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Mariana [72]
3 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]3 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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How long does it take light to travel 850 km in a vacuum? Answer in ms.(Express your answer to two significant figures.)
poizon [28]

Answer:

0.002833 sec

Explanation:

Speed of light in vacuum is 3\times 10^{8}m/sec

Given distance = 850 km = 850×1000=850000 m

We have to calculate the time that light take to travel the distance 850 km

Time T=\frac{distance }{speed}=\frac{850000}{3\times 10^8}=2.833\times 10^{-3}sec

So the time taken by light to travel 850 km is 0.002833 sec

5 0
3 years ago
atoms are the principle constituent of ______ A. all matter b. only solids liquids and gases c. all subatomic particles d. only
Soloha48 [4]
I think it is a. all matter

6 0
2 years ago
A go-cart is traveling at a rate of 25 m/sec for 20 seconds. How far will the go cart travel?
notsponge [240]

Answer:

Distance travel by go-cart = 500 meter

Explanation:

Given:

Speed of go cart = 25 m/s

Time travel = 20 seconds

Find:

Distance travel by go-cart

Computation:

Distance = Speed x time

Distance travel by go-cart = Speed of go cart x Time travel

Distance travel by go-cart = 25 x 20

Distance travel by go-cart = 500 meter

4 0
2 years ago
Two airplanes leave an airport at the same time. The velocity of the first airplane is 730 m/h at a heading of 44.3 ◦ . The velo
Goshia [24]

Answer:

So airplane will be 1324.9453 m apart after 2.9 hour

Explanation:

So if we draw the vectors of a 2d graph we see that the difference in angles is  = 83 - 44.3 = 83-44.3=38.7^{\circ}

Distance traveled by first plane = 730×2.9 = 2117 m

And distance traveled by second plane = 590×2.9 = 1711 m

We represent these distances as two sides of the triangle, and the distance between the planes as the side opposing the angle 38.7.

Using the law of cosine, d^2 representing the distance between the planes, we see that:

d^2=2117^2 + 1711^2 -2\times (2117)\times (1711)cos(38.7)=1755480.2482

d = 1324.9453 m

4 0
3 years ago
Jake is helping Fin push a box at a constant velocity up an incline that makes an angle of 30.0° above the horizontal by applyin
andre [41]

Given data

The angle of inclination of the plane is theta = 30 degree

The applied force in the inclined plane is F = 94 N

The distance moved in the inclined plane is d = 2.30 m

The coefficient of kinetic friction is u_k = 0.280

The free-body diagram of the above configuration is shown below:

Here, the normal reaction force on the box is N, the acceleration due to gravity is denoted as g, the friction force on the box is F_f, and the mass of the box is denoted as m.

(a)

The expression for the work done by the pushing force is given as:

W=Fd

Substitute the value in the above equation.

\begin{gathered} W=94\text{ N}\times2.30\text{ m} \\ W=216.2\text{ J} \end{gathered}

Thus, the work done by the pushing force is 216.2 J.

(b)

The box is moving at the constant velocity, therefore, the pushing force will be equal to the frictional force and the component of the gravitational force in the inclined plane.

\begin{gathered} F=F_f+mg\sin \theta \\ F=\mu_kN+mg\sin \theta \end{gathered}

The expression for the normal reaction force is given as:

N=mg\cos \theta

The expression for the mass of the box is given as:

\begin{gathered} F=\mu_k\times mg\cos \theta+mg\sin \theta \\ m=\frac{F}{\mu_kg\cos \theta+g\sin \theta} \end{gathered}

Substitute the value in the above equation.

\begin{gathered} m=\frac{94\text{ N}}{0.28\times9.8m/s^2\times\cos 30^o+9.8m/s^2\times\sin 30^0} \\ m=12.9\text{ kg} \end{gathered}

Thus, the mass of the box is 12.9 kg.

7 0
10 months ago
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