Rubidium is an element that belongs to Group 1. As such it will have physical properties similar to the other Group 1 elements. Rubidium is below
Potassium in the periodic table but above
Cesium. As such it would be most like one of those two elements.
<span>Answer
is: activation energy of this reaction is 212,01975 kJ/mol.
Arrhenius equation: ln(k</span>₁/k₂) = Ea/R (1/T₂ - 1/T₁<span>).
k</span>₁<span> = 0,000643
1/s.
k</span>₂ = 0,00828
1/s.
T₁ = 622 K.
T₂ = 666 K.
R = 8,3145 J/Kmol.
1/T₁<span> = 1/622 K = 0,0016 1/K.
1/T</span>₂<span> = 1/666 K =
0,0015 1/K.
ln(0,000643/0,00828) = Ea/8,3145 J/Kmol · (-0,0001 1/K).
-2,55 = Ea/8,3145 J/Kmol · (-0,0001 1/K).
Ea = 212019,75 J/mol = 212,01975 kJ/mol.</span>
The complete question is shown in the image attached to this answer.
Answer:
C
Explanation:
Let us quickly remember that the EMF of a cell under non standard conditions in given by the Nernst equation.
This equation states that;
E = E°cell - 0.592/n log Q
Where
E = EMF under non standard conditions
E°cell= standard EMF of the cell
n = number of electrons transferred
Q = reaction quotient
If the reaction quotient is greater than 1 then cell potential is less than the standard cell potential.
The cell that generates the lowest cell potential is the cell depicted in option C because Q has the greatest positive value(Q<1).
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:
