Answer is: cobalt(II) hydroxide.
Ksp(lead(II) chromate) = [Pb²⁺][CrO₄²⁻]. [Pb²⁺] = √2,8·10⁻¹³ = 5,3·10⁻⁷ M.<span>
Ksp(cobalt(II) hydroxide) = </span>[Co²⁺][OH⁻]². [Co²⁺] = ∛1,3·10⁻¹⁵ = 1,1·10⁻⁵ M.
Ksp(cobalt(II) sulfide) = [Co²⁺][S²⁻]. [Co²⁺] = √5·10⁻²² = 2,23·10⁻¹¹ M.
Ksp(chromium(III) hydroxide) = [Cr³⁺][OH⁻]³. [Cr³⁺] = √(√1,6·10⁻³⁰) = 3,6·10⁻⁸.
Ksp(silver sulfide) = [Ag⁺]²[S²⁻].
Answer:
4.00L
Explanation:
Using Charle's law, which have the following equation:
V1/T1 = V2/T2
Where;
V1 = initial volume (litres)
T1 = initial temperature (Kelvin)
V2 = final volume (litres)
T2 = final temperature (Kelvin)
According to the information provided, T1 = 275K, T2 = 400K, V1 = ?, V2 = 5.82L
Hence, using the formula;
V1/275 = 5.82/400
400 × V1 = 275 × 5.82
400V1 = 1600.5
V1 = 1600.5 ÷ 400
V1 = 4.001
Therefore, volume of the gas before it is heated is 4.00L
Explanation:
Kp remains constant (if T=const.).
If Q<Kp, more reactants are consumed (the direct reaction is in progress). If Q>Kp the reverse reaction is in progress (the products are consumed).
1. A reaction will occur in which PCl5 (g) is consumed
A. If Kp > Qp then 1,2,3 and 5 are F. 4 is T
B. Kc > Qc then 1,3 are T. 2,4,5 are F.
C. Qp > Kp then 1.3.4 are T. 2,5 are F.
