralization. The entire diluted flask still needs $10.00\mathrm{mL}$ of that base: dilution preserves the selected inventory but sampling selects a fraction of it.

Overfilling lowers the intended molar concentration. Removing some well-mixed chemical solution to restore the mark also removes solute, so it does not recover the intended inventory. Full transfer, known final volume, and mixing are separate requirements.

NASA's Wind Tunnel Experiment Details, slide 27, describes a practical manometer measurement: partially fill a Tygon tube with water and connect its two ends to total pressure and static pressure ports in a wind tunnel. The total pressure connection pushes its liquid meniscus downward; the vertical length difference supplies the pressure difference. The slide predicts increasing separation as the tunnel speed increases. It supplies an apparatus and method, rather than a table of observed readings.^[ref]

A control for this practical is to expose both connections to the same atmospheric pressure before the run: settled liquid meniscus heights should agree in a symmetric column. Read each liquid meniscus at eye level, reconnect sealed tubing to the identified ports, and wait for settling. Recheck the equal-pressure zero afterwards; reverse connections to check the sign. These checks establish neither mass density nor ruler accuracy.^[ref]

The inference is a differential pressure from hydrostatic pressure, with stagnation pressure on one side and static pressure on the other in NASA's practical. Converting that difference to a flow speed additionally needs the hypotheses discussed under dynamic pressure and Pitot-static tube. A changing pressure can leave the column oscillating; leaks, blocked pressure tappings or trapped liquid in a gas connection can alter or delay the readout. The column should be read after settling, and temperature should be recorded when choosing the measuring liquid's mass density.

Two corrections clarify the limits. If both gas ports are at the same height and carry the same gas of constant mass density ${\rho }_g$, keeping its hydrostatic pressure gives $p_L-p_R=(\rho -{\rho }_g)g\Delta h$. Different port heights require their separate gas column heights. Also, a curved liquid meniscus can sustain a capillary pressure jump through surface tension. Write $J_L,J_R$ for liquid pressure minus gas pressure at the two surfaces. Neglecting gas weight, the balance becomes $p_L-p_R=\rho g\Delta h+J_R-J_L$. Identical meniscus shapes in matching clean bores cancel those jumps; unequal bores or different wetting conditions need not. Thus the useful simplicity of $\rho g\Delta h$ rests on identifiable physical conditions.

Two water molecules associated by a donor O–H···O hydrogen bond. Gas-phase pair energy does not include the complete solvent and entropy changes governing association in solution.

Injecting water to displace oil through a porous reservoir. Spatial permeability contrasts cause preferential flow and early water breakthrough, while unswept slower regions retain oil. The recovery also depends on fractional flow, capillary pressure and phase mixing; a passive streamline model alone does not capture all two-phase effects.

Choose approximate atomic weights $A_{\mathrm{r}}(\mathrm{H})=1.008$ and $A_{\mathrm{r}}(\mathrm{O})=15.999$, and use $M_{\mathrm{u}}\approx 1\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$. The chemical formula ${\mathrm{H}}_2\mathrm{O}$ then gives water molar massThese are deliberately chosen rounded inputs, not exact atomic weights or a universal value for every water sample. Different isotopic abundances, notably those in heavy water, change the mean molar mass; chemical formula alone does not specify them. Adding atomic weights also neglects the tiny mass change associated with chemical bond energies, immaterial at this rounding precision.

For an illustrative pure water sample of mass $4.50375\mathrm{g}$, this convention givesThe count refers to water molecules. Their hydrogen chemical atom count is approximately $2N=3.01107038\times 10^{23}$, and their oxygen chemical atom count is approximately $N$. The mean mass per water molecule is $M/N_{\mathrm{A}}\approx 2.99146\times 10^{-26}\mathrm{kg}$. Extra displayed digits expose the arithmetic; they do not imply that the sample composition or mass was measured that precisely.

