Approximations significant figures

In each example, the initial values of the factors are expressed in U.S. customary units, and the dimensionless value is calculated. Then the factors are converted to SI units, and the dimensionless value is recalculated. The two dimensionless values will be approximately the same. (Small variations occur due to the number of significant figures carried in the solution.)... 

In applying the rules governing the use of significant figures, you should keep in mind that certain numbers involved in calculations are exact rather than approximate. To illustrate this situation, consider the equation relating Fahrenheit and Celsius temperatures ... 

This conversion factor is exact the inch is defined to be exactly 2.54 cm. The other factors listed in this column are approximate, quoted to four significant figures. Additional digits are available if needed for vary accurate calculations. For example, the pound is defined to be 453.59237 g. 

A two-liter plastic soft drink bottle can withstand a pressure of 5 atm. Half a cup (approximately 120 mL) of ethyl alcohol, C2H5OH (d = 0.789 g/mL), is poured into a soft drink bottle at room temperature. The bottle is then heated to 100°C (3 significant figures), changing the liquid alcohol to a gas. Will the soft drink bottle withstand the pressure, or will it explode ... 

In practice, in numerical calculations with a computer, both rational and imtiooal numbers are represented by a finite number of digits. In both cases, then, approximations are made and die errors introduced in the result depend on the number of significant figures carried by the computer - the machine precision. In die case of irrational numbers such errors cannot be avoided. 

Rather than solve this mathematically, one can solve this by successive approximation. Simply pick a value for x, plug into the above expression and see if you get the value of the Ksp. If your value is different (which it probably is), pick a higher or lower number. Keep going until you find x, in this case, to one significant figure. Start with x = 6 x 10-8. 

By definition, the mole fraction of NaCl in this solution is = 0 2 35 2 = 0.0036. (Note the number of significant figures [see Section 13.4] reported here taking the density of the solution as 1, and calculating the weight of water therefore as 1000 g would not have affected this calculation. The density correction could, therefore, have been omitted. Also, as a rule of thumb, 1 liter of water can be approximated as 55 mol F O.)... 

Similarly, to standardize a solution means to determine its concentration to three or more significant figures. For example, if a solution of NaOH is made up to be approximately 0.1 M, a standardization experiment may be performed and the concentration determined to be 0.1012 M. We will discuss the details of such an experiment in Section 4.6. 

The half-life and the abundance (in bold face from [7]) are shown followed by their units "%" symbol in the case of abundance) which are followed by the uncertainty, in italics,in the last significant figures. For example, 8.1 s 10 means 8.1 1,0 s. For some very short-lived nuclei, level widths rather than half-lives are given, There also, the width is followed by unite e.g., eV, keV, or MeV) which are followed by the uncertainty in italics, if known, As stated above when a limit or an approximate value is given it is based on systematics (sy), mostly from [5]. A in this field indicates that T is not known, For 2p and 2e decay only the lowest value of their several limits e.g., for Ov or 2v, etc.) is given,... 

By the approximate rule, the answer should be 1.1 (two significant figures). However, a difference of 1 in the last place of 9.3 (9.3 0.1) results in an error of about 1 percent, while a difference of 1 in the last place of 1.1 (1.1 0.1) yields an error of roughly 10 percent. Thus the answer 1.1 is of much lower percentage accuracy than 9.3. Hence in this case the answer should be 1.06, since a difference of 1 in the last place of the least exact factor used in the calculation (9.3) yields a percentage of error about the same (about 1 percent) as a difference of 1 in thelastplaceof 1.06 (1.06 0.01). Similarly, 0.92 x 1.13 = 1.04. 

The specific heat from Example 2 is 0.11 cal/g °C (to two significant figures). The approximate atomic weight is calculated to be... 

This procedure had converged in 4 or 5 iterations to four significant figures for all cases tried in this study. The accuracy of the calculations depends on the time increment At because the finite difference approximations become more accurate as At gets smaller. A summary of some iteration results and a comparison between this technique and the numerical integration with Gear s method will be presented after the following discussion on the stability of the temperature equation. 

A series of small stresses were applied to a 0.03 M sample with an equilibrium 4-value of 183 A, reducing the 4-value to 143 A over a period of about four hours. A larger stress was then applied, and the gel was allowed to come to equilibrium with the applied stress for 15 hours. The 4-value is plotted as a function of time in Figure 3.2. After 2 hours the 4-value was 118 A, and after 15 hours when the curve has obviously flattened out, the 4-value was 115 A. In the available neutron beam time, it was not possible to allow the gel to come fully to equilibrium with each applied stress. An equilibration time of two hours was chosen. After this time the voltage was always steady to two significant figures, and we may surmise that the error of approximately 3 A introduced into the 4-spacing is an absolute one, which does not affect the shape of the measured force-distance curve. 

The evaluated limit of detection should not be given with more than two significant figures as it has only an approximate character, because the normal distribution of results may be doubtful in the range of very small concentrations. Even a very large number of measurements cannot guarantee that the result bears no serious errors. 

In long-term clinical trials, latanoprost has been shown to be at least as effective as timolol maleate, a P-blocker, in reducing lOP. The ocular hypotensive effect of latanoprost is approximately 27% to 30%, whereas timolol reduces lOP approximately 20% (Figure 10-1). These results are of clinical significance because they reflect... 

Table 5.1 illustrates the frozen-core approximation for the case of sodium using a simple Slater (4.38) basis in the analytic-orbital representation. The core (Is 2s 2p ) is first calculated by Hartree—Fock for the state characterised by the 3s one-electron orbital, which we call the 3s state. The frozen-core calculation for the 3p state uses the same core orbitals and solves the 3p one-electron problem in the nonlocal potential (5.27) of the core. Comparison with the core and 3p orbitals from a 3p Hartree—Fock calculation illustrates the approximation. The overwhelming component of the 3p orbital agrees to almost five significant figures. 

Finally, within the Mott-Schottky approximation (Eq. 11), large values of or Ad can lead to the ratio Fh/ Fjc becoming significant. Figure 11 contains estimates of this ratio for several values of Aq for a semiconductor with a large s value (Ti02, s = 173) mapped as a function of the total potential drop across the interface [50]. Clearly, Fh can become a sizeable fraction of the total potential drop (approaching the situation for metals) under certain conditions. It has been shown [51] that the Mott-Schottky plots will still be linear but the intercept on the potential axis is shifted from the Fn, value. 

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