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Ksju [112]
3 years ago
14

Consider two electrochemical reaqctions. Reaction A results in the transfer of 2 mol of electrons per mole of reactant and gener

ates a current of 5 A on an electrode 2 cm2 in area. Reaction B results in the transfer of 3 mol of electrons per mole of reactant and generates a current of 15 A on an electrode 5 cm2 in area. What are the net reaction rates for reactions A and B (in moles of reactant per square centimeter per second)? Which reaction has the higher net reaction rate?
Engineering
2 answers:
poizon [28]3 years ago
8 0

Answer:

Reaction A has a higher net reaction rate

Explanation:

Rate of electrochemical reaction rate per unit area = \frac{i}{nFA}

F = 96500 Cmol^{-1} ( Faraday constant)

n = number of moles of electrons per mole of reactant

A = Area of electrode

i = current generated

convert the current from Amperes to Cs^{1}

To calculate net reaction of reaction A

i = 5 cs^{1}

n = 2

A = 2

back to the equation

=  5 / (2 * 96500 * 2) = (1.3 * 10 ^ -5)  mols^-1 cm^-2

To calculate net reaction of reaction B

i = 15 cs^{1}

n = 3

A = 5

back to the equation

= 15 / ( 3 * 96500 * 5) =  (1.036 * 10 ^ -5)  mols^-1 cm^-2

Delvig [45]3 years ago
3 0

Answer:

Reaction A has higher net reaction rate

Explanation:

Data:

The reaction rates:

Reaction A:

Number of electrons per area = 2 mol/ 2 cm²

                                                  = 1 mol/ cm²

Reaction B:

Number of electrons per area = 3 mol/ 5 cm²

                                                  = 0.6 mol/cm²

Based on the calculations above, the reaction B has a higher reaction rate.

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Determine the total condensation rate of water vapor onto the front surface of a vertical plate that is 10 mm high and 1 m in th
castortr0y [4]

Answer:

Q =  63,827.5 W

Explanation:

Given:-

- The dimensions of plate A = ( 10 mm x 1 m )

- The fluid comes at T_sat , 1 atm.

- The surface temperature, T_s = 75°C  

Find:-

Determine the total condensation rate of water vapor onto the front surface of a vertical plate

Solution:-

- Assuming drop-wise condensation the heat transfer coefficient for water is given by Griffith's empirical relation for T_sat = 100°C.

                            h = 255,310 W /m^2.K

- The rate of condensation (Q) is given by Newton's cooling law:

                           Q = h*As*( T_sat - Ts )

                           Q = (255,310)*( 0.01*1)*( 100 - 75 )

                           Q =  63,827.5 W

8 0
3 years ago
Read 2 more answers
What is a p-n junction? Show by the diagram.
Natalija [7]

Answer:

The p-n junction is a region formed when a p -type semiconductor material is joined to an n-type semiconductor material

Explanation:

The p type semiconductor has holes as its majority charge carriers making it positively charged while the n –types has an overall negative charge. At the junction the holes move towards the electron until such a time when there is a balance in charges from both materials, which leads to the formation of the depletion zone as shown in the attachment below

                           

5 0
4 years ago
2. The following segment of carotid artery has an inlet velocity of 50 cm/s (diameter of 15 mm). The outlet has a diameter of 11
ahrayia [7]

This question is incomplete, the missing diagram is uploaded along this answer below.

Answer:

the forces required to keep the artery in place is 1.65 N

Explanation:

Given the data in the question;

Inlet velocity V₁ = 50 cm/s = 0.5 m/s

diameter d₁ = 15 mm = 0.015 m

radius r₁ = 0.0075 m

diameter d₂ = 11 mm = 0.011 m

radius r₂ = 0.0055 m

A₁ = πr² = 3.14( 0.0075 )² =  1.76625 × 10⁻⁴ m²

A₂ = πr² = 3.14( 0.0055 )² =  9.4985 × 10⁻⁵ m²

pressure at inlet P₁ = 110 mm of Hg = 14665.5 pascal

pressure at outlet P₂ = 95 mm of Hg = 12665.6 pascal

Inlet volumetric flowrate = A₁V₁ = 1.76625 × 10⁻⁴ × 0.5 = 8.83125 × 10⁻⁵ m³/s

given that; blood density is 1050 kg/m³

mass going in m' = 8.83125 × 10⁻⁵ m³/s × 1050 kg/m³ = 0.092728 kg/s

Now, using continuity equation

A₁V₁ = A₂V₂

V₂ = A₁V₁ / A₂ = (d₁/d₂)² × V₁

we substitute

V₂ =  (0.015 / 0.011 )² × 0.5

V₂ = 0.92975 m/s

from the diagram, force balance in x-direction;

0 - P₂A₂ × cos(60°) + Rₓ = m'( V₂cos(60°) - 0 )    

so we substitute in our values

0 - (12665.6 × 9.4985 × 10⁻⁵)  × cos(60°) + Rₓ = 0.092728( 0.92975 cos(60°) - 0 )    

0 - 0.6014925 + Rₓ =  0.043106929 - 0

Rₓ = 0.043106929 + 0.6014925

Rₓ = 0.6446 N

Also, we do the same force balance in y-direction;

P₁A₁ - P₂A₂ × sin(60°) + R_y = m'( V₂sin(60°) - 0.5 )  

we substitute

⇒ (14665.5 × 1.76625 × 10⁻⁴) - (12665.6 × 9.4985 × 10⁻⁵) × sin(60°) + R_y = 0.092728( 0.92975sin(60°) - 0.5 )

⇒ 1.5484 + R_y = 0.092728( 0.305187 )

⇒ 1.5484 + R_y = 0.028299    

R_y = 0.028299 - 1.5484

R_y = -1.52 N

Hence reaction force required will be;

R = √( Rₓ² + R_y² )

we substitute

R = √( (0.6446)² + (-1.52)² )

R = √( 0.41550916 + 2.3104 )

R = √( 2.72590916 )

R = 1.65 N

Therefore, the forces required to keep the artery in place is 1.65 N

 

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You gently place several steel needles on the free surface of the water in a large tank. The needles come in two lengths: some a
Stolb23 [73]

Answer:

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Explanation:

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