Most properly it is 1-10 centimeters
The molarity of a solution is defined as the moles of solute dissolved per kilogram of solvent. Therefore, we first compute the moles of KOH using:
Moles = mass / Mr
Moles = 23 / 56
Moles = 0.41
The volume of solvent is 1.6 liters
The density is 1 gram/cm³ = 1 kg/L
Mass of solvent = density * volume
Mass of solvent = 1 * 1.6
Mass of solvent = 1.6 kg
Molality = moles / kilogram
Molarity = 0.41 / 1.6
Molarity = 0.26
The molality of the solution is 0.26 molal.
Answer:
Kc = 0.075
Explanation:
The dissociation (α) is the initial quantity that ionized divided by the total dissolved. So, let's calling x the ionized quantity, and M the initial one:
α = x/M
x = M*α
x = 0.354M
For the stoichiometry of the reaction (2:1:1), the concentration of H₂ and I₂ must be half of the acid. So the equilibrium table must be:
2HI(g) ⇄ H₂(g) + I₂(g)
M 0 0 <em> Initial</em>
-0.354M +0.177M +0.177M <em>Reacts</em>
0.646M 0.177M 0.177M <em>Equilibrium</em>
The equilibrium constant Kc is the multiplication of the products' concentrations (elevated by their coefficients) divided by the multiplication of the reactants' concentrations (elevated by their coefficients):
![Kc = \frac{[H2]*[I2]}{[HI]^2}](https://tex.z-dn.net/?f=Kc%20%3D%20%5Cfrac%7B%5BH2%5D%2A%5BI2%5D%7D%7B%5BHI%5D%5E2%7D)
Kc = 0.075
Answer:
1) ΔG°r(298 K) = - 28.619 KJ/mol
2) ΔG°r will decrease with decreasing temperature
Explanation:
- CO(g) + H2O(g) → H2(g) + CO2(g)
1) ΔG°r = ∑νiΔG°f,i
⇒ ΔG°r(298 K) = ΔG°CO2(g) + ΔG°H2(g) - ΔG°H2O(g) - ΔG°CO(g)
from literature, T = 298 K:
∴ ΔG°CO2(g) = - 394.359 KJ/mol
∴ ΔG°CO(g) = - 137.152 KJ/mol
∴ ΔG°H2(g) = 0 KJ/mol........pure substance
∴ ΔG°H2O(g) = - 228.588 KJ/mol
⇒ ΔG°r(298 K) = - 394.359 KJ/mol + 0 KJ/mol - ( - 228.588 KJ/mol ) - ( - 137.152 KJ7mol )
⇒ ΔG°r(298 K) = - 28.619 KJ/mol
2) K = e∧(-ΔG°/RT)
∴ R = 8.314 E-3 KJ/K.mol
∴ T = 298 K
⇒ K = e∧(-28.619/(8.314 E-3)(298) = 9.624 E-6
⇒ ΔG°r = - RTLnK
If T (↓) ⇒ ΔG°r (↓)
assuming T = 200 K
⇒ ΔG°r(200 K) = - (8.314 E-3)(200)Ln(9.624E-3)
⇒ ΔG°r (200K) = - 19.207 KJ/mol < ΔG°r(298 K) = - 28.619 KJ/mol
