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Strike441 [17]
3 years ago
10

How many grams of nickel metal are plated out when a constant current of 15.0 Ais passed through aqueous NICl2 for 80.0 minutes?

Physics
1 answer:
Alika [10]3 years ago
3 0

Answer:

21.89 g

Explanation:

i = 15 A

time = 80 minutes = 80 x 60 = 4800 second

Electro chemical equivalent of nickel, Z = 0.000304 gram per ampere second

Use the Faraday's law of electrolysis

m = z i t

where, m is the mass deposited, i be the current, z be the electro chemical equivalent and t be the time taken

m = 0.000304 x 15 x 4800

m = 21.89 g

Thus, the mass f nickel deposited is 21.89 g.

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Later research indicated that electric current is actually the flow of ____
iren [92.7K]

Answer:

Electrons.

Explanation:

Electricity was discovered before the discovery of electrons by J.J Thompson in 1896. Before the electron, it was thought that it is the positive ions that move through the wire and carry current—that's why today the conventional current represents the flow of positive charges.

After J.J Thompson's discovery of the electrons, it was realized that it is the electrons that actually carry the current through the conductor. But changing the direction of the conventional current didn't seem appropriate, and that's why the convention continues to be used to this day—reminding us that once it were the positive ions that were thought to carry the current.

6 0
3 years ago
And 8 kg bowling ball is rolling along the frictionless alley
VLD [36.1K]
It will stop eventually
8 0
3 years ago
A(n) 1.3 kg mass sliding on a frictionless surface has a velocity of 7.1 m/s east when it undergoes a one-dimensional elastic co
Oxana [17]

Answer: 2.12 kg

Explanation:

Since the 1.3 kg object moves to the west after the collision, the other object will move to the east after the collision.

In an elastic collision, the relative velocity after the collision is the opposite of the relative velocity before the collision. Since the 1.3 kg object’s velocity before the collision is 6.7 m/s greater than the other object, after the collision, its velocity will be 6.7 m/s less than the other object. To determine the other object’s velocity, use the following equation.

v = 1.7 – 7.1 = -5.4 m/s

The negative sign means it is moving eastward. Let’s use this number is a momentum equation to determine its mass.

Initial momentum = 1.3 * 7.1 = 9.23 east

For the 1.3 object, final momentum = 1.3 * 1.7 = 2.21 west

To determine the final momentum of the other object, add these two numbers.

Final momentum = 11.44 east

To determine its mass, use the following equation.

m * 5.4 = 11.44

m = 11.44 ÷ 5.4 = 2.12 kg

To make sure that kinetic energy is conserved, let’s round this number to 2 kg and determine the final kinetic energies.

For the 1.3 object, KE = 1/2 * 1/3* 1.7^2 = 0.48

For the 2 kg object, KE = 1/2* 2 * 5.4^2 = 29.64

Total final KE = 29.64

Initial KE = 0.5* 1.3 * 7.1^2 = 32.77

Since I rounded the mass up to 2kg, this proves that kinetic energy is conserved and the mass is correct!

3 0
3 years ago
Given the frequency of an electromagnetic wave, what else can you find immediately?
Oksana_A [137]

Answer:

wavelength

Explanation:

An electromagnetic waves is produced due to interaction of oscillating electric and magnetic field when they interacts perpendicular to each other. For example, Gamma rays, X rays , ultraviolet rays, visible radiations, infrared rays, micro waves, radio waves.

All the electromagnetic waves have velocity equal to velocity of light.

If the frequency of electromagnetic wave is unknown then we find the wavelength of wave. the formula used is

Wave speed = wavelength x frequency

3 0
3 years ago
Calculate the acceleration of gravity as a function of depth in the earth (assume it is a sphere). You may use an average densit
Ber [7]

Solution :

Acceleration due to gravity of the earth, g $=\frac{GM}{R^2}$

$g=\frac{G(4/3 \pi R^2 \rho)}{R^2}=G(4/3 \pi R \rho)$

Acceleration due to gravity at 1000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-1000) \times 5.5 \times 10^3\right)$

  $= 822486 \times 10^{-8}$

  $=0.822 \times 10^{-2} \ km/s$

 = 8.23 m/s

Acceleration due to gravity at 2000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-2000) \times 5.5 \times 10^3\right)$

  $= 673552 \times 10^{-8}$

  $=0.673 \times 10^{-2} \ km/s$

 = 6.73 m/s

Acceleration due to gravity at 3000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-3000) \times 5.5 \times 10^3\right)$

  $= 3371 \times 153.86 \times 10^{-8}$

  = 5.18 m/s

Acceleration due to gravity at 4000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-4000) \times 5.5 \times 10^3\right)$

  $= 153.84 \times 2371 \times 10^{-8}$

  $=0.364 \times 10^{-2} \ km/s$

 = 3.64 m/s

       

3 0
3 years ago
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