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Mashutka [201]
2 years ago
7

How many minutes will it take to plate out 16. 22 g of al metal from a solution of al 3 using a current of 12. 9 amps in an elec

trolytic cell?
Physics
1 answer:
MAXImum [283]2 years ago
8 0

It will take  103 minutes  to plate out 16. 22 g of al metal from a solution of al 3 using a current of 12. 9 amps in an electrolytic cell

Electrolytic cell, any device in which electrical energy is converted to chemical energy, or vice versa. Such a cell typically consists of two metallic or electronic conductors (electrodes) held apart from each other and in contact with an electrolyte (q.v.), usually a dissolved or fused ionic compound

First, look at the reaction to decided how many moles of electrons it will take:

Al3+ + 3e ------> Al(s)

It takes 3 electrons to convert 1 mole of Al3+

Next, figure out how many moles of Al metal you want:

I mole of Al mass = 26 g

1 g of Al will have = 1/26 moles

16. 22 g of al metal will have = 16.22 / 26 moles 0.624 moles

since , their are 3 electrons

0.624 x 3= 1.87 moles of electrons

using Faraday constant to find the number of coulombs (C) needed:

1 faraday = 96500 C per mole

1.87 x 96500 = 180455

Since,  1 amp = 1 C/sec,, time needed to supply a current of 12.9 amp will be

80455C  / 12.9 C = 6236.82 sec = 103 minutes

To learn more about electrolytic cell here

brainly.com/question/862677

#SPJ4

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An RC circuit consists of a resistor with resistance 1.0 kΩ, a 120-V battery, and two capacitors, C1 and C2, with capacitances o
Serhud [2]

Answer:

Q_t= 8.3 * 10^3 C

Explanation:

From the question we are told that:

Resistor R=1000ohms

Voltage v=120_V

Capacitance of c_1 c_1=20 \mu F

Capacitance of c_2 c_2=60 \mu F

Time t=0

Generally the equation for charges is mathematically given by

For C_1\\Charge\ on\ C_1 = CV = 20*120 = 2400 μC = 2.4 x 10^-3 C\\Charge\ on\ C_1 = 2400 μC = 2.4 x 10^-3 C\\Charge\ on\ C_1 = 2.4 x 10^-3 C\\

ForC_2\\Charge on C_2 = 60*120 =7200 μC =  7.2 x 10^-3\\Charge on C_2 =  7.2 x 10^-3

Generally the equation for voltage across capacitors is mathematically given by

V_c(t)=V(1-e^{-t/RC})

C=C_1+C_2=80 \mu f\\t=2RC=>160000s

V_c(t)=120(1-e^{-(160000)/1000*(80)})

V_c(t)=103.7598

Generally the equation for charges is mathematically given by

Q1(t) = C1Vc(t)\\Q1(t) = 20*103.7598\\Q1(t) = 2075.196\\\\Q2(t) = 60*103.7598\\Q2(t) = 6225.6\\

Generally the equation for total charges Q_t is mathematically given by

Q_t=Q1(t)+Q2(t)

Q_t= 8.3 * 10^3 C

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