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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.
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 ...
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 any ...
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
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).
The theoretical molar yield is 2.0 mol (the molar amount of the limiting compound, acetic acid). The molar yield of the product is calculated from its weight (132 g ÷ 88 g/mol = 1.5 mol). The % yield is calculated from the actual molar yield and the theoretical molar yield (1.5 mol ÷ 2.0 mol × 100% = 75%). [citation needed]
[1] [2] [3] It is defined as the number of millilitres (mL) of pure ethanol present in 100 mL (3.5 imp fl oz; 3.4 US fl oz) of solution at 20 °C (68 °F). The number of millilitres of pure ethanol is the mass of the ethanol divided by its density at 20 °C (68 °F), which is 0.78945 g/mL (0.82353 oz/US fl oz; 0.79122 oz/imp fl oz; 0.45633 oz ...
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).