Answer:
(c) only Ca2+(aq) and Hg2+(aq)
Explanation:
- In the first step, hydrochloric acid (HCl) is added to the solution. In this case the equilibrium that could take place is:
Ag⁺(aq) + Cl⁻(aq) ↔ AgCl(s)
But no precipitate was formed, so Ag⁺(aq) is absent.
- By adding H₂SO₄(aq) the next equilibrium that could take place is:
Ca⁺²(aq) + SO₄⁻²(aq) ↔ CaSO₄(s)
A white precipitate was formed, so Ca⁺² is present in the solution.
- The following could take place after adding H₂S(aq):
Hg²⁺(aq) + S⁻² ↔ HgS(s)
A black precipitate formed, so Hg⁺² is present as well.
The molecules are more loose and not as compact and bonded together by hydrogen bonds as solids. Liquid water also has an indefinite shape meaning it can shift into anything
Answer:
B = - 0.0326 dm³/mol
Explanation:
virial eq until second term:
∴ P = 10 bar * (atm/ 1.01325 bar) = 9.869 atm
∴ T = 200°C = 473 K
∴ Vm = 3.90 dm³/mol
∴ R = 0.08206 dm³.atm/K.mol
⇒ PVm / RT = 1 + B/Vm
⇒ ((9.869 atm)*(3.90 dm³/mol)) / ((0.08206 dm³.atm/mol.K)*(473K)) = 1 + B/Vm
⇒ 0.99164 = 1 + B/Vm
⇒ B/Vm = - 8.357 E-3
⇒ B = (3.90 dm³/mol)*( - 8.357 E-3 )
⇒ B = - 0.0326 dm³/mol
Yes, that is true. in order for it to be a redox reaction, both oxidation and reduction must be occurring.
