Answer:
C. effusion because there is a movement of a gas through a small opening into a larger volume
Explanation:
Effusion makes fluid/gas molecules move to the container with less pressure or larger volume. In diffusion, the movement should work two ways even though one side might receive more. But in effusion, the movement is rather one way.
This case shows how effusion work because its not the concentration that makes the balls moving to the bottom part of the container. No ball moving from bottom container to top either.
Answer : The results would show more amount of water in the hydrated sample.
Explanation :
The amount of water of crystallization can be found by taking the masses of hydrated copper sulfate and anhydrous copper sulfate.
The difference in masses indicates the mass of water lost during dehydration process.
If during dehydration process, some of the copper sulfate spatters out of the crucible, then this would give us less mass for anhydrous sample than the actual.
As a result, the difference in masses of hydrated sample and the anhydrous sample would be more.
Therefore the results would show more amount of water in the hydrated sample.
Answer: electrons
Explanation: moving electrons cause momentarily charge
Distribution on molecule. This distribution induces similar distribution to
Adjacent molecule.
Adding or removing protons from the nucleus changes the charge of the nucleus and changes that atom's atomic number. So, adding or removing protons from the nucleus changes what element that atom is! For example, adding a proton to the nucleus of an atom of hydrogen creates an atom of helium.
Answer: 
Explanation:
cM 0 0
So dissociation constant will be:
Given: c = 0.15 M
pH = 1.86
= ?
Putting in the values we get:
Also ![pH=-log[H^+]](https://tex.z-dn.net/?f=pH%3D-log%5BH%5E%2B%5D)
![1.86=-log[H^+]](https://tex.z-dn.net/?f=1.86%3D-log%5BH%5E%2B%5D)
![[H^+]=0.01](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D0.01)
![[H^+]=c\times \alpha](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3Dc%5Ctimes%20%5Calpha)
As ![[H^+]=[ClCH_2COO^-]=0.01](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D%5BClCH_2COO%5E-%5D%3D0.01)
![K_a=1.67\times 10^{-3]](https://tex.z-dn.net/?f=K_a%3D1.67%5Ctimes%2010%5E%7B-3%5D)
Thus the vale of for the acid is
