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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.
In chemistry, the most commonly used unit for molarity is the number of moles per liter, having the unit symbol mol/L or mol/dm 3 in SI units. A solution with a concentration of 1 mol/L is said to be 1 molar , commonly designated as 1 M or 1 M .
To convert from / to /, multiply by 100. To convert from / ... 1 dm 3 /mol = 1 L/mol = 1 m 3 /kmol = 0.001 m 3 /mol (where kmol is kilomoles = 1000 moles) References
The epoxy value is defined as the number of moles of epoxy group per 100g resin. So as an example using an epoxy resin with molar mass of 382 and that has 2 moles of epoxy groups per mole of resin, the EEW = 382/2 = 191, and the epoxy value is calculated as follows: 100/191 = 0.53 (i.e. the epoxy value of the resin is 0.53). [6]
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).
In chemistry, a mole map is a graphical representation of an algorithm that compares molar mass, number of particles per mole, and factors from balanced equations or other formulae. [1] They are often used in undergraduate -level chemistry courses as a tool to teach the basics of stoichiometry and unit conversion .
The ideal gas equation can be rearranged to give an expression for the molar volume of an ideal gas: = = Hence, for a given temperature and pressure, the molar volume is the same for all ideal gases and is based on the gas constant: R = 8.314 462 618 153 24 m 3 ⋅Pa⋅K −1 ⋅mol −1, or about 8.205 736 608 095 96 × 10 −5 m 3 ⋅atm⋅K ...
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.