Calculating amounts from equations

The formulae of some ionic compounds may be written as AB-nH20, where nH20 indicates that n molecules of water are associated with each AB unit within the crystal lattice. Because the water is within the lattice, the substance is perfectly dry to the touch but it contains water of crystallization and is said to be hydrated. The water may be driven off by strong heating (dehydration) and the residue, that now does not contain water of crystallization, is termed anhydrous. 

Blue copper(II) sulfate crystals (CUSO4-5H2O) become white, anhydrous copper(II) sulfate on heating  

If water is added to anhydrous copper(II) sulfate, it turns blue again. Other common substances that contain water of crystallization are 

The dehydration of copper(ll) sulfate by heating Is a stepwise process 

You have already seen that chemical equations give the following information  

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The second objection amounts to saying that our p+Aare not kp because we had calculated them from equation (4) on the assumptions that Pn+ is the only propagator and that its concentration is given by equation (8) ... 

Finally, after convergence is obtained and we have arrived to the final value of the spreading pressure we calculate x from lAST (Equation 14.9) and the adsorbed amounts from Equation 14.12a,b. 

In addition to the sulfuric acid required for pH adjustment, some amount of acid is consumed by the reduction reaction (Equation 8.15). If sulfur dioxide is used as the reducing agent, it will provide all the acid consumed by this reaction, and additional acid will not be required. However, if sodium bisulfite or sodium metabisulfite is used, additional acid must be supplied to satisfy the acid demand. This acid requirement is stoichiometric and can be calculated from Equations 6.19 to 6.22. 

Thus, log-log plots of S versus C provide an easy way to obtain the values for K (the intercept) and N (the slope of the line). The log-log plot can be used for graphic interpolation of adsorption at other concentrations, or, when values for K and N have been obtained, the amount of adsorption can be calculated from Equation 20.9. Figure 20.9 shows an example of adsorption isotherms for phenol adsorbed on Frio sandstone at two different temperatures. Note that when N = 1, Equation 20.9 simplifies to Equation 20.6 (i.e., adsorption is linear). 

Within each stage, the amount of catalyst, W, may be calculated from equation 21.5-4, together with an appropriate rate law, and the energy equation 21.5-8. Optimization problems relating to minimizing W may be considered in terms of choice of values of r and fAi. 

An important group of analytical methods is based on measurements of the change in isotopic ratio when active and non-active isotopes are mixed. In the simplest case, a known amount w1 of labelled analyte of known specific activity at is added to the sample. After isotopic mixing has been established sufficient of the analyte is separated (nor normally 100%) to allow the new specific activity a2 to be measured. Measurements of activity and the amount of the analyte separated are thus required. Subsequently the amount w2 of analyte in the sample may be calculated from equation (10.17). 

Figure 5.12 shows a graphic representation of a back titration. The long vertical block on the left (down arrow) represents the equivalents of the first titrant added. This amount actually exceeds the equivalents of the substance titrated present in the reaction flask. The short vertical block on the lower right (up arrow) represents the amount of the second titrant (the so-called back titrant) used to come back to the end point, to titrate the excess amount of the initial titrant. The difference between the total equivalents of the first titrant and the total equivalents of the second titrant is the number of equivalents that actually reacted with the analyte. It is this number of equivalents that is needed for the calculation. The calculation therefore uses the following equation, derived from Equation (4.40) in Chapter 4 ... 

As far as the velocity and the extent of the conversion are concerned, the two processes are, however, altogether different. Whereas an acid is practically instantaneously and completely converted into a salt by an equivalent amount of a sufficiently strong base (neutralisation), a process on which, indeed, alkalimetry and acidimetry depend, it is not possible to obtain from equimolecular amounts of acid and alcohol the theoretical (calculated) amount of ester. A certain maximal quantity of ester is formed, but always falls short of the theoretical, and it is impossible, even by indefinitely extending the duration of the reaction, to make the unchanged acid and alcohol produce ester in excess of that maximum. If, for example, equimolecular amounts of acetic acid and alcohol are allowed to interact in a closed system, only two-thirds of each enter into reaction, and it is impossible to induce the remaining third of acetic acid to react with that of alcohol. The maximum yield of ester therefore amounts to only two-thirds, or 66-7 per cent, of the theoretical quantity. The quantitative difference in the course of the two reactions mentioned above depends on the fact that esterification is a so-called reversible reaction , i.e. one in which the reaction products represented on the right-hand side of the equation (ester and water) also interact in the opposite direction ... 

The energy equivalent of the calorimeter, e, and the enthalpy of the isothermal calorimetric process, A//icp, were derived from equations 8.2 and 8.4, respectively. The standard enthalpy of reaction 8.5 was computed as Ar//°(8.5) = AZ/icp/n, where n is the amount of substance of Mo(ri5-C5H5)2(C2H4) used in the experiment. The data in table 8.1 lead to a mean value Ar//°(8.5) = — 186.0 2.1 kJ mol-1, where the uncertainty is twice the standard deviation of the mean (section 2.6). This value was used to calculate the enthalpy of reaction (8.6), where all reactants and products are in their standard reference states, at 298.15 K, from... 

