Answer:
0.0164 g
Explanation:
Let's consider the reduction of silver (I) to silver that occurs in the cathode during the electroplating.
Ag⁺(aq) + 1 e⁻ → Ag(s)
We can establish the following relations.
- 1 A = 1 C/s
- The charge of 1 mole of electrons is 96,468 C (Faraday's constant)
- 1 mole of Ag(s) is deposited when 1 mole of electrons circulate.
- The molar mass of silver is 107.87 g/mol
The mass of silver deposited when a current of 0.770 A circulates during 19.0 seconds is:

Answer: Option (e) is the correct answer.
Explanation:
A bond that is formed when an electron is transferred from one atom to another results in the formation of an ionic bond.
For example, NaBr will be an ionic compound as there is transfer of electron from Na to Br.
Whereas a bond that is formed by sharing of electrons is known as a covalent bond.
For example,
will be a covalent compound as there is sharing of electron between carbon and bromine atom.
Also, when electrons are shared between the combining atoms and there is large difference in electronegativity of these atoms then partial charges develop on these atoms. As a result, it forms a polar covalent bond.
For example, in a HBr compound there is sharing of electrons between H and Br. Also, due to difference in electronegativity there will be partial positive charge on H and partial negative charge on Br.
Thus, we can conclude that out of the given options HBr is the only compound that has polar covalent bonds.
Answer:
Kc = 2.145 × 10⁻⁸¹
Explanation:
Let's consider the following reaction:
O₂(g) ⇄ 2O(g)
The standard Gibbs free energy for the reaction (ΔG°) can be calculated using the following expression:
ΔG° = Σnp. ΔG°f(p) - Σnp. ΔG°f(p)
where,
ni are the moles of products and reactants
ΔG°f(p) are the standard Gibbs free energy of formation of products and reactants
In this case,
ΔG° = 2 × ΔG°f(O) - 1 × ΔG°f(O₂)
ΔG° = 2 × 230.1 kJ/mol - 1 × 0 kJ/mol
ΔG° = 460.2 kJ/mol
With this information, we can calculate the equilibrium constant (Kc) using the following expression:
