Moles finding

The saturated solution of silver iodate in water at 25°C has a molality equal to 0.00179, and the activity coefficient y in this saturated solution may be taken to be 0.989. The heat of solution tends at extreme dilution to the value +13,200 c d/mole Find in electron-volts per ion pair the value of L, and also the value of dL/dT in water at 25°C. 

When an electrostatic field of field strength E is applied, each type j of ion carrying charge ZjF (per mole) finds itself under the effect of an electric driving force... 

Obtain a general formula for the most probable three-dimensional translational quantum number j = jmax for a gas (assume a Boltzmann distribution). Evaluate this expression for NO2 at 1000 K (assume a cubic container 0.1 m on each side). Determine the translational energy that this corresonds to (J/mole). Find the fraction of molecules having a translational energy level greater than jmax. Hint Solution to this problem will involve the error function, erf(x). 

Recalling that the mole ratio figures are obtained by dividing each number of moles by the smallest number of moles, find those numbers and place them in the table. ... 

Fig. 1. The dependence of the stable stationary values of the adsorption and conformational variables on the control parameter, Xe. a-total adsorption per the mole of the nucleotides, b-the probability of finding of an arbitrary NA unit in the A form, c-the probability of finding of an arbitrary NA unit in the B-form. Param-(ders values used to obtain numerical results Vmi = 3,nL = 15.4, = 3.24,6° =...

Procedure. Calculate the heats of solution of the two species, KF and KF HOAc, at each of the four given molalities from a knowledge of the heat capacity. Calculate the enthalpy of solution per mole of solute at each concentration. Find... 

Strike couldn t find any decent nitroethane synths except for a couple of Chemical Abstract articles. One suggestion is to treat 1.5 moles of Na2C02 with 1 mole of sodium ethylsulfite and 0.0645 moles of K2CO3 at 125-130°C. Another route would be to use silver nitrate and ethyl iodide [8 p119]. This type of reaction has been used to nitrate other paraffins and would probably work. 

At the equivalence point, all the Cd initially present is now present as CdY -. The concentration of Cd, therefore, is determined by the dissociation of the CdY -complex. To find pCd we must first calculate the concentration of the complex, initial moles Cd McdVcd... 

Where Is the Equivalence Point In discussing acid-base titrations and com-plexometric titrations, we noted that the equivalence point is almost identical with the inflection point located in the sharply rising part of the titration curve. If you look back at Figures 9.8 and 9.28, you will see that for acid-base and com-plexometric titrations the inflection point is also in the middle of the titration curve s sharp rise (we call this a symmetrical equivalence point). This makes it relatively easy to find the equivalence point when you sketch these titration curves. When the stoichiometry of a redox titration is symmetrical (one mole analyte per mole of titrant), then the equivalence point also is symmetrical. If the stoichiometry is not symmetrical, then the equivalence point will lie closer to the top or bottom of the titration curve s sharp rise. In this case the equivalence point is said to be asymmetrical. Example 9.12 shows how to calculate the equivalence point potential in this situation. 

This experiment examines the effect of reaction time, temperature, and mole ratio of reactants on the synthetic yield of acetylferrocene by a Eriedel-Crafts acylation of ferrocene. A central composite experimental design is used to find the optimum conditions, but the experiment could be modified to use a factorial design. 

The derivatives are hydroxyethyl and hydroxypropyl cellulose. AH four derivatives find numerous appHcations and there are other reactants that can be added to ceUulose, including the mixed addition of reactants lea ding to adducts of commercial significance. In the commercial production of mixed ethers there are economic factors to consider that include the efficiency of adduct additions (ca 40%), waste product disposal, and the method of product recovery and drying on a commercial scale. The products produced by equation 2 require heat and produce NaCl, a corrosive by-product, with each mole of adduct added. These products are produced by a paste process and require corrosion-resistant production units. The oxirane additions (eq. 3) are exothermic, and with the explosive nature of the oxiranes, require a dispersion diluent in their synthesis (see Cellulose ethers). 

Equation (13-14) is solved iteratively for V/F, followed by the calculation of values ofx, andy from Eqs. (13-12) and (13-13) andL from the total mole balance. Any one of a number of numerical root-finding... 

The findings prove that peroxyacid oxidizes the reaction product from 2 to 3 moles of PMSA ai e spent per 1 mole of RP. If the components ai e mixed unintermptedly, the output of RP does not depend on the order of mixing. Value of (1.2-1.3)xl0 mole -cm -1 is close to values, given for other oxidants. At the sui plus of H Dm compai ed to nickel (6 1)3 moles of PMSA ai e spent per 1 mole of RP, while at the sui plus of PMSA (10 1) 3 moles of H Dm ai e spent per 1 mole of RP. The data obtained do not let us affirm what exactly oxidize in the RP - Ni(II) or H Dm. It is proved, that under conditions of formation, RP is not oxidized by PMSA. 

