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2 years ago

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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]2 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.

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Ryan and Becca are moving a folding table out of the sunlight. A cup of lemonade, with the message 0.44 kg is on the table. Becc

jarptica [38.1K]

Calculate the weight of the table through the equation,

W = mg

where W is the weight, m is the mass, and g is the acceleration due to gravity. Substituting the known values,
W = (0.44 kg)(9.8 m/s²)
<em>W = 4.312 N</em>

The components of this weight can be calculated through the equation,

Wx = W(sin θ)

and Wy = W(cos θ)

x - component:
Wx = W(sin θ)
Substituting,
Wx = (4.312 N)(sin 150°) = <em>2.156 N</em>

Wy = (4.312 N)(cos 150°) =<em> -3.734 N</em>

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2 years ago

Two identical conducting spheres, A and B, sit atop insulating stands. When they are touched, 1.51 × 1013 electrons flow from sp

erma4kov [3.2K]

Answer:

A = -0.576 μC

B = 4.256 μC

Explanation:

Suppose a single electron charge is $1.6\times10^{-19}C$ . Then the total charge that is flowing from B to A is:

$1.6\times10^{-19} * 1.51 \times 10^{13} = 2.416\times10^{-6}C = 2.416 \mu C$

Let A and B be the initial charge of spheres A and B, respectively. Since the net charge is 3.68μC we have the following equation

$A + B = 3.68$ (1)

When they touch 2.416μC flows from B to A, then they are equal, so we have the following equation

$A + 2.416 = B - 2.416$

$-A + B = 2.416 + 2.416 = 4.832$ (2)

Add equation (1) to equation (2) we have

$2B = 3.68 + 4.832 = 8.512$

$B = 8.512 / 2 = 4.256 \mu C$

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2 years ago

Let’s consider tunneling of an electron outside of a potential well. The formula for the transmission coefficient is T \simeq e^

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Answer:

L' = 1.231L

Explanation:

The transmission coefficient, in a tunneling process in which an electron is involved, can be approximated to the following expression:

$T \approx e^{-2CL}$

L: width of the barrier

C: constant that includes particle energy and barrier height

You have that the transmission coefficient for a specific value of L is T = 0.050. Furthermore, you have that for a new value of the width of the barrier, let's say, L', the value of the transmission coefficient is T'=0.025.

To find the new value of the L' you can write down both situation for T and T', as in the following:

$0.050=e^{-2CL}\ \ \ \ (1)\\\\0.025=e^{-2CL'}\ \ \ \ (2)$

Next, by properties of logarithms, you can apply Ln to both equations (1) and (2):

$ln(0.050)=ln(e^{-2CL})=-2CL\ \ \ \ (3)\\\\ln(0.025)=ln(e^{-2CL'})=-2CL'\ \ \ \ (4)$

Next, you divide the equation (3) into (4), and finally, you solve for L':

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8 0

2 years ago

A helicopter flies 250 km on a straight path in a direction 60° south of east. The east component of the helicopter’s displaceme

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East component = 125 km

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2 years ago

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A box rests on the (horizontal) back of a truck. The coefficient of static friction between the box and the surface on which it

vredina [299]

Answer:

The distance is 11 m.

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Given that,

Friction coefficient = 0.24

Time = 3.0 s

Initial velocity = 0

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Using newton's second law

$F = ma$ ...(I)

Using formula of friction force

$F= \mu m g$ ....(II)

Put the value of F in the equation (II) from equation (I)

$ma=\mu mg$ ....(III)

$a = \mu g$

Put the value in the equation (III)

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We need to calculate the distance,

Using equation of motion

$s = ut+\dfrac{1}{2}at^2$

$s=0+\dfrac{1}{2}2.352\times(3.0)^2$

$s=10.584\ m\ approx\ 11\ m$

Hence, The distance is 11 m.

3 0

2 years ago