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The equivalent weight of an element is the mass which combines with or displaces 1.008 gram of hydrogen or 8.0 grams of oxygen or 35.5 grams of chlorine. The equivalent weight of an element is the mass of a mole of the element divided by the element's valence. That is, in grams, the atomic weight of the element divided by the usual valence. [2]
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]
An equivalent (symbol: officially equiv; [1] unofficially but often Eq [2]) is the amount of a substance that reacts with (or is equivalent to) an arbitrary amount (typically one mole) of another substance in a given chemical reaction. It is an archaic quantity that was used in chemistry and the biological sciences (see Equivalent weight § In ...
It is the number of Nitrogens x 56.1 (Mwt of KOH) x 1000 (convert to milligrams) divided by molecular mass of the amine functional compound. So using Tetraethylenepentamine (TEPA) as an example: Mwt = 189, number of nitrogen atoms = 5 So 5 x 1000 x 56.1/189 = 1484. So the Amine Value of TEPA = 1484
Where HV is the hydroxyl value; V B is the amount (ml) potassium hydroxide solution required for the titration of the blank; V acet is the amount (ml) of potassium hydroxide solution required for the titration of the acetylated sample; W acet is the weight of the sample (in grams) used for acetylation; N is the normality of the titrant; 56.1 is ...
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:
Given two bodies, one with mass m 1 and the other with mass m 2, the equivalent one-body problem, with the position of one body with respect to the other as the unknown, is that of a single body of mass [1] [2] = = + = +, where the force on this mass is given by the force between the two bodies.
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.