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4 years ago

Why do atoms like carbon and nitrogen not like to make ions, while sodium and chlorine do?

Physics

2 answers:

Daniel [21]4 years ago

5 0

<u>Explanation:</u>

This can be very well explained with the help of reactivity.

Reactivity of an element is defined as the tendency to loose or gain electrons.

Sodium is an element belonging to group 1 and is considered as a metal. This element can easily loose electrons and hence is considered as reactive element. This element tends to form positive ions.

Chlorine is an element belonging to Group 17 and is considered as a non-metal. This element can easily gain electrons and hence is considered as a reactive element. This element tends to form negative ions.

Carbon and nitrogen belong to group 14 and group 15 respectively. They are not very reactive because they can't loose or gain electrons easily. Carbon tends to form covalent bonds by sharing of electrons. Nitrogen very rarely forms ionic compounds.

This is the reason why carbon and nitrogen does not form ions but sodium and chlorine forms ions.

krok68 [10]4 years ago

3 0

Atoms like carbon and nitrogen do not form ions because the electronegativity of these atoms are not that high nor very low which means electrons are fairly stable in the atom. While chlorine has very high electronegativity and for sodium very low, atoms tend to receive or release electrons.

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A 3 kg object is moving along a horizontal surface. The kinetic energy of the object is increasing at a constant rate of 6 J/m;

Whitepunk [10]

To solve this problem we will apply the concepts of energy conservation and Newton's second law that defines force as the product of the object's mass with its acceleration. Additionally we will apply concepts related to the kinematics equations of linear motion.

For conservation of energy we have that work is equal to kinetic energy therefore,

$W = KE$

$Fd = \frac{1}{2} mv^2$

Here,

F = Force

d = Displacement

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v= Velocity

At the same time we have the relation of

$F = \frac{W}{d}$

Therefore the value of the force can be interpreted as the rate of increase in energy per unit of distance, which makes it equivalent to

$F = \frac{W}{d} = 6J/m$

Applying Newton's Second Law

$F = ma$

$6J/m = (3kg)a$

$a = 2m/s^2$

In 4 seconds final velocity of the object becomes

$v = at$

$v= 2*4$

$v= 8m/s$

Then the work done is equal to,

$W = KE$

$W = \frac{1}{2} mv^2$

$W = \frac{1}{2} (3)(82)$

$W = 96J$

Then the displacement is,

$W = F*d$

$d = \frac{W}{F}$

$d = \frac{96}{6}$

$d = 16 m$

Therefore the distance moved is 16m

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4 years ago

A bullet is fired through a board 13.0 cm thick in such a way that the bullet's line of motion is perpendicular to the face of t

GREYUIT [131]

Answer:

392307.6923 m/s²

Explanation:

t = Time taken

u = Initial velocity = 560 m/s

v = Final velocity = 460 m/s

s = Displacement = 13 cm

a = Acceleration

From the equation of motion we have

$v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{460^2-560^2}{2\times 13\times 10^{-2}}\\\Rightarrow a=-392307.6923\ m/s^2$

The acceleration of the bullet as it passes through the board is -392307.6923 m/s²

5 0

3 years ago

The following is an example of a physical or a chemical change?? Alex made a beautiful origami swan out of a blue sheet of paper

REY [17]

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5 0

4 years ago

Two metal bars experience an equal change in volume due to an equal change in temperature. The first bar has a coefficient of ex

maksim [4K]

Answer:

The original volume of the first bar is half of the original volume of the second bar.

Explanation:

The coefficient of cubic expansivity of substances is given by;

γ = ΔV ÷ ( $V_{1}$ Δθ)

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Since they have equal change in volume,

Δ $V_{1}$ = Δ $V_{2}$ = ΔV

For the first metal bar,

2γ = ΔV ÷ ( $V_{1}$ Δθ)

⇒ Δθ = ΔV ÷ (2γ $V_{1}$ )

For the second metal bar,

γ = ΔV ÷ ( $V_{2}$ Δθ)

⇒ Δθ = ΔV ÷ ( $V_{2}$ γ)

Since they have equal change in temperature,

Δθ of first bar = Δθ of the second bar

ΔV ÷ (2γ $V_{1}$ ) = ΔV ÷ ( $V_{2}$ γ)

So that;

(1 ÷ 2 $V_{1}$ ) = (1 ÷ $V_{2}$ )

2 $V_{1}$ = $V_{2}$

$V_{1}$ = $\frac{V_{2} }{2}$

Thus, original volume of the first bar is half of the original volume of the second bar.

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