To calculate Ar//(I0.10), it is necessary to know n, the amount of substance of Cr(CO)6 consumed during period t. This is best done by analyzing the final reaction mixture, but as mentioned, it can also rely on the reaction quantum yield, [Pg.150]

The next step involved cooling the reaction mixture to -196°C, removing the H2 at low pressure, and sealing the tube. This sealed tube was then used in the equilibrium measurements. When it warmed up, a fraction of the hydride complex reacted with benzene, yielding H2 and the phenyl complex, according to equilibrium 14.12. Therefore, the total amount of substance of H2 in equation 14.18 is given by the sum of the initial amount of substance of H2 (no) and the amount of substance of Sc(Cp )2Ph in equilibrium. The latter is easily calculated from the relative concentrations of Sc(Cp )2Ph and Sc(Cp )2H determined by H NMR, and the known initial concentration of Sc(Cp )2H (5.4 x 10-5x 1000/0.5 = 0.108 mol dm-3). To evaluate the initial amount of substance of H2, consider the experimental procedure before and after reaction 14.19 takes place. When this reaction occurs (at 25 °C) a certain amount of H2 remains in solution, and it can be calculated by an equation similar to 14.17. This amount will be equal to no, by assuming that (1) there is no further H2 solubilization when the tube is rapidly cooled to — 196 °C, and (2) only the H2 dissolved in the frozen reaction mixture is not removed by the evacuation procedure. 

The previous description illustrates well the complications that may arise in second law studies when phase and reaction equilibria occur simultaneously. A number of assumptions are usually made, some of which may influence the final thermochemical results. For instance, it is possible that the equilibrium concentration of hydrogen obtained in the study by Bercaw and co-workers is not very accurate, because no may be underestimated (the cooling to — 196 °C will increase the amount of H2 in the condensed phase). Nevertheless, this error, which is constant for all the measurements at different temperatures (average T = 316 K), will probably have a negligible effect on the calculated Ar//j16 and Ar,V) l6 values, 28.3 1.9 kJ mol-1 and —5.3 6.1 J K-1 mol-1, respectively. These values were obtained from equation 14.20, which is a linear fit of the... 

From this equation the amount of liquid to be distilled in order to obtain a liquid of given concentration in the still may be calculated, and from this the average composition of the distillate may be found by a mass balance. 

Preparation from Borax. Dissolve 12 g of borax in 25 ml of water in a beaker with heating. What is the reaction of the solution to litmus and what is it due to Write the molecular and net-ionic equations of the borax hydrolysis reaction. How can the hydrolysis of borax be facilitated Calculate what amount of a 25% hydrochloric acid solution is needed to prepare boric acid from 12 g of borax. Measure off the calculated amount of acid, and taking a small excess amount, pour the acid into the hot borax solution. Let the solution cool slowly. What substance crystallizes Filter off the crystals on a Buchner funnel, dry them between filter paper sheets, and recrystallize them from hot water, guiding yourself by the table of solubilities. Determine the product yield (in per cent). Keep the prepared boric acid for the following experiments. 

In pure water we expect equal amounts of H+ ( hydrogen ion ) and OH- ( hydroxide ion ). From equation (3) we can calculate the concentration of H+ or OH in pure water to be 10 7 M. Therefore, a solution with an H+ concentration of 10-7 M is defined as neutral. An H+ concentration greater than IO-7 M indicates an acidic solution an H+ concentration less than IO-7 M indicates a basic solution. Rather than deal with exponentials, it is convenient to express the H+ concentration on a pH scale, the term pH being defined by the equation... 

Values of /JAB and AB have been calculated by applying equation (20), and often show striking catalytic effects of the additive i.e. very small amount of additive produces reductions in / A which are far greater than can be explained in terms of the ssh theory for vibration-translation transfer. Now that vibration-vibration transfer is better understood (see Section 5 below), it has become quite clear that many of these cases, where B is a polyatomic molecule, can be explained in terms of rapid vibration-vibration transfer from... 

The thermal time constant, defined in Equation 9.5, is useful for calculating heating and cooling times, which often take up a considerable amount of the cycle time of batch and semi-batch reactions. We start from Equation 9.4 that, after variable separation, combines with Equation 9.5, to become... 

As follows from equations (2.16) and (2.17), it is not obligatory to operate the amount of expended inducer in the calculations. Moreover, there are conjugated processes that proceed without inducer participation. Therefore, it is of much greater importance to account for the expenditure of compound A in both reactions and to determine the induction factor via it. From these positions, equation (2.17) is somehow universal, because it reflects chemical induction in any display. Moreover, it is shown below, how many important consequences follow from it, which may not be deduced from equation (1.3). 

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