Answer:
Filtration (The process of separating liquids and solids)
Explanation:
Filtration works best when the solute isn't dissolved in the solvent. Basically like how you can filter dirt or sand from water. The opposite of this would be if water and sugar where mixed together you cannot unmix this duo with filtration but with the example given you can use the process of filtration to rid the gasoline of sand.
Mass measures the amount of matter in an object
Explanation:
For the given reaction 
Now, expression for half-life of a second order reaction is as follows.
![t_{1/2} = \frac{1}{[A_{0}]k}](https://tex.z-dn.net/?f=t_%7B1%2F2%7D%20%3D%20%5Cfrac%7B1%7D%7B%5BA_%7B0%7D%5Dk%7D)
....... (1)
Second half life of this reaction will be 
. So, expression for this will be as follows.
= ![\frac{1}{k} [\frac{1}{[A]_{f}} - \frac{1}{[A_{0}]}]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bk%7D%20%5B%5Cfrac%7B1%7D%7B%5BA%5D_%7Bf%7D%7D%20-%20%5Cfrac%7B1%7D%7B%5BA_%7B0%7D%5D%7D%5D)
...(2)
where ![[A]_{f}](https://tex.z-dn.net/?f=%5BA%5D_%7Bf%7D)
is the final concentration that is, ![\frac{[A]_{0}}{4}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5D_%7B0%7D%7D%7B4%7D)
here and ![[A]_{i}](https://tex.z-dn.net/?f=%5BA%5D_%7Bi%7D)
is the initial concentration.
Hence, putting these values into equation (2) formula as follows.
= ![\frac{1}{k} [\frac{4}{[A]_{0}} - \frac{1}{[A_{0}]}]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bk%7D%20%5B%5Cfrac%7B4%7D%7B%5BA%5D_%7B0%7D%7D%20-%20%5Cfrac%7B1%7D%7B%5BA_%7B0%7D%5D%7D%5D)
= ![\frac{3}{[A_{0}]k}](https://tex.z-dn.net/?f=%5Cfrac%7B3%7D%7B%5BA_%7B0%7D%5Dk%7D)
...... (3)
Now, dividing equation (3) by equation (1) as follows.
= ![\frac{3}{[A_{0}]k} \times [A_{0}]k](https://tex.z-dn.net/?f=%5Cfrac%7B3%7D%7B%5BA_%7B0%7D%5Dk%7D%20%5Ctimes%20%5BA_%7B0%7D%5Dk)
= 3
or, = 3 
Thus, we can conclude that one would expect the second half-life of this reaction to be three times the first half-life of this reaction.
1.34*10^6
Move the decimal 6 times to the left, it is a number 1-10
Answer:
<span>Dipole-Dipole Forces are common </span><span>to all polar molecules but not non polar molecules.
Explanation:
An Asymmetrical molecule having a region of high electron density (partial negative) and lower density (partial positive) interacts with its neighbor molecules through Dipole-Dipole Interactions. The partial positive part of one molecule interacts with the partial negative part of another.
Example:
Acetone having a partial positive carbon (of carbonyl group) and partial negative oxygen interacts through Dipole-Dipole forces. Hence, acetone does not involves Hydrogen Bond interactions.</span>
