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
The carnivores are the hawk, owl, and fox.
Answer: A balanced equation for the given reaction is 
.
Explanation:
The reaction equation will be as follows.
Number of atoms on the reactant side is as follows.
Number of atoms on the product side is as follows.
Since number of atoms on both the reactant and product sides are equal. Hence, the reaction equation is balanced.
Thus, we can conclude that a balanced equation for the given reaction is .
Both are mainly composed of droplets of condensed water
Answer:
Mass of ring = 32 g
Volume of ring = 4 mL
Density of ring = 8 g/mL
Explanation:
From the question given above, the following data were obtained:
Mass of ring = 32 g
Volume of water = 64 mL
Volume of water + ring = 68 mL
Density of ring =?
Next, we shall determine the volume of the ring. This can be obtained as follow:
Volume of water = 64 mL
Volume of water + ring = 68 mL
Volume of ring =?
Volume of ring= (Volume of water + ring) – (Volume of water)
Volume of ring = 68 – 64
Volume of ring = 4 mL
Finally, we shall determine the density of the ring. This can be obtained as follow:
Mass of ring = 32 g
Volume of ring = 4 mL
Density of ring =?
Density = mass / volume
Density of ring = 32 / 4
Density of ring = 8 g/mL
