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Equating the Number of Atoms

Glaister [22] from the Department of Mathematics, University of Reading, showed a simple method to balance this equation by equating the number of atoms on each [Pg.382]


In a balanced chemical equation, the number of atoms of each element on the left-hand, or reactant, side will equal the number of atoms of eaeh element on the right-hand, or product, side. The above equation states that one mole of potassium perchlorate (KCIO 4, a reactant) will react with 4 moles of magnesium metal to produce one mole of potassinm chloride (KCl) and 4 moles of magnesinm oxide (MgO). [Pg.122]

Balancing a chemical equation requires an understanding of the Law of Conservation of Mass, which says that mass cannot be created or destroyed. The amount of mass in the reactants will be the amount of mass in the products. The credit for this discovery is given to Antoine Lavoisier, who took very careful measurements of the quantities of chemicals and equipment that he used. Conservation of mass also holds true when balancing equations. The number of atoms of each element in the reactants will be equal to the number of atoms of each element in the products. A useful mnemonic device for conservation of mass is What goes in, must come out. ... [Pg.102]

By use of these equations the number of atoms in any series of successive transformations can be calculated. For the daughter nuclide 2, cq. (4.17) is obtained. [Pg.44]

This is a balanced equation. The numbers of atoms of caibon. hydrogen and oxygen on the left-hand side are equal to the numbers of atoms of caibon, hydrogen and oxygen on the r t-band side. [Pg.27]

STEP Q Check the final equation to confirm it is balanced. In the final equation, the numbers of atoms of C, H, and O are the same in both the reactants and the products. The equation is balanced. [Pg.225]

In the matter of ring stability it is the number of atoms in the ring that is of central importance. In the following equations 1, m, and m represent the number of ring (1) and backbone (m, m )atoms in the species shown. Note that for linear species m and m are proportional to the degree of polymerization but are more suitable for present purposes than the latter. The following general reactions are pertinent to the present discussion ... [Pg.327]

The overall form of each of these equations is fairly simple, ie, energy = a constant times a displacement. In most cases the focus is on differences in energy, because these are the quantities which help discriminate reactivity among similar stmctures. The computational requirement for molecular mechanics calculations grows as where n is the number of atoms, not the number of electrons or basis functions. Immediately it can be seen that these calculations will be much faster than an equivalent quantum mechanical study. The size of the systems which can be studied can also substantially ecHpse those studied by quantum mechanics. [Pg.164]

It is extremely difficult to know values for all of these parameters precisely. Therefore, absolute quantitation is almost never attempted. The determination of relative atomic ratios is an inherently more tractable approach, however. This method is best illustrated by consideration of a binary material composed exclusively of atoms A and B that is perfectiy homogeneous up to the surface. In this case, independent equations can be developed relating the number of atoms sampled to the xps intensity for each atom as follows ... [Pg.278]

The solutions tothese equations give the number of atoms of each nucHde that are present at time t as... [Pg.446]

Formulate the constraining material-balance equations, based on conservation of the total number of atoms of each element in a system comprised of w elements. Let subscript k identify a particular atom, and define Ai as the total number of atomic masses of the /cth element in the feed. Further, let a be the number of atoms of the /cth element present in each molecule of chemical species i. The material balance for element k is then... [Pg.543]

The reactivities of the substrate and the nucleophilic reagent change vyhen fluorine atoms are introduced into their structures This perturbation becomes more impor tant when the number of atoms of this element increases A striking example is the reactivity of alkyl halides S l and mechanisms operate when few fluorine atoms are incorporated in the aliphatic chain, but perfluoroalkyl halides are usually resistant to these classical processes However, formal substitution at carbon can arise from other mecharasms For example nucleophilic attack at chlorine, bromine, or iodine (halogenophilic reaction, occurring either by a direct electron-pair transfer or by two successive one-electron transfers) gives carbanions These intermediates can then decompose to carbenes or olefins, which react further (see equations 15 and 47) Single-electron transfer (SET) from the nucleophile to the halide can produce intermediate radicals that react by an SrnI process (see equation 57) When these chain mechanisms can occur, they allow reactions that were previously unknown Perfluoroalkylation, which used to be very rare, can now be accomplished by new methods (see for example equations 48-56, 65-70, 79, 107-108, 110, 113-135, 138-141, and 145-146)... [Pg.446]

Further checks, uhich can readily be verified from the equations of balance, are (a) the number of atoms in a neutral borane molecule = 2(s + I + y + x), and (b) there are as many framework electrons as (here are atoms in a neutral borane B H , since each BH group supplies 2 electrons and each of the m — n) "extra H atoms supplies I electron, making n + m in all. [Pg.176]

