Answer:
The vapor pressure of the solution is 23.636 torr
Explanation:
Where;
is the vapor pressure of the solution
is the mole fraction of the solvent
is the vapor pressure of the pure solvent
Thus,
15.27 g of NaCl = [(15.27)/(58.5)]moles = 0.261 moles of NaCl
0.67 kg of water = [(0.67*1000)/(18)]moles = 37.222 moles of H₂O
Mole fraction of solvent (water) = (number of moles of water)/(total number of moles present in solution)
Mole fraction of solvent (water) = (37.222)/(37.222+0.261)
Mole fraction of solvent (water) = 0.993
<u>Note:</u> the vapor pressure of water at 25°C is 0.0313 atm
Therefore, the vapor pressure of the solution = 0.993 * 0.0313 atm
the vapor pressure of the solution = 0.0311 atm = 23.636 torr
When looking to power your equipment or vehicle with natural gas, the first question that springs to mind is: what is natural gas? A lot of people use natural gas in their homes for cooking and heating, but they don’t really give it some thought. So, let’s see what natural gas is and how it’s different from other forms of fossil fuels like oil and coal.
Natural Gas is a fossil fuel that exist in a gaseous state and is composed mainly of methane (CH4) a small percentage of other hydrocarbons (e.g. ethane). The use of natural gas is becoming more and more popular as it can be used with commercial, industrial, electric power generation and residential applications.
The answer for the following problem is described below.
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>
Explanation:
Given:
enthalpy of combustion of glucose(Δ
of 
) =-1275.0
enthalpy of combustion of oxygen(Δ of 
) = zero
enthalpy of combustion of carbon dioxide(Δ of 
) = -393.5
enthalpy of combustion of water(Δ of 
) = -285.8
To solve :
standard enthalpy of combustion
We know;
Δ = ∈Δ (products) - ∈Δ (reactants)
(s) +6 (g) → 6 (g)+ 6 (l)
Δ = [6 (-393.5) + 6(-285.8)] - [6 (0) + (-1275)]
Δ = [6 (-393.5) + 6(-285.8)] - [0 - 1275]
Δ = 6 (-393.5) + 6(-285.8) - 0 + 1275
Δ = -2361 - 1714 - 0 + 1275
Δ =-2800 kJ
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>
ΔG° at 450. K is -198.86kJ/mol
The following is the relationship between ΔG°, ΔH, and ΔS°:
ΔH-T ΔS = ΔG
where ΔG represents the common Gibbs free energy.
the enthalpy change, ΔH
The temperature in kelvin is T.
Entropy change is ΔS.
ΔG° = -206 kJ/mol
ΔH° equals -220 kJ/mol
T = 298 K
Using the formula, we obtain:
-220kJ/mol -T ΔS° = -206kJ/mol
220 kJ/mol +206 kJ/mol =T ΔS°.
-T ΔS = 14 kJ/mol
for ΔS-14/298
ΔS=0.047 kJ/mol.K
450K for the temperature Completing a formula with values
ΔG° = (450K)(-0.047kJ/mol)-220kJ/mol
ΔG° = -220 kJ/mol + 21.14 kJ/mol.
ΔG°=198.86 kJ/mol
Learn more about ΔG° here:
brainly.com/question/17214066
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Ideal gases are hypothetical gases whose molecules occupy negligible space and have no interactions, and that consequently obeys the gas laws exactly.
Not exactly sure about the amount...
I hope this helps! :)
