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The very ambiguous terms "percent solution" and "percentage solutions" with no other qualifiers, continue to occasionally be encountered. This common usage of % to mean m/v in biology is because of many biological solutions being dilute and water-based, an aqueous solution. Liquid water has a density of approximately 1 g/cm 3 (1 g/mL). Thus 100 ...
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).
Mass fraction can also be expressed, with a denominator of 100, as percentage by mass (in commercial contexts often called percentage by weight, abbreviated wt.% or % w/w; see mass versus weight). It is one way of expressing the composition of a mixture in a dimensionless size ; mole fraction (percentage by moles , mol%) and volume fraction ...
When expressed in percent, it is known as the mole percent or molar percentage (unit symbol %, sometimes "mol%", equivalent to cmol/mol for 10 −2). The mole fraction is called amount fraction by the International Union of Pure and Applied Chemistry (IUPAC) [ 1 ] and amount-of-substance fraction by the U.S. National Institute of Standards and ...
In chemistry, concentration is the abundance of a constituent divided by the total volume of a mixture. Several types of mathematical description can be distinguished: mass concentration, molar concentration, number concentration, and volume concentration. [1]
An example batch calculation may be demonstrated here. The desired glass composition in wt% is: 67 SiO 2, 12 Na 2 O, 10 CaO, 5 Al 2 O 3, 1 K 2 O, 2 MgO, 3 B 2 O 3, and as raw materials are used sand, trona, lime, albite, orthoclase, dolomite, and borax. The formulas and molar masses of the glass and batch components are listed in the following ...
In general, if an increase of x percent is followed by a decrease of x percent, and the initial amount was p, the final amount is p (1 + 0.01 x)(1 − 0.01 x) = p (1 − (0.01 x) 2); hence the net change is an overall decrease by x percent of x percent (the square of the original percent change when expressed as a decimal number).
As an example, a sample with 70 % of R isomer and 30 % of S will have a percent enantiomeric excess of 40. This can also be thought of as a mixture of 40 % pure R with 60 % of a racemic mixture (which contributes half 30 % R and the other half 30 % S to the overall composition).