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Partial equation

The term 4H20 is omitted, since the reaction is carried out in dilute solution, and the water concentration may be assumed constant. The hydrogen ion concentration is taken as molar. The complete reaction may be divided into two half-cell reactions corresponding to the partial equations ... [Pg.69]

The partial equation (36) may be multiplied by 5 in order to balance (35) electrically ... [Pg.69]

These partial equations must then be multiplied by coefficients which result in the number of electrons utilised in one reaction being equal to those liberated in the other. Thus equation (4) must be multiplied by two, and we have ... [Pg.849]

Now the gain and loss of electrons must be equal. One permanganate ion utilises 5 electrons, and one iron(II) ion liberates 1 electron hence the two partial equations must apply in the ratio of 1 5. [Pg.850]

The various stages in the deduction of the second partial equation are... [Pg.850]

One dichromate ion uses 6e, and two iodide ions liberate 2e hence the two partial equations apply in the ratio of 1 3 ... [Pg.850]

We can use the ideal gas equation to calculate the molar mass. Then we can use the molar mass to identify the correct molecular formula among a group of possible candidates, knowing that the products must contain the same elements as the reactants. The problem involves a chemical reaction, so we must make a connection between the gas measurements and the chemistry that takes place. Because the reactants and one product are known, we can write a partial equation that describes the chemical reaction CaC2(. ) +H2 0(/) Gas -I- OH" ((2 q) In any chemical reaction, atoms must be conserved, so the gas molecules can contain only H, O, C, and/or Ca atoms. To determine the chemical formula of the gas, we must find the combination of these elements that gives the observed molar mass. [Pg.302]

Reactions involving the creation, destruction, and elimination of defects can appear mysterious. In such cases it is useful to break the reaction down into hypothetical steps that can be represented by partial equations, rather akin to the half-reactions used to simplify redox reactions in chemistry. The complete defect formation equation is found by adding the partial equations together. The mles described above can be interpreted more flexibly in these partial equations but must be rigorously obeyed in the final equation. Finally, it is necessary to mention that a defect formation equation can often be written in terms of just structural (i.e., ionic) defects such as interstitials and vacancies or in terms of just electronic defects, electrons, and holes. Which of these alternatives is preferred will depend upon the physical properties of the solid. An insulator such as MgO is likely to utilize structural defects to compensate for the changes taking place, whereas a semiconducting transition-metal oxide with several easily accessible valence states is likely to prefer electronic compensation. [Pg.32]

Write partial equations for the oxidation and the reduction. Then (1) Balance charges by adding in acid solutions or OH in basic solutions. (2) Balance the number of O s by adding H O s to one side. (3) Balance the number of H s by adding H s to one side. The number added is the number of equivalents of oxidant or reductant. [Pg.278]

The change in oxidation state (number) method uses partial equations. One partial equation is used for the oxidation, and another partial equation is used for the reduction. [Pg.144]

First, the equation must be divided into two skeleton partial equations, showing the atoms that change their oxidation states. [Pg.145]

Incorporation of the superposition approximation leads inevitably to a closed set of several non-linear integro-differential equations. Their nonlinearity excludes the use of analytical methods, except for several cases of asymptotical automodel-like solutions at long reaction time. The kinetic equations derived are solved mainly by means of computers and this imposes limits on the approximations used. For instance, we could derive the kinetic equations for the A + B — C reaction employing the higher-order superposition approximation with mo = 3,4,... rather than mo = 2 for the Kirkwood one. (How to realize this for the simple reaction A + B —> B will be shown in Chapter 6.) However, even computer calculations involve great practical difficulties due to numerous coordinate variables entering these non-linear partial equations. [Pg.126]

In this section, you will begin with the reaction questions. Recall, you are given three partial equations. In each question, you are provided the reactants and any conditions of the reaction. You are then to write a balanced equation for the reaction and answer a question about reaction. The directions for this section tell you to assume that all solutions are aqueous unless otherwise indicated. You are also instructed to represent substances in solution as ions if the material is extensively ionized, and you should omit the formulas for any ions or molecules that are unchanged by the reaction (spectators, for instance). It is necessary to balance the equations in this section. Here is a sample problem that illustrates this portion of the test ... [Pg.16]

Let us take as a very simple case a reaction with a diagram as shown in Fig. 11 and the two sets of partial equations... [Pg.343]

The hydrt enation of carbon monoxide can be expressed by a number of partial equations. These reflect the complex parallel and ctmsecutive reactions whose relative rates depend on catalysts and reaction conditions. [Pg.50]

Examples 4.2 and 4.3 and the models from Section 4.4 show that the stochastic models can frequently be described mathematically by an assembly of differential partial equations. [Pg.247]

Useful in the application of cubic equations of state to mixtures are partial equation-of-state parameters. For the parameters of the generic cubic, represented by Eqs. (4-104), (4-105), and (4-106), the definitions are... [Pg.676]

Governing Conservation Partial Equations Used in Computational Fluid Dynamic Approach... [Pg.248]

Construct partial equation for reducing agent in the same way. [Pg.150]

Multiply each partial equation by a factor so that when the two are added the electrons just gets compensated. [Pg.150]

Add the partial equations and cancel out the substances which appear on both sides of the equation. [Pg.150]

Multiply each partial equation by a number chosen so that the total number of electrons lost by the reducing agent equals the number of electrons gained by the oxidizing agent. [Pg.178]

Add the two partial equations resulting from the multiplications. In the sum equation, cancel any terms common to both sides. All electrons should cancel. [Pg.178]


See other pages where Partial equation is mentioned: [Pg.848]    [Pg.849]    [Pg.849]    [Pg.849]    [Pg.849]    [Pg.849]    [Pg.849]    [Pg.131]    [Pg.98]    [Pg.105]    [Pg.204]    [Pg.329]    [Pg.98]    [Pg.149]    [Pg.150]    [Pg.150]    [Pg.150]    [Pg.178]    [Pg.178]    [Pg.179]    [Pg.179]   
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