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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 ...
For convenience in avoiding conversions in the imperial (or US customary units), some engineers adopted the pound-mole (notation lb-mol or lbmol), which is defined as the number of entities in 12 lb of 12 C. One lb-mol is equal to 453.592 37 g‑mol, [6] which is the same numerical value as the number of grams in an international avoirdupois pound.
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 ...
List of orders of magnitude for molar concentration; Factor (Molarity) SI prefix Value Item 10 −24: yM 1.66 yM: 1 elementary entity per litre [1]: 8.5 yM: airborne bacteria in the upper troposphere (5100/m 3) [2]
Normality is defined as the number of gram or mole equivalents of solute present in one liter of solution.The SI unit of normality is equivalents per liter (Eq/L). = where N is normality, m sol is the mass of solute in grams, EW sol is the equivalent weight of solute, and V soln is the volume of the entire solution in liters.
Unit conversion formula from mmol/L to mg/dL [5] m g / d L = m m o l / L × m o l e c u l a r w e i g h t ÷ 10 {\displaystyle mg/dL=mmol/L\times molecular\ weight\div 10} Since the molecular mass of glucose C 6 H 12 O 6 is 180.156 g/mol, the factor between the two units is about 18, so 1 mmol/L of glucose is equivalent to 18 mg/dL.
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potassium permanganate has a molar mass of 158.034(1) g mol −1, and reacts with five moles of electrons per mole of potassium permanganate, so its equivalent weight is 158.034(1) g mol −1 /5 eq mol −1 = 31.6068(3) g eq −1. Historically, the equivalent weights of the elements were often determined by studying their reactions with oxygen.