Bases equivalent mass

The results of this analysis are shown in Figure 5.16, which contains the UV trace, at 220 nm, and the equivalent mass spectral trace. In this case, the mass spectral data are represented by a modification of a reconstructed ion chromatogram (RIC) known as the base-peak trace, i.e. a plot of the intensity of the base peak in each scan (the m jz value of this may well vary from scan to scan) as a function of time. 

TNT equivalency is a simple method for equating a known energy of a combustible fuel to an equivalent mass of TNT. The approach is based on the assumption that an exploding fuel... 

In Part II, a weak acid is titrated with the strong base from Part I and the equivalence point will be somewhat higher than 7 (8 - 9). By titrating a known amount of solid acid with the standardized NaOH, it will be possible to determine the number of moles of ff the acid furnished. From this information, one can obtain the equivalent mass (EMa) of the acid ... 

Once the student had determined the exact concentration of the base, the student then proceeded to determine the equivalent mass of an unknown acid. To do this, the student measured out 0.500 grams of an unknown solid acid and titrated it with the standardized base, recording pH with a calibrated pH meter as the base was added. The student added 43.2 mL of the base but went too far past the end point and needed to back-titrate with 5.2 mL of the 0.100 M HC1 to exactly reach the end point. 

Today aspartame is used in more than 6,000 food products. Aspartame is 160 times as sweet as sucrose based on mass equivalents. Approximately 16,000 tons are consumed annually on a global basis, with approximately 8,000 tons used in the United States and 2,500 tons in Europe. In the body aspartame is metabolized into its three components aspartic acid, phenylalanine, and methanol (Figure 11.1). Aspartic acid is a nonessential amino acid and phenylalanine is an essential amino acid. The condition called phenylketonuria (PKU) is a genetic disorder that occurs when a person lacks the enzyme phenylalanine hydroxylase and cannot process phenylalanine. This results in high phenylalanine blood levels that are metabolized into products one of these is phenylpyruvate, which contains a ketone group and... 

Because atmospheric humidity must be avoided, the reaction flask is isolated from the atmosphere with drying tubes. Moreover, since the solvent is rarely perfectly anhydrous and will contain traces of water due to its hygroscopic nature, its water content must be measured prior to the determination. The equivalence point of the titration reaction is detected by an electrical method instead of a visual method. The current intensity that passes between two platinum electrodes inserted in the reaction medium is measured (see Fig. 19.10). The reagent, which is a mixture of sulphur dioxide, iodine and a base, is characterised by the number of mg of water that can be neutralised by 1 cm3 of this reagent. This is referred to as the equivalent mass concentration of water, or the titre T of the reagent. 

Since in-house facilities for handling a large volume of samples for routine analysis were not available, the analytical work was contracted out to three commercial laboratories. We will refer to them as Laboratories A, B, and C. The contractors were selected on the basis of qualification tests which were intended to serve also for interlaboratory calibration. The results were reported to NRDL as d.p.m. or equivalent 285U thermal-neutron fissions at detonation time. All of the radiochemical data obtained from the laboratories are reported in Ref. 5. These values were punched on cards and converted by computer to equivalent fissions of the device, based on mass-chain yield values supplied by the weapons laboratories. At the same time, the calibration factors derived from qualification-test analyses were applied. Values of the ratios, 95, were formed. All of the ratios for a given nuclide i were then selected along with the corresponding values of r89t95, and the data points were fitted... 

Normality is sometimes used in acid-base or ion-exchange chemistry. With respect to adds and bases, the equivalent mass of a reagent is the amount that can donate or accept 1 mole of H+. With respect to ion exchange, the equivalent mass is the mass of reagent containing 1 mole of charge. 

The equivalent mass of a base is that fraction of the molar mass which contains or can supply one mole of OH-, or can react with one mole of H+. 

Sometimes, the calculation involves a monoprotic acid and a dihydroxy base or another set of conditions in which the relationship is not 1 1. We have to keep track of the various concentrations so that the molarities do not get mixed up. However, stoichiometric calculations involving solutions of specified normalities are even simpler. By the definition of equivalent mass in Chapter 12, two solutions will react exactly with each other if... 

In a case like this problem, where a substance can have several equivalent masses, the normality determined by one type of reaction is not necessarily the normality in other reactions. For instance, if a weak base like NH3 were to be used instead of a strong base for neutralizing the acid, or if the method of detecting the point of neutralization were changed (a different indicator), the equivalent mass of phosphoric acid (and the normality) might well be different. 

The mass of 1.243 g of an acid is required to neutralize 31.72mL of 0.1923 N standard base. What is the equivalent mass of this acid ... 

In order to determine the presence of propagational qualities without any deviation, it may be desirable to use a sealed cell filled with a known amount of boundary propagational substance. Furthermore, a reference cell which is filled with a substance of known calorific (endothermic) value is also useful for accurately measuring the heat of decomposition. In order to obtain clear-cut exothermic peaks over a distinct base line, the heat capacity of the sample cell must be the same as the reference cell therefore, a sample cell of as close to equivalent mass as possible must be used. Moreover, reference cells of different masses need to be kept on hand so that the mass of a sample-containing cell may be matched with that of an available reference cell. 