The same conversion makes stoichiometry a mass calculation. For the chemical equation $2{\mathrm{H}}_2+{\mathrm{O}}_2\to 2{\mathrm{H}}_2\mathrm{O}$, complete conversion of $0.250000\mathrm{mol}$ of molecular hydrogen and $0.125000\mathrm{mol}$ of molecular oxygen gives $0.250000\mathrm{mol}$ of water. With the chosen approximate atomic weights, the reactant masses are $0.50400\mathrm{g}$ and $3.99975\mathrm{g}$, summing to $4.50375\mathrm{g}$. The $2:1:2$ ratio belongs to molecule counts and amount of substance, not to masses; this bookkeeping neglects the much smaller reaction energy contribution to mass.

A molecule containing two hydrogen chemical atoms and one oxygen chemical atom. An isolated ground-state water molecule has a bent equilibrium geometry; ordinary bulk water contains interacting water molecules rather than a collection of independent rigid objects.

Water potential expresses the chemical potential of water relative to a stated pure-water reference in units of pressure. For a nearly incompressible liquid at fixed temperature, use a constant reference molar volume ${\bar{V}}_w$ and writeThe reference is pure water at the same temperature and a chosen pressure $P_0$. Both sides must use the same reference. Neglecting height differences and interactions with solid surfaces, the approximation isHere $a_w$ is the dimensionless thermodynamic activity of water, $R$ the molar gas constant, and $T$ the absolute temperature. The natural logarithm is therefore applied to a dimensionless quantity. The pressure contribution ${\Psi }_p=P-P_0$ increases with pressure; the solute potential ${\Psi }_s$ is negative when $a_w<1$. Across an available passive water route, net water transfer favors higher to lower water potential. The familiar plant description adds effects of height and interactions with solid surfaces when they matter; they are omitted in the two liquid compartments here.^[ref]

Adding solute and raising pressure can thus cancel in water potential while each contribution remains nonzero. Equality of water potential does not imply equality of pressure or of every solute concentration. In a supported plant cell, the cell wall permits a positive turgor pressure to balance a negative solute potential; plasmolysis illustrates the opposite loss of support after water leaves.

Chemical and physical exchange between water and rock, including dissolution, precipitation, hydration and changes in oxidation state.

Water vapor is gaseous water. Its absorption and emission contribute to atmospheric radiative transfer and the greenhouse effect.

A weak acid establishes a chemical equilibrium between its proton-donating form and its conjugate base in water. For a neutral monoprotic acid $\mathrm{HA}$, the aqueous chemical equation isWater accepts a proton to become hydronium; ${\mathrm{A}}^{-}$ is the conjugate base. The double arrow denotes reversible exchange at chemical equilibrium, not equal populations. The acid dissociation constant and composition determine the hydronium concentration; substituting total chemical acid concentration into the pH formula is generally incorrect.

“Weak” describes the tendency to transfer a proton in a specified solvent, rather than how much chemical acid was dissolved. A weak acid can be concentrated, and a strong acid can be dilute. Even the same weak acid can have a large weak acid dissociation fraction after sufficient dilution, while its acid dissociation constant stays fixed at the same temperature and standard state conventions.^[ref]

For a monoprotic acid with analytical concentration $C_T=[\mathrm{HA}]+[{\mathrm{A}}^{-}]$, define the dimensionless weak acid dissociation fractionThis counts acid units in the conjugate base form. Identifying it with $[{\mathrm{H}}_3{\mathrm{O}}^{+}]/C_T$ additionally requires negligible self-ionization of water and no other sources of electric charge. In the ideal dilute chemical solution model, with $h=[{\mathrm{H}}_3{\mathrm{O}}^{+}]/c^{\circ }$, the acid dissociation constant gives $\alpha =K_a/(K_a+h)$. Thus lowering hydronium thermodynamic activity favors the conjugate base.

