2HF + CaCO₂H₂--> 2H₂O + CaF₂ In the equation above 2 moles of HF react with mole(s) of CaCO2H2. *

Answerthe half-reaction having the greatest tendency to occur at the anode:     i₂(aq) + 2e⁻  ⇄ 2i⁻  ;   e = +0.53 v======oxidation occurs at the anode. we must therefore find the half-reaction with the greatest tendency to oxidize.the given values of e are reduction potentials (the form of the half-reactions are of reduction).the lower the reduction potential, the higher the potential to oxidize. the half-cell with the lowest reduction potential will have the greatest tendency to oxidize at the anode.the half-reaction having the greatest tendency must be    i₂(aq) + 2e⁻  ⇄ 2i⁻    e = +0.53 vbecause its reduction potential is the lowest out of the three.

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Six base units were recommended: the metre, kilogram,second, ampere, degree kelvin (later renamed kelvin), and candela. in 1960, the 11th  cgpm  named the system the international system of units, abbreviated si from the french name, le système international  d'unités.