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 atomic mass of Cs is 132.9. So the mole number of 675 g cesium is 675/132.9 = 5.08 mole. Then the answer should be 5.08 mole.
Answer:
increased
Explanation:
Consuming a compound increases the concentration. When you increase the concentration, the rate constant for that reaction also increases.
Answer:
16 mol NaCl.
Explanation:
Do the train track method to cancel out all the units except moles of NaCl on top. Remember one mole of any gas occupies 22.4 L at STP.
179.2 L CO2 x 1 mol CO2/22.4 L CO2 x 2 mol NaCl/1 mol CO2
= 16 mol NaCl
Answer:
see calculations in explanation
Explanation:
percent = part/total x 100%
part = ∑ atomic mass of element
- hydrogen = 1.008 amu (atomic mass units)
- carbon = 12.011 amu
- nitrogen = 14.007 amu
total = ∑ molecular mass of compound
= H amu + C amu + Namu
= 1.008 amu + 12.011 amu + 14.007 amu
= 27.026 amu
%H = (1.008amu/27.026amu)100% = 3.730%
%C = (12.011amu/27.026amu)100% = 44.442%
%N = (14.007amu/27.026amu)100% = 51.827%
Check results ∑%values = 100%
3.730% + 44.442% + 51.827% = 99.999% ≅ 100%
