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siniylev [52]
3 years ago
6

Why is a 9-v transistor battery used as the source of electrical power for electrolysis rather than the power supply that plugs

into the wall electrical current?
Chemistry
1 answer:
deff fn [24]3 years ago
8 0

Answer:

The 9-v transistor battery is a direct current source, while the power supply from the wall socket is an alternating current source.

Explanation:

A direct current (DC) is one that flows continuously without any minimum and maximum values. This is majorly obtained from generators, batteries or cells etc.

An alternating current (AC) is a type that flows in a sinusoidal manner, having minimum and maximum value with respect to its flowing frequency. An example is the one obtained from the wall electrical source , generators etc.

An alternating current is not mainly used for electrolysis because of its nature. It would cause a switch of the electrodes in every cycle. Therefore, a direct current source is used.

The 9-v transistor battery is a direct current source, while the power supply from the wall is an alternating current source.

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Use the formation reactions below such that when added together, they match the balanced equation for the combustion of methane.
muminat

Answer:

ΔH of the reaction is -802.3kJ.

Explanation:

Using Hess's law, you can know ΔH of reaction by the sum of ΔH's of half-reactions.

Using the reactions:

<em>(1) </em>Cgraphite(s)+ 2H₂(g) → CH₄(g) ΔH₁ = −74.80kJ

<em>(2) </em>Cgraphite(s)+ O₂(g) → CO₂(g) ΔH₂ = −393.5k J

<em>(3) </em>H₂(g) + 1/2 O₂(g) → H₂O(g) ΔH₃ = −241.80kJ

The sum of (2) - (1) produce:

CH₄(g) + O₂(g) → CO₂(g) + 2H₂(g) ΔH' = -393.5kJ - (-74.80kJ) = -318.7kJ

And the sum of this reaction with 2×(3) produce:

CH₄(g) + 2 O₂(g) → CO₂(g) + 2H₂O(g) And ΔH = -318.7kJ + 2×(-241.80kJ) =

<em>-802.3kJ</em>

7 0
3 years ago
What unusual thing can the bull shark do?
kotegsom [21]
Bull sharks have the unique ability of keeping salt in their bodies even freshwater
3 0
3 years ago
Explain the difference in the boiling point of 2-methylpropane and 2-iodo-2-methylpropane in terms of both molecular polarity an
Len [333]
The Boiling Point of 2-methylpropane is approximately -11.7 °C, while, Boiling Point of <span>2-iodo-2-methylpropane is approximately 100 </span>°C.

As both compounds are Non-polar in nature, So there will be no dipole-dipole interactions between the molecules of said compounds.

The Interactions found in these compounds are London Dispersion Forces.

And among several factors at which London Dispersion Forces depends, one is the size of molecule.

Size of Molecule:
                          There is direct relation between size of molecule and London Dispersion forces. So, 2-iodo-2-methylpropane containing large atom (i.e. Iodine) experience greater interactions. So, due to greater interactions 2-iodo-2-methylpropane need more energy to separate from its partner molecules, Hence, high temperature is required to boil them.
6 0
3 years ago
Electrons in conductors, like copper wire, are free to roam throughout the metal in the presence of the ions that have released
Scrat [10]

Answer:

  • <u><em>No, I would not consider a metal to be a plasma because plasma is just another state of matter, and the copper wire is in solid state.</em></u>

Explanation:

Metal is not a state of matter. Metals can be solid or liquid (molten) depending on their melting point and the temperature at which they are.

Plasma is a state of matter, similar to gas, but it is reached only at very high temperatures like in the Sun. The particles in plasma state are not neutral atoms or molecules but negatively charged  ions and electrons.

The copper wire is yet a solid, thus it cannot be considered a plasma.

Metals can be in plasma state only if the temperature is too high, like the temperatures in the stars. In fact, the metals in the Sun and other hotter stars are in plasma state.

5 0
3 years ago
A certain metal M crystallizes in a lattice described by a body-centered cubic (bcc) unit cell. The lattice constant (a, the edg
Furkat [3]

Answer:

radius = 156 pm

Explanation:

The relation between radius and edge length of unit cell of BCC is

r=a\sqrt{3}/4

Given

a = 360 pm

Therefore

r = r = radius = 360\sqrt{3}/4= 155.88 pm

Or

156 pm

3 0
3 years ago
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