Two Half Cell reactions are as follow,
3e⁻ + Al⁺³ → Al (Reduction Reaction)
Mg → Mg⁺² + 2e⁻ (Oxidation Reaction)
In order to balance the number of electrons, multiply reduction reaction by 2 and oxidation reaction by 3.
So,
6e⁻ + 2 Al⁺³ → 2 Al
3 Mg → 3 Mg⁺² + 6e⁻ (e⁻ cancelled)
_______________________
2 Al⁺³ + 3 Mg → 2 Al + 3 Mg⁺²
Result:
At Anode Magnesium metal is Oxidized and At Cathode Aluminium Ions are reduced.
Answer:
Here's what I find.
Explanation:
a. Structure
Acebutolol is a secondary amine (basic). It forms a substituted ammonium salt when treated with hydrochloric acid.
The structure of the salt is shown below, with a red arrow pointing toward the positive charge on the N atom.
b. Solubility
The formula of acebutolol is C₁₈H₂₈N₂O₄.
The amide, acetyl, and ether groups confer little solubility to the molecule.
The alcohol and secondary amine do confer some solubility, because they can donate and accept hydrogen bonds.
However, they can each overcome the hydrophobic properties of only three to five carbons, and acebutolol has 18 of them.
The free amine would be preferentially soluble in lipid material (fats)
The protonated amine is ionic and therefore much more soluble in aqueous media (e.g., blood).
c. Marketing
The drug must be delivered to the tissues of the heart, where it blocks the effects of adrenalin. The best way to do this is through the blood, so acebutolol is marketed as the hydrochloride salt.
Some examples of malleable materials are gold, silver, iron, aluminum, copper and tin.
Answer:
I think the correct answer is a
Explanation:
that's the only part that could really be used to protect itself,
.22M because u convert 100g to mols which is around .56mol then divide by 2.5