Algebra answers

Even if H and S are functionally independent, one still might argue that the commutator is likely to be small, and, thus, the idea could be a useful approximation. The difficulty here is with the subtleties of the concept of smallness in this context. We will not attempt to address this question quantitatively, but satisfy ourselves by examining the commutators of H and S for three systems. The first of these is a simple 2x2 system for which we may obtain an algebraic answer. The other two are matrices from real VB calculations of CH4 and the 7r-system of naphthalene. 

Using the. results (roiri the previous two problems, evaluate the integrals in the answer to Problem 5 and find t as a dosed algebraic expression for the Gaussian trial runetion. 

Calculating the solubility of Pb(I03)2 in distilled water is a straightforward problem since the dissolution of the solid is the only source of Pb + or lOa. How is the solubility of Pb(I03)2 affected if we add Pb(I03)2 to a solution of 0.10 M Pb(N03)2 Before we set up and solve the problem algebraically, think about the chemistry occurring in this system, and decide whether the solubility of Pb(I03)2 will increase, decrease, or remain the same. This is a good habit to develop. Knowing what answers are reasonable will help you spot errors in your calculations and give you more confidence that your solution to a problem is correct. 

What do molecular orbitals and their nodes have to do with pericyclic reactions The answer is, everything. According to a series of rules formulated in the mid-1960s by JR. B. Woodward and Roald Hoffmann, a pericyclic reaction can take place only if the symmetries of the reactant MOs are the same as the symmetries of the product MOs. In other words, the lobes of reactant MOs must be of the correct algebraic sign for bonding to occur in the transition state leading to product. 

Many reactions with complicated rate laws proceed by bimolecular steps. The complexity often arises from attendant equilibria. Several instances have been cited where no clear-cut choice could be made between algebraically compatible alternatives. Thus, do Cr2+, Fe3+, and Cl- react via CrCl+ and Fe3+ orCr2+ and FeCl2+ Does the first term in Eq. (6-33) correspond to CrOH+ and Fe3+ or Cr2+ and FeOH2+ Does the iodide-peroxide reaction necessarily imply that H302+ reacts with I- could not H202 and HI be responsible The answers to these questions will not be found strictly from the kinetics. Other experiments must be devised. Some have been mentioned previously, and two more will be cited here. 

This is really an algebraic exercise in partial fractions. The answer hides in Table 2.1. 

In each section, you will find a few pre-algebra problems mixed in—problems that ask you to deal with variables (letters that stand for unknown numbers, such as x or y), exponents (those little numbers hanging above the other numbers, like 24), and the like. These problems are a warm-up for Section 5, Algebra. If they are too hard for you at first, just skip them. If you can answer them, you will be ahead of the game when you get to Section 5. 

You may need to answer questions directly about the GCF or the LCM, or you will need to find these numbers in your work with fractions and algebra. 

One may therefore wish to know what are the potential functions V(r) that correspond to a given algebraic model. The general answer to this question is provided by the solution of the inverse Schrodinger problem Since one knows the spectrum of the algebraic model, one finds the potential that reproduces the spectrum.1 A simple approach consists in expanding the potential V(r) into a set of functions with unknown coefficients, say... 

To answer this question one has to look at the way the rate equation is derived. A rate equation based on a certain reaction mechanism may have been derived after the introduction of some approximations valid at atmospheric pressure. If, at higher pressure these approximations are no longer valid, a continuous use of the rate equation may lead to erroneous results. As approximations usually are introduced to reduce the number of parameters, it should be evident that equations with differing numbers of parameters most probably have different algebraic forms. The omission of a critical, initially small, but increasingly more important rate constant with increasing pressure will unavoidably lead to suspect interpretations. 

The math word problems in this book span some basic problems (using arithmetic) to the more complex (requiring algebraic skills). Even though I like to make example problems come out with whole-number answers, sometimes... 

Math problems sometimes require solving algebraic equations. The algebra is wonderful, but you run the risk of creating some extraneous roots. Extraneous roots or solutions are answers that satisfy the algebraic equation but don t really mean a thing in the situation. 

A math word problem is different from other arithmetic and algebra problems, because you first have to translate from the words to the symbols before you do the operations or solve the equation for the answer. 

Accuracy and common sense go hand in hand. You can t have one without the other. The accuracy part goes to the arithmetic or algebra involved in solving the problem and whether the relationships hold. The common sense refers to whether the answer — even though it s the solution of an equation — really fits the problem and the real world. 

MM7ord problems in algebra are problems that seem to take on a life of their own. You start out with a simple sentence in English and end up with an equation that, you hope, will answer the question that s been posed. 

Changing from words to equations involves identifying what the variables (the x s or y s or f s) represent and how to arrange them in an equation. Solving an equation requires algebraic know-how, but, if your equation is nonsense or doesn t fit the problem, then the answer to the equation will get you no closer to the answer to the problem than you were before you started. 

When algebraic equations or inequalities are used to find the answer to a question or problem, you write the equation or inequality carefully, you solve the algebraic equation or inequality using all the correct rules, and then you check to be sure that the solution from the algebra really answers the question that s been posed. 

You get that feeling of satisfaction when an algebraic equation or inequality works out and you get one or more solutions. The next step in word problems is then, of course, to see if the solution of the equation or inequality is an answer to the problem. If the solution doesn t work, then you go back to see if you ve done some miscomputation. But sometimes, no amount of good mathematics is going to get you an answer. It could be that the question just doesn t have an answer. 

If the algebraic manipulations required by equation 6.21 are becoming too complicated, there is a spreadsheet method that gives the answer directly from the uncertainty components and the function for y. It relies on the fact that uncertainties are usually only a small fraction of the quantities (a few percent at most), and so the simplifying assumption may be made that, for... 

The final combined standard uncertainty, whether obtained using algebra or a spreadsheet or other software, is the answer, and can be quoted as such. I recommend using the wording Result x units [with a] standard uncertainty [of] uc units [where standard uncertainty is as defined in the International Vocabulary of Basic and General Terms in Metrology, 3rd edition, 2007, ISO, Geneva, and corresponds to one standard deviation]. ... 

There are many questions we wish to answer. Why and when does the steady evolution of the system become unstable Why do other values of the rate constants or of the initial conditions yield oscillations either immediately (without the induction period) or, in some cases, not at all How quickly will the oscillations grow How long will they last How many will there be Surprisingly, perhaps, the answers to these and other features emerge in simple analytic results and in many cases require only some simple algebra. 

Now we ask the parallel question—what is the new choice of basis functions for the function space (the one which produced rred) which will produce matrices in their fully reduced form Once again we are looking at the opposite side of the coin whose two faces are a similarity transformation and a change of basis functions. To answer the question we have posed, we will invoke the Great Orthogonality Theorem and carry out a certain amount of straightforward algebra. 

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