Answer: the percentage of acetic acid will be low.
Explanation: The major aim during titration of acids and bases is to determine the endpoint , that is exact point where the acid in the beaker changes colour, (in this case, pink )with an additional drop from the burette containing the base, since it is usually difficult to mark the equivalence point that tells us when all the substrate in the beaker has been neutralized completely with the buretted substance.
Overshooting the end point is an error which can occur when the person involved in the the titration accidently goes beyond this endpoint by adding too much of the substance(base) from the burette into the beaker missing the exact endpoint.
This implies that the person has added too much of the burreted liquid, ie the base than required , making the acid in the beaker to continue to react resulting to a lower concentration of the acid (acetic acid) with excess base.(NaOH)
Answer is: Ksp for strontium arsenate is 2.69·10⁻¹⁸.
Balanced chemical reaction (dissociation):
Sr₃(AsO₄)₂(s) → 3Sr²⁺(aq) + AsO₄³⁻(aq).
s(Sr₃(AsO₄)₂) = 0.0650 g/L.
s(Sr₃(AsO₄)₂) = 0.0650 g/L ÷ 540.7 g/mol = 1.2·10⁻⁴ mol/L.
s(Sr²⁺) = 3s(Sr₃(AsO₄)₂).
s(AsO₄³⁻) = 2s(Sr₃(AsO₄)₂).
Ksp = s(Sr²⁺)³ · s(AsO₄³⁻)².
Ksp = (3s)³ · (2s)².
Ksp = 108s⁵.
Ksp = 108 · (1.2·10⁻⁴ mol/L)⁵ = 2.69·10⁻¹⁸.
I don't understand what is (g).
Maybe the answer is 2H<span>(aq)S</span>₂<span>−2(aq) </span>⇒ <span>H</span>₂<span>S</span>₂.
Answer:
670.68°C
Explanation:
Given that:
volume of water = 50 ml but 1 g = 1 ml. Therefore the mass of water (m) = 50 ml × 1 g / ml = 50 g
specific heat (C) = 4.184 J/g˚C
Initial temperature = 20°C, final temperature = 22°C. Therefore the temperature change ΔT = final temperature - initial temperature = 22 - 20 = 2°C
The quantity of heat (Q) used to raise the temperature of a body is given by the equation:
Q = mCΔT
Substituting values:
Q = 50 g × 4.184 J/g˚C × 2°C = 418.4 J
Since the mass of lead = 5 g and specific heat = 0.129 J/g˚C. The heat used to raise the temperature of water is the same heat used to raise the temperature of lead.
-Q = mCΔT
-418.4 J = 5 g × 0.129 J/g˚C × ΔT
ΔT = -418.4 J / ( 5 g × 0.129 J/g˚C) = -648 .68°C
temperature change ΔT = final temperature - initial temperature
- 648 .68°C = 22°C - Initial Temperature
Initial Temperature = 22 + 648.68 = 670.68°C
