Sodium is a metal and bromine is a nonmetal so they form an ionic compound
nonmetals and nonmetals form covalent compounds
just divide the mass 49.2 grams by the amount of gummi bears 23 to get the answer of 2.1391 then it says to round to 2 decimals so 2.14 grams
covalent bond is firmed between two atoms
An orbital that penetrates into the region occupied by core electrons is less shielded from nuclear charge than an orbital that does not penetrate and therefore has a lower energy.
Explanation:
The only true statement from the given options is that "an orbital that penetrates into the region occupied by core electrons is less shielded from nuclear charge than an orbital that does not penetrate and therefore has a lower energy." Inner orbitals which are also known to contain core electrons feels the bulk of the nuclear pull on them compared to the outermost orbitals containing the valence electrons.
- The nuclear pull is the effect of the nucleus pulling and attracting the electrons in orbitals.
- This pull is stronger for inner orbitals and weak on the outer ones.
- The outer orbitals are said to be well shielded from the pull of the nuclear charge.
- Also, based on the quantum theory, electrons in the outer orbitals have higher energies because they occupy orbitals at having higher energy value.
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The statement is true in this situation is C. The size of Ffric is the same as the size of Fapp:
From the diagram, since the body is in equilibrium, the sum of vertical forces equals zero. Also, the sum of horizontal forces equal zero.
So, ∑Fx = 0 and ∑Fy = 0
Since Fapp acts in the negative x - direction and Ffric acts in the positive x - direction,
∑Fx = -Fapp + Ffric = 0
-Fapp + Ffric = 0
Fapp = Ffric
Also, since Fgrav acts in the negative y - direction and Fnorm acts in the positive y - direction,
∑Fy = Fnorm + (-Fgrav) = 0
Fnorm - Fgrav = 0
Fnorm = Fgrav
So, we see that the size of Fapp <u>equals</u> size of Ffric and the size of Fnorm <u>equals</u> the size of Fgrav.
So, the correct option is C
The statement which is true in this situation is C. The size of Ffric is the same as the size of Fapp.
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