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For example, the equivalent weight of oxygen is 16.0/2 = 8.0 grams. For acid–base reactions, the equivalent weight of an acid or base is the mass which supplies or reacts with one mole of hydrogen cations (H +). For redox reactions, the equivalent weight of each reactant supplies or reacts with one mole of electrons (e −) in a redox ...
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 ...
An example of a base being neutralized by an acid is as follows. Ba(OH) 2 + 2 H + → Ba 2+ + 2 H 2 O. The same equation relating the concentrations of acid and base applies. The concept of neutralization is not limited to reactions in solution. For example, the reaction of limestone with acid such as sulfuric acid is also a neutralization ...
Normality can be used for acid-base titrations. For example, sulfuric acid (H 2 SO 4) is a diprotic acid. Since only 0.5 mol of H 2 SO 4 are needed to neutralize 1 mol of OH −, the equivalence factor is: f eq (H 2 SO 4) = 0.5. If the concentration of a sulfuric acid solution is c(H 2 SO 4) = 1 mol/L, then its normality is 2 N. It can also be ...
The first solvation shell of a sodium ion dissolved in water. An aqueous solution is a solution in which the solvent is water. It is mostly shown in chemical equations by appending (aq) to the relevant chemical formula. For example, a solution of table salt, also known as sodium chloride (NaCl), in water would be represented as Na + (aq) + Cl ...
In chemistry and thermodynamics, the enthalpy of neutralization (ΔH n) is the change in enthalpy that occurs when one equivalent of an acid and a base undergo a neutralization reaction to form water and a salt. It is a special case of the enthalpy of reaction. It is defined as the energy released with the formation of 1 mole of water.
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:
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, the molar concentration of water is