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QveST [7]
3 years ago
8

What would the potential of a standard hydrogen electrode (s.h.e.) be under the given conditions? [h+]=0.70 mph2=2.4 atmt=298 k?

Physics
1 answer:
Andru [333]3 years ago
7 0
Constant = 8.314JK⁻¹mol⁻¹
T is for temperature which is 298K
Faraday constant value is 96500C/mol
n is the number of electrons which are transferred in the reaction.
Ecell = E₀cell - RT/nFiN [cathode]/[anode]
Ecell = E₀cell - RT/nF In [PH₂]/[H⁺]²
Ecell = 0.00-8.314 JK⁻¹ mol⁻¹ × 298k/ 2× 96500C/mol In [2.4atm]/ 0.70]²
Ecell = 0.00 - 0.0129 In (2.59)
Ecell = 0.00 - 0.0129 × 0.951
Ecell = -0.0122V
∴Ecell is = -o.0122v
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A pendulum clock depends on the period of a pendulum to keep correct time. Suppose a pendulum clock is keeping correct time and
Mashutka [201]

Answer:

time period is increased so the clock will become SLOW

Explanation:

As we know that the time period of the simple pendulum is given by the formula

T = 2\pi\sqrt{\frac{L}{g}}

here we know that

L = distance of the pendulum bob from the hinge

g = acceleration due to gravity

now here the bob slide down so that the length of the pendulum is being increased

so time period T of the pendulum is also increased

so here the pendulum will take more time to oscillate or to complete one oscillation

so clock will become SLOW

8 0
3 years ago
Two blocks are connected by a string over a pulley. The hanging block has a mass of 8-kg and the one on the plane has a mass of
jeka94

Answer:

Let f be force of friction on the blocks kept on inclined plane. T be tension in the string

For motion of block on the inclined plane in upward direction

T - m₁gsin40 - f = m₁a

f = μ m₁gcos40

For motion of hanging  block on  in downward direction

m₂g  - T = m₂ a

Adding to cancel T

m₂g - - m₁gsin40 -  μ m₁gcos40 = a ( m₁+m₂ )

a =   g (m₂ - - m₁sin40 -  μ m₁cos40) / ( m₁+m₂ )

Putting the values

a = 9.8 ( 4.75 - 2.12-1.045) / 7.6

2.04 m s⁻²

M₂ will go down and M₁ will go up with acceleration .

Explanation:

7 0
3 years ago
Two blocks, joined by a string, have masses of 6.0 kg and 9.0 kg. They rest on a frictionless horizontal surface. A 2nd string,
Gennadij [26K]

Answer:

Explanation:

30 N force is pulling total mass of 15 kg , so acceleration in the system of masses

= 30 / 15

= 2 m / s²

Let us now consider forces acting on 9 kg . 30 N is pulling it in forward direction . Tension T in the string attached to it is pulling it in reverse direction

so net force on it

30 - T

Applying Newton's law of motion on it

30 - T = mass x acceleration

30 - T = 9  x 2

30 - 18 = T

T = 12 N

4 0
4 years ago
What are the 3 Laws of Newtonian Mechanics
Katyanochek1 [597]

In the first law, an object will not change its motion unless a force acts on it. In the second law, the force on an object is equal to its mass times its acceleration. In the third law, when two objects interact, they apply forces to each other of equal magnitude and opposite direction

7 0
3 years ago
Read 2 more answers
A microscope has an objective lens with a focal length of 14.0mm . A small object is placed 0.80mm beyond the focal point of the
Nikolay [14]

Answer:

A. 260 mm

B. - 18

C. 175

Explanation:

A

The expression for the lens equation is

\frac{1}{f} = \frac{1}{u} + \frac{1}{v}

\frac{1}{14.0mm} = \frac{1}{14.80 mm} + \frac{1}{v}

\frac{1}{v} = \frac{1}{14.0} - \frac{1}{14.80}

v = 259 mm

   = 260 mm or 26 cm (to 2 s.f)

check the attached files for additional solution

3 0
3 years ago
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