Answer: Humans use energy to carry out their daily activities, like moving around, working, running etc.
Explanation:
Carbon and hydrogen are present in all
nitrogen is common
sulphur in proteins is common
oxygen is common in ethanoic acids
phosphorus is common for phosphate sugars
hope that helps
Answer:
C2H5O
Explanation:
In a 100 g sample we would have
53.31 g of C
11.18g of H
35.51g of O
First, we find the relative number of atoms of each element by dividing the number of grams the element has in the compound by its atomic mass.
Atomic mass of carbon is 12.011
Relative number of carbon atoms = 53.31 / 12.011 = 4.4
Atomic mass of hydrogen = 1.007
Relative number of hydrogen atoms : 11.18/1.007 = 11.1
Atomic mass of oxygen : 15.999
Relative number of oxygen atoms : 35.51 / 15.999 = 2.2
Now we find a ratio of the relative number of atoms by dividing the # of relative atoms of each element by the element's relative number of atoms that had the lowest number. ( oxygen which had 2.2 ) The outcome of each will be the subscript or number of atoms of each element.
Carbon : 4.4 / 2.2 = 2
Hydrogen : 11.1 / 2.2 = 5
Oxygen : 2.2 / 2.2 = 1
The answer is C2H5O
Given what we know, we can confirm that since Mr. Summers has to test a hypothesis, his next step should be to design an experiment.
<h3>Why design an Experiment?</h3>
- The next step is to design an experiment.
- This is because Mr. Summers has already made an observation and created a problem.
- He must now gather data to be analyzed.
- In order to do this, he must first design and perform an experiment.
Therefore, we can confirm that Mr. Summers must design an experiment given that this is the best way to gather data in order to be analyzed in the future and draw a valid conclusion.
To learn more about Hypothesis visit:
brainly.com/question/2695653?referrer=searchResults
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:
