Answer:
Cation present- Cr^3+
Cations absent- Al^3+,Fe^3+ and Ni^2+
Explanation:
The cations in group III are; Cr^3+,Al^3+,Fe^3+ and Ni^2+
The first step in the analysis of group III cations is the addition of sodium hypochlorite and sodium hydroxide. Chromium reacts as follows;
2Cr^3+(aq) + 3OCl^-(aq) + 10OH^-(aq) ---------> 2CrO4^2-(aq) + 3Cl^-(aq) + 5H2O(l)
In alkaline solution, Aluminum is separated from Chromium by precipitation. The yellow solution formed is unaffected by treatment with ammonia. However, in acid medium, a blue solution is formed which confirms the presence of Chromium III.
(150 grams)(45/100)=(x)(85.5/100)
67.5=.855(x)
x=78.947368 grams of stock solution
Answer:
The value of the mole is equal to the number of atoms in exactly 12 grams of pure carbon-12. 12.00 g C-12 = 1 mol C-12 atoms = 6.022 × 1023 atoms • The number of particles in 1 mole is called Avogadro's Number (6.0221421 x 1023).
Explanation: