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allochka39001 [22]
3 years ago
12

When electrons move closer to a more electronegative atom, what happens?

Physics
1 answer:
marissa [1.9K]3 years ago
4 0

Answer:

There will be a reduction of the more electro-negative atom and energy will be released

Explanation:

oxidation is the loss of electron while reduction is the gain of electron. in most cases electronegative elements acts as oxidizing agents and are reduced, while electropositive elements acts as reducing agents and are oxidized.

In a situation where we have two electronegative atoms, the less electronegative would tend to behave as electropositve and lose its electron.

When these electron is gained by the more electronegative atom, it will become reduced and energy will be released.

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Learning Check: Ranking Elements
loris [4]
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3 years ago
A small object with momentum 7.0 kg∙m/s approaches head-on a large object at rest. The small object bounces straight back with a
EastWind [94]

Answer:

The magnitude of the large object's momentum change is 3 kilogram-meters per second.

Explanation:

Under the assumption that no external forces are exerted on both the small object and the big object, whose situation is described by the Principle of Momentum Conservation:

p_{S,1}+p_{B,1} = p_{S,2}+p_{B,2} (1)

Where:

p_{S,1}, p_{S,2} - Initial and final momemtums of the small object, measured in kilogram-meters per second.

p_{B,1}, p_{B,2} - Initial and final momentums of the big object, measured in kilogram-meters per second.

If we know that p_{S,1} = 7\,\frac{kg\cdot m}{s}, p_{B,1} = 0\,\frac{kg\cdot m}{s} and p_{S, 2} = 4\,\frac{kg\cdot m}{s}, then the final momentum of the big object is:

7\,\frac{kg\cdot m}{s} + 0\,\frac{kg\cdot m}{s} = 4\,\frac{kg\cdot m}{s}+p_{B,2}

p_{B,2} = 3\,\frac{kg\cdot m}{s}

The magnitude of the large object's momentum change is:

p_{B,2}-p_{B,1} = 3\,\frac{kg\cdot m}{s}-0\,\frac{kg\cdot m}{s}

p_{B,2}-p_{B,1} = 3\,\frac{kg\cdot m}{s}

The magnitude of the large object's momentum change is 3 kilogram-meters per second.

4 0
3 years ago
E Describe the meaning of a line crossing the x-axis in a<br> velocity-time graph.
gizmo_the_mogwai [7]

Answer:

A line crossing the x-axis in a velocity-time graph means that the moving object has changed its direction.

Explanation:

Velocity-Time graph:

A velocity-time graph is a two dimensional graph with velocity at its y-axis and time at its x-axis. At any point, value of y represents the velocity and value of x represents the time. The slope of the graph gives us the acceleration or deceleration of the moving object.

In a velocity-time graph:

- A straight line represents constant velocity.

- A diagonal line means that the velocity of a body is changing.

*Referring to the figure attached with the answer*

The velocity of the moving object increases at a constant rate for the first 10 minutes. Then the velocity is 60 m/min for the next 5 minutes. After that the velocity is decreasing. Till 30th minute when the velocity is at 0 m/min.

What happens here?

Velocity is a vector quantity. It has some direction. In a velocity-time graph, we are only concerned with two directions of velocity:

- Forward direction

- Backward direction

So, the object stops at 30th minute and starts moving in the reverse direction after that with an increasing velocity. <u>The point where the line cuts the x-axis is basically the point where the object starts moving in the reverse direction.</u>

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3 years ago
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It takes about 24 hours for a sperm cell to fertilize an egg. When the sperm penetrates the egg, the surface of the egg changes so that no other sperm can enter. At the moment of fertilization, the baby's genetic makeup is complete, including whether it's a boy or girl.

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3 years ago
Read 2 more answers
Two charged particles are placed 2.0 meters apart. the first charge is +2.0 e-6 c and the second charge is +4.0 e-6
Yakvenalex [24]
The Coulomb force between two charges is defined as: Fc=(k*Q₁*Q₂)/r² where k=9*10^9 N m² C⁻², Q₁ and Q₂ are charges and r is the distance between those charges. 
In our case:

Q₁=2*10^-6 C
Q₂=4*10^-6 C
r=2 m

Now we simply plug in the numbers into the equation:

Fc={(9*10^9)*(2*10^-6)*(4*10^-6)}/2^2=0.018 N

The Coulombs force between the two positive charges is Fc=0.018 N and it is a repulsive force because both charges are positive.
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3 years ago
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