All categories

3 years ago

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

Physics

2 answers:

Daniel [21]3 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]3 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.

You might be interested in

Which one describes how much matter an object contains, using SI units?

ahrayia [7]

Si units or Systeme' de Internationale' is a widely adopted unit system in measuring basic and derived dimensions In this case, the SI units here are kilograms, meter and seconds. Pounds is an English unit. mass is the measure of <span>how much matter an object contains, hence the answer is A. 43 kg.</span>

5 0

3 years ago

Read 2 more answers

A quarterback passes a football from height h = 2.1 m above the field, with initial velocity v0 = 13.5 m/s at an angle θ = 32° a

alisha [4.7K]

Answer:

a) x = v₀² sin 2θ / g

b) t_total = 2 v₀ sin θ / g

c) x = 16.7 m

Explanation:

This is a projectile launching exercise, let's use trigonometry to find the components of the initial velocity

sin θ = $v_{oy}$ / vo

cos θ = v₀ₓ / vo

v_{oy} = v_{o} sin θ

v₀ₓ = v₀ cos θ

v_{oy} = 13.5 sin 32 = 7.15 m / s

v₀ₓ = 13.5 cos 32 = 11.45 m / s

a) In the x axis there is no acceleration so the velocity is constant

v₀ₓ = x / t

x = v₀ₓ t

the time the ball is in the air is twice the time to reach the maximum height, where the vertical speed is zero

v_{y} = v_{oy} - gt

0 = v₀ sin θ - gt

t = v_{o} sin θ / g

we substitute

x = v₀ cos θ (2 v_{o} sin θ / g)

x = v₀² /g 2 cos θ sin θ

x = v₀² sin 2θ / g

at the point where the receiver receives the ball is at the same height, so this coincides with the range of the projectile launch,

b) The acceleration to which the ball is subjected is equal in the rise and fall, therefore it takes the same time for both parties, let's find the rise time

at the highest point the vertical speed is zero

v_{y} = v_{oy} - gt

v_{y} = 0

t = v_{oy} / g

t = v₀ sin θ / g

as the time to get on and off is the same the total time or flight time is

t_total = 2 t

t_total = 2 v₀ sin θ / g

c) we calculate

x = 13.5 2 sin (2 32) / 9.8

x = 16.7 m

4 0

2 years ago

Consider a uniformly charged sphere of radius Rand total charge Q. The electric field Eout outsidethe sphere (r≥R) is simply tha

AlexFokin [52]

1) Electric potential inside the sphere: $\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})$

2) Ratio Vcenter/Vsurface: 3/2

3) Find graph in attachment

Explanation:

The electric field inside the sphere is given by

$E=\frac{1}{4\pi \epsilon_0}\frac{Qr}{R^3}$

where

$\epsilon_0=8.85\cdot 10^{-12}F/m$ is the vacuum permittivity

Q is the charge on the sphere

R is the radius of the sphere

r is the distance from the centre at which we compute the field

For a radial field,

$E(r)=-\frac{dV(r)}{dr}$

Therefore, we can find the potential at distance r by integrating the expression for the electric field. Calculating the difference between the potential at r and the potential at R,

$V(R)-V(r)=-\int\limits^R_r E(r)dr=-\frac{Q}{4\pi \epsilon_0 R^3}\int r dr = \frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)$

The potential at the surface, V(R), is that of a point charge, so

$V(R)=\frac{Q}{4\pi \epsilon_0 R}$

Therefore we can find the potential inside the sphere, V(r):

$V(r)=V(R)+\Delta V=\frac{Q}{4\pi \epsilon_0 R}+\frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})$

At the center,

r = 0

Therefore the potential at the center of the sphere is:

$V(r)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})\\V(0)=\frac{3Q}{8\pi \epsilon_0 R}$

On the other hand, the potential at the surface is

$V(R)=\frac{Q}{4\pi \epsilon_0 R}$

Therefore, the ratio V(center)/V(surface) is:

$\frac{V(0)}{V(R)}=\frac{\frac{3Q}{8\pi \epsilon_0 R}}{\frac{Q}{4\pi \epsilon_0 R}}=\frac{3}{2}$

The graph of V versus r can be found in attachment.

We observe the following:

- At r = 0, the value of the potential is $\frac{3}{2}V(R)$ , as found in part b) (where $V(R)=\frac{Q}{4\pi \epsilon_0 R}$ )

- Between r and R, the potential decreases as $-\frac{r^2}{R^2}$

- Then at r = R, the potential is $V(R)$

- Between r = R and r = 3R, the potential decreases as $\frac{1}{R}$ , therefore when the distance is tripled (r=3R), the potential as decreased to 1/3 ( $\frac{1}{3}V(R)$ )