а о
Explanation:
The given cation:
(Rf₂Al₂F₃)³⁺
The oxidation number gives the extent to which a specie is oxidized in a reaction.
This number is assigned based on some rules:
- Elements in combined state whose atoms combines with themselves have an oxidation number of zero.
- The charge carried on simple ions gives their oxidation number.
- Algebraic sum of all the oxidation numbers of atoms in neutral compound is zero. In an ion with more than one kind of atom, the charge on it is the oxidation number.
for the specie given;
Known:
oxidation number of Al = +3
F = -1
charge = +3
let the oxidation number of Rf = k
2k + 2(3) + 3(-1) = +3
2k + 6 - 3 = 3
2k = 0
k = 0
The oxidation state of rutherfodium is 0
learn more:
Oxidation state brainly.com/question/10017129
#learnwithBrainly
Xylene moles =\frac{17.12}{106.16×1000}=0.00016moles=
106.16×1000
17.12
=0.00016moles
Moles of CO_2 =\frac{56.77}{44.01×1000}=0.0013CO
2
=
44.01×1000
56.77
=0.0013
Moles of H_2O= =\frac{14.53}{18.02×1000}=0.0008H
2
O==
18.02×1000
14.53
=0.0008
Moles ratios
\frac{0.0013}{0.0008}=1.625
0.0008
0.0013
=1.625
\frac{0.0008}{0.0008}=1
0.0008
0.0008
=1
Hence molecular fomula
The empirical formula is C 4H 5.
The molecular formula C8H10
part 1 : the final volume : 1.404 L
part 2 : the initial concentration : 4.06 M
<h3>Further explanation
</h3>
Dilution is the process of adding a solvent to get a more dilute solution.
The moles(n) before and after dilution are the same.
Can be formulated :
M₁V₁=M₂V₂
M₁ = Molarity of the solution before dilution
V₁ = volume of the solution before dilution
M₂ = Molarity of the solution after dilution
V₂ = Molarity volume of the solution after dilution
part 1 :
M₁=44.8%
V₁=0.73 L
M₂=23.3%
part 2 :
V₁=739 ml=0.739 L
V₂=1.5 L
M₂=2
