Sodium metal forms at the cathode
Answer:
ruthenium (Ru), rhodium (Rh), palladium (Pd), osmium (Os), iridium (Ir), platinum (Pt), gold (Au), silver (Ag).
Explanation:
Answer:
Percentage dissociated = 0.41%
Explanation:
The chemical equation for the reaction is:

The ICE table is then shown as:

Initial (M) 1.8 0 0
Change (M) - x + x + x
Equilibrium (M) (1.8 -x) x x
![K_a = \frac{[C_3H_6ClCO^-_2][H^+]}{[C_3H_6ClCO_2H]}](https://tex.z-dn.net/?f=K_a%20%20%3D%20%5Cfrac%7B%5BC_3H_6ClCO%5E-_2%5D%5BH%5E%2B%5D%7D%7B%5BC_3H_6ClCO_2H%5D%7D)
where ;


Since the value for
is infinitesimally small; then 1.8 - x ≅ 1.8
Then;




Dissociated form of 4-chlorobutanoic acid = 
Percentage dissociated = 
Percentage dissociated = 
Percentage dissociated = 0.4096
Percentage dissociated = 0.41% (to two significant digits)
Answer:
-3.7771 × 10² kJ/mol
Explanation:
Let's consider the following equation.
3 Mg(s) + 2 Al³⁺(aq) ⇌ 3 Mg²⁺(aq) + 2 Al(s)
We can calculate the standard Gibbs free energy (ΔG°) using the following expression.
ΔG° = ∑np . ΔG°f(p) - ∑nr . ΔG°f(r)
where,
n: moles
ΔG°f(): standard Gibbs free energy of formation
p: products
r: reactants
ΔG° = 3 mol × ΔG°f(Mg²⁺(aq)) + 2 mol × ΔG°f(Al(s)) - 3 mol × ΔG°f(Mg(s)) - 2 mol × ΔG°f(Al³⁺(aq))
ΔG° = 3 mol × (-456.35 kJ/mol) + 2 mol × 0 kJ/mol - 3 mol × 0 kJ/mol - 2 mol × (-495.67 kJ/mol)
ΔG° = -377.71 kJ = -3.7771 × 10² kJ
This is the standard Gibbs free energy per mole of reaction.
Answer:
No
Explanation:
The solubility of a solid in water refers to the amount of that solid that dissolves in water.
It is not possible to calculate the solubility of the solid because the student threw away the first precipitate that formed. We already have the volume of water, but having lost some mass of precipitate, it has become impossible to accurately determine the solubility.
Hence the answer provided above.