Answer:
% of n-propyl chloride = 43.48 %
Explanation:
There are 2 secondary hydrogens and 6 primary hydrogens
The rate of abstraction of seondary hydrogen = 3.9 X rate of abstraction of primary hydrogen
probability of formation of isopropyl chloride = 3.9 X 1 (relative rate X relative number of secondary hydrogens)
Probability of formation of n-propyl chloride = 1 X 3 (relative rate X relative number of primary hydrogens)
Total probability = 3.9
% of n-propyl chloride = 3 X 100 / 6.9 = 43.48 %
Answer:
D. Ni²⁺
Explanation:
We know at once that the answer cannot be A or C, because Ni and Cu are already in their lowest oxidation states.
The correct answer must be either B or D.
An electrolytic cell is the opposite of a galvanic cell. In the former, the reaction proceeds spontaneously. In the latter, you must force the reaction to occur.
One strategy to solve this problem is:
- Look up the standard reduction potentials for the half reaction·
- Figure out the spontaneous direction.
- Write the equation in the reverse direction.
1. Standard reduction potentials
E°/V
Cu²⁺ + 2e⁻ ⟶ Cu; 0.3419
Ni²⁺ + 2e⁻ ⟶ Ni; -0.257
2. Galvanic Cell
We reverse the direction of the more negative half cell and add.
<u>E°/V
</u>
Ni ⟶ Ni²⁺ + 2e⁻; 0.257
<u>Cu²⁺ + 2e⁻ ⟶ Cu; </u> 0.3419
Ni + Cu²⁺ ⟶ Cu + Ni²⁺; 0.599
This is the spontaneous direction.
Cu²⁺ is reduced to Cu.
3. Electrochemical cell
<u>E°/V</u>
Ni²⁺ + 2e⁻ ⟶ Ni; -0.257
<u>Cu ⟶ Cu²⁺ + 2e⁻; </u> <u>-0.3419</u>
Cu + Ni²⁺ ⟶ Ni + Cu²⁺; -0.599
This is the non-spontaneous direction.
Ni²⁺ is reduced to Ni in the electrolytic cell.
Answer is: <span>1.0 mol X left over.
</span>Chemical reaction: X + 2Y → XY₂.
n(X) = 3,0 mol, excess reactant.
n(Y) = 4,0 mol, limiting reagent.
n - amount of substance.
from reaction: n(X) : n(Y) = 1 : 2.
n(X) : 4 mol = 1 : 2.
n(X) = 2 mol, that reacts.
excess of X: 3 mol - 2 mol = 1 mol.
1Na3PO4 + 3HCl -> 3NaCl + H3PO4
