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n a certain electroplating process gold is deposited by using a current of 14.0 A for 19 minutes. A gold ion, Au+, has a mass of approximately 3.3 × 10-22 g. How many grams of gold are deposited by this process?

Respuesta :

Answer:

32.9 g

Explanation:

Faraday's law: m = MQ / zF

  • M = 3.3x10^-22 x 6.02x10^23
  • Q = 19 x 14 x 60
  • z = 1
  • F =  96485

m = (3.3x10^-22 x 6.02x10^23) x  (19 x 14 x 60) / (1 x 96485) = 32.9 g

dsdrajlin

38 g of gold are produced in a certain electroplating process by using a current of 14.0 A for 19 minutes.

Let's consider the reduction of gold in an electroplating process.

Au⁺(aq) + 1 e⁻ ⇒ Au(s)

We can calculate the mass of gold deposited when the electroplating uses a current of 14.0 A for 19 minutes, using the following conversion factors.

  • 1 min = 60 s.
  • 1 A = 1 C/s.
  • 1 electron has a charge of 1.6 × 10⁻¹⁹ C.
  • 1 atom of Au is reduced by the gain of 1 electron.
  • 1 atom of Au has a mass of 3.3 × 10⁻²² g (approximately the same as Au⁺).

[tex]19 min \times \frac{60s}{1min} \times \frac{14.0C}{1s} \times \frac{1e^{-} }{1.6 \times 10^{-19}C } \times \frac{1atomAu}{1e^{-} } \times \frac{3.3 \times10^{-22}gAu } {1atomAu} = 38 g[/tex]

38 g of gold are produced in a certain electroplating process by using a current of 14.0 A for 19 minutes.

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