Question 616946
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First calculate the number of grams of mass in the key before it went in the acid.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2.03\ \text{oz/key}\ \times\ 28.35\ \text{gm/oz}\ =\ 57.5505\ \text{gm/key}]


Next, calculate the number of grams lost in 10 minutes:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 10\ \text{min}\ \times\ 60\ \text{sec/min}\ \times\ 0.00235\ \text{gm/sec}\ =\ 1.41\ \text{grams lost}]


Next, calculate the mass remaining after 10 minutes, i.e. what it <i>was</i> minus what it <i>lost</i>:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 57.5505\ \text{gm/key}\ -\ 1.41\ \text{gm}\ =\ 56.1405\ \text{gm/(acid eaten key)}]


Then divide the final mass by the density to calculate the volume:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{56.1405\ \text{gm}}{6.347\ \text{gm/cm^3}}\ \approx\ 8.85\ \text{cm}^3]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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