If by classical you mean formula name it is CrPO4<span>.</span>
<span>Percent yield is the measured recovery of a substance in grams divided by the theoretical amount of substance expected in grams (times 100 to convert to %). Undissolved Aluminum trihydroxide would add mass to the Aluminum sulphate one is making here. The measured mass would be inaccurately high, increasing the percent yield calculated.</span>
Answer:
figure no. A is right answer.......
Answer:
5x10⁻⁶ = [HTeH₄O₆⁺]
Explanation:
The first dissociation equilibrium of the telluric acid in water is:
H₂TeH₄O₆ + H₂O ⇄ HTeH₄O₆⁺ + H₃O⁺
Using H-H equation for telluric acid:
<em>pH = pKa + log₁₀ [HTeH₄O₆⁺] / [H₂TeH₄O₆]</em>
pKa of telluric acid is -logKa1
pKa = -log 2.0x10⁻⁸
pKa = 7.699
As concentration of [H₂TeH₄O₆] is 0.25M, replacing in H-H equation:
3.00 = 7.699+ log₁₀ [HTeH₄O₆⁺] / [0.25M]
-4.699 = log₁₀ [HTeH₄O₆⁺] / [0.25M]
2x10⁻⁵ = [HTeH₄O₆⁺] / [0.25M]
<h3>5x10⁻⁶ = [HTeH₄O₆⁺]</h3>
Answer:
The work done by the gas expansion is 5875 J,
Since the work done is positive, the work is done by the gas on the surroundings.
Explanation:
Given;
change in internal energy, ΔU = -4750 J
heat transferred to the system, Q = 1125 J
The change in internal energy is given by;
ΔU = Q - W
Where;
W is the work done by the system
The work done by the system is calculated as;
W = Q - ΔU
W = 1125 - (-4750)
W = 1125 + 4750
W = 5875 J
Since the work done is positive, the work is done by the gas on the surroundings (energy flows from the gas to the surroundings).
Therefore, the work done by the gas expansion is 5875 J
