Answer:
ya same here i hope so btw
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
Answer:
0.74M
Explanation:
Step 1 :
Data obtained from the question.
Initial concentration (C1) = 3M
Initial volume (V2) = 185mL
Final volume (V2) = 750mL
Final concentration (C2) =..?
Step 2:
Determination of the new concentration of the solution.
The new concentration of the solution can be obtained by using the dilution formula as shown below:
C1V1 = C2V2
3 x 185 = C2 x 750
Divide both side by 750
C2 = 3 x 185 / 750
C2 = 0.74M
Therefore, the new concentration of the solution is 0.74M
Butane is C₄H₁₀.
The balanced equation is 2 C₄H₁₀ + 13 O₂ <span>→</span> 8 CO₂ + 10 H₂O.
Answer:A
Explanation:2 phosphorus and 2 oxygen
