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4vir4ik [10]
3 years ago
6

I know the enthalpy of a reaction is 23kj/mol. Initially the reaction is taking place at 273 k. To what temperature do i need to

heat the reaction in order to double the equilibrium constant?

Physics
1 answer:
Vladimir79 [104]3 years ago
7 0

Answer:

293k

Explanation:

In this question, we are asked to calculate the temperature to which the reaction must be heated to double the equilibrium constant.

To find this value, we will need to use the Van’t Hoff equation.

Please check attachment for complete solution

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Answer: Your answer is<u> 1.36.</u>

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5 0
3 years ago
A 97.3 kg horizontal circular platform rotates freely with no friction about its center at an initial angular velocity of 1.67 r
Yuliya22 [10]

Answer:

w = 1.14 rad / s

Explanation:

This is an angular momentum exercise. Let's define a system formed by the three bodies, the platform, the bananas and the monkey, in such a way that the torques during the collision have been internal and the angular momentum is preserved.

Initial instant. The platform alone

        L₀ = I w₀

Final moment. When the bananas are on the shelf

we approximate the bananas as a point load and the distance is indicated

x = 0.45m

        L_f = (m x² + I ) w₁

angular momentum is conserved

         L₀ = L_f

         I w₀ = (m x² + I) w₁

         w₁ = \frac{I}{m x^2 + I}  \ w_o

Let's repeat for the platform with the bananas and the monkey is the one that falls for x₂ = 1.73 m

initial instant. The platform and bananas alone

        L₀ = I₁ w₁

         I₁ = (m x² + I)

           

final instant. After the crash

        L_f = I w

        L_f = (I₁ + M x₂²) w

the moment is preserved

        L₀ = L_f

        (m x² + I) w₁ = ((m x² + I) + M x₂²) w

         (m x² + I) w₁ = (I + m x² + M x₂²) w

we substitute

         w =  \frac{m x^2 +I}{I + m x^2 + M x_2^2} \ \frac{I}{m x^2 + I} \ w_o

         w =  \frac{I}{I + m x^2 + M x_2^2} \ w_o

the moment of inertia of a circular disk is

         I = ½ m_p x₂²

we substitute

         w =  \frac{ \frac{1}{2} m_p x_2^2 }{ \frac{1}{2} m_p x_2^2 + M x_2^2 + m x^2} \ \ w_o

let's calculate

          w =\frac{ \frac{1}{2} \ 97.3 \ 1.73^2 }{ \frac{1}{2} \ 97.3 \ 1.73^2 + 21.9 \ 1.73^2 + 9.67 \ 0.45^2 } \ \ 1.67

          w =  \frac{145.60 }{145.60 \ + 65.54 \ + 1.958} \ \ 1.67

          w = 1.14 rad / s

5 0
3 years ago
Use dimensional analysis to determine how the linear acceleration a in m/s2 of a particle traveling in a circle depends on some,
astraxan [27]

I have to say, i love this kind of problems.

So, we got the linear acceleration, as we all know, linear the acceleration its, in dimensional units:

[a]=[\frac{distance}{time^2}].

Now, we got the radius

[r] = [distance]

the angular frequency

[\omega] = \frac{1}{s}

and the mass

[m]=[mass].

Now, the acceleration doesn't have units of mass, so it can't depend on the mass of the particle.

The distance in the acceleration has exponent 1, and so does in the radius. As the radius its the only parameter that has units of distance, this means that the radius must appear with exponent 1. Lets write

a \propto r.

The time in the acceleration has exponent -2 As the angular frequency its the only parameter that has units of time, this means that the angular frequency must appear, but, the angular frequency has an exponent of -1, this means it must be squared

a \propto r \omega^2.

We are almost there. If this were any other problem, we would write:

a = A r \omega^2

where A its an dimensionless constant. Its common for this constants to appears if we need an conversion factor. If we wanted the acceleration in cm/s^2, for example. Luckily for us, the problem states that there is no dimensionless constant involved, so:

a = r \omega^2

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3 years ago
You toss a coin into a eishimg well full of a liquid denser than the coin which of the following could be true the coin will
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Then the coin will float on the surface of the liquid in the eishimg well.
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4 years ago
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