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Varvara68 [4.7K]
3 years ago
5

Electrolysis is a technique used to separate compounds, not mixtures. What has to happen in order to separate a compound? Is it

easier or harder to separate a compound than a mixture?
Physics
1 answer:
wel3 years ago
6 0

Answer:

In order to predict the products of electrolysis, we first need to understand what electrolysis is and how it works. Electrolysis is a method of separating bonded elements and compounds by passing an electric current through them. It uses a direct electric current (DC) to drive an otherwise non-spontaneous chemical reaction. Electrolysis is very important commercially as a stage in the separation of elements from naturally occurring sources, such as ores, using an electrolytic cell.

The main components required to achieve electrolysis are:

An electrolyte: a substance containing free ions, which are the carriers of electric current in the electrolyte. If the ions are not mobile, as in a solid salt, then electrolysis cannot occur.

A direct current (DC) supply: provides the energy necessary to create or discharge the ions in the electrolyte. Electric current is carried by electrons in the external circuit.

Two electrodes: an electrical conductor that provides the physical interface between the electrical circuit providing the energy and the electrolyte.

The Interchange of Atoms and Ions

The key process of electrolysis is the interchange of atoms and ions by the removal or addition of electrons to the external circuit. The required products of electrolysis are in a different physical state from the electrolyte and can be removed by some physical processes.

Each electrode attracts ions that are of the opposite charge. Positively charged ions, or cations, move toward the electron-providing cathode, which is negative; negatively charged ions, or anions, move toward the positive anode. You may have noticed that this is the opposite of a galvanic cell, where the anode is negative and the cathode is positive.

At the electrodes, electrons are absorbed or released by the atoms and ions. Those atoms that gain or lose electrons become charged ions that pass into the electrolyte. Those ions that gain or lose electrons to become uncharged atoms separate from the electrolyte. The formation of uncharged atoms from ions is called discharging. The energy required to cause the ions to migrate to the electrodes, and the energy to cause the change in ionic state, is provided by the external source.

Oxidation and Reduction

Oxidation of ions or neutral molecules occurs at the anode, and reduction of ions or neutral molecules occurs at the cathode. For example, it is possible to oxidize ferrous ions to ferric ions at the anode

It is also possible to reduce ferricyanide ions to ferrocyanide ions at the cathode

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What does it mean for waves to be polarized?
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I throw a football in the x direction with a fore of 124N. The football is 4 kilograms. The football accelerated for 5 seconds.
castortr0y [4]

31 (unit)

acceleration formula

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Consider a car travelling at 60 km/hr. If the radius of a tire is 25 cm, calculate the angular speed of a point on the outer edg
vlabodo [156]

To solve this problem it is necessary to apply the concepts given in the kinematic equations of movement description.

From the perspective of angular movement, we find the relationship with the tangential movement of velocity through

\omega = \frac{v}{R}

Where,

\omega =Angular velocity

v = Lineal Velocity

R = Radius

At the same time we know that the acceleration is given as the change of speed in a fraction of the time, that is

\alpha = \frac{\omega}{t}

Where

\alpha =Angular acceleration

\omega = Angular velocity

t = Time

Our values are

v = 60\frac{km}{h} (\frac{1h}{3600s})(\frac{1000m}{1km})

v = 16.67m/s

r = 0.25m

t=6s

Replacing at the previous equation we have that the angular velocity is

\omega = \frac{v}{R}

\omega = \frac{ 16.67}{0.25}

\omega = 66.67rad/s

Therefore the angular speed of a point on the outer edge of the tires is 66.67rad/s

At the same time the angular acceleration would be

\alpha = \frac{\omega}{t}

\alpha = \frac{66.67}{6}

\alpha = 11.11rad/s^2

Therefore the angular acceleration of a point on the outer edge of the tires is 11.11rad/s^2

5 0
3 years ago
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