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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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A 16.5-kg crate starts at rest at the top of a 60.0° incline. The coefficients of friction are μs = 0.400 and μk = 0.300. The cr
irga5000 [103]

Answer:

t = 1.62 s

Explanation:

given,

mass of the block m₁ = 16.5 Kg

m₂ = 8 Kg

angle of inclination = 60°

μs = 0.400 and μk = 0.300

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a) let a is the acceleration of the block m₁ downward.

Net force acting on m₂,

F₂ = T - m₂ g

m₂a = T - m₂ g

a = \dfrac{T}{m_2} - g.......(1)

net force acting on m₁

F₁ = m₁g sin(60°) - μ_k m₁g cos (60°) - T

m₁ a = m₁g sin(60°) - μ_k m₁g cos (60°) - T

a = g sin(60^0) - \mu_k g cos (60^0) - \dfrac{T}{m_1}.........(2)

from equations 1 and 2

\dfrac{T}{m_2} - g = g sin(60^0) - \mu_k g cos (60^0) - \dfrac{T}{m_1}

\dfrac{T}{m_2} +\dfrac{T}{m_1} = g+ g sin(60^0) - \mu_k g cos (60^0)

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 T = \dfrac{g+ g sin(60^0) - \mu_k g cos (60^0)}{\dfrac{m_1+m_2}{m_2\times m_1}}

 T = {m_2\times m_1}\dfrac{g+ g sin(60^0) - \mu_k g cos (60^0)}{{m_1+m_2}}  

T = {16.5\times 8}\dfrac{9.8 + 9.8 sin(60^0) - 0.3\times 9.8 cos (60^0)}{{16.5+8}}

T = 90.61 N

from equation (1)

a = \dfrac{90.61}{8} - 9.8.......(1)

a = 1.52 m/s²

let t is the time taken

Apply,

d = ut + 0.5 a t²

2 = 0 + 0.5 x 1.52 x t²

t = \sqrt{2.63}

t = 1.62 s

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How are the electric field lines around a positive charge affected when a second positive charge is near it?
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The given parameters:

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<h3>Maximum height of fall;</h3>
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