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Juli2301 [7.4K]
2 years ago
12

Consider a steady-state experiment in which the observed current due to reduction of Ox to R is 85 mA/cm2. What is the concentra

tion gradient at the electrode surface? Knowing that the concentration of the species being reduced is zero at the electrode surface, and 0.10 mM in the bulk, how thick is the diffusion layer in this experiment?
Engineering
1 answer:
Elanso [62]2 years ago
5 0

Answer: Yes

Explanation:

Surface Concentration: In electrochemistry, there is an important distinction between the concentration of a species at the electrode’s surface and its concentration at some distance from the electrode’s surface (in what we call the bulk solution). Suppose we place an electrode in a solution of Fe3+ and fix the potential at 1.00 V. At this potential Fe3+ is stable—the standard state reduction potential for Fe3+ to Fe2+ is +0.771 V, the concentration of Fe3+ remains the same at all distances from the electrode’s surface.

Bulk Concentraton: If we change the electrode’s potential to +0.500 V, the concentration of Fe3+ at the electrode’s surface decreases to approximately zero. The concentration of Fe3+ increases as we move away from the electrode’s surface until it equals the concentration of Fe3+ in bulk solution. The resulting concentration gradient causes additional Fe3+ from the bulk solution to diffuse to the electrode’s surface.

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A furnace wall is to be built of 20-cm firebrick and building (structural) brick of same thickness. The thermal conductivities o
Norma-Jean [14]

Answer:

q=2313.04W/m^2

T=690.86°C

Explanation:

Given that

Thickness t= 20 cm

Thermal conductivity of firebrick= 1.6 W/m.K

Thermal conductivity of structural brick= 0.7 W/m.K

Inner temperature of firebrick=980°C

Outer temperature of structural brick =30°C

We know that thermal resistance

R=\dfrac{t}{KA}

These are connect in series

R=\left(\dfrac{t}{KA}\right)_{fire}+\left(\dfrac{t}{KA}\right)_{struc}

R=\dfrac{0.2}{1.6A}+\dfrac{0.2}{0.7A}\ K/W

R=\dfrac{23}{56A}\ K/W

Heat transfer

Q=\dfrac{\Delta T}{R}

Q=56A\times \dfrac{980-30}{23}\ W

So heat flux

q=2313.04W/m^2

Lets temperature between interface is T

Now by equating heat in both bricks

\dfrac{980-T}{\dfrac{0.2}{1.6A}}=\dfrac{T-30}{\dfrac{0.2}{0.7A}}

So T=690.86°C

6 0
3 years ago
A well insulated rigid tank contains 4 kg of argon gas at 450 kPa and 30 C. A valve is opened, allowing the argon to escape unti
natima [27]

Answer:

Final mass of Argon=  2.46 kg

Explanation:

Initial mass of Argon gas ( M1 ) = 4 kg

P1 = 450 kPa

T1 = 30°C = 303 K

P2 = 200 kPa

k ( specific heat ratio of Argon ) = 1.667

assuming a reversible adiabatic process

<u>Calculate the value of the M2 </u>

Applying ideal gas equation ( PV = mRT )

P₁V / P₂V = m₁ RT₁ / m₂ RT₂

hence : m2 = P₂T₁ / P₁T₂ * m₁

                   = (200 * 303 ) / (450 * 219 ) * 4

                   = 2.46 kg

<em>Note: Calculation for T2 is attached below</em>

5 0
2 years ago
Write a class named FBoard for playing a game, where player x is trying to get her piece to row 7 and player o is trying to make
OverLord2011 [107]

Answer:

12345=V

Explanation:

7 0
3 years ago
3 facts about the Tokyo skytree tower aka the tallest tower in the world
UkoKoshka [18]

Answer:

1. Tokyo skytree tower is the tallest tower in the world, measuring 2080 feet. That's almost twice the size of the Eiffel Tower!

2.The process of building the tower began in 2008. The project was completed on 29 February 2012.

3. Pairing form with function, Skytree will serve as a TV and radio broadcast tower.

6 0
3 years ago
Read 2 more answers
• Suppose that a particular algorithm has time complexity T(n) = 10 ∗ 2n, and that execution of the algorithm on a particular ma
elena-s [515]

Answer:

The number of inputs processed by the new machine is 64

Solution:

As per the question:

The time complexity is given by:

T(n) = 10\times 2n

where

n = number of inputs

T = Time taken by the machine for 'n' inputs

Also

The new machine is 65 times faster than the one currently in use.

Let us assume that the new machine takes the same time to solve k operations.

Then

T(k) = 64 T(n)

\frac{T(k)}{T(n)} = 64

\frac{20k}{20n} = 64

k = 64n

Thus the new machine will process 64 inputs in the time duration T

8 0
3 years ago
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