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searching for experimental values of the consumption rate of $O_2$ by tumor cells I found an article that measures the rate in the units of measure

$$\frac{mol}{cell\cdot s}.$$

The actual measurement was

$$5.5\cdot 10^{-15}\frac{mol}{cell\cdot s},$$

I precise to convert the expression in terms of $kg,\;cm,\;s$ units, but I'm not sure of how to do that. When I tried the result where far away from the order of values that I was expecting. So, I appreciate any help.

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  • $\begingroup$ What is $cell$? Like what are the units / what is the size of it? $\endgroup$
    – SuperCiocia
    Commented Jul 5, 2020 at 3:45
  • $\begingroup$ @SuperCiocia I'm guessing it's a number? As in "per cell" or "per person"? $\endgroup$
    – Philip
    Commented Jul 5, 2020 at 4:01
  • $\begingroup$ Oxygen has a molar mass of 32 g, and there are 1000 g in a kg. Your answer should end up being g/cell-s or kg/cell-s, but chemists would normally go with g/cell-s. By the way, why would you want to convert to something besides mole/cell-s? $\endgroup$ Commented Jul 5, 2020 at 4:02
  • $\begingroup$ @Philip yeah but you need to give that in units of cm. $\endgroup$
    – SuperCiocia
    Commented Jul 5, 2020 at 4:02
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    $\begingroup$ I read the book @Philip (the part with the matter of interest), it was very good. Thanks! $\endgroup$
    – Fernando
    Commented Jul 6, 2020 at 19:25

1 Answer 1

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You have to write the old units in terms of the new units.

  • 1 mole of $0_2$ is an Avogradro's number ($N_A = 6.02214076 \times 10^{23}$) of particles. Each particle here is an $O_2$ molecule, so the mass of a single particle is $m_{O_2} = 2\times m_0$ (technically minus the binding energy, but it's small so let's ignore it). $m_0$ is the number of nucleons in an oxygen atom (neglecting the electrons since their mass contribution is so small) times by the mass of a nucleon. So $m_0 = 16\times u$ (for $^{16}O$) where $u$ is the atomic mass unit $1.66 \times 10^{-27}$ kg.

    Putting all together: $$ 1\,\mathrm{mol} = N_A\cdot m_{0_2} = N_A\cdot 2\cdot m_0 = N_A \cdot 2 \cdot 16 \cdot u \approx 0.032 \,\mathrm{kg}. $$

  • Whatever the dimension of a cell is, let's say it's $1 \, \mathrm{cell} = \alpha\, \mathrm{cm}$, where $\alpha$ is the dimension of the cell in centrimetres.

So, all together: $$ 5.5\cdot 10^{-15}\frac{\mathrm{mol}}{\mathrm{cell}\cdot \mathrm{s}} \Rightarrow 5.5\cdot 10^{-15}\frac{0.032 \,\mathrm{kg}}{\alpha\, \mathrm{cm}\cdot \mathrm{s}} = \frac{1.76\cdot 10^{-16}}{\alpha}\frac{\mathrm{kg}}{\mathrm{cm}\cdot \mathrm{s}}.$$

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  • $\begingroup$ @Fernando I am so sorry, $87$ is something i had in my head from something else. For oxygen, it should be $16$ (twice the $Z$ number) unless you are dealing with isotopes. Edited the answer. $\endgroup$
    – SuperCiocia
    Commented Jul 6, 2020 at 19:49
  • $\begingroup$ Thanks, I was looking a lot and always appear to be 16, then I decided to ask, jajajaja $\endgroup$
    – Fernando
    Commented Jul 6, 2020 at 19:50

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