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.
C. NaOH ammmonia is also an base but not as strong as NaOH
It would be potassium chloride or kcl
Answer:
26.0 g/mol is the molar mass of the gas
Explanation:
We have to combine density data with the Ideal Gases Law equation to solve this:
P . V = n . R .T
Let's convert the pressure mmHg to atm by a rule of three:
760 mmHg ____ 1 atm
752 mmHg ____ (752 . 1)/760 = 0.989 atm
In density we know that 1 L, occupies 1.053 grams of gas, but we don't know the moles.
Moles = Mass / molar mass.
We can replace density data as this in the equation:
0.989 atm . 1L = (1.053 g / x ) . 0.082 L.atm/mol.K . 298K
(0.989 atm . 1L) / (0.082 L.atm/mol.K . 298K) = 1.053 g / x
0.0405 mol = 1.053 g / x
x = 1.053 g / 0.0405 mol = 26 g/mol
I would say diamond, because it consists only of carbon
