of course migration
Explanation:
if you need help again i'll be there.
Answer:
I can't draw diagrams on this web site but I can do with numbers I think. So an electron is moved from n = 1 to n = 5. I'm assuming I've interpreted the problem correctly; if not you will need to make a correction. I'm assuming that you know the electron in the n = 1 state is the ground state so the 4th exited state moves it to the n = 5 level.
n = 5 4th excited state
n = 4 3rd excited state
n = 3 2nd excited state
n = 2 1st excited state
n = 1 ground state
Here are the possible spectral lines.
n = 5 to 4, n = 5 to 3, n = 5 to 2, n = 5 to 1 or 4 lines.
n = 4 to 3, 4 to 2, 4 to 1 = 3 lines
n = 3 to 2, 3 to 1 = 2 lines
n = 2 to 1 = 1 line. Add 'em up. I get 10.
b. The Lyman series is from whatever to n = 1. Count the above that end in n = 1.
c.The E for any level is -21.8E-19 Joules/n^2
To find the E for any transition (delta E) take E for upper n and subtract from the E for the lower n and that gives you delta E for the transition.
So for n = 5 to n = 1, use -Efor 5 -(-Efor 1) = + something which I'll leave for you. You could convert that to wavelength in meters with delta E = hc/wavelength. You might want to try it for the Balmer series (n ending in n = 2). I think the red line is about 650 nm.
Explanation:
Answer:
Ag⁺(aq) + Cl⁻(aq) ==> AgCl(s)
Explanation:
The net ionic equation can be described as the equation that contains only those species which would be participated in the chemical reaction. The spectator ions are the type of the ions that are present in both sides of the chemical equation these ions could not be present in the net ionic equation
First, it is easiest if you write the compete molecular equation:
AgNO₃(aq) + KCl(aq) ⇔ AgCl(s) + KNO₃(aq)
we look up which compounds are soluble (aq) and which are not (s). In this case, silver chloride (AgCl) is not soluble. Thus, the net ionic equation is...
Ag⁺(aq) + Cl⁻(aq) ==> AgCl(s)
