If 10.0 mL of a .600 M of HNO3 reacts with 31.0 mL of .700M Ba(OH)2 solution, what is the molarity of Ba(OH)2 after the reaction is complete

0.456M

Explanation:

1. Calculate the number of moles of each reactant:

Number of moles = molarity × volume in liter

HNO₃:

#moles = 0.600M × 10.0mL × 1L/1,000mL = 0.00600 moles

Ba(OH)₂:

#moles = 0.700M × 31.0mL × 1L/1,000mL = 0.0217 moles

2. Reaction

2HNO₃ + Ba(OH)₂ → Ba(NO₃)₂ + 2H₂O

3. Mole ratio

2 mol HNO₃ : 1 mol Ba(OH)₂

4. Excess reactant

As per the mole ratio 0.006 moles of HNO₃ will react with 0.003 moles of Ba(OH)₂.Since there are 0.0217 moles of Ba(OH)₂ available it is in excess.0.0217 moles - 0.003 moles = 0.0187 moles will be left after the reaction is complete

5. Molarity of Ba(OH)₂

Molarity = number of moles / volume of solution in litersMolarity = 0.0187mol / (0.010 + 0.0310)L = 0.456M

What do you need with it’s not a question your just stating what it says

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.