Sorry if I'm wrong but it's 1) learner engages in scientifically oriented questions. 2) learner gives priority to evidence in responding to questions. 3) learners formulate explanations from evidence. 4) learner connects explanations to scientific knowledge. 5) learner communicates explanations.
It is B: Fixed volume, takes the shape of the container
This is an incomplete question, here is a complete question.
A 0.130 mole quantity of NiCl₂ is added to a liter of 1.20 M NH₃ solution. What is the concentration of Ni²⁺ ions at equilibrium? Assume the formation constant of Ni(NH₃)₆²⁺ is 5.5 × 10⁸
Answer : The concentration of 
ions at equilibrium is, 
Explanation : Given,
Moles of 
= 0.130 mol
Volume of solution = 1 L
Concentration of = Concentration of = 0.130 M
Concentration of 
= 1.20 M
The equilibrium reaction will be:
![Ni^{2+}(aq)+6NH_3(aq)\rightarrow [Ni(NH_3)_6]^{2+}](https://tex.z-dn.net/?f=Ni%5E%7B2%2B%7D%28aq%29%2B6NH_3%28aq%29%5Crightarrow%20%5BNi%28NH_3%29_6%5D%5E%7B2%2B%7D)
Initial conc. 0.130 1.20 0
At eqm. x [1.20-6(0.130)] 0.130
= 0.42
The expression for equilibrium constant is:
![K_f=\frac{[Ni(NH_3)_6^{2+}]}{[Ni^{2+}][NH_3]^6}](https://tex.z-dn.net/?f=K_f%3D%5Cfrac%7B%5BNi%28NH_3%29_6%5E%7B2%2B%7D%5D%7D%7B%5BNi%5E%7B2%2B%7D%5D%5BNH_3%5D%5E6%7D)
Now put all the given values in this expression, we get:
Thus, the concentration of ions at equilibrium is,
No, they are both different
1 answer · Chemistry
Best Answer
Water steam condenses if its pressure is equal to vapor saturation vapor pressure.
Use the Clausius-Clapeyron relation.
I states the temperature gradient of the saturation pressure is equal to the quotient of molar enthalpy of phase change divided by molar volume change due to phase transition time temperature:
dp/dT = ΔH / (T·ΔV)
Because liquid volume is small compared to vapor volume
ΔV in vaporization is approximately equal to to the vapor volume. Further assume ideal gas phase:
ΔV ≈ V_v = R·T/p
Hence
dp/dT = ΔHv / (R·T²/p)
<=>
dlnp/dT = ΔHv / (R·T²)
If you solve this DE an apply boundary condition p(T₀)= p₀.
you get the common form:
ln(p/p₀) = (ΔHv/R)·(1/T₀ - 1/T)
<=>
p = p₀·exp{(ΔHv/R)·(1/T₀ - 1/T)}
For this problem use normal boiling point of water as reference point:
T₀ =100°C = 373.15K and p₀ = 1atm
Therefore the saturation vapor pressure at
T = 350°C = 623.15K
is
p = 1atm ·exp{(40700J / 8.314472kJ/mol)·(1/373.15K - 1/623.15K)} = 193 atm
hope this helps
