Answer:
0.057 M
Explanation:
Step 1: Given data
Solubility product constant (Ksp) for HgBr₂: 2.8 × 10⁻⁴
Concentration of mercury (II) ion: 0.085 M
Step 2: Write the reaction for the solution of HgBr₂
HgBr₂(s) ⇄ Hg²⁺(aq) + 2 Br⁻
Step 3: Calculate the bromide concentration needed for a precipitate to occur
The Ksp is:
Ksp = 2.8 × 10⁻⁴ = [Hg²⁺] × [Br⁻]²
[Br⁻] = √(2.8 × 10⁻⁴/0.085) = 0.057 M
Elements in the third row can break the octet rule
Answer:
1. The metal atom/ion in these compounds are Ni and Ni2+ respectively.
2. The electrons from s oribital will jump to d orbital and so I expect CO to donate electron pairs in 4p and 4s orbitals and form sp3 hybridisation.
Answer: Your Answer is carbon dioxide, water, nutrients, and energy from sunlight.
