Given:
K = 0.71 = Kp
The reaction of sulphur with oxygen is
S(s) + O2(g) ---> SO2(g)
initial Pressure 6.90 0
Change -x +x
Equilibrium 6.90-x x
Kp = pSO2 / pO2 = 0.71 = x / (6.90-x)
4.899 - 0.71x = x
4.899 = 1.71x
x = 2.86 atm = pressure of SO2 formed
temperature = 950 C = 950 + 273.15 K = 1223.15 K
Volume = 50 L
Let us calculate moles of SO2 formed using ideal gas equation as
PV = nRT
R = gas constant = 0.0821 L atm / mol K
putting other values
n = PV / RT = 2.86 X 50 / 1223.15 X 0.0821 = 1.42 moles
Moles of Sulphur required = 1.42 moles
Mass of sulphur required or consumed = moles X atomic mass of sulphur
mass of S = 1.42 X 32 = 45.57 grams or 0.04557 Kg of sulphur
Answer:
B. It is highly flammable.
Explanation:
Chemical properties deal with how the substance will behave when brought into contact with other substances, and usually have to do with it undergoing some kind of reaction. Flammability is a chemical property.
Density, melting point, and color are all considered physical properties.
Blank 1: nothing (to keep 2 total nitrogen)
blank 2: 3 (to make 6 total hydrogen)
blank 3: 2 (to make 2 total nitrogen and 6 total hydrogen)
hope this helps!! :)
