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
D = m / V
d = 5.0 / 45.0
d = 0.111 g/cm³
Chemical energy to thermal
Answer: Photosynthesis
Explanation:
<em>The process of photosynthesis</em> is used to help make food for plants. The inputs are sunlight, water (H2O), and carbon dioxide (CO2). Then the outputs are glucose (C6H12O6) which is food, and oxygen (O2) is released.
I hope this helps :)
Answer:
A. 266g/mol
Explanation:
A colligative property of matter is freezing point depression. The formula is:
ΔT = i×Kf×m <em>(1)</em>
Where:
ΔT is change in temperature (0°C - -0,14°C = 0,14°C)i is Van't Hoff factor (1 for a nonelectrolyte dissolved in water), kf is freezing point molar constant of solvent (1,86°Cm⁻¹) and m is molality of the solution (moles of solute per kg of solution). The mass of the solution is 816,0g
Replacing in (1):
0,14°C = 1×1,86°Cm⁻¹× mol Solute / 0,816kg
<em>0,0614 = mol of solute</em>.
As molar mass is defined as grams per mole of substance and the compound weights 16,0g:
16,0g / 0,0614 mol = 261 g/mol ≈ <em>A. 266g/mol</em>
I hope it helps!
