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2 years ago

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If the activation energy for a given compound is found to be 103 kJ/mol, with a frequency factor of 4.0 × 1013 s-1, what is the

rate constant for this reaction at 398 K?

2 answers:

Crank2 years ago

7 0

Answer: The Rate constant is 1.209

Explanation:

in the attachment

sasho [114]2 years ago

5 0

Explanation:

Below is an attachment containing the solution.

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In the design of a timing mechanism, the motion of pin P in the fixed circular slot is controlled by the guide A, which is being

german

Answer: Got It!

<em>Explanation:</em> Guide A Starts From Rest With Pin P At The Lowest Point In The Circular Slot, And Accelerates Upward At A Constant Rate Until It Reaches A Speed Of 175 Mm/s At The ... In the design of a timing mechanism, the motion of pin P in the fixed circular slot is controlled by the guide A, which is being elevated by its lead screw.

6 0

2 years ago

5.16 An insulated container, filled with 10 kg of liquid water at 20 C, is fitted with a stirrer. The stirrer is made to turn by

Anna007 [38]

Answer:

a) W=2.425kJ

b) $\Delta E=2.425kJ$

c) $T_f=20.06^{o}C$

d) Q=-2.425kJ

Explanation:

a)

First of all, we need to do a drawing of what the system looks like, this will help us visualize the problem better and take the best possible approach. (see attached picture)

The problem states that this will be an ideal system. This is, there will be no friction loss and all the work done by the object is transferred to the water. Therefore, we need to calculate the work done by the object when falling those 10m. Work done is calculated by using the following formula:

$W=Fd$

Where:

W=work done [J]

F= force applied [N]

d= distance [m]

In this case since it will be a vertical movement, the force is calculated like this:

F=mg

and the distance will be the height

d=h

so the formula gets the following shape:

$W=mgh$

so now e can substitute:

$W=(25kg)(9.7 m/s^{2})(10m)$

which yields:

W=2.425kJ

b) Since all the work is tansferred to the water, then the increase in internal energy will be the same as the work done by the object, so:

$\Delta E=2.425kJ$

c) In order to find the final temperature of the water after all the energy has been transferred we can make use of the following formula:

$\Delta Q=mC_{p}(T_{f}-T_{0})$

Where:

Q= heat transferred

m=mass

$C_{p}$ =specific heat

$T_{f}$ = Final temperature.

$T_{0}$ = initial temperature.

So we can solve the forula for the final temperature so we get:

$T_{f}=\frac{\Delta Q}{mC_{p}}+T_{0}$

So now we can substitute the data we know:

$T_{f}=\frac{2 425J}{(10000g)(4.1813\frac{J}{g-C})}+20^{o}C$

Which yields:

$T_{f}=20.06^{o}C$

d)

For part d, we know that the amount of heat to be removed for the water to reach its original temperature is the same amount of energy you inputed with the difference that since the energy is being removed this means that it will be negative.

$\Delta Q=-2.425kJ$

3 0

2 years ago

a 250 mH coil of negligible resistance is connected to an AC circuit in which as effective current of 5 mA is flowing. if the fr

mash [69]

Answer:

the inductive reactance of the coil is 1335.35 Ω

Explanation:

Given;

inductance of the coil, L = 250 mH = 0.25 H

effective current through the coil, I = 5 mA

frequency of the coil, f = 850 Hz

The inductive reactance of the coil is calculated as;

$X_l = \omega L = 2\pi f L\\\\X_l = 2\pi \times 850 \times 0.25\\\\X_l = 1335.35 \ ohms$

Therefore, the inductive reactance of the coil is 1335.35 Ω

6 0

1 year ago

An automobile accelerates from zero to 30 m/s in 6.0 s. The wheels have a diameter of 0.40 m. What is the average angular accele

leva [86]

To solve this problem we will use the concepts related to angular motion equations. Therefore we will have that the angular acceleration will be equivalent to the change in the angular velocity per unit of time.

Later we will use the relationship between linear velocity, radius and angular velocity to find said angular velocity and use it in the mathematical expression of angular acceleration.

The average angular acceleration

$\alpha = \frac{\omega_f - \omega_0}{t}$

Here

$\alpha$ = Angular acceleration

$\omega_{f,i} =$ Initial and final angular velocity

There is not initial angular velocity,then

$\alpha = \frac{\omega_f}{t}$

We know that the relation between the tangential velocity with the angular velocity is given by,

$v = r\omega$

Here,

r = Radius

$\omega$ = Angular velocity,

Rearranging to find the angular velocity

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

$\omega = \frac{30}{0.20} \rightarrow$ Remember that the radius is half te diameter.

Now replacing this expression at the first equation we have,

$\alpha = \frac{30}{0.20*6}$

$\alpha = 25 rad /s^2$

Therefore teh average angular acceleration of each wheel is $25rad/s^2$

3 0

2 years ago

Physics help, please? :)

Svetllana [295]

Answer:

Explanation:

3. Newton’s third law explains how every action has an equal but opposite reaction, meaning that forces comes in pairs. While the locomotive’s wheels are pushing back against the ground as the action force, the ground is producing a reaction force towards the locomotive, propelling it forward. Another pair of forces that act on the locomotive is gravity and normal force. While gravity is pulling the locomotive towards the ground, the normal force the ground exerts on the locomotive is why the locomotive doesn’t fall through the ground.

4. The force of Earth’s gravity on the Sun is weaker than the force of the Sun’s gravity on Earth. The Sun’s attraction affects the motion of Earth more than the Earth’s attraction affects the Sun’s motion because according to Newton’s second law, force has mass as one of its factors. The Sun has a significantly higher mass than Earth, meaning that its force of gravity would also be significantly higher. Newton’s third law is why the Earth doesn’t get marginally closer to the Sun, stating that every action has an equal and opposite reaction. As the Sun is pulling Earth towards itself, Earth is pulling away from the Sun.

8 0

1 year ago