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marta [7]
4 years ago
6

How do valence electrons relate to cation formation?

Physics
2 answers:
tatiyna4 years ago
5 0

Valence electrons relate to cation formation as atoms lose valence electrons to form cations.

Answer: Option B

<u>Explanation: </u>

Each atom has a specific set of electrons at the outermost orbit. These electrons are denoted as valence electrons. Now, to gain stability and become reclusive of reacting with other elements' atoms, each atom tries to sustain to eight valence electrons in its outermost orbit.

Thus, atoms either lose or gain valence electrons by interacting with other atoms and get ionised. Ions are formed by either gain or lose electrons. If an atom loses its valence electrons, it gets more positively charged and called Cation. Whereas, if atoms gain valence electrons, they become more negatively charged and called Anion.

Delicious77 [7]4 years ago
3 0

Answer:

B. Atoms lose valence electrons to form cations.

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Explain the process of refraction of light​
brilliants [131]

The process of refraction of light​ occurs when light rays bends when travelling between media of different densities.

What is refraction of light?

Refraction of light is the bending of light rays or the change in the direction of light rays as it travels between media of different densities.

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2 years ago
If I traveled 30 miles in 1 hour, then 20 miles in 1 hour, and 50 miles in 3
Harrizon [31]

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The answer is 20 miles/hr

Explanation:

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3 years ago
HELP! block is released from rest and slides down a frictionless ramp inclined at 30' from the horizontal. When the block reache
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Answer:

A

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5 0
3 years ago
A hollow cylinder with an inner radius of 4 mm and an outer radius of 25 mm conducts a 4-A current flowing parallel to the axis
ivann1987 [24]

Answer:

2.1 × 10⁻⁵ T

Explanation:

Given:

Inner radius, r = 4 mm = 0.004 m

Outer radius, R = 25 mm = 0.025 m

Current, I = 4 A

Distance of the point from the center, a = 17 mm = 0.017 m

μ₀ = 4π × 10⁻⁷ T·m/A

Now,

For the hollow cylinder magnetic field (B) is given as:

B=\frac{\mu_0 I(a^2-r^2)\textup{}}{2\pi a(R^2-r^2)\textup{}}

on substituting the respective values, we get

B=\frac{4\pi\times10^{-7}\times4(0.017^2-0.004^2)\textup{}}{2\pi\times0.017(0.025^2-0.004^2)\textup{}}

or

B = 2.1 × 10⁻⁵ T

5 0
4 years ago
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