Answer:
- 416 kJ/mol
Explanation:
The standard enthalpy of the reaction (Δ
H
∘
rxn) is independent of the pathway, so it can be calculated by the enthalpy of formation of the reactants and the products:
Δ
H
∘
rxn = ∑n*Δ
H
∘f products - ∑n*Δ
H
∘f reactants
Where n is the number of moles in the balanced reaction. So, for the reaction given:
Na₂O(s) + 1/2O₂(g) → Na₂O₂(s)
Because O₂ is formed by only one elements, its Δ
H
∘f is 0 kJ/mol:
-89.0 = (1*(-505) - (1*Δ
H
∘fNa₂O)
Δ
H
∘fNa₂O = -505 + 89
Δ
H
∘fNa₂O = - 416 kJ/mol
Density = mass / volume
Because the dimensions of the barometer are the same for each case, we can compare the heights of the liquids instead of the volume.
ρ(mercury) = 13.6ρ(water)
mass(mercury)/760 = 13.6 x mass(water)/Height(water)
Masses of the liquids will also be the same.
Height of water = 13.6 x 760
Height of water = 10,336 mm
a
The speed of molcules in gas mood are more than liquide
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Answer:
Final pressure is 1.42atm
Explanation:
Based on Gay-Lussac's law, pressure of a gas is directely proportional to its absolute temperature. The equation of this law is:
P₁T₂ = P₂T₁
<em>Where P is pressure and T is absolute temperature of 1, initial state and 2, final state of the gas.</em>
In the problem, initial conditions are Standard Temperature and Pressure, STP, that are 1 atm and 273.15K.
If the final temperature is 115°C = 388.15K (115°C + 273.15 = 388.15K), using Gay-Lussac's law:
P₁T₂ = P₂T₁
1atmₓ388.15K = P₂ₓ273.15K
1.42atm = P₂
<h3>Final pressure is 1.42atm</h3>
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