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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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Molten metal is poured into the pouring cup of a sand mold at a steady rate of 400 cm3/s. The molten metal overflows the pouring
umka2103 [35]

Answer:

the proper diameter at its base so as to maintain the same volume flow rate is 1.6034 cm

Explanation:

Given the data in the question ;

flowrate Q = 400 cm³/s

cross section of the sprue is round

Diameter of sprue at the top d_top = 3.4 cm

Height of sprue = 20 cm = 0.2 m³

the proper diameter at its base so as to maintain the same volume flow rate = ?

first we determine the velocity at the sprue base

V_base = √2gh = √( 2×9.81×0.2) = √3.924 = 1.980908 m = 198.0908 cm

so, diameter of the sprue at the bottom  will be

Q = AV = [ (( πd²_bottom)/4) × V_bottom ]

d_bottom =  √(4Q/πV_bottom)

we substitute

d_bottom =  √((4×400)/(π×198.0908 ))

d_bottom =  √( 1600/622.3206)

d_bottom =  √2.571022

d_bottom =  1.6034 cm

Therefore, the proper diameter at its base so as to maintain the same volume flow rate is 1.6034 cm

8 0
3 years ago
Technician A says some coolant recovery systems are pressurized. Technician B says the coolant recovery system is used to keep a
Digiron [165]

Answer: Technician A

Explanation: A coolant recovery system Is one of the most important part of a vehicles cooling system. When a coolant gets too hot, it forces it way out of the spring loaded radiator cap so as to relieve pressure. Any coolant which has escaped is recovered back through the discharge tube located in the recovery tank. The system is usually air free

8 0
3 years ago
How many volts of electricity would it take to power up an entire city? Take Tokyo for example. Please explain!
jek_recluse [69]

Answer:

maybe a couple million volts

Explanation:

because there are outlets you need to charge your phone maybe you car if you have a tesla and skyscrapers

3 0
3 years ago
A square steel bar has a length of 8.4 ft and a 2.1 in by 2.1 in cross section and is subjected to axial tension. The final leng
nikitadnepr [17]

Answer:

Poissons ratio = -0.3367

Explanation:

Poissons ratio = Lateral Strain / Longitudinal Strain

In this case, the longitudinal strain will be:

Strain (longitudinal) = Change in length / total length

Strain (longitudinal) = (8.40392 - 8.4) / 8.4

Strain (longitudinal) = 4.666 * 10^(-4)

While the lateral strain will be:

Strain (Lateral) = Change in length / total length

Strain (Lateral) = (2.09967 - 2.1) / 2.1

Strain (Lateral) = -1.571 * 10^(-4)

Solving the poisson equation at the top we get:

Poissons ratio = -1.571 / 4.666                                     <u>( 10^(-4) cancels out )</u>

Poissons ratio = -0.3367

6 0
3 years ago
A section of highway has a free-flow speed of 55 mph and a capacity of 3300 veh/hr. In a given hour, 2100 vehicles were counted
alexira [117]

Answer:

Space mean speed = 44 mi/h

Explanation:

Using Greenshield's linear model

q = Uf ( D - D^{2}/Dj )

qcap = capacity flow that gives Dcap

Dcap = Dj/2

qcap = Uf. Dj/4

Where

U = space mean speed

Uf =  free flow speed

D = density

Dj = jam density

now,

Dj = 4 × 3300/55

    = 240v/h

q = Dj ( U - U^{2}/Uf)

2100 = 240 ( U - U^{2}/55)

Solve for U

U = 44m/h

5 0
3 years ago
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