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Being dimensionless, its unit is 1; it is expressed as a number, e.g., 0.18. 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 ...
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).
1% m/v solutions are sometimes thought of as being gram/100 mL but this detracts from the fact that % m/v is g/mL; 1 g of water has a volume of approximately 1 mL (at standard temperature and pressure) and the mass concentration is said to be 100%. To make 10 mL of an aqueous 1% cholate solution, 0.1 grams of cholate are dissolved in 10 mL of ...
11.6 g of NaCl is dissolved in 100 g of water. The final mass concentration ρ(NaCl) is ρ(NaCl) = 11.6 g / 11.6 g + 100 g = 0.104 g/g = 10.4 %. The volume of such a solution is 104.3mL (volume is directly observable); its density is calculated to be 1.07 (111.6g/104.3mL) The molar concentration of NaCl in the solution is therefore
This improper name persists, especially in elementary textbooks. In biology, the unit "%" is sometimes (incorrectly) used to denote mass concentration, also called mass/volume percentage. A solution with 1 g of solute dissolved in a final volume of 100 mL of solution would be labeled as "1%" or "1% m/v" (mass/volume). This is incorrect because ...
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A tenfold dilution for each step is called a logarithmic dilution or log-dilution, a 3.16-fold (10 0.5-fold) dilution is called a half-logarithmic dilution or half-log dilution, and a 1.78-fold (10 0.25-fold) dilution is called a quarter-logarithmic dilution or quarter-log dilution.
For example, when d=4, the hash table for two occurrences of d would contain the key-value pair 8 and 4+4, and the one for three occurrences, the key-value pair 2 and (4+4)/4 (strings shown in bold). The task is then reduced to recursively computing these hash tables for increasing n , starting from n=1 and continuing up to e.g. n=4.