cite specific situation that you experienced at home or in school where magnet used to seperate mixtures?
1 answer:
You might be interested in
Answer: 35.72 % of Barium ions will be present in the original unknown compound.
Explanation: The reaction of Barium ions and sodium sulfate is:
Here, Sodium sulfate is present in excess, Barium ions are the limiting reagent because it limits the formation of product.
Now, 1 mole of barium sulfate is produced by 1 mole of Barium ions.
Molar mass of Barium sulfate = 233.38 g/mol
Molar mass of Barium ions = 137.327 g/mol
233.38 g/mol of barium sulfate will be produced by 137.323 g/mol of Barium ions, so
0.4105 grams of barium sulfate will be produced by = of Barium ions
Mass of barium ions = 0.2415 grams
To calculate percentage by mass, we use the formula:
Mass of the solution = 0.6760 grams
Putting the value in above equation, we get
% mass of Barium ions = 35.72%.
The Molecule of Sodium Formate along with Formal Charges (in blue) and lone pair electrons (in red) is attached below.
Sodium Formate is an ionic compound made up of a positive part (Sodium Ion) and a polyatomic anion (Formate).
Nomenclature:
In ionic compounds the positive part is named first. As sodium ion is the positive part hence, it is named first followed by the negative part i.e. formate.
Name of Formate:
Formate ion has been derived from formic acid ( the simplest carboxylic acid). When carboxylic acids looses the acidic proton of -COOH, they are converted into Carboxylate ions.
E.g.
HCOOH (formic acid) → HCOO⁻ (formate) + H⁺
H₃CCOOH (acetic acid) → H₃CCOO⁻ (acetate) + H⁺
Formal Charges:
Formal charges are calculated using following formula,
F.C = [# of Valence e⁻] - [e⁻ in lone pairs + 1/2 # of bonding electrons]
For Oxygen:
F.C = [6] - [6 + 2/2]
F.C = [6] - [6 + 1]
F.C = 6 - 7
F.C = -1
For Sodium:
F.C = [1] - [0 + 0/2]
F.C = [1] - [0]
F.C = 1 - 0
F.C = +1
