Answer is: an oxybromate compound is KBrO₄ (x = 4).
ω(Br) = 43.66% ÷ 100%.
ω(Br) = 0.4366; mass percentage of bromine.
If we take 100 grams of compound:
m(Br) = ω(Br) · 100 g.
m(Br) = 0.4366 · 100 g.
m(Br) = 43.66 g; mass of bromine.
n(Br) = m(Br) ÷ M(Br).
n(Br) = 43.66 g ÷ 79.9 g/mol,
n(Br) = 0.55 mol; amoun of bromine.
From chemical formula (KBrOₓ), amount of potassium is equal to amount of bromine: n(Br) = n(K).
m(K) = 0.55 mol · 39.1 g/mol.
m(K) = 21.365 g; mass of potassium in the compound.
m(O) = 100 g - 21.365 g - 43.66 g.
m(O) =34.97 g; mass of oxygen.
n(O) = 34.97 g ÷ 16 g/mol.
n(O) = 2.185 mol.
n(K) : n(Br) : n(O) = 0.55 mol : 0.55 mol : 2.185 mol /÷ 0.55 mol.
n(K) : n(Br) : n(O) = 1 : 1 : 4.
Explanation:
Ionic equation
NaCl(aq) --> Na+(aq) + Cl-(aq)
Na2SO4(aq) --> 2Na+(aq) + SO4^2-(aq)
In NaCl solution, 1 mole of Na+ is dissociated in 1 liter of solution while in Na2SO4, 2 moles of Na+ is dissociated in 1 liter of solution.
Molecular weight of NA2SO4 = (23*2) + 32 + (16*4)
= 142 g/mol
Molecular weight of NaCl = 23 + 35.5
= 58.5 g/mol
Masses
% Mass of NA+ in Na2SO4 = mass of Na+/total mass of Na2SO4 * 100
= 46/142 * 100
= 32.4%
% Mass of NA+ in NaCl = mass of Na+/total mass of NaCl * 100
= 23/58.5 * 100
= 39.3%
Therefore, the % mass of Na+ in NaCl and Na2SO4 are different so it cannot be used.
The 3d sublevel is not filled until after the 4s sublevel, because the 3d sublevel has more energy than the 4s sublevel, and less energy than the 4p sublevel. (:
Explanation:
All matter has mass and occupies space. Volume is a measure of how much space an object occupies. ... "Mass" only depends on how much matter is in an object. "Weight", on the other hand, depends on how strongly gravity pulls on an object.