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3 years ago

Which atom is most likely to form a metallic bond ?

Physics

1 answer:

Nana76 [90]3 years ago

5 0

<u>Answer:</u>

"Aluminium (Al)" atom is most likely to form a metallic bond.

<u>Explanation:</u>

A kind of chemical bonding which generates between conduction electrons and positively charged metal ions from the attractive electrostatic force is understood as "metallic bonding". It may be defined as sharing free electrons among a positively charged ion structure.

A sheet of foil made from aluminum is constructed from metal bonds. Metallic relations are formed by strong forces of attraction. This property leads to low volatility, high melting and boiling points, and most metals are high density.

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An x-ray photon is scattered by an originally stationary electron. how does the frequency of the scattered photon compare relati

Viefleur [7K]

The frequency of the scattered photon decreases or it will be lower compare to the frequency of incident photon. An x-ray photon scatters in one direction after a collision and some energy is transferred to the electron as it recoils in another direction resulting to have less energy in the scattered photon. In addition, the frequencies will also depend on the differences of the angle at which the scattered photon leaves the collision and this incident is called Compton Effect.

8 0

3 years ago

At constant pressure, the volume of a fixed mass of gas and its kelvin temperature are said to be

juin [17]

They are said to be directly related.

a) directly related.

This is Charles' Law.

3 0

3 years ago

Read 2 more answers

Two charges, each of 2.9 microC are placed at two corners of a square 50cm on a side, If the charges are on one side of the squa

anyanavicka [17]

Answer:

The magnitude of the electric field and direction of electric field are $146.03\times10^{3}\ N/C$ and 75.36°.

Explanation:

Given that,

First charge $q_{1}= 2.9\mu C$

Second charge $q_{2}= 2.9\mu C$

Distance between two corners r= 50 cm

We need to calculate the electric field due to other charges at one corner

For E₁

Using formula of electric field

$E_{1}=\dfrac{kq}{r'^2}$

Put the value into the formula

$E_{1}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\sqrt{2}\times10^{-2})^2}$

$E_{1}=52200=52.2\times10^{3}\ N/C$

For E₂,

Using formula of electric field

$E_{1}=\dfrac{kq}{r^2}$

Put the value into the formula

$E_{2}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\times10^{-2})^2}$

$E_{2}=104400=104.4\times10^{3}\ N/C$

We need to calculate the horizontal electric field

$E_{x}=E_{1}\cos\theta$

$E_{x}=52.2\times10^{3}\times\cos45$

$E_{x}=36910.97=36.9\times10^{3}\ N/C$

We need to calculate the vertical electric field

$E_{y}=E_{2}+E_{1}\sin\theta$

$E_{y}=104.4\times10^{3}+52.2\times10^{3}\sin45$

$E_{y}=141310.97=141.3\times10^{3}\ N/C$

We need to calculate the net electric field

$E_{net}=\sqrt{E_{x}^2+E_{y}^2}$

Put the value into the formula

$E_{net}=\sqrt{(36.9\times10^{3})^2+(141.3\times10^{3})^2}$

$E_{net}=146038.69\ N/C$

$E_{net}=146.03\times10^{3}\ N/C$

We need to calculate the direction of electric field

Using formula of direction

$\tan\theta=\dfrac{141.3\times10^{3}}{36.9\times10^{3}}$

$\theta=\tan^{-1}(\dfrac{141.3\times10^{3}}{36.9\times10^{3}})$

$\theta=75.36^{\circ}$

Hence, The magnitude of the electric field and direction of electric field are $146.03\times10^{3}\ N/C$ and 75.36°.

4 0

3 years ago

During a softball game, a player running from second base to third base reaches a speed

andrey2020 [161]

Answer: I would say it would be 3.9

but i believe there is something missing shorty.

4 0

3 years ago

What are the complementary colors of magenta? Blue? And cyan?

zysi [14]

Answer:

Magneta is a mix of blue and red and is a secondary colour.

Explanation:

As we can see green is the complementary colour of Magneta.Complementary colours are the pairs of colours which when combined cancel each other out. This means that when combined they produce a grayscale colour like white or black .

4 0

3 years ago