Answer: option D. the ability of a base to react with a soluble metal salt.
Justification:
NaOH is a strong base, which means that in water it will dissociate according to this reaction:
- NaOH(aq) → Na⁺ (aq) + OH⁻ (aq)
On the other hand, CuSO₄ is a soluble ionic salt which in water will dissociate into its ions according to this other reaction:
Hence, in solution, the sodium ion (Na⁺) will react with the metal salt in a double replacement reaction, where the highly reactive sodium ion (Na⁺) will substitute the Cu²⁺ in the CuSO₄ to form the sodium sulfate salt, Na₂SO₄ (water soluble), and the copper(II) hydroxide, Cu(OH)₂ (insoluble).
That is what the given reaction represents:
CuSO₄ (aq) + 2NaOH(aq) → Cu(OH)₂(s) + Na₂SO₄(aq)
↑ ↑ ↑ ↑
soluble metal salt strong base insoluble base solube salt
Unsaturation (IHD) 2 hydrogen Needed
IHD = [(2n+2) -H]/2
(H: X=1, N=-1, O= zero)
Unsaturation:
Double bonds = 1
Rings = 1
Triple Bonds = 2
The degrees of unsaturation in a molecule are additive — a
molecule with one double bond has one degree of unsaturation, a molecule with
two double bonds has two degrees of unsaturation, and so forth.
The micromoles of mercury(II) iodide : 0.013 μ moles
<h3>Further explanation</h3>
Given
215.0mL of a 6.0x10⁻⁵mmol/L HgI₂
Required
micromoles of HgI₂
Solution
Molarity(M) = moles of solute per liters of solution
Can be formulated :
M = n : V
n = moles
V = volume of solution
V = 215 mL = 0.215 L
so moles of solution :
n = M x V
n = 6.10 mmol/L x 0.215 L
n = 1.312 . 10⁻⁵ mmol
mmol = 10³ micromol
so 1.312 mmol = 1.312.10⁻⁵ x 10³ = 0.01312 micromoles ⇒ 2 sif fig = 0.013 μ moles
