Answer:
The transition from lower energy level to higher energy level require a gain of energy.
Explanation:
When transition occur from lower energy level to higher energy level require a gain of energy. Electron could not jump unto higher energy level without gaining thew energy.
When electron jump into lower energy level from high energy level it loses the energy.
For example electron when jumped from 2nd to 3rd shell it gain energy and when in return back to 2nd shell from 3rd shell it loses energy.
The process is called excitation and de-excitation.
Excitation:
When the energy is provided to the atom the electrons by absorbing the energy jump to the higher energy levels. This process is called excitation. The amount of energy absorbed by the electron is exactly equal to the energy difference of orbits.
De-excitation:
When the excited electron fall back to the lower energy levels the energy is released in the form of radiations. this energy is exactly equal to the energy difference between the orbits. The characteristics bright colors are due to the these emitted radiations. These emitted radiations can be seen if they are fall in the visible region of spectrum.
3. Eukarya...is the answer
Answer:
Any of the six chemical elements that markup group1
of the periodic table.
Explanation:
The idea here is that you need to figure out how many moles of magnesium chloride,
MgCl
2
, you need to have in the target solution, then use this value to determine what volume of the stock solution would contain this many moles.
As you know, molarity is defined as the number of moles of solute, which in your case is magnesium chloride, divided by liters of solution.
c
=
n
V
So, how many moles of magnesium chloride must be present in the target solution?
c
=
n
V
⇒
n
=
c
⋅
V
n
=
0.158 M
⋅
250.0
⋅
10
−
3
L
=
0.0395 moles MgCl
2
Now determine what volume of the target solution would contain this many moles of magnesium chloride
c
=
n
V
⇒
V
=
n
c
V
=
0.0395
moles
3.15
moles
L
=
0.01254 L
Rounded to three sig figs and expressed in mililiters, the volume will be
V
=
12.5 mL
So, to prepare your target solution, use a
12.5-mL
sample of the stock solution and add enough water to make the volume of the total solution equal to
250.0 mL
.
This is equivalent to diluting the
12.5-mL
sample of the stock solution by a dilution factor of
20
.
