It is often said that stomach acid is as acidic as a car battery. If this is the case, why doesn't the acid in your stomach doesn't eat through your stomach walls

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Jrstx333

11.04.2022

Chemistry

Approximately $279\; \rm mg$ of fluorine would be produced.

Explanation:

Look up the relative atomic mass of these calcium and fluorine on a modern periodic table:

$\rm Ca$ : 40.078

. $\rm F$ : 18.998

In other words, the mass of each mole of calcium atoms is (approximately) $40.078\; \rm g$ . Similarly, the mass of each mole of fluorine atoms is (approximately) $18.998\; \rm g$ .

Calculate the number of moles of calcium atom that were produced:

$\begin{aligned} &n({\rm Ca}) \\ &= \frac{m({\rm Ca})}{M({\rm Ca})} \\ &= \frac{294\; \rm mg}{40.078\; \rm g \cdot mol^{-1}} \\ &= \frac{0.294\; \rm mg}{40.08 \; \rm g \cdot mol^{-1}}\\ & \approx 7.3357 \times 10^{-3}\; \rm mol\end{aligned}$ .

Every formula unit of calcium fluoride, $\rm CaF_2$ , contains twice as many fluorine atoms as calcium atoms.

Hence, if the sample contained approximately $7.33570 \times 10^{-3}\; \rm mol$ calcium atoms, it would contain twice as many fluorine atoms:

$\begin{aligned}n({\rm F}) &= 2\, n({\rm Ca}) \\ &\approx 2 \times 7.33570\times 10^{-3}\; \rm mol \\ &\approx 1.4671 \times 10^{-2}\; \rm mol\end{aligned}$ .

Calculate the mass of that many fluorine atoms:

$\begin{aligned}& m({\rm F}) \\ &= n({\rm F}) \cdot M({\rm F}) \\ &\approx 1.4671 \times 10^{-2}\; \rm mol \times 18.998\; \rm g \cdot mol^{-1} \\ &\approx 278.72 \times 10^{-3}\; \rm g \\&\approx 279\; \rm mg\end{aligned}$ .

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