Conversion factors involving volume

From the back inside cover, we find 1 L = 10 m, but there is no relationship listed involving km. From our knowledge of SI prefixes, however, we know 1 km = 10 m and we can use this relationship between lengths to write the desired conversion factor between volumes ... 

As pointed out in Chapter 3, a balanced equation can be used to relate moles or grams of substances taking part in a reaction. Where gases are involved, these relations can be ex tended to include volumes. To do this, we use the ideal gas law and the conversion factor approach described in Chapter 3. 

The various kinetic and thermodynamic factors involved in vinyl free radical polymerization have been considered for the case of a batch (or semi-batch) polymerization being carried out to very high conversion. In particular, computations have been done for the final stage of the reaction when monomer concentration is reduced from approximately 5 volume % to 0.5 volume %. 

You are probably well aware that the pressures at great depths under the ocean are extremely large, and the deeper you go, the greater the pressure that is exerted. Let s start small and look at a beaker filled with water. The beaker has a diameter of 10 cm, and the height of the water is 20 cm. How much pressure does the water exert on the bottom of the beaker To answer this question we simply need to recall that P = FI A. The force involved is the weight of the water contained in the beaker. The weight of the water can be calculated from the mass of the water, and the mass of the water can be calculated from the volume, using density as a conversion factor. 

Because virtually all stoichiometric calculations involve moles (abbreviated mol) of material, molarity is probably the most common concentration unit in chemistry. If we dissolved 1.0 mol of glucose in enough water to give a total volume of 1.0 L, we would obtain a 1.0 molar solution of glucose. Molarity is abbreviated with a capital M. Notice that, because molarity has units of moles per liter, molar concentrations are conversion factors between moles of material and liters of solution. 

When reactants are liquids, they are almost always measured by volume. So, to do calculations involving liquids, you add two more steps to the sequence of mass-mass problems—the conversions of volume to mass and of mass to volume. Five conversion factors—two densities, two molar masses, and a mole ratio—are needed for this type of calculation, as shown in Skills Toolkit 4. 

CDP4-P Fairly straight forward California registration problem where you must carry out a number of calculations involving conversion factors to calculate CSTR and PFR reactor volumes and a batch reaction time. 

Stoichiometric problems involving gas volumes can be solved by the general mole-ratio method outlined in Chapter 9. The factors 1 mol/22.4 L and 22.4 L/1 mol are used for converting volume to moles and moles to volume, respectively. (See Figure 12.16.) These conversion factors are used under the assumption that the gases are at STP and that they behave as ideal gases. In actual practice, gases are measured at other than STP conditions, and the volumes are converted to STP for stoichiometric calculations. 

Gases in Chemical Reactions Stoichiometric calculations involving gases are similar to those that do not involve gases in that the coefficients in a balanced chemical equation provide conversion factors among moles of reactants and products in the reaction. For gases, the amoimt of a reactant or product is often specified by the volume of reactant or product at a given temperature and pressiue. The ideal gas law is then used to convert from these quantities to moles of reactant or product. Alternatively at standard temperature and pressure, volume can be converted directly to moles with ffie equality ... 

As you saw earlier, you can use molarity as a conversion factor, and in this way you can calculate the volume of solution that is equivalent to a given mass of solute (see Example 4.10). This means that you can replace mass measurements in solution reactions by volume measurements. In the next example, we look at the volumes of solutions involved in a given reaction. 

Let s say we want to determine what volume of O2 is required to react completely with 65.8 mL of CO at STP. We could use the ideal gas equation to convert the volume of CO to moles, use the stoichiometric conversion factor to convert to moles O2, and then use the ideal gas equation again to convert moles O2 to volume. But this method involves several unnecessary steps. We get the same result simply by using the conversion factor expressed in miUihters ... 

Examples 1-2 and 1-3 further illustrate that numerical calculations involving density are generally of two types determining density from mass and volume measurements and using density as a conversion factor to relate mass and volume. 

There are circumstances where weight is an important factor (Example 3), but the calculations involving gases may be in terms of volumes of gases involved. The conversion from volumes of gas to mass is done through the numbers of moles. The methods used in these problem solutions are as in Chapter 4 except that the numbers of moles converted to mass (g, lb, etc.) must be determined from the volume, temperature, and pressure of the gases. 

Figure 10.1. Comparison of normal (top) and surface-enhanced (bottom) Raman scattering. The top panel shows the conversion of incident laser light of intensity /(vl) into Stokes scattered light /NRS, which is proportional to the Raman cross section and the number of target molecules N in the probed volume. In the bottom panel

The whole of the internal surface area of a porous catalyst will be available for the catalytic reaction if the rates of diffusion of reactant into the pores, and of product out of them, are fast compared with the rate of the surface reaction. In contrast, if the reactant diffuses slowly but reacts rapidly, conversion to product will occur near the pore entrances and the interior of the pores will play no role in the catalysis. Ion exchange resins are typical examples of catalysts for which such considerations are important (cf. Sect. 2.3). The detailed mathematics of this problem have been treated in several texts [49-51] and we shall now quote some of the main theoretical results derived for isothermal conditions. The parameters involved tend to be those employed by chemical engineers and differ somewhat from those used elsewhere in this chapter. In particular, the catalyst material (active + support) is present in the form of pellets of volume Vp and the catalytic rates vv are given per unit volume of pellet (mols m 3). The decrease in vv brought about by pore diffusion is then expressed by an effectiveness factor, rj, defined by... 

Sources of error can be introduced in each conversion from volume to moles and back to weight, although for simple examples such as the one above it does not really matter which method of calculation is employed as long as the correct answer for the purity of citric acid is obtained. However, for more complicated calculations, involving the use of back and blank titrations, this author believes that factors and equivalents simplify volumetric analysis and they will be used for that reason (rather than any reason of dogma) in the remainder of this book. 

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