Answer is: the compound is B₂O₃.
ω(O) = 68.94% ÷ 100%.
ω(O) = 0.6894; percentage of oxygen in the compound.
ω(X) = 31.06% ÷ 100%.
ω(X) = 0.3106; percentage of unknown element in the compound.
If we take 69.7 grams of the compound:
M(compound) = 69.7 g/mol.
n(compound) = 69.7 g ÷ 69.7 g/mol.
n(compound) = 1 mol.
n(O) = (69.7 g · 0.6894) ÷ 16 g/mol.
n(O) = 3 mol.
M(compound) = n(O) · M(O) + n(X) · M(X).
n(X) = 1 mol ⇒ M(X) = 21.7 g/mol; there is no element with this molecular weight.
n(X) = 2 mol ⇒ M(X) = 10.85 g/mol; this element is boron (B).
It's testable
(adding random stuff right now to make the answer longer but the answer is it's testable
Which of the following is not a disaccharide:
<span>d)</span>
