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As an example, a measured NO x concentration of 45 ppmv in a dry gas having 5 volume % O 2 is: 45 × ( 20.9 - 3 ) ÷ ( 20.9 - 5 ) = 50.7 ppmv of NO x. when corrected to a dry gas having a specified reference O 2 content of 3 volume %. Note: The measured gas concentration C m must first be corrected to a dry basis before using the above equation.
1 Nm 3 of any gas (measured at 0 °C and 1 atmosphere of absolute pressure) equals 37.326 scf of that gas (measured at 60 °F and 1 atmosphere of absolute pressure). 1 kmol of any ideal gas equals 22.414 Nm 3 of that gas at 0 °C and 1 atmosphere of absolute pressure ... and 1 lbmol of any ideal gas equals 379.482 scf of that gas at 60 °F and ...
As a more complex example, the concentration of nitrogen oxides (NO x) in the flue gas from an industrial furnace can be converted to a mass flow rate expressed in grams per hour (g/h) of NO x by using the following information as shown below: NO x concentration = 10 parts per million by volume = 10 ppmv = 10 volumes/10 6 volumes NO x molar mass
For example, the conversion factor between a mass fraction of 1 ppb and a mole fraction of 1 ppb is about 4.7 for the greenhouse gas CFC-11 in air (Molar mass of CFC-11 / Mean molar mass of air = 137.368 / 28.97 = 4.74). For volume fraction, the suffix "V" or "v" is sometimes appended to the parts-per notation (e.g. ppmV, ppbv, pptv).
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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.
10 −1 g dg decigram 10 1 g dag decagram 10 −2 g cg: centigram: 10 2 g hg hectogram 10 −3 g mg: milligram: 10 3 g kg: kilogram: 10 −6 g μg: microgram (mcg) 10 6 g Mg megagram 10 −9 g ng: nanogram: 10 9 g Gg gigagram 10 −12 g pg picogram 10 12 g Tg teragram 10 −15 g fg femtogram 10 15 g Pg petagram 10 −18 g ag attogram 10 18 g Eg ...
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: