You start by using proportions to find the number of liters of solution:
180 g of glucose / 1 liter of solution = 18 g of glucose / x liter of solution
=> x = 18 g of glucose * 1 liter of solution / 180 g of glucose = 0.1 liter of solution.
If you assume that the 18 grams of glucose does not apport volume to the solution but that the volume of the solution is the same volumen of water added (which is the best assumption you can do given that you do not know the how much the 18 g of glucose affect the volume of the solution) then you should add 0.1 liter of water.
Answer: 0.1 liter of water.
Look it up, it’s not that hard.
(a) One form of the Clausius-Clapeyron equation is
ln(P₂/P₁) = (ΔHv/R) * (1/T₁ - 1/T₂); where in this case:
Solving for ΔHv:
- ΔHv = R * ln(P₂/P₁) / (1/T₁ - 1/T₂)
- ΔHv = 8.31 J/molK * ln(5.3/1.3) / (1/358.96 - 1/392.46)
(b) <em>Normal boiling point means</em> that P = 1 atm = 101.325 kPa. We use the same formula, using the same values for P₁ and T₁, and replacing P₂ with atmosferic pressure, <u>solving for T₂</u>:
- ln(P₂/P₁) = (ΔHv/R) * (1/T₁ - 1/T₂)
- 1/T₂ = 1/T₁ - [ ln(P₂/P₁) / (ΔHv/R) ]
- 1/T₂ = 1/358.96 K - [ ln(101.325/1.3) / (49111.12/8.31) ]
(c)<em> The enthalpy of vaporization</em> was calculated in part (a), and it does not vary depending on temperature, meaning <u>that at the boiling point the enthalpy of vaporization ΔHv is still 49111.12 J/molK</u>.
Answer:
In my opinion, water is a compound.
Explanation:
Water is mixed from Oxygen(O) and Hydrogen(H) and they cannot be splitted.
Sorry if there any mistakes
Answer:
Yes
Explanation:
Based on the graph, nuclear energy is one of the least contributors of CO2 emissions.
You would support this source of energy .
