Answer:
-3
Explanation:
The oxidation state or oxidation number of an atom is the total number of electrons that an atom either gains or loses in order to form a chemical bond with another atom.
The complex anion here is [Cr(CN)6]3-.
Now, as the oxidation state of CN or cyanide ligand is -1, and if we suppose the oxidation state of Cr to be 'x', then; x - 6 = -3 (overall charge on the anion),
so x= +3. Hence the oxidation state of Chromium in this complex hexacyanochromium (III) anion comes out to be -3.
.
Answer:
molarity= 0.238 mol L-
Explanation:
The idea here is that you need to use the fact that all the moles of sodium phosphate that you dissolve to make this solution will dissociate to produce sodium cations to calculate the concentration of the sodium cations.
Na 3 PO 4 (aq) → Na + (aq) + PO3−4 (aq)
Use the molar mass of sodium phosphate to calculate the number of moles of salt used to make this solution.
3.25g⋅1 mole N 3PO4 163.9g = 0.01983 moles Na3 PO 4
Now, notice that every
1 mole of sodium phosphate that you dissolve in water dissociates to produce
3bmoles of sodium cations in aqueous solution.
Answer:
25.35%
Explanation:
Again let me restate the the equation of the reaction;
H2O (ℓ) + 2 MnO4 - (aq) + 3 CN- (aq) → 2 MnO2 (s) + 3 CNO- (aq) + 2 OH- (aq)
Amount of potassium permanganate reacted = 10.2/1000 * 0.08035 = 8.1957 * 10^-4 moles
If 2 moles of MnO4 - reacts with 3 moles of CN-
8.1957 * 10^-4 moles of MnO4 - reacts with 8.1957 * 10^-4 * 3/2
= 1.229 * 10^-3 moles of CN-
Mass of CN- reacted = 1.229 * 10^-3 moles of CN- * 26.02 g/mol
= 0.03 g
Hence, percentage of the cyanide = 0.03 g/0.1183 g * 100
= 25.35%
Answer:bubuvuvuvftctcrchj
Explanation:
Answer:
2Li(s) + 2H₂O(ℓ) ⟶ 2Li⁺(aq) + 2OH⁻(aq) + H₂(g)
Explanation:
An ionic equation uses the symbols (aq) [aqueous] to indicate molecules and ions that are soluble in water, (s) [solid] to indicate insoluble solids, and (ℓ) to indicate substances (usually water) in the liquid state.
In this reaction, solid lithium reacts with liquid water to form soluble lithium hydroxide and gaseous hydrogen
.
1. Molecular equation
2Li(s) + 2H₂O(ℓ) ⟶ 2LiOH(aq) + H₂(g)
2. Ionic equation
Lithium hydroxide is a soluble ionic compound, so we write it as hydrated ions.
2Li(s) + 2H₂O(ℓ) ⟶ 2Li⁺(aq) + 2OH⁻(aq) + H₂(g)
