Use Arrhenius equation:
k = A*exp(-Ea/RT)
We have:
1.35x10^2/s = A*exp(-85600/(8.314*298.15))
or: A = 1.342x10^17/s
It is a piece of cake to calculate:
k = 1.342x10^17*exp(-85600/(8.314*348.15))
= 1.92x10^4/s
Glucose is used by intestinal cells and red blood cells, while the rest reaches the liver, adipose tissue and muscle cells, where it is absorbed and stored as glycogen.
(it is saved to be used later)
<span>1.16 moles/liter
The equation for freezing point depression in an ideal solution is
ΔTF = KF * b * i
where
ΔTF = depression in freezing point, defined as TF (pure) ⒠TF (solution). So in this case ΔTF = 2.15
KF = cryoscopic constant of the solvent (given as 1.86 âc/m)
b = molality of solute
i = van 't Hoff factor (number of ions of solute produced per molecule of solute). For glucose, that will be 1.
Solving for b, we get
ΔTF = KF * b * i
ΔTF/KF = b * i
ΔTF/(KF*i) = b
And substuting known values.
ΔTF/(KF*i) = b
2.15âc/(1.86âc/m * 1) = b
2.15/(1.86 1/m) = b
1.155913978 m = b
So the molarity of the solution is 1.16 moles/liter to 3 significant figures.</span>
Answer:
When n = 1, the reaction is of the First Order
Explanation:
Find attach the solution
