Answer:
The equilibrium partial pressure of O2 is 0.545 atm
Explanation:
Step 1: Data given
Partial pressure of SO2 = 0.409 atm
Partial pressure of O2 = 0.601 atm
At equilibrium, the partial pressure of SO2 was 0.297 atm.
Step 2: The balanced equation
2SO2 + O2 ⇆ 2SO3
Step 3: The initial pressure
pSO2 = 0.409 atm
pO2 = 0.601 atm
pSO3 = 0 atm
Step 4: Calculate the pressure at the equilibrium
pSO2 = 0.409 - 2X atm
pO2 = 0.601 - X atm
pSO3 = 2X
pSO2 = 0.409 - 2X atm = 0.297
X = 0.056 atm
pO2 = 0.601 - 0.056 = 0.545 atm
pSO3 = 2*0.056 = 0.112 atm
Step 5: Calculate Kp
Kp = (pSO3)²/((pO2)*(pSO2)²)
Kp = (0.112²) / (0.545 * 0.297²)
Kp = 0.261
The equilibrium partial pressure of O2 is 0.545 atm
<span>The composition of a fertilizer is usually express in NPK number. NPK number is in terms of Percent by mass of the said element which are Nitrogen, Phosphorus and Potassium. A 15-35-15 fertilizer has 15%
Nitrogen, 35% Phosphorous, and 15% Potassium by mass. If you have 10 g of this
fertilizer, to get the number of moles of phosphorus, you multiply the mass by
35%, which is equal to 10*0.35 or 3.5 g phosphorus. Then you divide the
calculated mass of phosphorous by its molar mass which is 30.97 g/mol.
Therefore, you have 3.5/30.97 which is equal to 0.1130 mol Phosphorus. This is the amount of Phosphorus in moles in the fertilizer.</span>
Answer:
6
Explanation:
the value is 6 because its an even number
Hello!
At
Standard Pressure and Temperature, an ideal gas has a molar density of
0,04464 mol/L.So, we need to apply a simple conversion factor to calculate the density of Sulfur Dioxide using the molar mass of Sulfur Dioxide.
So, the Density of Sulfur Dioxide (SO₂) at STP is
2,8599 g/LHave a nice day!
2NH₂ + O₂ → N₂ + 2H₂O
<u>Explanation:</u>
Balancing the equation means, the number of atoms on both sides of the equation must be the same.
In the case of the given equation, we have to find out whether it is balanced or not.
2NH₂ + O₂ → N₂ + 2H₂O
Atoms Number of atoms before balancing after balancing
LHS RHS LHS RHS
N 1 2 2 2
H 2 2 4 4
O 2 1 2 2
To balance the N atoms, we have to put 2 in front of NH₂, and then to balance the H, O atoms, we have to put 2 in front of H₂O, so that each atom in left hand as well as right hand side of the equation was balanced.
