Answer:
Ammonia is an Arrhenius base and a Brønsted-Lowry base.
Explanation:
An Arrhenius base is any substance which, when it is dissolved in an aqueous solution, produces hydroxide (OH^-), ions in solution. An aqueous solution is a solution that has water present in it.
A Bronsted-Lowry base is a substance that accepts a proton, that is, a hydrogen ion (H^+).
Looking at the equation above, ammonia satisfies both characteristics. We can see that when ammonia is dissolved in water, hydroxide ions is produced in the solution. Hence it is an Arrhenius base. Similarly, the hydroxide ion is formed when ammonia accepts a proton. This is a characteristic of a Brownstead-Lowry base. Hence ammonia is both an Arrhenius base and a Brownstead-Lowry base.
There are two possible products from this elimination:
-2,3-dimethylbut-1-ene
-2,3-dimethylbut-2-ene
As the base is relatively unhindered, the reaction will form the Saytzeff product as the major product. The Saytzeff product is the most substituted alkene which is more stable due to hyperconjugation. In this reaction the Saytzeff product is 2,3-dimethylbut-2-ene.
Total number of atoms = 7
Total number of H atom = 5
% of H in ammonium hydroxide = 5/7 ×100 = 71.4 %
Given:
0.060 mol of NiC2O4
Ksp = 4 x 10⁻¹⁰
1.0 L of solution
Kf of Ni(NH3)6 2⁺ = 1.2 x 10⁹
<span>NiC2O4 + 6NH3 ⇋ Ni(NH3)6 2+ + 2O4 2- </span>
<span>NiC2O4 ⇋ Ni 2+ + C2O4 2- ...Ksp </span>
<span>Ni2+ + 6NH3 ⇋ Ni(NH3)6 2+...Kf </span>
Ksp * Kf = (4 x 10⁻¹⁰) * (1.2 x 10⁹) = 0.48
K = 0.48 = [Ni(NH3)6 2+][C2O4 2-] / [NH3]⁶<span>
</span>0.48 = (0.060)² / [NH3]⁶<span> ... (dissolved C2O4 2- = 0.060M)
</span><span>[NH3]</span>⁶<span> = (0.060)</span>²<span> / 0.48 = </span>0.0036 / 0.48 = 0.0075
NH3 = ⁶√0.0075
NH3 = 0.44 M
