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[1] [2] Fractions are collected based on differences in a specific property of the individual components. A common trait in fractionations is the need to find an optimum between the amount of fractions collected and the desired purity in each fraction. Fractionation makes it possible to isolate more than two components in a mixture in a single run.
A fraction in chemistry is a quantity collected from a batch of a substance in a fractionating separation process. In such a process, a mixture is separated into fractions, which have compositions that vary according to a gradient. A fraction can be defined as a group of chemicals that have similar boiling points.
Volume percent is the concentration of a certain solute, measured by volume, in a solution.It has as a denominator the volume of the mixture itself, as usual for expressions of concentration, [2] rather than the total of all the individual components’ volumes prior to mixing:
A second way to show division is to use the division sign (÷, also known as obelus though the term has additional meanings), common in arithmetic, in this manner: This form is infrequent except in elementary arithmetic. ISO 80000-2-9.6 states it should not be used.
It is one way of expressing the composition of a mixture in a dimensionless size; mole fraction (percentage by moles, mol%) and volume fraction (percentage by volume, vol%) are others. When the prevalences of interest are those of individual chemical elements , rather than of compounds or other substances, the term mass fraction can also refer ...
The calculations involved in the design of petroleum fractionation columns require in the usual practice the use of numerable charts, tables, and complex empirical equations. In recent years, however, a considerable amount of work has been done to develop efficient and reliable computer-aided design procedures for fractional distillation.
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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: