Ask question

Ask question

All categories

3 years ago

6

For a particular reaction, the change in enthalpy is 51kJmole and the activation energy is 109kJmole. Which of the following cou

ld NOT be the change in enthalpy (ΔH) and the activation energy (Ea), respectively, for the catalyzed reaction? Select all that apply.

1 answer:

Ronch [10]3 years ago

5 0

Answer

given,

change in enthalpy = 51 kJ/mole

change in activation energy = 109 kJ/mole

when a reaction is catalysed change in enthalpy between the product and the reactant does not change it remain constant.

where as activation energy of the product and the reactant decreases.

example:

ΔH = 51 kJ/mole

E_a= 83 kJ/mole

here activation energy decrease whereas change in enthalpy remains same.

You might be interested in

Two identical metal balls a and b are mounted on insulating rods. Ball a has a charge of +q/2 and ball b is initially uncharged.

oksano4ka [1.4K]

Answer:

Help me please?

Explanation:

Did you get the answer? I believe it’s either C. +q or D. 0

6 0

3 years ago

Read 2 more answers

a ray of light strikes a plane mirror at an angle of 53 degrees to the normal. what is the angle of reflection?

Alexxx [7]

So for an incident rag from vacuum to a medium with n=1.6, and an incident angle of 53 degrees, the the angle of refraction will be .

4 0

3 years ago

Three ideal polarizing filters are stacked, with the polarizing axis of the second and third filters at 21 degrees and 61 degree

kvv77 [185]

Answer:

1

When second polarizer is removed the intensity after it passes through the stack is

$I_f_3 = 27.57 W/cm^2$

2 When third polarizer is removed the intensity after it passes through the stack is

$I_f_2 = 102.24 W/cm^2$

Explanation:

From the question we are told that

The angle of the second polarizing to the first is $\theta_2 = 21^o$

The angle of the third polarizing to the first is $\theta_3 = 61^o$

The unpolarized light after it pass through the polarizing stack $I_u = 60 W/cm^2$

Let the initial intensity of the beam of light before polarization be $I_p$

Generally when the unpolarized light passes through the first polarizing filter the intensity of light that emerges is mathematically evaluated as

$I_1 = \frac{I_p}{2}$

Now according to Malus’ law the intensity of light that would emerge from the second polarizing filter is mathematically represented as

$I_2 = I_1 cos^2 \theta_1$

$= \frac{I_p}{2} cos ^2 \theta_1$

The intensity of light that will emerge from the third filter is mathematically represented as

$I_3 = I_2 cos^2(\theta_2 - \theta_1 )$

$I_3= \frac{I_p}{2}(cos^2 \theta_1)[cos^2(\theta_2 - \theta_1)]$

making $I_p$ the subject of the formula

$I_p = \frac{2L_3}{(cos^2 \theta [cos^2 (\theta_2 - \theta_1)])}$

Note that $I_u = I_3$ as $I_3$ is the last emerging intensity of light after it has pass through the polarizing stack

Substituting values

$I_p = \frac{2 * 60 }{(cos^2(21) [cos^2 (61-21)])}$

$I_p = \frac{2 * 60 }{(cos^2(21) [cos^2 (40)])}$

$=234.622W/cm^2$

When the second is removed the third polarizer becomes the second and final polarizer so the intensity of light would be mathematically evaluated as

$I_f_3 = \frac{I_p}{2} cos ^2 \theta_2$

$I_f_3$ is the intensity of the light emerging from the stack

substituting values

$I_f_3 = \frac{234.622}{2} * cos^2(61)$

$I_f_3 = 27.57 W/cm^2$

When the third polarizer is removed the second polarizer becomes the

the final polarizer and the intensity of light emerging from the stack would be

$I_f_2 = \frac{I_p}{2} cos ^2 \theta_1$

$I_f_2$ is the intensity of the light emerging from the stack

Substituting values

$I_f_2 = \frac{234.622}{2} cos^2 (21)$

$I_f_2 = 102.24 W/cm^2$

7 0

3 years ago

Please help! If the instantaneous speed of an object remains constant, can its instantaneous velocity change? If the instantaneo

kondaur [170]

<span>
I think the the answer would be no. </span> If the instantaneous speed of an object remains constant, then its instantaneous velocity would not change. S<span>ince v = c for c is some constant, then |v| = |c|. This is regardless of the sign of c. Hope this answers the question.</span>

4 0

4 years ago

Match the measurements with the proper SI unit.

Ivan

Answer:

Acceleration:

C. Meters per second squared

Velocity:

B. Meters per second

Distance:

A. Meters

Explanation:

We must remember that the international system of measures (SI) takes into account for the length as the main unit the meter, for the mass the kilogram, for the time the second.

The acceleration is calculated using the following expression

a = v/t = (m/s/s) = (m/s^2]

The velocity is calculated using the following expression

v = x/t = (m)/(s) = (m/s)

The distance for the SI system is given in meters

7 0

3 years ago

Read 2 more answers