Answer:
energy
Explanation:
The photon of light that is emitted as an electron drops back to its original orbit is energy and this energy is released during de-excitation process.
The electron is jumped into higher level and back into lower level by absorbing and releasing the 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. For example if electron jumped from K to L it must absorbed the energy which is equal the energy difference of these two level. The excited electron thus move back to lower energy level which is K by releasing the energy because electron can not stay longer in higher energy level and comes to ground state.
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
A. the distance between towns
Answer:
0.444 mol/L
Explanation:
First step is to find the number of moles of oxalic acid.
n(oxalic acid) = 
Now use the molar ratio to find how many moles of NaOH would be required to neutralize 
of oxalic acid.
n(oxalic acid): n(potassium hydroxide)
1 : 2 (we get this from the balanced equation)
: x
x = 0.0111 mol
Now to calculate what concentration of KOH that would be in 25 mL of water:
Answer:
The specific heat of the metal is 2.09899 J/g℃.
Explanation:
Given,
For Metal sample,
mass = 13 grams
T = 73°C
For Water sample,
mass = 60 grams
T = 22°C.
When the metal sample and water sample are mixed,
The addition of metal increases the temperature of the water, as the metal is at higher temperature, and the addition of water decreases the temperature of metal. Therefore, heat lost by metal is equal to the heat gained by water.
Since, heat lost by metal is equal to the heat gained by water,
Qlost = Qgain
However,
Q = (mass) (ΔT) (Cp)
(mass) (ΔT) (Cp) = (mass) (ΔT) (Cp)
After mixing both samples, their temperature changes to 27°C.
It implies that
, water sample temperature changed from 22°C to 27°C and metal sample temperature changed from 73°C to 27°C.
Since, Specific heat of water = 4.184 J/g°C
Let Cp be the specific heat of the metal.
Substituting values,
(13)(73°C - 27°C)(Cp) = (60)(27°C - 22℃)(4.184)
By solving, we get Cp =
Therefore, specific heat of the metal sample is 2.09899 J/g℃.
Reactants would be the missing word in the sentence because they make up the products in a chemical reaction
