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Olegator [25]
3 years ago
15

The average kinetic energy of uf6(g) is 4.98 kj/mol at some temperature. what is the average kinetic energy of sf6(g) at that sa

me temperature?
Physics
2 answers:
pashok25 [27]3 years ago
6 0

Answer:

The average kinetic energy of SF₆(g) is 4.98 kj/mol.

Explanation:

Given that,

Kinetic energy  of UF₆(g)=4.98\ kj/mol

We need to calculate the average kinetic energy of SF₆(g)

Using formula of average kinetic energy

E_{avg}=\dfrac{3}{2}kT

E_{avg}=\dfrac{3}{2}\dfrac{RT}{N_{A}}

Where, k = Boltzmann constant

N_{A} =Avogadro number

R =gas constant

Here, average kinetic energy depends only on temperature not on special molar mass

So, The SF₆(g) average kinetic energy is same as UF₆(g

Hence, The average kinetic energy of SF₆(g) is 4.98 kj/mol.

Darya [45]3 years ago
5 0
4.98 kJ/mol would be the answer I believe :) hope this helped
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A pole AB of length 10.0m and weight 600N has its center of gravity 4.0m from the end A, and lies on horizontal ground .Calculat
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Answer:

The force required to begin to lift the pole from the end 'A' is 240 N

Explanation:

The given parameters for the pole AB are;

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The weight of the pole, W = 600 N ↓

The distance of the center of gravity of the pole from the side 'A' = 4.0 m

Let 'F_A' represent the force required to begin to lift the pole from the end 'A' and let a force applied in the upwards direction be positive

For equilibrium, the sum of moment about the point 'B' = 0, therefore, taking moment about 'B', we have

F_A × 10.0 m - W × 4.0 m = 0

∴ F_A × 10.0 m = W × 4.0 m = 600 N × 4.0 m

F_A × 10.0 m = 600 N × 4.0 m

∴  F_A = 600 N × 4.0 m/(10.0 m) = 240 N

The force required to begin to lift the pole from the end 'A', F_A = 240 N.

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Read 2 more answers
You need to focus a 10 mW, 632.8 nm Gaussian laser beam that is 5.0 mm in diameter into a sample. You have access to a lens with
Anna [14]

Answer:

ee that the lens with the shortest focal length has a smaller object

           

Explanation:

For this exercise we use the constructor equation or Gaussian equation

        \frac{1}{f}  = \frac{1}{p} + \frac{1}{q}

where f is the focal length, p and q are the distance to the object and the image respectively.

Magnification a lens system is

          m = \frac{h'}{h} = - \frac{q}{p}

             h ’= -\frac{h q}{p}

In the exercise give the value of the height of the object h = 0.50cm and the position of the object p =∞

Let's calculate the distance to the image for each lens

f = 6.0 cm

           \frac{1}{q} = \frac{1}{f }  - \frac{1}{p}

as they indicate that the light fills the entire lens, this indicates that the object is at infinity, remember that the light of the laser rays is almost parallel, therefore p = inf

          q = f = 6.0 cm

for the lens of f = 12.0 cm q = 12.0 cn

to find the size of the image we use

           h ’= h q / p

where p has a high value and is the same for all systems

           h ’= h / p q

Thus

f = 6 cm h ’= fo 6 cm

 

f = 12 cm h ’= fo 12  cm

therefore we see that the lens with the shortest focal length has a smaller object

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