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3 years ago

The electron cloud model describes which of the following?

Physics

2 answers:

son4ous [18]3 years ago

7 0

Answer:

1.B.The trials left by an electron as it moves around the nucleus .

2.8

3.Its mass is lowered, but it is still the same element.

Explanation:

1.Electron cloud model: It says that we cannot know exactly where an electron is at any given time, but the electrons are more likely to be in specific areas.The electron model says that electron have no fixed position so, it traces their path.

Therefore, the trials left by an electron as it moves around the nucleus .

Option B is true.

2.We are given that

Number of protons=8

Number of neutrons=10

We have to find the atomic number

We know that atomic number=Number of protons=8

Hence, atomic number of oxygen=8

3.When an atom loses a neutron then mass decreases but the element does not change.Because when mass number different and atomic number are same then the element are called isotopes of an element.In isotopes , neutron numbers are different.

If an atom loses an neutron then atomic number remain same and element remain same.

Answer:Its mass is lowered, but it is still the same element.

GalinKa [24]3 years ago

3 0

1. I think it's the trails left by an electron as it moves around the nucleus.

2. The atomic number is the number of protons so it is 8.

3. It's mass is lowered but it is still the same element.

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A motorcycle is traveling with a velocity of 4 m/s and has a mass of 112 kg

Zarrin [17]

Answer: 896J

Explanation:

Given that

Mass of motorcycle = 112kg

kinetic energy = ?

velocity of the motorcycle = 4m/s

Since kinetic energy is the energy possessed by a moving object, and it depends on the mass (m) of the object and the velocity (v) by which it moves. Therefore, the motorcycle has kinetic energy

i.e K.E = 1/2mv^2

KE = 1/2 x 112kg x (4m/s)^2

KE = 0.5 x 112 x 16

KE = 896J

Thus, the motorcycle has 896 joules of kinetic energy.

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4 years ago

How much heat is released to freeze 47.30 grams of copper at its freezing point of 1,085°C? The latent heat of fusion of copper

krek1111 [17]

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3 years ago

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6.6 kg block initially at rest is pulled to theright along a horizontal, frictionless surfaceby a constant, horizontal force of

neonofarm [45]

Answer:

v = 3.04 m/s

Explanation:

given,

mass of the block, M = 6.6 Kg

horizontal force, F = 12.2 N

distance, L = 2.5 m

initial speed = 0 m/s

speed of the block,v = ?

we now

Work done is equal to change in Kinetic energy.

Work done = Force x displacement

W = Δ K E

Δ K E = Force x displacement

$\dfrac{1}{2}mv^2 - \dfrac{1}{2}mu^2= F .s$

$\dfrac{1}{2}\times 6.6 \times v^2 - 0= 12.2\times 2.5$

3.3 v² = 30.5

v² = 9.242

v = 3.04 m/s

speed of the block is equal to 3.04 m/s

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3 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 ( $\frac{1}{3}V(R)$ )