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Paraphin [41]
3 years ago
14

What mass of silver can be plated onto an object in 33.5 minutes at 8.70 a of current? ag (aq e- ? ag(s what mass of silver can

be plated onto an object in 33.5 minutes at 8.70 a of current? ag (aq e- ? ag(s 9.78 g 0.102 g 3.07 g 0.326 g 19.6 g?
Physics
2 answers:
Allisa [31]3 years ago
5 0

Answer:

Last option 19.6 g

Explanation:

In order to do this, we need to apply the Faraday law which is the following:

q = I*t (1)

Where:

I: current in Amperius

t: time in seconds

q: charge in Coulombs

the time in seconds is:

t = 33.5 min * 60 s/min = 2,010 s

Replacing in (1):

q = 8.7 * 2,010 = 17,487 C

With this value, we need to calculate the moles of Ag. This can be done dividing this value with the respective charge in one mole:

moles = q/n

n: value of 1 mole of electron / C = 96500 C/mol e

Replacing we have:

moles = 17.487 / 96,500

moles = 0,1812 moles e

Remember now that each mole of Ag+ can only catch one electron at a time, so, 0.1812 moles e, would be 0.1812 moles.

Finally, we need the molecular mass of Ag, which is 107.87 g/mol so:

m = 0.1812 * 107.87

m = 19.55 g or 19.6 g

Schach [20]3 years ago
4 0
To determine the mass plated, we use Faraday's Law of Electrolysis. We calculate as follows:

q = It
q = 8.70 (33.5) (60)
q = 17487 C

mass = 17487 C ( 1 mol e- / 96500 C) ( 1 mol / 2 mol e-) (107.9 g /mol)
mass = 9.78 g

Hope this helps.

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Complete Question

An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field in the wire changes with time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is measured in seconds.

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Answer:

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Explanation:

From the question we are told that

    The diameter of the wire is  d =  0.205cm = 0.00205 \ m

     The radius of  the wire is  r =  \frac{0.00205}{2} = 0.001025  \ m

     The resistivity of aluminum is 2.75*10^{-8} \ ohm-meters.

       The electric field change is mathematically defied as

         E (t) =  0.0004t^2 - 0.0001 +0.0004

     

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       Q = \int\limits^{t}_{0} {\frac{A}{\rho} E(t) } \, dt

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       A =  \pi r^2 =  (3.142 * (0.001025^2)) = 3.30*10^{-6} \ m^2

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       \frac{A}{\rho} =  \frac{3.3 *10^{-6}}{2.75 *10^{-8}} =  120.03 \ m / \Omega

Therefore

      Q = 120 \int\limits^{t}_{0} { E(t) } \, dt

substituting values

      Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt

     Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] }  \left | t} \atop {0}} \right.

From the question we are told that t =  5 sec

           Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] }  \left | 5} \atop {0}} \right.

          Q = 120 [ \frac{0.0004(5)^3 }{3} - \frac{0.0001 (5)^2}{2} +0.0004(5)] }

         Q =2.094 C

     

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