The free energy change(Gibbs free energy-ΔG)=-8.698 kJ/mol
<h3>Further explanation</h3>Given
Ratio of the concentrations of the products to the concentrations of the reactants is 22.3
Temperature = 37 C = 310 K
ΔG°=-16.7 kJ/mol
Required
the free energy change
Solution
Ratio of the concentration : equilbrium constant = K = 22.3
We can use Gibbs free energy :
ΔG = ΔG°+ RT ln K
R=8.314 .10⁻³ kJ/mol K
0.127 mol Au
<h3>General Formulas and Concepts:</h3><u>Math</u>
Order of Operations: BPEMDAS
<u>Chemistry</u>
<u>Atomic Structure</u>
<u>Stoichiometry</u>
<u>Step 1: Define</u>
[Given] 25.0 g Au
[Solve] moles Au
<u>Step 2: Identify Conversions</u>
[PT] Molar Mass of Au - 196.97 g/mol
<u>Step 3: Convert</u>
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
0.126923 mol Au ≈ 0.127 mol Au
Answer: pH of resulting solution will be 13
Explanation:
pH is the measure of acidity or alkalinity of a solution.
Moles of ion =
Moles of ion =
For neutralization:
1 mole of ion will react with 1 mole of
ion
0.01 mol of ion will react with =
of
ion
Thus (0.012-0.01)= 0.002 moles of are left in 20 ml or 0.02 L of solution.
Thus the pH of resulting solution will be 13