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Ratling [72]
2 years ago
12

Which of these atoms is the most electronegative? select one: a. si b. cl c. p d. f e. c

Physics
1 answer:
katrin [286]2 years ago
4 0

The atom that is the most electronegative is fluorine (F).

<h3>What is electronegative?</h3>

Electronegativity, is the tendency for an atom of a given chemical element to attract shared electrons when forming a chemical bond.

Electronegativity increases across the groups from left to right of the periodic table and decreases down the group.

Examples of electronegative elements arranged in decreasing order;

  • fluorine,
  • oxygen,
  • nitrogen,
  • chlorine,
  • bromine,
  • iodine,
  • sulfur,
  • carbon, and
  • hydrogen.

Thus,  the atom that is the most electronegative is fluorine (F).

Learn more about electronegativity here: brainly.com/question/24977425

#SPJ1

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If a person lives in china in which two hemispheres does he or she live
Delvig [45]

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In figure 1, charge q2 experiences no net electric force. What is q1?
lukranit [14]

By using Coulomb's law, we want to find the value of q₁ given that q₂ experiences no net electric force. We will find that q₁ = 8nC

<h3>Working with Coulomb's law.</h3>

Coulomb's law says that for two charges q₁ and q₂ separated by a distance r, the force that each one experiences is:

F = k\frac{q_1*q_2}{r^2}

Where k is a constant

Here we can see that q₂ interacts with two charges, then the total force on q₂ will be:

F = k\frac{q_1*q_2}{(20cm)^2} + k\frac{-2nC*q_2}{(10cm)^2}

And we know that it must be equal to zero, so we can write it as:

F = k\frac{q_1*q_2}{(20cm)^2} + k\frac{-2nC*q_2}{(10cm)^2} = 0\\\\k*q_2*(\frac{q_1}{(20cm)^2} + \frac{-2nC}{(10cm)^2}) = 0\\

The parenthesis must be equal to zero, so we can write:

\frac{q_1}{(20cm)^2} + \frac{-2nC}{(10cm)^2} = 0

And now we can solve this for q₁ to get:

q_1  = 2nC*(\frac{(20cm)^2}{(10cm)^2} ) = 8nC

If you want to learn more about Coulomb's law, you can read:

brainly.com/question/24743340

3 0
3 years ago
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