1. Review of Fundamentals

Michael Mombourquette

1 Calculate the following mathematical expressions and give the answer to the correct number of sig figs. (Answers to practice problems are below)

    \[\begin{array}{|c|c|c|c|} \hline a.\; 1.2343 \times 43.21 & b.\; 23.554 \times 21.5 & c.\; 1250 + 2.54 & d.\; 1.546 \times 33.1 \\ \hline e.\; 23.9 \times 452.1& f.\; \frac{23.665 \times 20.34}{44.13} & g.\; \frac{128.325 \times 0.34}{289.4} & h.\; \frac{35.2 + 75.3}{1.4532} \\ \hline i.\; \frac{103.2 \times 103.1}{0.0023} & j.\; log(134)& k.\; log(0.0023) \times 2.43 & l.\; antilog(3.554) \\ \hline m.\; log(14.35^3)&&&\\ \hline \end{array}\]

2 An American vehicle’s speedometer is calibrated in units of miles per hour (mph). The vehicle is travelling in Canada, where the speed limits are posted in metric units (km/h).  What speed must the driver of the American vehicle maintain in order to remain in compliance with a posted speed limit of 65 km/h?
(Use the equalities: 1 in = 2.54 cm, 12 in = 1 ft, 1 mile = 5280 ft)

3 An industrial plant produces 1.4 kg/s of product.  How much product will the plant produce in one week (7 days) if the plant operates 24hours–a–day with no stoppages or slowdowns?

4 Two solutions are sampled and their pH is found to be 4.3 and 5.8, respectively. Find an expression to determine the ratio of the [H^+] concentrations for the two solutions and then calculate the ratio.

Answers to practice problems

1

  1. 1.2343 \times 43.21 = 53.33
  2. 23.554 \times \;21.5 = 506 or 5.06\times 10^2
  3. 1250 + 2.54 = 1253
  4. 1.546 \times 33.1 = 51.2
  5. 23.9 \times 452.1 = 10800 more correctly 1.08\times 10^4
  6. \frac{23.665 \times 20.34}{44.13} = 10.91 Go with the smallest sig figs.
  7. \frac{128.325 \times 0.34}{289.4} = .15 or 1.5\times 10^{-1}
  8. \frac{35.2 + 75.3}{1.4532} = \frac{110.5}{1.4532} = 76.04 or 7.603\times 10^1
  9. \frac{103.2 \times 103.1}{0.0023} = 4600000 more correctly 4.6\times 10^6
  10. log(134) = 2.127 the 2 is the power of 10, not a sig fig.
  11. log(0.0023) \times 2.43 = -2.63_{83} \times 2.43 = -6.41 See how you have to work your sig figs step by step.
  12. antilog(3.554) = 3580 more correctly 3.58\times 10^2
  13. log(14.35^3) =\; 3\;\times log(14.35) = 3\times 1.1569 = 3.4706 Note that the 1 in the 1.1549 is the power of 10. You’ve ‘gained’ a sig fig.

2
\frac{65\;km}{1\;h}\times \frac{1000\;m}{1\;km}\times \frac{100\;cm}{1\;m}\times \frac{1\;in}{2.54\;cm}\times \frac{1\;ft}{12\;in}\times \frac{1\;mile}{5280\;ft}\times = 40\; miles/h or 40\; mph.

3 Using the extended fraction formalism:
\left{|}\frac{1.4\;kg}{1 s}\right{|}\frac{3600\;s}{1\;h}\left{|}\frac{24\;h}{1\;day}\right{|}\frac{7\;days}{} = 8.5\times10^5 kg

4 We know that we can use the concentration of the hydronium ion to determine pH=-log([H^+]).  Or, in exponential form, 10^{-pH}=[H^+].  So to get an expression for the ratio of the concentrations of the two solutions, we take divide two equations (one for each solution).

\frac{[H^+]_1}{[H^+]_2} = \frac{10^{-pH_1}}{10^{-pH_2}}

So, we can calculate the concentration ratio:

\frac{[H^+]_1}{[H^+]_2} = \frac{10^{-4.3}}{10^{-5.8}} = 32

So although the pH values are only separated by a difference of 1.5, solution 1 is 32 times more concentrated than solution 2.

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To the extent possible under law, Michael Mombourquette has waived all copyright and related or neighboring rights to 1. Review of Fundamentals, except where otherwise noted.

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