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
Na because its a metal. Metals are the best conductors. S and Ne are nonmetals. and Ge is a metalloid. (Metalloids are semi conductors)
Violet, indigo, blue, green, yellow, orange, red
1.07 mol would be your answer
Answer:
3.0 L of NH₃
Solution:
The equation is as follow,
N₂ + 3 H₂ → 2 NH₃
According to equation,
67.2 L (3 mole) H₂ at STP produces = 44.8 L (3 mole) of NH₃
So,
4.50 L of H₂ will produce = X L of NH₃
Solving for X,
X = (4.50 L × 44.8 L) ÷ 67.2 L
X = 3.0 L of NH₃
