Search results
Results from the WOW.Com Content Network
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.
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
The nonideality of the solution is reflected by a slight decrease (roughly 2.2%, 1.0326 rather than 1.055 L/kg) in the volume of the combined system upon mixing. As the percent ethanol goes up toward 100%, the apparent molar volume rises to the molar volume of pure ethanol.
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).
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 ...