Answer:
0,2,8,3
Explanation:
k shell takes 2,L shell takes 8 and M shell takes the remaining which is 3
<span>1.40 x 10^5 kilograms of calcium oxide
The reaction looks like
SO2 + CaO => CaSO3
First, determine the mass of sulfur in the coal
5.00 x 10^6 * 1.60 x 10^-2 = 8.00 x 10^4
Now lookup the atomic weights of Sulfur, Calcium, and Oxygen.
Sulfur = 32.065
Calcium = 40.078
Oxygen = 15.999
Calculate the molar mass of CaO
CaO = 40.078 + 15.999 = 56.077
Since 1 atom of sulfur makes 1 atom of sulfur dioxide, we don't need the molar mass of sulfur dioxide. We merely need the number of moles of sulfur we're burning. divide the mass of sulfur by the atomic weight.
8.00 x 10^4 / 32.065 = 2.49 x 10^3 moles
Since 1 molecule of sulfur dioxide is reacted with 1 molecule of calcium oxide, just multiply the number of moles needed by the molar mass
2.49 x 10^3 * 56.077 = 1.40 x 10^5
So you need to use 1.40 x 10^5 kilograms of calcium oxide per day to treat the sulfur dioxide generated by burning 5.00 x 10^6 kilograms of coal with 1.60% sulfur.</span>
Answer:
0.99mol/L
Explanation:
Molarity ( M) = # of moles of solute / volume of solution (L)
Volume of Solution = 100 mL or 0.1 L
Molarity ( M) = # of moles of solute / volume of solution (L)
Molarity (M) = 0.099 mol / 0.1 L = 0.99mol/ L
Answer:
A its a step by step equation the answer is A
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
