Answer: the electrons remain around the atomic nuclei due to the existence of a positive charge on the nuclei that, of course, atract the negative charged electrons. The protons are the paricles in the nuclei that hold the positive charge.
Justification:
First, I wish to explaing the sense of the question. The question arises because given that the electrons have negative electric charge how is that they do not repeal each other to the point that they end leaving the nucleous of the atom alone.
This is you know that equal charges repel each other, so how is it that the electrons stand around the nucleous instead of separateing and levaing the atomic nucleous alone.
The answer is due to the existence of a positive charge on the nuclei that, of course, atract the negative charged electrons. That positive charge is the protons.
The protons are particles in the atomic nuclei that are positive charged and they exert the right attractive force upon the electrons to permit them stay in the orbitals (regions of the space around the nuclei of the atoms where the electrons are found).
Answer:
18.45 g of C
Explanation:
This is a problem of rules of three:
1 mol of C₃H₈ contains 3 moles of C and 8 moles of H
If 8 moles of H are contained in 1 mol of propane
4.10 moles of H are contained in (4.1 . 1) /8 = 0.5125 moles
Now, If 1 mol of propane contains 3 moles of C
0.5125 moles of propane may contain (0.5125 . 3) / 1 = 1.5375 moles of C
Let's convert the moles to mass:
1.5375 mol . 12 g /mol = 18.45 g
Answer:
V = 36.7L of 
Explanation:
1. Write the chemical reaction for the decomposition of sodium azide:

2. Find the number of moles of
produced by the decomposition of 71.4g of
:

3. Use the ideal gas equation to find the Volume of
occupied by 1.65 moles of
, at the temperature and pressure given by the problem:

Solving for V:

Converting the temperature from ◦C to K:
25◦C+273.15=298.15K
Replacing values:

V=36.7L of 
Answer:
no because the suns light is way more bright than lightbulbs.
Explanation:
The question is as follows: What is the% m / m of a solution in which 22 g of solute are dissolved in 44 g of solvent?
Answer: The% m/m of a solution in which 22 g of solute are dissolved in 44 g of solvent is 50%.
Explanation:
Given: Mass of solute = 22 g
Mass of solvent = 44 g
The percentage m/m is calculated using the following formula.

Substitute the values into above formula as follows.

Thus, we can conclude that the% m/m of a solution in which 22 g of solute are dissolved in 44 g of solvent is 50%.