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).
C6H15, C2H5 has a molar mass of 29g/mol. 87 divided by 29 is 3. Then multiply each element subscript by 3
Answer:
Explanation:
Due to Coulomb´s law electric force can be described by the formula 
, where K is the Coulomb´s constant (
), 
= Charge 1 (Na+ in this case), 
is the charge 2 (Cl-) and r is the distance between both charges.
Work made by a force is W=F.d and total work produced is the change in energy between final and initial state. this is 
.
so we have ![W=W_{f} -W_{i} =(K\frac{q_{(Na+)}q_{(Cl-)}rf}{r_{f} ^{2}})-(K\frac{q_{(Na+)}q_{(Cl-)}ri}{r_{i} ^{2}})=Kq_{(Na+)}q_{(Cl-)[\frac{1}{{r_{f}}} -\frac{1}{{r_{i}}}]](https://tex.z-dn.net/?f=W%3DW_%7Bf%7D%20-W_%7Bi%7D%20%3D%28K%5Cfrac%7Bq_%7B%28Na%2B%29%7Dq_%7B%28Cl-%29%7Drf%7D%7Br_%7Bf%7D%20%5E%7B2%7D%7D%29-%28K%5Cfrac%7Bq_%7B%28Na%2B%29%7Dq_%7B%28Cl-%29%7Dri%7D%7Br_%7Bi%7D%20%5E%7B2%7D%7D%29%3DKq_%7B%28Na%2B%29%7Dq_%7B%28Cl-%29%5B%5Cfrac%7B1%7D%7B%7Br_%7Bf%7D%7D%7D%20-%5Cfrac%7B1%7D%7B%7Br_%7Bi%7D%7D%7D%5D)
Given that ri= 1.1nm= 
and rf= infinite distance
![W=(9x10^{9})(1.6x10^{-19})(-1.6x10^{-19})[\frac{1}{\alpha }-\frac{1}{(1.1x10^{-9})}]=2.1x10^{-19}J](https://tex.z-dn.net/?f=W%3D%289x10%5E%7B9%7D%29%281.6x10%5E%7B-19%7D%29%28-1.6x10%5E%7B-19%7D%29%5B%5Cfrac%7B1%7D%7B%5Calpha%20%7D-%5Cfrac%7B1%7D%7B%281.1x10%5E%7B-9%7D%29%7D%5D%3D2.1x10%5E%7B-19%7DJ)
Answer:
15 people x 4 slices so you’ll need 60 pieces
60/12(the amount of slices per pizza) =5
You need 5 pizzas. $14.78 x 5 = 73.90.
Explanation:
the answer is that one 73.90 is rounded nearest cent : ) took me a while to get it
