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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.
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 ...
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
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 ...
An increase of $0.15 on a price of $2.50 is an increase by a fraction of 0.15 / 2.50 = 0.06. Expressed as a percentage, this is a 6% increase. While many percentage values are between 0 and 100, there is no mathematical restriction and percentages may take on other values. [4]
Thus 100 mL of water is equal to approximately 100 g. Therefore, a solution with 1 g of solute dissolved in final volume of 100 mL aqueous solution may also be considered 1% m/m (1 g solute in 99 g water). This approximation breaks down as the solute concentration is increased (for example, in water–NaCl mixtures). High solute concentrations ...
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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).