It’s easy, if the PH of any acidic solution = -Log[H+], where [H+] is hydrogen ion concentration, multiply each term by (-1) then raise each term as a power to (10), so it will become like this:-
[H+] = 10^(-PH)
It will have traveled 0.78 m. You find this by multiplying .013 by 60
Find the number of moles
C = n / V
C(Concentration) = 0.30 moles / L
V ( Volume) = 2 L
n = ??
n = C * V
n = 0.30 mol / L * 2 L
n = 0.60 mol
Find the molar mass
2Na = 23 * 2 = 46 grams
1S = 32 * 1 = 32 grams
O4 = 16 * 4 = 64 grams
Total = 142 grams / mol
Find the mass
n = given mass / molar mass
n = 0.06 mol
molar Mass = 142 grams / mol
given mass = ???
given mass = molar mass * mols
given mass = 142 * 0.6
given mass = 85.2 grams.
85.2 are in a 2 L solution that has a concentration of 0.6 mol/L
They exist in the outer orbitals
I believe it's answer #3. Logically, at least.
You can test #1 through trial and error.
You can experiment #2 also through trial and error.
You cannot test #3 through trial and error, because that would be catastrophic.
You can test #4 through a survey and individual study and data collection.
