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It is the same concept as volume percent (vol%) except that the latter is expressed with a denominator of 100, e.g., 18%. The volume fraction coincides with the volume concentration in ideal solutions where the volumes of the constituents are additive (the volume of the solution is equal to the sum of the volumes of its ingredients).
The ideal gas equation can be rearranged to give an expression for the molar volume of an ideal gas: = = Hence, for a given temperature and pressure, the molar volume is the same for all ideal gases and is based on the gas constant: R = 8.314 462 618 153 24 m 3 ⋅Pa⋅K −1 ⋅mol −1, or about 8.205 736 608 095 96 × 10 −5 m 3 ⋅atm⋅K ...
Some SI units of volume to scale and approximate corresponding mass of water. To ease calculations, a unit of volume is equal to the volume occupied by a unit cube (with a side length of one). Because the volume occupies three dimensions, if the metre (m) is chosen as a unit of length, the corresponding unit of volume is the cubic metre (m 3).
For example, if there are 10 grams of salt (the solute) dissolved in 1 litre of water (the solvent), this solution has a certain salt concentration . If one adds 1 litre of water to this solution, the salt concentration is reduced. The diluted solution still contains 10 grams of salt (0.171 moles of NaCl).
The number density (symbol: n or ρ N) is an intensive quantity used to describe the degree of concentration of countable objects (particles, molecules, phonons, cells, galaxies, etc.) in physical space: three-dimensional volumetric number density, two-dimensional areal number density, or one-dimensional linear number density.
Gas stoichiometry calculations solve for the unknown volume or mass of a gaseous product or reactant. For example, if we wanted to calculate the volume of gaseous NO 2 produced from the combustion of 100 g of NH 3, by the reaction: 4 NH 3 (g) + 7 O 2 (g) → 4 NO 2 (g) + 6 H 2 O (l) we would carry out the following calculations:
The condition to get a partially ideal solution on mixing is that the volume of the resulting mixture V to equal double the volume V s of each solution mixed in equal volumes due to the additivity of volumes. The resulting volume can be found from the mass balance equation involving densities of the mixed and resulting solutions and equalising ...
For a substance X with a specific volume of 0.657 cm 3 /g and a substance Y with a specific volume 0.374 cm 3 /g, the density of each substance can be found by taking the inverse of the specific volume; therefore, substance X has a density of 1.522 g/cm 3 and substance Y has a density of 2.673 g/cm 3. With this information, the specific ...