Answer:
λ = 6.5604 x 1016 nm
Explanation:
Given Data:
The energy of the red line in Hydrogen Spectra = 3.03 x 10-19
Formula to calculate Wave length
E= hv
Where E is Energy
h is Planks Constant = 6.626 x 10–34 J s
v is frequency
In turn
v= c/ λ
where c is speed of light = 3.00 x 108 m s–1
λ is wavelength = to find
Solution:
Formula to be Used:
E= hv………………………… (1)
Putting the value v in equation 1
E= h c/ λ…………………… (2)
Put the value in equation 2
3.03 x 10-19 J = (6.626 x 10–34 J s) x (3.00 x 108 m s–1) / λ ……………………….(3)
By rearranging equation 3
λ = (6.626 x 10–34 J s) x (3.00 x 108 m s–1) /3.03 x 10-19 J
λ = 6.5604 x 107 m
The answer is in “m”
So we have to convert it into nm
So for this to convert “m” to “nm” multiply the answer with 109
λ = 6.5604 x 107 x 109
λ = 6.5604 x 1016 nm
It is by looking on the dates of the paper.
Hoped this helped.
~Bob Ross®
In an oxidation-reduction reaction there is an exchange of electrons.
The exchange of electrons implies change in the oxidation states: at least one element increases its oxidation number while other reduces it.
By simple ispection you can predict that in the equation b. there is a change in oxidation states of Cl and Mn.
Now you can check it:
Equation 4H Cl + Mn O2 -> Mn Cl2 + 2H2 O + Cl2
oxidation sates 1+ 1- 4+ 2- 2+ 1- 1+ 2- 0
The oxidation state of Cl in HCl is 1- and it changed to 0 in Cl2
The oxidation state of Mn in MnO2 is 4+ and it changed to 2+ in MnCl2
Answer b.
Onization energy is the energy required to lose an electron and form an ion. The stronger is the attraction of the atom and the electron the higher the ionization energy, and the weaker is the attraction of the atom and the electron the higher the ionization energy. This leads to a clear trend in the periodic table. Given that the larger the atom the weaker the attraction of the atom to the valence electrons, the easier they will be released, and the lower the ionization energy. This is, as you go downward in a group, the ionization energy decreases. So, the element at the top of the group will exhibit the largest ionization energy. <span>Therefore, the answer is that of the four elements of group 7A, fluorine will have the largest first ionization energy.</span>
Explanation:
When chlorine bonds and becomes stable, (had a full outer electron shell), because we know that chlorine has 7 electrons in it’s outer shell, it requires 1 to gain noble gas structure. So, when if bonds with an element and then becomes stable, the gaining of an electron means it becomes a negatively charged ion. (an ion being a charged particle.)
