Answer:
1.634 molL-1
Explanation:
The mol ration between NH3 and HCl is 1 : 1
Using Ca Va / Cb Vb = Na / Nb where a = acid and b = base
Na = 1
Nb = 1
Ca = 0.208 molL-1
Cb = ?
Va = 19.64 mL
Vb = 25.00mL
Solving for Cb
Cb = Ca Va / Vb
Cb = 0.208 * 19.64 / 25.0
Cb = 0.1634 molL-1 (Concentration of diluted ammonia solution)
Using the dilution equation;
C1V1 = C2V2
Initial Concentration, C1 = ?
Initial Volume, V1 = 25.00 mL
Final Volume, V2 = 250 mL
Final Concentration, C2 = 0.1634 molL-1
Solving for C1;
C1 = C2 * V2 / V1
C1 = 0.1634 * 250 / 25.00
C1 = 1.634 molL-1
<span>Based on your information 1000 times greater than pH 13 is the best I can come up </span>with.
Answer:
A. 2.45 L - Answer my answers is right.
Answer:
Hg(NO₃)₂(aq) + Na₂SO₄(aq) → 2NaNO₃(aq) + HgSO₄(s)
Moles of Hg(NO₃)₂ = 55.42 / 324.7 ==> 0.1707 moles
Moles of Na₂SO₄ = 16.642 / 142.04 ==> 0.1172 moles
Limiting reagent is Na₂SO₄ as it controls product formation
Moles of HgSO₄ formed = 0.1172 moles
= 0.1172 x 296.65
= 34.757g
Explanation:
