POH = - log [ OH⁻ ]
pOH = - log [ 1 x 10⁻⁹ ]
pOH = 9
Answer C
hope this helps!
Using the exponential decay model; we calculate "k"
We know that "A" is half of A0
A = A0 e^(k× 5050)
A/A0 = e^(5050k)
0.5 = e^(5055k)
In (0.5) = 5055k
-0.69315 = 5055k
k = -0.0001371
To calculate how long it will take to decay to 86% of the original mass
0.86 = e^(-0.0001371t)
In (0.86) = -0.0001371t
-0.150823 = -0.0001371 t
t = 1100 hours
When the balanced equation for this reaction is:
2Fe + 3H2O → Fe2O3 + 3H2
and according to the vapour pressure formula:
PV= nRT
when we have P is the vapor pressure of H2O= 0.121 atm
and V is the volume of H2O = 4.5 L
and T in Kelvin = 52.5 +273 = 325.5 K
R= 0.08205 atm-L/g mol-K
So we can get n H2O
So, by substitution:
n H2O = PV/RT
= (0.121*4.5)/(0.08205 * 325.5) = 0.02038 gmol
n Fe2O3 = 0.02038 * (1Fe2O3/ 3H2O) = 0.00679 gmol
Note: we get (1FeO3/3H2O) ratio from the balanced equation.
we can get the Mass of Fe2O3 from this formula:
Mass = number of moles * molecular weight
when we have a molecular weight of Fe2O3 = 159.7
= 0.00679 * 159.7 = 1.084 g
∴ 1.084 gm of Fe2O3 will produced
Answer:
This is a coal combustion process and we will assume
Inlet coal amount = 100kg
It means that there are
15kg of H2O, 2kg of Sulphur and 83kg of Carbon
Now to find the mole fraction of SO2(g) in the exhaust?
Molar mass of S = 32kg/kmol
Initial moles n of S = 2/32 = 0.0625kmols
Reaction: S + O₂ = SO₂
That is 1 mole of S reacts with 1 mole of O₂ to give 1 mole of SO₂
Then, it means for 0.0625 kmoles of S, we will have 0.0625 kmole of SO2 coming out of the exhaust
The mole fraction of SO2(g) in the exhaust=0.0625kmols
Explanation:
