Answer: The product from the reduction reaction is
CH3-CH2-CH(CH3)-CH2-CH2OH
IUPAC name; 3- Methylpentan-1-ol
Explanation:
Since oxidation is simply the addition of oxygen to a compound and reduction is likewise the addition of hydrogen to a compound.
Therefore, hydrogen is added onto the carbon atom adjacent to oxygen in 3- methyl pentanal
CH3 CH2 CHCH3 CH2 CHO thereby -CHO( aldehyde functional group) are reduced to CH2OH ( Primary alcohol) which gives;
3-methylpenta-1-ol .
The structure of the product is:
CH3-CH2-CH(CH3)-CH2-CH2OH
They occur between polar molecules
Ka is the acid dissociation equilibrium constant. The larger the value of the Ka, the stronger is the acid. To find Ka from pKa, the equation is:
pKa = -log[Ka]
@pKa = 7
7 = -log[Ka]
Ka = 1×10⁻⁷
@pKa = 10
10 = -log[Ka]
Ka = 1×10⁻¹⁰
This, pKa 7 is more acidic than pKa 10. The scale factor would be:
1×10⁻⁷/1×10⁻¹⁰ = 1,000
<em>Therefore, Compound A is 1,000 times more acidic than Compound B.</em>
Answer:
<h3>1. 10 e⁻</h3>
Oxidation numbers
I₂O₅(s): I (5+); O(2-)
CO(g): C(2+); O(2-)
I₂(s): I(0)
CO₂(g): C(4+); O(2-)
<h3>2. 4 e⁻</h3>
Oxidation numbers
Hg²⁺(aq): Hg(2+)
N₂H₄(aq): N(2-); H(1+)
Hg(l): Hg(0)
N₂(g): N(0)
H⁺(aq): H(1+)
<h3>3. 6 e⁻</h3>
Oxidation numbers
H₂S(aq): H(1+); S(2-)
H⁺(aq): H(1+)
NO₃⁻(aq): N(5+); O(2-)
S(s): S(0)
NO(g): N(2+); O(2-)
H₂O(l): H(1+); O(2-)
Explanation:
In order to state the total number of electrons transferred we have to identify both half-reactions for each redox reaction.
1. I₂O₅(s) + 5 CO(g) → I₂(s) + 5 CO₂(g)
Oxidation: 10 e⁻ + 10 H⁺(aq) + I₂O₅(s) → I₂(s) + 5 H₂O(l)
Reduction: 5 H₂O(l) + 5 CO(g) → 5 CO₂(g) + 10 H⁺(aq) + 10 e⁻
2. 2 Hg²⁺(aq) + N₂H₄(aq) → 2 Hg(l) + N₂(g) + 4 H⁺(aq)
Oxidation: N₂H₄(aq) → N₂(g) + 4 H⁺(aq) + 4 e⁻
Reduction: 2 Hg²⁺(aq) + 4 e⁻ → 2 Hg(l)
3. 3 H₂S(aq) + 2H⁺(aq) + 2 NO₃⁻(aq) → 3 S(s) + 2 NO(g) + 4H₂O(l)
Oxidation: 3 H₂S(aq) → 3 S(s) + 6 H⁺(aq) + 6 e⁻
Reduction: 8 H⁺(aq) + 2 NO₃⁻(aq) + 6 e⁻ → 2 NO(g) + 4 H₂O
