Answer: 0.050M urea, 0.10M glucose, 0.2M sucrose, pure water
Explanation:
Vapor pressure refers to the ease with which a liquid substance is transformed into vapour. High vapour density implies that the liquid is easily transformed into gas. Pure water is expected to have the lowest vapour density since it is held by strong intermolecular forces in the liquid state. Urea is an organic liquid held by weak Van der Waals forces hence its extremely high vapor pressure.
Answer is: 6,16 kJ.
1) changing temperature of ice from -25°C to 0°C.
Q₁ = m·C·ΔT
Q₁ = 18 g · 2 J/g·°C · 25°C
Q₁ = 900 J.
m(H₂O) = 1mol · 18 g/mol = 18 g.
C - <span>specific heat of ice.
</span>2) changing temperature of water from 0°C to 70°C.
Q₁ = m·C·ΔT
Q₁ = 18 g · 4,18 J/g·°C · 70°C
Q₁ = 5266,8 J.
C - specific heat of water.
Q = Q₁ + Q₂ = 900 J + 5266,8 J
Q = 6166,8 J = 6,16 kJ.
Answer:
<span>The energy required to go from liquid to gas is called as Latent Heat of Vaporization.
Explanation:
The process of conversion of liquid into gas phase is known as vaporization while the conversion of gas into liquid state is called as condensation. The liquid having stronger intermolecular forces than gases require some energy to break those interactions hence, the heat provided to break these interactions and convert it into gas phase is called as heat of vaporization. Remember, heat of vaporization and heat of condensation are same for a given substance but with different signs.
Example:
Heat of Vaporization of Water = 40.65 kJ/mol
Heat of Condensation of Water = - 40.65 Kj/mol</span>
Answer:
5.88atm
Explanation:
First, we obtain the number of mole of CO2 present in the vessel. This is illustrated below:
Molar Mass of CO2 = 12 + (2x16) = 12 + 32 = 44g/mol
Mass of CO2 from the question = 345.1g
Number of mole of CO2 =?
Number of mole = Mass/Molar Mass
Number of mole of CO2 = 345.1/44
= 7.84moles
Now we can easily calculate the pressure by doing the following:
Data obtained from the question include:
V (volume) = 32.1 L
T (temperature) = 20°C = 20 + 273 = 293K
R (gas constant) = 0.0821atm*L/mole*K
n (number of mole) = 7.84moles
P (pressure) =?
We will be making use of the ideal gas equation PV = nRT to calculate the pressure
PV = nRT
P = nRT/V
P = 7.84 x 0.0821 x 293/32.1
P = 5.88atm
Therefore, the pressure is 5.88atm
Your best guess for the boiling point of any version of Coke would be 100 C, the boiling point of water.
Diet Coke is mostly water (the flavourings are a very small amount relative to the amount of water). The largest ingredient will be the sweetener but there will be only a fraction of a gram of that. It is unlikely you will notice any deviation from the properties of water.
Standard Coke has quite a lot of sugar in it. A standard can (~300ml) contains about 40g of sugar. To put it another way, the contents are more than 10% sugar by weight and the solution is about 1/3 mol/L of sucrose (other sugars will be slightly different). A standard calculation using the ebullioscopic constant for water suggests the elevation of the boiling point will be barely 0.2 C, so small you'd struggle to measure it without good instruments and a good experimental setup.
