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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?

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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A piano wire with mass 2.60g and length 84.0 cm is stretched with a tension of 25.0 N. A wave with frequency 120.0 Hz and amplit

likoan [24]

Answer:

Power will be 0.2023 watt

And when amplitude is halved then power will be 0.0505 watt

Explanation:

We have given mass of the Piano wire m = 2.60 gram = 0.0026 kg

Length of wire l = 84 cm = 0.84 m

So mass density $\mu =\frac{m}{l}=\frac{0.0026}{0.84}=0.0031kg/m$

Tension in the wire T = 25 N

Frequency f = 120 Hz

So angular frequency $\omega =2\pi f=2\times 3.14\times 120=753.6rad/sec$

And amplitude A = 1.6 mm = 0.0016 m

We have to find the generated power

Power is given by $P=\frac{1}{2}\sqrt{\mu T}\omega ^2A^2=\frac{1}{2}\times \sqrt{0.0031\times 25}\times 753.6^2\times 0.0016^2=0.2023watt$

From the relation we can see that power $P\ \propto\ A^2$

So if amplitude is halved then power will be $\frac{1}{4}$ times

So power will be equal to $\frac{0.2023}{2}=0.0505watt$

4 0

2 years ago

1. Indica cuáles son los reactivos y productos de la reacción química representada por la ecuación siguiente, y señala en que es

bearhunter [10]

Answer:

Los reactivos, con su correspondiente estado de agregación, son:

Mg: estado sólido
HCl: estado acuoso

Los productos, con su correspondiente estado de agregación, son:

MgCl₂: estado acuoso
H₂: estado gaseoso

Explanation:

Una reacción química es un proceso por el cual una o más sustancias se transforman en otra u otras sustancias con propiedades diferentes. En otras palabras, las reacciones químicas son procesos en los que los átomos rompen sus enlaces químicos y forman otros nuevos, cambiando la naturaleza de las sustancias que intervienen.

En una ecuación química se representan dos términos. En el primero, el de la derecha, se escriben los reactivos, expresados mediante sus fórmulas químicas correspondientes y separados por un signo más. Por otro lado, el término de la izquierda, en el que aparecen los productos, también representados por sus formulas químicas correspondientes y con signos más entre ellos. Entre ambos términos suele ponerse una flecha que indica que se ha producido la reacción química correspondiente.

Las sustancias que hay antes de producirse el cambio y que desaparecen se llaman reactivos. Las sustancias que hay después de producirse el cambio y que aparecen o se generan se llaman productos.

El estado de agregación en una reacción química se indica poniendo detrás del nombre o la fórmula de la sustancia y entre paréntesis la notación siguiente:

Sustancias sólidas (s)
Sustancias líquidas (l)
Sustancias gaseosas (g)
Sustancias disueltas en agua (aq)

En este caso tienes la reacción química:

Mg (s) + 2 HCl (aq) -> MgCl₂ (aq) + H₂ (g)

Entonces, <u><em>en este caso los reactivos, con su correspondiente estado de agregación, son:</em></u>

<u><em>Mg: estado sólido</em></u>
<u><em>HCl: estado acuoso</em></u>

<u><em>Por otro lado, los productos, con su correspondiente estado de agregación, son:</em></u>

<u><em>MgCl₂: estado acuoso</em></u>
<u><em>H₂: estado gaseoso</em></u>

6 0

2 years ago

A wave has a wavelength of 9 mm and a frequency of 14 hertz. What is its speed?

lesantik [10]

Wavespeed = frequency x wavelength
= 14 x 9
= 126 mm/s
= 0.126 m/s

5 0

2 years ago

Red blood cells can be modeled as spheres of 6.53 μm diameter with −2.55×10−12 C excess charge uniformly distributed over the su

yKpoI14uk [10]

Complete Question

Red blood cells can be modeled as spheres of 6.53 μm diameter with −2.55×10−12 C excess charge uniformly distributed over the surface. Find the electric field at the following locations, with radially outward defined as the positive direction and radially inward defined as the negative direction. The permittivity of free space ????0 is 8.85×10−12 C/(V⋅m). What is the electric field

E⃗ 1 inside the cell at a distance of 3.05 μm from the center?

E⃗ 2 Just inside the surface of the cell

E⃗ 3 Just outside the surface of the cell

E⃗ 4 At a point outside the cell 3.05 μm from the surface

Answer:

E⃗ 1

0 V/m

E⃗ 2

0 V/m

E⃗ 3

$E_3 = 2.153 *10^{9} \ V/m$

E⃗ 4

$E_4 = 5.754 *10^ {8} \ V/m$

Explanation:

From the question we are told that

The diameter is $d = 6.53 \mu m = 6.53*10^{-6}\ m$

The charge is $Q = -.2.55 *10^{-12} \ C$

The permittivity of free space is $\epsilon_o = 8.85* 10^{-12}\ C / V.m$

The distance considered is $d = 3.05 \mu m = 3.05 *10^{-6} \ m$

Generally the electric field inside the cell at a distance of 3.05 μm from the center is

0 V/m

This because there is no electric field felt inside the cell according Gauss the cell is taken as a point charge

Generally the electric field just inside the surface of the cell is 0 V/m

This because there is no electric field felt inside the cell according Gauss the cell is taken as a point charge

Generally the electric field just outside the cell is mathematically represented as

$E_3 = \frac{ k * |Q|}{ r^2 }$

Here $k$ is the coulomb constant with value

$k = 9*10^{9}\ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}$

r is the radius of the sphere which is mathematically as

$r = \frac{d}{2} = \frac{6.53*10^{-6}}{2} = 3.265 *10^{-6} \ m$

$E_3 = \frac{ 9*10^{9} * |-2.55 *10^{-12} |}{ [3.265 *10^{-6} ]^2 }$

$E_3 = 2.153 *10^{9} \ V/m$

Generally the electric field at a point outside the cell 3.05 μm from the surface is mathematically represented as

$E_4 = \frac{ k * |Q|}{ R^2 }$

Here R is mathematically represented as

$R = 3.265 *10^{-6} + 3.05 *10^{-6}$

=> $R = 6.315 *10^{-6}$

So

$E_4 = \frac{ 9*10^{9} * |-2.55 *10^{-12} |}{ [ 6.315 *10^{-6} ]^2 }$

$E_4 = 5.754 *10^ {8} \ V/m$

3 0

2 years ago

Displacement is which of the following types of quantities?A. vectorB. scalarC. magnitudeD. dimensional

agasfer [191]

Answer: A. vector

Explanation:

Movement is the change of position of a body at a certain time. The body that experiences this change of position is called mobile.

In this sense, the <u>displacement</u> is the distance in a straight line between the initial and final position of this mobile, which differentiates it from <u>the trajectory, </u> which is the <u>path followed by the body in its movement and a scalar magnitude.</u>

Hence, the displacement is a vector.

6 0

2 years ago