The diagram uses stipulated dimensionless acid dissociation constant $K_a=10^{-5}$ and self-ionization of water ion product $K_w=10^{-14}$ at fixed temperature; these are model inputs, not measurements of a named substance. It solves the weak acid mass and charge balances. At $C_T=0.010\mathrm{mol}{\mathrm{L}}^{-1}$, the calculated weak acid dissociation fraction is about $3.11\%$ and pH about $3.51$; at $C_T=10^{-4}\mathrm{mol}{\mathrm{L}}^{-1}$, they are about $27.0\%$ and $4.57$. More acid units are dissociated as a fraction of the total, yet fewer hydronium ions occur per volume.

At extreme dilution, self-ionization of water makes $h$ approach $\sqrt{K_w}$, so pH approaches $7$ for the chosen model. The limiting weak acid dissociation fraction is $K_a/(K_a+\sqrt{K_w})\approx 0.9901$, rather than exactly one. The calculation assumes pure added water, a neutral monoprotic acid, and no competing chemical reactions; actual contamination becomes increasingly important as $C_T$ falls.

A pH measurement becomes evidence for a weak acid chemical equilibrium when paired with a known analytical concentration and a calibrated glass electrode. For the pure-weak acid model, measured pH estimates $h=10^{-\mathrm{pH}}$ only in the ideal thermodynamic activity approximation. Then $o=K_w/h$, $x=h-o$, $y=c-x$, and $K_a=hx/y$. Require $0<x<c$; values outside that range signal inconsistent measurements or an inadequate model. Repeating this reconstruction across a dilution series tests whether one acid dissociation constant describes the data, rather than merely fitting one point.

The Analytical Sciences Digital Library's Potentiometric Titration of an Unknown Weak Monoprotic Acid provides a complementary acid-base titration experiment. Its apparatus includes a glass electrode and meter, a stirring magnet, a volumetric flask, pipettes, and a burette containing standardized sodium hydroxide. Prepare an aqueous acid stock of known analytical concentration, transfer a measured portion, and stir gently with the glass electrode clear of the magnet. Record added volume and stable pH after each addition; use a preliminary run to locate the steep region and smaller additions there on a fresh portion. The readout is the pH versus added volume curve, not a visual count of dissociated molecules.

Calibrate with fresh standard buffer solutions bracketing the pH range being interpreted. Use their assigned values at the actual temperature; a pair near pH $4$ and $7$ can cover the midpoint of many weak acid examples, but does not cover an entire alkaline endpoint region. Rinse between standards and samples to limit carryover, keep the sensing bulb and reference electrode junction immersed, and wait for a stable reading. These instrument practices are documented in HI5221 & HI5222 Instruction Manual: pH and mV and ISE and Temperature Bench Meters by Hanna Instruments (pp. 45–47, 49–50 and 73–74: calibration, measurement, rinsing and electrode conditioning).

Locate the acid-base titration equivalence region from the steepest change of pH with added volume. For one-to-one neutralization, standardized sodium hydroxide concentration times equivalence volume estimates the initial amount of substance of acid. Half that volume gives approximately equal $[\mathrm{HA}]$ and $[{\mathrm{A}}^{-}]$ only when subsequent chemical equilibrium redistribution is small. The Henderson-Hasselbalch equation then gives $\mathrm{pH}\approx \mathrm{p}{K}_a$. Unequal activity coefficients or appreciable hydronium or hydroxide inventories can shift this estimate. The full chemical equilibrium curve supplies a stronger consistency check than its midpoint alone.

Useful controls for this model test are a fresh independently prepared acid portion, a water blank, a check standard after measurement to detect instrument drift, and fixed recorded temperature. A blank diagnoses contamination but should not be subtracted directly from pH, which is a logarithm. Keep exposure to atmospheric carbon dioxide comparable, especially at high dilution, and record any background salt that changes ionic strength. These are experimental-design controls suggested by the model's assumptions. The observed curve and its uncertainty determine whether the model is adequate; the diagrams here contain no experimental readouts. An aqueous acid dissociation constant cannot be inferred unchanged from a sample requiring a different solvent mixture, and a nearly neutral dilute sample makes the difference $h-K_w/h$ particularly sensitive to pH error.