Consider the isomerization of butane with equilibrium constant is 2.5 .The system is originally at equilibrium with :
[butane]=1.0 M , [isobutane]=2.5 M
If 0.50 mol/L of butane is added to the original equilibrium mixture and the system shifts to a new equilibrium position, what is the equilibrium concentration of each gas?
Answer:
The equilibrium concentration of each gas:
[Butane] = 1.14 M
[isobutane] = 2.86 M
Explanation:
Butane ⇄ Isobutane
At equilibrium
1.0 M 2.5 M
After addition of 0.50 M of butane:
(1.0 + 0.50) M -
After equilibrium reestablishes:
(1.50-x)M (2.5+x)
The equilibrium expression will wriiten as:
![K_c=\frac{[Isobutane]}{[Butane]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BIsobutane%5D%7D%7B%5BButane%5D%7D)
x = 0.36 M
The equilibrium concentration of each gas:
[Butane]= (1.50-x) = 1.50 M - 0.36M = 1.14 M
[isobutane]= (2.5+x) = 2.50 M + 0.36 M = 2.86 M
Answer:
Coefficient of 
Coefficient of 
=8
Explanation:
We are given that a reaction in which 
reacts with
We have to find the coefficient of each reactants in balanced reaction
Coefficient is defined the constant value multiplied with a reactant in a reaction.
Coefficient of =3
Coefficient of 
Coefficient of 
Coefficient of 
Coefficient of 
Coefficient of KOH=2
Hence, Coefficient of and coefficient of
Answer is B- F has a smaller radius than F− because an additional electron causes greater repulsion in F− is the correct choice and the nuclear charge the radius of the anion increases.
Explanation: I hoped that helped!
There would be a 2 infront of the NH3 and and 2 Infront of the H2O
Answer:
a) 1,6%
b) 64,775 g/mol
c) 3,6×10⁻² M
d) 2,3×10⁻³ g/mL
Explanation:
a) The mass fractium of helium is obtained converting the moles of the four gases to grams with molar weight and then caculating of the total of grams how many are of helium, thus:
- Helium: 0,25 moles ×
= 1 g of Helium - Argon: 0,25 moles ×
= 10 g of Argon - Krypton: 0,25 moles ×
= 20,95 g of krypton - Xenon: 0,25 moles ×
= 32,825 g of Xenon
Total grams: 1g+10g+20,85g+30,825g= 62,675 g
Mass fraction of helium: 
× 100 = <em>1,6%</em>
<em />
<em>The mass fraction of Helium is 1,6%</em>
<em />
<em>b)</em><em> </em>Because the mole fraction of all gases is the same the average molecular weight of the mixture is:
= 64,775 g/mol
c) The molar concentration is possible to know ussing ideal gas law, thus:
= M
Where:
P is pressure: 150 kPa
R is gas constant: 8,3145
T is temperature: 500 K
And M is molar concentration. Replacing:
M = 3,6×10⁻² M
d) The mass density is possible to know converting the moles of molarity to grams with average molecular weight and liters to mililiters, thus:
3,6×10⁻² 
× 
× 
=
2,3×10⁻³ g/mL
I hope it helps!
