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3 years ago

In both ionic and molecular bonds, the resulting compound is stabilized because

1 answer:

7 0

Answer: In both ionic and molecular bonds, the resulting compound is stabilized because each atom's outer electronic orbital is full.

Explanation:

Molecular bonds are also called covalent bonds. A covalent bond is formed by sharing of electrons between two or more atoms.

For example, atomic number of hydrogen is 1 and atomic number of nitrogen is 7 (2, 5). In order to attain stability hydrogen atom needs to gain one electron whereas nitrogen needs to gain 3 electrons.

Hence, 3 atoms of hydrogen chemically combine with one atom of nitrogen by sharing electrons and thus it forms the compound $NH_{3}$ .

Ionic bonds are the bonds formed by transfer of electrons from one atom to another.

For example, atomic number of sodium is 11 (2, 8, 1) and atomic number of chlorine is 17 (2, 8, 7). In order to attain stability sodium needs to lose one electron whereas chlorine needs to gain one electron.

Hence, when sodium combines chemically with chloride then sodium will transfer its 1 valence electron to the chlorine atom and thus it forms the compound NaCl.

Therefore, we can conclude that in both ionic and molecular bonds, the resulting compound is stabilized because each atom's outer electronic orbital is full.

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A diver is swimming underneath an oil slick with a thickness of 200 nm and an index of refraction of 1.50. A white light shines

Tcecarenko [31]

Answer:

Explanation:

thickness of oil t = 200 nm

index of refraction μ = 1.5

For transmitted light :---

path difference = 2μ t

For constructive interference

path difference = n λ , λ is wavelength of light

2μ t = n λ

λ = 2μ t / n

For longest λ , n = 1

λ = 2μ t

= 2 x 1.5 x 200 nm

= 600 nm

Wavelength in water

= 600 / refractive index of water

= 600 / 1.33

= 451.1 nm Ans

4 0

3 years ago

1. If a car travels 400m in 20 seconds how fast is it going?

Marrrta [24]

Answer:

1) 20 m/s

2) 5 m/s

3) 2 m/s

4) 395,000m/9000s

5) 16 km/0.25h

Explanation:

i dunno

5 0

3 years ago

A pendulum is made up of a small sphere of mass 0.500 kg attached to a string of length 0.950 m. The sphere is swinging back and

Semenov [28]

Answer:

W = 0.842 J

Explanation:

To solve this exercise we can use the relationship between work and kinetic energy

W = ΔK

In this case the kinetic energy at point A is zero since the system is stopped

W = K_f (1)

now let's use conservation of energy

starting point. Highest point A

Em₀ = U = m g h

Final point. Lowest point B

Em_f = K = ½ m v²

energy is conserved

Em₀ = Em_f

mg h = K

to find the height let's use trigonometry

at point A

cos 35 = x / L

x = L cos 35

so at the height is

h = L - L cos 35

h = L (1-cos 35)

we substitute

K = m g L (1 -cos 35)

we substitute in equation 1

W = m g L (1 -cos 35)

let's calculate

W = 0.500 9.8 0.950 (1 - cos 35)

W = 0.842 J

7 0

3 years ago

A thin spherical spherical shell of radius R which carried a uniform surface charge density σ. Write an expression for the volum

ozzi

Answer:

Explanation:

From the given information:

We know that the thin spherical shell is on a uniform surface which implies that both the inside and outside the charge of the sphere are equal, Then

The volume charge distribution relates to the radial direction at r = R

∴

$\rho (r) \ \alpha \ \delta (r -R)$

$\rho (r) = k \ \delta (r -R) \ \ at \ \ (r = R)$

$\rho (r) = 0\ \ since \ r< R \ \ or \ \ r>R---- (1)$

To find the constant k, we examine the total charge Q which is:

$Q = \int \rho (r) \ dV = \int \sigma \times dA$

$Q = \int \rho (r) \ dV = \sigma \times4 \pi R^2$

∴

$\int ^{2 \pi}_{0} \int ^{\pi}_{0} \int ^{R}_{0} \rho (r) r^2sin \theta \ dr \ d\theta \ d\phi = \sigma \times 4 \pi R^2$

$\int^{2 \pi}_{0} d \phi* \int ^{\pi}_{0} \ sin \theta d \theta * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2$

$(2 \pi)(2) * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2$

Thus;

$k * 4 \pi \int ^{R}_{0} \delta (r -R) * r^2dr = \sigma \times R^2$

$k * \int ^{R}_{0} \delta (r -R) r^2dr = \sigma \times R^2$

$k * R^2= \sigma \times R^2$

$k = R^2 --- (2)$

Hence, from equation (1), if k = $\sigma$

$\mathbf{\rho (r) = \delta* \delta (r -R) \ \ at \ \ (r=R)}$

$\mathbf{\rho (r) =0 \ \ at \ \ rR}$

To verify the units:

$\mathbf{\rho (r) =\sigma \ * \ \delta (r-R)}$

↓ ↓ ↓

c/m³ c/m³ × 1/m

Thus, the units are verified.

The integrated charge Q

$Q = \int \rho (r) \ dV \\ \\ Q = \int ^{2 \ \pi}_{0} \int ^{\pi}_{0} \int ^R_0 \rho (r) \ \ r^2 \ \ sin \theta \ dr \ d\theta \ d \phi \\ \\ Q = \int ^{2 \pi}_{0} \ d \phi \int ^{\pi}_{0} \ sin \theta \int ^R_{0} \rho (r) r^2 \ dr$