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4 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?

2 answers:

poizon [28]4 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 C $mol^{-1}$ ( 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 C $s^{1}$

To calculate net reaction of reaction A

i = 5 c $s^{1}$

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 c $s^{1}$

n = 3

A = 5

back to the equation

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

Delvig [45]4 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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True or false If 2 is a factor of y and 3 is a factor of y, then 6 is a factor of y

melamori03 [73]

Answer:

True

Explanation:

The product of any subset of the factors of y is also a factor of y. If 2 and 3 are factors of y, then 2·3 = 6 is a factor of y.

The statement is true.

4 0

2 years ago

Determine whether or not it is possible to cold work steel so as to give a minimum Brinell hardness of 225 and at the same time

rodikova [14]

Answer:

First we determine the tensile strength using the equation;

Tₓ (MPa) = 3.45 × HB

{ Tₓ is tensile strength, HB is Brinell hardness = 225 }

therefore

Tₓ = 3.45 × 225

Tₓ = 775 Mpa

From Conclusions, It is stated that in order to achieve a tensile strength of 775 MPa for a steel, the percentage of the cold work should be 10

When the percentage of cold work for steel is up to 10,the ductility is 16% EL.

And 16% EL is greater than 12% EL

Therefore, it is possible to cold work steel to a given minimum Brinell hardness of 225 and at the same time a ductility of at least 12% EL

7 0

3 years ago

A car generator turns at 400 rpm (revolutions per minute) when the engine is idling. It has a rectangular coil with 300 turns of

larisa86 [58]

Answer:

The field strength needed is 0.625 T

Explanation:

Given;

angular frequency, ω = 400 rpm = (2π /60) x (400) = 41.893 rad/s

area of the rectangular coil, A = L x B = 0.0611 x 0.05 = 0.003055 m²

number of tuns of the coil, N = 300 turns

peak emf = 24 V

The peak emf is given by;

emf₀ = NABω

B = (emf₀ ) / (NA ω)

B = (24) / (300 x 0.003055 x 41.893)

B = 0.625 T

Therefore, the field strength needed is 0.625 T

4 0

3 years ago

Find the derivative of y = sin(ln(5x2 − 2x))

pickupchik [31]

Answer:

$y = \cos[\ln x + \ln (5\cdot x - 2)]\cdot \left(\frac{1}{x} + \frac{5}{5\cdot x-2} \right)$

Explanation:

Let $y = \sin[\ln(5\cdot x^{2}-2\cdot x)]$ and we proceed to find the derivative by the following steps:

1) $y = \sin[\ln(5\cdot x^{2}-2\cdot x)]$ Given

2) $y = \sin [\ln[x\cdot (5\cdot x - 2)]]$ Distributive property

3) $y = \sin[\ln x + \ln (5\cdot x - 2 )]$ $\ln (a\cdot b) = \ln a + \ln b$

4) $y = \cos[\ln x + \ln (5\cdot x - 2)]\cdot \left(\frac{1}{x} + \frac{5}{5\cdot x-2} \right)$ $\frac{d}{dx} (\sin x) = \cos x$ / $\frac{d}{dx}(\ln x) = \frac{1}{x}$ / $\frac{d}{dx}(c\cdot x^{n}) = n\cdot c\cdot x^{n-1}$ /Rule of chain/Result

3 0

3 years ago

Consider a Carnot refrigeration cycle executed in a closed system in the saturated liquid-vapor mixture region using 0.96 kg of

Eduardwww [97]

Answer:

$P_m_i_n= 356.9 KPa$

Explanation:

Coefficient of performance (COP) of refrigeration cycle is given by:

$COP=\frac{1}{\frac{T_H}{T_L}-1 }$

We are given:

$T_H=1.2T_L$

$COP=\frac{1}{1.2-1 }$

COP= 5

We can also write Coefficient of performance (COP) of refrigeration cycle as:

$COP_R=\frac{Q_L}{W_i_n}$

Amount of heat absorbed by low temperature reservoir can be found as:

$Q_L=COP_R * W_i_n$

$Q_L=5 * 22 KJ$

$Q_L=110 KJ$

According to first law of thermodynamics amount of heat rejected by hot reservoir is given by:

$Q_H=Q_L + W_i_n$

$Q_H=110 KJ + 22 KJ$

$Q_H=132 KJ$

We are given the mass of 0.96 kg. So,

$q_H=\frac{Q_H}{m}$

$q_H=\frac{132 KJ}{0.96Kg}$

$q_H=137.5 KJ/Kg$

Since it is a saturated liquid-vapour mixture $q_H=h_f_g$ .

$q_H=h_f_g=137.5 KJ/Kg$

From Refrigerant 134-a tables $T_H$ at $h_f_g=137.5 KJ/Kg$ is 61.3 C. (We calculated this by interpolation)

Converting $T_H$ from Celsius to Kelvin:

$61.3^{o} C+273 = 334.3^{o} K$

$T_H= 334.3^{o} K$

We are given:

$T_H=1.2T_L$

$T_L=\frac{T_H}{1.2}$

$T_L=\frac{334.3}{1.2}$

$T_L=278.58^{o} K$

Converting $T_L$ from Kelvin to Celsius:

$278.58^{o} K-273 = 5.58^{o} C$

$T_L= 5.58^{o} C$

From Refrigerant 134-a tables $P_m_i_n$ at $T_L=5.58^{o} C$ is 356.9 KPa. (We calculated this by interpolation).

$P_m_i_n= 356.9 KPa$

8 0

4 years ago