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:
"Spherical: Bacteria shaped like a ball are called cocci, and a single bacterium is a coccus. Examples include the streptococcus group, responsible for “strep throat.”
Rod-shaped: These are known as bacilli (singular bacillus). ...
Spiral: These are known as spirilla (singular spirillus).
" - Medical News Today
Answer:
½O 2 + 2e - + H 2O → 2OH.
Explanation:
Redox reactions - Higher
In terms of electrons:
oxidation is loss of electrons
reduction is gain of electrons
Rusting is a complex process. The example below show why both water and oxygen are needed for rusting to occur. They are interesting examples of oxidation, reduction and the use of half equations:
iron loses electrons and is oxidised to iron(II) ions: Fe → Fe2+ + 2e-
oxygen gains electrons in the presence of water and is reduced: ½O2 + 2e- + H2O → 2OH-
iron(II) ions lose electrons and are oxidised to iron(III) ions by oxygen: 2Fe2+ + ½O2 → 2Fe3+ + O2-
<u>Given information:</u>
Mass of H2 = 2 g
Mass of O2 = 32 g
<u>To determine:</u>
Mass of H2O2 produced
<u>Explanation:</u>
The reaction between H2 and O2 can be given as:
H2 + O2 → H2O2
Based on the reaction stoichiometry:
1 mole of H2 reacts with 1 mole of O2 to form 1 mole of H2O2
# moles of H2 = mass of H2 / molar mass of H2 = 2 g/ 2 g.mol-1 = 1 mole
# moles of O2 = mass of O2/ molar mass of O2 = 32 g/ 32 g.mol-1 = 1 mole
Hence for the given reactant conditions, moles of H2O2 produced = 1
Mass of H2O2 = moles of H2O2 * molar mass H2O2 = 1 mole * 34 g.mole-1 = 34 g
<u>Ans</u>: 34 g of H2O2 is produced in this reaction
