how long does it take to deposit a coating of gold 1.00 μm thick on a disk - shaped medallion 4.00 cm in…

Explanation:

Step1: Calculate the surface - area of the disk

The formula for the surface - area of a disk is $A=\pi r^{2}$, where $r = \frac{d}{2}$. Given $d = 4.00\ cm$, so $r=2.00\ cm$. Then $A=\pi\times(2.00\ cm)^{2}=4\pi\ cm^{2}$.

Step2: Calculate the volume of the gold coating

The thickness of the gold coating $t = 1.00\ \mu m=1.00\times10^{- 4}\ cm$. The volume of the gold coating $V = A\times t=4\pi\ cm^{2}\times1.00\times10^{-4}\ cm = 4\pi\times10^{-4}\ cm^{3}$.

Step3: Calculate the mass of the gold coating

Using the density formula $\rho=\frac{m}{V}$, where $\rho = 19.3\ g/cm^{3}$ and $V = 4\pi\times10^{-4}\ cm^{3}$. Then $m=\rho V=19.3\ g/cm^{3}\times4\pi\times10^{-4}\ cm^{3}\approx2.42\times10^{-2}\ g$.

Step4: Calculate the number of moles of gold

The molar mass of gold $Au$ is $M = 197\ g/mol$. The number of moles $n=\frac{m}{M}=\frac{2.42\times10^{-2}\ g}{197\ g/mol}\approx1.23\times10^{-4}\ mol$.

Step5: Determine the number of electrons transferred

For $Au^{3 +}+3e^{-}\rightarrow Au$, 3 moles of electrons are transferred per mole of $Au$ deposited. So the number of moles of electrons $n_{e}=3n = 3\times1.23\times10^{-4}\ mol=3.69\times10^{-4}\ mol$.

Step6: Calculate the charge transferred

Using Faraday's constant $F = 96485\ C/mol$. The charge $Q=n_{e}F=3.69\times10^{-4}\ mol\times96485\ C/mol\approx35.6\ C$.

Step7: Calculate the time

Using the formula $I=\frac{Q}{t}$, where $I = 81\ A$. Then $t=\frac{Q}{I}=\frac{35.6\ C}{81\ A}\approx0.44\ s$.

Answer:

$0.44\ s$