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Source: Basic Chemistry, Steven S. Zumdahl, Donald J. DeCoste

Referring to the above, rule it seems $(105-32)\times\frac{5}{9}=40.5555\ldots$ has infinitely many significant figures. Am I correct?

M.A.R.
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buzzee
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  • You should base the number of significant digits in your answer on the number of significant digits in your starting numbers. – Shafter Apr 02 '16 at 19:11
  • @Shafter How should I base the number of significant digits? – buzzee Apr 02 '16 at 19:35
  • Your example is pure math, where everything is exact. For a real world chemistry problem, you have to give a context, and the error of each starting number. And then do the proper math to calculate the error propagation. – Karl Apr 03 '16 at 21:14

2 Answers2

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As GumpyCede notes in his answer, if you are calculating the result from the pure numbers $105$, $32$, $5$ and $9$, then yes, the result has an infinite number of significant figures.

If one or more of those values is derived from some sort of measurement(s), though, then the significant figures of those measured value(s) would enter into consideration and the repeating decimal would need to be truncated appropriately.

hBy2Py
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  • 1457 has four nonzero integers, and captive zeros always count as significant figures. 105 is not defined as 105 in = 266.7 cm, so 105 has three significant numbers, 1, 0, 5. – buzzee Apr 04 '16 at 13:42
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All numbers (105, 32, $\frac{5}{9}$, and 40.555...) have infinitely many significant figures because there is no uncertain digit.

michaelchang64
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    As for, 105 and 32, when considering captive zero and nonzero integers they have three and two significant figures; the 1 and the 0 and the 5 in the fist, the 3 and the 2 in the second. – buzzee Apr 02 '16 at 19:47