Answer:
Explanation:
The expression for Clausius-Clapeyron Equation is shown below as:
Where,
P is the vapor pressure
ΔHvap is the Enthalpy of Vaporization
R is the gas constant (8.314×10⁻³ kJ /mol K)
c is the constant.
For two situations and phases, the equation becomes:
Given:
= 13.95 torr
= 144.78 torr
= 25°C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T = (25 + 273.15) K = 298.15 K
= 298.15 K
= 75°C = 348.15 K
So,
Answer: Option (C) is the correct answer.
Explanation:
Biopower means production of electricity with the help of biomass. The unused wood, crops, agricultural residue, animal residue etc when burned then they help in generating biopower.
It is a renewable energy. Since biomass is readily available in most of the places. Therefore, the advantage of biopower is that it is also readily available in most places.
For example, bacterial decay produces a certain amount of heat which further helps in the production of energy.
• Take a look at the steps below to see how to balance this equation. Let's start by writing the unbalanced equation given the information.
Unbalanced Equation : C₃H₈ (g) + O₂ (g) → CO₂ (g) + H₂O (g)
,
Start by Balancing the Carbons : C₃H₈ (g) + O₂ (g) → 3CO₂ (g) + H₂O (g)
Now let's balance the Hydrogen : C₃H₈ (g) + O₂ (g) → 3CO₂ (g) + 4H₂O (g)
Balancing the Oxygen : C₃H₈ (g) + 5O₂ (g) → 3CO₂ (g) + 4H₂O (g)
Balanced Equation : C₃H₈ (g) + 5O₂ (g) → 3CO₂ (g) + 4H₂O (g)
• Let's apply dimensional analysis here,
0.7 L propane × (5 liters Oxygen / 1 liter Propane) = 3.5 Liters of Oxygen
• Similarly we can identify the liters of carbon dioxide produced in the reaction,
0.7 L propane × (3 liters Carbon Dioxide / 1 liter Propane) = 2.1 Liters of Carbon Dioxide
• 0.7 L propane × (4 liter water vapor / 1 liter propane ) = 2.8 Liters of Water Vapor
Answer:
= 46.06 grams
Explanation:
P=cRT
25 = c (0.0821)(298)
concentration = c = 1.021 M
number of moles = (1.021)(0.75) = 0.7664 moles
mass = (0.7664)(60.10)=40.06 grams
Answer:
No
Explanation:
No, his mass remains the same no matter where he is in the universe.
But then again the moon has less gravitational pull, therefore your weight and mass will be smaller in space and on the moon than on earth
I hope this was helpful! ;)
