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The following formulas can be used to calculate the volumes of solute (V solute) and solvent (V solvent) to be used: [1] = = where V total is the desired total volume, and F is the desired dilution factor number (the number in the position of F if expressed as "1/F dilution factor" or "xF dilution"). However, some solutions and mixtures take up ...
For example, sulfuric acid (H 2 SO 4) is a diprotic acid. Since only 0.5 mol of H 2 SO 4 are needed to neutralize 1 mol of OH −, the equivalence factor is: f eq (H 2 SO 4) = 0.5. If the concentration of a sulfuric acid solution is c(H 2 SO 4) = 1 mol/L, then its normality is 2 N. It can also be called a "2 normal" solution.
In fractions like "2 nanometers per meter" (2 n m / m = 2 nano = 2×10 −9 = 2 ppb = 2 × 0.000 000 001), so the quotients are pure-number coefficients with positive values less than or equal to 1. When parts-per notations, including the percent symbol (%), are used in regular prose (as opposed to mathematical expressions), they are still pure ...
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).
m(NaCl) = 2 mol/L × 0.1 L × 58 g/mol = 11.6 g. To create the solution, 11.6 g NaCl is placed in a volumetric flask, dissolved in some water, then followed by the addition of more water until the total volume reaches 100 mL. The density of water is approximately 1000 g/L and its molar mass is 18.02 g/mol (or 1/18.02 = 0.055 mol/g). Therefore ...
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 ...
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The solution has 1 mole or 1 equiv Na +, 1 mole or 2 equiv Ca 2+, and 3 mole or 3 equiv Cl −. An earlier definition, used especially for chemical elements , holds that an equivalent is the amount of a substance that will react with 1 g (0.035 oz) of hydrogen , 8 g (0.28 oz) of oxygen , or 35.5 g (1.25 oz) of chlorine —or that will displace ...