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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, the molar concentration of water is
The formula from milligrams (mg) to milli-equivalent (mEq) ... (0.001 mol) of Na + is equal to 1 meq, while 1 mmol of Ca 2+ is equal to 2 meq. References
V eq is the volume of titrant (ml) consumed by the crude oil sample and 1 ml of spiking solution at the equivalent point, b eq is the volume of titrant (ml) consumed by 1 ml of spiking solution at the equivalent point, 56.1 g/mol is the molecular weight of KOH, W oil is the mass of the sample in grams. The normality (N) of titrant is calculated as:
Calculated osmolarity = 2 Na + Glucose + Urea (all in mmol/L) As Na+ is the major extracellular cation, the sum of osmolarity of all other anions can be assumed to be equal to natremia, hence [Na+]x2 ≈ [Na+] + [anions] To calculate plasma osmolality use the following equation (typical in the US): = 2[Na +
Composition per 1 mL (once preparation complete): sodium chloride 7.14 mg (122.17 mmol), potassium chloride 0.38 mg (5.097 mmol), calcium chloride dihydrate 0.154 mg (1.04754 mmol), magnesium chloride hexahydrate 0.2 mg (0.983767 mmol), dibasic sodium phosphate 0.42 mg (2.95858 mmol), sodium bicarbonate 2.1 mg (24.998 mmol), dextrose 0.92 mg (5 ...
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.
The osmol gap is typically calculated with the following formula (all values in mmol/L): = = ([+] + [] + []) In non-SI laboratory units: Calculated osmolality = 2 x [Na mmol/L] + [glucose mg/dL] / 18 + [BUN mg/dL] / 2.8 + [ethanol/3.7] [3] (note: the values 18 and 2.8 convert mg/dL into mmol/L; the molecular weight of ethanol is 46, but empiric data shows that it does not act as an ideal ...
Parts-per notation is often used describing dilute solutions in chemistry, for instance, the relative abundance of dissolved minerals or pollutants in water.The quantity "1 ppm" can be used for a mass fraction if a water-borne pollutant is present at one-millionth of a gram per gram of sample solution.