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:
Idk tbh srry man am in quarantine so I have a big packet to do
Solid- particles are packed tightly together so they don’t move much
Liquid- particles are still close together but move freely
Gas- particles are neither close together nor fixed in place
Answer: Option (A) is the correct answer.
Explanation:
A covalent bond is formed when there occurs sharing of electrons between atoms.
For example, in hydrogen atom there is one electron in its orbit and in a chlorine atom there is 7 valence electron. So, in order to attain stability both hydrogen and chlorine share electrons when they come close to each other.
Whereas except hydrogen and chlorine rest of the given atoms will form ionic bond, that is, bond formed by transfer of electrons.
Thus, we can conclude that a pair of hydrogen and chlorine will form a covalent bond.
Answer:
The solutions should be added in this order NaCl > Na2SO4 > Na2S
Explanation:
Silver is insoluble as a chloride, so the silver ions get precipitated on addition of chloride ion as silver chloride. This means Ag+ would be removed the first.
So we will add NaCl in the first step.
The following reaction will occur.
Ag+ + Cl- → AgCl(s)
Both, Pb2 and Ni are soluble as chlorides. (lead chloride is soluble as a hot solution but will ppt when colder).
When we add Na2SO4, Pb2+ will get precipitated (because it's insoluble) as PbSO4 and Ni will remain soluble as NiSO4 is soluble in water.
The reaction that will occur is:
Pb^2+ + SO4^2- → PbSO4(s)
Nickel is insoluble as a sulfide. So when we will add Na2S, nickel will be precipitated as sulfide and be able to separate and be collected.
The solutions should be added in this order NaCl > Na2SO4 > Na2S
