The first step in the two-step process of making a solution is the breakdown of the solute source into Atomic particles
for a solution to break the solute must be dissociated and break into the atomic particles
so correct option is D
hope it helps
Balanced equation for the above reaction is as follows;
Mg(OH)₂ + 2HCl ---> MgCl₂ + 2H₂O
stoichiometry of Mg(OH)₂ to MgCl₂ is 1:1
mass of Mg(OH)₂ reacted - 1.82 g
number of moles of Mg(OH)₂ - 1.82 g/ 58.3 g/mol = 0.0312 mol
number of Mg(OH)₂ moles reacted - number of MgCl₂ moles formed
number of MgCl₂ moles formed - 0.0312 mol
mass of MgCl₂ formed - 0.0312 mol x 95.2 g/mol = 2.97 g
mass of MgCl₂ formed - 2.97 g
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:
I think the best answer is B. Even this is the broadest case for the Conservation of matter and the one for Energy, the only way this can be applied is in nuclear rxns.