Sinee there are six unknowns and three equations, there are three independent variables. We ean associate these with any three elementary independent modes of point defect formation which conserve the numbers of atoms. These are like basis vectors for representing arbitrary point defect concentrations. Let us define them as follows ... [Pg.341]

By differentiating G with respect to Na, the number of atoms of type A, we obtain the chemical potential Pa- In doing so care must be taken to include the dependence of N on the number of atoms via the equation ... [Pg.344]

Because of the way in which rate of decay is measured (Figure 19.3), it is often described by the activity (A) of the sample, which expresses the number of atoms decaying in unit time. The first equation above can be written... [Pg.518]

From his equation Planck was able to calculate the number of atoms in a gram-molecule ... [Pg.525]

For a polyatomic molecule, the complex vibrational motion of the atoms can be resolved into a set of fundamental vibrations. Each fundamental vibration, called a normal mode, describes how the atoms move relative to each other. Every normal mode has its own set of energy levels that can be represented by equation (10.11). A linear molecule has (hr) - 5) such fundamental vibrations, where r) is the number of atoms in the molecule. For a nonlinear molecule, the number of fundamental vibrations is (3-q — 6). [Pg.502]

We have seen earlier that for a linear polyatomic molecule, the vibrational motions can be divided into (3rj — 5) fundamentals, where rj is the number of atoms. For a nonlinear molecule (3rj - 6) fundamentals are present. In either case, each fundamental vibration can be treated as a harmonic oscillator with a partition function given by equations (10.100) and (10.101). Thus. [Pg.541]

Using models in learning about ehemieal equations has proved a successful tool, espeeially by students with diffieulties in eoneept understanding. By eounting the number of atoms in partiele representations they better understood the meaning of a balanced equation. Some students still have problems with balaneing chemical equations when models are not avail-... [Pg.324]

Students stressed that it is mueh easier to write down ehemieal equation and check whether it is balaneed with the use of models, beeause they ean eoneretely count the number of atoms on eaeh side of the ehemieal equation. [Teacher from Sehool N° 3, Section 1 ]... [Pg.326]

Balanced equation An equation using formulae where the number of atoms of each element involved in the reaction is the same on each side of the arrow. [Pg.11]

Identify a process to solve the problem. The question asks about the volume of one silver atom. Mass and volume are related through density p — mj V. From this equation, we can calculate the total volume of the silver atoms. The problem also gives the total number of silver atoms transferred from the wire to the spoon. The volume of a single atom is the total volume divided by the number of atoms. Oftentimes, a flow chart helps to summarize the process ... [Pg.47]

Moles and atoms of mercury are the goal of the calculation. We need to solve Equation for the number of atoms atoms = n... [Pg.102]

C03-0042. Diagram the process for converting from the mass of a compound of a known chemical formula to the number of atoms of one of its constituent elements. Include all necessary equations and conversion factors. [Pg.183]

The stoichiometric coefficients in a balanced chemical equation must be chosen so that the atoms of each element are conserved. Many chemical equations can be balanced by inspection. Balancing by inspection means changing stoichiometric coefficients until the number of atoms of each element is the same on each side of the arrow. Usually, we can tell what changes need to be made by looking closely at the reaction and matching the numbers of atoms of each element on both sides of the equation. Consider the following example. [Pg.202]

The equation is not balanced, because there are too many carbon and hydrogen atoms on the left and too many oxygen atoms on the right. We need to change the numbers of molecules by changing stoichiometric coefficients until the numbers of atoms of each element are equal. [Pg.203]

Our task is to estimate the volume occupied by one atom of lithium. As usual, the mole is a convenient place to begin the calculations. Visualize a piece of lithium containing one mole of atoms. The molar mass, taken from the periodic table, tells us the number of grams of Li in one mole. The density equation can be used to convert from mass to volume. Once we have the volume of one mole of lithium, we divide by the number of atoms per mole to find the volume of a single atom. [Pg.435]


See other pages where Equating the Number of Atoms is mentioned: [Pg.85]    [Pg.153]    [Pg.382]    [Pg.599]    [Pg.85]    [Pg.153]    [Pg.382]    [Pg.599]    [Pg.120]    [Pg.396]    [Pg.203]    [Pg.19]    [Pg.95]    [Pg.215]    [Pg.247]    [Pg.302]    [Pg.312]    [Pg.375]    [Pg.956]    [Pg.19]    [Pg.162]    [Pg.167]    [Pg.536]    [Pg.153]    [Pg.16]   


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