<span>i think its Uranium Dating </span>
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: All organic compound depends on H-bonding with water. more stronger H-bonding with water more will be soluble.
Explanation:
1. It depends primarily upon the function groups of that compound. It also depends on the size of the compound.
2. some organic compound which soluble in water for example: alcohols, ethers, carboxylic acids. Because of the functional groups attached to the organic structure (the C-H backbone) are what effect the solubilities.Like carboxylic acids and alcohols form hydrogen bonds with the water, helping to solubilize it.
3. Take alcohols for example: methanol, ethanol, and isopropanol are all completely soluble in water. By the time you get to butanol and some of the larger alcohols, including those with more complex structures, they tend to be less soluble.
heat flows<span> from the </span>warmer<span> body to the </span>cooler<span> body </span>