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zepelin [54]
2 years ago
4

Everything else being equal, write a general expression to show how the exchange current density for a reaction changes as a fun

ction of temperature. In other words, write an expression for j0(T) at an arbitrary temperature T as a function of j0(T0) at a reference temperature T0. If a reaction has j0 = 10-8 A/cm2 at 300K and j0 = 10-4 A/cm2 at 600K, what is ∆G1 + + for the reaction? Assume that the T0 is 300K.

Engineering
2 answers:
Norma-Jean [14]2 years ago
8 0

Answer:

The Gibbs free energy for the reaction is 45944.8619 J/mol

Explanation:

the solution is in the attached Word document

Download docx
Alex2 years ago
7 0

Answer:

Gibbs free energy is 45944.86 J/mol

Explanation:

In Engineering, thermodynamics, the Gibbs free energy is a potential that is used to calculate the maximum of reversible work that may be performed by a thermodynamic system at a constant temperature and pressure.

Please kindly go through the attached file for a rundown of how the answer is gotten.

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You work in Madison, Wisconsin. It is January and the area has been hit with bad weather. Another weather front is expected to a
Lelu [443]

Answer:

The best first sentence of your e-mail will be-

<u>Weather forecasters are predicting a blizzard this afternoon, so, as a result of this news, our supervisor has decided to close the office at noon so people can travel home safely.</u>

<u />

5 0
2 years ago
A 150-lbm astronaut took his bathroom scale (a spring scale) and a beam scale (compares masses) to the moon where the local grav
kozerog [31]

Answer:

a)Wt =25.68 lbf

b)Wt = 150 lbf

F= 899.59 N

Explanation:

Given that

g = 5.48 ft/s^2.

m= 150 lbm

a)

Weight on the spring scale(Wt) = m g

We know that

1\ lbf=32.17 \ lmb.ft/s^2

Wt = 150 x 5.48/32 lbf

Wt =25.68 lbf

b)

On the beam scale

This is scale which does not affects by gravitational acceleration.So the wight on the beam scale will be 150 lbf.

Wt = 150 lbf

If the plane is moving upward with acceleration 6 g's then the for F

F = m a

We know that

1\ ft/s^2= 0.304\ m/s^2

5.48\ ft/s^2= 1.66\ m/s^2

a=6 g's

a=9.99\ m/s^2

So

F = 90 x 9.99 N

F= 899.59 N

3 0
2 years ago
It is said that Archimedes discovered his principle during a bath while thinking about how he could determine if King Hiero’s cr
arlik [135]

Answer: The crown is not made of pure gold.

Explanation:

Archimedes discovered that any solid, of any shape, when submerged in a liquid receives an upward force, equal to the weight of the volume of the liquid removed by the solid, which is equal to the solid volume.

So, if any body is weighed in air, the normal force will be equal to the gravity force (which we call weight) which can be expressed as follows:

Fg = m g = δ V g = 34.7 N

When submerged in water, the normal force is equal to the difference between the actual weight, and the upward force due to Archimedes' principle, called buoyant force, as follows:

Fn = Fg - Ep = δx. V. g - δH20 . V. g = 31.5 N

Dividing Fg between Fn, and simplifying common terms, we have:

δx / (δx - δh20) = 34.7 / 31.5 = 1.10

Solving for δx, we get the following value:

δx = 11,000 Kg/m3, less dense than pure gold, so we can conclude that the crown was not made of pure gold.

3 0
2 years ago
Water flows at a rate of 10 gallons per minute in a new horizontal 0.75?in. diameter galvanized iron pipe. Determine the pressur
ruslelena [56]

Answer:

\frac{\delta p }{l} = 30.4 lb/ft^3

Explanation:

Given data:

flow rate = 10 gallon per  minute = 0.0223 ft^3/sec

diameter = 0.75 inch

we know discharge is given as

Q =  VA

solve for velocity V = \frac{Q}{A}[/tex]

V = \frac{0.223}{\frac{\pi}{4} \frac{0.75}{12}}

V = 7.27 ft/sec

we know that Reynold number

Re = \frac{VD}{\nu}

Re = \frac{7.27 \times \frac{0.75}{12}}{1.21\times 10^{-5}}

Re = 3.76 \times 10^4

calculate the \frac{\epsilon }{D}ratio to determine the fanning friction f

\frac{\epsilon }{D} = \frac{0.0005}{\frac{0.75}{12}} = 0.008

from moody diagram f value corresonding to Re and \frac{\epsilon }{D}is 0.037

for horizontal pipe

\delta p = \frac{f l \rho v^2}{2D}

\frac{\delta p }{l} = \frac{1 \times 0.037 \times 1.94 \times 7.27}{\frac{0.75}{12}}

where 1.94 slug/ft^3is density of  water

\frac{\delta p }{l} = 30.4 lb/ft^3

3 0
2 years ago
A conical enlargement in a vertical pipeline is 5 ft long and enlarges the pipe diameter from 12 in. to 24 in. diameter. Calcula
makkiz [27]

Answer:

F_y = 151319.01N = 15.132 KN

Explanation:

From the linear momentum equation theory, since flow is steady, the y components would be;

-V1•ρ1•V1•A1 + V2•ρ2•V2•A2 = P1•A1 - P2•A2 - F_y

We are given;

Length; L = 5ft = 1.52.

Initial diameter;d1 = 12in = 0.3m

Exit diameter; d2 = 24 in = 0.6m

Volume flow rate of water; Q2 = 10 ft³/s = 0.28 m³/s

Initial pressure;p1 = 30 psi = 206843 pa

Thus,

initial Area;A1 = π•d1²/4 = π•0.3²/4 = 0.07 m²

Exit area;A2 = π•d2²/4 = π•0.6²/4 = 0.28m²

Now, we know that volume flow rate of water is given by; Q = A•V

Thus,

At exit, Q2 = A2•V2

So, 0.28 = 0.28•V2

So,V2 = 1 m/s

When flow is incompressible, we often say that ;

Initial mass flow rate = exit mass flow rate.

Thus,

ρ1 = ρ2 = 1000 kg/m³

Density of water is 1000 kg/m³

And A1•V1 = A2•V2

So, V1 = A2•V2/A1

So, V1 = 0.28 x 1/0.07

V1 = 4 m/s

So, from initial equation of y components;

-V1•ρ1•V1•A1 + V2•ρ2•V2•A2 = P1•A1 - P2•A2 - F_y

Where F_y is vertical force of enlargement pressure and P2 = 0

Thus, making F_y the subject;

F_y = P1•A1 + V1•ρ1•V1•A1 - V2•ρ2•V2•A2

Plugging in the relevant values to get;

F_y = (206843 x 0.07) + (1² x 1000 x 0.07) - (4² x 1000 x 0.28)

F_y = 151319.01N = 15.132 KN

6 0
2 years ago
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