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n200080 [17]
3 years ago
9

The efficiency of a machine can be increased by decreasing distance. decreasing force. reducing friction. reducing work.

Chemistry
2 answers:
crimeas [40]3 years ago
5 0

Answer:

Reducing friction

Explanation:

The efficiency of a machine is a how the input energy is converted in output work

Then in the middle you have things consuming the input energy and not going to the output work, the best you can do is reducre those things consuming, for a machine it can be friction, heat, vibration or loose parts.

For that reason Reducing friction is the answer

xeze [42]3 years ago
5 0

Answer: Reducing friction

Explanation:

Friction refers to the resistance that a surface encounters while moving on the surface of another. Machines often posses moving parts such as rollers. Those parts move on other surfaces generating friction. Friction in machine parts leads to useless work and reduces the amount of input work used by the machine to do useful work. The efficiency of a machine has to do with its ability to convert input work into output work. This efficiency is greatly reduced by friction in the moving parts of the machine. Hence friction decreases the efficiency (work done by a machine)

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The materials in the flux covering on an electrode determine the electrical characteristics of the electrode.
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Answer:

a. true

Explanation:

It is true that, the materials in the flux covering on an electrode determine the electrical characteristics of the electrode.

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2 years ago
Which substance used during photosynthesis does a plant take in from the soil?
Simora [160]

Answer:

the answer is water

Explanation:

During photosynthesis, plants take in carbon dioxide (CO2) and water (H2O) from the air and soil. Within the plant cell, the water is oxidized, meaning it loses electrons, while the carbon dioxide is reduced, meaning it gains electrons. This transforms the water into oxygen and the carbon dioxide into glucose.

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2 years ago
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4 0
3 years ago
What is the E°cell for the cell represented by the combination of the following half-reactions? ClO4–(aq) + 8H+(aq) + 8e– Cl–(aq
Vinvika [58]

Answer:

The E°cell for the cell represented by the combination of the given half-reactions is 0.398 V

Explanation:

Oxide-reduction reactions, also called redox, involve the transfer or transfer of electrons between two or more chemical species. In these reactions two substances interact: the reducing agent and the oxidizing agent.

The gain of electrons is called reduction and the loss of electrons oxidation. That is to say, there is oxidation whenever an atom or group of atoms loses electrons (or increases its positive charges) and in the reduction an atom or group of atoms gains electrons, increasing its negative charges or decreasing the positive ones.

The species that supplies electrons is the reducing agent (that is, it is that species that oxidizes, yielding electrons and increasing its positive charge, or decreasing the negative one causing the reduction of the other species) and the one that gains them is the oxidizing agent ( that is, it is that species that is reduced, capturing electrons and increasing its negative charge, or decreasing its positive charge, causing oxidation of the other species).

This is the type of reaction that occurs in this case.

ClO₄⁻(aq) + 8 H⁺ (aq) + 8 e⁻ ⇔ Cl⁻ (aq) + 4 H₂O(l) E° = 1.389 V

VO₂⁺(aq) + 2 H⁺(aq) + e⁻ ⇔ VO⁺(aq) + H₂O(l) E° = 0.991 V

In this case both are written as reductions, and their E ° values ​​as well. The species that has the greatest potential for reduction will be the species that will be reduced, that is, it will be the oxidizing agent. In this case, it will be the first half-reaction expressed.  Therefore, to obtain a reaction, the second semi-reaction must be reversed to be an oxidation, maintaining its constant value. Then:

Reduction: ClO₄⁻(aq) + 8 H⁺ (aq) + 8 e⁻ ⇔ Cl⁻ (aq) + 4 H₂O(l) E° = 1.389 V

Oxidation: VO⁺(aq) + H₂O(l) ⇔ VO₂⁺(aq) + 2 H⁺(aq) + e⁻ E° = 0.991 V

<em>E°cell=Ereduction - Eoxidation</em>

E°cell=1.389 V - 0.991 V

<em>E°cell=0.398 V</em>

Then <u><em>the E°cell for the cell represented by the combination of the given half-reactions is 0.398 V.</em></u>

Another way of thinking is that, by inverting the second semi-reaction to be an oxidation, the value of E ° is reversed in the sign, unlike the previous case in which it was constant. Then:

Reduction: ClO₄⁻(aq) + 8 H⁺ (aq) + 8 e⁻ ⇔ Cl⁻ (aq) + 4 H₂O(l) E° = 1.389 V

Oxidation: VO⁺(aq) + H₂O(l) ⇔ VO₂⁺(aq) + 2 H⁺(aq) + e⁻ E° = -0.991 V

In this case:

E°cell=Ereduction + Eoxidation=

E°cell=1.389 V + (-0.991 V)=1.389 V-0.991 V

<em>E°cell=0.398 V</em>

Note that the result obtained is the same. This indicates that either of the two ways proposed is correct, and you will use the one that is most comfortable for you.

4 0
3 years ago
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