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Elenna [48]
3 years ago
14

The first-order decomposition of cyclopropane has a rate constant of 6.7 × 10–4 s–1. If the initial concentration of cyclopropan

e is 1.33 M, what is the concentration of cyclopropane after 644 s?
Physics
1 answer:
Ket [755]3 years ago
7 0

Answer:

.864 M

Explanation:

For first order decomposition,

rate constant k = 1/t x ln a / (a - x )

given , a = 1.33 M , t = 644 s , k = 6.7 x 10⁻⁴ , a - x  = ? = b( let )

6.7 x 10 ⁻⁴ = 1/644 x ln 1.33/b

ln 1.33/b = 6.7 x 10⁻⁴ x 644 = .4315

1.33 / b = e⁰ ⁴³¹⁵ = 1.5395

b = 1.33 / 1.5395 = .864 M.

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Complete Question

The complete question shown on the first uploaded image

Answer:

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The force on Q due to dipole is Attractive

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The charge Q exerts attractive force on the dipole

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2 years ago
What do you think will happen to Charlie now that he is smart? Explain.
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A parallel-plate capacitor initially has a potential difference of 400 V and is then disconnected from the charging battery. If
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The capacitance of a capacitor is the ratio of the stored charge to its potential difference, i.e.

C = Q/ΔV

C is the capacitance

Q is the stored charge

ΔV is the potential difference

Rearrange the equation:

ΔV = Q/C

We also know the capacitance of a parallel-plate capacitor is given by:

C = κε₀A/d

C is the capacitance

κ is the capacitor's dielectric constant

ε₀ is the electric constant

A is the area of the plates

d is the plate separation

If we substitute C:

ΔV = Qd/(κε₀A)

We assume the stored charge and the area of the plates don't change. Then if we double the plate spacing, i.e. we double the value of d, then the potential difference ΔV is also doubled.

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3 years ago
un avión aterriza en la superficie de un portaaviones a 50 m/s y se detiene por completo en 120 metros, ¿cuál es la aceleración
sergejj [24]

Answer:

La aceleración necesaria para detener el avión es - 10.42 m/s².

Explanation:

Un movimiento uniformemente acelerado (M.U.A) es aquél cuya aceleración es constante y la velocidad de un objeto cambia a medida que el movimiento evoluciona.

Siendo la aceleración "a" el cambio de velocidad al tiempo transcurrido en un punto A a B, la velocidad inicial la velocidad que tiene un cuerpo al iniciar su movimiento en un período de tiempo y la velocidad final la velocidad que tiene un cuerpo al finalizar su movimiento en un período de tiempo, entonces en  M.U.A se cumple:

Vf² - Vo² = 2*a*d

donde:

  • Vf: Velocidad final
  • Vo: Velocidad inicial
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  • d: Distancia recorrida

En este  caso:

  • Vf: 0 m/s, porque el avión se detiene
  • Vo: 50 m/s
  • a: ?
  • d: 120 m

Reemplazando:

(0 m/s)² - (50 m/s)² = 2*a*120 m

Resolviendo:

a=\frac{(0 m/s)^{2} -(50 m/s)^{2} }{2*120 m}

a= - 10.42 m/s²

<u><em>La aceleración necesaria para detener el avión es - 10.42 m/s².</em></u>

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3 years ago
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