432g divided by 200cm^3 = 2.16g/cm^3
To answer this question, you need to know the aluminum acetate weight and formula. Molecular weight of aluminum acetate is 204.11g/mol, so the concentration should be: (25.20/204.11 )/(0.185l) = 0.1234 mol/0.185l= 0.6673 M<span>
</span>The aluminum acetate Al(C2H3O2)3 which mean for one aluminum acetate there will be 3 acetate ion. The concentration of acetate should be: 0.6673 M= 2M
The empirical formula CH₂O has a mass [(12 × 1) + (1 × 2) + (16 × 1)] = 30 g/mol
If the empirical formula is 30 g/mol,
and the molecular formula is 60 g/mol
Then the multiple is = 60 g/mol ÷ 30 g/mol
= 2
Therefor the molecular formula is 2(CH₂O) = C₂H₄O₂ (OPTION 2)
The given complex ion is as follow,
[Ru (CN) (CO)₄]⁻
Where;
[ ] = Coordination Sphere
Ru = Central Metal Atom = <span>Ruthenium
CN = Cyanide Ligand
CO = Carbonyl Ligand
The charge on Ru is calculated as follow,
Ru + (CN) + (CO)</span>₄ = -1
Where;
-1 = overall charge on sphere
0 = Charge on neutral CO
-1 = Charge on CN
So, Putting values,
Ru + (-1) + (0)₄ = -1
Ru - 1 + 0 = -1
Ru - 1 = -1
Ru = -1 + 1
Ru = 0
Result:
<span>Oxidation state of the metal species in each complex [Ru(CN)(CO)</span>₄]⁻ is zero.
