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Elemental mEq to compound weight Potassium (reference) K 39.098 g/mol 1 (K +) 20 mEq potassium 20*39.098/1=782 mg Potassium citrate monohydrate C 6 H 7 K 3 O 8: 324.41 g/mol 3 (K +) Liquid potassium citrate/gluconate therapy for adults and teenagers taken two to four times a day [3] 20 mEq potassium 20*324/3=2160 mg Potassium gluconate ...
Molecular weight (M.W.) (for molecular compounds) and formula weight (F.W.) (for non-molecular compounds), are older terms for what is now more correctly called the relative molar mass (M r). [8] This is a dimensionless quantity (i.e., a pure number, without units) equal to the molar mass divided by the molar mass constant .
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 standard atomic weight takes into account the isotopic distribution of the element in a given sample (usually assumed to be "normal"). For example, water has a molar mass of 18.0153(3) g/mol, but individual water molecules have molecular masses which range between 18.010 564 6863(15) Da ( 1 H
The volume and normality of the sodium hydroxide are used, along with the weight of the sample, to calculate the free fatty acid value. [3] Acid value is usually measured as milligrams of KOH per gram of sample (mg KOH/g fat/oil), or grams of KOH per gram of sample (g KOH/g fat/oil). [5]
Mass fraction can also be expressed, with a denominator of 100, as percentage by mass (in commercial contexts often called percentage by weight, abbreviated wt.% or % w/w; see mass versus weight). It is one way of expressing the composition of a mixture in a dimensionless size ; mole fraction (percentage by moles , mol%) and volume fraction ...
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 ...
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