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This formula expects weight to be measured in kilograms and creatinine to be measured in mg/dL, as is standard in the US. The resulting value is multiplied by a constant of 0.85 if the patient is female. This formula is useful because the calculations are simple and can often be performed without the aid of a calculator.
l 3 n −1 In chemistry and related fields, the molar volume , symbol V m , [ 1 ] or V ~ {\displaystyle {\tilde {V}}} of a substance is the ratio of the volume ( V ) occupied by a substance to the amount of substance ( n ), usually at a given temperature and pressure .
This article uses the standard notation ISO 80000-2, which supersedes ISO 31-11, for spherical coordinates (other sources may reverse the definitions of θ and φ): . The polar angle is denoted by [,]: it is the angle between the z-axis and the radial vector connecting the origin to the point in question.
Whereas osmolality (with an "l") is defined as the number of osmoles (Osm) of solute per kilogram of solvent (osmol/kg or Osm/kg), osmolarity (with an "r") is defined as the number of osmoles of solute per liter (L) of solution (osmol/L or Osm/L). As such, larger numbers indicate a greater concentration of solutes in the plasma.
The equation can only be applied when the purged volume of vapor or gas is replaced with "clean" air or gas. For example, the equation can be used to calculate the time required at a certain ventilation rate to reduce a high carbon monoxide concentration in a room.
In chemistry, the mass concentration ρ i (or γ i) is defined as the mass of a constituent m i divided by the volume of the mixture V. [1]= For a pure chemical the mass concentration equals its density (mass divided by volume); thus the mass concentration of a component in a mixture can be called the density of a component in a mixture.
The osmol gap is typically calculated with the following formula (all values in mmol/L): = = ([+] + [] + []) In non-SI laboratory units: Calculated osmolality = 2 x [Na mmol/L] + [glucose mg/dL] / 18 + [BUN mg/dL] / 2.8 + [ethanol/3.7] [3] (note: the values 18 and 2.8 convert mg/dL into mmol/L; the molecular weight of ethanol is 46, but empiric data shows that it does not act as an ideal ...
This gives a = 100 μg/mL if the drug stays in the blood stream only, and thus its volume of distribution is the same as that is = 0.08 L/kg. If the drug distributes into all body water the volume of distribution would increase to approximately V D = {\displaystyle V_{D}=} 0.57 L/kg [ 8 ]