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3 years ago

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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.

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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A tin can collapses if all air inside it is taken out why

Veseljchak [2.6K]

That only happens when the tin can is IN air.

In the familiar, comfy part of Earth's atmosphere where we live, the normal pressure of air is around 14.6 pounds on every square inch of everything. That's a big part of the reason why we're built with bodies that generate that same amount of pressure on the INSIDE pressing OUT. That way, we always have the same pressure pushing in both directions, so we know that we won't get crushed or blow up like balloons.

But we have to be careful with our bodies or other things when they're in places where the atmospheric pressure on the outside is NOT normal.

-- When a deep-sea diver goes hundreds of feet down in the ocean, and the pressure of the water is much GREATER than normal air.

-- When an astronaut has to go outside ... where there's NO air ... and fix something on the International Space Station.

When the pressure on the outside becomes very unusual, we have to wear special suits to protect our bodies from the unusual conditions.

The tin can in the story is a lot like our bodies. As long as it has air inside and air outside, the pressure is the same in both directions, so there's no particular force trying to deform the can. But ...

-- If you seal the can with the air inside it, take the can into a vacuum chamber, and pump the air out of the vacuum chamber, then the can only has pressure inside. It'll expand, and eventually spring a little hole in the metal, and all the air inside will blow out.

-- If you take all the air OUT of the can (so the can is REALLY 'empty'), then the pressure on it is all from the outside. In that situation, the can simply collapses, because there's nothing inside to provide pressure in the outward direction.

One more little thing to think about:

When you want some toothpaste to come drizzling out of the tube onto your brush, what do you do ? Do you perhaps squeeze the tube, and increase the pressure on the outside ?

4 0

3 years ago

F = 2.0*10^20 N,

natka813 [3]

$F=\dfrac{Gm_1m_2}{r^2}$

With the given values of $F,G,m_1,r$ , we have

$2.0\times10^{20}\,\mathrm N=\dfrac{\left(6.67\times10^{-11}\,\frac{\mathrm{Nm}^2}{\mathrm{kg}^2}\right)\left(5.98\times10^{24}\,\mathrm{kg}\right)m_2}{\left(3.8\times10^8\,\mathrm m\right)^2}$

Try dealing with the powers of 10 first: On the right, we have

$\dfrac{10^{-11}\times10^{24}}{(10^8)^2}=\dfrac{10^{24-11}}{10^{16}}=10^{-3}$

Meanwhile, the other values on the right reduce to

$\dfrac{6.67\times5.98}{3.8^2}\approx2.76$

Then taking units into account, we end up with the equation

$2.0\times10^{20}\,\mathrm N=\left(2.76\times10^{-3}\,\dfrac{\mathrm N}{\mathrm{kg}}\right)m_2$

Now we solve for $m_2$ :

$m_2=\dfrac{2.0\times10^{20}\,\mathrm N}{2.76\times10^{-3}\,\frac{\mathrm N}{\mathrm{kg}}}\approx0.725\times10^{20-(-3)}\,\mathrm{kg}$

$m_2=7.25\times10^{22}\,\mathrm{kg}$

or, if taking significant digits into account,

$m_2=7.3\times10^{22}\,\mathrm{kg}$

5 0

3 years ago

A solid, uniform disk of radius 0.250 m and mass 45.2 kg rolls down a ramp of length 5.40 m that makes an angle of 17.0° with th

serious [3.7K]

Explanation:

Given that,

Radius of the disk, r = 0.25 m

Mass, m = 45.2 kg

Length of the ramp, l = 5.4 m

Angle made by the ramp with horizontal, $\theta=17^{\circ}$

Solution,

As the disk starts from rest from the top of the ramp, the potential energy is equal to the sum of translational kinetic energy and the rotational kinetic energy or by using the law of conservation of energy as :

(a) $mgh=\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2$

h is the height of the ramp

$sin\theta=\dfrac{h}{5.4}$

$h=sin(17)\times 5.4=1.57\ m$

v is the speed of the disk's center

I is the moment of inertia of the disk,

$I=\dfrac{1}{2}mr^2$

$\omega=\dfrac{v}{r}$

$mgh=\dfrac{1}{2}mv^2+\dfrac{1}{2}\dfrac{1}{2}mr^2\times (\dfrac{v}{r})^2$

$gh=\dfrac{1}{2}v^2+\dfrac{1}{4}v^2$

$gh=\dfrac{3}{4}v^2$

$9.8\times 1.57=\dfrac{3}{4}v^2$

v = 4.52 m/s

(b) At the bottom of the ramp, the angular speed of the disk is given by :

$\omega=\dfrac{v}{r}$

$\omega=\dfrac{4.52}{0.25}$

$\omega=18.08\ rad/s$

Hence, this is the required solution.

4 0

3 years ago

What is 64 nanometers to m?

defon

$\sf Hello!$

$\sf We\: know \:that,$
$\sf 1\: meter$ = $\sf 10^{9} nm$

$\sf Then,$

$\sf Distance\: in \:m$ = $\sf Distance\: in\: nm$ × $\dfrac{\sf 1}{\sf 10^{9}}\: \sf m$

⇒ $\sf Distance\: in \:m$ = $\sf 64 × 10^{-9} \:m$

⇒ $\sf Distance\: in\: m$ = $\sf 6.4 × 10^{-8} \:m$

~ $\sf iCarl$

3 0

3 years ago

Read 2 more answers

Help me its emergency I need a correct answer please if you don't know then please dont guess

maxonik [38]

Answer:

True.

Explanation:

Total internal reflection is bound to occur when light travels from dense medium to less dense.

From the table given in the question above, we obtained:

Refractive Index of Crown glass = 1.52

Refractive Index of Asphalt = 1.635

From the above, Crown glass has a lower refractive index when compared to Asphalt. This means that Crown glass is less dense than Asphalt.

Therefore, total internal reflection will not occur when light travels from Crown glass to Asphalt because Crown glass is less dense than Asphalt.

7 0

3 years ago