One step in the isolation of pure rhodium metal (Rh) is the precipitation of rhodium(III) hydroxide from a solution containing rhodium(III) sulfate according to the following balanced chemical equation:
Rh2(SO4)3(aq) + NaOH(aq)
If 8.40g of rhodium(III) sulfate reacts with excess sodium hydroxide, what mass of rhodium(III) hydroxide may be produced?
Oa. 5.23 g
Ob. 16.8 g
Oc. 8.48
Od. 1.3 g
Oe. 10.4 g

Answer is: the equilibrium  shifts toward the reactants, to the left side of chemical reaction.it is because lead(ii) nitrate dissociates in water: pb(no₃)₂ → pb²⁺(aq) + 2no₃⁻(aq).according to lechatelier's principle, the position of equilibrium moves to counteract thechange (more lead cations, common ion), the position of equilibrium will move so that the concentration ofproducts decrease and concentration of reactants of chemical reaction increase.read more on -

It's potassium i believe

Tell me the answer choices and i would be able to answer. : )

Can you post list or organisms