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3 years ago

Which statement best describes metallic bonding?

Physics

2 answers:

aksik [14]3 years ago

3 0

THE ANSWER IS D. A NONMETAL ATOM TRANSFERS ELCTRONS TO A METAL ATOM

ITS A TYPE OF ENERGY BONDING THAT ARISES FROM ELECTROSTATIC. ATTRACTIVE FORCE BETWEEN CONDUCTION ELCTRONS (A NONMETAL ATOM) AND A POSITIVELY CHARGED METAL ATOM ( WHO ATTRACTS THE NONMETALS ELECTRONS BECAUSE ITS POSITIVELY CHARGED)

mario62 [17]3 years ago

3 0

Answer:

D.A nonmetal atom transfers electrons to a metal atom.

Explanation:

Metals are chemical elements that have as their main physical characteristic the ability to lose electrons and, consequently, form metal cations. For this reason, they can perform two types of chemical bonds: the ionic bond and the metallic bond.

The metal bond is established between atoms of a single metal element. This type of bonding occurs only between atoms of a single metal and exclusively because a metal cannot establish chemical bond with another different metallic element.

In the metal bond, the crystalline lattices that form the metals are actually an ionic cluster (composed only of cations and electrons). The electrons present in the valence layer of the metal atoms are delocalized, that is, they leave the valence layer, causing the atom to become a cation (electron deficient).

After being delocalized, the electrons from the metal atoms surround the cations, forming a true "sea of electrons". Each of the electrons in this sea has the ability to move through the crystal lattice of the metal freely.

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Genrish500 [490]

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3 years ago

Red light of wavelength 633 nm from a helium-neon laser passes through a slit 0.400mm wide. The diffraction pattern is observed

kogti [31]

Answer:

a) $y_{first}=5.3mm$

b) $y_{second}=10.6-5.3 =5.3 mm$

Explanation:

The width of the central bright in this diffraction pattern is given by:

$y=\frac{m\lambda D}{a}$ when m is a natural number.

here:

m is 1 (to find the central bright fringe)
D is the distance from the slit to the screen
a is the slit wide
λ is the wavelength

So we have:

$y_{first}=\frac{633*10^{9}*3.35}{0.0004}$

$y_{first}=5.3mm$

Now, if we do m=2 we can find the distance to the second minima.

$y_{2}=\frac{2*633*10^{9}*3.35}{0.0004}$

$y_{2}=10.6 mm$

Now we need to subtract these distance, to get the width of the first bright fringe :

$y_{second}=10.6-5.3 =5.3 mm$

I hope it heps you!

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2 years ago

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TiliK225 [7]

Answer:

Explanation:

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3 years ago

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Vlad [161]

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3 years ago