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Liono4ka [1.6K]
3 years ago
9

Explain how metallic atoms bond to form metallic compounds.

Physics
2 answers:
AlekseyPX3 years ago
8 0

<u>Answer:</u>

Metals possess low ionisation energy hence it loses electrons easily. The nucleus and the inner shell electrons is called the kernel.

The electron which gets lost leaves the influence of one kernel and moves towards the other.

Thus the electrons will be in constant movement and gets delocalized.

The electrostatic forces between the positive kernel and the sea of mobile electrons is called as metallic bond.

Metal atoms possess this type of attractive forces in between them.

For example the metal atoms of a sheet of copper possess this type of chemical bond between the atoms.

Thus, Metallic bonding is a type of chemical bonding that rises from the electrostatic attractive force between conduction electrons and positively charged metal ions

xz_007 [3.2K]3 years ago
4 0

In metallic bonds, the valence electrons from the s and p orbitals of the interacting metal atoms delocalize. That is to say, instead of orbiting their respective metal atoms, they form a “sea” of electrons that surrounds the positively charged atomic nuclei of the interacting metal ions

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Assume that in the Stern-Gerlach experiment for neutral silver atoms, the magnetic field has a magnitude of B = 0.21 T. (a) What
Marat540 [252]

Answer:

Explanation

Assume that in the Stern-Gerlach experiment for neutral silver atoms, the magnetic field has a magnitude of B = 0.21 T

A. To calculate the energy difference in the magnetic moment orientation

∆E = 2μB

For example, any electron's magnetic moment is measured to be 9.284764×10^−24 J/T

Then

μ = 9.284764 × 10^-24 J/T

∆E = 2μB

∆E = 2 × 9.284764 × 10^-24 × 0.21

∆E = 3.8996 × 10^-24 J

Then, to eV

1eV = 1.602 × 10^-19J

∆E = 3.8996 × 10^-24 J × 1eV / 1.602 × 10^-19J

∆E = 2.43 × 10^-5 eV

B. Frequency?

To determine the frequency of radiation hitch would induce the transition between the two states is,

∆E = hf

Where h is plank constant

h = 6.626 × 10-34 Js

Then, f = ∆E / h

f = 3.8996 × 10^-24 / 6.626 × 10^-34

f = 5.885 × 10^9 Hz

f ≈ 5.89 GHz

C. The wavelength of the radiation

From wave equation

v = fλ

In electromagnetic, we deal with speed of light, v = c

And the speed of light in vacuum is

c = 3 × 10^8 m/s

c = fλ

λ = c / f

λ = 3 × 10^8 / 5.885 × 10^9

λ = 0.051 m

λ = 5.1 cm

λ = 51 mm

D. It belongs to the microwave

From table

Micro waves ranges from

•Wavelength 10 to 0.01cm

Then we got λ = 5.1 cm, which is in the range.

•Frequency 3GHz to 3 Thz

Then, we got f ≈ 5.89 GHz, which is in the range

•Energy 10^-5 to 0.01 eV

We got ∆E = 2.43 × 10^-5 eV, which is in the range of the microwave

The value above is in microwave range

5 0
3 years ago
A container of gas molecules is at a pressure of 2 atm and has amass density of 1.7 grams per liter. All of the molecules in the
Tpy6a [65]

Answer:

The speed of nitrogen molecule is 1.87 m/s.

Explanation:

Given that,

Pressure = 2 atm

Density = 1.7 grams/liter

Atomic weight = 28 grams

We need to calculate the temperature

Using formula of idea gas

PV=nRT

P=\dfrac{WRT}{VM}

P=\dfrac{\rho RT}{M}

T=\dfrac{PM}{\rho R}

Put the value into the formula

T=\dfrac{2\times28}{1.7\times0.0821}

T=401.2\ K

We need to calculate the speed of nitrogen molecule

Using formula of RMS speed

V_{rms}=\sqrt{\dfrac{3RT}{M}}

V_{rms}=\sqrt{\dfrac{3\times0.0821\times401.2}{28}}

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Hence, The speed of nitrogen molecule is 1.87 m/s.

5 0
3 years ago
Which process is not required for an animal to obtain energy from food?
Romashka [77]

Complete Question:

Which process is not required for an animal to obtain energy from food?

a. Secretion.

b. Digestion.

c. Excretion.

d. Ingestion.

Answer:

c. Excretion.

Explanation:

Excretion is a metabolic process in which living organisms gets their system rid of metabolic waste, water and excess ions. This is usually done with the aid of kidney, skin and lungs in all vertebrates.

Example of excretions are urination, sweating and defecation.

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