A period of the periodic table ends when the highest energy level of an elements atoms has __ electrons

1 answer:

8 0

This is dependent on how many shells/layers/energy levels the element has. The first shell can only hold 2 electrons however every shell beyond that can hold 8 electrons

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What is the gravitational acceleration close to the surface of a planet with a mass of 9ME and radius of 3RE, where ME and RE ar

Papessa [141]

Answer:

9.78 m/s²

Explanation:

To solve this, we use the gravitational formula

g = GM/r², where

g = acceleration due to gravity

G = gravitational constant

M = mass of the planet

r = radius of the planet

From the question, we got that the mass of the planet is

M = 9ME, where ME = 5.95*10^24

M = 9 * 5.95*10^24

M = 5.355*10^25 kg

Also, the Radius of the planet, R = 3RE, where RE = 6.37*10^6

R = 3 * 6.37*10^6

R = 1.911*10^7 m

On applying the values of both R and M to the equation, we get

g = GM/r²

g = (6.67*10^-11 * 5.355*10^25) / (1.911*10^7)²

g = 3.57*10^15/3.65*10^14

g = 9.78 m/s²

Therefore, the acceleration due to gravity on the planet is 9.78 m/s²

Please vote brainliest if it helped you <3

5 0

2 years ago

Write an essay about the things an individual must do in order to adapt and survive the changes that are happening in his/her en

Scorpion4ik [409]

Sorry Man If you ask this question no one wants to write an essay and waste their time. You are smart enough to write an essay.

7 0

2 years ago

Two identical conducting spheres, A and B, carry equal charge. They are stationary and are separated by a distance much larger t

podryga [215]

Answer:

8F_i = 3F_f

Explanation:

When two identical spheres are touched to each other, they equally share the total charge. Therefore, When neutral C is first touch to A, they share the initial charge of A equally.

Let us denote that the initial charge of A and B are Q. Then after C is touched to A, their respective charges are Q/2.

Then, C is touched to B, and they share the total charge of Q + Q/2 = 3Q/2. Their respective charges afterwards is 3Q/4 each.

The electrostatic force, Fi, in the initial configuration can be calculated as follows.

$F_i = \frac{1}{4\pi\epsilon_0}\frac{q_Aq_B}{r^2} = \frac{1}{4\pi\epsilon_0}\frac{Q^2}{r^2}[/tex}The electrostatic force, Ff, in the final configuration is [tex]F_f = \frac{1}{4\pi\epsilon_0}\frac{q_Aq_B}{r^2} = \frac{1}{4\pi\epsilon_0}\frac{3Q^2/8}{r^2}[/tex}Therefore, the relation between Fi and Ff is as follows[tex]F_i = F_f\frac{3}{8}\\8F_i = 3F_f$