In preparing diluted solutions from concentrated solutions we can use the following formula
c1v1 = c2v2
c1 and v1 are the concentration and volume of the concentrated solution respectively
c2 and v2 are the concentrations and volume of the diluted solution respectively
Substituting these values ,
20 mL x 1.0 M = C x 60 mL
C = 0.33 M
The concentration of the resulting diluted solutions is 0.33 M
Explanation:
Objects or substances with their density greater than that of water will sink in it whiles those less than water will float on it.
From the question the object has a density of 3.4 g/mL.
Since it's density is greater than that of water the object will sink.
Hope this helps you
Because metallic bonding is non-localized, and extends throughout the metallic lattice. Metal nuclei can move with respect to other metal nuclei without disrupting the forces of attraction.
Based on the given scenario, the rate of increase in water pressure in atm/km is known to be 992 atm/km.
<h3>What is water pressure?</h3>
Pressure is known to be that kind of force that is known to often pushes water via pipes.
Note that Water pressure is a tool that is often used to know or identify the flow of water from any given tap. The amount of pressure at one's tap is said to be often dependent on how high the service tank or water tower is above one's home.
Therefore since 1 atm/m = 1000 atm/km
Hence 0.992 atm/m :
= 0.992 x 1000
= 992 atm/km.
Therefore, Based on the given scenario, the rate of increase in water pressure in atm/km is known to be 992 atm/km.
Learn more about water pressure from
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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.
