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Mariana [72]
4 years ago
15

An oxygen molecule is adsorbed onto a small patch of the surface of a catalyst. It's known that the molecule is adsorbed on of p

ossible sites for adsorption (see sketch at right). Calculate the entropy of this system. Round your answer to significant digits, and be sure it has the correct unit symbol.
Physics
1 answer:
FromTheMoon [43]4 years ago
4 0

The given question is incomplete. The complete question is as follows.

An oxygen molecule is adsorbed onto a small patch of the surface of a catalyst. It's known that the molecule is adsorbed on 1 of 36 possible sites for adsorption. Calculate the entropy of this system.

Explanation:

It is known that Boltzmann formula of entropy is as follows.

             s = k ln W

where,   k = Boltzmann constant

              W = number of energetically equivalent possible microstates or configuration of the system

In the given case, W = 36. Now, we will put the given values into the above formula as follows.

                  s = k ln W

                    = 1.38 \times 10^{-23} ln (36)        

                    = 4.945 \times 10^{-23} J/K

Thus, we can conclude that the entropy of this system is 4.945 \times 10^{-23} J/K.

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jeka94

Answer:

Lol, yes

Explanation:

5 0
3 years ago
The floor of a railroad flatcar is loaded with loose crates having a coefficient of static friction of 0.320 with the floor. If
mart [117]

Answer:  29.50 m

Explanation: In order to calculate the higher accelation to stop a train  without moving the crates inside the wagon which is traveling at constat speed we have to use the second Newton law so that:

f=μ*N the friction force is equal to coefficient of static friction  multiply the normal force (m*g).

f=m.a=μ*N= m*a= μ*m*g= m*a

then

a=μ*g=0.32*9.8m/s^2= 3.14 m/s^2

With this value we can determine the short distance to stop the train

as follows:

x= vo*t- (a/2)* t^2

Vf=0= vo-a*t then t=vo/a

Finally; x=vo*vo/a-a/2*(vo/a)^2=vo^2/2a= (49*1000/3600)^2/(2*3.14)=29.50 m

5 0
3 years ago
An initially stationary object experiences an acceleration of 6 m/s2 for a time of 15 s. How far will it travel during that time
andrew-mc [135]

Answer:

The object will travel 675 m during that time.

Explanation:

A body moves with constant acceleration motion or uniformly accelerated rectilinear motion (u.a.r.m) when the path is a straight line, but the velocity is not necessarily constant because there is an acceleration.

In other words, a body performs a u.a.r.m when its path is a straight line and its acceleration is constant. This implies that the speed increases or decreases uniformly.

In this case, the position is calculated using the expression:

x = xo + vo*t + ½*a*t²

where:

  • x0 is the initial position.
  • v0 is the initial velocity.
  • a is the acceleration.
  • t is the time interval in which the motion is studied.

In this case:

  • x0= 0
  • v0= 0  because the object is initially stationary
  • a= 6 \frac{m}{s^{2} }
  • t= 15 s

Replacing:

x= 0 + 0*15 s + ½*6 \frac{m}{s^{2} }*(15s)²

Solving:

x=½*6 \frac{m}{s^{2} }*(15s)²

x=½*6 \frac{m}{s^{2} }*225 s²

x= 675 m

<u><em> The object will travel 675 m during that time.</em></u>

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3 years ago
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