Explanation:
Soaps attach to both water and grease molecules.
The grease molecules are attracted more strongly towards each other as compared to water molecules. Also, water molecules are smaller in size hence, strong intermolecular force is required to break the hydrogen bonds of water molecule so that grease or oil molecules can enter the water molecule.
A soap molecule goes in between water and grease molecule and helps them to bind. The force for linkage between water and grease molecule through the soap molecule is weak london dispersion force.
The soap molecule has its salt end as ionic and water soluble. When grease or oil is added to the soap and water solution then the soap acts as an emulsifier. The soap forms miscelles of the non-polar tails and grease molecules are trapped between these miscelles. This miscelle is easily soluble in water hence, the grease is washed away.
Thus, it can be concluded that the nonpolar end of a soap molecule attaches itself to grease.
Answer : The mole fraction of nitrogen will be 0.4615.
Explanation : When nitrogen (
)and hydrogen (
)are mixed, the mole ratio becomes 1 : 1.5,
Now we know that () is acting as a limiting agent.
So at the time of when 0.4 moles of (
) is been formed it requires 0.4 moles of () and 3.4 moles of ()
So, we find the the remaining () will be 0.6 and
() will be 0.3 mole present in mixture.
So, the mole fraction of () becomes = 0.6 / (0.6 + 0.4 + 0.3) Which becomes = 0.4615
solution:
Weight of caffeine is W = 0.170 gm.
Volume of water is V= 10 ml
Volume of methylene chloride which extracted caffeine is v= 5ml
No of portions n=3
Distribution co-efficient= 4.6
Total amount of caffeine that can be unextracted is given by
![w_{n}=w\times[\frac{k_{Dx}v}{k_{Dx}v+v}]^n\\w_{3}=0.170[\frac{4.6\times10}{(4.6\times10+5)}]^3\\=0.170[\frac{46}{46+5}]^3\\=0.170[\frac{46}{51}]^3\\=0.170[\frac{97336}{132651}]\\=0.170\times0.734=0.125gms](https://tex.z-dn.net/?f=w_%7Bn%7D%3Dw%5Ctimes%5B%5Cfrac%7Bk_%7BDx%7Dv%7D%7Bk_%7BDx%7Dv%2Bv%7D%5D%5En%5C%5C%3C%2Fp%3E%3Cp%3Ew_%7B3%7D%3D0.170%5B%5Cfrac%7B4.6%5Ctimes10%7D%7B%284.6%5Ctimes10%2B5%29%7D%5D%5E3%5C%5C%3C%2Fp%3E%3Cp%3E%3D0.170%5B%5Cfrac%7B46%7D%7B46%2B5%7D%5D%5E3%5C%5C%3C%2Fp%3E%3Cp%3E%3D0.170%5B%5Cfrac%7B46%7D%7B51%7D%5D%5E3%5C%5C%3C%2Fp%3E%3Cp%3E%3D0.170%5B%5Cfrac%7B97336%7D%7B132651%7D%5D%5C%5C%3C%2Fp%3E%3Cp%3E%3D0.170%5Ctimes0.734%3D0.125gms)
amount of caffeine un extracted is 0.125gms
amount of caffeine extracted=0.170-0.125
=0.045 gms
1. What do they have in common?
As mentioned in the problem, these gases are present in equal amounts. So, that would infer that they are common in terms of their mass. Also, it is specified that the temperature is 25°C. Connected to that is the average kinetic energy, which is directly proportional. Hence, they are also common in temperature and average kinetic energy.
2. What are the differences?
They differ in type, of course. Also, they differ in average velocities which is a factor of temperature of molar mass. Since they are 3 different types of gases with different molar masses, they would also differ in their average velocities.
