Answer:
1. 2 M
2. 2 M
Explanation:
1. Determination of the final concentration.
Initial Volume (V₁) = 2 L
Initial concentration (C₁) = 6 M
Final volume (V₂) = 6 L
Final concentration (C₂) =?
The final concentration can be obtained as follow:
C₁V₁ = C₂V₂
6 × 2 = C₂ × 6
12 = C₂ × 6
Divide both side by 6
C₂ = 12 / 6
C₂ = 2 M
Therefore, the final concentration of the solution is 2 M
2. Determination of the final concentration.
Initial Volume (V₁) = 0.5 L
Initial concentration (C₁) = 12 M
Final volume (V₂) = 3 L
Final concentration (C₂) =?
The final concentration can be obtained as follow:
C₁V₁ = C₂V₂
12 × 0.5 = C₂ × 3
6 = C₂ × 3
Divide both side by 3
C₂ = 6 / 3
C₂ = 2 M
Therefore, the final concentration of the solution is 2 M
Answer:
Lead(II) sulfate
Explanation:
This looks like a double displacement reaction, in which the cations change partners with the anions.
The possible products are
Pb(NO₃)₂ (aq)+ Na₂SO₄(aq) ⟶PbSO₄(?) + 2NaNO₃(?)
To predict the product, we must use the solubility rules. Two important ones for this question are:
- Salts containing Group 1 elements are soluble.
- Most sulfates are soluble, but PbSO₄ is an important exception.
Thus, NaNO₃ is soluble and PbSO₄ is the precipitate.
Answer:
Please explain the question nicely first
