If one mole of aluminum chloride dissociates into one mole of aluminum and three moles of chloride ions, how many chloride ions will be produced when 5.22 moles of aluminum chloride dissociates?

The answer is: 15.66 moles of chloride ions.

Dissociation of aluminium chloride in water: AlCl₃(aq) → Al³⁺(aq) + 3Cl⁻(aq).

n(AlCl₃) = 5.22 mol; amount of aluminium chloride.

From chemical reaction: n(AlCl₃) : n(Cl⁻) = 1 : 3.

n(Cl⁻) = 3 · 5.22 mol.

n(Cl⁻) = 15.66 mol; amount of chloride ions.

N(Cl⁻) = n(Cl⁻) · Na.

N(Cl⁻) = 15.66 mol · 6.022·10²³ 1/mol.

N(Cl⁻) = 9.43·10²⁴; number of chloride ions.

the answer is   c.) 1575 j.

explanation:

a). in this case, dh> 0 and sd> 0. if t becomes larger, the term "-tds" will become more negative. at some point, it is possible that -dts will be more negative than the dh is positive, and the reaction will become spontaneous. this is probably the correct answer, but you should look at all of the other possibilities and make sure of that.