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lianna [129]
3 years ago
14

The relative density of oxygen and carbon dioxide are 16, 12 respectively. If 25cm3 of carbon dioxide effuse out in 75 sec what

volume of oxygen will effuse out in 96 sec under similar conditions?​
Physics
1 answer:
Mrac [35]3 years ago
8 0

Answer:

32 cm³

Explanation:

The given gas data are;

The relative density of oxygen = 16

The relative density of carbon dioxide = 12

The time it takes 25 cm³ of carbon dioxide to effuse out = 75 seconds'

The duration of effusion of the oxygen = 96 seconds

The rate of effusion of carbon dioxide, R1 = 25 cm³/(75 sec) = (1/3) cm³/sec

According to Graham's law of diffusion and effusion of a gas, we have;

\dfrac{Rate \ of \ effudion \ of \ gas \ 1}{Rate \ of \ effudion \ of \ gas \ 2} =\dfrac{The \ relative \ density \ of \ gas \ 2}{The \ relative \ density \ of \ gas \ 1}

Therefore, we have;

\dfrac{Rate \ of \ effudion \ of \ oxygen}{(1/3)} =\dfrac{12}{16}

The \ rate \ of \ effudion \ of \ oxygen}=\dfrac{12}{16} \times \left(\dfrac{1}{3 } \ cm^3/sec\right ) = \dfrac{1}{4} \ cm^3/sec

The volume of effusion = The rate of effusion × Time

The volume of the oxygen that will effuse in 96 seconds is given as follows;

The rate of effusion of a gas × Time

V = The rate of effusion of oxygen × Time = (1/3) cm³/sec × 96 sec = 32 cm³

The volume of oxygen that will effuse in 96 seconds, V = 32 cm³.

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4 years ago
Consider a particle of mass m which can move freely along the x axis from -a/2 to a/2, but which is strictly prohibited from bei
nalin [4]

Answer:

  φ = B sin (2π n/a   x)

Explanation:

In quantum mechanics when a particle moves freely it implies that the potential is zero (V = 0), so its wave function is

     φ = A cos kx + B sin kx

we must place the boundary conditions to determine the value of the constants A and B.

In our case we are told that the particle cannot be outside the boundary given by x = ± a / 2

therefore we must make the cosine part zero, for this the constant A = 0, the wave function remains

    φ = B sin kx

the wave vector is

      k = 2π /λ

now let's adjust the period, in the border fi = 0 therefore the sine function must be zero

         φ (a /2) = 0

          0 = A sin (2π/λ  a/2)

therefore the sine argument is

          2π /λ   a/2 = n π

          λ= a / n

we substitute

          φ = B sin (2π n/a   x)

7 0
3 years ago
charge, q1 =5.00μC, is at the origin, a second charge, q2= -3μC, is on the x-axis 0.800m from the origin. find the electric fiel
IRISSAK [1]

Answer:

Explanation:

Electric field due to charge at origin

= k Q / r²

k is a constant , Q is charge and r is distance

= 9 x 10⁹ x 5 x 10⁻⁶ / .5²

= 180 x 10³ N /C

In vector form

E₁ = 180 x 10³ j

Electric field due to q₂ charge

= 9 x 10⁹ x 3 x 10⁻⁶ /.5² + .8²

= 30.33 x 10³ N / C

It will have negative slope θ with x axis

Tan θ = .5 / √.5² + .8²

= .5 / .94

θ = 28°

E₂ = 30.33 x 10³ cos 28 i - 30.33 x 10³ sin28j

= 26.78 x 10³ i - 14.24 x 10³ j

Total electric field

E = E₁  + E₂

= 180 x 10³ j +26.78 x 10³ i - 14.24 x 10³ j

= 26.78 x 10³ i + 165.76 X 10³ j

magnitude

= √(26.78² + 165.76² ) x 10³ N /C

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3 0
4 years ago
A long coaxial cable consists of an inner cylindrical conductor with radius a and an outer coaxial cylinder with inner radius b
Tcecarenko [31]

Answer:

(a) E=λ/(2\pi e0 r)

(b) E = 0

Explanation:

(a) We can use the Gaussian's Law to calculate the electric field at any distance r from the axis. By using a cylindrical Gaussian surface we have:

\int \vec{E}\cdot d\vec{r}=\frac{\lambda}{\epsilon_o}

where  λ is the total charge per unit length inside the Gaussian surface. In this case we have that the Electric field vector is perpendicular to the r vector. Hence:

E\int dr=E2\pi r=\frac{\lambda}{\epsilon_o}\\\\E=\frac{\lambda}{2\pi r \epsilon_o}

(b) outside of the outer cylinder there is no net charge inside the Gaussian surface, because charge of the inner radius cancel out with the inner surface of the cylindrical conductor.

Hence, we  have that E is zero.

hope this helps!!

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