A sample of 5.00 g of a solid acid was treated with 50.00 mL of 2.500 N NaOH, which dissolved it completely (by reacting with it). There was enough excess NaOH to require 9.13 mL of 1.000 N HCl to neutralize the excess base. What is the equivalent mass of the acid ... 

Chemistry is a very wide field however, only a very small portion, indeed, of this seemingly complex subject is used in this book. These include equivalents and equivalent mass, methods of expressing concentrations, activity and active concentration, equilibrium and solubility product constants, and acids and bases. This knowledge of chemistry will be used under the unit processes part of this book. 

The literature shows confused definitions of equivalents and equivalent masses and no universal definition exists. They are defined based on specific sifuafions and are never unified. For example, in wafer chemistry, three methods of defining equivalenf mass are used equivalenf mass based on ionic charge, equivalenf mass based on... 

Equivalent mass based on ionic charge. In this method, the equivalent mass is dehned as (Snoeyink and Jenkins, 1980) ... 

In the case of the calcium hydroxide, since calcium has a charge of +2 and the coefficient of the term is 2, the number of reference species is 4. Thus, we have now two possible values for the same reaction. In this situation, there are two alternatives the 2 or the 4 as the number of reference species. As mentioned previously, either can be used provided, when one is chosen, all subsequent calculations are based on the one particular choice however, adopt the convention wherein the number of reference species to be chosen should be the smallest value. Thus, the number of reference species in the second reaction is 2, not 4— and all the equivalent masses of the participating substances are obtained by dividing each balanced term of the reaction by 2 Fe(HC03)2/2, 2Ca(OH)2/2, Fe(OH)2/2, 2CaCOf2 and 2H2O/2. 

In Equation (12.56), the equivalent masses could have been based on the number of electrons involved. The ferrous ion is oxidized to the ferric ion thus, the number of reference species is 1 mole of electrons, for which the equivalent masses would have been divided by 1. If this were done, however. Equation (12.56) would stand... 

Similar statements also hold with respect to bases added to raise the pH. They need to react first with existing acidities before the pH can increase, and then-equivalent masses also need to be determined by referring to a single unifying species. This single unifying species is the hydrogen ion. We will discuss the hydroxide ion first. 

To find the equivalent masses of the caustic soda and soda ash, they must be reacted with an acid, which may be represented by the hydrogen ion. Again, remember that the bases specified previously must first consume all the existing acidity of the water represented in the overall by H before they can raise the pH. Thus,... 

To make the previous equations compatible, use 2 as the number of reference species, based on Ca(OH)2. Therefore the equivalent masses are lime = CaO/2 = 37.05 calcium bicarbonate = Ca(HC03)2/2 = 81.05 and phosphorus = 3/5P/2 = 9.3. In this particular case, because the equivalent mass of calcium bicarbonate is also equal to 81.05, [ Calgeq . ... 

If the previous equations are simply added without modifications, electrons will remain free to roam in solution this is not possible. In actual reactions, the previous equations must be modified to make the electrons given up by Equation (15.13) be balanced by the electrons accepted by Equations (15.14) and (15.15). Starting with one gram-equivalent of NH4-N, based on Equation (15.15), assume m equivalents are incorporated into the cell of Nitrosomonas, C5H7NO2. Thus, the NH4-N equivalent remaiiung for the donor reaction of Equation (15.13) is 1 - m, equals (1 - m) [((1/20)N)/1] (1/N) = (1 - m) moles. [The 1/20 came from Equation (15.15) used to compute the equivalent mass of NH4-N.] Thus, (1 - m) moles of NH4-N is available for the donor reaction to donate electrons. By Equation (15.13), the modified donor reaction is then... 

As active substances are often salts of acids or bases, their content in the formulation must be given as an acid or base equivalent to make products with different counter ions comparable. Glyphosate, for instance, is supplied as glyphosate sodium, glyphosate ammonium, glyphosate isopropyl ammonium, and glyphosate trimethylsulfonium (trimesium) salts, each of which has a different molar mass and consequently a different content of free glyphosate per kg formulated product. 

An alternative approach based on a mass spring model (MSM) has been proposed by Williams [ 1 ]. The test system is represented by a lumped mass model with the contact stifBiess between the striker and specimen being ki, this acts on an equivalent mass m which is 17/35 of the specimen mass, and the specimen stiffness is k2- A vital factor is the contact stiffness ki, which controls the dynamics of the system. In reality, this factor is not linear but for a case of the contact of a finite cylinder on plane used here, it can be approximated as linear. From this it is possible to calculate the natural frequency of the system o) where... 

Equivalent mass distribution is derived based on the assumption that the particles are spherical and of equal density. 

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