Answer:
3s and 3p
Explanation:
From the question given above, the following data were obtained:
Electronic configuration =>
1s² 2s²2p⁶ 3s²3p⁴
Location of valence electron =?
From the electronic configuration given above, we can see clearly that the atom has three (3) shells.
Valence electron(s) are located at the outer most shell of an atom.
The outer most shell of the atom above is 3s and 3p.
Therefore, 3s and 3p will contain the valence electron(s)
Because CH4 is the limiting reagent, we must find how many moles we have.
1. 5.14 g CH4 * (1 mol CH4 / 16.04 g/mol) = .32125 moles CH4
Now we know that there are .32125 moles of CCl4 produced as well.
Next, take .32125 moles and multiply it by the atomic mass of CCl4, which is 153.82 grams = 49.29 g CCl4 produced by the reaction.
Answer :
The complete equation for the reaction of sulfuric acid and sugar is,
By the stoichiometry of the reaction,
1 mole of sucrose react with the 11 moles of sulfuric acid to give 12 moles of carbon and 11 moles of water.
In this reaction, sulfuric acid react with sucrose (sugar). It dehydrates the sugar molecules which means it eliminates the water.
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:
A pure substance has "one set of universal properties". This means they have some of the universal properties in common.
<h3>The definition of universal property</h3>
A characteristic that describes some structures up to an isomorphism is known as a universal property in mathematics, more specifically in category theory.
As a result, independent of the construction technique used, some objects can be described using universal properties. For example, one can define polynomial rings as derived from the field of their coefficients, rational numbers as derived from integers, real numbers as derived from integers, and rational numbers as derived from real numbers.
All of these definitions can be made in terms of universal properties. In particular, the concept of universal property offers a simple demonstration of the equality of any real number structures, requiring only that they satisfy the same universal property.
<h3>
What is the universal property of all substances?</h3>
Diamagnetism is a feature that all substances share.
To learn more about Diamagnetism click on the link below:
brainly.com/question/22078990
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