To find the moles of diethyl ether needed to reach tlie 1.9% LFL. the gmol of air in the refrigerator is multiplied by 0.019 ... 

Find the standard heat of formation of benzene (() given the following heats of combustion data (in kcal/g-mole) at 1 atm and 25°C ... 

Consider 1 mole of a completely dissociated uni-univalent solute in aqueous solution at extreme dilution at 25°C, each positive and each negative ion having a diameter equal to 3.0 angstroms. Find from (19), in calories per mole per degree, what would be the total amount of entropy lost by the solvent in the fields of all these ions, if (19) could correctly be used for a sphere as small as 3 angstroms. 

If a small additional number of particles of any species is added to the solution to which (47) applies, the mole fraction of every species will thereby be changed. Taking this into account, we find from (47) that if, for example, dnB particles of species B are added, the increment in In Wef per particle is given by... 

Since the saturated solutions of AgT and AgCl are both very dilute, it is of interest to examine their partial molal entropies, to see whether we can make a comparison between the values of the unitary terms. As mentioned above, the heat of precipitation of silver iodide was found by calorimetric measurement to be 1.16 electron-volts per ion pair, or 26,710 cal/mole. Dividing this by the temperature, we find for the entropy of solution of the crystal in the saturated solution the value... 

The heat of solution of silver bromide in water at 25°C is 20,150 cal/mole. Taking the value of the entropy and the solubility of the crystalline solid from Tables 44 and 33, find by the method of Secs. 48 and 49 the difference between the unitary part of the partial inolal entropy of the bromide ion Br and that of the iodide ion I-. 

Turning next to AgBr, we see from Table 33 that the value of L increases from 0.931 electron-volt at 15° to 0.935 at 35°, a difference of 0.004. Dividing by 20, we find that the average value of dL/dT in the neighborhood of 25° is 2 X 10 1 electron-volt/deg. Multiplying by 23,060, we find this is equivalent to 4.6 cal/mole. It follows that the value of the conventional entropy of solution AS0 in the neighborhood of 25°C is approximately... 

The value for the equilibrium constant at 25°C given in the last column of Table 10 is 4.52 X 10 7. Multiplying logoff by RT, we find for AF° the value 8683 cal/mole. The heat of dissociation has been measured calorimetrically1 and found to be 1843 cal/mole. By subtraction we find... 

Values for the equilibrium constant at temperatures between 0 and 60°C were included in Table 9. From the value at 25° we find —log,K = 23.8. Multiplying by RT, we obtain AF° = 13,983 cal/mole. From the temperature coefficient of AF° we obtain for the heat of dissociation the value AH° = 3500 cal. Hence... 

In each case, from the value of r we can at once calculate the value of J for the proton transfer. At 25°C a liter of ethanol weighs 786 grams and contains 17.0 moles. We find then for the proton transfer (43)... 

The macroscopic dielectric constant of liquid formic acid at 25° has the value 64, not much lower than that of water. Hence, from the simple electrostatic point of view, we should expect. /c for the proton transfer (211) carried out in formic acid solution, to have a value somewhat greater, but not much greater, than when the same proton transfer is carried out in water as solvent. In Table 12 we found that, in aqueous solution, the value of (./ + Jenv) rises from 0.3197 at 20°C to 0.3425 at 40°C. Measurements in formic acid at 25°C yielded for the equilibrium of (211) the value — kT log K = 4.70. Since for formic acid the number of moles in the b.q.s. is M = we find... 

Notice that the formula of a substance must be known to find its molar mass. It would be ambiguous, to say the least, to refer to the molar mass of hydrogen. One mole of hydrogen atoms, represented by the symbol H, weighs 1.008 g the molar mass of H is 1.008 g/mol. One mole of hydrogen molecules, represented by the formula H weighs 2.016 g the molar mass of H2 is 2.016 g/mol. 

Strategy Find the molar mass of CgH T and use it to convert moles to grams in (a), grams to moles in (b), and (with Avogadro s number) grams to molecules in (c). 

Knowing the formula of a compound, Fe203> you can readily calculate the mass percents of its constituent elements. It is convenient to start with one mole of compound (Example 3.4a). The formula of a compound can also be used in a straightforward way to find the mass of an element in a known mass of the compound (Example 3.4b). 

---