Answer:
Q = 1.08x10⁻¹⁰
Yes, precipitate is formed.
Explanation:
The reaction of CoF₂ with NaOH is:
CoF₂(aq) + 2 NaOH(aq) ⇄ Co(OH)₂(s) + 2 NaF(aq).
The solubility product of the precipitate produced, Co(OH)₂, is:
Co(OH)₂(s) ⇄ Co²⁺(aq) + 2OH⁻(aq)
And Ksp is:
Ksp = 3x10⁻¹⁶= [Co²⁺][OH⁻]²
Molar concentration of both ions is:
[Co²⁺] = 0.018Lₓ (8.43x10⁻⁴mol / L) / (0.018 + 0.022)L = <em>3.79x10⁻⁴M</em>
[OH⁻] = 0.022Lₓ (9.72x10⁻⁴mol / L) / (0.018 + 0.022)L = <em>5.35x10⁻⁴M</em>
Reaction quotient under these concentrations is:
Q = [3.79x10⁻⁴M] [5.35x10⁻⁴M]²
<em>Q = 1.08x10⁻¹⁰</em>
As Q > Ksp, <em>the equilibrium will shift to the left producing Co(OH)₂(s) </em>the precipitate
Explanation:
As it is given that stream A is super heated. From stream tables, we get that specific enthalpy, (
) is 3054.29 kJ/kg.
For stream B, it is saturated water at 25 degree celsius. It's
is 2546.54 kJ/kg.
Stream C, data will be as follows.
P = 200 kPa,
= 0.9
So,
=
= 504.47 + 0.9 \times 2201.7
= 2486 kJ/kg
Now, energy balance formula will be as follows.
= 
= 
= 28.16 kg/s
Hence, inflow of saturated liquid is 28.16 kg/s. According to steam table, temperature of wet steam is 
Mass flow rate of out flow is 31.16 kg/s.
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