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:
36.92 mg of oxygen required for bio-degradation.
Explanation:
Mass of benzene = 30 mg = 0.03 g (1000 mg = 1 g )
Moles benzene =
According to reaction 5 moles of benzene reacts with 15 moles of oxygen gas.
Then 0.0003846 mol of benzene will react with:
of oxygen gas
Mass of 0.0011538 moles of oxygen gas:
0.0011538 mol × 32 g/mol = 0.03692 g = 36.92 mg
36.92 mg of oxygen required for bio-degradation.
Answer:
Molarity = 0.24 M
Explanation:
Given data:
Number of moles of Mg(NO₃)₂ = 0.181 mol
Volume of solution = 0.750 L
Molarity of solution = ?
Solution:
Molarity is used to describe the concentration of solution. It tells how many moles are dissolve in per litter of solution.
Formula:
Molarity = number of moles of solute / L of solution
by putting values,
Molarity = 0.181 mol / 0.750 L
Molarity = 0.24 M
Answer:
in 1983, Lake Mead had a recorded water level of 1,225 feet above sea level. The closest it's been to that level in modern times was in 2000. Explanation:
<span>
Phenobarbital is derivative of
Barbituric Acid and Barbituric Acid is derivative of
Urea. (structures shown in Fig below)
Urea has H</span>₂N- group attached to Carbonyl Group (C=O), and such class of comounds conataining H₂N-C=O bond are called as Amides.
Result: So, <span>Phenobarbital belongs to
Amides.</span>
