An intensive property would be something like color. You could have 1 mol or 20 mols of something and the color would be the same. Think intense color like silver or gold. Another example is density. 1 mol of iron has the same density as 28 mol of iron.
An extensive property would be something like mass. How much mass you have will determine the extensive property of the material.
Complete Question
The complete question is shown on the first uploaded image
Answer:
The partial pressure is
Explanation:
The Partial pressure of is mathematically represented as
Where is the total pressure of water with a value of 15.5 mm of Hg
is the partial pressure of water with a value 753 mm of Hg
Now substituting values
1.the butter would begin to soften and melt
2. I would put it back in the fridge
3. If you’re done using a food item that’s able to spoil, put it back in the condition where it can live it’s best life ... ( I don’t know if that makes sense but...?)
Answer:
1.9 x 10⁻¹⁴
Explanation:
The Gibbs-Helmholtz equation which gives us the dependendence of the equilibrium constant, k, at the different temperatures, T₁ and T₂, will be used to solve this question:
ln(k₂/k₁) = - ΔHº/R ( 1/T₂ - 1/T₁ )
Here we are to assume ΔHº is constant over the temperature range 25-100 ºC.
We have all the data input required, so lets substitute and solve for k₂
T₁ = (25 + 273) K = 298 K
T₂ = (53 +273) K = 326 K
R = 8.314 J
ln( k₂/1.0 x 10⁻¹⁴ ) = - 5.58 x 10⁴ J/K x (8.314 J/K ( 1/326 K - 1/298 K ))
ln( k₂/1.0 x 10⁻¹⁴ ) = 1.9
Notice the units cancel each other as we would expect it to be since k₂/k₁ is unitless.
Now take inverse ln to both sides of the equation:
k₂/ 1.0 x 10⁻¹⁴ = 1.93 ⇒ k₂ = 1.9 x 10⁻¹⁴
This result is logical because the reaction is endothermic given ΔH° is positive at 25 ºC, so at 53 ºC we would expect k₂ greater than k₁.
according to this reaction:
by using the ICE table:
CO(OH)2 → CO2+ + 2OH-
intial 0 0
change +X +2X
Equ X 2X
when Ksp = [CO2][OH-]^2
by substitution:
3.6 x 10^-16 = X * (2X)^2
3.6 x 10^-16 = 4 X^3
∴X = 3.88 x 10^-6 M