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3 years ago

What does the potential energy diagram of a chemical reaction tell you

Physics

2 answers:

alexdok [17]3 years ago

5 0

Whether the reaction is endothermic or exothermic. Apex

sattari [20]3 years ago

4 0

By looking at the potential energies before and after the reaction, we can tell that the reaction is exothermic (final < initial) or endodermic (final > initial).

Also, the amount of activation energy gives an idea of the external energy required to initiate the reaction (for example, by heating the reactants).
Furthermore, by the same principle, we can also deduce the activation energy for the reverse reaction.

If a catalyst is available, the diagram will show a reduced activation energy, compared to a reaction without catalyst. However, it will also show that the catalyst does not alter the initial and final energies of the reaction.

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A cornea transplant involves the grafting of a donor cornea into a recipient’s anterior eye. The sutures to hold the graft in pl

Sladkaya [172]

Answer:

The cornea of an eye is avascular

Explanation:

The healing process in case of a cornea transplant is slow, because the cornea is of an eye is avascular, i.e., it does not contain blood vessels and in the absence of the blood vessels in the cornea tissue the nutrients and oxygen is not able to get delivered there.

Thus the immune system, without the blood vessels in the cornea is weak and hence healing takes time in case of cornea transplant.

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2 years ago

A uniform, spherical, 1900.0 kg shell has a radius of 5.00 m. Find the gravitational force this shell exerts on a 1.80 kg point

Mandarinka [93]

Answer:

$F=9.09\times 10^{-9} N$

Explanation:

We are given that

Mass of spherical shell, $m_1$ =1900 kg

Mass= $m_2=1.80 kg$

Radius of shell=r=5 m

Distance between two masses=r=5.01 m

Because distance measure from center .

Gravitational force

$F=G\frac{m_1m_2}{r^2}$

$G=6.67\times 10^{-11} Nm^2/kg^2$

Using the formula

$F=6.67\times 10^{-11}\times \frac{1900\times 1.80}{(5.01)^2}$

$F=9.09\times 10^{-9} N$

Hence,the gravitational force = $F=9.09\times 10^{-9} N$

6 0

2 years ago

If the sun is 400 times bigger than the moon, how couild the moon possibly cover the sun during a solar eclipse?

AlekseyPX

Explanation:

the Moon passes between Earth and the Sun Even though the Moon is much smaller than the Sun, because it is just the right distance away from Earth, the Moon can fully block the Sun's light from Earth's perspective This completely blocks out the Sun's light

3 0

2 years ago

Read 2 more answers

The experimental apparatus shown in the figure above contains a pendulum consisting of a 0.66 kg ball attached to a string of le

lara31 [8.8K]

The problem is solved and the questions are answered below.

Explanation:

a. To calculate the speed of the 0.66 kg ball just before the collision

V₀ + K₀ = V₁ + K₁

= mgh₀ = 1/2 mv₁²

where, h= r - r cosθ

V = $\sqrt{2gh}$

V = 2.42 m/s

b. Calculate the speed of the 0.22 kg ball immediately after the collision

y = y₀ + Vy₀t - 1/2 gt²

0 = 1.2 - 1/2 gt²

t = 0.495 s

x = x₀ + Vx₀t

1.4 = 0 + vx₀ (0.495)

Vx₀ = 2.83 m/s

C. To Calculate the speed of the 0.66 kg ball immediately after the collision

m₁ v₁ = m₁ v₃ + m₂ v₄

(0.66)(2.42) = (0.66) v₃ + (0.22)(2.83)

V₃ = 1.48 m/s

D. To Indicate the direction of motion of the 0.66 kg ball immediately after the collision is to the right.

E. To Calculate the height to which the 0.66 kg ball rises after the collision

V₀ + k₀ = V₁ + k₁

1/2 mv₀² = mgh₁

h₁ = v₀²/2 g

= 0.112 m

F. Based on your data, No the collision is not elastic.

Δk = 1/2 m₁v₃² =1/2 m₂v₄² - 1/2 m₁v₁²

= 1/2 (0.66)(1.48)² + 1/2 (0.22)(2.83)² - 1/2 (0.66)(2.42)²

= - 0.329 J

Hence, kinetic energy is not conserved.

8 0

2 years ago

Two uniform solid cylinders, each rotating about its central (longitudinal) axis, have the same mass of 3.56 kg and rotate with

Natali5045456 [20]

Answer:

(a) 3107.98 J

(b) 14530.6 J

Explanation:

mass, m = 3.56 kg

angular speed, ω = 179 rad/s

Moment of inertia of solid cylinder, I = 1/2 mr^2

where, m is the mass and r be the radius of the cylinder.

(a) radius, r = 0.330 m

I = 0.5 x 3.56 x 0.330 x 0.330 = 0.194 kgm^2

The formula for the rotational kinetic energy is given by

$K = \frac{1}{2}I\omega ^{2}$

K = 0.5 x 0.194 x 179 x 179 = 3107.98 J

(b) radius, r = 0.714 m

I = 0.5 x 3.56 x 0.714 x 0.714 = 0.907 kgm^2

The formula for the rotational kinetic energy is given by

$K = \frac{1}{2}I\omega ^{2}$

K = 0.5 x 0.907 x 179 x 179 = 14530.6 J

3 0

3 years ago