oWhat is the change in free energy at 298 K when 70.0 mL of 0.704 M calcium chloride is combined with 47.9 mL of 0.859 M sodium carbonate? (R = 8.31 J/(K ∙ mol) )
Ca²⁺(aq) + CO₃^2-(aq) → CaCO₃(s)

Question

oWhat is the change in free energy at 298 K when 70.0 mL of 0.704 M calcium chloride is combined with 47.9 mL of 0.859 M sodium carbonate? (R = 8.31 J/(K ∙ mol) ) Ca<sup>2+</sup>(aq) + CO<sub>3</sub><sup>2-</sup>(aq) → CaCO<sub>3</sub>(s) A) 42.6 kJ B) -2.21 × 10<sup>3</sup> kJ C) 46.1 kJ D) 47.4 kJ E) 2.56 × 10<sup>3</sup> kJ

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The Answer of oWhat is the change in free energy at 298 K when 70.0 mL of 0.704 M calcium chloride is combined with 47.9 mL of 0.859 M sodium carbonate? (R = 8.31 J/(K ∙ mol)) Ca²⁺(aq)+ CO₃^2-(aq)→ CaCO₃(s) A)42.6 kJ B)-2.21 × 10³ kJ C)46.1 kJ D)47.4 kJ E)2.56 × 10³ kJ

OWhat Is the Change in Free Energy at 298 K