Answer:
ΔG° = -533.64 kJ
Explanation:
Let's consider the following reaction.
Hg₂Cl₂(s) ⇄ Hg₂²⁺(aq) + 2 Cl⁻(aq)
The standard Gibbs free energy (ΔG°) can be calculated using the following expression:
ΔG° = ∑np × ΔG°f(products) - ∑nr × ΔG°f(reactants)
where,
ni are the moles of reactants and products
ΔG°f(i) are the standard Gibbs free energies of formation of reactants and products
ΔG° = 1 mol × ΔG°f(Hg₂²⁺) + 2 mol × ΔG°f(Cl⁻) - 1 mol × ΔG°f(Hg₂Cl₂)
ΔG° = 1 mol × 148.85 kJ/mol + 2 mol × (-182.43 kJ/mol) - 1 mol × (-317.63 kJ/mol)
ΔG° = -533.64 kJ
The greatest molar amount of metal product is obtained from silver (1) nitrate.
Let us determine the mass of metal obtained in each case.
For La^3+;
La^3+ + 3e ----> La
1 mole of La is deposited by 3F
x moles of La is deposited by 10 F
x = 1 × 10/3
x = 3.33 moles
For Zn^2+;
1 mole Zn^2+ is deposited by 2F
x moles of Zn^2+ is deposited by 10 F
x = 5 F
For Ag^+
1 mole of Ag^+ is deposited by 1 F
x moles of Ag^+ is deposited by 10 F
x = 10 moles
For Ba^2+;
1 mole of Ba^2+ is deposited by 2 F
x moles of Ba^2+ is deposited by 10 F
x = 5 moles
Learn more: brainly.com/question/967776
Answer:
D
Explanation:
The high jump of ionization energy indicates that we are trying to remove electron from noble gas configuration state.
The ionization energy data specifies that the Elements are from group 1 at period 3 or greater.
Removing the first electron require 496 kJ and the second ionization energy jump significantly due to the removal of electron from the noble gas configuration which is logical because electron try to maintain the especially stable state.
Answer:
O2
Explanation:
A compound is made of two or more chemical elements. With that in mind, O2 is only one element: oxygen.
