Answer:
just use the tongs and put it on a plate
Explanation:
Answer:
C) The compound is largely ionic with A as the cation.
Explanation:
Pulings proposed the method to determine if the compound is ionic in nature or covalent in nature , by finding the difference between the electronegativity of the respective cation and anion .
The ion with higher electronegativity is the anion and the ion with lower electronegativity is the cation.
The electronegativity difference above 1.7 make the compound ionic in nature.
Hence, from the question ,
A is the cation and B is the anion.
And the electronegativity difference above 1.7 so the compound is ionic in nature.
Answer:
It would be 5.8 times 10^9cm
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
