The total pressure when the new equilibrium is stabilized is half of the initial pressure of the system.
The given chemical reaction at a stable equilibrium is,
2H₂O(g)+O₂(g) = 2H₂O₂(g)
According to the ideal gas equation,
PV = nRT
P is pressure,
V is volume,
n is moles
R is gas constant,
T is temperature.
Assuming the temperature is constant.
If the volume of the system is twice the initial volume then the total pressure at the new equilibrium can be found out as,
P₁V₁ = P₂V₂
Where, P₁ and V₁ are initial volume and pressure while P₂ and V₂ are final pressure and volume.
If V₂ = 2V₁,
P₂ = P₁/2
So, the final total pressure will be half of the initial pressure.
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Answer:
The answer is: <u>Al2O3</u>
Explanation:
The data they give us is:
To find the empirical formula without knowing the grams of the compound, we find it per mole:
- 0.545 g Al * 1 mol Al / 27 g Al = 0.02 mol Al
- 0.485 g O * 1 mol O / 16 g O = 0.03 mol O
Then we must divide the results obtained by the lowest result, which in this case is 0.02:
- 0.02 mol Al / 0.02 = 1 Al
- 0.03 mol O / 0.02 = 1.5 O
Since both numbers have to give an integer, multiply by 2 until both remain integers:
Now the answer is given correctly:
Answer:
ΔS° = - 47.2 J/mol.K
Explanation:
ΔS°= 4(S°mH3PO4) - 6(S°mH2O) - S°mP4O10
∴ S°mH2O(l) = 69.9 J/mol.K
∴ S°mP4O10 = 231 J/mol.K
∴ S°mH3PO4 = 150.8 J/mol.K
⇒ ΔS° = 4*(150.8) - 6*(69.9) - 231
⇒ ΔS° = - 47.2 J/mol.K
Answer:
40.4 kJ
Explanation:
Step 1: Given data
- Heat of sublimation of CO₂ (ΔH°sub): 32.3 kJ/mol
Step 2: Calculate the moles corresponding to 55.0 g of CO₂
The molar mass of CO₂ is 44.01 g/mol.
n = 55.0 g × 1 mol/44.01 g = 1.25 mol
Step 3: Calculate the heat (Q) required to sublimate 1.25 moles of CO₂
We will use the following expression.
Q = n × ΔH°sub
Q = 1.25 mol × 32.3 kJ/mol = 40.4 kJ
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