Answer:
The answer is 5.7 minutes
Explanation:
A first-order reaction follow the law of
. Where <em>[A]</em> is the concentration of the reactant at any <em>t</em> time of the reaction,
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
. 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:

Answer:
The energy transferred between samples of matter because of a difference in their temperatures is called a. heat.
Explanation:
The first law of thermodynamics establishes that when two bodies with different temperatures are put in contact they will find thermic equilibrium to a final temperature by transferring heat. Thus the correct answer is (a).
Thermochemistry is the study of the transformations of heat energy on the chemical reactions. Chemical kinetics is the study of the rate of chemical reactions. And temperature is the measure of the heat.
Explanation:
its hard to explain unless we know what the question fully asks..
Answer:
There is 50.2 kJ heat need to heat 300 gram of water from 10° to 50°C
Explanation:
<u>Step 1: </u>Data given
mass of water = 300 grams
initial temperature = 10°C
final temperature = 50°C
Temperature rise = 50 °C - 10 °C = 40 °C
Specific heat capacity of water = 4.184 J/g °C
<u>Step 2:</u> Calculate the heat
Q = m*c*ΔT
Q = 300 grams * 4.184 J/g °C * (50°C - 10 °C)
Q = 50208 Joule = 50.2 kJ
There is 50.2 kJ heat need to heat 300 gram of water from 10° to 50°C
Hello!
When something is a liquid then turns to a gas, this is known as evaporating so we would call it the heat of evaporation or also known as latent heat. Just remember liquid to gas is vapor!
I hope this helped!
I am, yours most sincerely,
SuperHelperThingy