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).
I’m pretty one of the answers is water, but I don’t know if there is more
this is a lot sorry i can't help
Ionic bonds are formed when there is complete transfer of valence electrons between two atoms.
Electronegativity tells the trend of an atom to atract electrons.
You should search for the complete set of rules that indicate whether an ionic or covalent bond happens.
There are two relevant rules to state if whether an ionic bond will happen:
- When the difference of electronegativities between the two atoms is greater than 2.0, then the bond is ionic.
- When the difference is between 1.6 and 2.0, the bond is ionic if one of the elements is a metal.
You need to list the electronegativities of the five elements (there are tables with this information)
Element electronegativity
Cu: 1.9
H: 2.2
Cl 3.16
I: 2.66
S: 2.58
Differences:
Cu / S: 2.58 - 1.9 = 0.68
H / S: 2.58 - 2.2 = 0.38
Cl / S: 3.16 - 2.58 =0.58
I / S: 2.66 - 2.58 = 0.08
Those differences are too low to consider that the bond is ionic.
Then the answer is that none of those atoms forms an ionic bond with sulfur.
