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3 years ago

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A jeweler needs to electroplate gold onto a bracelet using an ionic solution. He knows that the charge carriers in the ionic sol

ution are gold ions of charge e, and that each gold ion has a mass of 22g. The gold ions move through the solution, and are deposited on the bracelet. He has calculated that he must deposit 0.20 g of 901d to reach the necessary thickness. If each gold ion has a mass of g, what current should he use to electroplate the bracelet in three hours

1 answer:

Studentka2010 [4]3 years ago

7 0

Explanation:

First, we will calculate the total charge as follows.

$q_{total} = \frac{\text{weight of gold to be deposited}}{\text{atomic mass of gold}} \times N_{A} \times \text{charge on electron}$

= $\frac{0.2 g}{196.97 g/mol} \times 6.02 \times 10^{23} \times 1.6 \times 10^{-19} C$

= 97.80 C

Now, we will calculate the current required by the jeweler to plate the bracelet as follows.

i = $\frac{q_{total}}{t_{total}}$

= $\frac{97.80 C}{3 \times 60 \times 60 sec}$

= $9.05 \times 10^{-3}$ A

or, = 9.05 mA

thus, we can conclude that 9.05 mA current should he use to electroplate the bracelet in three hours.

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$\begin{aligned}\text{GPE change} &= \text{Initial GPE} - \text{Final GPE} \\ &= 0 - \left(-\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\right) = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \end{aligned}$

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