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To convert from / to /, divide by 1000. a (L ... 11.77 0.07685 Carbon monoxide: 1.505 ... (where kmol is kilomoles = 1000 moles) References
Atmospheric pollutant concentrations expressed as mass per unit volume of atmospheric air (e.g., mg/m 3, μg/m 3, etc.) at sea level will decrease with increasing altitude because the atmospheric pressure decreases with increasing altitude. The change of atmospheric pressure with altitude can be obtained from this equation: [2]
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.
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 ...
It is commonly expressed in mass of oxygen consumed over volume of solution, which in SI units is milligrams per liter (mg/L). A COD test can be used to quickly quantify the amount of organics in water. The most common application of COD is in quantifying the amount of oxidizable pollutants found in surface water (e.g. lakes and rivers) or ...
In chemistry, the mole fraction or molar fraction, also called mole proportion or molar proportion, is a quantity defined as the ratio between the amount of a constituent substance, n i (expressed in unit of moles, symbol mol), and the total amount of all constituents in a mixture, n tot (also expressed in moles): [1]
l 3 n −1 In chemistry and related fields, the molar volume , symbol V m , [ 1 ] or V ~ {\displaystyle {\tilde {V}}} of a substance is the ratio of the volume ( V ) occupied by a substance to the amount of substance ( n ), usually at a given temperature and pressure .
For example, 50 g of zinc will react with oxygen to produce 62.24 g of zinc oxide, implying that the zinc has reacted with 12.24 g of oxygen (from the Law of conservation of mass): the equivalent weight of zinc is the mass which will react with eight grams of oxygen, hence 50 g × 8 g/12.24 g = 32.7 g.