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yanalaym [24]
3 years ago
7

The rate constant for this second‑order reaction is 0.760 M−1⋅s−1 at 300 ∘C. A⟶products How long, in seconds, would it take for

the concentration of A to decrease from 0.750 M to 0.330 M?
Physics
1 answer:
Mashcka [7]3 years ago
3 0

Answer:

2.23 s

Explanation:

For a second-order reaction:

1 / [A] = 1 / [A]₀ + kt

Given [A] = 0.330 M, [A]₀ = 0.750 M, and k = 0.760 M⁻¹s⁻¹:

1 / 0.330 = 1 / 0.750 + 0.760t

t = 2.23

It would take 2.23 seconds.

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A. The potential energy stored in the compressed spring of a dart gun, with a spring constant of 62.00 N/m, is 0.940 J. Find by
valina [46]

a. The spring is compressed  by  0.174 m.

b.  The vertical distance the dart travels from its position when the spring is compressed to its highest position is 9.6 m.

c. The horizontal velocity of the dart at that time is  13.74 m/s.

d. The horizontal distance from the equilibrium position at which the dart hits the ground is 19.236 m.

<h3>What is potential energy?</h3>

The energy by virtue of its position is called the potential energy.

a. Given is the potential energy stored in the compressed spring of a dart gun, with a spring constant of 62.00 N/m, is 0.940 J.

PE  of spring = 1/2 kx²

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b. Given is a 0.010 kg dart is fired straight up.

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v = √2gh

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v = √2 x 9.8x 9.6

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d.  Time that dart spends in air, t = √2h/g

t = √(2x9.6)/9.81

t = 1.4 s

The horizontal distance from the equilibrium position at which the dart hits the ground.

Horizontal distance = (Velocity on x direction) x time

Horizontal distance = 13.74 m/s x 1.4s

Horizontal distance = 19.236 m

Thus, the horizontal distance is 19.236 m.

Learn more about potential energy.

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