Answer:
Intensive properties can be used to help identify a sample because these characteristics do not depend on the amount of sample, nor do they change according to conditions.
Explanation:
Intensive properties are bulk properties, which means they do not depend on the amount of matter that is present. Examples of intensive properties include:
Boiling Point
Density
State of Matter
Color
Melting Point
Odor
Temperature
Refractive Index
Luster
Hardness
Ductility
Malleability
Silicon shapes our technological reality because computer chips are made out of silicon. Different computer parts, are made out of silicon chips, tablets, phones, different electronic devices, everything has some silicon in them. Without silicon we would (currently) not have any of those. That is how silicon shapes our technological reality :)
Inhalation is the process of taking air into the lungs. For this to occur, the air pressure inside the lungs must be lower than that of the external atmosphere as air flows from areas of higher pressure to lower pressure. <span>Exhalation is the process of expelling air out of the lungs. For this to occur, the air pressure inside the lungs must be higher than that of the external atmosphere as air flows from areas of higher pressure to ones of lower pressure.</span>
Answer:
The answer is 5.7 minutes
Explanation:
A first-order reaction follow the law of ![Ln [A] = -k.t + Ln [A]_{0}](https://tex.z-dn.net/?f=Ln%20%5BA%5D%20%3D%20-k.t%20%2B%20Ln%20%5BA%5D_%7B0%7D)
. Where <em>[A]</em> is the concentration of the reactant at any <em>t</em> time of the reaction, ![[A]_{0}](https://tex.z-dn.net/?f=%5BA%5D_%7B0%7D)
is the concentration of the reactant at the beginning of the reaction and <em>k</em> is the rate constant.
Dropping the concentration of the reactant to 6.25% means the concentration of A at the end of the reaction has to be ![[A]=\frac{6.25}{100}.[A]_{0}](https://tex.z-dn.net/?f=%5BA%5D%3D%5Cfrac%7B6.25%7D%7B100%7D.%5BA%5D_%7B0%7D)
. And the rate constant (<em>k</em>) is 8.10×10−3 s−1
Replacing the equation of the law:
![Ln \frac{6.25}{100}.[A]_{0} = -8.10x10^{-3}s^{-1}.t + Ln[A]_{0}](https://tex.z-dn.net/?f=Ln%20%5Cfrac%7B6.25%7D%7B100%7D.%5BA%5D_%7B0%7D%20%3D%20-8.10x10%5E%7B-3%7Ds%5E%7B-1%7D.t%20%2B%20Ln%5BA%5D_%7B0%7D)
Clearing the equation:
![Ln [A]_{0}.\frac{6.25}{100} - Ln [A]_{0} = -8.10x10^{-3}s^{-1}.t](https://tex.z-dn.net/?f=Ln%20%5BA%5D_%7B0%7D.%5Cfrac%7B6.25%7D%7B100%7D%20-%20Ln%20%5BA%5D_%7B0%7D%20%3D%20-8.10x10%5E%7B-3%7Ds%5E%7B-1%7D.t)
<em>Considering the property of logarithms: </em>
Using the property:
![Ln \frac{[A]_{0}}{[A]_{0}}.\frac{6.25}{100} = -8.10x10^{-3}s^{-1}.t](https://tex.z-dn.net/?f=Ln%20%5Cfrac%7B%5BA%5D_%7B0%7D%7D%7B%5BA%5D_%7B0%7D%7D.%5Cfrac%7B6.25%7D%7B100%7D%20%3D%20-8.10x10%5E%7B-3%7Ds%5E%7B-1%7D.t)
Clearing <em>t </em>and solving:
The answer is in the unit of seconds, but every minute contains 60 seconds, converting the units:
